terminology - What is so "spectral" about spectra?

It seems reasonable to me that in operator theory the term "spectrum" comes from the Latin verb spectare (paradigm: specto, -as, -avi, -atum, -are), which means "to observe". After all in quantum mechanics the spectrum of an observable, i.e. the eigenvalues of a self adjoint operator, is what you can actually see (measure) experimentally.

Edit: after having a look to an online etymological dictionary, it seems the relevant Latin verb is another: spècere (or interchangeably spicere)= "to see", from which comes the root spec- of the latin word spectrum= "something that appears, that manifests itself, vision". Furthermore, spec- = "to see", -trum = "instrument" (like in spec-trum). Also the term "spectrum" in astronomy and optics has the same origin.

In algebraic geometry, I believe the term "spectrum", and the corresponding concept, has been introduced after the development of quantum mechanics became well known. In this context, the concept of spectrum as a space made of ideals is perfectly analogous of that in operator theory (think of Gelfand-Naimark theory, and that the Gelfand spectrum of the abelian C-star algebra generated by one operator is nothing but the spectrum of that operator).

I wouldn't be surprised if the term "spectral sequence" had something to do with "inspecting" [b.t.w. also "to inspect" comes from in + spècere...] step by step the deep properties of some cohomological constructions.

Maybe the term "spectrum" in homotopy theory and generalized (co)homology -but I don't know almost anything about these- has to do with "probing", "testing", a space via maps from (or to?) certain standard spaces such as the Eilenberg-MacLane spaces or the spheres. Does it sound reasonable?

Edit: The following paragraph from the wikipedia article on "primon gas" seems to support my guess:

"The connections between number theory and quantum field theory can be somewhat further extended into connections between topological field theory and K-theory, where, corresponding to the example above, the spectrum of a ring takes the role of the spectrum of energy eigenvalues, the prime ideals take the role of the prime numbers, the group representations take the role of integers, group characters taking the place the Dirichlet characters, and so on"

ag.algebraic geometry - Finding n points that are equidistant around the circumference of an ellipse

I'll attempt to answer your question by misinterpreting it.

As pointed out in the comments, computing elliptic integrals is not going to be easy. But what if you wanted to find $n$ points arranged around an ellipse which form the vertices of an equilateral polygon? Now the answer to the question is given by a real algebraic variety. It's possible that this question may be computationally more tractable.

There are $2n$ variables $(x_i,y_i)$, $i=1,ldots,n$, and $2n-1$ equations:
$$ frac{x_i^2}{a^2}+frac{y_i^2}{b^2}=1, i=1,ldots, n,$$
$$(x_i-x_{i+1})^2+(y_i-y_{i+1})^2=(x_{i+1}-x_{i+2})^2+(y_{i+1}-y_{i+2})^2, i=1,ldots, n-1,$$
indices taken $(mod n)$.

Also, for geometric reasons, one expects $n-1$ components to this variety, each of which is a circle. For example, if $n=5$, one would expect two solutions which are oriented in different directions, and two solutions which are star shaped.
As one moves around the circle, the solution should move around.

Thus, one expects this variety to be a complete intersection defined by quadratic equations. There are methods from algebraic geometry to find solutions to such equations. There are versions of Newton's recursion which may be effective for finding a numerical solution. The dihedral symmetry might further constrain the solutions. Maybe someone could point you to some references if this sort of solution would suffice for your application?

lo.logic - Generalized Cox Theorems, valuations on boolean sets, bayesian probabilities and posets

Bayesian probabilities are usually justified by the Cox theorems, that can be written this way:

Under some technical assumptions (continuity, etc, etc...), given a set $P$ of objects $A, B, C, ldots$, with a boolean algebra defined over it with operations $A wedge B$ (and) and $A | B$ (or) such that:

1) $A wedge B = B wedge A$

2) $A wedge (B wedge C) = (A wedge B) wedge C$

3) $A | (B wedge C) = (A|B) wedge (A|C)$

and a "valuation":

$f : P rightarrow mathcal{R}$

there is a strictly monotonic "regraduation function" $R : mathcal{R} rightarrow mathcal{R}$ such that, for:

$R(f(Awedge B)) = R(f(A)) + R(f(B))$ (sum rule)

and

$R(f(A|B)) = R(f(A) ) R(f(B))$ (product rule)

This theorem allows one to show that any system designed to "evaluate" boolean expressions consistently with a single real number redunds in the laws of classical probability (this can be seen shortly here: arxiv:physics/0403089 and more thoroughly here: arxiv:abs/0808.0012)

Recently this has been extended for valuations of the type $f : P rightarrow mathcal{R}^2$ in http://arxiv.org/abs/0907.0909 and they proved that there are just 5 canonical valuations compatible with the underlying Boolean algebra (one of them giving a complex field structure to the "valuation" field).

My question/proposal is: is it possible/interesting/feasible to classify at least a class of valuations of the type:

$f : P rightarrow W$

where W is a continuous manifold? If we retrict our attention to $mathcal{R}^n$ for example, is there, for each n, a set of canonical valuations to which all others can be reduced after a regraduation?

If this can be done, are those nice rules for inference in some sense? Are they useful as inference tools in specific situations?

ag.algebraic geometry - Can a singular Deligne-Mumford stack have a smooth coarse space?

I think if the coarse moduli space is smooth, so is the DM stack, because XX --> X is a gerbe, which is always smooth (since smoothness can be checked fppf locally on X, and B(G/X) is smooth over X). A stack (or a morphism of stacks, not necessarily representable) is defined to be smooth if one can find a presentation which is smooth over the base. And if it is smooth, then any presentation is smooth. That's why I got confused on Anton's example. Maybe someone can explain this to me. Thanks in advance.

lo.logic - Ultimate limits of Tennenbaum's Theorem

This is version 2 of a question about the ultimate limits of Tennenbaum's Theorem. The attempt to find these limits by moving up the induction heirarchy, as in Wilmer's Theorem, seems somehow indecisive. I suggested that maybe there is a Theory $T$ extending open induction such that

1) $T$ has a recursively presentable nonstandard model.

2) If the sentence $phi$ is not provable from $T$, then
$T+phi$ has no recursively presentable nonstandard model.

François G. Dorais immediately replied that this just amounts to $T$ being complete.

