nt.number theory - Intuition behind the Eichler-Shimura relation?

(1) Short answer to first question: $T_p$ is about $p$-isogenies, and in char. $p$ there is a canonical $p$-isogeny, namely Frobenius.

Details:

The Hecke correspondence $T_p$ has the following definition, in modular terms:
Let $(E,C)$ be a point of $X_0(N)$, i.e. a modular curve together with a cyclic subgroup
of order $N$. Now $T_p$ (for $p$ not dividing $N$) is a correspondence (multi-valued function) which maps
$(E,C)$ to $sum_D (E/D, (C+D)/D)$, where $D$ runs over all subgroups of $E$ of degree $p$.
(There are $p+1$ of these.)

Here is another way to write this, which will work better in char. $p$:
map $(E,C)$ to $sum_{phi:E rightarrow E'}(E',phi(C)),$
where the sum is over all degree $p$ isogenies $phi:Erightarrow E'.$ Giving a degree
$p$ isogeny in char. 0 is the same as choosing a subgroup of order $D$ of $E$ (its kernel),
but in char. $p$ the kernel of an isogeny can be a subgroup scheme which is non-reduced,
and so has no points, and hence can't be described simply in terms of subgroups of points.
Thus this latter description is the better one to use to compute the reduction of the
correspondence $T_p$ mod $p$.

Now if $E$ is an elliptic curve in char. $p$, any $p$-isogeny $E to E'$ is either
Frobenius $Fr$, or the dual isogeny to Frobenius (often called Vershiebung).
Now Frobenius takes an elliptic curve $E$ with $j$-invariant $j$ to the elliptic curve
$E^{(p)}$ with $j$-invariant
$j^p$. So the correspondence on $X_0(N)$ in char. $p$ which maps $(E,C)$ to $(E^{(p)},
Fr(C))$ is itself the Frobenius correspondence on $X_0(N)$. And the correspondence
which maps $(E,C)$ to its image under the dual to Frobenius is the transpose
to Frobenius (domain and codomain are switched). Since there are no other $p$-isogenies in char. $p$
we see that $T_p$ mod $p = Fr + Fr'$ as correspondences on $X_0(N)$ in char. $p$;
this is the Eichler--Shimura relation.

(2) Note that only weight 2 eigenforms with rational Hecke eigenvalues give elliptic curves;
more general eigenforms give abelian varieties.

An easy computation shows that if $f$ is a Hecke eigenform, than the $L$-funcion
$L(f,s)$, obtained by Mellin transform, has a degree 2 Euler product. A more conceptual answer would probably involve describing how automorphic representations
factor as a tensor product of local factors, but that it a very different topic from Eichler--Shimura, and I won't say more here.

co.combinatorics - Does War have infinite expected length?

Dear Joel David,
I will try to explain it, but I have to note that article is quite primitive, and is written in a readable English. Moreover there are many figures. But I will try:
I will make a list of statements and then You can mention the number of the non clear one:

1. By our assumption (players do not have strategy and do not have fixed rules how to return cards) the game is a Markov chain.




2. Absorbing (final) state is a state where you stay forever :)
   For us it means the end of the game i.e. the state when one of players has got all cards.

3A. In finite Markov chain, assuming arbitrary initial state, you are absorbed with probability ONE If And Only If "for each vertex of the Markov chain graph there is a way to an absorbing state."

3B. So we have to prove that for the graph of our game of war, there no exists such initial state that players do not have any chance to reach the end.

1. To prove it we should consider first the simplification. Consider the game with cards {1,...,n} i.e. every value meets only once.

We call a vertex attaining if it has got terminal states as its descendants, and wandering otherwise. It is obvious that a descendant of a wandering vertex is again wandering, and an ancestor
of attaining is again attaining.
For an arbitrary oriented graph it is possible that an attaining vertex has got wandering vertices among its descendants. We show that for our graph G it is not so. For that, we need to understand some properties of the graph G.

LEMMA 1.
A: Let state be such that one of the players has got only one
card in his hand, then this state has got exactly one ancestor.

B: If both players have got at least two cards, then this state has got exactly two ancestors.

LEMMA 2. For the graph of the game it holds that a descendant of an attaining vertex is again an attaining vertex.
(Page 5 of the article)

Lemma 3. The states in which one of the players has got only one card are attaining. (page 6)

Lemma 4. Every vertex has got an ancestor that corresponds to the state in which one of the players has got only one card.

Therefore, we have shown that each vertex has got an ancestor that corresponds to the state in which each player has got exactly one card. This state is attaining by Lemma 3. By Lemma 2 descendants of attaining vertices are again attaining, therefore, the initial state is again attaining, and we have proved

Theorem: Graph G does not have any wandering vertices.

---

Now how to apply it to the standard GAME:
We use the following obvious fact: If a subgraph of an oriented graph does not have wandering vertices, then the original graph does not have any wandering vertices either.

Now the proof is similar.

I hope it is better to read the article, I am sorry.
and I want to note once more time, that question of strategy is never been discussed.

---

[Added by J.O'Rourke:]
The paper has appeared: "On Finiteness in the Card Game of War,"
Evgeny Lakshtanov and Vera Roshchina,
The American Mathematical Monthly,
Vol. 119, No. 4 (April 2012) (pp. 318-323).
JSTOR link.

soft question - Analogies between analogies

It is unfortunate Jan Weidner asked his question -- "Can someone give an example of an analogy between analogies?" -- in such a way that he is unlikely to receive very many insightful responses of the kind offered by Joel David Hamkins and philip314. I don't know how insightful my own response will be, but I can provide a context for the topic of analogy, as I was the contributor who added the quotations by Stefan Banach and Stanislaw M. Ulam to the list of famous mathematical quotes.

It is true that Banach is the original source of the quotation, but I first encountered the topic in Ulam's memoir Adventures of a Mathematician (1976; 1991). From there I learned that the University of California Press published a collection of papers by Ulam titled Analogies between Analogies: The Mathematical Reports of S.M. Ulam and His Los Alamos Collaborators (1991). The collection includes the paper "On the Notion of Analogy and Complexity in Some Constructive Mathematical Schemata" (1981). Rather than attempting to summarize Ulam's results, I'll simply provide an excerpt from the paper's introduction so that interested readers can investigate the topic on their own:

Throughout the development of
mathematics and with the growth of new
concepts and more complicated notions,
a cohesive tendency and organic
structure have been guided by a
feeling of analogy between the old and
new ideas.

Historically, problems posed by the
development of a new mathematical
discipline, which originally was only
metamathematical, coalesced into new
parts of mathematics itself. One could
cite, as obvious examples, the study
of transformations of a space as
points of a new space of such
transformations, and the study of
algorithms for solving equations as
entities per se (group theory, for
instance).

The increasing proliferation of
notions in pure mathematics may
suggest that the idea of analogy
itself is amenable to mathematical
discussion. One finds that old and
elementary formulations of this idea
are, in special cases, present in the
definitions of the similarity of
geometrical figures, more generally in
the equivalence of figures-sets,
through the elements of a group of
transformations, or, more generally
yet, through the identity of proximity
of such sets in spaces which encompass
them.

Two abstract sets of elements may be
felt to be "analogous" if the
difference between their cardinalities
(in the finite case) is small compared
to the cardinalities themselves. Two
classes of such sets may be deemed to
be analogous if the numbers of sets in
the two classes differ by "little" and
if the cardinalities of the
corresponding sets also do not differ
by much. Obviously, one needs to
attempt to formulate a quantitative
criterion, and it is clear a priori
that the notion of analogy will not
be, in general, transitive.

In this report we merely want to
discuss some of the salient features
of analogy and exhibit them on a class
of examples where we shall attempt to
define it, at first, as proximity in
the sense of a metric distance in
suitably defined spaces. (Ulam 1991; 514)

Hopefully, Ulam's paper will provide some tools to make the topic of analogy less "far out."

ag.algebraic geometry - The canonical line bundle of a normal variety

Edit (11/12/12): I added an explanation of the phrase "this is essentially equivalent to $X$ being $S_2$" at the end to answer aglearner's question in the comments.
[See also here and here]

Dear Jesus,

I think there are several problems with your question/desire to define a canonical divisor on any algebraic variety.

First of all, what is any algebraic variety? Perhaps you mean a quasi-projective variety (=reduced and of finite type) defined over some (algebraically closed) field.

OK, let's assume that $X$ is such a variety. Then what is a divisor on $X$? Of course, you could just say it is a formal linear combination of prime divisors, where a prime divisor is just a codimension 1 irreducible subvariety.

OK, but what if $X$ is not equidimensional? Well, let's assume it is, or even that it is irreducible.

Still, if you want to talk about divisors, you would surely want to say when two divisors are linearly equivalent. OK, we know what that is, $D_1$ and $D_2$ are linearly equivalent iff $D_1-D_2$ is a principal divisor.

But, what is a principal divisor? Here it starts to become clear why one usually assumes that $X$ is normal even to just talk about divisors, let alone defining the canonical divisor. In order to define principal divisors, one would need to define something like the order of vanishing of a regular function along a prime divisor. It's not obvious how to define this unless the local ring of the general point of any prime divisor is a DVR. Well, then this leads to one to want to assume that $X$ is $R_1$, that is, regular in codimension $1$ which is equivalent to those local rings being DVRs.

OK, now once we have this we might also want another property: If $f$ is a regular function, we would expect, that the zero set of $f$ should be 1-codimensional in $X$. In other words, we would expect that if $Zsubset X$ is a closed subset of codimension at least $2$, then if $f$ is nowhere zero on $Xsetminus Z$, then it is nowhere zero on $X$. In (yet) other words, if $1/f$ is a regular function on $Xsetminus Z$, then we expect that it is a regular function on $X$. This in the language of sheaves means that we expect that the push-forward of $mathscr O_{Xsetminus Z}$ to $X$ is isomorphic to $mathscr O_X$. Now this is essentially equivalent to $X$ being $S_2$.

So we get that in order to define divisors as we are used to them, we would need that $X$ be $R_1$ and $S_2$, that is, normal.

Now, actually, one can work with objects that behave very much like divisors even on non-normal varieties/schemes, but one has to be very careful what properties work for them.

