at.algebraic topology - Equivariant maps inducing isomorphism in integral cohomology

I don't know of a reference off-hand but here's one way to think about it. First, one can think of $H^i(X;mathbb{Z})$ as $[X,K(mathbb{Z},n)]$, the set of homotopy classes of maps. Notice that a cellular model for $K(mathbb{Z},n)$ can be taken to be $S^n$ union higher cells that kill off the higher homotopy groups. Second, any map $f:Xto Y$ can be replaced by an inclusion $iota:Xto M_f$, where $M_f$ is the mapping cylinder and it has the same homotopy type as that of $Y$. This works in the equivariant setting also. The third fact is that if any equivariant map $f:Xto Y$ induces an isomorphism in cellular cohomology and $f$ acts freely on both $X$ and $Y$ then $f$ induces an isomorphism on equivariant cohomology as well. The equivariant cohomology can be thought of as maps from spaces to $K(mathbb{Z},n)$ up to equivariant homotopy.

Now think of $f:Xto Y$ as in inclusion and there is a long exact sequence in cohomology
$cdotsto H^ast(X;mathbb{Z})to H^{ast+1}(Y,X;mathbb{Z})to H^{ast+1}(Y;mathbb{Z})to H^{ast+1}(X;mathbb{Z})to H^{ast+2}(X;mathbb{Z})tocdots$
which tells you in your case that $H^{i}(Y,X;mathbb{Z})=0$ if $i>i_0$. The kernel of $H^{i_0}(Y;mathbb{Z})to H^{i_0}(X;mathbb{Z})$ is just the image of $H^{i_0}(Y,X;mathbb{Z})$ in $H^{i_0}(Y;mathbb{Z})$. Thinking of $H^{i_0}(Y,X;mathbb{Z})$ as relative homotopy classes of maps from $(Y,X)$ to $K(mathbb{Z},i_0)$. These maps only probe the $i_0$ skeleton of $(Y,X)$ because $H^i(Y,X;mathbb{Z})=0$ if $i>i_0$. And since any map can be made homotopic to a cellular map we need only study homotopy classes of maps from the $i_0$-skeleton of $(Y,X)$ to the $i_0$-skeleton of $K(mathbb{Z},i_0)$ which is $S^{i_0}$. These are precisely the different ways of factoring a given equivariant map $Xto S^{i_0}$ via $Xstackrel{f}{to} Yto S^{i_0}$ (all upto equivariant homotopy).

evolution - How do archaea relate to eukaryotes and bacteria?

The relationship between the three is so overly complicated because of horizontal gene transfer, i.e. cells sharing parts of their genome with others instead of the normal passing-down to the next generation. This works between eukaryotes, archaea and bacteria and causes the mixing of the three.

An overview (originally proposed by Woese et al):

• Bacteria are simplest considering cellular organisation. They have no nucleus or other internal membranes and no cytoskeleton. Their DNA is generally "junk-free"* (i.e. no introns), not bound by histones ("organising" proteins) and the organisation of regulatory sequences is relatively straightforward, typically with one 'operator' sequence before a gene.


• Eukaryotes are complex cells in comparison. They have a nucleus and are full of internal membranes (ER, golgi, vesicles,...). Their complex cytoskeleton allows them to grow very large in comparison. Their DNA is usually full of more or less "junk"*, organised into tight bundles by histones and they tend to have a ridiculous amount of regulatory sequences way before, just at the start and even inside genes. Very importantly, they possess endosymbionts as you mentioned - mitochondria or plasmids, derived probably from some protobacterium.


• Archaea at first appear to be a mixture between the two. A nice way to sum it up (though not very accurate) is "eukaryote in a bacterium's clothing". Their cells look a lot like prokaryotes because they are similar size, have no nucleus, endomembranes or cytoskeleton. However, some archaeas' DNA is bound by histones and they use similar machinery as eukaryotes for DNA replication, transcription and translation.

One possible explanation for this (theory by Margulis & Schwartz as explained by my tutor) is: From the last universal common ancestor, first prokaryotes and archaea diverged. After this branching, the differences in genetic machinery evolved. Archaea then branched, producing a protoeukaryote line which went on to endosymbiosis with some protobacteria. That would explain the similarities I outlined above: similar cell structure between archaea and prokaryotes but similar genetic machinery between archaea and eukaryotes.

It gets a bit more complicated though, because archaea and bacteria can exchange genetic material, and eukaryotes incorporated a lot of their endosymbionts' DNA in their own genome, so you end up with all of them having a huge mix of genes from the others.

Edit: Genetic machinery refers to the enzymes used for DNA replication and transcription (DNA and RNA polymerase and the transcription initiation factors) and translation (ribosomes and several other translation factors involved). For example: Prokaryotes only have one type of RNA polymerase which uses sigma factor to bind the initiaion site. Eukaryotes and archaea both have three types of RNA polymerase which use TATA-binding protein for initiation. Or: In Prokaryotes, the translation start codon codes for formyl-methionine while in eukaryotes and archaea it codes for "normal" methionine. (Source)

*) Unrelated to your question, but in my own defence: I am aware that the usage of the word "junk" is somewhat unaccepted nowadays, because it supposes that there is "useful" and "unuseful" genetic material. Also, some of what was thought to be junk has been identified to have regulatory function etc. However, it's also a fact that a lot has been identified to have no function to the cell whatsoever, most of it parasitic DNA which is dead or still happily replicating away (LINEs and SINEs). That is, "junk" to the cell.

co.combinatorics - Asymptotics of q-Catalan numbers

It's not hard to compute numerical values. If you do this, in the regime $0 < q < 1$ it looks like $C_n$ grows exponentially, i. e. $C_n sim alpha_q beta_q^n$ for some constants $alpha_q$ and $beta_q$ which depend on q.