So... What about asking for the maximum $n$ such that the theory of all true (in the integers) all-2 sentences with n existential quantifiers has a recursive nonstandard model?
What is known about this? Is it known that $n<2$?

homological algebra - Does Ext commute with direct limit?

A short comment, which I can't post as a comment as I've just opened a new account (I apologize).
The $R^1varprojlim mathrm{Hom}_R(M_{alpha}, Q)$ contribution can be taken care by arranging the transition maps in the directed system $(M_{alpha})$ to be injective.

Suppose we show $mathrm{Ext}^1_R(cdot, Q)$ vanishes on all finitely generated $R$-modules ($R$ Noetherian: we'll be using that the category of finitely generated $R$-modules is abelian when $R$ is Noetherian). We can write $M$ as the directed union of its finitely generated $R$-submodules and arrange all transition maps to be injective. Applying $mathrm{Hom}_R(cdot, Q)$ to the injection $M_{alpha}to M_{alpha'}$, $alpha'gealpha$, we have an exact-in-the-middle seq

$$mathrm{Hom}_R(M_{alpha'}, Q)to mathrm{Hom}_R(M_{alpha}, Q) to mathrm{Ext}^1_R(A, Q)$$

for $A$ a finitely generated $R$-module. That is, the inverse system $(X_{alpha})$, $X_{alpha} := mathrm{Hom}_R(M_{alpha}, Q)$, satisfies the Mittag-Leffler condition (because $mathrm{Ext}^1_R(A, Q) = 0$) and therefore it has vanishing $R^1varprojlim (cdot)$.

Torsten's answer shows that in the case $R$ is Noetherian, one reduces to check injectivity of an $R$-module $Q$ to computing $mathrm{Ext}^1_R(R/mathfrak{p}, Q)$ to be trivial for all prime ideals of $R$ (as for $M$ finitely generated, given ses's of finitely generated $R$-modules:

$$0to M'to Mto M''to 0$$

and by functoriality of Ext's, we get exact-in-the-middle sequences

$$mathrm{Ext}^1_R(M'', Q)to mathrm{Ext}^1_R(M, Q)to mathrm{Ext}^1_R(M', Q)$$

so if we show vanishing of the outer terms, we show vanishing of the middle one. This reduces consideration to the case of simple $R$-modules (again, here $R$ is Noetherian!), ie. of the form $R/mathfrak{p}$, $mathfrak{p}$ a prime ideal).

Eg. Let $R = mathbf{Z}/p^2mathbf{Z}$. Showing $R$ is an injective $R$-module is equivalent to showing $mathrm{Ext}^1_R(R/p, R) = 0$. More in general, can show $mathbf{Z}/nmathbf{Z}$ is injective as a module over itself this way.

Best

lo.logic - What does it mean for a mathematical statement to be true?

This is a philosophical question, rather than a matehmatical one. Anyway personally (it's a metter of personal taste!) I totally agree that mathematics is more about correctness than about truth.

In the following paragraphs I will try to (partially) answer your specific doubts about Goedel incompleteness in a down to earth way, with the caveat that I'm no expert in logic nor I am a philosopher. (See also this MO question, from which I will borrow a piece of notation). I had some doubts about whether to post this answer, as it resulted being a bit too verbose, but in the end I thought it may help to clarify the related philosophical questions to a non-mathematician, and also to myself.

---

The point is that there are several "levels" in which you can "state" a certain mathematical statement; more: in theory, in order to make clear what you formally want to state, along with the informal "verbal" mathematical statement itself (such as $2+2=4$) you should specify in which "level" it sits. (Of course, as mathematicians don't want to get crazy, in everyday practice all of this is left completely as understood, even in mathematical logic).

For example, suppose we work in the framework of Zermelo-Frenkel set theory ZF (plus a formal logical deduction system, such as Hilbert-Frege HF): let's call it Set1. In this setting, you can talk formally about sets and draw correct (relative to the deduction system) inferences about sets from the axioms. You can also formally talk and prove things about other mathematical entities (such as $mathbb{N}$, $mathbb{R}$, algebraic varieties or operators on Hilbert spaces), but everything always boils down to sets.

Still in this framework (that we called Set1) you can also play the game that logicians play: talking, and proving things, about theories $T$. How? Well, you only have sets, and in terms of sets alone you can define "logical symbols", the "language" $L$ of the theory you want to talk about, the "well formed formulae" in $L$, and also the set of "axioms" of your theory. Examples of such theories are Peano arithmetic PA (that in this incarnation we should perhaps call PA2), group theory, and (which is the reason of your perplexity) a version of Zermelo-Frenkel set theory ZF as well (that we will call Set2). Note that every piece of Set2 "is" a set of Set1: even the "$in$" symbol, or the "$=$" symbol, of Set2 is itself a set (e.g. a string of 0's and 1's specifying it's ascii character code...) of which we can formally talk within Set1, likewise every logical formula regardless of its "truth" or even well-formedness. Stating that a certain formula can be deduced from the axioms in Set2 reduces to a certain "combinatorial" (syntactical) assertion in Set1 about sets that describe sentences of Set2.

The good think about having a meta-theory Set1 in which to construct (or from which to see) other formal theories $T$ is that you can compare different theories, and the good thing of this meta-theory being a set theory is that you can talk of models of these theories: you have a notion of semantics.

In the light of what we've said so far, you can think of the statement "$2+2=4$" either as a statement about natural numbers (elements of $mathbb{N}$, constructed as "finite von Neumann ordinals" within Set1, for which $0:=emptyset$, $1:=${$emptyset$} etc.); or as a sentence of PA2 (which is actually itself a bare set, of which Set1 can talk).

An interesting (or quite obvious?) thing is that in some cases it makes sense to go on to "construct theories" also within the lower levels. For example, within Set2 you can easily mimick what you did at the above level and have formal theories, such as ZF set theory itself, again (which we can call Set3)!
A crucial observation of Goedel's is that you can construct a version of Peano arithmetic not only within Set2 but even within PA2 itself (not surprisingly we'll call such a theory PA3). So you have natural numbers (of which PA2 formulae talk of) codifying sentences of Peano arithmetic!

So, if we loosely write "$A-triangleright B$" to indicate that the theory or structure $B$ can be "constructed" (or "formalized") within the theory $A$, we have a picture like this:

Set1 $-triangleright$ ($mathbb{N}$; PA2 $-triangleright$ PA3; Set2 $-triangleright$ Set3; T2 $-triangleright$ T3; ...).

