ac.commutative algebra - What properties define open loci in excellent schemes?

Long,

For the $mathbb{Q}$-Gorenstein question, I don't know of a reference either but it should be easy, if $omega_R^{(n)}$ is locally free at a point, then it's locally free in a neighborhood of that point (of course, I'm probably assuming normal or G1 + S2 to make sense of $omega_R^{(n)}$).

A couple other that jump to mind are the following.

I. Seminormality / weak normality.

II. Being $F$-split in characteristic $p$ (at least in the $F$-finite case, $F$-finite is another decent one on its own). Of course, all the sings of the MMP (canonical, terminal, lc, klt, slc, rational, Du Bois, etc.) Most of the singularities of tight closure theory (strong F-regularity, F-purity, F-injectivity, F-rationality, some of these requiring again the $F$-finite case).

With reguards to your second question: The general common thing that virtually all the singulraity classes mentioned above possess, which makes them open is the following:

Almost all of these singularities are checked by either showing that a particular module $M$ is isomorphic to $R$ or $omega_R$. Alternately, that a particular module is zero. For example, klt and log canonical singularities can both be checked by this (is the multiplier ideal / non-lc ideal equal to $R$). This also holds in the characteristic $p$ world, although its not the usual way things are phrased.

For example, strong $F$-regularity can be checked by looking at the ideal
$$
J = sum_{e geq 0} sum_{phi} phi(F^e_* cR)
$$
where $phi$ runs over all elements of $Hom_R(F^e_* R, R)$ and $c$ is chosen such that $R_c$ is regular. If this ideal equals $R$, then $R$ is strongly $F$-regular.

lo.logic - cardinal of a quotient space

Let me first treat the case where the
underlying set is infinite.

In the infinite case, your cardinal $beta$ is either $0$
or equal to $alpha$, depending on whether all points are
equivalent or not. The reason is that if the relation is
not trivial, then every point is inequivalent to some other
point, so $alphaleqbeta$, and conversely
$betaleqalphacdotalpha=alpha$ by infinite cardinal
arithmetic.

For question 1, the answer is therefore that $kappa$ is
not determined by $alpha$ and $beta$. As you observed,
$kappa$ is the number of classes, and the same infinite
set of size $alpha$ can be partitioned into any number
$kappa$ of classes, provided $1leqkappaleqalpha$.

For question 2, when $alpha$ is infinite, then since
$beta=alpha$ (unless there is only one class, in which
case $beta=0$), the bound $gamma$ is not very helpful. But the largest $kappa$ can be is $alpha$. (This is under AC; without AC, then it is possible that $kappa$ could be strictly larger than $alpha$, as I expain at the bottom.)

Similarly, for question 3, the smallest $kappa$ can be is
$1$, when $beta=0$, and otherwise, $kappa=2$ is possible,
since you can divide $alpha$ into $2$ classes, each of
size $alpha$.

In the infinite case, there are some interesting
issues that arise with the Axiom of Choice in this
question. Your observation that the quotient has size
$kappa$ seems to rely on AC, since the chains are
essentially choice functions. More generally, it was
observed in a previous MO answer by Dr. Strangechoice that
$kappa$ can actually be strictly larger than $alpha$! That is, one can partition a set into strictly more classes than there are points! For
example, consider the relation $E$ on the reals, where
$xEy$ if $x=y$ or if both $x$ and $y$ code a well order on
the natural numbers having the same order type. This is an
equivalence relation on the reals, but it is consistent
with ZF that there is no $omega_1$-sequence of reals, and
in this case there can be no injection from the $E$-classes
into the reals, since this would provide such an
$omega_1$-sequence. But there is a converse injection,
since we can injectively map reals to reals that don't code
well-orders. So this is a situation where the number of
equivalence classes is a strictly larger cardinality than
the underlying set.

---

Update. In the finite case, I happened to observe that again $kappa$ is not a function of $alpha$ and $beta$. To see this, let $sim_1$ and $sim_2$ be two relations on 6 points, the first partitioning it as $2+2+2$, with three classes, and the second partitiioning it as $3+1+1+1$, with four classes. In each case, we have $alpha=6$. But unless I have made a counting mistake, it seems we also have $beta=24$ in each case, since each equivalence relation adds 3 equivalent (unordered) pairs beyond the identity pairs, making for 12 equivalent pairs (a,b), and hence $beta=36-12=24$ inequivalent pairs in each case. But the first relation has $kappa=3$ and the second has $kappa=4$; so $kappa$ is not determined by $alpha$ and $beta$.

at.algebraic topology - What is a principal refinement of a Postnikov system?

The key idea is that not all fibrations $E to B$ with fibre an Eilenberg-MacLane space $K(pi,n)$ can be constructed by pulling the principal path fibration $K(pi,n) to PK(pi,n+1) to K(pi,n+1)$ along a classifying map $B to K(pi,n+1)$. If you can construct the fibration in this way then the classifying map is the Postnikov $k$-invariant. Clearly this at least requires that the group $pi$ is abelian.

Now, it is a nice little exercise to check that existence of a principal refinement of the Postnikov tower is equivalent to $pi_1$ being nilpotent and acting nilpotently on all of the higher homotopy groups.

(Recall that a group $G$ acts nilpotently on a group $H$ if $H$ has a finite sequence of $G$-invariant subgroups $H supset H_1 supset H_2 supset cdots H_k = 1$ such that $H_i/H_{i+1}$ is abelian and the action of $G$ on it is trivial.)

The general idea of Hilton, Mislin, and Roitberg is that is it obvious how to localise abelian groups, and nilpotent groups are those which can be assembled from abelian groups one layer at a time. So we can localise nilpotent groups by working one layer at a time, and then we can localise nilpotent spaces by working up the refined Postnikov tower one stage at a time.

at.algebraic topology - tangent sphere bundle over sphere

If you like clutching maps descriptions of bundles the sphere has a nice one. Think of $S^n$ as the union of two discs corresponding to an upper and lower hemi-sphere. Then the tangent bundle trivializes over both hemispheres. You can write down the trivializations explicitly with some linear algebra constructions. Think of the intersection of the two hemi-spheres as an $S^{n-1}$, this allows you to think of the tangent bundle as a union $D^n times mathbb R^n cup D^n times mathbb R^n$ along the common boundary $S^{n-1} times mathbb R^n$. The clutching (gluing) map is then a map of the form:

$$ c: S^{n-1} to SO_n $$

and it is explicitly the map $c(v) = M(v)M(x_{n+1})$

where $v in S^{n-1} subset mathbb R^n subset mathbb R^{n+1}$ where we think of $mathbb R^n$ as the orthogonal complement of the $(n+1)$-st coordinate vector $x_{n+1}$ in $mathbb R^{n+1}$. $M(v)$ denotes mirror reflection fixing the orthogonal complement to $v$.

The basic idea in this construction is that if one takes a geodesic between two points on a sphere, parallel transport from one point to the other can be written as a composite of two reflections, the latter reflection corresponding to the mid-point of the geodesic, the initial reflection corresponding to the initial point of the geodesic.

co.combinatorics - Is every k-edge-connected graph also k-trail-ordered?

This is an old question of Aradhana Narula-Tam and Philip Lin that I think deserves wider circulation. It appeared in Discrete Math. 257 (2002), page 613, but not many people have looked at it and it might have a simple solution.

A graph is $k$-trail-ordered if for every list $v_0, ldots, v_k$ of $k + 1$ distinct vertices, there is a trail (i.e., a walk with no repeated edges) that visits
$v_0, ldots, v_k$ in order (among a possibly longer trail). The question is, is every
$k$-edge-connected graph also $k$-trail-ordered? If not, then what is
the least edge-connectivity $f(k)$ such that every $f(k)$-edge-connected graph is $k$-trail-ordered?

Note that in the desired trail, each $v_i$ may appear repeatedly, as long as
the list of vertices visited has $v_0, ldots, v_k$ as a sublist in order. The problem arose in the context of a wavelength assignment problem in a WDM optical network.
Whitney’s Theorem on edge-disjoint paths yields $f(2)=2$, as desired. In [1] it was
proved that $f(3)=3$. Florian Pfender points out that a result of Okamura [2] yields
a proof that $f(k)=k$ when $k$ is even, and consequently $f(k) le k + 1$ when $k$ is odd.
Therefore, it remains to decide between $k$ and $k + 1$ when $k$ is odd.

[1] A. Narula-Tam, P. J. Lin, E. Modiano, Efficient routing and wavelength assignment for reconfigurable WDM networks, IEEE J. Selected Areas Comm. 20 (2002) 75–88.

[2] H. Okamura, Paths in $k$-edge-connected graphs, J. Combin. Theory Ser. B 45 (1988) 345–355.

vector bundles - Harmonic Expansion

I'd look in books on representation theory rather than books on homogeneous vector bundles for this sort of thing. Perhaps George Mackey's famous book "Unitary group representations in physics, probability, and number theory" would be of interest, and cover such facts.

To answer your question, it's probably better to consider the finite group case, before the more general Lie group case. So let $H$ be a subgroup of a finite group $G$. Let $(tau, W)$ be a finite-dimensional irreducible representation of $H$ (it's helpful to give the vector space its own name). Let $psi$ be a section of the homogeneous bundle over $G/H$ associated to $tau$. In this finite context, this simply means that $psi: G rightarrow W$ is a function which satisfies the identity:
$$psi(gh) = [ tau(h^{-1})] (psi(g)).$$

In other words, $psi$ is simply a vector in the induced representation $Ind_H^G W$. One may decompose this induced representation into irreducible representations of $G$:
$$Ind_H^G W = bigoplus_{lambda in hat G} V_lambda^{m_lambda},$$
where $(lambda, V_lambda)$ runs over the irreducible representations of $G$, and
$$m_lambda = dim(Hom_G(V_lambda, Ind_H^G W)) = dim(Hom_H(V_lambda, W))$$
is the multiplicity of $V_lambda$ in the induced representation, or equivalently (by Frobenius reciprocity) $m_lambda$ is the multiplicity of $W$ in the restriction of $V_lambda$.

Given such a decomposition, we may decompose the vector $psi in Ind_H^G W$ as a tuple (or finite sequence of tuples):
$$psi = (v_lambda^{(1)}, ldots, v_lambda^{(m_lambda)})_{lambda in hat G}.$$

This is the same, in spirit, to the formula you mentioned. The main difference is that I am averse to choosing bases (unlike many physicists). In fact, I think my dislike of choosing bases (or slight laziness) prevents me from deriving your formula on the nose. Sorry!

But to finish my answer, I should say that the entire argument above can be carried out with few changes, when working with a compact Lie group $G$ and closed subgroup $H$, and unitary representations throughout. (This situation, I would bet, is covered by Mackey). A classic example would be when $G = SO(n)$ and $H = SO(n-1)$, so that $G/H$ is homeomorphic to the $(n-1)$-dimensional sphere $S^{n-1}$.

There, when one works with finite-dimensional unitary representations, the Lie group action yields a natural action of the Lie algebra, and hence the universal enveloping algebra. In other words, the representation spaces also have actions of differential operators arising from the groups.

