Saturday 26 January 2008

genetics - Are there dog breeds that are so far apart genetically that they can't produce viable offspring?

I actually found some sort of reference for this. Apparently in the case of a beagle and irish setter pairing, they had a lot of difficulty in producing pups, but as breeds have a lot of genetic quirks, this might be due to a genetic accident; this is probably a case of mutual infertility rather than speciation. There are probably mutations in their genomes which are causing non viability in the pups the way that it does with some human parents who have difficulty having children - an immunological incompatibility or traits which cause most offspring non-viable.



In fact the discussion points out that the species designation does not always mean they cannot produce offspring, but simply do not. Coyotes and wolves will produce offspring too if they are encouraged to do so, but are competing and antagonistic in the wild and so never do mate in practice.



So the answer is probably no. (the beagle/setter pups did show up, but with a much smaller litter than usual with just 2 pups).

Wednesday 23 January 2008

biochemistry - Does caffeine increase the speed at which sperm travels?

The cell response for caffeine, is at the molecular level of RyR (Ryanodine receptor), and the main effect could be a temporal rise in Ca+2 cytosolic concentration, so it can change the electrical behavior for the whole cell, and then affect their mobility. But the main area for RyR research is at spermatogenesis, there are a lot of papers like this one, for that topic

co.combinatorics - Hamilton cycle decompositions of the complete graph

Just reporting that I wrote another algorithm for this and found the following values:



3 1
5 6
7 960
9 40037760


I ran this through the superseeker on Sloane and it came up with nothing (so perhaps nobody has counted these before).



Here's my code below (it uses GAP). We generate a (n-1) x n Latin rectangle where each row is an n-cycle and the i-th and (i+(n-1)/2)-th rows are inverses.



EnumerateHamiltonDecompositionsBacktrackingAlgorithm:=function(n,L,step)
local i,j,k,count,A;
i:=Int((step-1)/n)+1;
j:=(step-1) mod n+1;
count:=0;

if(n mod 2=0 or n<3) then return fail; fi;
if(j=1) then A:=[Minimum(Filtered([2..n],i->ForAll([1..n-1],t->L[t][1]<>i)))]; else A:=Filtered([1..n],s->ForAll([1..n-1],t->L[t][j]<>s) and ForAll([1..n],t->L[i][t]<>s)); fi;
for k in A do
L[i][j]:=k;
L[i+(n-1)/2][k]:=j;
if((j=n and CycleLengths(PermList(L[i]),[1..n])=[n]) or j<n) then
if(i=(n-1)/2 and j=n) then count:=count+1;
else count:=count+EnumerateHamiltonDecompositionsBacktrackingAlgorithm(n,L,step+1); fi;
fi;
L[i][j]:=0;
L[i+(n-1)/2][k]:=0;
od;
return count;
end;;

EnumerateHamiltonDecompositions:=function(n)
local L;
if(n mod 2=0 or n<3) then return fail; fi;
if(n=3) then return 1; fi;
L:=List([1..n-1],i->List([1..n],j->0));
L[1]:=List([1..n],i->i mod n+1);
L[1+(n-1)/2]:=ListPerm(Inverse(PermList(List([1..n],i->i mod n+1))));
return Factorial(n-2)*EnumerateHamiltonDecompositionsBacktrackingAlgorithm(n,L,n+1);
end;;


The extra data point comes from assuming that (12..n) is one of the cycles, then multiplying the result by (n-2)!. This is legitimate since each decomposition contains a unique cycle with the edge 12, and by permuting the remaining n-2 edges, we generate a unique decomposition with the cycle (12..n). There are no automorphisms under this group action, so each orbit has cardinality (n-2)!.

Tuesday 22 January 2008

fa.functional analysis - L_p norm balls for 12?

The L_1 ball in 2D is shaped like a diamond (L_1 is also known as the Manhattan norm). The L_∞ ball is shaped like a square (L_∞ is also known as the supremum norm). They are similar, i.e. have same shape. The L2 ball is shaped like a circle.



Hypothesis: For all p in the interval (1,2), there is q>2 such that the q-ball and the p-ball are similar. One further hypothesis is that this occurs precisely when p,q are Hölder conjugates.



I wasn't sure how to tag this.

microscopy - Online repositories of scanning electron microscope photographs?

First off I'd like to reccomend the University of Dartmouth's publicly available collection located here. They have both SEM and TEM images of a wide range of organisms and cells from algae to see urchins through everything from cholera to mammalian cells. Images are high quality, fully captioned and properly attributed.



I'm a little confused as to the problem you're having with google - see my original comment on your question. However I have looked at the photos of Wikimedia Commons in the SEM category and see that they are either images taken with, images of SEM Microscopes or technical drawings regarding their function. I assume that you're looking for the images taken with SEM.



I was able to find a great number of similar images to those in the commons with a Google Image search - I read above that you were having problems with this approach so perhaps try the search term "SEM images" in image search (or this search to avoid duplications from images you have already seen on Wiki Commons)?



Examples found on just the first page:
Mushroom Spores (Uni. Dartmouth Ash Grain (USGS) - not biological but interesting nonetheless

cell biology - Is "exhaustion" of the Hodgkin-Huxley membrane at constant stimulation a real phenomenon?

(I probably ought to have a pat answer to this on the tip of my mind, but since I don't I'm going to wing it. This is probably just an opportunity to make an utter fool of myself. Please treat everything that follows with extreme suspicion.)



I think this is effectively an artefact of the model. That may not be true in the strictest sense -- it is possible such behaviour could be produced in real experimental preparations -- but it would require driving them in drastically non-physiological ways. I am not aware of this having been done, but I'm sure that's just my ignorance -- I would be surprised if nobody has tried.



In physiological terms, though: where would such a large constant current come from in a real cell? Where would the charges go? How would that be sustained with realistic boundary conditions?



While it is possible that such an effect can come into play very transiently in living systems, it seems unlikely that a true steady state of this kind could ever be reached.

Monday 21 January 2008

at.algebraic topology - Does homology detect chain homotopy equivalence?

Yes, this is true. Suppose $C_*$ is such a chain complex of free abelian groups.



For each $n$, choose a splitting of the boundary map $C_n to B_{n-1}$, so that $C_n cong Z_n oplus B_{n-1}$. (You can do this because $B_{n-1}$, as a subgroup of a free group, is free.) For all $n$, you then have a sub-chain-complex $cdots to 0 to B_n to Z_n to 0 to cdots$ concentrated in degrees $n$ and $n+1$, and $C_*$ is the direct sum of these chain complexes.



Given two such chain complexes $C_*$ and $D_*$, you get a direct sum decomposition of each, and so it suffices to show that any two complexes $cdots to 0 to R_i to F_i to 0 to cdots$, concentrated in degrees $n$ and $n+1$, which are resolutions of the same module $M$ are chain homotopy equivalent; but this is some variant of the fundamental theorem of homological algebra.



This is special to abelian groups and is false for modules over a general ring.

Algebra / unital associative algebra: better terminology?

If your work requires you to work mostly with unital assosiative $k$-algebras with unital morphisms, define somewhere prominent in your work 'algebra' and 'morphism of algebras' to mean precisely that, and say 'Lie algebras', 'not necessarily associative algebra', 'possibly non unital morphism of algebra', and so on in the meagre 5% of the remaining cases.

nt.number theory - Irrational logs and the harmonic series

Consider the series
$$ S_f = sum_{x=1}^infty frac{f}{x^2+fx}. $$
Goldbach showed that, for integers $f ge 1$,
$$ S_f = 1 + frac12 + frac13 + ldots + frac1f $$
(this follows easily by writing $S_f$ as a telescoping series).
Thus $S_f$ is rational for all natural numbers $f ge 1$.
Goldbach claimed that, for all nonintegral (rational) numbers $f$,
the sum $S_f$ would be irrational.



Euler showed, by using the substitution
$$ frac1k = int_0^1 x^{k-1} dx, $$
that
$$ S_f = int_0^1 frac{1-x^f}{1-x} dx. $$
He evaluated this integral for $f = frac12$ and found
that $S_{1/2} = 2(1 - ln 2)$ (this also follows easily from
Goldbach's series for $S_f$). Thus Goldbach's claim holds for all
$f equiv frac12 bmod 1$ since $S_{f+1} = S_f + frac1{f+1}$.



Here are my questions:



  1. The irrationality of $ln 2$ was established by Lambert, who
    proved that $e^r$ is irrational for all rational numbers
    $r ne 0$. Are there any (simple) direct proofs?