Unfortunately, I don't know what $alpha_q$ and $beta_q$ are. For example, when q = 1/2 the ratio $C_n/C_{n-1}$ approaches a constant which is approximately 1.6022827223; I claim this is $beta_{1/2}$. Then $C_{50}/beta_{1/2}^{50} = 0.5757566503$, which I claim is $alpha_{1/2}$. Neither of these constants appears in the inverse symbolic calculator.

The generating function $C(q,z) = C_0 + C_1 z + C_2 z^2 + ldots$, where the $C_n$ are $q$-Catalan numbers, ought to satisfy some functional equation, and then one could use techniques from singularity analysis (see, for example, Analytic Combinatorics by Flajolet and Sedgewick). But I am having trouble finding that functional equation.

qa.quantum algebra - What happened to the Vacuum Hypothesis in TQFT?

I remember that in the beginning, there was an axiom for $(n+1)$-dimensional
TQFT that said that the state space $V(Sigma)$ assigned to an $n$-dimensional
oriented manifold is spanned by the invariants of all $n+1$-dimensional oriented manifolds
$M$ with $partial M=Sigma$. If we call the invariant of $M$, $Z(M)in V(Sigma)$, this
just says that $V(Sigma)$ is spanned by all $Z(M)$. Maybe it was in Segal's Swansea notes, maybe it was in an early version of Atiyah's axioms. It doesn't seem to have made it
into "The Geometry and Physics of Knots"

For instance if you read "Topological Quantum Field Theories Derived from the Kauffman Bracket" by Blanchet, Habegger, Masbaum and Vogel, the vacuum hypothesis is implicit in their constructions.

There is a theorem that the category of Frobenius algebras is equivalent to the category of
$1+1$-dimensional TQFT's. For instance, let $A=mathbb{C}[x]/(x^3)$, with Frobenius map
$epsilon(1)=epsilon(x)=0$ and $epsilon(x^2)=-1$. This is the choice that Khovanov made to construct his $sl_3$-invariant of links. At this point, no one would deny that this gives rise to a TQFT.

The state space associated to a circle is just $A$. The $2$-manifolds with boundary
the circle are classified by genus. Using TQFT to compute them, I get that a disk has invariant $1$, a surface of genus one with one boundary component has invariant $3x^2$, and any other surface has invariant $0$. The invariants don't span $A$.

The problem as I learned from Chris French is that $A$ is not semisimple. In fact, $A$ being spanned by the invariants of surfaces with one boundary component is equivalent to the semisimplicity of $A$.

Here is my question. At what point, and why was the vacuum hypothesis abandoned?

at.algebraic topology - Graded commutativity of cup in Hochschild cohomology

The cup product in Hochschild cohomology$H^bullet(A,A)$ is graded commutative for all unitary algebras. If $M$ is an $A$-bimodule, then the cohomology $H^bullet(A,M)$ with values in $M$ is a symmetric graded bimodule over $H^bullet(A,A)$.

(If $M$ itself is also an algebra such that its multiplication map $Motimes Mto M$ is a map of $A$-bimodules, then in general $H^bullet(A,M)$ is not commutative (for example, take $A=k$ to be the ground field, and $M$ to be an arbitrary non-commutative algebra! I do not know of a criterion for commutatitivity in this case)

These results originally appeared in [M. Gerstenhaber, The cohomology structure of an associative ring, Ann. of Math. (2) 78 (1963), 267–288.], and they are discussed at length in [Gerstenhaber, Murray; Schack, Samuel D. Algebraic cohomology and deformation theory. Deformation theory of algebras and structures and applications (Il Ciocco, 1986), 11--264, NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., 247, Kluwer Acad. Publ., Dordrecht, 1988.] Both references give proofs of a rather computational nature.

You can find an element-free proof of the graded commutativity in this paper, which moreover applies to the cup-products of many other cohomologies.

As for what happens with non-unital algebras, I do not know. But they are very much used today as they were before. One particular context in which they show up constantly is in the intersection of K-theory and functional analysis, where people study `algebras' which are really just ideals in rings of operators of functional spaces---one egregius example is the algebra of compact operators in a Hilbert space.

By the way, you say that group cohomology is a special case of Hochschild cohomology: it is only in a sense... There is a close relationship between the group cohomology $H^bullet(G,mathord-)$ of a group $G$, and the Hochschild chomology $H^bullet(kG,mathord-)$ of the group algebra, but they are not the same. You can use the second to compute the first (because more generally if $M$ and $N$ are $G$-modules, then $mathord{Ext}_{kG}^bullet(M,N)=H^bullet(kG, hom(M,N))$, where on the right we have Hochschild cohomology of the group algebra $kG$ with values in the $kG$-bimodule $hom(M,N)$), but the «principal» group cohomology $H^bullet(G,k)$ is only a little part of the «principal» Hochschild cohomology $H^bullet(kG,kG)$.