In particular, if one takes an irreducible subrepresentation of $Ind_H^G W$ as before, one gets a space of functions from $G$ to $W$, which is an irreducible representation of $G$ by translation. It follows that the center of the universal enveloping algebra of $G$ acts via scalars on this space of functions. Hence, if $G$ is a simple compact connected Lie group, the Casimir operator (generalization of the Laplacian) acts via a scalar on this space of functions -- i.e., the functions satisfy a nice second order differential equation. This is the source of "spherical harmonics", for example, and the reason for the word harmonic in this context.

ag.algebraic geometry - algebraic group G vs. algebraic stack BG

If G is a group scheme over k (algebraic closed), then let me talk through how to get G back by looking at the stack BG. The k points of BG (which is a groupoid) consist of one point whose automorphisms are the k points of G. The pullback of this point to Spec A for any k-algebra A has automorphisms given by the A points of G. If you think of BG points as principal bundles, I'm saying the automorphsims of the trivial bundle on Spec A are the A points of the group.

So what happens if you deform BG? You still have this one point, you can't deform that to anything, so you can only change its morphisms. That's your G' (you get an algebraic group since you can pullback to all the Spec A's). How you see it's BG' is a little trickier, so maybe I should leave it to a real algebraic geometer, but I think that the idea is that BG is distinguished by being the sheafication of the trivial bundles in the smooth/fppf topology, and this won't change when you deform.

pr.probability - Random projection and finite fields

Suppose the vectors are $e_1,dots,e_n$. The kernel of projection onto a random subspace of dimension $n+r$ is a random subspace of dimension $n-r$, so you want the probability that such a subspace has trivial intersection with the span of $e_1,dots, e_n$. Now just count the number of choices for a basis $v_1,dots, v_{n-r}$ of such a space: $2^{2n} - 2^n$ for the first vector, then $2^{2n} - 2^{n+1}$ for the second, and so on. This is to be compared with $2^{2n} - 1$ choices for the first vector if one doesn't have an restriction, $2^{2n}-2$ for the second and so on.

So the probability of this happening is the ratio of these two quantities, which you need to find a good approximation for; a very brief back-of-an-envelope calculation suggested it's about $1 - c2^{-r}$, at least if $r$ is largeish. For your specific needs, then, $d - n$ should be about $Clog n$.

st.statistics - E[log(Z_t^2)], proof of convergence with Law of Large Numbers

For every $t$, let $Y_t=log(Z_t^2)$. Fix some $t$. The sequence $(Y_{t-k})_{kgeqslant0}$ is i.i.d. with $E[Y_t]lt0$ hence the usual law of large numbers yields $frac1jsumlimits_{k=0}^{j-1}Y_{t-k}to E[Y_1]$. Fix some negative $mgt E[Y_1]$.

Then $frac1jsumlimits_{k=0}^{j-1}Y_{t-k}leqslant m$ for every $j$ large enough, that is, for every $jgeqslant J$ where $J$ is random and almost surely finite. In particular, for every $jgeqslant0$, $sumlimits_{k=0}^{j}Y_{t-k}leqslant mj+X$, for some almost surely finite random $X$. This implies the pointwise convergence of the series since
$$
sum_{jgeqslant0}expleft(sumlimits_{k=0}^{j}Y_{t-k}right)leqslantsum_{jgeqslant0}mathrm e^Xmathrm e^{mj}=mathrm e^X(1-mathrm e^m)^{-1}.
$$
Note that the RHS above is almost surely finite since $mlt0$ but is not (a priori) uniformly bounded since $X$ may be unbounded.

ag.algebraic geometry - What do heat kernels have to do with the Riemann-Roch theorem and the Gauss-Bonnet theorem?

Added 2 June:

Since the summary below is already a bit long, I thought I'd add a few lines at the beginning as a guide. The proofs all proceed as follows:

1. Identify the quantity of interest (like the Euler characteristic) as the index of an operator going from an 'even' bundle to an 'odd' bundle.




2. Use Hodge theory to write the index in terms of the dimensions of harmonic sections, i.e., kernels of Laplacians.




3. Use the heat evolution operator for the Laplacians and 'supersymmetry' to rewrite this as a 'supertrace.'




4. Write the heat evolution operator in terms of the heat kernel to express the supertrace as the integral of a local density.




5. Use the eigenfunction expansion of the heat kernel to identify the constant (in time) part of the local density.

Most of this is general nonsense, and the difficult step is 5. By and large, the advances made after the seventies all had to do with finding interpretations of this last step that employed intuition arising from physics.

---

I suffered over this proof quite a bit in my pre-arithmetic youth and wrote up
a number of summaries. A condensed and extremely superficial version is given here, mostly for my own review.
If by chance someone finds it at all useful, of course I will be delighted. I apologize that I don't say anything about physical
intuition (because I have none), and for repeating parts of the previous nice answers.
It's been years since I've thought about these matters, so I will forgo
all attempts at even a semblance of analytic rigor. In fact, the main pedagogical reason for posting is that a basic outline of the proof is possible to understand with almost no analysis.

The usual setting has a compact Riemannian manifold $M$, two hermitian bundles $E^+$ and $E^-$, and a linear
operator
$$P:H^+rightarrow H^-,$$
where $H^{pm}:=L^2(E^{pm})$.
With suitable assumptions (ellipticity), $ker(P)$ and $coker(P)$
have finite dimension, and the number of interest is the index:
$$Ind(P)=dim(ker(P))-dim(coker(P)).$$
This can also be expressed as
$$dim(ker(P))-dim(ker(P^{*})),$$ where
$$P^{*}:H^-rightarrow H^+$$
is the Hilbert space adjoint. A straightforward generalization of the Hodge theorem allows us also to write this in terms of Laplacians
$Delta^+=P^* P$ and $Delta^-=PP^*$
as
$$dim(ker(Delta^+))-dim(ker(Delta^-)).$$
Things get a bit more tricky when we try to identify the index with the expression ('supertrace,' so-called)
$$Tr(e^{-tDelta^+})-Tr(e^{-tDelta^-}).$$
The operator
$$e^{-tDelta^{pm}}:H^{pm}rightarrow H^{pm}$$
sends a section $f$ to the solution of the heat equation
$$frac{partial}{partial t} F(t,x)+Delta^{pm}F(t,x)=0$$
($x$ denoting a point of $M$) at time $t$ with intial condition $F(0,x)=f(x).$
One important part of this is that there are discrete Hilbert direct sum decompositions
$$H^+=oplus_{lambda} H^+(lambda)$$
and $$H^-=oplus_{mu} H^-(mu)$$
in terms of finite-dimensional eigenspaces for the Laplacians with non-negative eigenvalues. And then, the identities
$$Delta^-P=PP^{*}P=PDelta^+$$
and
$$Delta^+P^{*}=P^{*}PP^{*}=P^{*}Delta^-$$
show that the (supersymmetry) operators $P$ and $P^{*}$ can be used to define isomorphisms between all non-zero eigenspaces of the two Laplacians with
a correspondence for eigenvalues as well.
Thus, once you believe that the exponential operators are trace class,
it's easy to see that the only contributions to the trace are from the kernels of the plus and minus Laplacians. This is the 'easy cancellation' that occurs in this proof.
But on the zero eigenspaces, the heat evolution operators are clearly the identity, allowing us to identify the supertrace with the index.
To summarize up to here, we have
$$Ind(P)=Tr(e^{-tDelta^+})-Tr(e^{-tDelta^-}).$$
This identity also makes it obvious that the supertrace is in fact independent of $t>0$.

The proofs under discussion all have to do with identifying this supertrace in terms of local expressions that
relate naturally to characteristic classes. The beginning of this process involves first writing the operator
$e^{-tDelta^+}$ in terms of an integral kernel
$$K^+_t(x,y)=sum_i e^{-tlambda_i } phi^+_i(x)otimes phi^+_i(y)$$
where the $phi^+_i$ make up an orthonormal basis of eigenvectors for the Laplacian.
That is,
$$[e^{-tDelta^+}f](x)=int_M K^+_t(x,y)f(y)dvol(y)=sum_i e^{-tlambda_i } int_M phi^+_i(x) langle phi^+_i(y),f(y)rangle dvol(y).$$
Formally, this identity is obvious, and the real work consists of the global analysis necessary to justify the formal computation.
Obviously, there is a parallel discussion for $Delta^-$. Now, by an infinite-dimensional version of the formula
that expresses the trace of a matrix as a sum of diagonals, we get that
$$Tr(e^{-tDelta^+})=int_M Tr(K^+_t(x,x))dvol(x)=int_M sum_ie^{-tlambda_i}||phi^+_i(x)||^2 dvol(x),$$
an integral of local (point-wise) traces, and similarly for $Tr(e^{-tDelta^-})$. One needs therefore, techniques to evaluate the density

$$sum_ie^{-tlambda_i}||phi^+_i(x)||^2 dvol(x)-sum_ie^{-tmu_i}||phi^-_i(x)||^2 dvol(x).$$

More analysis gives an asymptotic expansion for the plus and minus densities of the form
$$ a^{ pm }_{-d/2}(x) t^{-d/2}+a^{ pm }_{d/2+1}(x) t^{-d/2+1}+cdots $$
where $d$ is the dimension of $M$.

Up to here the discussion was completely general, but then the proof begins to involve special cases, or
at least, broad division into classes of cases. But note that even for the special cases mentioned in the original question,
one would essentially carry out the procedure outlined above for a specific operator $P$.

The breakthrough in this line of thinking
came from Patodi's incredibly complicated computations for the operator $d+d^*$
going from even to odd differential forms,
where one saw that the
$$a^{+}_i(x)$$
and
$$a^{-}_i(x)$$
canceled each other out locally, that is, for each point $x$, for all the
terms with negative $i$. I think it was fashionable to refer to this cancellation as 'miraculous,' which it is, compared to the easy cancellation above.
At this point, Patodi could take a limit
$$lim_{trightarrow 0}[sum_ie^{-tlambda_i}||phi^+_i(x)||^2 dvol(x)-sum_ie^{-tmu_i}||phi^-_i(x)||^2 dvol(x)],$$
that he identified with the Euler form. This important calculation set a pattern that recurred in all other versions of
the heat kernel approach to index theorems. One proves the existence of an analogous
limit as $trightarrow 0$ and identifies it. The identification
as a precise differential form representative for a characteristic class is referred to sometimes as a local index theorem, a statement
more refined than the topological formula for the global index. There is even a beautiful version of a local families index theorem
that relates eventually to deep work in arithmetic intersection theory and Vojta's proof of the Mordell conjecture.

As I understand it, Gilkey's contribution was an invariant theory
argument that tremendously simplified the calculation and allowed a differential form representative for
the $hat{A}$ genus to emerge naturally
in the case of the Dirac operator. And then, I believe there is a $K$-theory argument that deduces the index theorem for a general elliptic operator
from the one for the twisted Dirac operator.

Experts can correct me if I'm wrong, but
from a purely mathematical point of view, essentially all the work on the heat kernel proof was done at this point.
Subsequent interpretations of the proof (more precisely, the supertrace), in terms of supersymmetry, path integrals, loop spaces, etc., were tremendously
influential to many areas of mathematics and physics, but the mathematical core of the index theorem itself appears to have remained largely unchanged for almost forty years. In particular, the terminology I've used myself above, the super- things, didn't occur at all in the original papers of Patodi, Atiyah-Bott-Patodi, or Gilkey.