So, you see that in some cases a theory can "talk about itself": PA2 talks about sentences of PA3 (as they are just natural numbers!), and there is a formally precise way of stating and proving, within Set1, that "PA3 is essentially the same thing as PA2 in disguise".

Furthermore, you can make sense of otherwise loose questions such as "Can the theory $T$ prove it's own consistency?". How? Well, you construct (within Set1) a version of $T$, say T2, and within T2 formalize another theory T3 that also "works exatly as $T$". Then you have to formalize the notion of proof.

So, the Goedel incompleteness result stating that

"Peano arithmetic cannot prove its own consistency"

is really a theorem of Set1 asserting that "PA2 cannot prove the consistency of PA3". This means: however you've codified the axioms and formulae of PA as natural numbers and the deduction rules as sentences about natural numbers (all within PA2), there is no way, manipulating correctly the formulae of PA2, to obtain a formula (expressed of course in terms of logical relations between natural numbers, according to your codification) that reads like "It is not true that axioms of PA3 imply $1neq 1$".

You can say an exactly analogous thing about Set2 $-triangleright$ Set3, and likewise about every theory "at least compliceted as PA".

Now, about truth. First of all, the distinction between provability a and truth, as far as I understand it. It is easy to say what being "provable" means for a formula in a formal theory $T$: it means that you can obtain it applying correct inferences starting from the axioms of $T$. This is a purely syntactical notion.

The concept of "truth", as understood in the semantic sense, poses some problems, as it depends on a set-theory-like meta-theory within which you are supposed to work (say, Set1). Saying that a certain formula of $T$ is true means that it holds true once interpreted in every model of $T$ (Of course for this definition to be of any use, $T$ must have models!).

If we could convince ourselves in a rigorous way that ZF was a consistent theory (and hence had "models"), it would be great because then we could simply define a sentence to be "true" if it holds in every model. This was Hilbert's program. Unfortunately, as said above, it is impossible to rigorously (within ZF itself for example) prove the consistency of ZF.

---

About true undecidable statements.

The assertion of Goedel's that

"There is a property of natural numbers that is true but unprovable from the axioms of Peano arithmetic"

is a theorem of Set1 stating that there is a sentence of PA2 that holds true* in any model of PA2 (such as $mathbb{N}$) but is not obtainable as the conclusion of a finite set of correct logical inference steps from the axioms of PA2.

*(that a sentence of PA2 is "true in any model" here means: "the corresponding interpretation of that sentence in each model, which is a sentence of Set1, is a consequence of the axioms of Set1")

According to Goedel's theorems, you can find undecidable statements in any consistent theory which is rich enough to describe elementary arithmetic. That is, such a theory is either inconsistent or incomplete.

---

About meaning of "truth".

Foundational problems about the absolute meaning of truth arise in the "zeroth" level, i.e. about sentences expressed in what is supposed to be the foundational theory Th0 for all of mathematics According to some, this Th0 ought to be itself a formal theory, such as ZF or some theory of classes or something weaker or different; and according to others it cannot be prescribed but in an informal way and reflect some ontological -or psychological- entity such as the "real universe of sets".
I would roughly classify the former viewpoint as "formalism" and the second as "platonism".

One point in favour of the platonism is that you have an absolute concept of truth in mathematics. One drawback is that you have to commit an act of faith about the existence of some "true universe of sets" on which you have no rigorous control (and hence the absolute concept of truth is not formally well defined). According to platonism, the Goedel incompleteness results say that

"Logic cannot capture all of mathematical truth".

On the other hand, one point in favour of "formalism" (in my sense) is that you don't need any ontological commitment about mathematics, but you still have a perfectly rigorous -though relative- control of your statements via checking the correctness of their derivation from some set of axioms (axioms that vary according to what you want to do). One consequence (not necessarily a drawback in my opinion) is that the Goedel incompleteness results assume the meaning:

"There is no place for an absolute concept of truth: you must accept that mathematics (unlike the natural sciences) is more a science about correctness than a science about truth".

ca.analysis and odes - Universal covers of domains in complex projective space

The Uniformization Theorem states that the universal cover of a Riemann surface is biholomorphic to the extended complex plane, the complex plane or the open unit disk. Each of these three is a domain in the extended complex plane. In particular, then, the universal cover of a domain in the extended complex plane is biholomorphic to a domain in the extended complex plane. This leads to an analogous question in higher dimensions: Is the universal cover of a domain in complex projective space biholomorphic to a domain in complex projective space? More precisely, I am asking for a counterexample. Many results in one complex variable break in several complex variables, and the Uniformization Theorem is fairly delicate, so it seems reasonable to expect it to break. Perhaps there is a counterexample that one can see just by topology?

gr.group theory - A question about iterated quotients in riemannian geometry

The group $D$ is the preimage of $E$ in $N(F)$, so it is as you expect. The finiteness hypothesis can be weakened, which is important for many applications. Things become clearer if one thinks categorically in terms of the universal properties.

Say an arbitrary group $G$ acts freely and properly discontinuously and isometrically on $X$ and $H$ is a normal subgroup of $G$. Then $G/H$ acts freely and properly discontinuously and isometrically on $X/H$ with $X rightarrow (X/H)(G/H)$ a $G$-invariant map. The induced map $f:X/G rightarrow (X/H)(G/H)$ is an isomorphism. Indeed, both sides composed back with the natural map from $X$ satisfy the same universal property, and $f$ respects the maps from $X$, so $f$ is an isomorphism. QED

In fact, with more work this can all be done more generally with $G$ and Lie group and $H$ a closed normal Lie subgroup, under suitable "niceness" hypotheses for the orbit maps (which are satisfied in the above situation): see Proposition 13 in section 1.6 of Chapter III of Bourbaki LIE.

lie groups - Non-Lie Subgroups

A result of Borel and Lichnerowicz states that the holonomy group of a connection on a principal $G$-bundle is a Lie subgroup of $G$ (Cartan had earlier asserted this, but apparently without proof).
This restriction, that it be a Lie subgroup, allows for a lot of poorly-behaved subgroups, for example a line with irrational slope on a torus. This subgroup comes from a perfectly fine immersion of the Lie group $mathbb{R}$, but it's not closed in the induced topology of the torus.