Finally, in the paper by Sletsjøe the definition for the boundary is given as you say because he only considers commutative algebras and only principal coefficients, that is $H^bullet(A,A)$.

ag.algebraic geometry - Minimal size of an open affine cover

This is not a complete answer by any means, but here are the two most basic arguments. First of all, you have that every projective scheme that can be embedded in $mathbb{P}^n$ can be covered by $n+1$ open affines, namely the closed subschemes of the affines $U_i = lbrace z_i neq 0 rbrace cong mathbb{A}^n$.

For a lower bound, think Cech-cohomologically: if $X$ can be covered by $k$ affine opens, then $check{H}^l(X) = 0$ for every $l > k$. If $X$ is Noetherian separated, then Cech cohomology coincides with sheaf cohomology, which indicated that you need at least $maxlbrace l ;|; H^l(X) neq 0 rbrace$ open affines to cover it.

ag.algebraic geometry - Relation between characteristic variety and support of D-Module

I'm not a $D$-module person. I'm hoping someone else can give a slightly more insightful explanation. (It looks like YBL gives a clear picture of the holonomic case.)

By definition $Ch(M)$ is the support of the associated graded of $M$, for a suitable filtration, so it should lie in the preimage of the support of $M$. However, in many interesting cases the inclusion would be strict. If $M$ is holonomic, $Ch(M)$ is Lagrangian, so it wouldn't coincide with the preimage of $supp(M)$. The simplest case where this
happens is when $M$ is a flat connection, then $Ch(M)$ is the zero section of $T^*X$.

Continuation: Perhaps it's worthwhile making this a little more explicit.
The simplest example is $X=mathbb{A}^n$. Then (the global sections of) $D_X$
is the Weyl algebra with generators $x_1,ldots, x_n, partial_1,ldots, partial_n$ with
commutation relations $[x_i,partial_j]= delta_{ij}$.
If we force these to commute, by passing to the associated graded with respect to
the filtration by order of operators, we obtain the polynomial ring in $2n$ variables
or in other words the coordinate ring of $T^*mathbb{A}^n$. Any finitely generated $D_X$-module $M$, carries a (noncanonical) compatible filtration, so can define $Ch(M)= Supp(GrM)subset T^*mathbb{A}^n$ (it is independent of the filtration). Now let
$M= D/sum Dpartial_i$, which corresponds to $mathcal{O}_X$. I'll omit the
details, but one can see that
$Ch(M) = V(partial_1,ldots partial_n)$ (the zero section), and $pi^*supp(M) = T^*mathbb{A}^n$ where
$pi:T^*mathbb{A}^nto mathbb{A}^n$.

For a nonholonomic example, take $M= D_X$. Then $Ch(M)= pi^{-1}supp(M) = T^*mathbb{A}^n$.

ag.algebraic geometry - Rank of a linear combination of quadratic forms

Suppose we have a set of quadratic forms $Q_i (x_1, dots, x_n)$ for $1 leq i leq k$ in $n$ variables, defined over $mathbb{R}$. We suppose these are 'collectively nondegenerate' in the sense that there does not exist a change of variables which takes us into a set of quadratic forms with less than $n$ variables.

I am looking at linear combinations of these forms: $$ Q_{boldsymbol{lambda}}(textbf{x})=sum_i lambda_i Q_i(x_1, dots, x_n)$$ for $boldsymbol{lambda} = (lambda_1, dots , lambda_k) in mathbb{R}^k$. My question is whether we are guaranteed a set of $lambda$s which gives us a quadratic form of full rank i.e. $n$? Edit:: this has been shown to be untrue, so...

Is there anything we can do to guarantee a 'high' rank, say bigger than 5? For example by taking $n gg k$?

enumerative geometry - Chow Ring of Moduli Space of Abelian Varieties

Van der Geer has written a paper computing what he calls the tautological subring of the chow ring of $mathcal{A}_g$. He also computes the tautological ring for a smooth toroidal compactification.

G. van der Geer, Cycles on the Moduli Space of Abelian Varieties, in "Moduli of Curves and Abelian Varieties (The Dutch Intercity Seminar on Moduli)", p. 65-89 (Carel Faber and Eduard Looijenga, editors), Aspects of Mathematics, Vieweg, Wiesbaden 1999.

It is available on the van der Geer's website here

Regarding intersection theory, Erdenberger, Grushevsky, and Hulek have been working on this for the toroidal compactifications, mostly for small values of $g$. For example, see the following references.

C. Erdenberger, S. Grushevsky, K. Hulek, Intersection theory of toroidal compactifications of $mathcal{A}_4$. Bull. London Math. Soc. 38 (2006), no. 3, 396--400.

C. Erdenberger, S. Grushevsky, K. Hulek, Some intersection numbers of divisors on toroidal compactifications of $mathcal{A}_g$. J. Algebraic Geom. 19 (2010), no. 1, 99--132.