Added:

Here is just a little bit of geometric-physical intuition regarding the heat kernel in the Gauss-Bonnet case, which I'm sure is completely banal for most people. The density
$$sum_ie^{-tlambda_i}||phi^+_i(x)||^2 dvol(x)-sum_ie^{-tmu_i}||phi^-_i(x)||^2 dvol(x)$$
expresses the heat kernel in terms of orthonormal bases for the even and odd forms. When $trightarrow infty$ all terms involving the positive eigenvalues decay to zero, leaving only contributions from orthonormal harmonic forms. This is one way to to see that the integral of this density, which is independent of $t$, must be the Euler characteristic. On the other hand, as $trightarrow 0$, the operator
$$K^+_t(x,y)dvol(y)=[sum_i e^{-tlambda_i } phi^+_i(x)otimes phi^+_i(y)]dvol(y)$$
literally approaches the identity operator on all even forms (except for the fact that it diverges). That is, the heat kernel interpolates between the identity and the projection to the harmonic forms, in some genuine sense expressing the diffusion of heat from a point distribution to a harmonic steady-state. A similar discussion holds for the odd forms as well. I can't justify this next point even vaguely at the moment, but one should therefore think of $$[K^+_t(x,y)-K^-_t(x,y)]dvol(y)$$ as regularizing the current on $Mtimes M$ given by the diagonal $Msubset Mtimes M$. Thus, the integral of $$[TrK^{+}_t(x,x)-TrK^-_t(x,x)]dvol(x)$$ ends up computing a deformed self-intersection number of the diagonal in $Mtimes M$. From this perspective, it shouldn't be too surprising that the Euler class, representing exactly this self-intersection, shows up.

Added:

I forgot to mention that the Riemann-Roch case is where $$P=bar{partial}+bar{partial}^*$$
going from the even to the odd part of the Dolbeault resolution associated to a holomorphic vector bundle. The limit of the local density is a differential form representing the top degree portion of the Chern character of the bundle multiplied by the Todd class of the tangent bundle. Perhaps it's worth stressing that these special cases all go through the general argument outlined above.

ag.algebraic geometry - Comparison of Local and Global Duality

In the case of varieties over a perfect field this question is explained with great detail in the book by J. Lipman:

Dualizing sheaves, differentials and residues on algebraic varieties. (French summary)
Astérisque No. 117 (1984)

known by some people as "Lipman's blue book". If you want to get rid of the base field then you should look at Greenlees May duality for schemes and Grothedieck duality for formal schemes, see joint work by Alonso, Jeremías & Lipman.

ds.dynamical systems - topologically mixing subshifts without ergodic measures

Let me go a little further: I believe that there also exists a topologically mixing subshift on two symbols with no fully-supported invariant measure. I don't know of a reference for this result, but I think that a direct construction should be "not too hard", in the sense that a completely detailed proof would take up perhaps five pages or so.

Here is a very rough sketch. We want to find a sequence $x=(a_n)_{n=-infty}^infty in {1,2}^{mathbb{Z}}$ such that the orbit closure of $x$ is a subshift with the desired properties. Let us denote this orbit closure by $A$, and the shift by $T$. Our criteria mean the following:

1) Topological mixing. This means that for every pair of subwords $u$ and $v$ of $(a_n)_{n=-infty}^infty$, there exists an integer $N(u,v)>0$ such that for all $n geq N(u,v)$, we can find a word $omega$ of length $n$ such that $uomega v$ is a subword of $(a_n)_{n=-infty}^infty$. Since every nonempty open set contains a cylinder I think it is not too hard to see that this implies topological mixing.

2) There is no fully-supported invariant probability measure on $A$. In particular, every invariant measure on the orbit closure gives zero measure to the cylinder set ${(b_n)_{n=-infty}^{infty} colon b_0=2}$. To achieve this we want $(a_n)_{n=-infty}^infty$ to have the following property: there is a sequence of positive real numbers $(varepsilon_n)_{n=1}^infty$, which decreases to zero, such that for every $n in mathbb{N}$, every subword of $(a_n)_{n=-infty}^infty$ with length $n$ contains at most $nvarepsilon_n$ instances of the symbol $2$. If this property holds, then it is not difficult to see that
[lim_{ntoinfty}sup_{x in A} frac{1}{n}sum_{i=0}^{n-1}chi_{{x_0=2}}left(T^ixright)=0]
and by the Birkhoff ergodic theorem combined with the ergodic decomposition theorem, this forces every invariant measure on $A$ to give zero measure to the aforementioned cylinder set.

I am not really going to say much more, except that I think that these requirements can be seen to be compatible, in the sense that such a sequence $(a_n)_{n=-infty}^infty$ exists. To give a very rough idea of how it might be constructed, we could start like this. Begin by defining $a_0:=2$, and $a_{pm1}:=1$. We will build the sequence outward from the origin in a sequence of inductive stages, roughly as follows. At each stage, we consider the set of all subwords of the finite sequence defined in the previous stage. We construct the new stage by appending copies of these subwords to the sequence defined in the previous stage in a careful manner. To ensure that (1) holds for the limit construction, we will need to include an awful lot of copies of every pair of subwords from the previous stage, separated by big lists of ones, once each for each pair of subwords and each different separating length in a certain range depending on the pair of subwords. On the other hand, to ensure that (2) holds, we need to make sure that every time a subword containing some twos is appended, it is padded out on both sides with big blocks of ones to a sufficient extent that we do not create a subword with too high a proportion of ones in it. The easiest way to achieve this is to ensure that $N(u,v)$ is always very, very, very large relative to the lengths of $u$ and $v$.

I think that this program can be followed, although I haven't gone to the significant effort of trying to write down all of the details. If there is a general moral, then I think it is this: there are an awful lot of subshifts, and if a small collection of criteria on a possible subshift does not lead to an obvious contradiction, then the desired subshift can probably be constructed by this kind of combinatorial procedure.

ac.commutative algebra - What is the set of possible values of the degree of the sum of two algebraic numbers with fixed degrees?

This isn't a solution, just a comment that got too long for the comment box: Assuming that all finite groups occur as Galois groups over $k$, the answer to this question should only depend on the characteristic of $k$.

Consider the more detailed question:

For a finite group $G$, and subgroups
$H_1$, $H_2$ and $H_3$, is there a
Galois extension $L$ of $k$ with
Galois group $G$, and elements $v_i
> in L$ such that the stabilizer of
$v_i$ is $H_i$ and $v_1+v_2+v_3=0$.

I claim that, for given $(G, H_1, H_2, H_3)$, assuming that there is some $G$-extension of $k$, the answer to this question only depends on the characteristic of $k$. Proof: As a $G$-representation, $L$ is the permutation representation on $X:=G/(H_1 cap H_2 cap H_3)$. The question, then, is whether we can find $L$-valued functions, $f_i$ on $X$, such that $f_i$ is constant on $H_i$ orbits (but not for any larger subgroup) and $f_1+f_2+f_3=0$. This is a collection of linear equalities and inequalities in $3 |X|$ variables, with integer coefficients. So whether or not they have a solution depends only on the characteristic of $k$ (we are using that $k$ is infinite). To see that the characteristic can matter, take $G=S_3$, $H_1$ and $H_2$ two different subgroups of order $2$ and $H_3$ the subgroup of order $3$. You should get a solution in characteristic $3$, and not otherwise.

Of course, answering the original question just means answering this question for all $(G, H_1, H_2, H_3)$ with $|G/H_i| = d_i$. (One can immediately make two reductions. First, a necessary condition is that $H_1 cap H_2 = H_1 cap H_3= H_2 cap H_3$. Second, one can immediately reduce to the case that $H_1 cap H_2$ contains no nontrivial normal subgroup. The latter means that $|G| leq (d_1 d_2)!$, so the problem is finite.)

I see no reason to believe that you will get a nicer answer by forgetting the groups and only remembering the degrees, but of course I haven't thought very hard about the problem.

ag.algebraic geometry - Compactifiable morphisms

Let's say a morphism $f:Xto Y$ is compactifiable if it admits a factorization $f = pj$ with $j:Xto P$ an open immersion and $p:Pto Y$ proper.

In SGA 4 Exp. XVII, Deligne says that Nagata proved that any morphism of separated integral northerian schemes is compactifiable but that he didn't understand the proof.

My questions:

• Where can I find a proof of Nagata's theorem?


• What about the complex analytic setting?

ho.history overview - Archive of the Work of J Sutherland Frame

The short answer is "no". Frame did interesting work, though usually
outside the conceptual mainstream of representation theory. Some of his
calculations of character tables (such as that of the most exceptional
Weyl group) have permanent value, I think.

co.combinatorics - partition of a set

First, we see this example.
Suppose we have a set of 6 elements, we can get 3 subsets of it, each of which has 2 elements, but no two sets overlap.
But if our set has 5 elements, we want to get 3 subsets of it each of which has 2 elements
. Then two of them need to overlap on 1 element.
Now generally suppose we have a set of #a elements and we want 3 subsets of it each of which has #a' elements.
We also want each two of them have the same but least # in common.
Could you get the relationship between a and a'? How many does any two of them in common?
How many does three of them has in common?
For example, if a=3a', the we can have no pair of the three subsets overlap.

ag.algebraic geometry - Intersection numbers and theorem of the cube

On a smooth projective surface, the intersection number of two divisors $C,D$ may be defined as $chi(O)-chi(O(C))-chi(O(D))+chi(O(C+D))$ (Hartshorne, ex. V.1.1). Bilinearity of the intersection number translates to the formula $sum_{Isubset{1,2,3}}(-1)^{|I|}chi(O(sum_{iin I}C_i))=0$ which is somewhat similar to the theorem of the cube in the form $bigotimes_{Isubset{1,2,3}}m_I^*mathcal L^{(-1)^{|I|}}cong O$. Is this just a coincidence, or is there any deeper reason for this similarity?

ca.analysis and odes - Factoring and solving trinomials

One can of course apply general algorithms for irreducibility testing and factorization, so I presume you are asking if there is something more efficient or more explicit that can be said in the case of trinomials. Except for special cases I don't believe that is the case.

While it is known that every binomial in $mathbb{Q}(x)$ must have an irreducible factor that is either binomial or trinomial, no analogous bound is known in the trinomial case. It is at least 8 terms due to the known [1] example

$$f(x)f(-x) = - x^{14} - 27180501562500 x^2 + 1244325625000000$$

for $f(x) = x^7 + 20 x^6 + 200 x^5 + 2450 x^4 + 29000 x^3 + 545000 x^2 + 8101250 x + 35275000$

1 Choudhry and A. Schinzel (1992).
On the number of terms in the irreducible factors of a polynomial over $mathbb Q$.
Glasgow Mathematical Journal (1992), 34, 11-15.

To dig deeper I suggest starting with the work of Schinzel - who has studied these and related factorization problems intensively for almost half a century, e.g. see

MR1254093 (95d:11146) 11R09 (12E05 12E10)
Schinzel, Andrzej.
On reducible trinomials.
Dissertationes Math. (Rozprawy Mat.) 329 (1993), 83 pp.