As far as I can tell, the best way is to work with Weil divisorial sheaves which are really reflexive sheaves of rank $1$. On a normal variety, the sheaf associated to a Weil divisor $D$, usually denoted by $mathcal O_X(D)$, is indeed a reflexive sheaf of rank $1$, and conversely every reflexive sheaf of rank $1$ on a normal variety is the sheaf associated to a Weil divisor (in particular a reflexive sheaf of rank $1$ on a regular variety is an invertible sheaf) so this is indeed a direct generalization. One word of caution here: $mathcal O_X(D)$ may be defined for Weil divisors that are not Cartier, but then this is (obviously) not an invertible sheaf.

Finally, to answer your original question about canonical divisors. Indeed it is possible to define a canonical divisor (=Weil divisorial sheaf) for all quasi-projective varieties. If $Xsubseteq mathbb P^N$ and $overline X$ denotes the closure of $X$ in $mathbb P^N$, then the dualizing complex of $overline X$ is
$$
omega_{overline X}^bullet=R{mathscr H}om_{mathbb P^N}(mathscr O_{overline X}, omega_{mathbb P^N}[N])
$$
and the canonical sheaf of $X$ is
$$
omega_X=h^{-n}(omega_{overline X}^bullet)|_X=mathscr Ext^{N-n}_{mathbb P^N}(mathscr O_{overline X},omega_{mathbb P^N})|_X
$$
where $n=dim X$. (Notice that you may disregard the derived category stuff and the dualizing complex, and just make the definition using $mathscr Ext$.) Notice further, that
if $X$ is normal, this is the same as the one you are used to and otherwise it is a reflexive sheaf of rank $1$.

As for your formula, I am not entirely sure what you mean by "where the $D_i$ are representatives of all divisors in the Class Group". For toric varieties this can be made sense as in Josh's answer, but otherwise I am not sure what you had in mind.

(Added on 11/12/12):

Lemma A scheme $X$ is $S_2$ if and only if for any $iota:Zto X$ closed subset of codimension at least $2$, the natural
map $mathscr O_Xto iota_*mathscr O_{Xsetminus Z}$ is an isomorphism.

Proof
Since both statements are local we may assume that $X$ is affine.
Let $xin X$ be a point and $Zsubseteq X$ its closure in $X$. If $x$ is a codimension at most $1$ point, there is nothing to prove, so we may assume that $Z$ is of codimension at least $2$.

Considering the exact sequence (recall that $X$ is affine):
$$
0to H^0_Z(X,mathscr O_X) to H^0(X,mathscr O_X) to H^0(Xsetminus Z,mathscr O_X) to H^1_Z(X,mathscr O_X) to 0
$$
shows that $mathscr O_Xto iota_*mathscr O_{Xsetminus Z}$ is an isomorphism
if and only if
$H^0_Z(X,mathscr O_X)=H^1_Z(X,mathscr O_X)=0$ the latter condition is equivalent to
$$
mathrm{depth}mathscr O_{X,x}geq 2,
$$
which given the assumption on the codimension is exactly the condition that $X$ is $S_2$ at $xin X$. $qquadsquare$

ag.algebraic geometry - A GAGA question

A GAGA question.

Say I have a ``quasi-projective'' (*) subvariety X over the complex
numbers within a smooth complex algebraic variety Z.

True or False: The analytic and algebraic closure of X (within Z) coincide.

I guess the answer must be `True' and is contained somewhere within Serre's GAGA,
or some elucidation thereof. If I'm right
could someone point me to a precise reference, either within GAGA or
elsewhere? If I'm wrong I'd love to hear about it.

Elucidation:

(*) By `quasi-projective'' I mean X is defined by
a finite number ofalgebraic equations' $f_i = 0$
and inequalities $g_a ne 0$, As is the case when Z is projective
space, the $f_i$ may not be globally defined;
same for the $g_a$. In my situation,
the Zariski open set defined by intersecting the sets $g_a ne 0$
is an affine set (in the usual schemy sense) and the $f_i$ are polynomials
on this affine set.

My Z is probably projective -- I'm not positive here, just pretty sure.
(My Z is obtained by iterating the construction of taking the
bundle over a smooth projective variety whose
fiber is the Grassmannian of d-planes within said variety's tangent space. )

graph theory - Why is edge-coloring less interesting than vertex-coloring?

I do not know whether edge coloring is more or less interesting than vertex coloring,
this is probably someting that could only be settled by a poll.
The chief reasons why edge coloring receives less attention than vertex coloriong would,
if I had to guess, be the third and fourth you offer.

Note though that if $X$ is $k$-regular graph, then Vizing's theorm tells us that
its edge-chromatic number is $k$ or $k+1$. If it is $k$, then $X$ has a 1-factorization --
we can partition its edges into $k$ pairwise edge-disjoint perfect matchings.
There is a very large literature on problems related to 1-factorizations.

Vertex coloring is also arguably more important in practice, since it arises in
connection with important scheduling and register allocation problems. Although many of
us are pure mathematicians and may feel no need to consider practical problems,
the influence of the ``real world'' on pure mathematics is nonetheless very strong.

st.statistics - Has the Lie group preserving a probability distribution been used in Bayesian statistics?

For a (possibly signed) nondegenerate probability measure $pi$ on ${1,dots,n}$ define
$$langle pi rangle := {R in operatorname{STO}(n): pi R = pi }.$$
Here $operatorname{STO}(n)$ denotes the group of invertible stochastic matrices, i.e., the set of matrices (with entries of either sign) whose row sums are unity. Also, write $langle R rangle equiv pi$ for $R in langle pi rangle$. $langle pi rangle$ is a $(n-1)^2$-dimensional Lie group under matrix multiplication. Its Lie algebra $mathfrak{lie}(langle pi rangle)$ has a basis of the form
$$e_{(j,k)}^{langle pi rangle} := e_{(j,k)} - frac{pi_j}{pi_n}e_{(n,k)}, quad 1 le j, k le n-1$$
where $e_{(j,k)} = e_j(e_k^* - e_n^*)$ and we use typical notation for the standard basis of $mathbb{R}^n$. A block matrix decomposition shows that $mathfrak{lie}(langle pi rangle) cong mathfrak{gl}(n-1)$, and it is easy to show that
$$exp Ce_{(j,k)}^{langle pi rangle} = I + frac{e^{C(pi_j/pi_n + delta_{jk})} - 1}{pi_j/pi_n + delta_{jk}} e_{(j,k)}^{langle pi rangle}.$$
In particular, this object seems pretty tractable (although I haven't bothered to think about the Haar measure).

My question is, has it been used in
Bayesian statistics (or elsewhere)? It seems
naturally suited for such an
application.

linear algebra - Sequence of constant rank matrices

I think it is best to settle this problem geometrically, that is if you think of matrices as linear maps from $mathbb R^m$ to $mathbb R^n$. The images of these maps are $r$-dimensional linear subspaces of $mathbb R^n$. Let $X_k$ denote the image of $A_k$, then $u_kin X_k$, and you want to prove that the limit vector $u$ belongs to the image of $A$.

Let $X$ denote the set of all possible limits of sequences such that the $k$th element of the sequence belongs to $X_k$ for every $k$ (for example, ${u_k}$ is one such sequence, hence $uin X$). Clearly $X$ is a linear subspace of $mathbb R^n$, and it contains the image of $A$. And the dimension of $X$ is no greater than $r$. Indeed, suppose that $X$ contains $r+1$ linearly independent vectors $w_1,dots,w_{r+1}$. Each $w_i$ is a limit of a sequence of vectors $u_k^{(i)}in X_k$. For a sufficiently large $k$, the vectors $u_k^{(1)},dots,u_k^{(r+1)}$ are linearly independent because the set of linearly independent $(r+1)$-tuples is open. This contradicts the fact that $dim X_k=r$.

Since $X$ is a linear subspace of dimension at most $r$ and it contains the ($r$-dimensional) image of $A$, it must coincide with that image. Since $uin X$ by definition, it follows that $u$ belongs to the image of $A$, q.e.d.

human biology - Can Naegleria fowleri enter through wounds into the bloodstream?

No, Naegleria fowleri is a free-living excavate form of protist that lives in warm fresh water. Fowleri finds its way into the brain by eating through the olfactory neurons in the nose where it multiplies itself greatly by feeding on nerve tissue.

Once it penetrates the nervous tissue, fowleri's feeding results in significant necrosis of and hemorrhaging in the olfactory bulbs. The protist then climbs along nerve fibers through the floor of the cranium via the cribriform plate and into the brain.

Since the protist travels by feeding on nervous tissue, fowleri could not enter the brain via bloodstream.

Just some more facts about Naegleria:

• Death usually occurs within 14 days of exposure due to destruction the autonomic nerve cells of the medulla oblongata.


• Luckily for everyone this disease is rare. There were only 300 cases as of 2008

genetics - Autosomal Recessive Trait when skipping one generation

What you are saying is very similar to the statement that:

Absence of evidence is not evidence of absence

Suppose you have an autosomic gene A, which has a mutant allele a, causing an illness (or any other phenotype, for that matters). The particular nature of a makes its associated phenotype recessive.

You have three possibilities:

AA -> no illness  
Aa (or aA) -> carrier, with no illness 
aa -> ill

Now, let's take an aa individual (ill) and cross it with a sane AA individual:

aa x AA

All of the individuals of the first generation (F1) will be carriers (Aa) because the father will always give a and the mother always A.

Now if we breed them together

aA x aA

both parents have 1/2 probability of giving a and 1/2 of giving A. We will then have a proportion of:

1/4 aa
1/2 Aa (1/4 A from mother and a from father, 1/4 vice-versa) 
1/4 AA

But this is just down to statistics. You have only 1/4 chances of seeing the trait reappearing, so you are much less likely to see it if they have 1 offspring then if they have 20.
Moreover, you have Aa offsprings in the progeny... this means that these could pass the mutant allele to the third or fourth generation.

Another important thing is that all we said is true if the first cross is

aa * AA

If it was

aa * aa

or

aa * Aa

The trait will (surely in the first case, and with 1/2 chance in the second) appear directly in the first generation.

gr.group theory - Finite-dimensional version of the word problem for groups

The (uniform) word problem for groups can be stated in several equivalent ways:

Word Problem for Groups (WP)
Instance: A finite presentation of a group G and an element w of G as a product of generators and their inverses.
Question: Does every linear representation of G in a (not necessarily finite-dimensional) Hilbert space map w to the identity operator?
Equivalently: Does every unitary representation of G in a Hilbert space map w to the identity operator?
Equivalently: Does every normal subgroup of G contain w?
Equivalently: Is w=1 in G?