S. Grushevsky, Geometry of $mathcal{A}_g$ and its compactifications. Algebraic geometry---Seattle 2005. Part 1, 193--234, Proc. Sympos. Pure Math., 80, Part 1, Amer. Math. Soc., Providence, RI, 2009.

What is the difference between matrix theory and linear algebra?

Let me elaborate a little on what Steve Huntsman is talking about. A matrix is just a list of numbers, and you're allowed to add and multiply matrices by combining those numbers in a certain way. When you talk about matrices, you're allowed to talk about things like the entry in the 3rd row and 4th column, and so forth. In this setting, matrices are useful for representing things like transition probabilities in a Markov chain, where each entry indicates the probability of transitioning from one state to another. You can do lots of interesting numerical things with matrices, and these interesting numerical things are very important because matrices show up a lot in engineering and the sciences.

In linear algebra, however, you instead talk about linear transformations, which are not (I cannot emphasize this enough) a list of numbers, although sometimes it is convenient to use a particular matrix to write down a linear transformation. The difference between a linear transformation and a matrix is not easy to grasp the first time you see it, and most people would be fine with conflating the two points of view. However, when you're given a linear transformation, you're not allowed to ask for things like the entry in its 3rd row and 4th column because questions like these depend on a choice of basis. Instead, you're only allowed to ask for things that don't depend on the basis, such as the rank, the trace, the determinant, or the set of eigenvalues. This point of view may seem unnecessarily restrictive, but it is fundamental to a deeper understanding of pure mathematics.

mg.metric geometry - Four Dimensional Origami Axioms

In the $n$-dimensional origami question, you start with a generic set of hyperplanes and their intersections, which can then be some collection of $k$-dimensional planes. An "axiom" is a set of incidence constraints that determines a unique reflection hyperplane, or conceivably a reflection hyperplane that is an isolated solution even if it is not unique. The space of available hyperplanes is also $n$-dimensional. Each incidence constraint has a codimension. A set of independent constraints makes an axiom when their codimensions add to $n$.

It is easy to write down a simple incidence constraint on a reflection $R$ and compute its codimension. If $n=2$, then the simple constraints are as follows:

1. $R(L) = L$ for a line $L$,


2. $R(x) = x$ for a point $x$,


3. $R(x) in L$,


4. $R(L_1) = L_2$, and


5. $R(x_1) = x_2$

The first three constraints have codimension 1 and the last two have codimension 2. Formally there are 8 ways that to combine these constraints to make axioms. However, 1 cannot be used twice in Euclidean geometry, so that leaves 7 others. These are the seven axioms listed on Robert Lang's page. As it happens, Huzita missed combining 1 and 3. If you could have hyperbolic origami, you would have 8 axioms.

You can go through the same reasoning in $n=3$ dimensions. The simple constraints can again be written down:

1. $R(x) = x$ for a point $x$ (1)


2. $R$ fixes a line $L$ pointwise (2)


3. $R(L) = L$ by reflecting it (2)


4. $R(P) = P$ for a plane $P$ (1)


5. $R(x) in L$ (2)


6. $R(x) in P$ (1)


7. $R(L) subset P$ (2)


8. $R(x_1) = x_2$ (3)


9. $R(L_1) cap L_2 ne emptyset$ (1)


10. $R(P_1) = P_2$ (3)

I've written the codimensions of these constraints in parentheses. As before, you can combine these constraints. Certain pairs, such as 3 and 4, can't combine. You also can't use 4 three times. If you go through all of this properly, it is not all that hard to make a list of axioms that resembles the ones in 2 dimensions. (Possibly I made a mistake in this list or missed something, but it is not hard to go through this properly.)

However, there is a possible subtlety that I don't know how to address. Namely, suppose that you do make some complicated configuration using combinations of these constraints, at first. Can you create a configuration with the property that the codimensions don't simply add? For instance, ordinarily you can't use condition 7 twice to define the reflection $R$, because the total codimension is 4, which is too large. However, if the lines and planes in this condition are related to each other, is the true codimension sometimes 3? I would guess that you can make this happen. If so, then potentially you'd have to add "unstable" construction axioms to the list. But then it is not clear whether an unstable construction axiom is actually needed, or whether an unstable axiom can always be replaced by a sequence of stable axioms.

nt.number theory - Why do congruence conditions not suffice to determine which primes split in non-abelian extensions?

A prime splits completely in $L$ over $K$ (an extension of number fields) if and only if
it splits completely in the Galois closure of $L$ over $K$. Thus to answer the question we may assume that $L$ is Galois over $K$.

Ben Linowitz's argument then holds in generality: suppose that all $wp$ congruent to $1$ modulo some conductor $mathfrak m$ split in $L$. Then by the Lemma in Ben's answer, $L$ is contained in the ray class field of conductor $mathfrak m$ over $K$, and hence is abelian.

(As far as I can tell, this is not at all obvious without class field theory, and in fact, a big part of the development of class field theory involved the realization that class fields
--- which were defined in terms of splitting conditiosn described by congruences --- were the same things as abelian extensions. In some sense, the equivalence of these two conditions is the essence of class field theory.)