Let $K$ be a field. It is well known
that a binomial $x^n+ain K[x]$ is
reducible iff it has the form
$x^{pk}-b^p$ ($p$ prime) or
$x^{4k}+4b^4$. In this treatise the
reducibility of trinomials
$x^n+ax^m+b$ $(a,bneq 0)$ is
investigated. It turns out that the
situation is very complicated. A
satisfactory answer is obtained if $K$
is a rational function field. For
algebraic function fields in one
variable and for algebraic number
fields, less complete results are
proved. It is assumed throughout that
the characteristic of $K$ does not
divide $mn(n-m)$.

It is easy to find trinomials with
linear or quadratic factors. Table 1
of this paper provides additional
families of reducible trinomials if
$(n,m)$ belongs to a list of 12 pairs,
the largest being $(15,5)$. Perhaps
the simplest example is
$$x^6+4(v+1)x^2-v^2=(x^3+2x^2+2x-v)(x^3-2x^2+2x+v).$$ Every reducible trinomial
$f(x)=x^n+ax^m+b$ gives rise to
additional examples by considering
$u^nf(x^l/u)$ (with $uin K^times$
and $lgeq 1$) or $x^nf(1/x)/b$.
Theorem 1 essentially states that
every reducible trinomial arises in
this manner from the examples
indicated before if $K$ is a rational
function field. (More precisely, it is
assumed that $a^{-n}b^{n-m}$ is not a
constant.) Table 2 lists $7$ families
of reducible trinomials
$x^n+A(v,w)x^m+B(v,w)$ with $(v,w)in
> E(K)$, where $A$, $B$ are polynomials
over $mathbb Z$ and $E$ is an
elliptic curve defined by an equation
$z^2=C(w)$, where $C$ is a monic
polynomial over $mathbb Z$. The
polynomials $A,B$ and the
corresponding factorizations of the
trinomials are too complicated to be
included in this review. (For the
largest pair $(n,m)=(21,7)$ the
corresponding $A$ fills 10 lines in
the paper.) In Theorem 2 it is assumed
that $K$ is a finite extension of a
rational function field $F(t)$ such
that $overline FK$ has genus $g>0$
and $a^{-n}b^{n-m}notin
> overline{F}$. If $g=1$ then there are
no additional examples of reducible
trinomials. If $g>1$ then essentially
new examples with $n<24g$ may exist.
Theorem 3 reduces the case where $K$
is a finite separable extension of
$F(t)$ and $a^{-n}b^{n-m}inoverline
> F$ to studying reducibility over
$Kcapoverline F$. If $K$ is an
algebraic number field then for fixed
$n$, $m$ a finite number of
essentially new examples of reducible
trinomials $x^n+ax^m+b$ may exist
(Theorem 6). The author conjectures
that for every $K$ there is only a
finite number of these ``sporadic
trinomials''. If the conjecture holds
then there exists a constant $c(K)$
such that every trinomial over $K$ has
an irreducible factor with at most
$c(K)$ nonzero coefficients
(Consequence 2). Table 5 contains all
52 sporadic trinomials over $mathbb
> Q$ known to the author. Their degrees
lie in the range from $8$ to $52$. The
rest of the paper is devoted to
studying the reducibility of
$ax^n+bx^m+cinmathbb Z[x]$. Theorem
9 (refining a result of Nagell)
derives necessary conditions, which in
the case $(m,n)=1$ yield an explicit
bound for $b$ in terms of $a,c,m,n$.
For every positive integer $d$ there
exist only finitely many $n,m,b$ with
$n/(m,n)>d$ and $|b|>2$ such that
$x^n+bx^mpm 1$ has a factor of degree
$d$; and these can be effectively
computed. Theorem 10 derives necessary
conditions from the existence of a
factor (of $ax^n+bx^m+c)$ of given
degree $d$. These imply that there
exists $n_0(d)$ such that $x^n+bx^m+1$
is irreducible if $ngeq n_0(d)$,
$nneq 2m$, $|b|>2$. By Theorem 8, for
every $n$ there exist only finitely
many reducible trinomials $x^n+bx^m+1$
with $nneq 2m$.

The proof of Theorem 10 does not
depend on the other results of the
paper. The same applies to Theorem 9.
All other theorems except for Theorem
3 are based on lower estimates for the
genus of certain function fields.
These estimates show that the
existence of a factor of degree $k$ of
$x^n+ax^m+bin K[x]$ imposes severe
restrictions on $k,m,n,a,b$ provided
$K$ is a function field. The remaining
cases are treated in a long series of
lemmas applying to every field $K$
whose characteristic does not divide
$mn(n-m)$. In several cases the proofs
require extensive manipulations (with
polynomials in several variables)
which were performed by means of
computer algebra systems. Faltings'
theorem (solving Mordell's conjecture)
is invoked in the proof of Theorem 6
(dealing with number fields). Theorems
7 and 8 (concerning
$ax^n+bx^m+cinmathbb Z[x])$ are
proved by using the corresponding
theorems for rational function fields
together with a lemma which may be
viewed as a refinement of Hilbert's
irreducibility theorem. The proof of
this lemma is based on Siegel's
theorem (on integral points of curves
of positive genus) and on a result of
Maillet (1919) dealing with rational
functions over $mathbb Q$ taking
infinitely many integral values at
rational points.

{Reviewer's remarks: In Theorem 2 the
term $u^{nu-mu}$ in the expression
for $B$ has to be replaced by $u^nu$.
The proof of Lemma 27 employs Lemma
2(c) although this lemma only applies
to separable extensions. In order to
prove Lemma 49 one has to know that
every finite separable extension $L$
of $K(t)$ with $Lsubseteq overline
> K(t)$ is contained in $K'(t)$ for some
separable extension $K'$ of $K$. (One
can in fact prove that $L=K'(t)$ for
suitable $K'$. This need not be true
for inseparable $L$.) The proof of
Theorem 6 is apparently based on the
incorrect assumption that a divisor
$P$ of a function field $L=K(t,y)$ has
degree $1$ or is ramified with respect
to $K(t)$ if $t$ and $y$ are congruent
to elements of $Kbmod P$.}

REVISED (1995)

Reviewed by G. Turnwald

soft question - Your favorite surprising connections in Mathematics

Here is one of my favorite, that I learned from A. G. Khovanskii: let $f$ be a univariate rational function with real coefficients. Then, you can think of $f$ as inducing a continuous self-map of $mathbb{RP}^1 cong S^1$, in particular, it has a topological degree, say $[f]$, and if $f$ happens to be a polynomial, it is obvious that $[f]=0$ if $deg(f)$ is even, and that $[f]=pm 1$ if $deg(f)$ is odd (depending on the sign of the main coefficient).

If the decomposition of $f$ in continued fraction is
$$ f=P_0+frac{1}{P_1+frac{1}{P_2+ddots}}$$
Then one can prove easily that $[f]$ is the (finite) sum: $[f]=sum_{i geq 0} (-1)^i[P_i]$.
(Khovanskii himself taught this to high-schoolers in Moscow.)

The interesting connection for me follows: for any real polynomial $P$, the topological degree of the fraction $P'/P$ is clearly the (negative of the) number of real roots of $P$. Thus, the computation formula above applied to $[P'/P]$ allows us to recover Sturm's theorem.

I don't know if it really qualifies as a new proof of the theorem, but it's definitely a different point of view on that proof.

soft question - What happens to the boundary conditions as a PDE is approximated by a lesser order PDE?

The question seems to contain a misprint, as Harald Hanche-Olsen pointed out above. I am assuming that the real situation is as follows: $ epsilonnabla^4u + cnabla^2u-lambda u=F(x,y)$, where $ (x,y)inGamma $, with a pair of boundary conditions, say $ u=u'=0$, where $ (x,y)indeltaGamma $.

When $epsilon$ is "small" it is natural to assume that solutions of this PDE should be "close" in some sense to solutions of the reduced PDE $ cnabla^2u'-lambda u'=F(x,y) $.
This is a typical singular perturbation problem (see Van Dyke's "Perturbation methods in fluid mechanics" or any other text on asymptotic expansions that contains a discussion of singular perturbation problems). It is clear that since the order of the original equation has been reduced, one can now only satisfy one boundary condition, not two of the original problem. This basically indicates the presence of a boundary layer. This boundary layer is a solution of another asymptotic limit of the problem, the one for which $epsilonnabla^4u$ can no longer be omitted (see the comment by Willie Wong). The problem can generally be non-trivial in 2D; let me just show the idea how this type of problems is solved in 1D.

Consider for $xge 0$
$$ epsilon^2u_{,xxxx}-c^2u_{,xx}-lambda u=F(x) quadmbox{subject to}quad u(0)=u_{,x}(0)=0 $$
(I made some parameter changes to simplify the forthcoming algebra and changed the sign of $c^2$ as it actually makes more sense to pre-tension beam anyway). The "regular" solution $u_r$, as you pointed out, satisfies $c^2u_{r,xx}+lambda u_r=-F(x)$. The boundary layer solution is typically extracted by "zooming" into the vicinity of the boundary by using a coordinate transformation $xi=x/epsilon^{alpha}$ (so that $partial/partial x=epsilon^{-alpha}partial/partialxi$). How do we find $alpha$? Intuitively, we want to keep the previously vanishing term $epsilon^2u_{,xxxx}$ in the equation and the non-trivial answer should balance it with at least one other term of the original equation. If $alpha=1/2$ then first term is balanced by the third at $O(1)$, but the second term becomes $O(epsilon^{-1})$. Hence, one needs to take $alpha=1$ to balance the first two terms with the resulting equation
$$ epsilon^{-2}u_{,xixixixi}-c^2epsilon^{-2}u_{,xixi}-lambda u=F(x). $$
Clearly, when $epsilon$ is "small", the solution of this equation is approximated by the solution of boundary layer equation $u_{bl,xixi}-c^2u_{bl}=0$. What happens next depends on the sign of $c^2$. I shall take $c^2$ to be strictly positive. Then the boundary layer solution is a rapidly decaying exponent $u_{bl}=U_{bl}exp(-cxi)=U_{bl}exp(-cx/epsilon)$.

Let us now assume that the general solution of our problem is given by sum $u=u_r+u_{bl}$ (this is so-called Vishik-Lyusternik method). This means that when you were posing boundary condition $u_{,x}(0)=0$, you effectively posed
$$ u_{r,x}(0)+u_{bl,x}(0)=0 $$
Since $u_{bl}=U_{bl}exp(-cx/epsilon)$, then $u_{bl,x}=-cU_{bl}exp(-cx/epsilon)/epsilon=-cu_{bl}/epsilon$. Therefore, we can conclude that
$$ u_{r,x}(0)+u_{bl,x}(0)=u_{r,x}(0)-cu_{bl}(0)/epsilon=0 quadmbox{so that}quad u_{bl}(0)=epsilon u_{r,x}(0)/c. $$
If we now turn our attention to the original first boundary condition, the same line of argument gives
$$ u(0)=u_r(0)+u_{bl}(0)=u_r(0)+epsilon u_{r,x}(0)/c=0. $$

The conclusion then is: instead of solving the original fourth-order ODE with clamped boundary conditions, for sufficiently small $epsilon$ one can simply solve the following second order boundary value problem
$$ c^2u_{r,xx}+lambda u_r=-F(x) quadmbox{subject to the "effective" condition}quad u_r(0)+epsilon u_{r,x}(0)/c=0. $$
The resulting solution error will be of the order $O(epsilon^2)$ (except in the close vicinity of the boundary where we made error $O(epsilon)$). Your own suggested answer $u_r(0)=0$ is only accurate to within $O(epsilon)$ everywhere. My explanation is quite loose here, but I hope it should give you the idea of how to tackle this kind of problems.