  2. Has Goldbach's claim about the irrationality of $S_f$ for
    nonintegral rational values of $f$ been settled in other cases?


Sunday 20 January 2008

ag.algebraic geometry - Geometric Intuition for Big Monodromy

In various contexts, I have come across results referred to as "big monodromy." A standard arithmetic example is the open image theorem for the image of Galois action on non-CM elliptic curves. A general setup for such a result in algebraic geometry is:



Given a proper, generically smooth map $pi:X rightarrow S$ of relative dimension d, say S is connected. This gives rise to an $l$-adic representations of the etale fundamental group $pi_1(U)$ where $U$ is smooth locus of $pi$ corresponding to higher pushforward $R^d pi_* Q_l$. One might say it has "big monodromy" if the Zariski closure of the image is as big as it can be given that it has to respect cup-product, etc.



My specific question is what are the geometric consequences of big monodromy? If we know such a result for $pi$, what does that say about the geometry of the fibration or at the very least is there geometric intuition for what it should mean?



I welcome intuition from number theory, algebraic geometry, or complex geometry.



I have also heard that "one should expect big monodromy unless there is a reason not to" (for example, complex multiplication). What are other examples of things which inhibit big monodromy?

linear algebra - Spheres over rational numbers and other fields

Any orthonormal set extends to an orthonormal basis, over any field
of characteristic not $2$. This is a special case of Witt's theorem.



EDIT: In response to Vipul's comment: The proof of Witt's theorem is constructive, and leads to the following recursive algorithm for extending an orthonormal set $lbrace v_1,ldots,v_r rbrace$ to an orthonormal basis.



Let $e_1,ldots,e_n$ be the standard basis of $K^n$, where $e_i$ has $1$ in the $i^{operatorname{th}}$ coordinate and $0$ elsewhere. It suffices to find a sequence of reflections defined over $K$ whose composition maps $v_i$ to $e_i$ for $i=1,ldots,r$, since then the inverse sequence maps $e_1,ldots,e_n$ to an orthonormal basis extending $v_1,ldots,v_r$. In fact, it suffices to find such a sequence mapping just $v_1$ to $e_1$, since after that we are reduced to an $(n-1)$-dimensional problem in $e_1^perp$, and can use recursion.



Case 1: $q(v_1-e_1) ne 0$, where $q$ is the quadratic form. Then reflection in the hyperplane $(v_1-e_1)^perp$ maps $v_1$ to $e_1$.



Case 2: $q(v_1+e_1) ne 0$. Then reflection in $(v_1+e_1)^perp$ maps $v_1$ to $-e_1$, so follow this with reflection in the coordinate hyperplane $e_1^perp$.



Case 3: $q(v_1-e_1)=q(v_1+e_1)=0$. Summing yields $0=2q(v_1)+2q(e_1)=2+2=4$, a contradiction, so this case does not actually arise.

Saturday 19 January 2008

evolution - Genetic Models for Natural Selection?

The rate of mutation has been studied since the early 1900s, which is when your question was put to rest in a prescient paper by JBS Haldane (1949). Essentially the answer is that it varies a lot between organisms and environmental conditions, but for any given gene a mutation occurs once in about 50,000 sexual replications. That is equivalent to, in a human, several hundred new mutations in each person born. Impressively, Haldane's calculations have since been confirmed by a variety of other methods (e.g. Nachman & Crowell, 2000; Xue et al., 2009).



Faster rates of evolution have been recorded in other organisms, and there have been many, many studies looking at the rate of increase in sizes of the whole body or of particular organs of a wide variety of animals and plants.



For example, Azevedo et al. (2002) look at the rate of mutations affecting size (trying to select for decrease in body size) in C. elegans, and found mutations affecting size occured at a rate of ~0.0025 per haploid genome per generation.



In another example, with a result more definite regarding your question, Keightley (1998) investigated the rate of change in average body weight under conditions of selection for increased body size, over 50 generations in mice. They recorded an average increase of 0.23-0.57% body weight per generation from new mutations. That's pretty huge, and demonstrates how rapidly a complex trait can change by mutation under selection.



I wrote a small piece of python code to calculate the increase after x generations at the same rate (the formula is very simple)...



initial = 1
rate = 0.35
generations = 1000
result = initial * (1.0 + rate) ** generations
print result
>>>20.10655586861808


Thus if we assume that approximately the average rate found in the mouse study (0.35%) could be the rate at which giraffes' necks might increase under strong selection, after 1000 generations the mean neck length would be over 20 times the original length. A low estimate for wild giraffe longevity is around 20 years, thus for giraffe necks to increase around 20x might take 20,000 years. Of course this basic estimate is making a whole lot of assumptions, but the point is that rates of mutation are actually very high, easily enough to explain the diversity of traits we see in the world.



References



Friday 18 January 2008

zoology - Examples of animals that co-opt toxins?

Glaucus atlanticus consumes and reuses the nematocysts of jellyfish siphonophores. Perhaps not quite what you were looking for, as the animal doesn't simply concentrate the toxin, but actually co-opts entire stinging cells.



Further to your mention of dinoflagellate toxins being concentrated by fish, I seem to recall reading something about some toxins being modified in vivo into more toxic derivatives, but cannot find any mention of it.

Thursday 17 January 2008

What can be said about the homotopy groups of a CW-complex in terms of its (co)homology?

Try looking up some references on rational homotopy theory. Rational homotopy theory studies the homotopy groups tensor Q, so basically you kill all torsion information. If we focus only on homotopy groups tensor Q, the question you ask becomes easier. As Steven Sam mentions in the comments, the homotopy groups of spheres are really crazy. But the rational homotopy groups of spheres are quite tractable (in fact completely known, by a theorem of Serre) and can be more or less obtained from cohomology, if I recall correctly.



One particularly impressive theorem, of Deligne-Griffiths-Morgan-Sullivan, says that if your space is a compact Kähler manifold (e.g. a smooth complex projective variety), and if you know its rational cohomology ring, then you can compute for instance the ranks of all of its homotopy groups (maybe you need an extra assumption that the space is simply connected or has nilpotent fundamental group).

Wednesday 16 January 2008

gr.group theory - Abelianization of a semidirect product

I agree with Ryan and Victor, except that you don't need presentations. The subgroup $[G ltimes H,G ltimes H]$ is generated by $[H,H] cup [G,H] cup [G,G]$, so you can write
$$(G ltimes H)^{ab} = (G ltimes H) / langle [H,H] cup [G,H] cup [G,G] rangle.$$
If you apply the relators $[H,H]$, you get $G ltimes H^{ab}$; then if you apply the relators $[G,H]$, you get $G times (H^{ab})_G$; then finally if you apply $[G,G]$, you get $G^{ab} times (H^{ab})_G$. You can add this as an extra half-paragraph or footnote rather than giving a citation.



I don't think that the referee has the right to demand a longer explanation than this, unless maybe you are writing a textbook.

evolution - Why is 'Grudger' an evolutionary stable strategy?

I am currently reading 'The Selfish Gene' by Richard Dawkins, which I am sure many here have read. The topic are evolutionary stable strategies (ESS) regarding cooperation.



I apologise for the long question. If you are already familiar with the topic and Dawkins' model of Cheat, Sucker and Grudger: my question is, how can Grudger be an ESS if it could be invaded both by Suckers (because they have no disadvantage against Grudger) and Cheats (because a Cheat minority is unlikely to meet the same Grudger twice, turning Grudger into Sucker effectively)?



More detailed:



The model



Near the end of chapter 10 (p 185 in my version), Dawkins uses a model of birds who clean each other of parasites, therefore helping in survival (as cleaning themselves they cannot reach every spot of their body). He defines three different behaviours for the model:



  • Sucker - birds who indiscriminately help and clean other birds

  • Cheat - birds who let others help them but never do so themselves

  • Grudger - birds who help others and remember who they helped. If the same bird does not help them later (reciprocate), they will not help that bird again.

Claim: Cheat and Grudger are ESS



He claims that both Cheat and Grudger in themselves are ESS - that is, if all birds behave this way, none of the other behaviours can develop because they will be immediately penalised by lower chances of reproducing.



The part that makes sense: Suckers is not an ESS, Cheat is



Sucker is of course not an ESS. If all birds were Suckers, any Cheat that developed would have a huge reproductive advantage and Cheat genes would overtake the population.



Being an ESS makes sense for Cheat. If all birds cheat, nobody will ever be helping each other. A minority of Suckers would be spending all their time helping and not getting anything in return, Cheats have the advantage and Suckers die out again. Grudger would be unlikely to meet a Cheat who they helped before again, so they too will spend all their time helping and die out again.