[EDIT:] This edit is in response to Buzzard's comments on the original question, and also his answer and subsequent comments.

Suppose that $L$ over $K$ is Galois (as we may) and that for some non-empty subset $S$ in some ray
class group $Cl_{mathfrak m}$ we know that all (but finitely many) primes lying in $S$ mod $mathfrak m$ split in $L.$

If we furthermore assume if and only if in the preceding statement, then Buzzard's answer shows that $S$ must contain the trivial class, and hence $L$ is an abelian extension contained in the ray class field of conductor $mathfrak m$; class field theory then takes over to show that $S$ is in fact a subgroup.

But what if we don't assume if and only if (i.e. we allow that other primes besides those
lying in $S$ split)? Can we still argue that $L$ is abelian over $K$?

Let $L'$ be the compositum of $L$ and the ray class field of conductor $mathfrak m$
over $K$, let $G = Gal(L/K)$, and let $G' = Gal(L'/K)$.

Then $G' hookrightarrow G times Cl_{mathfrak m}$ (via the Galois action on $L'$
and the ray class field resp.); let $p$ and $q$ be the first and second projections
(note that they are both surjective).

Our assumption translates into the statement that $p (q^{-1}(S)) = {1}$, i.e.
$q^{-1}(S) subset 1 times Cl_{mathfrak m}.$

Now choose $s in S$, and suppose that $(g,1) in G'$. The previous paragraph together with the surjectivity of $q$ shows that also $(1,s) in G'$. Then $(g,1) (1,s) = (g,s) in G',$
since $G'$ is a subgroup of the product. But $(g,s)$ lies in $q^{-1}(S)$, hence $g = 1$.
In other words, if the second coordinate of an element of $G'$ is trivial, so is
the first. Thus in fact $L'$ equals the ray class field of conductor $mathfrak m,$
i.e. $L$ is contained in the latter field. This is what had to be shown.

ag.algebraic geometry - Family of Enriques surfaces and GRR, Part 2

Q1: It is the other way round. For a smooth family the differential $T_Y to f^ast T_T$ is surjective and the relative tangent is the kernel, so you have an exact sequence

$0 to T_f to T_Y to f^ast T_T to 0$.

In this way the tangent to $f$ actually restricts to the tangent of the fibers.

Q2: I don't think that the classes $c_i(T_f)$ are determined by the fibers alone; they depend on the family. It does not even make sense to say that $c_i(T_f)$ are determined by the fibers since these classes live in $H^{2i}(Y)$ anyway, so you have to know at least the total space.

But since $T_f$ restricts to the tangent of the fibers, you know, by naturality of the Chern classes, that if $i colon E to Y$ is the inclusion of a fiber $i^ast c_i(T_f) = c_i(T_E)$.

And these you can compute using the fact that $E$ is Enriques. Namely $2 c_1(T_E) = 0$ since twice the canonical is trivial and $c_2(T_E) = chi(E) = 12$.

Q3: Surely it is not injective in the top degree, for trivial dimensional reasons. I do not see any reason why it should be in other degrees.

Q4: As is written in the article, this follows from $f_ast c_2 = 12$. This is more or less clear in cohomology. In this case $f_*$ is the integration along fibers, and since $c_2(T_E)$ is $12$ times the fundamental class of $E$ for all fibers $E$ (see Q2), that integral is $12$.

To translate this in the Chow language, I think the folllowing will do. Let $D$ be a cycle representing $c_2(T_f)$. Since $Y$ is smooth, we can compute the intersection number $D cdot E = c_2(T_f) cap E = c_2(T_E) cap E = 12$. So $D$ intersectts the generic fiber in $12$ points, and the morphism $D to T$ has degree $12$. In follows that $f_ast D = 12 [T]$, which is what you want.

dna - Do mitochondria simply automatically convert glucose to ATP?

Mitochondria are comprised of ~3000 proteins. However, the mitochondrial genome has only 13-14 protein-encoding genes. The remaining 99.6% of mitochondrial proteins are encoded by genes in the nuclear genome. (Wikipedia) Chloroplast genomes are only slightly larger (~100 genes).

Gene regulation and signaling between the nucleus and mitochondria (and between nucleus & chloroplast in plants/algae) occurs in both directions. Anterograde regulation is the signalling from nucleus to mitochondria and was once thought to be the only method of regulating organelle function. We now also know that Retrograde regulation occurs, in which the mitochondria sends signals to the nucleus.

To answer your question directly (but not thoroughly), both the mitochondria and the nucleus are sensing the environment and needs of the cell and signaling to each other to regulate ATP production.

If you want to learn more about retrograde signaling, googling "retrograde regulation" will bring up lots of papers on this topic in different organisms. This paper also has some diagrams of signalling pathways.

ca.analysis and odes - convergence of a series involving cosines

Question:

1) How to determine the convergence of

$displaystyle sum_{k=1}^{infty} frac{cos(k^{alpha} x)}{k^{alpha}} (-1)^k $

where $x in mathbb{R}$ and $alpha in (0,1]$. I am especially interested in the case of $alpha = 1/2$.

2) For a fixed $alpha$, if the above series converges for every $x$, is the convergence uniform? Is the resulting sum bounded in $x$?