2 knots - Alexander polynomial or Reidemeister torsion for knotted surfaces?

Yes for $S^2$, and more generally, depending on what you are after.

Given any torsion module $M$ over the PID $Q[t,t^{-1}]$, the order of $M$ is well defined in
$Q[t, t^{-1}]$ up to units, in the usual way. Moreover $Motimes Q(t)=0$. These two facts are at the heart of why the Alexander polynomial is related to Reidemeister torsion. The way it works is that if $C$ is a f.g. chain complex over $Q[t, t^{-1}]$ its homology is torsion iff $Cotimes Q(t)$ is acyclic. Then the two notions of torsion are related by

$$prod_k order(H_k(C))^{(-1)^k}=tau(C)$$

If $S^nsubset S^{n+2}$ and $X$ its complement, then $X$ has a $Z$ cover $tilde{X}$ and hence an exact sequence of chain complexes $0to C(tilde{X})to C(tilde{X})to C(X)to 0.$ The corresponding long exact sequence shows that $H_k(tilde{X})$ is torsion for all $k$, since $H_k(X)=H_k(S^1)$.

Setting $Delta_k$ to be the order of $H_k(tilde{X})$ gives the $k$-th Alexander polynomial. You get Reidemeister torsion equals the multiplicative Euler characteristic:

$$tau= prod Delta_k^{(-1)^k}$$

In the case of $S^1subset S^3$ Poincare duality relates $Delta_2$ and $Delta_1$, and $Delta_0$ is independent of the knot, so you recover Alexander poly "=" Reidemeister torsion. But in general $tau$ combines all the $Delta_k$.

All this (including how to pick the basis of the acyclic complex) is explained in Milnor's article "infinite cyclic covers."

As far as what happens more generally, all this extends, but you have to be careful when the homology is not torsion, which can happen for surfaces in $S^4$. Then you have to view Reidemeister torsion as a function of the homology , etc. Milnor's other Torsion articles are worthwhile reading for these topics.

rt.representation theory - Representations of finite Coxeter groups

More context is needed for this question. I am going to address the case of generic representations in the spherical case.

It is an old theorem of Lusztig that over the ring $mathbb{C}[v,v^{-1}]$, the generic Hecke algebra $H_v$ with generators $T_i$ and relations

$$T_i T_j ldots = T_j T_i ldots (m_{ij} text{ factors}), qquad (T_i+v^2)(T_i-v^2)=0$$

is isomorphic to the group ring of the corresponding finite Coxeter group $W.$ Thus every representation of $W$ can be canonically deformed to a representation of $H_v.$ In the course of developing representation theory of reductive groups over a finite field, Lusztig developed quite a bit of machinery describing these representations (fake degrees, etc). This is described in his book

G. Lusztig, Characters of reductive groups over a finite field. Annals of Mathematics Studies, 107. Princeton University Press, Princeton, NJ, 1984

A more recent source, in a more general situation and with improved proofs, is

G. Lusztig, Hecke algebras with unequal parameters. CRM Monograph Series, 18. American Mathematical Society, Providence, RI, 2003

For specific non-zero $v$ that are not roots of unity, the same result holds. The case of roots of unity and of more general Hecke algebras (in particular, for Coxeter system of affine type) has also been studied.

dg.differential geometry - existence of Morse functions satisfying the Palais-Smale condition

Well, I have been away from this kind of question for a long while, so please don't the following remarks as definitive in any way, but I am not aware of any counter-example and in the infinite dimensional case I also do not know any positive result. So, I think it is a nice question you are asking and if you can prove something in this direction it would be interesting and probably publishable.

A relatively minor point---I assume that you mean a differentiable Hilbert manifold, and you probably want to assume the manifold is separable. With those assumption I believe it is not hard to show that the manifold can be smoothly embedded as a closed submanifold of Hilbert space, so it gets a complete Riemannian metric, and it would be really nice if you could show that for some such embedding there was a ``height function'' (i.e., a continuous linear functional on the Hilbert space) that when restricted to the manifold satisfied Condition C. If I had to bet, I would guess this is so.

at.algebraic topology - Maslov index of a pullback bundle

When you have a vector bundle on a manifold $X$ with boundary, trivialised over $partial X$, there are characteristic classes valued in $H^ast (X,partial X)$. Here, when $L$ is orientable, the Maslov index is twice the first Chern class of $u^ast TM$ relative to the trivialisation on the boundary induced by $L$, evaluated on $[Sigma,partial Sigma]$.

When $Sigma$ is closed, the Chern number of $u^ast TM$ is the signed count of zeroes of a transversely-vanishing section $s$ of $u^astLambda^{max}_{mathbb{C}}TM$.

When there is an orientable boundary condition, the relative Chern number is the same thing, but you choose $s$ non-vanishing along the boundary and tangent to the real line sub-bundle $Lambda^{max}_{mathbb{R}} TL$.

This doesn't make sense when $u^*|_{partial Sigma} TL to partial Sigma$ is not orientable: its top exterior power then has no non-vanishing section. Besides, the Maslov index is odd in this case.

ADDED: Here's a proof using the method of Robbin's appendix to McDuff-Salamon ("$J$-holomorphic curves and symplectic topology"). Robbin characterises the boundary Maslov index as an invariant of bundle pairs (complex vector $E$ bundle over a surface, totally real sub-bundle $F$ over the boundary) which is additive under direct sum and under sewing boundaries and is suitably normalised for line bundles over the disc. The uniqueness proof, by "pair-of pants induction", still applies when $F$ is assumed orientable. The invariant "twice the relative Chern number" evidently satisfies the direct sum and sewing properties, and the section $zmapsto z$ of the trivial line bundle over the disc satisfies the standard Maslov-index 2 boundary condition. Done!

nt.number theory - Isolated quadratic residues in integers mod p

In addition to the character sum approach explained by Charles and Robin, which gives you a precise formula for the number of NRN patterns $mod p$ for any $p$, in the case $pequiv 3(mod 4)$ there is an elementary proof that avoids elliptic curves or Gauss sums. (Notation: N is quadratic non-residue, R is quadratic residue.)

I claim that it is sufficient to find an arithmetic progression $mod p$ of the form $-a^2,1,-b^2$ with non-zero $a,b.$ Indeed, $-a^2$ and $-b^2$ are N and 1 is R. Choose the sign so that $n=pm(1+a^2)$ is R: this is possible since $-1$ is a quadratic non-residue $(mod p),$ by the assumption on $p$. Dividing by $n$, we'll get three consecutive non-zero numbers $mod p$ in pattern NRN. Thus it remains to show that $a^2+b^2=-2 (mod p)$ has a solution with $a,bne 0,$ which is a special case of the well known fact that for $pgeq 7,$ any non-zero element of $mathbb{Z}/pmathbb{Z}$ is the sum of two non-zero squares. Conversely, all NRN patterns are obtained in this way and with a tiny bit of extra work, one can count that there are $(p+1)/8$ or $(p-3)/8$ of them, depending on whether $pequiv 7(mod 8)$ or $pequiv 3(mod 8).$

books - Matsumura: "Commutative Algebra" versus "Commutative Ring Theory"

By comparing the tables of contents, the two books seem to contain almost the same material, with similar organization, with perhaps the omission of the chapter on excellent rings from the first, but the second book is considerably more user friendly for learners. There are about the same number of pages but almost twice as many words per page. The first book was almost like a set of class lecture notes from Professor Matsumura's 1967 course at Brandeis. Compared to the second book, the first had few exercises, relatively few references, and a short index. Chapters often began with definitions instead of a summary of results. Numerous definitions and basic ring theoretic concepts were taken for granted that are defined and discussed in the second. E.g. the fact that a power series ring over a noetherian ring is also noetherian is stated in the first book and proved in the second. The freeness of any projective modules over a local ring is stated in book one, proved in the finite case, and proved in general in book two. Derived functors such as Ext and Tor are assumed in the first book, while there is an appendix reviewing them in the second. Possibly the second book benefited from the input of the translator Miles Reid, at least Matsumura says so, and the difference in ease of reading between the two books is noticeable. Some arguments in the second are changed and adapted from the well written book by Atiyah and Macdonald. More than one of Matsumura's former students from his course at Brandeis which gave rise to the first book, including me, themselves prefer the second one. Thus, while experts may prefer book one, for many people who are reading Hartshorne, and are also learning commutative algebra, I would suggest the second book may be preferable.

Edit: Note there are also two editions of the earlier book Commutative algebra, and apparently only the second edition (according to its preface) includes the appendix with Matsumura's theory of excellent rings.

big picture - Intuitive and/or philosophical explanation for set theory paradoxes

The history of axiomatic set theory did not proceed in the way that you suggest here. Zermelo, for example, was motivated to form his axiomatic system primarily in order to give a careful proof of his well-ordering theorem, and not to avoid the set-theoretic antinomies. If trying to avoid contradiction were our primary goal, then we would almost surely scale way back and work with much weaker axiomatic systems such as those described in Simpson's book Subsystems of Second-Order Arithmetic, which suffice for a huge fraction of mathematics. Thus your "if this axiom system does not work, let us just toy with it until we get something that looks consistent" is a straw man. I don't know of anybody who has done serious work in foundations that has taken anything even remotely resembling that attitude.

That said, you can still ask for some justification for why we should expect, say, ZFC to be consistent. Michael Greinecker has given a good answer. I would like to add one key word to his account that you may find helpful if you want to do more reading on this subject: impredicativity. Intuitively, the set-theoretic paradoxes all arise because of "self-reference" in some sense. We define something by quantifying over a set that contains the thing being defined. The intuition is that if we avoid such "impredicative" definitions, by defining new sets only in terms of sets that we have already constructed, we should block the paradoxes.

I should note, however, that ZFC is generally regarded as being impredicative, so this doesn't fully answer your question in the case of ZFC. Nevertheless, the consistency of ZFC is typically justified by the describing the so-called cumulative hierarchy of sets, which is built from the ground up, and thus is based on the same intuition that if you define things in stages with each stage building on the previous stage, then self-referential loops should not have any way of arising.

sheaf theory - What books should I read before beginning Masaki Kashiwara's "Sheaves on Manifolds"

The Kashiwara's book is quite focused and technical. I won't recommend it as an introduction to sheaves, since the abstract language of sheaves and homological algebra is most useful when you already know a big class of examples.

If you're planning on hitting algebraic geometry one day, it could be a good idea to start with reading about it now. Any technical book, e.g. Hartshorne or others suggested in this MO question
will contain such material as sheaves, functors, derived functors, Verdier duality, etc.