The part that confuses: Grudger is an ESS?



But Dawkins also claims that Grudger is an ESS, and he seems very confident in that. Now I don't consider myself enough of a smartypants to claim that he's wrong, but I don't understand how Grudger can be an ESS. If all birds behave in this way, and for any reason some Sucker developed - the Sucker would have no disadvantage. All birds would still always be helping each other, so nothing would stop the Suckers from propagating equally well as the Grudgers, invading the gene pool. That's already the ESS broken, but even further, the presence of Suckers would mean that if Cheats came up, they would have a realistic chance of surviving - Grudgers would shun them after having helped once, but if the number of Suckers is large enough, Cheats will have an advantage.



Moreover, back to the initial setting of Grudgers only - if a Cheat developed, he would be unlikely to meet the same Grudger twice, receiving the benefit all the time but never paying the cost. He would have an advantage and spread Cheat genes.



The problem



I'm not familiar enough with how these kinds of models are calculated in order to state chances that Cheats will take over completely, but however I think of it Grudger does not seem to be an ESS to me.



Does anyone have an explanation why Dawkins is so sure that it is? Seeing as in nature we do see patterns like Sucker and Grudger all the time, I must be missing something important here.

Tuesday 15 January 2008

Protein Biology Cheat Sheet - Biology

For 95% of the work I do with protein, I find this chart very helpful, showing common post-translational modifications, for each amino acid, with molecular weights, associated codons, pKa values, 1- and 3-letter amino acid codes, chemical structure and chemical properties (polar, non-polar, aromatic, acidic, basic).

ct.category theory - What precisely Is "Categorification"?

The Wikipedia answer is one answer that is commonly used: replace sets with categories, replace functions with functors, and replace identities among functions with natural transformations (or isomorphisms) among functors. One hopes for newer deeper results along the way.



In the case of work of Lauda and Khovanov, they often start with an algebra (for example ${bf C}[x]$ with operators $d (x^n)= n x^{n-1}$ and $x cdot x^n = x^{n+1}$ subject to the relation $d circ x = x circ d +1$) and replace this with a category of projective $R$-modules and functors defined thereupon in such a way that the associated Grothendieck group is isomorphic to the original algebra.



Khovanov's categorification of the Jones polynomial can be thought of in a different way even though, from his point of view, there is a central motivating idea between this paragraph and the preceding one. The Khovanov homology of a knot constructs from the set of $2^n$ Kauffman bracket smoothing of the diagram ($n$ is the crossing number) a homology theory whose graded Euler characteristic is the Jones polynomial. In this case, we can think of taking a polynomial formula and replacing it with a formula that inter-relates certain homology groups.



Crane's original motivation was to define a Hopf category (which he did) as a generalization of a Hopf algebra in order to use this to define invariants of $4$-dimensional manifolds. The story gets a little complicated here, but goes roughly like this. Frobenius algebras give invariants of surfaces via TQFTs. More precisely, a TQFT on the $(1+1)$ cobordism category (e.g. three circles connected by a pair of pants) gives a Frobenius algebra. Hopf algebras give invariants of 3-manifolds. What algebraic structure gives rise to a $4$-dimensional manifold invariant, or a $4$-dimensional TQFT? Crane showed that a Hopf category was the underlying structure.



So a goal from Crane's point of view, would be to construct interesting examples of Hopf categories. Similarly, in my question below, a goal is to give interesting examples of braided monoidal 2-categories with duals.



In the last sense of categorification, we start from a category in which certain equalities hold. For example, a braided monoidal category has a set of axioms that mimic the braid relations. Then we replace those equalities by
$2$-morphisms that are isomorphisms and that satisfy certain coherence conditions. The resulting $2$-category may be structurally similar to another known entity. In this case, $2$-functors (objects to objects, morphisms to morphisms, and $2$-morphisms to $2$-morphisms in which equalities are preserved) can be shown to give invariants.



The most important categorifications in terms of applications to date are
(in my own opinion) the Khovanov homology, Oszvath-Szabo's invariants of knots, and Crane's original insight. The former two items are important since they are giving new and interesting results.

Monday 14 January 2008

mathematics education - effective teaching

I just watched the video -- I didn't know of it and it is certainly very interesting and provides much food for thought.



From the perspective of someone who teaches, I could relate to parts of it: particularly the decreasing ability to understand student's questions. From the perspective of someone who was a student, though, I do remember learning during lectures: perhaps not in all courses, but certainly in some which I still remember fondly. These courses were usually lecture-based; although in one case the lecturer would ask questions to the students all the time: picking a starting person and then moving systematically along the audience. This used to instill the "fear of God" in some people, but it meant one had to be on top of the material. I enjoyed that and, in fact, it boosted my confidence.



Now to answer the question, in the University of Edinburgh (where I am based) we started a few years ago to teach some of the introductory courses incorporating some element of Peer Instruction. I personally have not taught introductory courses for a while, so I cannot say how this is panning out. The School of Physics (I'm in Maths) has been teaching the first-year introductory course using Peer Instruction for some time now and they seem to be very happy with the result.



I wonder whether some variant of this method can work for final-year courses, though.

pr.probability - Is there an introduction to probability theory from a structuralist/categorical perspective?

One can argue that an object of the right category of spaces in measure theory is not a set equipped
with a $sigma$-algebra of measurable sets, but rather a set $S$ equipped with a $sigma$-algebra $M$ of measurable
sets and a $sigma$-ideal $N$ of $M$ consisting of sets of measure $0$.
The reason for this is that you can hardly state any theorem of measure theory or probability
theory without referring to sets of measure $0$.
However, objects of this category contain less data than the usual measured spaces,
because they are not equipped with a measure.
Therefore I prefer to call them measurable spaces.
A morphism of measurable spaces $(S,M,N)to(T,P,Q)$ is a map $Sto T$ such that
the preimage of every element of $P$ is a union of an element of $M$ and a subset of an element of $N$
and the preimage of every element of $Q$ is a subset of an element of $N$.



Irving Segal proved that for a measurable space the following properties are equivalent:



  1. The Boolean algebra $M/N$ of equivalence classes of measurable sets is complete;

  2. The space of equivalence classes of all bounded (or unbounded) real-valued functions on $S$ is Dedekind-complete;

  3. Radon-Nikodym theorem is true for $(S,M,N)$;

  4. Riesz theorem is true for $(S,M,N)$;

  5. Equivalence classes of bounded functions on $S$ form a von Neumann algebra (aka $W^*$-algebra).

  6. $(S,M,N)$ is a coproduct (disjoint union) of points and real lines.

A measurable space that satisfies these conditions is called localizable.



This theorem tells us that if we want to prove anything nontrivial about measurable
spaces, we better restrict ourselves to localizable measurable spaces.
We also have a nice illustration of the claim I made in the first paragraph:
None of these statements would be true without identifying objects that differ
on a set of measure $0$.
For example, take a non-measurable set $G$
and a family of one-element subsets of $G$ indexed by themselves.
This family of measurable sets does not have a supremum in the Boolean algebra
of measurable sets, thus disproving a naïve version of (1).



Another argument for restricting to localizable spaces is the following version of
Gelfand-Neumark theorem:




The category of localizable measurable spaces is equivalent to the category of commutative von Neumann algebras (aka $W^*$-algebras) and their morphisms (normal unital homomorphisms of $^*$-algebras).




I actually prefer to define the category of localizable measurable spaces
as the opposite category of the category of commutative $W^*$-algebras.
The reason for this is that the classical definition of measurable space
exhibits immediate connections only to descriptive set theory (and with additional
effort to Boolean algebras), which are mostly
irrelevant for the central core of mathematics,
whereas the description in terms of operator algebras immediately connects
measure theory to other areas of the central core (noncommutative geometry,
algebraic geometry, complex geometry, differential geometry etc.).
Also it is easier to use in practice. Let me illustrate this statement
with just one example: When we try to define measurable bundles of Hilbert spaces
on a localizable measurable space set-theoretically, we run into all sorts of problems
if the fibers can be non-separable, and I do not know how to fix this problem in the set-theoretic framework.
On the other hand, in the algebraic framework we can simply say that a bundle
of Hilbert spaces is a Hilbert module over the corresponding $W^*$-algebra.