I found the series tests (alternating test,etc.) I learned not useful in this situation, except that the convergence is clear for $x = 0$...

oc.optimization control - Hardness of combinatorial optimization after adding one constraint

Sure. We'll construct a normally trivial problem that turns into a question of four-coloring when we restrict one parameter.

For a graph $G$ on $n$ vertices, consider binary words of length $2n+1$. This will represent an assignment of colors 1-4 on the $n$ vertices, with the last bit telling us what the restriction on neighboring colors is. Namely, if the last digit is $i$, then $f$ spits out a 0 (calls it an improper coloring) if some pair of adjacent vertices have colors that are $i$ apart. When it is proper, $f$ spits out the reciprocal of the number of colors used.

Well, this is easy to maximize: just make every vertex the same color and choose the last digit to be 1. Congrats, you only used one color. But if our constraint is that the last digit is 0, then now you're asking whether the graph needs 1, 2, 3, 4, or more colors to properly color (in the usual sense), which you can't answer in polynomial time.

cohomology - Classifying Algebra Extensions over a fixed extension?

I think your problem is not constrained enough to have an interesting answer. Notice that your intertwining condition can be rephrased by saying that $g: B to A$ is a homomorphism of $R$-algebras, where $A$ is given the structure of $R$-algebra given by $f$. In these terms, what you are looking for is the comma category $mathcal{C} = (mathbf{Alg}_R downarrow A)$, whose objects are precisely the pairs $(B, g)$ as above, and whose morphisms $operatorname{Hom}_{mathcal{C}}((B, g), (B', g'))$ are the $R$-algebra homomorphisms $h: B to B'$ such that $g = g' circ h$. I am not sure it is possible to capture this beast with a cohomology group of any sort.

What you can do is restrict the class of objects that you are looking at. For example, you can classify square-zero extensions: fixing an $A$-module $I$, you can look at $R$-algebras $B$ such that you have a short exact sequence $0 to I to B to A to 0$; the name square-zero comes from the fact that $I$ is an ideal of $B$ with $I^2 = 0$. You can read about them in the first chapter of Sernesi's Deformations of Algebraic Schemes.

ag.algebraic geometry - When is a homogeneous space a variety?

I'll try to answer both questions, though I will change the first question somewhat. Let's work in the setting of a real reductive algebraic group $G$ and a closed subgroup $H subset G$.

Your first question asks when $G/H$ is an open subset of some (presumably complex) variety. I think that this question should be modified in a few ways.

You can't really say that $G/H$ "is a subset" of a variety, since $G/H$ is not a priori endowed with a complex structure. So you need a bit more data to go with the question -- a complex structure on the homogeneous space $G/H$. Such a complex structure can be given by an embedding of the circle group $U(1)$ as a subgroup of the center of $H$. Let $phi: U(1) rightarrow G$ be such an embedding, and let $iota = phi(i)$ be the image of $e^{pi i} in U(1)$ under this map. Such an embedding yields an integrable complex structure on the real manifold $G/H$, I believe (though I haven't seen this stated in this degree of generality).

So now one can ask if $G/H$, endowed with such a complex structure, is an open subset of a complex algebraic variety. But again, I have some objection to this question -- it's not really the right one to ask. Indeed, it's very interesting when one finds that some quotients $Gamma backslash G /H$ are (quasiprojective) varieties -- but such quotients are not obtained as quotients in a category of varieties, from $G/H$ to $Gamma backslash G / H$. They are complex analytic quotients, but not quotient varieties in any sense that I know.

So what's the point of knowing whether $G/H$ is an open subset of a variety? Really, one needs to know properties of $G/H$ as a Riemannian manifold and complex analytic space (e.g. curvature, whether it's a Stein space). That's the most important thing!

As Kevin Buzzard and his commentators note, under the assumption that $G$ comes from a reductive group over $Q$, and under the assumption that $H$ is a maximal compact subgroup of $G$, and under the assumption that there is a "Shimura datum" giving the quotient $G/H$ a complex structure, the quotient $G/H$ is a period domain for Hodge structures, and the quotients $Gamma backslash G / H$ are quasiprojective varieties when $Gamma$ is an arithmetic subgroup of $G$.

But these are quite strong conditions, on $G$ and on $H$! I have also wondered about other situations when $X = Gamma backslash G / H$ might have a natural structure of a quasiprojective variety. A general technique to prove such a thing is to use a differential-geometric argument. A great theorem along this line is due to Mok-Zhong (Compactifying complete Kähler-Einstein manifolds of finite topological type and bounded curvature, Ann. of Math 1989). The theorem, as quoted from MathSciNet, reads:

"Let $X$ be a complex manifold of finite topological type. Let $g$ be a complete Kähler metric on $X$ of finite volume and negative Ricci curvature. Suppose furthermore that the sectional curvatures are bounded. Then $X$ is biholomorphic to a Zariski-open subset $X'$ of a projective algebraic variety $M$."

Such results can be applied to prove quasiprojectivity of Shimura varieties of Hodge type. I believe I first learned this by reading J. Milne's notes on Shimura varieties.