There are also better places to learn about D-modules and related stuff; e.g. note Kashiwara's book says:

(p.411) Although perverse sheaves have a short history ...

and, indeed, 30 years later there are quite a few introductions to perverse sheaves that are easier to read.

I don't know about microlocalization, perhaps this topic should be indeed read from Kashiwara.

Now we'll be able to recommend a more specific text if you tell us what exactly you planned on reading Kashiwara for and where you get stuck!

gr.group theory - Finite groups with a character having very few nonzero values?

For n=2, your question is addressed in:

Gagola, Stephen M., Jr. "Characters vanishing on all but two conjugacy classes."
Pacific J. Math. 109 (1983), no. 2, 363–385.
MR721927
euclid.pjm/1102720107

In it, he shows that if such a nearly-zero character exists, it is unique, and the unique faithful irreducible character of G (similar to the extra-special 2 and 3 groups mentioned by Kevin Buzzard). A doubly transitive Frobenius group has such a character, but the non-solvable doubly transitive Frobenius groups are in short supply (five or so). The possible forms of more general non-solvable examples are restricted in Theorem 5.6 (basically SL2).

---

Berkovich and Zhmud have results for more general cases. See chapter 27 of their book on character theory (volume 2) which answers a broader question for n=2 and n=3:

Which groups have all irreducible non-linear characters taking on only n distinct values?

The list of groups is fairly short, but I haven't had time to verify it; they are all far from simple, usually with a normal Sylow and nilpotent quotient. They have a result for n=4 too, on p. 239, but it is much less complete.

Since your character is nonzero except on n classes, it can take on at most n nonzero values. It is fairly likely to be faithful, since every class contained in the kernel is a wasted class that cannot be zero. This allows some of the lemmas to be used. If Gagola's case is indicative, then the "all" versus "one" will not actually be a huge difference, at least mod the kernel of your single character.

Here are the paper references:

Berkovich, Yakov; Chillag, David; Zhmud, Emmanuel. "Finite groups in which all nonlinear irreducible characters have three distinct values."
Houston J. Math. 21 (1995), no. 1, 17–28.
MR1331241

Zhmudʹ, È. M. "On finite groups, all of whose irreducible characters take at most two nonzero values."
Ukraïn. Mat. Zh. 47 (1995), no. 8, 1144–1148;^{translation in
Ukrainian Math. J. 47 (1995), no. 8, 1308–1313 (1996)}MR1367729
DOI:10.1007/BF01057720

---

Berkovich and Zhmud also pose an exercise that if all the irreducible characters of degree coprime to p take on only 3 values, then the group is p-solvable. The same is true if "coprime to" is replaced by "divisible by". This is page 238.

mathematical writing - Capitalization of theorem names

In English, proper nouns are capitalized. The numbered instances you mention are all usages as proper nouns, but merely refering to a lemma or corollary not by its name is not using a proper noun, and so is uncapitalized.

Thus, for example, one should write about the lemma before Theorem 1.2 having a proof similar to Lemma 5, while the main corollary of Section 2 does not.

---

Edit. Well, I've become conflicted. The Chicago Manual of Style, which I have always taken as my guide in such matters, asserts in item 7.136 that "the word chapter is lowercased and spelled out in text". And in 7.141 they favor act 3 and scene 5 in words denoting parts of poems and plays. This would seem to speak against Section 2 and possibly against Theorem 1.2. In 7.135 they say that common titles such as foreward, preface, introduction, contents, etc. are lowercased, as in "Allan Nevins wrote the foreward to...". This may also be evidence against Theorem 1.2. But in 7.147 they favor Piano Sonata no. 2, which may be evidence in favor of Theorem 1.2. But they don't treat mathematical writing explicitly, and now I am less sure of what I have always believed, above. I do note that the CSM text itself refers to "fig. 1.2" and "figure 9.3", and not Figure 1.2, which would clearly speak against Theorem 1.2. So I am afraid that I may have to change my mind about this.

lo.logic - Why worry about the axiom of choice?

The best answer I've ever heard --- and I think I heard it here on MathOverflow from Mike Shulman, which suggests that this question is roughly duplicated somewhere else --- is that you should care about constructions "internal" to other categories:

1. For many, many applications, one wants "topological" objects: topological vector spaces, topological rings, topological groups, etc. In general, for any algebraic gadget, there's a corresponding topological gadget, by writing the original definition (a la Bourbaki) entirely in terms of sets and functions, and then replacing every set by a topological space and requiring that every function be continuous.


2. A closely related example is that you might want "Lie" objects: sets are replaced by smooth manifolds and functions by smooth maps.


3. Another closely related example is to work entirely within the "algebraic" category.

In all of these cases, the "axiom of choice" fails. In fact, from the internal-category perspective, the axiom of choice is the following simple statement: every surjection ("epimorphism") splits, i.e. if $f: Xto Y$ is a surjection, then there exists $g: Y to X$ so that $fcirc g = {rm id}_Y$. But this is simply false in the topological, Lie, and algebraic categories.

This leads to all sorts of extra rich structure if you do algebra internal to these categories. You have to start thinking about bundles rather than products, there can be "anomalies", etc.

Update:

In the comments, there was a request for a totally explicit example, where Axiom of Choice is commonly used but not necessary. Here's one that I needed recently. Let $mathcal C$ be an abelian tensor category, by which I mean that it is abelian, has a monoidal structure $otimes$ that is biadditive on hom-sets, and that has a distinguished natural isomorphism $text{flip}: Xotimes Y oversetsimto Yotimes X$ which is a "symmetry" in the sense that $text{flip}^2 = text{id}$. Then in $mathcal C$ is makes sense to talk about "Lie algebra objects" and "associative algebra objects", and given an associative algebra $A$ you can define a Lie algebra by "$[x,y] = xy - yx$", where this is short-hand for $[,] = (cdot) - (cdot circ text{flip})$ — $x,y$ should not be read as elements, but as some sort of generalization. So we can makes sense of the categories of $text{LIE}_{mathcal C} = $"Lie algebras in $mathcal C$" and $text{ASSOC}_{mathcal C} = $"associative algebras in $mathcal C$", and we have a forgetful functor $text{Forget}: text{ASSOC}_{mathcal C} to text{LIE}_{mathcal C}$.

Then one can ask whether $text{Forget}$ has a left adjoint $U: text{LIE}_{mathcal C} to text{ASSOC}_{mathcal C}$. If $mathcal C$ admits arbitrary countable direct sums, then the answer is yes: the tensor algebra is thence well-defined, and so just form the quotient as you normally would do, being careful to write everything in terms of objects and morphisms rather than elements. In particular, if $mathfrak g in text{LIE}_{mathcal C}$, then $Umathfrak g in text{ASSOC}_{mathcal C}$ and it is universal with respect to the property that there is a Lie algebra homomorphism $mathfrak g to Umathfrak g$.

Let's say that $mathfrak g$ is representable if the map $mathfrak g to Umathfrak g$ is a monomorphism in $text{LIE}_{mathcal C}$. By universality, if there is any associative algebra $A$ and a monomorphism $mathfrak g to A$, then $mathfrak g to Umathfrak g$ is mono, so this really is the condition that $mathfrak g$ has some faithful representation. The statement that "Every Lie algebra is representable" is normally known as the Poincare-Birkoff-Witt theorem.

The important point is that the usual proof — the one that Birkoff and Witt gave — requires the Axiom of Choice, because it requires picking a vector-space basis, and so it works only when $mathcal C$ is the category of $mathbb K$ vector spaces for $mathbb K$ a field, or more generally when $mathcal C$ is the category of $R$-modules for $R$ a commutative ring and $mathfrak g$ is a free $R$-module, or actually the proof can be made to work for arbitrary Dedekind domains $R$. But in many abelian categories of interest this approach is untenable: not every abelian category is semisimple, and even those that are you often don't have access to bases. So you need other proofs. Provided that $mathcal C$ is "over $mathbb Q$" (hom sets are $mathbb Q$-vector spaces, etc.), a proof that works constructively with no other restrictions on $mathcal C$ is available in

• Deligne, Pierre; Morgan, John W.
  Notes on supersymmetry (following Joseph Bernstein). Quantum fields and strings: a course for mathematicians, Vol. 1, 2 (Princeton, NJ, 1996/1997), 41--97, Amer. Math. Soc., Providence, RI, 1999. MR1701597.

They give a reference to

• Corwin, L.; Ne'eman, Y.; Sternberg, S.
  Graded Lie algebras in mathematics and physics (Bose-Fermi symmetry).
  Rev. Modern Phys. 47 (1975), 573--603. MR0438925.

in which the proof is given when $mathcal C$ is the category of modules of a (super)commutative ring $R$, with $otimes = otimes_R$, and, importantly, $2$ and $3$ are both invertible in $R$.

Finally, with $R$ an arbitrary commutative ring and $mathcal C$ the category of $R$-modules, if $mathfrak g$ is torsion-free as a $mathbb Z$-module then it is representable. This is proved in:

• Cohn, P. M.
  A remark on the Birkhoff-Witt theorem.
  J. London Math. Soc. 38 1963 197--203. MR0148717

So it seems that almost all Lie algebras are representable. But notably Cohn gives examples in characteristic $p$ for which PBW fails. His example is as follows. Let $mathbb K$ be some field of characteristic $pneq 0$; then in the free associative algebra $mathbb Klangle x,yrangle$ on two generators we have $(x+y)^p - x^p - y^p = Lambda_p(x,y)$ is some non-zero Lie series. Let $R = mathbb K[alpha,beta,gamma] / (alpha^p,beta^p,gamma^p)$ be a commutative ring, and define $mathfrak g$ the Lie algebra over $R$ to be generated by $x,y,z$ with the only defining relation being that $alpha x = beta y + gamma z$. Then $mathfrak g$ is not representable in the category of $R$-modules: $Lambda_p(beta y,gamma z)neq 0$ in $mathfrak g$, but $Lambda_p(beta y,gamma z)= 0$ in $Umathfrak g$.

soft question - eBook readers for mathematics

For math books you probably need to look at larger devices (8" or 9"):

• iRex DR1000s and DR800SG (am not if sure the company still exists)


• The new Kindle DX Graphite


• The iPad (not a reader, but it's here)

However, I find that 6" devices are more portable, unlike the (bulkier?) 8" or larger screens. But having a smaller screen means less screen estate and the PDF format cannot re-flow correctly because the page layout is designed for a specified size and is constrained by graphics and positioning of other elements. This means that what you see on the screen is a smaller version of the page, sometimes not easy on the eyes or downright illegible. The solution is to read in landscape mode and, if possible, to cut the page margins in the PDF.

For PDF and Dejaview formats take a look at Pocketbook 301+. In my opinion, this is the best PDF reader in the 6" class. There is a newer model (Pocketbook 302) but it has a glossy touch screen, so I would avoid that. I use it mostly for technical documents, heavy mathematics with graphics and it's great because:

1. Zoom to arbitrary level and keep the zoom between page switches (persistent zoom)


2. Landscape mode with automatic margins cut


3. Multiple-column mode (great for two columns papers in A4 or Letter size)


4. Two PDF viewers (XPDF and Adobe with DRM support) and an excellent DejaView viewer.

The PB301+ is a EB600EM clone with 8 gray scale levels (revised in 2009). Do not mistake it with other EB clones (there is a gazillion of them): the firmware is what makes Pocketbook a great PDF viewer and you cannot install it on other clones.