As an example of something that's not a Lie subgroup, let $G= mathbb{R}$, consider an uncountable set of $mathbb{Q}$-independent points, none of which are rational, and consider the subgroup they generate. If this were a Lie subgroup it would be the image of an uncountable discrete space (there can't be anything $1$-dimensional, since we left out the rationals), which wouldn't be second countable, hence not a manifold and not a Lie group.

This seems like a pretty contrived example, and I suspect there is more content to 'being a Lie subgroup' than having countably many components. However, I can't seem to pin down something that would illustrate this. Can anyone give me an example of a connected subgroup of a Lie group that is not a Lie subgroup?

Algorithm for determining if a path exists in a graph or if not, the closest edit distance.

Given a directed acyclic graph G and a path made up from its set of nodes N, what is the closest approximate match to N, equipped with an intuitive notion of distance?

A path in a directed acyclic graph is essentially a string (that uses the set of nodes as the alphabet), so when comparing paths one can make use of edit distances developed for approximate string matching:

There are many algorithms for approximate string matching:

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

This string matching answers the question when the G itself is also a path.. then we're merely asking to compare two strings.

Asking if an arbitrary graph A built from the same set of nodes is a sub-graph of G is the general case of the problem, but I'm only interested in the case where A is a path and G is directed & acyclic.

Any general pointers are also welcome.

--

Edit: I ended up using a dynamic programming algorithm, independently also suggested in the accepted answer. Good call! It is probably the most accurate option as well, and barely more "complex" than the string-to-string case when the average # of edges per node is low.

quantum field theory - How to calculate partition function of a QFT by summing over irreducible representations of the symmetry group?

The following article by A.B. Balantekin describes in section III the structure of thermal grand canonical partition functions (i.e., with chemical potentials) for systems with Lie group symmetries (which is the partition function of the type described in the reference given in the question).

The sum over the group charcters originates from the presence of the chemical potentials,
which are the (analytical continuations of the) coordinates of the maximal torus of the group.

The Balantekin article describes the "sum over states" formulas of the partition functions in which the summation over the group representations is explicit, in contrast to the Feynman's "sum over paths" (mentioned in the question) in which the summation over the representations is implicit.

I'll describe here a simple example of an explicit calculation of the grand canonical partition function over the two-sphere.

Let $-H$ be the scalar Laplacian on the two sphere. The state Hilbert space is spanned the spherical representations (with multiplicity 1),
corresponding to integer spins only. The symmetry group $SU(2)$ is generated by the usual set of generators $[J_i, J_j] = epsilon_{ijk} J_k$

The grand canonical partition function is given by:

$Z = textrm{Tr}(exp(-beta H + mu J_3)) = sum_{j=0}^{infty}chi_j(i mu)exp(-beta j(j+1))$

$= sum_{j=0}^{infty}frac{sinh((2j+1) mu)}{sinh(mu)}exp(-beta j(j+1))$

$= frac{e^{frac{beta}{4}}}{2sinh(mu) }theta_1(mu, beta)$

The actual computation (of the N=4 SYM) referred to in the question needs much more work and it relies on several assumptions and approximations.
The introduction section of the following article by Yamada and Yaffe and references therein may be helpful.

gn.general topology - Applications of Brouwer's fixed point theorem

In many nonlinear equations, the existence of a solution (but not its uniqueness) follows from a topological argument in the vein of BFP Theorem. The BFP is at work especially when the equation is posed in some finite dimensional vector space, and you can establish an a priori estimate of the size of the solution. This means that you know a ball $B_R$ containing all the solutions.

The most important example of this situation is the stationary Navier-Stokes equation, with Dirichlet condition $u=0$ on the boundary of the domain. Of course, the ambient space is infinite dimensional, so you first establish the existence of an approximate solution in a subspace of dimension $n$ (Galerkin procedure); this is where you use the BFP Theorem, or its equivalent form that a continuous vector field over $B_R$ which is outgoing on $partial B_R$ must vanish somewhere. Then passing to the limit as $nrightarrowinfty$ is pedestrian.

The BFP Theorem is a consequence of the fact that the Euler-Poincaré characteristic of the ball is non-zero. There are counterparts when you work on a compact manifold (with boundary) whose EPC is non-zero. This happened to me in a very interesting way. I considered the free fall of a rigid body in water filling the entire space. The mathematical problem is a coupling between Navier-Stokes and the Euler equation for the top. I looked at a permanent regime, in which the solid body has a time-independent velocity field, and that of the fluid is time-independent as well, once you consider it in the moving frame attached to the solid. The difficulty is that you don't know a priori the direction of the vertical axis (the direction of gravity) in this frame. After a Galerkin procedure, the problem reduces to the search of a zero of a tangent vector field over $B_Rtimes S^2$. This vector field is outgoing on the boundary $partial B_Rtimes S^2$. Because
$$EP(B_Rtimes S^2)=EP(B_R)cdot EP(S^2)=1cdot2ne0,$$
such a zero exists. Therefore the permanent regime does exist. Remark that because $EP=2$, we even expect an even number of solutions when counting multiplicities, at least at each level of the Galerkin approximation.

na.numerical analysis - How well does a truncated fourier expansion of a stepfunction perform near the expansionpoint

Hey, i'am currently trying to make a stability analysis of a binary fluid at its phase border. As the governing equations in this double-diffusive problem are rather complicated i have to do a series expansion of my density profile. My problem is, that my PDEs contain a hevyside step function right at the border and i want to do a fourier expansion (i can then use symmetries of my problem and do some horrible algebra to actually get a "solution" which can be computed for up to 5 or 6 modes).

Now i know that fourier expansion performs quite horribly at a discontinous jump if only the first few modes are kept. Is it nervertheless possible to obtain a "margin of error"? Numerical precision is not of utmost concern, but the physics behind my problem should not all be thrown out by such an expansion. Do you know of any criteria or "better" expansion where its possible to make heavy use of symmetry/antisymmetry?

all the best,
jan

dg.differential geometry - Does a smooth, constant-rank, integrable distribution have a basis in which the traces of the structure constants are the divergences of the corresponding basis elements?

In a previous question, I asked an utterly trivial question, which Deane Yang correctly pointed out was utterly trivial. I will now ask a similar question, which is the one I meant to ask last time; I hope it's not similarly trivial.

I am working on $mathbb R^n$, although in fact any manifold with volume form is good enough. And my question is local: I have an open neighborhood $U ni 0$, and you are allowed to make it smaller if you want.