I tried once to apply this to an arithmetic quotient of $G/H$, where $H$ was a bit smaller than a maximal compact (when $G/H$ was the twistor covering of a quaternionic symmetric space) -- I couldn't prove Mok-Zhong's conditions for quasiprojectivity, and I still don't know whether such quotients are quasiprojective.

co.combinatorics - On the Bell Numbers

It's easy to see that $B_n ge 2 B_{n-1}$ since we always have a choice of whether to add $n$ to the same part as $n-1$ or not. Since the number of parts in a typical set partition of size $n-1$ grows, the choices for adding $n$ to a new or existing part grow, so

$$lim_{ntoinfty} B_{n-1}/B_n = 0.$$
There are asymptotics in the Wikipedia article on the Bell numbers, but it may not be obvious how to work with the Lambert $W$-function in that expression, or how to bound $B_{n-1}/B_n$. A faster proof that the limit is $0$ can be obtained from Dobrinski's formula, that $B_n$ is the $n$th moment of a Poisson distribution with mean $1$:

For any $c in mathbb R$, the Poisson distribution has positive probability of being greater than $c$. So, for large enough $n$, the $n$th moment $B_n$ is at least $c^n$. By Jensen's inequality, the moments satisfy

$$B_n^{frac{n+1}{n}} le B_{n+1}$$

$$c le sqrt[n]{B_n} le frac {B_{n+1}}{B_n}$$

ag.algebraic geometry - What is the Hirzebruch-Riemann-Roch formula for the flag variety of a Lie algebra?

Riemann-Roch for the flag variety is the Weyl Character formula!

More specifically, let $L$ be an ample line bundle on $G/B$, corresponding to the weight $lambda$. According to Borel-Weil-Bott, $H^0(G/B,L)$ is $V_{lambda}$, the irreducible representation of $G$ with highest weight $lambda$, and $H^i(G/B,L)=0$ for $i>0$. So the holomorphic Euler characteristic of $L$ is $mathrm{dim} V_{lambda}$.

As we will see, computing the holomorphic Euler characteristic of $L$ by Hirzebruch-Riemann-Roch gives the Weyl character formula for $mathrm{dim} V_{lambda}$.

Notation:

$G$ is a simply-connected semi-simple algebraic group, $B$ a Borel and $T$ the maximal torus in $B$. The corresponding Lie algebras are $mathfrak{g}$, $mathfrak{b}$, $mathfrak{t}$. The Weyl group is $W$, the length function on $W$ is $ell$ and the positive roots are $Phi^{+}$. It will simplify many signs later to take $B$ to be a lower Borel, so the weights of $T$ acting on $mathfrak{b}$ are $Phi^{-}$.

We will need notations for the following objects:
$$rho = (1/2) sum_{alpha in Phi^{+}} alpha.$$
$$Delta = prod_{alpha in Phi^{+}} alpha.$$
$$delta = prod_{alpha in Phi^{+}} (e^{alpha/2}-e^{-alpha/2}).$$

They respectively live in $mathfrak{t}^*$, in the polynomial ring $mathbb{C}[mathfrak{t}^*]$ and in the power series ring $mathbb{C}[[mathfrak{t}^*]]$.

Geometry of flag varieties

Every line bundle $L$ on $G/B$ can be made $G$-equivariant in a unique way. Writing $x$ for the point $B/B$, the Borel $B$ acts on the fiber $L_x$ by some character of $T$. This is a bijection between line bundles on $G/B$ and characters of $T$. Taking chern classes of line bundles gives classes in $H^2(G/B)$. This extends to an isomorphism $mathfrak{t}^* to H^2(G/B, mathbb{C})$ and a surjection $mathbb{C}[[mathfrak{t}^*]] to H^*(G/B, mathbb{C})$. We will often abuse notation by identifiying a power series in $mathbb{C}[[mathfrak{t}^*]]$ with its image in $H^*(G/B)$.

We will need to know the Chern roots of the cotangent bundle to $G/B$.
Again writing $x$ for the point $B/B$, the Borel $B$ acts on the tangent space $T_x(G/B)$ by the adjoint action of $B$ on $mathfrak{g}/mathfrak{b}$. As a $T$-representation, $mathfrak{g}/mathfrak{b}$ breaks into a sum of one dimensional representations, with characters the positive roots. We can order these summands to give a $B$-equivariant filtration of $mathfrak{g}/mathfrak{b}$ whose quotients are the corresponding characters of $B$. Translating this filtration around $G/B$, we get a filtration on the tangent bundle whose associated graded is the direct sum of line bundles indexed by the negative roots. So the Chern roots of the tangent bundle are $Phi^{+}$. (The signs in this paragraph would be reversed if $B$ were an upper Borel.)

The Weyl group $W$ acts on $mathfrak{t}^*$. This extends to an action of $W$ on $H^*(G/B)$. The easiest way to see this is to use the diffeomorphism between $G/B$ and $K/(K cap T)$, where $K$ is a maximal compact subgroup of $G$; the Weyl group normalizes $K$ and $T$ so it gives an action on $K/(K cap T)$.