Concerning larger device: there is a number of them arriving soon on the market, including the Pocketbook 901 (9.7" screen size) expected in September 2010.

Beware of the touch screens: they reflect a lot (screen glare). I owned a Sony PRS600 but I had to send it back. I really wanted this to be a non-issue, but the glare and the reflections ended by getting in my way when reading: I needed a constant conscious effort to not see the reflection of my face in the device, all this combined with a sort of "fuzzy display" due probably to the "through the glass" effect induced by the screen.

On the web, the mother-lode site of mobile readers is http://wiki.mobileread.com (the technical descriptions are here: http://wiki.mobileread.com/wiki/E-book_devices). The forums are a good source of personal opinions.

Now, surprisingly, reading mathematics on a 6" screen turned out to be a much better experience than I have thought. This is subjective, but again, I expected worse. Here are some of my impressions (for a PB301+, 6" screen):

• Landscape mode is OK: you will see one third or half a page at a time but you can switch quite fast, up and down, in the same page


• Page turn is slower (1 sec), so turning pages back and forth is not practical. You can jump directly to any page but you will have to type the page number... not very practical on a device without keyboard


• Bookmarks are great and allows you to jump anywhere in the document.


• Reading is just like paper, not very stressful for the eyes


• Some documents are just too small even on maximum zoom in landscape mode

Good luck with your choice!

constructive mathematics - Direct construction of the integers

I would say: the free group on one element. I guess you can translate this into a series of first-order axioms. Notice that multiplication comes for free as composition between automorphisms of the group with itself.

Addendum: Prompted by the comment below, I am not thinking about the usual description of the free group through a chain of $1$'s and $-1$'s but on the universal property.

Let me give some specifics. A group is a tuple $(G,m,e,i)$ with $G$ a set, $m colon G times G to G$ a map $e in G$ and $i colon G to G$ satisfying certain commutativities that amount to the defining properties of group (associativity, $e$ is the neutral element and $i(g)$ is the inverse of the element $g in G$). A free group in one element is such a tuple $(F, dot , 1, op)$ satisfying that for any choice of a $g in G$ from a group $(G,m,e,i)$ there is one and only one homomorphism $(F, dot , 1, op) to (G,m,e,i)$ taking $1$ to $g$. I propose to translate this description into a series of first order formulas, that was my suggestion.

Addendum 2: I have just realized that this way the description is second-order.

schemes - Can one check formal smoothness using only one-variable Artin rings?

Let $f:Xrightarrow Y$ be a morphism of schemes over a field $k$. Can one check that $f$ is formally smooth using only Artin rings of the form $k^{prime}left[tright]/t^{n}$, where $k^{prime}$ is also a field?

Considering cuspidal curves one can show that you do at least need arbitrarily large $n$.

Suppose the independent number of a graph is bounded. How small the clique number can be?

This is basically a question in Ramsey theory. The Ramsey number $R(s, t)$ is the minimum integer $n$ for which every red-blue coloring of the edges of a complete $n$-vertex graph induces either a red complete graph of order $s$ or a blue complete graph of order $t$. So for example Kim (Random Structures and Algorithms 7 (1995), 173-207) showed that $R(3,t)asymp t^2/log t$. Roughly speaking this means that a graph with $t^2/log t$ vertices with independence number at most 2 must have a clique of size $t$. So in this case the clique number grows like the square root of the number of vertices of the graph. I'm not sure what the best known results are in general but this should point you to the right literature.

qa.quantum algebra - Weyl Character Formula for Quantum Groups

The question still feels a bit vague to me, but this started to get too long to be a comment. There are a number of issues:

There's the question of the definition of $mathbb C[G_q]$, and say an analogue of the Peter-Weyl theorem, and there is also the issue of doing this at a root of unity case, or better studying things integrally. For this say, a recent paper of Lusztig gives a definition of a quantum coordinate ring for any (finite type) root datum, which specializes to the Kostant-Chevalley form.

Andersen-Polo-Wen and others have studied studied quantum induction functors which correspond to taking global sections on the classical flag variety, and these might be what you want (Ryom-Hansen also proved a version of Kempf vanishing in this context for example). This was also more recently taken up by Kumar and Littelmann in the context of studying Frobenius splitting. Finally there's the issue of understand quantum flag varieties as noncommutative spaces as in the previous comment, for which along with the Lunts-Rosenberg paper, there is also more recent work of Backelin and Kremnizer.

homotopy theory - What is the "intuition" behind "brave new algebra"?

This is a general phrase that refers to the direction of

• higher category theory, per Lurie (you know references)


• scheme homotopy theory, per Voevodsky


• derived spaces, per Ben-Zvi and Nadler (0706.0322, 0805.0157)

The idea is that we're again changing the fundamental nature of space — first it was something easily drawn, then topology, then schemes, then stacks. Now we're doing some infinity versions of spaces, e.g. space --> $infty$-category, ring --> $E_infty$ category and that's brave new (the person who wrote this was quoting somebody from the 80s — below I explain that this person may very well be not Manin). In one sentence, we're not just taking functions now, but also forms etc.

Later he actually explains that "the homotopy picture becomes more important, and if you want discrete, you need to factorize".

---

Note that the "brave new" phrase is absent from the Russian version of the interview linked from AMS:

И поэтому я не предвижу ничего такого экстраординарного в ближайшие двадцать лет. Происходит перестройка того, что я называю основаниями математики, не в нормативном смысле слова, а как свод подчас даже не эксплицитных правил, критериев ценности, способов представления результатов, который присутствует в мозгу у работающего математика здесь и сейчас, в каждое конкретное время.
Вот это я называю основаниями математики. Их можно делать эксплицитными, при этом в нескольких вариантах, и представители разных вариантов могут начать спорить, но, поскольку это существует в мозгах работающего поколения математиков, там всегда есть нечто общее. Так вот, после Кантора и Бурбаков в мозгах, что бы там ни говорили, сидит теоретико-множественная математика.

which was translated to

And so I don’t foresee anything extraordinary
in the next twenty years. Probably, a rebuilding of
what I call the “pragmatic foundations of math-
ematics” will continue. By this I mean simply a
codification of efficient new intuitive tools, such
as Feynman path integrals, higher categories, the
“brave new algebra” of homotopy theorists, as
well as emerging new value systems and accepted
forms of presenting results that exist in the minds
and research papers of working mathematicians
here and now, at each particular time.
When “pragmatic foundations” of mathematics
are made explicit, usually in several variants, the
advocates of different versions may start quarrel-
ing, but to the extent that it all exists in the brains
of the working generation of mathematicians,
there is always something they have in common.
So, after Cantor and Bourbaki, no matter what
we say, set theoretic mathematics resides in our
brains.

The translation is accurate except for the italicized phrase. That phrase should be translated as

The things that I call the foundation of math are being rebuilt; not in the normative meaning of that word, but rather as the codex of — not even explicit rules, but rather values, ways to represent the results that exist in the brain of a working mathematician, here and now, at every given moment of time.

(I'm going for more literal translation: the original uses present tense, "brain" rather then "mind" and there is no "codification of mathematics", rather there are "values and ways" that are "being rebuilt")

Interesting, but as you see this is referring to the general idea of change in the "homotopy" direction rather then to the specific papers. In particular, "codification" should refer to the process when this "homotopy-think" becomes firmly established in the textbooks, rather then in the recent research articles.

---

It's a mystery to me as to why highly intelligent people didn't notice the discrepancy when preparing the interview for publication. In some other places the words are changed, e.g. "then you factorize..." --> "then you pass to the set of connected components of a
space defined only up to homotopy", and it appears this was made to make the interview more readable and unambiguous in English — it's very informal, though understandable, in the source.

A possibility, of course, would be that Manin himself edited the English version after it was translated.

ag.algebraic geometry - Can a coequalizer of schemes fail to be surjective?

This might lead to a pedestrian example: Let $R$ be a local ring with maximal ideal $m$, $X$ a scheme, and $f:Spec(R) to X$ a map. If $U$ is an open subscheme of $X$ containing $f(m)$ then $f$ factors through a map $Spec(R) to U$. Thus if we have two arrows $g,h:Y to Spec(R)$, the coequalizer $c: Spec(R)to C$ must be an affine scheme (else $c$ would factor strictly through an affine neighborhood $U$ of $c(m)$, and $U$ would be a "better coequalizer" than $C$). So If $Y=Spec(B)$ is also affine, then the coequalizer of $g,h$ is just $Spec(A)$ where $A$ is the equalizer of $g^sharp,h^sharp$.

let R be $k[x_i]_{(x_i)}$ and $S'=k[y_i,z_i]_{(y_i,z_i)}.$ there are two maps $g',h':R to S$ given by

$$g'(x_i) = y_i$$

$$h'(x_i) = z_i$$

suppose that $I$ is an ideal of $R$. let $I_y, I_z$ be the ideals generated by $g(I)$ and $h(I)$ respectively and let $S = S'/(I_y + I_z)$. write $g$ and $h$ for the induced maps $R to S$. the equalizer of $g,h$ is just $A = k + I subset R$. it seems unlikely to me that $spec R to spec A$ is surjective for all choices of $I$.

Relation between Sheaf and Group Cohomology

I doubt that in general one can construct a reasonable sheaf on $U$ with the required properties. To see what kind of bad things can happen, let us try to understand why this works for $X$ an elliptic curve and the sheaf $cal{O}^{times}$ on it.

We have the derived global sections functor from the $D^b$ of sheaves on $X$ to the $D^b$ of sheaves on a point, i.e. graded vector spaces. But given a sheaf $F$ on $X$ we can compute its global sections in a roundabout way: we can first take the pullback to $U$, then take global sections and then take the $G$-invariants where $G=pi_1(X)$. Passing to the derived categories we get $$RGamma (F)=R(RGamma f^{-1}(F))^G$$ where $Fin D^b(X)$, $f:Uto X$ is the projection, $RGamma f^{-1}$ is the right derived functor of the left exact functor $Gamma f^{-1}$ and $R(cdot)^G$ is the right derived functor of the functor of $G$-invariants (this functor goes from the $D^b$ of $G$-modules to graded vector spaces).

We have the Grothendieck spectral sequence that converges to $H^ast(X,F)$ with the $E_2$ sheet given by $$E_2^{p,q}=H^p(G,H^q(U,f^{-1}(F)).$$

Now if $X$ is an elliptic curve, $F=cal{O}^{times}$ and $U=mathbf{C}$, then it follows from the exponential exact sequence that $H^q(U,f^{-1}(F))=0$ for $qneq 0$, so the above spectral sequence collapses and we get the required isomorphism $H^ast (X,F)=H^ast(G,H^0(U,f^{-1}(F))$. This also happens when say $F$ is locally constant and $U$ is contractible. But in general there seems no reason to expect the spectral sequence to collapse, let alone to be concentrated in one row only.

soft question - Making sure that you have comprehended a concept

There is a famous quote from John Von Neumann:

“In mathematics you don't understand things. You just get used to them.”