Categorical properties of $W^*$-algebras (hence of localizable measurable spaces)
were investigated by Guichardet.
Electronic version of this paper is available here.
Let me mention some of his results.
The category of localizable measurable spaces admits equalizers and coequalizers,
arbitrary coproducts and hence arbitrary colimits.
It also admits products, although they are quite different from what one might think.
For example, the product of two real lines is not $Bbb R^2$ with the two obvious projections.
The product contains $Bbb R^2$, but it also has a lot of other stuff, for example, the diagonal of $Bbb R^2$,
which is needed to satisfy the universal property for the two identity maps on $Bbb R$.
The more intuitive product of measurable spaces ($Bbb Rtimes Bbb R=Bbb R^2$) corresponds to the spatial
tensor product of von Neumann algebras and forms a part of a symmetric monoidal
structure on the category of measurable spaces.
See Guichardet's paper for other categorical properties (monoidal structures
on measurable spaces, flatness, existence of filtered limits, etc.).



Finally let me mention pushforward and pullback properties of measures on measurable spaces.
I will talk about more general case of $L^p$ spaces instead of just measures (i.e., $L^1$ spaces).
For the sake of convenience let $L_p(M) := L^{1/p}(M)$, where $M$ is a measurable space.
Here $p$ can be an arbitrary complex number with a nonnegative real part.
Note that you don't need a measure on $M$ to define $L_p(M)$.
In particular, $L_0$ is the space of all bounded functions (i.e., the $W^*$-algebra itself),
$L_1$ is the space of finite complex-valued measures (the dual of
$L_0$ in the $sigma$-weak topology), and $L_{1/2}$ is the Hilbert space of half-densities.
I will also talk about extended positive part $E^+L_p$ of $L_p$ for real $p$.
In particular, $E^+L_1$ is the space of all (not necessarily finite) positive measures.



Pushforward for $L_p$ spaces: Suppose we have a morphism of measurable spaces $Mto N$.
If $p=1$, then we have a canonical map $L_{1}(M) to L_{1}(N)$, which just the dual of $L_{0}(N) to L_{0}(M)$
in the $sigma$-weak topology. Geometrically, this is the fiberwise integration map.
If $pneq 1$, then we only have a pushforward map of the extended positive parts:
$E^+L_p(M) to E^+L_p(N)$, which is non-additive unless $p=1$.
Geometrically, this is the fiberwise $L_p$ norm.
Thus $L_1$ is a functor from the category of measurable spaces to the category of Banach spaces
and $E^+L_p$ is a functor to the category of "positive homogeneous $p$-cones".
The pushforward map preserves the trace on $L_1$ and hence
sends a probability measure to a probability measure.



To define pullback of $L_p$ spaces (in particular, $L_1$ spaces) one needs to pass to a different
category of measurable spaces.
In algebraic language, if we have two $W^*$-algebras $A$ and $B$, then
a morphism from $A$ to $B$ is a usual morphism of $W^*$-algebras $f: Ato B$
together with an operator valued weight $T: E^+(B)→E^+(A)$ associated to $f$.
Here $E^+(A)$ denotes the extended positive part of A (think of positive functions on $mathrm{Spec} A$
that can take infinite values).
Geometrically, this is a morphism $mathrm{Spec} f:  mathrm{Spec} B to mathrm{Spec} A $ between the corresponding
measurable spaces and a choice of measure on each fiber of $mathrm{Spec} f$.
Now we have a canonical additive map $E^+L_p(mathrm{Spec} A) to E^+L_p(mathrm{Spec} B)$,
which makes $E^+L_p$ into a contravariant functor from the category of measurable spaces
equipped with a fiberwise measure to the category of “positive homogeneous additive cones”.



If we want to have a pullback of $L_p$ spaces themselves and not just their extended positive parts,
we need to replace operator valued weights in the above definition
by finite complex-valued operator valued weights $T: B to A$ (think of fiberwise complex-valued measure).
Then $L_p$ becomes a functor from the category of measurable spaces to the category
of Banach spaces (if the real part of $p$ is at most $1$) or quasi-Banach spaces (if the real part of $p$ is greater than $1$).
Here $p$ is an arbitrary complex number with a nonnegative real part.
Notice that for $p=0$ we get the original map $f: Ato B$ and in this (and only this) case we don't need $T$.



Finally, if we restrict ourselves to an even smaller subcategory of measurable
spaces equipped with a finite operator valued weight T such that $T(1)=1$
(i.e., T is a conditional expectation; think of fiberwise probability measure), then the pullback map preserves
the trace on $L_1$ and in this case the pullback of a probability measure is a probability measure.



There is also a smooth analog of the theory described above: The category of measurable spaces
and their morphisms is replaced by the category of smooth manifolds and submersions,
$L_p$ spaces are replaced by bundles of $p$-densities, operator valued weights are
replaced by sections of the bundle of relative $1$-densities,
integration map on $1$-densities is defined via Poincaré duality (to avoid circular dependence
on measure theory) etc.
The forgetful functor that sends a smooth manifold to its underlying measurable space
commutes with everything and preserves everything.

evolution - Did we first have swimming birds or flying birds?

Flying came first, as far as we know. The earliest known bird (currently), Archaeopteryx lithographica, already had aerodynamic feathers (Feduccia and Tordoff, 1979). The Solnhofen Limestone, where it was discovered, is ~145 million years old, so we can place the "when" flight evolved to greater than or equal to that time.



Gansus yumenensis is regarded as the earliest aquatic bird and is dated from ~120 million years ago. The Hesperornithiformes are the sister taxon of Gansus (less derived) and also aquatic. However, they date from ~85 million years ago. It is not clear whether the ancestors of Hesperornithiformes were also aquatic. Because they branched off the main line of birds before Gansus, these ancestors would likely predate Gansus, but those relatives have not yet been discovered.

ag.algebraic geometry - Stacks and sheaves

Let me see if I understand your example correctly: you are fixing $X$ and $Y$, families
of curves over $S$, and now you are considering the functor which maps an $S$-scheme $T$
to the set of $T$-isomorphisms $f^*X to f^*Y$ (where $f$ is the map from $T$ to $S$).



If I have things straight, then this functor shouldn't be so bad to think about, because it is actually representable, by an Isom scheme. In other words, there is an $S$-scheme
$Isom_S(X,Y)$ whose $T$-valued points, for any $f:T to S$, are precisely the $T$-isomorphisms
from $f^*X$ to $f^*Y$. (One can construct the Isom scheme by looking inside a
certain well-chosen Hilbert scheme.)



One way to think about this geometrically is as follows: one can imagine that two
curves over $k$ (a field) are isomorphic precisely when certain invariants coincide
(e.g. for elliptic curves, the $j$-invariant). (Of course this is a simplification,
and the whole point of the theory of moduli spaces/schemes/stacks is to make it precise,
but it is a helpful intuition.) Now if we have a family $X$ over $S$, these invariants
vary over $S$ to give a collections of functions on $S$ (e.g. a function $j$ in the
genus $1$ case), and similarly with $Y$. Now $X$ and $Y$ will have isomorphic
fibres precisely at those points where the invariants coincide, so if we look
at the subscheme $Z$ of $S$ defined by the coincidence of the invariants,
we expect that $f^*X$ and $f^*Y$ will be isomorphic precisely if the map $f$
factors through $Z$. Thus $Z$ is a rough approximation to the Isom scheme.



It is not precisely the Isom scheme, because curves sometimes have non-trivial
automorphisms, and so even if we know that $X_s$ and $Y_s$ are isomorphic for
some $s in S$, they may be isomorphic in more than one way. So actually the
Isom scheme will be some kind of (possibly ramified) finite cover of $Z$.



Of course, if one pursues this line of intuition much more seriously, one will
recover the notions of moduli stack, coarse moduli space, and so on.



Added: The following additional remark might help:



The families $X$ and $Y$ over $S$ correspond to a map $phi:S to {mathcal M}_g
times {mathcal M}_g$. The stack which maps a $T$-scheme to $Isom_T(f^*X,
f^*Y)$ can then seen to be the fibre product of the map $phi$ and the diagonal
$Delta:{mathcal M}_g to {mathcal M}_g times {mathcal M}_g$.



In the particular case of ${mathcal M}_g$ the fact that this fibre product is representable is part of the condition that ${mathcal M}_g$ be an algebraic stack.



But in general, the construction you describe is the construction of a fibre product
with the diagonal. This might help with the geometric picture, and make the relationship to Mike's answer clearer. (For the latter:note that the path space into $X$ has a natural
projection to $Xtimes X$ (take the two endpoints), and the loop space is the fibre product
of the path space with the diagonal $Xto Xtimes X$.)