We need the following formula, valid for any $h in mathbb{C}[[mathfrak{t}^*]]$:
$$int h = mbox{constant term of}left( (sum_{w in W} (-1)^{ell(w)} w^*h)/Delta right). quad (*)$$
Two comments: on the left hand side, we are considering $h in H^*(G/B)$ and using the standard notation that $int$ means "discard all components not in top degree and integrate." On the right hand side, we are working in $mathbb{C}[[mathfrak{t}^*]]$, as $Delta$ is a zero divisor in $H^*(G/B)$.

Sketch of proof of (*): The action of $w$ is orientation reversing or preserving according to the sign of $ell(w)$. So $int h = int (sum_{w in W} (-1)^{ell(w)} w^*h) / |W|$. Since the power series $sum_{w in W} (-1)^{ell(w)} w^*h$ is alternating, it is divisible by $Delta$ and must be of the form $Delta(k + (mbox{higher order terms}))$ for some constant $k$. The higher order terms, multiplied by $Delta$, all vanish in $H^*(G/B)$, so we have $int h = k int Delta/|W|$. The right hand side of $(*)$ is just $k$.

By the Chern root computation above, the top chern class of the tangent bundle is $Delta$. So $int Delta$ is the (topological) Euler characteristic of $G/B$. The Bruhat decomposition of $G/B$ has one even-dimensional cell for every element of $W$, and no odd cells, so $int Delta = |W|$ and we have proved formula $(*)$.

The computation

We now have all the ingredients. Consider an ample line bundle $L$ on $G/B$, corresponding to the weight $lambda$ of $T$. The Chern character is $e^{lambda}$. HRR tells us that the holomorphic Euler characteristic of $L$ is
$$int e^{lambda} prod_{alpha in Phi^{+}} frac{alpha}{1 - e^{- alpha}}.$$
Elementary manipulations show that this is
$$int frac{ e^{lambda + rho} Delta}{delta}.$$

Applying $(*)$, and noticing that $Delta/delta$ is fixed by $W$, this is
$$mbox{Constant term of} left( frac{1}{Delta} frac{Delta}{delta} sum_{w in W} (-1)^{ell(w)} w^* e^{lambda + rho} right)= $$
$$mbox{Constant term of} left( frac{sum_{w in W} (-1)^{ell(w)} e^{w(lambda + rho)}}{delta} right).$$

Let $s_{lambda}$ be the character of the $G$-irrep with highest weight $lambda$. By the Weyl character formula, the term in parentheses is $s_{lambda}$ as an element of $mathbb{C}[[mathfrak{t}^*]]$. More precisely, a character is a function on $G$. Restrict to $T$, and pull back by the exponential to get an analytic function on $mathfrak{t}$. The power series of this function is the expression in parentheses. Taking the constant term means evaluating this character at the origin, so we get $dim V_{lambda}$, as desired.

dna - Altering the human genome

First of all it is important to note that, within certain limits, human DNA is not much different than, say, a mouse DNA: it has the same structure, it is constituted by the same bases etc etc. Therefore it is teorethically possible (leaving aside ethical issues, of course) to selectively modify it as you would do for a mouse. There are, however, some technical issues and obviously many ethical ones.

To answer specifically to your questions

Is it even theoretically possible to alter human genetics using only an injected substance?

Yes it is. Aside from the good example of radioactive substances given by Alexander, another good one (and probably one that is more common) is that of chemical mutagens.

It is important to understand that the mutation caused by these substances are random, or only partially selective.
For instance, certain substances, called intercalating agents can interpose themselves in between the bases of the DNA, interfering and possibly giving rise to mutations during DNA replication.

Other substances such as alkilating agents are a bit more specific, although still not gene- or locus- specific.
For instance ethylnitrosourea (ENU) favours A->T base transversions and AT->GC transitions.

I'd imagine trying to get a single DNA change to propogate through the entire body without being killed off by the immune system would be near impossible.

The immune system would not necessarily kill cells with mutations. Think of what happens during cancer developement.
Also, an extremely important point: mutating a somatic cell (i.e. not a sperm or an egg) can induce a mutant phenotype in the individual but it will not be transmitted to his/her progeny.

Since we are aware of genetic markers that are related to predispositions to certain medical conditions (e.g. heart problems, cancers, high blood pressure, strokes, etc), is it possible to modify human DNA to create these issues in a living person?

As Alexander says, there is much research in this field.
Some attempts of gene therapy have indeed been made, and some have a big potential. We are, however, still far away for gene therapy to become a routine treatment.

Also, remember that:

1. not all diseases are genetic diseases


2. not all genetic diseases are monogenic: some derive from a complex interplay of different factors and hence curing them is not as simple as providing the correct version of the mutated gene.

Has much research gone into how genetics affects thought and opinion, or the way we perceive and react to situations at an emotional level? If so, is it theoretically possible to alter this at a biological level?

Yes, but I would say that we are very very far away from a complete understanding of how thought and emotions are coded in our brain. The very difficult point is that one needs to study these events at multiple scales. We can study a big region of the brain and say that it is involved in fear coding, for instance. But then if you want to go deeper and study what happens at the single cell level you need to look at a lot of different cells to see how they interact between each other.
From the other side of the "size scale", it is possible to study biochemical processes in the single neuron, but then again it is difficult to study how these processes are integrated in big networks of cells, especially in vivo.