Recall that a constant-rank distribution $D$ on $U$ is a vector subbundle of the tangent bundle ${rm T}U$. Let's fix the rank to be $kleq n$, and suppose that everything is smooth: $Gamma(D)$ is a $C^infty$-submodule of $Gamma({rm T}U)$. The distribution is involutive if $Gamma(D)$ is a Lie subalgebra of $Gamma({rm T}U)$ (over $mathbb R$ — the Lie bracket on $Gamma({rm T}U)$ is not $C^infty$-linear). The distribution $D$ is smooth if $Gamma(D)$ is a free rank-$k$ module over $C^infty$, i.e. if $Gamma(D)$ has a basis ${v_1,dots,v_k}$ so that $Gamma(D) = operatorname{Span}_{C^infty}{v_1,dots,v_k}$.

So, suppose that on $U subseteq mathbb R^n$ I have a smooth involutive rank-$k$ distribution. Given a basis $v_1,dots,v_k in Gamma(D)$ (and I will use coordinates $x^1,dots,x^n$ on $mathbb R^n$, so I will write $v_a = sum_i v_a^i(x) frac{partial}{partial x^i}$), then I can define the structure coefficients $f_{a,b}^c(x)$, $a,b,c = 1,dots,k$, via $[v_a,v_b] = sum_c f_{a,b}^c v_c$, or, in coordinates:
$$ sum_i left( v_a^i(x),frac{partial v_b^j}{partial x^i} - v_b^i(x),frac{partial v_a^j}{partial x^i}right) = sum_c f_{a,b}^c(x),v_c^j(x) $$

Question: Does there exist a basis ${v_a}$ for a given smooth involutive constant-rank distribution so that for each $a=1,dots,k$ and each $xin U$, we have $displaystyle sum_b f^b_{a,b}(x) = sum_i frac{partial v^i_a(x)}{partial x^i}$?

For example, by Frobenius theorem (my utterly trivial question), I can find a basis so that the LHS vanishes for each $a$. Or, by another of my trivial questions, I could make the basis entirely divergence-free. But I don't think I can simultaneously make the basis consist of divergence-free vector fields.

Question: If so, how many choices of such a basis do I have? Clearly ${rm GL}(k,mathbb R)$ acts on the set of choices; are there more?

reference request - Inverse limit in metric geometry

Question. Did you ever see inverse limits to be used (or even seriousely considered) anywhere in metric geometry (but NOT in topology)?

The definition of inverse limit for metric spaces is given below. (It is usual inverse limit in the category with class of objects formed by metric spaces and class of morphisms formed by short maps.)

Definition.
Consider an inverse system of metric spaces $X_n$ and short maps $phi_{m,n}:X_mto X_n$ for $mge n$;
i.e.,(1) $phi_{m,n}circ phi_{k,m}=phi_{k,n}$ for any triple $kge mge n$ and (2) for any $n$, the map $phi_{n,n}$ is identity map of $X_n$.

A metric space $X$ is called inverse limit of the system $(phi_{m,n}, X_n)$ if its underlying space consists of all sequences $x_nin X_n$ such that $phi_{m,n}(x_m)=x_n$ for all $mge n$ and for any two such sequences $(x_n)$ and $(y_n)$ the distance is defined by

$$ | (x_n) (y_n)| = lim_{ntoinfty} | x_n y_n | .$$

Why: I have a theorem, with little cheating you can stated it this way: The class of metric spaces which admit path-isometries to Euclidean $d$-spaces coincides with class of inverse limits of $d$-polyhedral spaces.
In the paper I write: it seems to be the first case when inverse limits help to solve a natural problem in metric geometry. But I can not be 100% sure, and if I'm wrong I still have time to change this sentence.

pr.probability - Birkhoff ergodic theorem for dynamical systems driven by a Wiener process

At the risk of asking a stupid question I have the following problem.

Suppose I have a measure preserving dynamical system $(X, mathcal{F}, mu, T_s)$, where

• $X$ is a set


• $mathcal{F}$ is a sigma-algebra on $X$,


• $mu$ is a probability measure on $X$,


• $T_s:X rightarrow X$, is a group of measure preserving transformations parametrized by $s in mathbb{R}$.

Suppose that this dynamical system is ergodic, so that for any $f in L^1(mu)$,

$lim_{trightarrow infty}frac{1}{2t}int_{-t}^t f(T_s x) ds = int f(x)dmu(x)$.

Now let $B_s$ be a real valued Wiener process such that $B_0 = 0$, then I can define the following process:

$frac{1}{t}int_{0}^t f(T_{B_s} x) ds$

Does anybody know how this process would behave as $trightarrow infty$? Intuitively I would expect it to converge to a similar constant for a.e realisation of the brownian motion, but I can't find a convincing argument.

Thanks for your help.

ag.algebraic geometry - Principal bundle for contractible group is weak homotopy equivalence for ind schemes

My recollection is that when you turn these into analytic spaces you get something which is locally contractible topologically. In this case what you are describing is a principal bundle for locally contractible spaces in which the fiber is contractible. If the base is paracompact then this will indeed be a weak equivalence, in fact the space corresponding to X will be a topological product space $X = U times Y$.

This follows because you can build a global section (trivialization). How do you do this? You start with you local trivializations, and you choose a partition of unity subordinate to this cover. You also choose a contraction of U. You can then patch these together to obtain a global section. The exact method and formula is explained, for example, in the appendix of this paper. (This is probably not the only/first/best reference).

Segal, G. Cohomology of topological groups. 1970 Symposia Mathematica, Vol. IV (INDAM, Rome, 1968/69) pp. 377--387 Academic Press, London

So the real question is whether your space Y is paracompact. I'm pretty sure that your conditions (that Y is quasi-projective) ensure that this is the case.

at.algebraic topology - Betti Cohomology of singular Kummer Surface

I missed that the question concerned the singular Kummer surface (which I think
historically was what was what was called the Kummer surface but our current fixation on
non-singularity has changed that) so one needs a few more steps than Barth,
Peters, van de Ven: Compact complex surfaces (which will be my reference below).