Sunday 13 January 2008

set theory - Does every non-empty set admit a group structure (in ZF)?

You cannot in general put a group structure on a set. There is a model of ZF with a set A that has no infinite countable subset and cannot be partitioned into finite sets; such a set has no group structure.



See e.g at http://groups.google.com/group/sci.math/msg/06eba700dfacb6ed




Sketch of proof that in standard Cohen model the set $A={a_n:ninomega}$ of adjoined Cohen reals cannot be partitioned into finite sets:



Let $mathbb{P}=Fn(omegatimesomega,2)$ which is the poset we force with. The model is the symmetric submodel whose permutation group on $mathbb{P}$ is all permutations of the form $pi(p)(pi(m),n)=p(m,n)$ where $pi$ varies over all permutations of $omega$, (that is we are extending each $pi$ to a permutation of $mathbb{P}$ which I also refer to as $pi$) and the relevant filter is generated by all the finite support subgroups.



Suppose for contradiction that $pVdash " bigcup_{iin I}dot{A_i}=A$ is a partition into finite pieces"; let $E$ (a finite set) be the support of this partition. Take some $a_{i_0}notin E$ and extend $p$ to a $q$ such that $qVdash ``{a_{i_0},ldots a_{i_n}}$ is the piece of the partition containing $a_{i_0}$". Then pick some $j$ which is not in $E$ nor the domain of $q$ nor equal to any of the $a_{i_0},ldots a_{i_l}$. If $pi$ is a permutation fixing $E$ and each of $a_{i_1},ldots a_{i_n}$ and sending $a_{i_0}$ to $a_j$, it follows that $pi(q) Vdash " {a_j,a_{i_1},ldots a_{i_n}}$ is the piece of the partition containing a_j". But also $q$ and $pi(q)$ are compatible and here we run into trouble, because $q$ forces that $a_{i_0}$ and $a_{i_1}$ are in the same piece of the partition, and $pi(q)$ forces that this is not the case (and they are talking about the same partition we started with because $pi$ fixes $E$). Contradiction.

Saturday 12 January 2008

biochemistry - Single hormone opposite effects

An hormone is not different from most other molecules. To have an effect on a cell it binds to a (more or less specific) receptor, located either on the plasma membrane or inside the cell, and it initiates an intracellular cascade of events1.



There are several ways an hormone can have different effects:



  1. there can be multiple receptors for the same hormone. For instance, prolactin can bind to two receptors, called prolactin receptor (PRL-R) short and long form. The short form of the receptor has been shown to lack the ability to promote milk protein genes transcription (See Lesueur et al., PNAS - 1991).


  2. the same receptor can be coupled to different intracellular pathways in different cell types / physiological conditions, thus resulting in different effects.


  3. each cell type/tissue expresses a set of protein that will interact in a different manner with the intracellular cascade promoted by the hormone.


  4. an hormone can interact with receptors for other molecules. For instance allopregnanolone, a metabolite of progesterone, is a potent agonist of the GABA-A receptor, giving it anxiolitic properties.


An interesting example is that of estrogenic compound. Several receptors exist for estradiol (E2). The "classical" receptors are called ER-alpha and ER-beta, and they are located in the cytoplasm. The binding of E2 to the ER promotes their dimerization and entrance into the nucleus where they can promote the transcription of various genes. ERs can also bind to other transcription factors and modulate their activity. So, depending on which transcription factors are present different genes will be transcribed in response to E2. Moreover, ER-beta can have opposite effects then ER-alpha (see for instance Weihua et al., PNAS - 2000). In addition, receptors like ERs can be activated also in absence of the endogenous hormone: for instance, dietary amino acids activate ER-alpha in liver by a mTOR-dependent phosphorylation (Della Torre et al., Cell Metabolism 2011)



To make things more complicated, membrane receptor for estrogen have been described, such as GPR30 (Revankar et al., Science - 2005), membrane "versions" of the classical ERs, plus various splicing variants of the ERs (mostly expressed in tumoral tissue).



GPR30 in breasts has been shown to activate certain molecular pathways (notably Erk1/2) that contribute to cellular growth (Filardo et al., Mol Endocrinol. - 2000), possibly linking it to proliferation of ER-negative breast tumours. Other tissues that do not express GPR30 will lack this E2-induced proliferative stimulus.



So, tissue- or time- dependent modulation of the receptors and of the intracellular pathways associated with it can allow the body to respond to the same hormone (or, more in general, to the same molecule) in opposite ways.




1 This is a simplification, not every substance (and not every hormone) acts by binding to receptors, but let's keep things simple.

gt.geometric topology - Triangulations coming from a poset. Or: What conditions are necessary and sufficient for a finite simplicial complex to be the order complex of a poset?

Here are necessary and sufficient conditions for an abstract, finite simplicial complex $mathcal{S}$ to be the order complex of some partially ordered set.



(i) $mathcal{S}$ has no missing faces of cardinality $geq 3$; and



(ii) The graph given by the edges (=$1$-dimensional simplices) of $mathcal{S}$ is a comparability.



[Definitions.
(a) A missing face of $mathcal{S}$ is a subset $M$ of its vertices (=$0$-dimensional simplices) such that
$M not in mathcal{S}$, but all proper subsets $Psubseteq M$ satisfy $Pin mathcal{S}$.
(b) A graph (=undirected graph with no loops nor multiple edges) is a comparability if its
edges can be transitively oriented, meaning that whenever edges ${p, r_1}, {r_1, r_2},ldots, {r_{u−1}, r_u}, {r_u, q}$ are oriented as $(p, r_1), (r_1, r_2),ldots, (r_{u−1}, r_u), (r_u, q)$, then there exists an edge ${p, q}$ oriented as $(p, q)$.]



This characterisation appears with a sketch of proof $-$ which is not hard, anyway $-$ in




M. M. Bayer, Barycentric subdivisions. Pacific J. Math. 135 (1988), no. 1, pp. 1-16.




As Bayer points out, the result was first observed in




R. Stanley, Balanced Cohen-Macaulay complexes, Trans. Amer. Math. Soc, 249 (1979), pp. 139-157.




@Rasmus and @Gwyn: The characterisation might perhaps disappoint you if you were expecting something more topological. However, it's easy to prove that no topological characterisation of order complexes is possible, and therefore a combinatorial condition such as the one on comparabilities must be used. For this, first check that the barycentric subdivision of any simplicial complex indeed is an order complex. Next observe that barycentric subdivision of a simplicial complex does not change the homeomorphism type of the underlying polyhedron of the complex. Finally, conclude that for any topological space $T$ that is homeomorphic to a compact polyhedron, there is an order complex whose underlying polyhedron is homeomorphic to $T$.



I hope this helps.

Friday 11 January 2008

pr.probability - What is known about the Gaussian measure of the unit ball in a Hilbert Space?

You can't talk about "the" Gaussian measure on an infinite-dimensional Hilbert space, for the same reason that you can't talk about a uniform probability distribution over all integers. It doesn't exist; see Richard's answer. However, there are a lot of non-uniform Gaussian measures on infinite dimensional Hilbert spaces.



Consider the measure on $mathbb{R}^infty$ where the $j$th coordinate is a Gaussian with mean 0 and variance $sigma_j^2$, where $sum_{j=1}^{infty} sigma_j^2 < infty$ (and different coordinates are independent). This is almost surely bounded in the $ell_2$ metric, and any projection onto a finite-dimensional space has a Gaussian distribution. The squared length of a vector drawn from this measure is a sum of squares of Gaussians, and so follows some kind of generalized $chi$-square distribution. If I knew more about generalized $chi$-square distributions, I might be able to tell you what the measure of the unit ball was.



This kind of Gaussian distribution is very important in quantum optics. In fact, in quantum optics, a thermal state is Gaussian, so "the" Gaussian measure actually makes some sense.

Thursday 10 January 2008

lo.logic - What is Realistic Mathematics?

This is more a philosophical question, and therefore hasn't a definite answer.



But if you want to make plausible that AC isn't realistic mathematics, you might reason something like the following:



There is a set of mathematical sentences, that has a direct (so, not indirect yet) relation with physics. Call this set B (of Basic). Personally, I think they are in the following four areas:

  • Computation
  • Probability
  • Geometry
  • Topology

Note, I count arithmetic as part of computation, since numbers are not a physical entity, but computation is. But, likely many people will disagree.



Also note, that the sentences in B, might be far simpler than the mathematical sentences suggested in your question or in one of the answers.