As is well-known, WP is undecidable; more specifically, it belongs to RE∖coRE, where RE is the class of recursively enumerable languages and coRE is the class of their complements. (In fact, it is known that WP remains undecidable even if we fix the group G and its finite presentation suitably, but that is not the topic of this question.)

We can consider the finite-dimensional version of WP.

Finite-Dimensional Word Problem for Groups (FWP)
Instance: Same as WP.
Question: Does every matrix representation of G map w to the identity matrix?
Equivalently: Does every unitary matrix representation of G map w to the identity matrix?
Equivalently: Does every normal subgroup of G of finite index contain w?
(The equivalence is based on a result by Malcev; see this post by Greg Kuperberg and the comments attached to it.)

The two problems have different answers for some instances; that is, sometimes w≠1 in G but every matrix representation maps w to the identity matrix. There even exists a finitely presented infinite group which does not have a nontrivial matrix representation. See the answers to the question “Finitely presented sub-groups of GL(n,C)” by Dmitri.

In fact, FWP is in coRE unlike WP: if we also give the dimension as part of the input, the problem becomes decidable (because it is a system of algebraic equations over ℂ), and therefore FWP can be put in coRE by trying all dimensions.

Question: Is FWP decidable?

I do not even know whether FWP is NP-hard or not. As far as I know, FWP can be in P or undecidable, or anywhere in between!

(Note: FWP can be viewed as a (very) special case of the problem discussed in the question “Decidability of matrix algebra” by Ricky Demer.)

at.algebraic topology - Maps inducing zero on homotopy groups but are not null-homotopic

Even if you ask that $f$ induces trivial maps on all (singular) homology and cohomology groups, there are still easy manifold examples. (This actually arises as an exercise in Hatcher's AT).

For instance, let $f:T^3rightarrow S^2$ be the composition $T^3rightarrow S^3rightarrow S^2$, where the map from $T^3$ to $S^3$ is simply collapsing the 2-skeleton to a point, and the map from $S^3$ to $S^2$ is the Hopf map.

As others have mentioned, since $T^3$ is a $K(mathbb{Z}^3, 1)$, if follows that $f$ induces trivial maps on homotopy groups.

Since the Hopf map induces trivial maps on homology and cohomology, it follows that $f$ does as well.

Finally, to see that $f$ is NOT nullhomotopic, assume it is. Since the map from $S^3$ to $S^2$ is a fiber bundle, it has the homotopy lifting property. Hence, we can lift the homotopy of $f$ to a homotopy $G:Itimes T^3rightarrow S^3$ where $G_0$ is the above map from $T^3$ to $S^3$ and $G_1$ is is a map from $T^3$ to $S^1subseteq S^3$, the preimage of a point in $S^2$ under the Hopf map.

But $G_0$ has degree 1, while $G_1$ has degree 0, a contradiction.

rt.representation theory - Decomposing a tensor product

For $SO(n)$ a calculation using LiE gives:
(using partition notation so $W$ is [2])
and assuming $n$ is not small

For $Wotimes W$, [4],[3,1],[2,2],[2],[1,1],[]
(all with multiplicity one)

and for $Wotimes Wotimes W$,
1.[6] 2.[5,1] 3.[4,2] 1.[3,3] 1.[4,1,1] 2.[3,2,1] 1.[2,2,2] 3.[4] 6.[3,1] 2.[2,2] 3.[2,1,1] 6.[2] 3.[1,1] 1.[]

The same works for $Sp(n)$ by taking conjugate partitions.

There is also a relationship with $SL(n)$.

This is taking your question at face value. If it is understanding you're after instead then the best approach is to use crystal graphs.

The notation I have used denotes a representation by a partition. I have put $m.$ in front to denote multiplicity is $m$. A partition is $[a_1,a_2,a_3,...]$ where $a_ige a_j$ if $i$ less than $j$. To convert to a highest weight vector add the appropriate number of $0$s to the end.
Then take $[a_1-a_2,a_2-a_3,a_3-a_4,...]$. This gives a dominant integral weight. The fundamental weights are the partitions $[1,,,,1]$. If this has length $k$ this corresponds to the $k$-th exterior power of the vector representation (provided $2k-1$ less than $n$).

In particular the trivial representation is $[]$, the vector representation $V$ is $[1]$, the exterior square of $V$ is $[1,1]$, the symmetric square is $[2]+[]$.

For the $k$-th tensor power of $W$ you will see partitions of $2k-2p$ for $0le ple k$ only and it remains to determine the multiplicities (possibly $0$). For $SL(n)$ just take the partitions of $2k$ (with their multiplicities) and ignore the rest.

ag.algebraic geometry - Why do flag manifolds, in the P(V_rho) embedding, look like products of P^1s?

Your first question is about two objects becoming isomorphic after quantization, and you're asking "Why?"

Here, the relevant quantum object is the spin representation of $mathfrak g$, which is a representation of $mathfrak gltimes mathit{Cliff}(mathfrak g)$,
where $mathit{Cliff}(mathfrak g)$ is the Clifford algebra of (the underlying vector space of) $mathfrak g$, with respect to some invariant inner product; a $mathbb Z/2$-graded algebra.

Let $S$ be the unique (up to grading reversal) irreducible $mathbb Z/2$-graded representation of $mathit{Cliff}(mathfrak g)$. It has a graded-commuting action of
$$C:=begin{cases}mathbb C&text{ if }quad dim(mathfrak g) text{ is even}
\
mathit{Cliff}(1)&text{ if }quad dim(mathfrak g) text{ is odd.}
end{cases}
$$
Let $V$ be the category of modules of the above algebra, so that $Sin V$.
To make things a bit more canonical, one can use the graded Morita equivalence between $C$ and $mathit{Cliff}(mathfrak h)$ to identify $V$ with the category of $mathit{Cliff}(mathfrak h)$-modules.

Let $alpha$ denote the adjoint action of $G$ on $mathit{Cliff}(mathfrak g)$. For any element $gin G$, we can pre-compose the action of $mathit{Cliff}(mathfrak g)$ on $S$ by $alpha_g$ to get a new, isomorphic $mathit{Cliff}(mathfrak g)$-module in $V$ [here, I'm using that $G$ is connected].
The map that sends $g$ to such an isomorphism is unique up to scalar, and so we get a projective representation of $G$ on $S$.

If $G$ is simply connected, this lifts to an honest action of $G$ on $S$, and so we get an action of $Gltimesmathit{Cliff}(mathfrak g)$ on the object $Sin V$.

All in all, (the underlying vector space of) $S$ has actions of $G$, of $mathit{Cliff}(mathfrak g)$, and of $mathit{Cliff}(mathfrak h)$.
Now, we can "cancel" two $mathit{Cliff}(mathfrak h)$ actions to get a vector space with actions of $G$ and of $mathit{Cliff}(mathfrak gominus h)$.
That's the irreducible $G$-rep with highest weight $rho$.

This vector space has two descriptions:
(1) The irreducible $G$-rep with highest weight $rho$.
(2) The irreducible $mathit{Cliff}(mathfrak gominus h)$-module.

Note however that this operation of "canceling" the two $mathit{Cliff}(mathfrak h)$ actions is a bit unnatural.
For example, the vector space (1) is purely even, while (2) is genuinely $mathbb Z/2$-graded. You probably see that same weirdness on the classical side of the problem when you try to identify the flag variety with a product of $mathbb P^1$s.

gr.group theory - Recursive presentations

The answer is a simple trick. Essentially no group theory is involved.

Suppose that we are given a group presentation with a set of generators, and relations R_0, R_1, etc. that have been given by a computably enumerable procedure. Let us view each relation as a word in the generators that is to become trivial in the group.

Now, the trick. Introduce a new generator x. Also, add the relation x, which means that x will be the identity. Now, let S_i be the relation (R_i)x^(t_i), where t_i is the time it takes for the word R_i to be enumerated in the enumeration algorithm. That is, we simply pad R_i with an enormous number of x's, depending on how long it takes for R_i to be enumerated into the set of relations. Clearly, S_i and R_i are going to give the same group, once we have said that x is trivial, since the enormous number of copies of x in S_i will all cancel out. But the point is that the presentation of the group with this new presentation becomes computably decidable (rather than merely enumerable), because given a relation, we look at it to see if it has the form Sx^t for some t, then we run the enumeration algorithm for t steps, and see if S has been added. If not, then we reject; if so, then we accept.

One can get rid of the extra generator x simply by using the relation R_0 from the original presentation. This gives a computable set of relations in the same generating set that generates the same group.

The essence of this trick is that every relation is equivalent to a ridiculously long relation, and you make the length long enough so that one can check that it really should be there.

examples - Request: A Serre fibration that is not a Dold fibration

You've already answered your own question, but here is another example.

Let $f: mathbb{Q}^delta to mathbb{Q}$ be the obvious map from the rational numbers with the discrete topology to the rational numbers with the usual topology. Let $M_f$ be the mapping cylinder. Then the projection
$$p:M_f to [0,1]$$
is a Serre fibration. This follows because any map of a disc into $mathbb{Q}$ factors through f. Hence as far as discs are concerned, $M_f$ might as well be $mathbb{Q}^delta times [0,1]$.

However this projection is not a Dold fibration. It is easy to construct a diagram using $Y = mathbb{Q}$ which will have no weak homotopy lift. Indeed consider the projection map $mathbb{Q} times [0,1] to [0,1]$ with the obvious initial lift. Any other initial lift vertically homotopic to this one in fact coincides with this one, hence it is easy to see that there is no weak lift of this map.

By replacing $mathbb{Q}^delta$ and $mathbb{Q}$ with their cones, we get a similar example where now the base, total space, and fibers are contractible. Hence they are path-connected and also have the homotopy type of CW complexes. So this also answers Ronnie Browns question (but surely the answer to that has been known for some time).

at.algebraic topology - Why do finite homotopy groups imply finite homology groups?

(This answer has been edited to give more details.)

Finitely generated homotopy groups do not imply finitely generated homology groups. Stallings gave an example of a finitely presented group $G$ such that $H_3(G;Z)$ is not finitely generated. A $K(G,1)$ space then has finitely generated homotopy groups but not finitely generated homology groups. Stallings' paper appeared in Amer. J. Math 83 (1963), 541-543. Footnote in small print: Stallings' example can also be found in my algebraic topology book, pp.423-426, as part of a more general family of examples due to Bestvina and Brady.