Let $picolontilde Xrightarrow X$ be the minimal resolution of singularities
and consider the Leray spectral sequence for $pi$. We have $pi_astmathbb
Z=mathbb Z$ and $R^2pi_astmathbb Z$ the skyscraper sheaf with one $mathbb
Z$ at each of the 16 singular points. The Leray s.s. thus gives that
$H^i(X,mathbb Z)=H^i(tilde X,mathbb Z)$ for $ineq2,3$ and hence
$H^i(X,mathbb Z)=mathbb Z$ for $i=0,4$ and $H^1(X,mathbb Z)=0$ as well as a
short exact sequence
$$
0rightarrow H^2(X,mathbb Z)rightarrow H^2(tilde X,mathbb Z)rightarrow
bigoplus_{vin V}mathbb Zvrightarrow H^3(X,mathbb Z)rightarrow0,
$$
where $V$ is the set of singular points. Now, it is easy to see that $H^2(tilde
X,mathbb Z)rightarrow mathbb Zv$ is given by $fmapsto deg(f_{E_v})$, where
$E_v:=pi^{-1}(v)$. We have $deg(f_{E_v})=langle e_v,frangle$, where $e_vin
H^2(tilde X,mathbb Z)$ is the fundamental class of $E_v$. Hence, we get to
begin with that $H^2(X,mathbb Z)$ is the orthogonal complement in $H^2(tilde
X,mathbb Z)$ of the $e_v$. By Cor. 5.6 (of BPV) this can be identified with
$H^2(A,mathbb Z)$. On the other hand, the image of $H^2(tilde X,mathbb Z)$ in
$bigoplus_{vin V}mathbb Zv$ contains the linear functions given by the $e_v$
and $e_v(v')=-2delta_{v,v'}$ so that we may consider the image of $H^2(tilde
X,mathbb Z)$ in $bigoplus_{vin V}mathbb Z/2v$. By the fact that the cup
product pairing on $H^2(tilde
X,mathbb Z)$ is perfect (by Poincaré duality) and by Prop. 5.5 we get that this
image is dual to the subspace of affine functions of $bigoplus_{vin V}mathbb
Z/2v$ (where $V$ is identified by the kernel of multiplication by $2$ in $A$)
and hence we get an identification of $H^3(X,mathbb Z)$ with the dual of the
$mathbb Z/2$-space of affine functions of $V$, in particular it has dimension
$5$.

Remark: It is interesting to note that while the quotient $A/sigma$ as a
topological space does not use the complex structure of $A$ it still seems
easier to use it (in a very weak form, the blowing up only uses that a conical
neighbourhood has a certain form) as we consider the complex blow up of the
singular points. Indeed, the use of Mayer-Vietoris tried by the poser does look
more difficult (of course that would also use the local form of the singularity
but somehow in a less complex fashion).

ct.category theory - How can I understand the "groupoid" quotient of a group action as some sort of "product"?

Recall the notion of groupoid (Wikipedia, nLab). An important construction of groupoids is as "action groupoids" for group actions. Namely, let $X$ be a groupoid and $G$ a group, and suppose that $G$ acts on $X$ by groupoid automorphisms. Then we can form a new groupoid $X//G$, which has as objects the objects of $X$, but the morphisms include, in addition to the original morphisms of $X$, a morphism $x overset g to gx$ for each $gin G$ and $xin X$. The composition of morphisms is well-defined if the action is by groupoid automorphisms. (When $X$ is a set, then $X//G$ is equivalent to the skeletal groupoid whose objects are the elements of the "coarse" quotient $X/G$, and with ${rm Aut}(bar x) = {rm Stab}_G(x)$.)

(Probably there is a fancier construction, in which the conditions on the word "group action" be relaxed to an "action" up to specified natural isomorphism, and then $G$ could act on $X$ by autoequivalences, rather than autoisomorphisms, but this generalization won't concern me.)

Let $1$ denote the one-point set, thought of as a groupoid with only identity morphisms. Then any group $G$ acts uniquely on $1$, and so we have the groupoid $1//G$. In general, although $Xtimes 1 cong X$, we do not have $X times (1//G) cong X//G$ for arbitrary $G$-actions on $X$ unless the action is trivial. (Here $times$ denotes the groupoid product, which is just what you think it is.) However, the construction provides natural bijections between the objects of $X//G$ and the objects of $X times (1//G)$, and between the morphisms of $X//G$ and the morphisms of $X times (1//G)$.

Question: Is there some sort of "semidirect" or "crossed" product of groupoids, which presumably depends on extra data, so that we do have $X//G cong X rtimes (1//G)$? By which I mean, what is the correct notion of "action" of a groupoid $Y$ on a groupoid $X$ and what is the corresponding correct notion of $X rtimes Y$?

I see that the page semidirect product in nLab defines $X rtimes G$ as something closely related to $X//G$. But clearly this ought to be called $Xrtimes (1//G)$, but then I do not know what the right definition for $Xrtimes Y$ is, hence the question. And really I'd like to know about a "double crossed product" $Xbowtie Y$.

My motivation for this question is from my answer to Do rational numbers admit a categorification which respects the following “duality”?.

ac.commutative algebra - Elements of trace zero in a field extension

In characteristic $3$, yes.

Proof: Let $y=1$.

In characteristic $2$, yes. In all other cases, no.

Proof: The symmetric bilinear form $Tr(yz)$ on the vector space $F$ is nondegenerate (for any finite separable field extension). Its restriction to the codimension one subspace $A$ is also nondegenerate, since the orthogonal complement of $A$ is spanned by $1$, which is not in $A$. Count the nonzero solutions of $x=yz$ with $x,y,z$ all in $A$: the number is $(q^2-1)(q-1)$, because $y$ can be any nonzero vector in $A$ and $z$ can be any nonzero vector in the (one-dimensional) orthogonal complement of $Ky$ in $A$. If we identify the solutions $(x,y,z)$ and $(x,cy,c^{-1}z)$ for $cin F$ then the count becomes $q^2-1$. This set is mapped to another set $A-0$ of the same size by $(x,y,z)mapsto x$. To decide whether the map is surjective, look at whether it's injective.

In characteristic different from $2$, nondegeneracy of the form implies that there is a solution $(x,y,z)$ with $y$ and $z$ linearly independent, so that $(x,z,y)$ is an inequivalent solution.