Now, we have more complex mathematical sentences, that are still "realistic", if they can be converted or "instantiated" to sentences of B. Call this extended set be E. These more complex sentences capture a higher principle, which can be powerful in the science of physics. Still, there is no direct link with physics, the mathematic sentence first needs to be instantiated, to make a direct link with physics. Example, any sentence with a real number, is not be part of B, because we can not observe real numbers in physics, but they can be part of E.



Consider that there is a sentence s1 ∈ E and s2 ∈ E. Furthermore, that s2 follows from s1, but with a rather difficult proof. Suppose there is a proposed axiom a, such that s1 + a leads more directly to s2 (a simpler proof). However, a ∉ E. So, axiom a is independent from sentences in B.



From above concept it follows that axioms can exist that are "useful", because they make proofs shorter, but have nevertheless no "meaning". I do believe that AC is such axiom.



About CH, I think it is not useful and not having a meaning.



But again, this is more an opinion.

Wednesday 9 January 2008

computer science - Reconstructing an ordering of a multiset from its consecutive submultisets

I have some unpublished notes that I think are relevant to this problem, and I am cherry-picking from them. (See also here for a related question.) Because the notes are unpublished I hope it will not be considered poor form to provide a rather lengthy response.



To save myself trouble while picking the relevant parts from them, I will write $q$ in place of $r$ and $k$ in place of $j$. Also, let $B = q^k$ and let $N$ be the length of a generic cyclic word (or necklace) over a $q$-ary alphabet $mathcal{A}_q$.



As Gerhard suggests, the problem can be recast in terms of necklaces and their subwords of length $k$. We introduce the appellation dimgraph for “directed multigraph”. Let $[w]_k$ denote the generalized order $(k – 1)$ de Bruijn dimgraph of the $q$-ary word $w$: i.e., vertices correspond to the $(k – 1)$-tuples, and directed edges correspond to the $k$-tuples in $w$ in the usual way. If $[w′]_k = [w]_k$, we say that $w$ and $w′$ belong to the same order $k$ de Bruijn homology class. It is easy to see that this really is an equivalence relation, which we denote $sim_k$ (suppressing subscripts when there is no risk of confusion). If the size of an equivalence class is greater than unity, it will be necessary to take into account the additional data associated to $p$, but I will not touch on that (or on the formula to determine the multiplicity of a nontrivial equivalence class, though if anyone wants this please leave a comment and I will try to reply with a PDF) here.



For convenience, assume first that $q = 2$ and define the frequency tuple $alpha = alpha(w)$ as the $B$-tuple of natural numbers indicating the frequency of occurrence in $w$ of the (lexicographically ordered) $k$-tuples over $mathcal{A}_q$. For later convenience we take the frequency tuple to be an integral column vector. For example,



$w = 00011 in mathcal{A}_2^5, k = 3 Rightarrow alpha(w) = (alpha_{000},dots,alpha_{111})^T equiv (alpha_0,dots,alpha_7)^T = (1,1,0,1,1,0,1,0)$.



Notice that we start the index at zero: this will be our default convention for all indices relating to $k$-tuples, and context should generally suffice to determine the indexing scheme for a particular object.



The Eulerian property of generalized de Bruijn graphs means that $alpha$ is completely specified by the $B/2 + 1$ components $alpha_0,alpha_1,dots,alpha_{B/2-1},alpha_{B-1}$: conversely, any such assignment completely specifies the corresponding generalized de Bruijn dimgraph—though notice here that we make no implicit requirement that such a dimgraph must as a matter of course be strongly connected (i.e., correspond to a word). We could make other choices for the “free” components: the essential thing is that the two “loop” components $alpha_0$ and $alpha_{B-1}$ must be free, and any other $B/2 – 1$ components will suffice to complete a specification. The reason for this is basically that the Eulerian property (besides its requirement of strong connectivity) provides $B/2$ independent equations in $B$ unknowns.



More explicitly, for k > 1, write the frequency tuple as a column vector:



$alpha = (alpha_0,alpha_{[0]}^{(k)},alpha_{[1]}^{(k)},alpha_{B-1})^T; quad alpha_{[0]}^{(k)} = (alpha_1,dots,alpha_{B/2-1})^T, quad alpha_{[1]}^{(k)} = (alpha_{B/2},dots,alpha_{B-2})^T$.



Now there is an integral matrix—actually there are many integral matrices—satisfying



$alpha_{[1]}^{(k)} = M_{[0 rightarrow 1]}^{(k)} alpha_{[1]}^{(k)}$.



There is an algorithm for computing these matrices for arbitrary $k$: it is just an exercise in index-juggling, however, so we omit the details. The point is that we can dispense with the unnecessary components of frequency tuples for the binary case, although some care must be taken. Also, it turns out to simplify matters considerably if we omit the words of all zeros and all ones from consideration: we will do so without comment when convenient. More generally, it is convenient to require that none of the components of the frequency tuple are zero (which also implies that none of them equal $N$), but we shall be explicit when making such an assumption.



We can treat some of the components of a frequency tuple in the manner just outlined as functions into the natural numbers. In turn, the matrix-tree and BEST theorems can be jointly recast as a recipe for another function—the BEST function—from this space into the natural numbers that gives the cardinality of the de Bruijn equivalence class.



...



With the binary case in hand, we seek to generalize to $k$-tuples over $mathcal{A}_q$. The approach will be somewhat different here. Define ancestor and descendant matrices by



$M_leftarrow := 1_{1 times q} otimes I_{B/q}, quad M_rightarrow := I_{B/q} otimes 1_{1 times q}$.



Now the indegree/outdegree (I/O) or weak Eulerian property reduces to



$M_leftarrow alpha = M_rightarrow alpha$.



Nothing is lost here by forming the augmented ancestor and descendant matrices



$tilde M_leftarrow := delta_1 otimes M_leftarrow + I_B, quad tilde M_rightarrow := delta_1 otimes M_rightarrow + I_B$



and characterizing the I/O property instead by



$tilde M_leftarrow alpha = tilde M_rightarrow alpha$.



The augmented matrices are invertible, and so we can rewrite the I/O property once more as



$M_{(I/O)}alpha equiv left( tilde M_rightarrow^{-1} tilde M_leftarrow -I right)alpha = 0$.



In other words, a proper frequency tuple must belong to the kernel of the consistency matrix on the LHS above. The dimension $d$ of this in/out (I/O) kernel $ker_{I/O}(q,k)$ is just the number of independent variables required to determine a frequency tuple.



This number is $d = B – B/q + 1$. A sketch of a proof is as follows: of the $B$ edges (including any with multiplicity zero) in an order $(k – 1)$ generalized de Bruijn dimgraph, $q$ are loops: the corresponding frequency tuple components obviously need to be treated as free variables. For the remaining edges, we have B/q independent linear equations in $B – q$ variables. This gives a net of $B – B/q$ independent variables for a minimal integral solution to the linear equations. The +1 is accounted for by scaling.



Accordingly, if the word length is fixed to $N$ then evidently there are $B – B/q$ free components.



For small $q, k$ an exact/integral basis for the in/out kernel can be formed easily (e.g., by using the null command in MATLAB with the r option), and since these are the only cases we can realistically address, we will not bother with an algorithm.



We have two comments at this point. The first is that the reader should be convinced (or take it as an exercise to prove) that the I/O kernel contains an integer lattice whose primitive cells are relatively small (a precise formulation and attendant proof are probably rather technical, and certainly uninteresting). The second is that the lattice can be generated by a positive integral basis: the few negative entries above can clearly be made to go away with a few elementary linear operations.



That is, we consider the I/O lattice



$Lambda_{I/O}(q,k) :- ker_{I/O}(q,k) cap mathbb{Z}^B$



and the associated lattice cone



$Lambda_{I/O}^{0+}(q,k) := ker_{I/O}(q,k) cap mathbb{N}^B$.



Except for some degenerate cases corresponding to disconnected generalized de Bruijn dimgraphs (to be addressed in the sequel), points in this cone correspond to valid frequency tuples. For computational purposes we want to be able to efficiently generate the frequency tuple simplices



$Lambda_{I/O}^{(N)}(q,k) := Lambda_{I/O}^{0+}(q,k) cap N Delta_B$



where a dilation of the standard simplex is indicated on the RHS. (It will be an implicit corollary of our construction of these sets that the I/O lattice is unimodular, but we do not use this.)



We first define a bucketspace (“$s$ balls in $r$ buckets”) $X_{r,s} := {alpha in mathbb{N}^r : sum_{j=1}^r alpha_j = s}$. Now it is clear that there exists a $B times d$ integer matrix $M^{(Lambda)}$ s.t.