I don't know for sure what this means but, perhaps it means that asking if you understand something is a question that does not really make sense, in fact your question is really asking: What is the definition of understanding. Such a question may be doomed from the start. Even if we could agree on some definition and you were to apply it to some mathematical concept I am sure that at some later time, after you have learned other things, your new perspective will make you feel like you never understood that concept anyway. So my advice is to take your concept and get used to it. Here are some ideas on how.

1)If it's a definition try to think up some examples and non-examples. The non-examples are especially useful when your concept is adding an adjective to something you already know; like adding "prime" to "number"

2) I it's a theorem try to identify exactly where the hypotheses are used in the proof, and try to think up counterexamples to the statement as you remove those hypotheses.

3) If there are exercises, ie if you are reading a textbook, try them. Even if you don't get to them all at least read the statements.

4) After you think you are used to your idea try to explain it to someone. This is a really good way to see if you have overlooked something.

5) As to the remark about moving on to the next few pages: Don't expect to go through a book line by line and not have to go back. The stuff later in the book may help.

In short if you work a lot with your concept it will cease to intimidate you. Don't worry about achieving understanding (equivalently seeing the true meaning), This will never happen, which is fine since it's the trying that is important anyway. So keep trying, and have fun doing so!

gr.group theory - on the computation of decomposition groups

Let $L/K$ be a finite Galois extension of function fields, with Galois groups $G$. I want to look at the ramification of primes in the extension, i.e. to get $e_p$ and $f_p$ for a prime $p$ in the base field $K$ (since the extension is Galois, the ramification index and inertia degree are independent of the choice of the prime lying above $p$). From Serre's 'Local Fields', it is clear that if we fix a prime $q$ in $L$ which lies over $p$, then we can look at the decomposition group associated to $q$, say $G_q$, and its inertia group, say $(G_q)_0$ (please forgive me for the notation :P), and an immediate result is that $e_p = left|{(G_q)_0} right|$ and $f_p = left| {G_q/(G_q)_0} right|$.

And here is my problem. Is there any nice way to compute the decomposition groups, inertia groups or just the cardinalities? If not, can we do something in some special cases? For example, when $G$ is cyclic?

ag.algebraic geometry - For which hypersurfaces in projective space does the complement admit an algebraic group structure?

Not an answer, but here's a Hodge-theoretic restriction on subvarieties $Z subset mathbf{P}^n$ such that $G = mathbf{P}^n - Z$ admits the structure of a linear algebraic group.

Under the above conditions, the natural mixed Hodge structure (MHS) on $RGamma(G,mathbf{Z})$ is of mixed Tate type (by the Bruhat decomposition, say); this condition on a Hodge structure means, roughly, that only (n,n) classes show up. As $G$ is smooth as a variety, it follows by duality that the same is true for compactly supported cohomology $RGamma_c(G,mathbf{Z})$. On the other hand, there is an exact triangle of MHSs

$RGamma_c(G,mathbf{Z}) to RGamma(mathbf{P}^n,mathbf{Z}) to RGamma(Z,mathbf{Z})$

This means that the MHS on $RGamma(Z,mathbf{Z})$ also has to be of mixed Tate type. This restriction rules out any Z with "interesting" cohomology.

Now if one further assumes that Z is smooth hypersurface, then the mixed Tate condition forces $h^{p,q}(Z) = 0$ unless p=q. Standard calculatons with Hodge numbers of hypersurfaces (see, eg, page 126 of "A Survey of the Hodge Conjecture" by Lewis) then show that Z has degree 1 or 2, at least when the ambient projective space is at least 6 dimensional.

(Edited to include degree restrictions in last paragraph.)

lo.logic - Reducing ACA₀ proof to First Order PA

According to the Wikipedia ACA₀ is a conservative extension of First Order logic + PA.

http://en.wikipedia.org/wiki/Reverse_Mathematics

First of all I have a few questions about the proof:
a - What is the general sketch of this proof, is it based on models?
b - Consider the theorem that ACA₀ is a conservative extension of First Order + PA, and the proof of that theorem is proven in a formal system, what kind of logic is needed? If the proof is based on models, then it requires second order logic. However, the theorem itself is a ∏⁰₂ question as far as I understand, and can be expressed in First Order logic + PA. Is there also a proof in First Order logic + PA?

Then I am interested in the following:
c - Given an ACA₀ formal proof that ends in a theorem that is part of First Order logic + PA, is there an algorithm that reduces the ACA₀ proof to First Order + PA proof?

One could just do a breath first search on First Order logic + PA and given the fact that ACA₀ is a conservative extension, it is guaranteed to end. So, the answer to question c is definitely "yes", but I am looking for something more clever.

I am struggling with this algorithm for months. In general an ACA₀ proof, with a First Order + PA end theorem reduces rather easier. However, there are some non-trivial cases. If the answer to question b is "yes", then that proof might give hints for constructing the algorithm.

I want to use this algorithm to reduce proofs of full second order, such that the reduced proof is First Order logic + PA, or contains the use of the induction scheme with a second order induction hypothesis.

In many cases the use of second order induction hypothesis, can be reduced by using the "Constructive Omega Rule". I want to understand the limitations of this (if any).

Thanks in advance,

Lucas

soft question - What are the most misleading alternate definitions in taught mathematics?

Many topics in linear algebra suffer from the issue in the
question. For example:

In linear algebra, one often sees the determinant of a
matrix defined by some ungodly formula, often even with
special diagrams and mnemonics given for how to compute it
in the 3x3 case, say.

det(A) = some horrible mess of a formula

Even relatively sophisticated people will insist that
det(A) is the sum over permutations, etc. with a sign for
the parity, etc. Students trapped in this way of thinking
do not understand the determinant.

The right definition is that det(A) is the volume of the
image of the unit cube after applying the transformation
determined by A. From this alone, everything follows. One
sees immediately the importance of det(A)=0, the reason why
elementary operations have the corresponding determinant,
why diagonal and triangular matrices have their
determinants.

Even matrix multiplication, if defined by the usual
formula, seems arbitrary and even crazy, without some
background understanding of why the definition is that way.

The larger point here is that although the question asked about having a single wrong definition, really the problem is that a limiting perspective can infect one's entire approach to a subject. Theorems,
questions, exercises, examples as well as definitions can be coming
from an incorrect view of a subject!

Too often, (undergraduate) linear algebra is taught as a
subject about static objects---matrices sitting there,
having complicated formulas associated with them and
complex procedures carried out with the, often for no
immediately discernible reason. From this perspective, many
matrix rules seem completely arbitrary.

The right way to teach and to understand linear algebra is as a fully dynamic
subject. The purpose is to understand transformations of
space. It is exciting! We want to stretch space, skew it,
reflect it, rotate it around. How can we represent these
transformations? If they are linear, then we are led to
consider the action on unit basis vectors, so we are led
naturally to matrices. Multiplying matrices should mean
composing the transformations, and from this one derives
the multiplication rules. All the usual topics in
elementary linear algebra have deep connection with
essentially geometric concepts connected with the
corresponding transformations.

pr.probability - Google question: In a country in which people only want boys

I think this is already implicit in the heavily up-voted answer, but it may be worth clarifying: there are two kinds of expectations that we can talk about.

The first is the distribution of G/B, G/(G + B), B/G, B/(B + G), values for the entire population (along with its expected value, standard deviation, etc.). Here, the distribution is over all possible "runs of history", so to speak, in the sense that we average over all possible ways history could turn out. If the population is large enough (thousands? millions?), then the expected values of all these quantities are what you would expect from a 50:50 split, and the standard deviations are near zero. Thus, as far as demographic estimations of the overall population are concerned, 50:50 is the way to go. In fact, at the population level, the ratio of girls to boys cannot be influenced by stopping strategies; any influence must either (i) affect the relative probability of conception of male versus female fetus (ii) adopt a post-conception filtering mechanism, such as induced abortion or infanticide).

The second is the expected G/B, G/(G + B), B/G, B/(B + G), etc., values over families. More generally, we may be interested in the distribution of different (G,B) values for different families. If we are interested in understanding family dynamics more thoroughly, we may also be interested in the birth orders, i.e., in what order girls and boys arise. Here, family stopping strategies could affect the distribution of (G,B) values and also of the birth orders. In particular, the strategy here ("stop as soon as you have a boy") gives 50% of the families with a single boy, 25% with one (older) girl and one (younger) boy, 12.5% with two older girls and one younger boy, and so on (assuming the complication of twins and triplets does not arise). This could have important demographic implications in the long term, when mating is done for the next generation (since birth order and the age gaps between children and their parents all play a role in mating and the creation of chlidren). However, that is getting beyond the current question.

For this second sense, it is not just the expected value per family that matters, but rather, the specific distribution of families. As already pointed out, since the variables are not independent, E[G/B] is not the same thing as E[G]/E[B], so what variable we choose to average over affects what answer we get. Looking at the whole distribution conveys more information.

When demographers are making short-term population estimates, it is the first sense (expected values for the population over runs of history) that is relevant, so stopping strategies can be discounted unless they are accompanied by post-conception selective strategies or strategies that affect conception probabilities. A deeper understanding of society would require knowing things in both the first and the second sense.

ct.category theory - Internal hom of sheaves

Well, here is a partial answer. The category of abelian group-valued sheaves is not a topos, the category of set-valued sheaves is. And I think you should look at set-valued presheaves, at least $hom(-,U)$ is one:

When the site comes from a topological space, you can see as follows that your two definitions coincide: When you insert into the $hom$ in your second expression an open set $V$, you either get $hom(V,U)=$the one-element-set containing only the inclusion, if $V subseteq U$, or $hom(V,U)=$empty set, if not, so taking a product with the set-valued presheaf $hom(-,U)$ either leaves X as it is or "deletes" it (i.e. transforms it into the (presheaf with value the) empty set).

So a natural transformation in $hom_{PreShv}(X times hom(-,U), Y)$ is given just by its components for all $V subseteq U$ and for other $V$ it extends to the unique map from the initial object into Y. This is the same as a natural transformation of restricted presheaves.

Edit: It now occurred to me that you were probably actually speaking of the sheaves of group homomorphisms, not just any natural transformations. You get that, when you apply the "free group"-functor to the set-valued sheaf "hom(-,U)" and take tensor product instead of product. The free group over the empty set ist the trivial group, thus the initial object in groups and tensoring with it trivializes $X$. Tensoring with the free group over one element doesn't change anything, so the same things happen as in the set case...

For more general sites, the expression $X| _U$ probably means that you look at $X$ as a functor defined on the slice category given by maps on your site into its object $U$. But you can also see $X$ as a sheaf not just on the site, but on all of its ambient sheaf category (as a hom-functor, and where you give to topos an appropriate topology - you sometimes do this in topos theory). Then looking at restricted sheaves is the same as passing to the slice category $Shv / hom(-,U)$, and $X times hom(-,U)$ is the image of $X$ under the canonical functor $Shv rightarrow Shv/hom(-,U)$ (which is given by taking product with $hom(-,U)$), so it actually $is$ $X| _U$. This still doesn't make it clear to me why the two functors in question coincide, but I guess it may be a way to look at it to find it out...