As Stallings noted, it follows that any finite complex $K$ with $pi_1(K)=G$ has $pi_2(K)$ nonfinitely generated, even as a module over $pi_1(K)$. This is in contrast to the example of $S^1 vee S^2$.

Finite homotopy groups do imply finite homology groups, however. In the simply-connected case this is a consequence of Serre's mod C theory, but for the nonsimply-connected case I don't know a reference in the literature. I asked about this on Don Davis' algebraic topology listserv in 2001 and got answers from Bill Browder and Tom Goodwillie. Here's the link to their answers:

http://www.lehigh.edu/~dmd1/tg39.txt

The argument goes as follows. First consider the special case that the given space $X$ is $BG$ for a finite group $G$. The standard model for $BG$ has finite skeleta when $G$ is finite so the homology is finitely generated. A standard transfer argument using the contractible universal cover shows that the homology is annihilated by $|G|$, so it must then be finite.

For a general $X$ with finite homotopy groups one uses the fibration $E to X to BG$ where $G=pi_1(X)$ and $E$ is the universal cover of $X$. The Serre spectral sequence for this fibration has $E^2_{pq}=H_p(BG;H_q(E))$ where the coefficients may be twisted, so a little care is needed. From the simply-connected case we know that $H_q(E)$ is finite for $q>0$. Since $BG$ has finite skeleta this implies $E_{pq}^2$ is finite for $q>0$, even with twisted coefficients. To see this one could for example go back to the $E^1$ page where $E_{pq}^1=C_p(BG;H_q(E))$, the cellular chain group, a finite abelian group when $q>0$, which implies finiteness of $E_{pq}^2$ for $q>0$. When $q=0$ we have $E_{p0}^2=H_p(BG;Z)$ with untwisted coefficients, so this is finite for $p>0$ by the earlier special case. Now we have $E_{pq}^2$ finite for $p+q>0$, so the same must be true for $E^infty$ and hence $H_n(X)$ is finite for $n>0$.

Sorry for the length of this answer and for the multiple edits, but it seemed worthwhile to get this argument on record.

Descriptive complexity theoretic-characterizations of P and NP

Prompted by Vinay Deolalikar's purported proof of P != NP, I've been reading up on Descriptive Complexity for some background material.

The major successes of Descriptive Complexity include Fagin's result that $NP=SOexists$ (that is, the class NP is equal to the class of models of a second-order existential query over some vocabulary), and also that $P = FO(LFP)$ (that the class P is equal to the class of models of first-order queries that might use a Least-Fixed-Point operator), and also $PH = SO$.

My understanding of mathematical logic is quite shaky, but from what I understand, second-order formulas are not expressible in first order logic - how does this fact stand in relation to the results I mentioned above? Why does it not separate NP from P, or PH from P?

ag.algebraic geometry - Why and how are moduli spaces of (semi)stable vector bundles well-behaved?

From a topological viewpoint, I believe the idea is that one wants to have a Hausdorff quotient space. In other words, consider the space of all holomorphic structures on a fixed (topological) vector bundle on a curve. Holomorphic structures can be viewed as differential operators on sections of the bundle, such that a section is holomorphic if and only if this operator evaluates to zero on the section. (See, for example, sections 5 and 7 of Atiyah and Bott's "The Yang-Mills equations over Riemann surfaces.") This makes the space of holomorphic structures (i.e. the space of bundles with a fixed topological type) into an affine space. The group of complex automorphisms of the bundle acts on this space, and the quotient is the moduli space of holomorphic bundles. If you don't restrict to stable bundles, this quotient space fails to be Hausdorff. Atiyah and Bott reference this to Mumford's 1965 GIT book. Actually, they just say that the moduli space of stable bundles is Hausdorff, due to the fact that the orbits of stable bundles are closed. (Hmmm... that really just says points are closed in the quotient...) I don't know how much of this is spelled out in Mumford; in particular, I don't know whether there's a proof in the literature that the full quotient space fails to be Hausdorff.

What are the various types of protein-protein interactions

There are so many types of protein-protein interactions via various domains, such as SH2 binding (in RTK signaling), Pleckstrin Homology domain (involved in signaling) among others. This site gives a nice list: http://pawsonlab.mshri.on.ca/index.php?option=com_content&task=view&id=30&Itemid=63

Of course, protein-protein interactions rely on the premises of basic biochemistry:

• van-der waals interactions (at the most basic level)


• Electrostatic interactions


• "Lock and Key" model: Some proteins have specific binding pockets for domains of other proteins


• Induced fit

There are many ways to study protein-protein interactions, including, but not limited to

• Pull-down assays


• 2D SDS-PAGE / MS


• Yeast two-hybrid


• Immunoprecipitation

nt.number theory - Can every finite graph be represented by an arithmetic sequence of natural numbers?

(This is a follow-up to my previous questions Natural models of graphs?.)

Erdös in The Representation of a Graph by Set Intersections (1966) states:

Theorem. Let $G$ be an arbitrary
graph. Then there is a set $S$ and a
family of subsets $S_1, S_2, ...$ of
$S$ which can be put into one-to-one
correspondence with the vertices of
$G$ in such a way that $x_i$ and $x_j$ are joined by an
edge of $G$ iff $i neq j$
and $S_i cap S_j neq emptyset$.

If we identify $S$ with a set of prime numbers and each $S_i$ with the product of its members we get the following:

Corollary. Let $G$ be an arbitrary finite
graph. Then there is a sequence of natural numbers $(n_1, n_2, ..., n_k)$
which can be put into one-to-one
correspondence with the vertices of
$G$ in such a way that $x_i$ and $x_j$ are joined by an edge iff $i neq j$ and GCD$(n_i, n_j) > 1$.

We can choose the prime numbers (the elements of $S$, from which the $n_i$ are built) arbitrarily, and so the question arises, whether they can always be choosen in such a way, that the set $(n_1, n_2, ..., n_k)$ is an arithmetic sequence.

Of course every complete graph on $k$ nodes can be represented by an arithmetic sequence: just take some consecutive sequence of even numbers. Green-Tao's Theorem guarantees that also every empty graph on $k$ nodes can be represented by an arithmetic sequence $(p_1, p_2, ..., p_k)$ of primes.

Question: Can every graph on $k$ nodes be represented by an arithmetic sequence
of natural numbers such that $n_i$ and $n_j$ are joined by an edge iff $n_i neq n_j$ and GCD$(n_i, n_j) > 1$

This would be one kind of natural model of a graph, that I was looking for, originally.

Maybe some references?

Added: Due to Kevin's concise answer and Thomas' comment, I'd like to add the following question:

Question: If not every graph on $k$ nodes can be represented by an arithmetic sequence
of natural numbers such that $n_i$ and $n_j$ are joined by an edge iff $n_i neq n_j$ and GCD$(n_i, n_j) > 1$: Are there interesting classes of graphs with this property?

fa.functional analysis - Need help understanding Riesz representation theorem for Reproducing Kernel Hilbert Spaces

This really isn't an answer. Like Pietro's, it's a comment that got out of hand.

I've been reading a number of books on and offline (thanks to Google books), and I now understand what the kernel of a linear operator is as well as the orthogonal projection theorem, but an understanding of the proof still eludes me. (By the way I've noted that almost all the proofs I've found are versions of each other.) Nevertheless, reading so much about the proof has shed some light on the nature of RKHS, such as:

• any linear evaluation function $f(x) = ; lt x , x_0 gt$ is an inner product ($x_0$ is the representer of the evaluation function)


• for each evaluation function there exists only one $x_0 in H$


• $parallel f parallel ; = ; parallel x_0 parallel$

Furthermore, according to "Smoothing Spline ANOVA Models" Gu, Chong 2002 (page 27)

"For every $g$ in a Hilbert space $mathcal{H}$, $L_gf ; = ; lt g , f gt $ defines a continuous linear functional $L_g$. Conversely, every continuous linear functional $L$ in $mathcal{H}$ has a representation $Lf ; = ; lt g_L , f gt$ for some $g_L in mathcal{H}$, called the representer of the evaluation."

This statement demystifies RKHS by the assertion that: every linear evaluation functional is (merely) an inner product of the representer and an element of the RKHS, with the result that the Riesz representation is increasingly seems to to be a definition i.e. something to be accepted and not a result that must be derived.

rt.representation theory - Intersection cohomology of flag varieties/Schubert varieties

First, let me rephrase your question in a slightly pedantic manner.

To establish some notation, for a point $p$ on the flag variety $G/B$, let $V_1(p)subsetcdots V_{n-1}(p)$ be the flag in $mathbb{C}^n$ that it corresponds to. (Be careful. There are no flags actually in the flag variety, just points. Rather, the points in the flag variety correspond to flags. If this confuses you you need a live person to straighten you out.)

You are asking for the intersection cohomology of the subvariety $Xsubset G/B times G/B$ consisting of points $(p,q)$ such that $dim(V_i(p)cap V_j(q))=a_{ij}$ (for some specified $a_{ij}$).

Now an answer:

Your variety $X$ has a projection onto the second factor, and this map is a fiber bundle whose base space is smooth (since it is the entire flag variety). Therefore, the local intersection cohomology for the whole space is determined entirely by the local intersection cohomology of the fibers.

If the conditions $a_{ij}$ are conditions that determine a Schubert variety, then the fibers are Schubert varieties, and hence local intersection cohomolgy Betti numbers are precisely given by Kazhdan--Lusztig polynomials.

If the conditions $a_{ij}$ are not conditions determining a Schubert variety, then your fibers will be unions of Schubert varieties. I don't know if anyone has bothered to do this, but I would think that if you take any of the definitions of Kazhdan--Lusztig polynomials $P_{u,v}(q)$ and modify it in the obvious way (if there is one) to allow $v$ to be an arbitrary lower ideal in Bruhat order rather than a principal lower ideal you should get the right thing.

big picture - Does the presence of cocycle conditions indicate the existence of an underlying cohomology theory?

I had lots of thoughts on that kind of question, and feel uneasy to speak as my answer can range from a tautology, through systematic and positive, but somewhat ignorant toward not-well understood cases, to mere impressions and (seeming?) "counterexample" oriented answer. The basic question is what you mean by a cocycle. Usually one talks on expressions of some higher categorical coherence, or about some notion of homotopy behind it. In such cases the answer is normally yes: the equivalent or homotopic cocycles will form cohomology classes and this can be in all understood cases done naturally and systematically. Higher nonabelian cohomology can be done for all $n$, as now many frameworks know (Brown, Jardine, Toen, Street...) and cohomology boils down to take homotopy classes into certain suspension of the coefficient object. For one recent framework we can advertise our own work (pdf).