$M^{(Lambda)}X_{B,N} = Lambda_{I/O}^{(N)}(q,k)$



(where the LHS is interpreted in the obvious way). In fact there may be several such integer matrices—each reflecting a particular choice of basis—but all we need is one. We will call it a bucketspace-I/O lattice matrix.



The problem of its construction appears subtle (but perhaps that is because we have focused on explicit examples). A useful heuristic is to invoke the LLL lattice basis reduction algorithm on an integer basis for the I/O kernel, then use the structure of the de Bruijn graph—in particular, a table of its cycles—to improve on LLL. Note that the cycle enumeration for de Bruijn graphs has only been effected for small values of $k$.



In practice there are some subtle considerations, which we sketch in the context of the two specific cases $q = 2$ and $k = 3, 4$. Since the I/O kernel is the span of the dimgraph cycle space (see the sequel), it makes sense to look for a set of cycles corresponding to a “nice” basis. A natural candidate is a reduced basis of positive lattice vectors with minimal length.



Once the simple cycles of the de Bruijn graph are enumerated, it is easy to produce such a positive basis with a greedy algorithm. Start with the cycle representatives as a proto-basis. Now pick one of the remaining simple cycles of minimal weight uniformly at random: if it is linearly independent of the existing proto-basis, add it. Keep going until a basis is formed.



For $q = 2, k = 3$ the 6 simple cycles and corresponding simple cycle vectors are



$(0) leftrightarrow (1,0,0,0,0,0,0,0)^T$



$(001) leftrightarrow (0,1,1,0,1,0,0,0)^T$



$(0011) leftrightarrow (0,1,0,1,1,0,1,0)^T$



$(01) leftrightarrow (0,0,1,0,0,1,0,0)^T$



$(011) leftrightarrow (0,0,0,1,0,1,1,0)^T$



$(1) leftrightarrow (0,0,0,0,0,0,0,1)^T$



and the greedy algorithm produces the unique minimal positive basis



${(0),(001),(01),(011),(1)}$.



Similarly for $q=2, k=4$ we have the 19 simple cycles



$(0),(0001),(0001011),(00010111),(00011),(0001101),(000111),(00011101),(001)$,
$(001011),(0010111),(0011),(001101),(00111),(0011101),(01),(011),(0111),(1)$



and the greedy algorithm produces either of the two minimal positive bases



${(0),(0001),(001),(001011),(0011),(01),(011),(0111),(1)}$



${(0),(0001),(001),(0011),(001101),(01),(011),(0111),(1)}$



Note that the simple vectors of length 5 are rejected by the greedy algorithm, as is the simple vector $(000111)$ of weight 6.



(Notice that Gordan’s lemma implies that the monoid [commutative semigroup with identity and cancellation] of integral points in a lattice cone is finitely generated. This in turn implies that we can use one of the minimal positive bases to produce frequency tuple simplices if we are willing to pay a large overhead factor over a more efficient [if less well-understood] method outlined below.)



By performing elementary column operations it is possible to find a workable bucketspace-I/O lattice matrix by trial and error (of course an algorithm would be preferable, but the cycle decomposition of de Bruijn graphs is known only for small values of $k$, suggesting that such a goal is too lofty). We find by direct experiment with $N = 16$ that the following matrices work:



alt text



The columns of the $k = 4$ matrix can be indicated graphically:



alt text



Finally, notice that the method outlined essentially boils down to finding the vertices of a frequency tuple simplex—but this appears to be more easily said than done.



...



Now that the I/O condition has been suitably developed, we move on to the question of (strong) connectivity. Let $B$ (with an abusive resue of notation) denote the incidence matrix of a dimgraph $G$. Without bothering to restate the common definition, we simple provide a relevant example: for $q = 2$ and $k = 4$ we have



alt text



Notice by way of passing that we can recast the I/O equations as



$B1 = 0$.



It can be shown that $ker B$ is the cycle space of $G$.



In any event it is a standard result for dimgraphs that



$mbox{rank}(B(G)) = |V(G)| - c(G)$



where on the RHS we indicate the number of vertices minus the number of components. In short, the rank of the incidence matrix allows us to judge whether or not the corresponding dimgraph is (strongly) connected.



Our intent here is that a generalized de Bruijn dimgraph has only a single component by definition: although we have not really produced a proper mathematical definition, this intent should suffice for the remainder. Now the requirement on a frequency tuple is that



$|V(alpha)| - mbox{rank}(B(alpha)) = 1$.



The number of vertices is just the number of $(k – 1)$-tuples making an appearance; the rank can be computed by Gaussian elimination. For $q = 2$, it can quickly be seen that the rank is just the number of nonzero components in $(alpha_1,dots,alpha_{B/2-1})$.



With that in hand, define the tuple canonical ensemble (TCE) by



$hat Lambda_{I/O}^{(N)}(q,k) := Lambda_{I/O}^{(N)}(q,k) cap {alpha: |V(alpha)|-mbox{rank}(B(alpha)) = 1}$.



The TCE gives all the frequency tuples for a given length, and no garbage points, although some care is required in its construction.



In conclusion: check to see if a given point is in the TCE. If it is, then use the matrix-tree and BEST theorems (again, ask me how and I'll provide a PDF if I see the comment) to determine its cardinality. Providing an enumeration and selecting $p$ minimal completes the solution to the problem, although I'm not sure about the value of $p$, complexity, etc.

Tuesday 8 January 2008

ca.analysis and odes - Are two probability distributions uniquely constrained by the sum of their p-norms?

Here is a proof that Steve's rescaling gives you all solutions, together with the trivial operation of permuting the components of $A$, $B$, and $C$ if you view them as vectors with positive coeifficients. (If you view them this way, then Steve's notation $||A||_p$ is just the usual $p$-norm.)



I first tried what Alekk tried: You can take the limit as $p to infty$ and eventually obtain certain power series expansions in $1/p$. Or you can take the limit $p to 0$ and obtain certain power series expansions in $p$. The problem with both approaches is that the information in the terms of these expansions is complicated. To help understand the second limit, I observed that the two sides of Steve's equation are analytic in $p$, but it only helped so much.



Then I realized that when you have a complex analytic function of one variable, you can get a lot of information from looking at singularities. So let's look at that. Let
$alpha_k = ln a_k$, so that
$$||A||_p = expleft( frac{ln bigl[exp(alpha_1 p) + exp(alpha_2 p) + cdots + exp(alpha_d p) bigr]}{p} right).$$
The expression inside the logarithm has been called an exponential polynomial in the literature, which I'll call $a(p)$. As indicated, $||A||_p$ has a logarithmic singularity when $a(p) = 0$. $||A||_p$ has another kind of singularity when $p = 0$, but won't matter for anything. Also $a(p)$ is an entire function, which means in particular that it is univalent and has isolated zeroes. Also, none of the zeroes of $a(p)$ are on the real axis. Let $b(p)$ and $c(p)$ be the corresponding exponential polynomials for $B$ and $C$.



Suppose that you follow a loop that starts on the positive real axis, encircles an $m$-fold zero of $a(p)$ at $p_0$, and then retraces to its starting point. Then the value of $||A||_p$, which is non-zero for $p > 0$, gains a factor of $exp(2mpi i/p_0)$. Thus Steve's equation is not consistent unless all three of $a(p)$, $b(p)$ and, $c(p)$ have the same zeroes with the same multiplicity. (Since $exp(2mpi i/p_0)$ cannot have norm 1, geometric sequences with this ratio but with different values of $m$ are linearly independent.)



At this point, the problem is solved by a very interesting paper of Ritt, On the zeros of exponential polynomials. Ritt reviews certain results of Tamarkin, Polya, and Schwengler, which imply in particular that if an exponential polynomial $f(z)$ does not have any zeroes, then it is a monomial $f_alpha exp(alpha z)$. Ritt's own theorem is that if $f(z)$ and $g(z)$ are exponential polynomials, and if the roots of $f(z)$ are all roots of $g(z)$ (with multiplicity), then their ratio is another exponential polynomial. Thus in our situation $a(p)$, $b(p)$, and $c(p)$ are all proportional up to a constant and an exponential factor. Thus, $A$, $B$, and $C$ must be the same vectors up to permutation, repetition, and rescaling of the coordinates. Repetition is an operation that hasn't yet been analyzed. If $A^{oplus n}$ denotes the $n$-fold repetition of $A$, then $||A^{oplus n}||_p = n^{1/p}||A||_p$. Again, since geometric sequences with distinct ratios are linearly independent, Steve's equation is not consistent if $A$, $B$, and $C$ are repetitions of the same vector by different amounts.