In characteristic $2$, since $Tr(y^2)$ is universally congruent to $Tr(y)^2$ mod $2$, $A$ is closed under squaring; thus every solution has the form $(x,y,ay)$ with $ain K$. And injectivity holds because if $(x,y_1,a_1y_1)$ and $(x,y_2,a_2y_2)$ are solutions then $frac{y_1}{y_2}$ is a square root of $frac{a_2}{a_1}$, so an element of $K$.

mathematics education - Relating Category Theory to Programming Language Theory

The most immediately obvious relation to category theory is that we have a category consisting of types as objects and functions as arrows. We have identity functions and can compose functions with the usual axioms holding (with various caveats). That's just the starting point.

One place where it starts getting deeper is when you consider polymorphic functions. A polymorphic function is essentially a family of functions, parameterised by types. Or categorically, a family of arrows, parameterised by objects. This is similar to what a natural transformation is. By introducing some reasonable restrictions we find that a large class of polymorphic functions are in fact natural transformations and lots of category theory now applies. The standard examples to give here are the free theorems.

Category theory also meshes nicely with the notion of an 'interface' in programming. Category theory encourages us not to look at what an object is made of but how it interacts with other objects, and itself. By separating an interface from an implementation a programmer doesn't need to know anything about the implementation. Similarly category encourages to think about objects up to isomorphism - it doesn't precisely what sets our groups are made of, it just matters what the operations on our groups are. Category theory precisely captures this notion of interface.

There is also a beautiful relationship between pure typed lambda calculus and cartesian closed categories. Any expression in the calculus can be interpreted as the composition of the standard functions that come with a CCC: like the projection onto the factors of a product, or the evaluation function. So lambda expressions can be interpreted as applying to any CCC. In other words, lambda calculus is an internal language for CCCs. This is explicated in Lambek and Scott. This means that the theory of CCCs is deeply embedded in Haskell, say, because Haskell is essentially pure typed lambda calculus with a bunch of extensions.

Another example is the way structurally recursing over recursive datatypes can be nicely described in terms of initial objects in categories of F-algebras. You can find some details here.

And one last example: dualising (in the categorical sense) definitions turns out to be very useful in the programming languages world. For example, in the previous paragraph I mentioned structural recursion. Dualising this gives the notions of F-coalgebras and guarded recursion and leads to a nice way to work with 'infinite' data types such as streams. Working with streams is tricky because how do you guard against inadvertently trying to walk the entire length of a stream causing an infinite loop? The appropriate dual of structural recursion leads to a powerful way to deal with streams that is guaranteed to be well behaved. Bart Jacobs, for example, has many nice papers in this area.

ac.commutative algebra - Pushouts of noetherian rings

Does the category of noetherian rings has pushouts?

Background: If $X/S$ is an abelian scheme, then the relative Picard functor $Pic_{X/S}$, is only defined on the category of locally noetherian $S$-schemes (as far as I know). It is a group functor and in some situations it is representable. We then get a group object in the category of locally noetherian $S$-schemes, and I ask myself if it has a multiplication morphism.

Observe that the tensor product of noetherian rings does not have to be noetherian (isn't this ugly?). Even for fields there is a counterexample: Let $L/K$ be a purely transcendental field extension of infinite transcendence degree. Then $Omega^1_{L/K}$ is infinite-dimensional, from which you can concluce that the kernel of $L otimes_K L to L, a otimes b mapsto ab$ is not finitely generated. Thus $L otimes_K L$ is not noetherian.

Of course, this does not disprove that $L leftarrow K rightarrow L$ has a pushout in the category of noetherian rings. How can this be done? The question has a similar spirit (you may call it pathological) as this one.

set theory - Does ZF prove that all PIDs are UFDs?

ZF alone does not prove that every PID is a UFD, according to this paper: Hodges, Wilfrid. Läuchli's algebraic closure of $Q$. Math. Proc. Cambridge Philos. Soc. 79 (1976), no. 2, 289--297. MR 422022.

One result in this paper is the following:

COROLLARY 10. Neither (a) nor (b) is provable from ZF alone:
(a) Every principal ideal domain is a unique factorization domain.
(b) Every principal ideal domain has a maximal ideal.

By the way, I didn't know the answer to this question until today. To find the answer, I consulted Howard and Rubin's book Consequences of the Axiom of Choice. (Actually, I did a search for "principal ideal domain" of their book using Google Books.)

Indeterminate "x" in Abstract Algebra/Ring Theory

Among infinitely many other places, this issue is discussed in Sections 4.3 and 4.4 of my notes on commutative algebra:

http://math.uga.edu/~pete/integral.pdf

In Section 4.3 I give Mariano's definition, with some commentary. A slight drawback to this definition is that it makes the associativity of the product look mysterious. In Section 4.4, I mention that this may be viewed as a special case of the semigroup algebra construction, namely we may define $R[x]$ to be the set of all finitely nonzero functions $t: mathbb{N} rightarrow R$ -- this is algebra, so $mathbb{N} = {{bf 0},1,2,ldots}$ -- with pointwise addition and convolution product. The associativity still has to be checked, but it is relatively satisfying to do this once and for all in this level of generality. (And this will probably come in handy elsewhere, e.g. the associativity of the convolution product is precisely the content of the Möbius Inversion Formula.)

On the other hand -- when $R$ is commutative, as I assume from now on -- of course it does make sense to plug in a polynomial at any element of $R$: in other words, a polynomial determines a function from $R$ to $R$. Indeed, the evaluation map gives a homomorphism of rings from $R[t]$ to the ring of all functions from $R$ to $R$ under pointwise addition and pointwise multiplication. As I mention in my notes, when $R$ is an infinite integral domain, this evaluation map is injective and one can use this to deduce the associativity of the multiplication in $R[t]$ for free.

However, when $R$ is finite it is important to distinguish between polynomials in the formal sense and polynomial functions. In particular, your definition of a polynomial is not correct when $R$ is e.g. the finite field $mathbb{Z}/pmathbb{Z}$, because it does not distinguish between the polynomial $x^p - x$ and the zero polynomial: both induce the zero function. (That multiple polynomials may determine the same function has some positive aspects as well; it can be used to give a proof of the Chevalley-Warning theorem.)

oa.operator algebras - Ideals in Factors

One can easily prove that factors have no nontrivial ultraweakly closed 2-sided ideals as these are equivalent to nontrivial central projections. One can also show type $I_n$, type $II_1$, and type $III$ factors are algebraically simple (any 2-sided ideal must contain a projection. All projections are comparable in a factor, so you can show 1 is in the ideal). Ideals in $B(H)$ ($dim(H)=infty$, $H$ separable) have been studied extensively. What about ideals in $II_infty$ factors?