I slightly believe anyway that some algebraic cases can be outside of the current homotopy categorical framework and I discussed that much on the n-category cafe, nforum and elsewhere. Namely model categories treat on equal footing homology and cohomology, while the minimal conditions on a setup to be able to do cohomology of homology is less than both simultaneously (cf. work of Rosenberg on "right exact structures" on a category, pdf).

Finally, we can imagine more complicated category-like structures where one can do much of the usual combinatorics but can not properly do the equivalence classes when needed for cohomology. There is one example which is maybe repairable, due Shahn Majid, namely he has a notion of bialgebra cocycles for a noncommutative and noncocommutative bialgebra. Now in special cocommutative or commutative cases like Lie algebras and/or abelian coefficients he recovers some known cohomology theories like Chevalley-Eilenberg cohomology for Lie algebras. In low dimensional cases he also gets some interesting nonabelian cocycles of much usage like Drinfel'd 2-twist and Drinfel'd 3-associator which are used in the study of monoidal categories, CFT, knot theory and quantum groups. In this example the differential and cocycles are defined for every $n$ but 17 years after the discovery, there is still no known way to define well the cohomology classes, for dimension 3 or more, for general bialgebra, despite the special cases and despite the cocycles and the differential. See the nlab page bialgebra cocycle for the basics (and the references therein).

nt.number theory - Is there an "elementary" proof of the infinitude of completely split primes?

Let K be a Galois extension of the rationals with degree n. The Chebotarev Density Theorem guarantees that the rational primes that split completely in K have density 1/n and thus there are infinitely many such primes. As Kevin Buzzard pointed out to me in a comment, there is a simpler way to see that there are infinitely many rational primes that split completely in K, namely that the Dedekind zeta-function ζ_K(s) has a simple pole at s = 1. While this result is certainly much easier to prove than Chebotarev's Theorem, it is still not an elementary proof.

Is there a known elementary proof of the fact that there are infinitely many rational primes that split completely in K?

Selberg's elementary proof of Dirichlet's Theorem for primes in arithmetic progressions handles the case where Gal(K/Q) is Abelian. I don't know anything about the general case. Since Dirichlet's Theorem is stronger than required, it is possible that an simpler proof exists even in the Abelian case.

Remarks on the meaning of elementary. I am aware that there is no uniformly recognized definition of "elementary proof" in number theory. While I am not opposed to alternate definitions, my personal definition is a proof which can be carried out in first-order arithmetic, i.e. without quantification over real numbers or higher-type objects. Obviously, I don't require it to be explicitly formulated in that way — even logicians don't do that! Odds are that whatever you believe is elementary is also elementary in my sense.

Kurt Gödel observed that proofs of (first-order) arithmetical facts can be much, much shorter in second-order arithmetic than in first-order arithmetic. This observation explains some of the effectiveness of analytic number theory, which is implicitly second-order. In view of Gödel's observation, it is possible that we have encountered arithmetical facts with a reasonably short second-order proof (i.e. could be found in an analytic number theory textbook) but no reasonable first-order proof (i.e. the production of any such proof would necessarily exhaust all of our natural resources). The above is unlikely to be such, but it is interesting to know that beasts of this type could exist...

evolution - Do men have more extreme variations than women?

_{This question was considered unsuitable for Skeptics and I think it is more suited to BIology than Cognitive Sciences}

I was reading this article which I found interesting. It is not supported with references but was an invited address for the American Psychological Association.

It makes the claim that men go to more extremes than women.

People can point to plenty of data that the average IQ of adult men is about the same as the average for women. So to suggest that men are smarter than women is wrong. No wonder some women were offended.

There are more males than females with really low IQs. Indeed, the
pattern with mental retardation is the same as with genius, namely
that as you go from mild to medium to extreme, the preponderance of
males gets bigger.

[...]

Almost certainly, it is something biological and genetic. And my guess
is that the greater proportion of men at both extremes of the IQ
distribution is part of the same pattern.

[...]

Nature rolls the dice with men more than women. Men go to extremes
more than women. It’s true not just with IQ but also with other
things, even height: The male distribution of height is flatter, with
more really tall and really short men.

---

Another notable example of a similar claim was Lawrence Summer's speech regarding the possible reasons for dearth of women in tenured positions in science and engineering at top schools.

Summers outlined 3 possible reasons, with a major kerfuffle erupting - resulting in his firing from the job as a Harvard University President - over his presented second reason [ quotes from Wiki ]:

The second hypothesis, different availability of aptitude at the high end, caused the most controversy. In his discussion of this hypothesis, Summers said that "even small differences in the standard deviation [between genders] will translate into very large differences in the available pool substantially out [from the mean]

Summers referenced research that implied differences between the standard deviations of males and females in the top 5% of twelfth graders under various tests. He then went on to argue that, if this research were to be accepted, then "whatever the set of attributes... that are precisely defined to correlate with being an aeronautical engineer at MIT or being a chemist at Berkeley... are probably different in their standard deviations as well".

---

QUESTION: Are there any studies or evidence to support the claim that men have more extreme variations than women (in other words, that the trait's distribution in males have fatter tails) in a variety of traits?

Possible areas of research can be:

• Intelligence (general IQ or specific linguistic or spacial abilities)


• Physical attributes (strength, height, weight)


• personality traits
  
  • for lack of anything better, let's take the FFM's Five:
  
  
  • openness
  
  
  • conscientiousness
  
  
  • extroversion
  
  
  • agreeableness
  
  
  • neuroticism
  
  

---

In order for the answer to be "yes", the research needs to show fat tails in at least 5-6 traits from all 3 of the buckets above (intelligence, physical, and personality), although precise trait mix may be arbitrary and not necessarily limited to my examples, as long as 5-6 somewhat independent traits fit.

In order for the answer to be "no", at least 8 of the 11 traits above should have no fatter tails in men.

Anything else means "inconclusive".

histology - What is the proper quick freezing /snap freezing protocol for pancreatic tissue?

I would like to do a quick freeze on pancreas from mice. I want to then make sections (30 µm thick). The idea is to preserve a fluorescent staining done in vivo.

I do not know if I should place my fresh tissue directly into liquid nitrogen, then embed the tissue in OCT, and then freeze everything in cold isopentane?

Or do I have to embed the fresh tissue in OCT first, prior to putting everything in nitrogen?

Or is there another protocol to do quick freezing in order to do sectioning?

complex geometry - what is the motivation of Shimura variety?

The theory of Shimura varieties was begun by Shimura, and further developed by Langlands (who introduced the name), and is now a central part of arithemtic geometry and of the theories
of automorphic forms, Galois representations, and motives.

Shimura varieties are certain moduli of Hodge structures; but that is perhaps not the best point of view to understand why people study them. Rather, the primary motivation is the following: Shimura varieties are attached to (certain) reductive linear algebraic groups over $mathbb Q$, and the geometry of the Shimura variety is closely linked to the theory of automorphic forms over the corresponding reductive group.

Thus Shimura varieties make a natural test-case for investigating the conjectural relations between motives and automorphic forms, since they are geometric objects with a direct link to the theory of automorphic forms. (I'm not sure that it's useful to be more specific about this in this particular answer, but for those to whom it is meaningful: on the one hand, one has an analogue, for any Shimura variety, of Eichler--Shimura theory, relating the cohomology of modular curves to modular forms; and on the other hand, by thinking of cohomology as being etale cohomology, one obtains Galois representations which one would like to show (and in many cases can show) are related to automorphic forms in a manner analogous to the relationship between the Galois representations on the etale cohomology of modular curves and Hecke eigenforms that was established by Deligne.)

matrices - Determinant of a 3x3 Magic Square

I don't have an explanation, but here is an outline of a proof (I've checked all the details myself, but it's laborious to write up correctly) that what you claim to happen actually does happen.

Result: Let $M$ be a 3x3 integer matrix whose columns, rows, diagonal, and anti-diagonal each total 15. The following are equivalent:

1. The entries of $M$ are distinct and positive;


2. The determinant of $M$ is $pm 360$;


3. Calling $h,i$ the (3,2) and (3,3) entries, $$(h,i)in {(1,6), (1,8), (3,4), (3,8), (7,2), (7,6), (9,2), (9,4)}.$$

The proof is just writing down the equations for the sums, rows, etc., to be 15 and solving, and then writing down the formula for the determinant and simplifying (it turns out $det = 45(h-5)(h+2i-15)$) and solving the resulting diophantine equation.

The explanation may well be just that there aren't very many 3x3 magic squares (basically, it looks like there's just one that we rotate and flip to get eight). For 4x4, the solution space will be 7d instead of 2d, and that is a lot of extra freedom. If you find a simply-described structure in the determinants of those, I'd be surprised.

dg.differential geometry - Local Triviality of an Associated Bundle

@Scott: I'm going to add this as an answer because it a bit long for a comment: Sorry, but I'm a bit slow and want to be certain that I understand exactly what you're saying: If $(p,v) in P times_{rho} mathbb{C}^n$, then $$(alpha_u times id)(p,v) = (pi(p),g_p,v) in U times G times mathbb{C}^n,$$ for some $g_p in G$, depending on the choice of $alpha_U$. This is not in $U times mathbb{C}^n$ as we would want, moreover it's not even well defined. So by $alpha_U times_{rho} id$ do you mean the composition
$$
(id times m) circ (id times rho times id)circ (alpha_U times id),
$$
that maps
$$
(p,v) mapsto (pi(p), rho(g_p)v)?
$$

gn.general topology - A good place to read about uniform spaces

I would motivate them as follows: if topological spaces were invented to give a general meaning to "continuous function", then uniform spaces were invented to give a general meaning to "uniformly continuous function". It is clear what "uniformly continuous" should mean for metric spaces and topological groups, but how should the general notion be formalized?

Once this is formalized, one can define the notion of Cauchy net in a uniform space (which is something you cannot do for general topological spaces). This leads to the notion of completeness of course (every Cauchy net converges to at least one point), although the theory is much cleaner for complete Hausdorff uniform spaces, where you have convergence to at most one point as well.