The same argument works for the generalized equation
$$x_1||A_1||_p + x_2||A_2||_p + cdots + x_n||A_n||_p = 0.$$
The result is that any such linear dependence trivializes, after rescaling the vectors and permuting their coordinates.



Update (by J.O'Rourke): Greg's paper based on this solution was just published:




"Norms as a function of $p$ are linearly
independent in finite dimensions," Amer. Math. Monthly, Vol. 119, No. 7, Aug-Sep 2012, pp. 601-3
(JSTOR link).


fourier analysis - A sum involving sines

Consider a scaled sine function, $sin(2pi x/2^n)$, for some positive integer $n$. For this, I have the following linear combination.



$$ sum_{x=1}^{2^{n-2}} c_x sin(2pi x/2^n).$$
(The upper limit to the sum is $2^{n-2}$.)



The question is whether there exist $c_x in {0, pm 1, pm 2}$, not all $0$, that make the above expression $0$, for infinitely many $n$?



If it helps, the above came up in a computation concerning the discrete Fourier Transform.

Monday 7 January 2008

cell culture - Hoechst or DAPI for nuclear staining?

The Hoechst 33342 dye is similar to DAPI in that both are UV-excited, minor groove-binding, and emit signals proportional to total DNA content. Both are maximally excited around 355 nm and emit around 460 nm. A UV light source is required, which may harm the cell. However, this is only a risk with Hoechst, as DAPI requires the cells to be fixed and/or permeabilized, which is incompatible with life. DAPI is more stable, but Hoechst is brighter. Both are subject to photo-bleaching after long exposure.



There is another newer DNA dye called DRAQ5 that makes up for all of the issues I mentioned earlier - it is bright, quite photo-stable, can be used in live cell imaging, and its excitation and emission wavelengths are in the far red end of the spectrum, meaning that no UV is needed. I found a good comparison chart at http://www.biostatus.com/product/draq5/live_cell_comparison_chart/ if you're interested. I have no connection to them, by the way, I've just used DRAQ5 a lot in the past.

Thursday 3 January 2008

ag.algebraic geometry - Standard reduction to the artinian local case?

Dear Workitout: The list of comments above is getting unwieldy, so let me post an answer here, now that you have finally identified 1.10.1 in Katz-Mazur as (at least one) source of the question. As I predicted, you'll see that the basic technique to be used adapts to many other settings, and that it is very hard to formulate a "meta-theorem" to cover all cases. The only sure-fire method I know is to read all of sections 8, 9, 11, 12 in EGA IV$_3$ and sections 17 and 18 in EGA IV$_4$, and then it becomes really routine. Maybe there's a better method (well I can think of one, but I won't say it here). I will use the terminology and notation around 1.10.1 from Katz-Mazur without explanation below.



Being a "full set of sections" of $Z/S$ is something which is sufficient to check using the constituents of a single open affine cover of $S$, and also a finite generating set of the coordinate ring of $Z$ over each such open. Thus, by working Zariski-locally on $S$ we may assume $S = {rm{Spec}}(R)$ and that both sides of (1) in KM 1.10.1 have $R$-free coordinate rings (when viewed as finite $R$-schemes), and likewise for their sum (as effective Cartier divisors). Now the assertions (1) and (2) in KM 1.10.1 are identities among finitely many elements of some finite free $R$-modules. For instance, (1) asserts that certain elements in a finite free $R$-module have vanishing image in a finite free $R$-module quotient. This is all now a bunch of identities among finitely many elements of $R$.



OK, finally we come to the part with a real idea. Consider the subring $R_0$ of $R$ generated over $mathbf{Z}$ generated by those finitely many elements. It is noetherian. Now unfortunately your initial algebro-geometric setup over $R$ (the smooth separated finite-type curve $C$, the various effective relative Cartier divisors, etc.) probably does not arise via base change from the exact same setup (including flatness properties!) over $R_0$. But that doesn't matter: what would really be swell is if some noetherian subring of $R$ which contains $R_0$ permits such a descent of the situation. Now express $R$ as the direct limit of its finitely generated $R_0$-subalgebras (all of which are noetherian). Does the entire situation descend to one of those? If it did, we'd be in great shape, since it would then suffice to solve the problem in the case of a noetherian base ring (as that would imply the result over $R$ by suitable base change of the descent from the noetherian subring back up to $R$).



How to implement this strategy of reduction to the noetherian case (after which we'll need to deal with the passage to artin local base with algebraically closed residue field)? OK, so that's where we are all very fortunate that Grothendieck actually wrote out the entire formalism in total detail to handle basically every such situation one could ever want to handle. So it becomes kind of a game in finding the references in EGA (which I admit is hard to do if one doesn't know where to look, but is really easy if one has read the right parts). For your particular situation with some smooth separated curves and some relative effective Cartier divisors, etc., the results you need are: EGA IV$_3$ 8.3.4, 8.9.1, 8.10.5 (e.g., (v)), 9.2.6.1, 11.2.6(ii), and IV$_4$ 17.7.9.



I'm not going to say more about how you combine those references here; that is where I again remind you of your pseudonym.



OK, now $R$ is noetherian (even finite type over $mathbf{Z}$, which is very useful extra stuff to have for other arguments with excellence later in life), and you're trying to prove some finite collection of identities among elements of $R$. To do that it suffices to check in the local rings of $R$, so you can assume $R$ is local. Now a pair of elements of a local noetherian ring are equal if and only if they have equal images in each artinian quotient (Krull intersection theorem). So it suffices to prove the general result over arbitrary artin local rings (really just the artinian quotients of finitely generated $mathbf{Z}$-algebra, so there's no set-theoretic quantification nonsense going on). Now $R$ is artin local. To check an identity in a ring it suffices to do so in a faithfully flat extension ring. So finally we haul out EGA 0$_{rm{III}}$ 10.3.1 to find a faithfully flat artin local extension with an algebraically closed residue field. That's it!



I presume you can now see why anyone who knows how to fill in such arguments never actually writes them out in papers: it is much simpler to say "by standard limit methods from EGA IV$_3$, sections 8,9, etc." (maybe even to be a bit more specific, as K-M are at the beginning of their 1.8.1), and to trust that the reader will pick up the clue that they should read those parts of EGA if they want to understand what is going in such arguments for themself. It is bad when people don't at least mention the relevance of sections 8, 9, etc., but things could be worse (e.g., Grothendieck could have not written EGA, leaving stuff in a complete mess reference-wise).

What roles do "base change" play in algebraic geometry?

It might be a not very specific problem. I just wanna know how much do we rely on the property of "base change closed". In the definition of Grothendieck pretopology, we require a collection of morphisms satisfied the "invariant under base change"



As far as I know, in noncommutative algebra. The flat morphism does not respect to base change in general. I heard from some guys in category theory telling me that the "topology" which satisfied the "base change closed" is called "square topology"(I am not sure whether I spell it in correct way).



My question is: if we do not have the property of "invariant under base change". How much algebraic geometry can we recover? Or Is there a big difference from classcial algebraic geometry?



In fact, I know some work of Kontsevich-Rosenberg on noncommutative algebraic geometry dealing with the topology which is lack of base change property. And they called "quasi-topology". I just wonder know if there are other people in category theory or in algebraic geometry ever considered this question.



Thanks in advance!

examples - Do good math jokes exist?

Test to tell the difference between a Physicist or a Mathematician



Consider the following scenario: A room with a sink at the far end with a working cold water faucet plus a table with the following items on top – small bucket, ring stand, Bunsen burner, and a pack of matches. The problem is to boil water.



If the individual picks up the bucket from the table, walks to the sink and fills the bucket from the faucet, brings it back to the table, sets it on the ring stand, puts the Bunsen burner under the stand, and then lights the burner and waits for the water to boil … this establishes the base line but does not separate which it the Physicist and which is the Mathematician.



Test scenario 2: The bucket is now sitting on the floor under the table and the problem is again to boil water.



If the individual picks up the bucket from under the table, walks directly to the sink and fills the bucket from the faucet, brings it back to the table, sets it on the ring stand, puts the Bunsen burner under the stand, and then lights the burner and waits for the water to boil … this proves that this individual is the Physicist.



However, if the individual picks up the bucket from under the table and places it back on top of the table thus reducing the current problem to a form that they have previously solved … this proves that this individual is the Mathematician.