To illustrate this: the Cauchy completion of a uniform space $X$ can be defined in the usual way via equivalence classes of Cauchy nets. It is a complete Hausdorff uniform space $bar{X}$ together with a map $i: X to bar{X}$ which satisfies a universal property: given a complete Hausdorff uniform space $Y$ and a uniformly continuous function $f: X to Y$, there is a unique uniformly continuous map $bar{f}: bar{X} to Y$ such that $bar{f} circ i = f$. (If you omit "Hausdorff" or "uniformly", you lose the universal property, which is arguably the point of the completion.)

The nLab has an article on uniform spaces with some material not included in the Wikipedia article.

botany - Is there a reason why human eyesight and plants make use of the same wavelength of light?

This question is related to the question: Why are some things transparent and others opaque?

Being able to see something requires that it is opaque and that sufficient light illuminates it.

UV and shorter wavelengths are not as prevalent as visible light on earth. The world would appear too dark to see if we used UV and shorter wavelengths. This is because our atmosphere absorbs most high energy light.

Infrared and longer wavelengths of light pass through many objects which would make vision difficult. There's less light here reaching the earth and even less being refracted.

Think about how much our vision relies on indirect light. The frequencies at which most objects are opaque makes those frequencies useful for vision due to the accumulation of refracted light.

Why are many objects opaque in the visible spectrum of light? Longer wavelengths of light have less energy than the valence electrons on most matter. Shorter wavelengths have too much energy, they cause chemical reactions, in addition to not being very prevalent on the surface of the earth.

Electrons are what absorb then remit light and have thresholds based on their chemistry for what they can absorb. No absorption = transparent. Too much energy and chemical reactions start to happen, which may be undesirable, or desirable in the synthesis of vitamin D by UV light.

Plants extract energy for chemical reactions from wavelengths shorter than infrared, which is too weak to drive photosynthesis, and not as abundant as visible light. But also absorb wavelengths longer than ionizing frequencies, which are not very prevalent, and usually cause damage.

Visible light is the spectrum of light which is prevalent enough on earth to see, but is not so energetic it would harm biological systems. The qualities for optimal light frequencies in sight and photosynthesis overlap because they have similar mechanisms for interacting with light. What is this mechanism? The chemistry of carbon based life.

set theory - Does Cantor-Bernstein hold for classes?

This is an interesting question. I think there are some issues which the others did not mention yet. But I'm not an expert at all, I might be wrong. Please leave me a comment!

In the following, I work in $ZF$. Thus, a class is just a formula. There are some constructions and relations of sets which directly carry over to classes. For example $A=B$ means that the formulas of $A,B$ are equivalent.

Let's sketch the proof for classes. Let $A,B$ classes, $f : A to B$, $g : B to A$ injective maps. Define the classes $A_n subseteq A, B_n subseteq B$ recursively by $A_0=A, B_0 = B, A_{n+1} = g[B_n], B_{n+1}=f[A_n]$. Then $h : A to B$, defined by $f$ on $cap_n A_n cup (cup_n A_{2n} setminus A_{2n+1})$ and by $g^{-1}$ in the rest, is well-defined and a bijection.

Unions, cuts, images of functions etc. are not the problem. But what about the recursion? What we really need here is a recursion scheme for classes. Actually there is a theorem which might be called the transfinite recursion scheme for classes:

Let $R$ be a well-founded and set-like relation of the class $A$ and $F : A times V to V$ a function. Then there is a function $G : A to V$, such that for all $x in A$

$G(x)=F(x,G|_{{y in A : y R x}})$.

However, note that the images of $G$ are sets, not proper classes. We can't use that theorem here.

I think we need a meta-theorem stating that the above also holds when $V$ is replaced by the set of formulas and $R=mathbb{N}$. Also, the meta-world should be able to talk about functions. But this is not plausible at all, since the resulting formula won't have to be finite, right?

For example, try to define a formula $G(n)$ recursively by $G(0)=phi_0$ (doesn't matter what $phi_n$ is) and $G(n)=G(n-1) wedge phi_n$. Why should there be a formula $psi(n,x)$ such that $psi(n,x) Leftrightarrow wedge_{i=0}^{n} phi_i$? I think we need a rather mighty logical calculus for that.

Also note that Francois' great answer here (proving Schröder-Bernstein without the existence of the set $omega$) also causes problems when you want to write down the formula for the part "...$exists s : {0,...,n} to A$ ...". Perhaps there is really no bijection between the two classes mentioned above (class of all singletons, and class of all 2-element sets)?

Does every gland type have stem cells for regeneration?

Stem cells are not all 'unipotent' - they cannot necessarily differentiate into any type of cell. For instance, resident stem cells in tissues such as muscle - myo-satellite cells - are partially differentiated and during cell division one daughter differentiates further to become a myocyte (for example), and the other daughter the replacement myosatellite cell. As far as I'm aware, the majority of tissues have resident stem cells in varying degrees of differentiation (exceptions might include neurons and cardiac muscle, where cells are not replaced over the lifespan of the organism).

Here is some evidence for resident stem cells in glands;

• The adult pituitary gland shows stem/progenitor cell activation in response to injury and is capable of regeneration (Fu et al, 2012),


• The pancreas has been heavily studied from the perspective of diabetes research, with many studies looking to implant stem cells from elsewhere (Rezania et al, 2012). However there is some evidence for resident stem cells of the pancreas that may turn out to be more promising as treatment strategies in the future (Xu et al, 2004, Venkatesan et al, 2011),


• Sebaceous glands are found throughout the skin, and secrete 'sebum' which keeps the skin waxy and waterproof. There is plenty of evidence for epidermal stem cells (Eckert et al, 2012),


• There is also recent evidence for the existence of thyroid stem cells too (Fierabracci, 2012, Malaguarnera et al, 2012),


• Saliva is mainly produced by the submandibular glands also contain resident stem cells, with this paper stating that stem cells have been isolted from salivary glands in humans (Okumura et al, 2012),

Given the limited (and very recent) evidence for the above (relatively well characterizd) glands, it seems likely that other glands will also have resident stem cells that remain to be identified.

Ageing: The so-called stem-cell pool gets depleted over a lifetime, so tissues lose the regenerative potential. However rejuvenation medicine is becoming a heavily invested in area of research (see www.SENS.org).

at.algebraic topology - Why can't the Klein bottle embed in $mathbb{R}^3$?

Assuming that K is smooth or locally flat, there is a nice geometric reason based on transversality.

The Klein bottle is not orientable. If it embedded in R3 so as to separate a small epsilon neighborhood of its image, then the neighborhood would be homeomorphic to a trivial line bundle over the Klein bottle, K times (-1,1), and therefore the neighborhood would also be non-orientable. But since any 3-dimensional submanifold of R3 is orientable this cannnot happen.

So the Klein bottle embedding would have to be non separating in this neighborhood. But this means there is a curve C in the neighborhood that intersects the embedded Klein bottle once, transversally.

Now C bounds a disk D in R3 since R3 is simply connected. (Bounding any surface is enough).
One can make D transverse to the Klein bottle K, leaving boundary D fixed. One then see that K intersects D in a union of arcs and closed curves, with a single boundary point. But any such collection has an even number of boundary points, a contradiction.

This argument works more generally to show a closed nonorientable surface cannot be embedded in a 3-manifold M with H1(M;Z2) = 0.

linear programming - Sorting a binary matrix diagonal in polynomial time while preserving rows

Is there a polynomial time solution to sort an arbitrary binary square matrix in polynomial time by rows so that the diagonal contains a 1 if any row contains a 1 in that column?

For example given matrix:

0 1 1 1 0  r0
1 0 0 1 0  r1
1 1 1 0 1  r2
0 0 0 0 1  r3
0 0 0 1 1  r4

A solution would be:

1 0 0 1 0  r1
1 1 1 0 1  r2
0 1 1 1 0  r0
0 0 0 1 1  r4
0 0 0 0 1  r3

Given a matrix:

1 0 0   r0
0 0 1   r1
1 0 1   r2

There could be multiple solutions:

1 0 0   r0    1 0 1   r2
0 0 1   r1    1 0 0   r0
1 0 1   r2    0 0 1   r1

gel electrophoresis - How often can you reuse coomassie stain?

Since there's a lot of methanol in Coomassie stain, a significant amount would probably evaporate off each time you microwave it. Therefore, it'd be probably a good idea not to reuse it because (1) you're losing methanol, and (2) the concentration of everything else is thrown off.

Why do you boil the stain? I just pour mine at RT right on the gel and it works fine.

differential equations - Does every ODE comes from something in physics?

A fairly silly answer is that the answer is obviously "Yes", since one can build a computer to integrate numerical solutions to your ODE. (now that I think about it, Sigfpe's answer is essentially the same as mine.)

Going along these lines I guess one can find more "physical" models of my suggestion (in the sense that doing physics is often finding toy models that contain the essence of the phenomena etc) by proposing various lattice models or cellular automata which are known to have universal constructors. Or by designing circuits made out of balls and springs.

I spent a little bit of time trying to put down the right words which would make your question more precise and more in line with your intent, but I think ultimately it boils down to what kind of physical models you'd be satisfied with.

As much of physics can be described in terms of ODEs, any sufficiently powerful type of model is going to contain the sort of answer I described above. I think the right question is what's the "simplest" (or perhaps "weakest") known physical model for Painlevé VI.

One kind of answer to that question would be finding a physical system for which some solution of Painlevé VI gives some physically measurable function - along these lines, I know that Painlevé functions are highly useful in various integrable models / lattice models, e.g. famously, the spin-spin correlation function in the 2D Ising model in the scaling limit is a solution to Painlevé III - thus, my guess is that Painlevé VI shows up in one of these contexts, but the literature is pretty vast.

at.algebraic topology - Stable ∞-categories as spectral categories

Let C be a stable ∞-category in the sense of Lurie's DAG I. (In particular I do not assume that C has all colimits.) Then C does have all finite colimits, the suspension functor on C is an equivalence, and C is enriched in Spectra in a way I don't want to make too precise (basically the Hom functor C^op × C → Spaces factors through Spectra and there are composition maps on the level of spectra).

Now suppose instead that C is an ∞-category which has all finite colimits and comes equipped with an enrichment in Spectra in the above sense. One can show easily that C then has a zero object which allows us to define a suspension on C. Suppose it is an equivalence. Is C then a stable ∞-category? Moreover, is the enrichment on C the one which comes from the fact that it is a stable ∞-category?

gr.group theory - Two finite groups with the same identical relations?

Two finite-dimensional prime algebras (in a certain extended sense, what includes the usual algebras with binary multiplication over a field) with the same identities are isomorphic over the algebraic closure of the ground field (Yu.P. Razmyslov, Identities of Algebras and Their Representations, AMS, 1994, p. 30 onwards). This suggests that the proper condition one should impose on finite groups to guarantee that the same identities imply isomorphism, would be something related to primeness.

Unfortunately, Razmyslov's reasonings seem not be extendible to a broader class of algebraic systems, in particular, to groups: the linear structure is crucial there (a relatively free algebra in the corresponding variety is enlarged, via the action of a suitable extension of the ground field, to an algebra which is isomorphic to an extension of the initial algebra). Also, it is not clear what the group-theoretic analog of primeness should be in this context.

However, there is an old result saying that two finite simple groups with the same identities are isomorphic (H. Neumann, Varieties of Groups, p. 166, Corollary 53.35). I believe that the machinery developed in that book (critical groups, etc.) would allow to answer this question for any reasonably defined class of finite groups.

nt.number theory - Is the product of first $n$ prime numbers $+1$ another prime number?

Here's a possible intended solution to show that $30031 = 2 cdot 3 cdot 5 cdot 7 cdot 11 cdot 13 + 1$ is composite without factoring it. Recall the Fermat primality test: if $a^{n-1} not equiv 1 bmod n$, then $n$ cannot be prime. It turns out that $2^{30030} equiv 21335 bmod 30031$, so $30031$ must indeed be composite. There is a well-known algorithm called binary exponentiation that is reasonably fast to implement by hand and that could conceivably be done on an exam. (I am not totally convinced that this would be faster than trial division until $p = 59$, though. And if you followed Leonid's suggestion in the comments your life would be even easier.)

Edit: Here is the solution by trial division. It suffices to check the primes from $17$ up. Note that the fact that we know the prime factorization of $30030$ helps a lot. All of the arithmetic necessary beyond what I wrote down was mental.

For $17$ write $30031 = 30 cdot 77 cdot 13 + 1 equiv -4 cdot 9 cdot -4 + 1 equiv -8 bmod 17$.

For $19$ write $30031 equiv -8 cdot 1 cdot -6 + 1 equiv 11 bmod 19$.

For $23$ write $30031 equiv 7 cdot 8 cdot -10 + 1 equiv -99 bmod 23$.

For $29$ write $30031 equiv 1 cdot -10 cdot 13 + 1 equiv -100 bmod 29$.

For $31$ write $30031 equiv 30000 bmod 31$.

For $37$ write $30031 equiv -7 cdot 3 cdot 13 + 1 equiv -13 bmod 37$.

For $41$ write $30031 equiv -11 cdot 5 cdot 13 + 1 equiv -14 cdot 13 + 1 equiv -140 bmod 41$.

For $43$ write $30031 equiv -13 cdot -9 cdot 13 + 1 equiv -26 bmod 43$.

For $47$ write $30031 = 210 cdot 143 + 1 equiv 25 cdot 2 + 1 equiv 4 bmod 47$.

For $53$ write $30031 equiv -2 cdot -16 + 1 equiv 33 bmod 53$.

Finally, for $59$ write $30031 equiv 33 cdot 25 + 1 equiv 11 cdot 16 + 1 equiv 0 bmod 59$.

nt.number theory - Reference for Deligne's construction of Galois representations attached to modular forms

I absolutely agree that Deligne is very terse. One thing that I ultimately found very helpful is Carayol's two papers where he proves the analogous theorem for Hilbert modular forms. I say "ultimately" because it took me a long time to read those papers. I would come back to them every few years and learn more, as I matured mathematically. The big problems with using Carayol to understand Deligne will be: (1) Carayol has to work much harder in places than Deligne, because the Shimura curves he uses are not the solution to a moduli problem of abelian varieties plus extra structure, so he has to use extra tricks which Deligne did not have to get into, and this will obfuscate things (I guess perhaps this is only when analysing the bad reduction of the curves, which is perhaps not the paper you'd be wanting to read anyway) and (2) Deligne had to deal with the fact that modular curves need compactifying, so he had to work with parabolic cohomology, which is a technicality he has to deal with and Carayol doesn't. But for the main part, the techniques are the same.

structural biology - How do you turn a minimal CIF description into a complete one?

I have a CIF file that I downloaded from the PDB, but if I try to use it in Coot, it complains that it is not a complete CIF definition. This page provides a batch script that I may be able to tease apart to suit my needs, but I thought there was a way to do this from within the CCP4 Program Suite GUI/job manager.

Maddeningly, I see in my project job history a couple "dictionary" jobs where I did this, but I utterly forgot how I set them up (forgive the asinine names):

When I try to use the Monomer Library Sketcher (under Refinement/Restraint Preparation) and open a CIF ("mmCIF") file, nothing happens.

Also under the sketcher, I can severely abuse the "Load Monomer from Library" and it looks like I get the right thing:

However, when setting up the above, it throws "Error: can't read "monomer_lib(code,n..." warnings, then when running it generates the Geometry file (seemingly successfully) then says that Libcheck failed when trying to create the coordinate file.

---

If there are other solutions, I'd be interested as well. I've used PRODRG to create CIF files before with its draw tool, but it can be a bit clumsy, it ignores how I drew my double bonds, and it creates new names for all the atoms.

A learning roadmap for Representation Theory

I second the suggestion of Fulton and Harris. It's a funny book, and definitely you want to keep going after you finish it, but it's a good introduction to the basic ideas.

You specifically might be happier reading a book on algebraic groups.

While I third the suggestion of Ginzburg and Chriss, I wouldn't call it a "second course." Maybe if what you really wanted to do was serious, Russian-style geometric representation theory, but otherwise you might want to try something a little less focused, like Knapp's "Lie Groups Beyond an Introduction."

If you want Langlandsy stuff, then Ginzburg and Chriss is actually a bit of a tangent; good enrichment, but not directly what you want, since it skips over all the good stuff with D-modules. Look in the background reading for the graduate student seminar we're having in Boston this year: http://www.math.harvard.edu/~gaitsgde/grad_2009/

reference request - Books on reductive groups using scheme theory

Personally, I find the "classical" books (Borel, Humphreys, Springer) unpleasant to read because they work in the wrong category, namely, that of reduced algebraic group schemes rather than all algebraic group schemes. In that category, the isomorphism theorems in group theory fail, so you never know what is true. For example, the map $H/Hcap Nrightarrow HN/N$ needn't be an isomorphism (take $G=GL_{p}$, $H=SL_{p}$, $N=mathbb{G}_{m}$ embedded diagonally). Moreover, since the terminology they use goes back to Weil's Foundations, there are strange statements like "the kernel of a homomorphism of algebraic groups defined over $k$ need not be defined over $k$". Also I don't agree with Brian that if you don't know descent theory, EGA, etc. then you don't "know scheme theory well enough to be asking for a scheme-theoretic treatment'.

Which explains why I've been working on a book whose goal is to allow people to learn the theory of algebraic group schemes (including the structure of reductive algebraic group schemes) without first reading the classical books and with only the minimum of prerequisites (for what's currently available, see my website under course notes). In a sense, my aim is to complete what Waterhouse started with his book.

So my answer to the question is, no, there is no such book, but I'm working on it....

ag.algebraic geometry - A book on locally ringed spaces?

A locally ringed space is nothing but a local ring object (in the internal sense) in a category of sheaves over a topological space, which happens to be an example of a topos.

So in order to study general properties of locally ringed spaces, one could proceed by studying properties of local ring objects in arbitrary topoi. As any proposition (in the internal language) on a local ring (a commutative ring with $0 neq 1$ and $s + t = 1 implies s in R^times lor t in R^times$) is true for any topos if and only if it can be derived intuitionistically (which is more or less the same as constructively), the original questions seems to boil down to:

"What are the constructively valid properties and constructions for a local ring?"

For example, the construction of Kähler differentials makes also sense constructively, which immediately implies (by the above reasoning) that every morphism $X to Y$ of locally ringed spaces has an associated module $Omega_{X/Y}$ of Kähler differentials.

And there is a lot of literature on constructive algebra. The book of Mines, Richman and Ruitenburg as well as many of the preprints on Fred Richman's homepage are a start. Some material can also be found in "Sheaves in Geometry and Logic" by Mac Lane and Moerdijk.

gr.group theory - Are the inner automorphisms the only ones that extend to every overgroup?

I struggled with the same question for quite some time and solved it for finite groups, only to then discover that it had already been solved. Schupp solved it for arbitrary groups. Martin Pettet later solved it for finite groups, and his proof works for p-groups, finite p-groups, finite pi-groups, solvable groups, etc.

1. On inner automorphisms of finite groups by Martin R. Pettet, Proceedings of the American Mathematical Society, Volume 106,Number 1, Page 87 - 90(May 1989)




2. Characterizing inner automorphisms of groups by Martin R. Pettet, Archiv der Mathematik, Volume 55,Number 5, Page 422 - 428(Year 1990)

These proofs also show that analogous statements are true if we replace injective embeddings with quotient maps (i.e., the only automorphisms that can be pulled back over all quotient maps are the inner ones).

I also have some notes on this and similar problems here: extensible automorphisms problem.

Is a polynomial with 1 very large coefficient irreducible?

Maybe this should have gone in the comments, but I couldn't see the button.

In any case, I'm wondering what you hope to be true. There are some obvious 'bad' examples (i.e. $10^{20} x^{2} - 1$, or $x^{2} - 10^{20}$, or $x^{2} - 2 10^{10} x + 10^{20}$) s.t. some coefficients can be arbitrarily larger than (any function of) the others, while the polynomial remains reducible. This isn't terrible (I can't come up with examples like this for all coefficients and all degrees), but certainly means you can't get a condition which just involves the largest coefficient, without regards to spacing.

There are also some 'nice' examples. In the same paper that he proves the criterion you mention above, Perron also proves that a polynomial is irreducible if $a_{n-2}$ is sufficiently larger than the rest.

The paper 'irreducibility of polynomials' by Dorwart (from the monthly in 1935 (!)) came up on a quick google search, and may be worth looking at.

For the last question, playing around with the various divisibility criteria (and Maple) seems to give many, many examples for moderate degree, but my algebra is not strong enough to turn this into a theorem. Of course, if you are only interested in infinitely many n (not all n, since this only works for n being a prime - 1), the cyclotomic polynomials seem like good examples, with all coefficients 1! Is there any reason that you believe a restriction on the size of the coefficients would do something?

human biology - Could an "overactive" brain increase the chances of Alzheimer's Disease?

I think the article that quoted(1) is different to what the link shows now(2). However, both the article quoted and the one in the link are relevant to the question.

Alzheimer's Disease : In a nutshell, AD is the most common form of dementia. Although the exact cause is still under investigation, several genes have been implicated in the development of AD. One of the things you see in the brain with disease progression is beta amyloid plaque deposition which disrupts brain activity.

Brain Activity : While the brain consumes around 20% of our body's energy at a baseline, an increase in brain activity only increases this consumption by around 5%. Hans Berger, the first to record an EEG stated that "mental work, as I explained elsewhere, adds only a small increment to the cortical work which is going on continuously and not only in the waking state".(2)

However, there are a few 'intrinsic' networks or regions that perform functions such as "information processing for interpreting, responding to and predicting environmental demands". One of the areas active in this state of the brain is referred to as the 'default mode network' (DMN). It's thought that DMN is the main target of the amyloid plaques responsible for Alzheimer's Disease.(2)

However, there are other multimodal regions of the brain that perform processing of various types of information. It's now suggested that "regions vulnerable to Ab deposition do not simply involve the DMN, but rather comprise multimodal brain regions that are highly interconnected, plastic, and capable of rapid ATP generation"(1).

So it's a variation of activity in these particular areas - not all - that are implicated with Alzheimer's Disease. When you are performing a task, activity in these areas diminish.

Why keep the brain intentionally active : The reason for the hypothesis that actively engaging in challenging activity may reduce the risk of AD is that individuals with a high cognitive reserve have been shown to cope with the pathological changes such as beta amyloid deposition better. Cognitive reserve has been linked to a bunch of factors including "education, occupation, socioeconomic status, social networks and lifelong participation in cognitive and physical activity" and reflects "lifelong patterns of behaviors, endogenous factors (including genetics) and exposure to environmental factors"(1).

Moreover, it's suggested that perhaps higher cognitive reserve increases neuronal efficiency which has been shown to reduce plaque deposition.(1).

So InquilineKea's question and jp89's answer aren't really contradictory.

1. August, et al. Lifespan brain activity, β-amyloid, and Alzheimer's disease.
   Trends in Cognitive Sciences. 2011. 15(11):520-526.




2. Raiche. Two views of the brain.
   Trends in Cognitive Sciences. 2010. 14(4):180-190.




3. Hafkemeijer, et al. Imaging the default mode network in aging and dementia.
   Biochimica et Biophysica Acta (BBA) - Molecular Basis of Disease. 2012. 1822(3):431-441.




4. Harrison's Principles of Internal Medicine. 16th Ed. McGraw Hill.

evolution - How many times did life emerge from the ocean?

I presume you mean how many times did life emerge from the ocean? ("how often" implies you want to know the regularity). Anyway, great question. I really enjoyed reading and thinking about it.

I doubt we know the precise number, or even anywhere near it. But there are several well-supported theorised colonisations which might interest you and help to build up a picture of just how common it was for life to transition to land. We can also use known facts about when different evolutionary lineages diverged, along with knowledge about the earlier colonisations of land, to work some events out for ourselves. I've done it here for broad taxonomic clades at different scales - if interested you could do the same thing again for lower sub-clades.

As you rightly point out, there must have been at least one colonisation event for each lineage present on land which diverged from other land-present lineages before the colonisation of land. Using the evidence and reasoning I give below, at the very least, the following 9 independent colonisations occurred:

• bacteria


• cyanobacteria


• archaea


• protists


• fungi


• algae


• plants


• nematodes


• arthropods


• vertebrates

Bacterial and archaean colonisation
The first evidence of life on land seems to originate from 2.6 (Watanabe et al., 2000) to 3.1 (Battistuzzi et al., 2004) billion years ago. Since molecular evidence points to bacteria and archaea diverging between 3.2-3.8 billion years ago (Feng et al.,1997 - a classic paper), and since both bacteria and archaea are found on land (e.g. Taketani & Tsai, 2010), they must have colonised land independently. I would suggest there would have been many different bacterial colonisations, too. One at least is certain - cyanobacteria must have colonised independently from some other forms, since they evolved after the first bacterial colonisation (Tomitani et al., 2006), and are now found on land, e.g. in lichens.

Protistan, fungal, algal, plant and animal colonisation
Protists are a polyphyletic group of simple eukaryotes, and since fungal divergence from them (Wang et al., 1999 - another classic) predates fungal emergence from the ocean (Taylor & Osborn, 1996), they must have emerged separately. Then, since plants and fungi diverged whilst fungi were still in the ocean (Wang et al., 1999), plants must have colonised separately. Actually, it has been explicitly discovered in various ways (e.g. molecular clock methods, Heckman et al., 2001) that plants must have left the ocean separately to fungi, but probably relied upon them to be able to do it (Brundrett, 2002 - see note at bottom about this paper). Next, simple animals... Arthropods colonised the land independently (Pisani et al, 2004), and since nematodes diverged before arthropods (Wang et al., 1999), they too must have independently found land. Then, lumbering along at the end, came the tetrapods (Long & Gordon, 2004).

Note about the Brundrett paper: it has OVER 300 REFERENCES! That guy must have been hoping for some sort of prize.

References

• Battistuzzi FU, Feijao A, Hedges SB. 2004. A genomic timescale of prokaryote evolution: insights into the origin of methanogenesis, phototrophy, and the colonization of land. BMC Evol Biol 4: 44.


• Brundrett MC. 2002. Coevolution of roots and mycorrhizas of land plants. New Phytologist 154: 275–304.


• Feng D-F, Cho G, Doolittle RF. 1997. Determining divergence times with a protein clock: Update and reevaluation. Proceedings of the National Academy of Sciences 94: 13028 –13033.


• Heckman DS, Geiser DM, Eidell BR, Stauffer RL, Kardos NL, Hedges SB. 2001. Molecular Evidence for the Early Colonization of Land by Fungi and Plants. Science 293: 1129 –1133.


• Long JA, Gordon MS. 2004. The Greatest Step in Vertebrate History: A Paleobiological Review of the Fish‐Tetrapod Transition. Physiological and Biochemical Zoology 77: 700–719.


• Pisani D, Poling LL, Lyons-Weiler M, Hedges SB. 2004. The colonization of land by animals: molecular phylogeny and divergence times among arthropods. BMC Biol 2: 1.


• Taketani RG, Tsai SM. 2010. The influence of different land uses on the structure of archaeal communities in Amazonian anthrosols based on 16S rRNA and amoA genes. Microb Ecol 59: 734–743.


• Taylor TN, Osborn JM. 1996. The importance of fungi in shaping the paleoecosystem. Review of Palaeobotany and Palynology 90: 249–262.


• Wang DY, Kumar S, Hedges SB. 1999. Divergence time estimates for the early history of animal phyla and the origin of plants, animals and fungi. Proc Biol Sci 266: 163–171.


• Watanabe Y, Martini JEJ, Ohmoto H. 2000. Geochemical evidence for terrestrial ecosystems 2.6 billion years ago. Nature 408: 574–578.

gn.general topology - What is a monoidal metric space?

For the general question about how categorical concepts look when applied to metric spaces, one place to look is Lawvere's paper 'Taking categories seriously', section 6 onwards.

Aside from that, here are a few examples.

Functor categories became function spaces with the uniform or sup metric. That is, if $A$ and $B$ are metric spaces construed as enriched categories, then the functor category $B^A$ is the set of distance-decreasing maps $A to B$ with the sup metric. (I use "decreasing" in the non-strict sense.)

The (cartesian) product $A times B$ of two metric spaces --- that is, their product in the category of metric spaces --- has the '$infty$-metric':
$$
d((a, b), (a', b')) = max{d(a, a'), d(b, b')}.
$$
The same goes for infinite products --- remembering that $infty$ is allowed as a distance. Once you know this, limits in general work in the obvious way.

The coproduct $A + B$ of two metric spaces $A$ and $B$ is their disjoint union, with $d(a, b) = d(b, a) = infty$ for all $a in A, b in B$. Again, it's crucial here to allow $infty$ as a distance. Otherwise, your category of metric spaces will lack lots of limits and colimits. The coequalizer of two maps $f, g: A to B$ is $B$ quotiented out by the usual equivalence relation $sim$ (as in the category of sets), and metrized by
$$
d([b], [b']) = inf{ d(y, y'): y sim b, y' sim b'}
$$
where $[b]$ denotes the equivalence class of $b$. General colimits work similarly.

I mentioned the cartesian product, but there's another kind of product. Generally, if $mathbf{V}$ is a monoidal category then any two $mathbf{V}$-enriched categories, $A$ and $B$, have a tensor product $A otimes B$. Its set of objects is the product of the sets of objects of $A$ and $B$. Its hom-objects are given by
$$
(A otimes B)((a, b), (a', b')) = A(a, a') otimes B(b, b').
$$
This gives us a tensor product of metric spaces. Given metric spaces $A$ and $B$, the point-set of $A otimes B$ is the product of the point-sets of $A$ and $B$. The distance is given by
$$
d((a, b), (a', b')) = d(a, a') + d(b, b').
$$
In other words, it's the '$1$-metric', also known as the taxicab metric, Manhattan metric, etc.

So, Andrew, when you ask 'What is a monoidal metric space?', you have to say which product you want to be monoidal with respect to. That is, are you asking about (weak) monoids in $(mathbf{Met}, times)$ or in $(mathbf{Met}, otimes)$?

From the tone of your question, I would guess: both. So here goes.

The answer for cartesian product $times$ doesn't seem so interesting. Assuming that your metric spaces satisfy the classical skeletality axiom ($d(a, b) = 0 Rightarrow a = b$), a monoidal category for $times$ is a metric space $A$ equipped with a monoid structure on its set of points such that
$$
d(a cdot b, a' cdot b') leq max{d(a, a'), d(b, b')}.
$$
I can't think of anything more to say about that.

The answer for tensor product $otimes$ seems more interesting. A monoid in $(mathbf{Met}, otimes)$ is a metric space $A$ equipped with a monoid structure on its set of points such that for all $a$, the maps $acdot -$ and $- cdot a$ are distance-decreasing. For example, if it's a group then this says that left or right translation is always an isometry. This often happens: consider the underlying additive group of a normed vector space, for instance.