One might think, since every $II_infty$ factor $M$ can be written as $Noverline{otimes} B(H)$ for $N$ a $II_1$ factor, if $Isubset B(H)$ is an ideal, then $Notimes I$ is a 2-sided ideal. This is false. One needs to take the ideal generated by $Notimes I$. What does that mean from a von Neumann algebra viewpoint? Is it the same as taking the norm closure?

We can also describe some ideals in terms of the trace. One has the equivalent of the Hilbert-Schmidt operators: $$I_2={xin M | tr(x^ast x)<infty}$$ and the trace class operators:
$$I_1={xin M | tr(|x|)<infty}=I_2^ast I_2 =left{sum^n_{i=1} x_i^ast y_i | x_i, y_iin I_2right}.$$
What is the relation of $I_j$ to $Notimes L^j(H)$ for $j=1,2$ (where $L^2(H)$ is the Hilbert-Schmidt operators and $L^1(H)$ is the trace class operators in $B(H)$)? Is $I_j$ the norm closure of $Notimes L^j(H)$?

differential topology - Homologically trivial submanifolds

Unuseful prequel

Let $M$ be a (compact, oriented, differentiable) manifold. Before knowing anything about homology theory a naif but clever mathematician may want to measure the holes in $M$ by the following procedure. One lets $B_k(M)$ be the abelian group generated by embedded submanifolds of $M$ of dimension $k$ with border. There is an obvious border operator $delta colon B_{k}(M) to B_{k-1}(M)$ and trivially $delta^2 = 0$ since the border of a manifold with border is a closed manifold. So one may define $H^K_{naif}(M)$ by taking the quotient of cycles modulo boundaries.

Ok, this does not really work, for at least two reasons. The first there is no way to make this functorial without considering more degenerate objects (singular simplices or currents being two possible choices). The second is that even if one adjusts the definition to get, for instance, the bordism groups of $M$, these will be highly nontrivial for differential geometric reasons, rather than for the topology of $M$ (for instance they will be nontrivial for a point).

Anyway, one finally chooses a working (co)homology theory, for instance singular, and then can use the fundamental class of submanifolds to link this theory to the naif one. Two questions naturally arise:

1. Are fundamental classes of manifolds enough to generate the cohomology? This is nicely answered here.


2. Are submanifolds with border enough to generate homology relations? This is the purpose of the present question.

The actual question

Let $M$ be a (compact, oriented, differentiable) manifold and let $N subset M$ be a (closed) submanifold. The problem is to test whether $N$ is the boundary of submanifold with boundary of $M$. There are two obvious obstructions:

1. $N$ should be bordant as an abstract variety


2. The class $[N] in H_{*}(M, mathbb{Z})$ should be $0$.

Assume 1 and 2 hold. Are there any conditions on $M$, $N$ or the embedding $N to M$ which guarantee that $N$ is the boundary of submanifold with boundary of $M$?

There are a few classical cases which come to mind:

1. If $N$ and $M$ are spheres, 1 and 2 are vacuous and the thesis is true by the (generalized) Jordan curve theorem.


2. If $N = S^1$ and $M = mathbb{R}^3$, the thesis is still true by the existence of Seifert surfaces.


3. If $dim N = 1$ and $dim M = 2$ the result seems true to me by the classification of surfaces as quotients of polygons and the usual Jordan curve theorem (but I did not check the details).

On the other hand I'm sure there are plenty of negative example even in $3$-dimensional topology. Is there anything nontrivial which can be said about this question, whether on the positive or negative side?

latex - Tools for collaborative paper-writing

I use bzr (any particular reason why git, by the way?) and I've ended up using it for just about everything: papers, seminars, teaching, configuration files, my entire website, just about everything I do on a computer is in a bzr repository. Although I've yet to convince any collaborator to use it as well, I still find that it makes life easier since I can easily keep a record of when I sent something to someone else and merge in changes against that particular revision. I can also publish a repository and make it easy for a collaborator to have access to the files without needing to use bzr themselves.

Within a paper, I use the changes.sty package for sharing comments back and forth between myself and a collaborator.

Bzr has "nice" frontends so it might be possible to persuade a non-technologically minded person to use it (I'm a commandline junkie so have no experience of the available GUIs).

I also use a wiki (nlab, naturally) but that is (at the moment) for less focussed projects than a specific paper. However, when writing anything substantial there then I do it "offline" (even so far as to "compiling" and viewing it) and only sending it into the ether when I'm happy with it.

I find it completely incomprehensible that people want everything to be "in the cloud". I have access to several high-powered computers which are capable of running whatever software I'm using incredibly fast. Why would I swap that for a slow, crackly internet link which is guaranteed to be down the one time that I really need it? By using a DVCS (distributed version control system), I only need to be connected to the internet at the start of a given session and I can get my files off any one of a number of machines so it doesn't matter if one or other is down. In the worst case scenario that I can't connect, I can work offline on something and then merge my changes back again later. Indeed, my entire DVCS currently takes up a mere 71Mb (of which 25Mb consists of my local copies of Instiki and xournal) so I could easily carry it around with me on a memory stick (encrypted, of course).

If I really did want to do some "real time" collaboration, I would use either gobby (for working on files or papers) or jarnal (for working on maths). Gobby has real-time editing (and has had for quite some time) whilst with jarnal I can use my graphics tablet to actually write the mathematics for the other person to see just as if we were at a blackboard together. After all, if I'm doing real-time collaboration then I don't want to bother with getting the LaTeX syntax exactly right. I'm not bad at TeX, but if I'm in "Math Mode" then I don't want to be bothering with it.

lo.logic - "Requires axiom of choice" vs. "explicitly constructible"

I think I'm a bit confused about the relationship between some concepts in mathematical logic, namely constructions that require the axiom of choice and "explicit" results.

For example, let's take the existence of well-orderings on $mathbb{R}$. As we all know after reading this answer by Ori Gurel-Gurevich, this is independent of ZF, so it "requires the axiom of choice." However, the proof of the well-ordering theorem that I (and probably others) have seen using the axiom of choice is nonconstructive: it doesn't produce an explicit well-ordering. By an explicit well-ordering, I simply mean a formal predicate $P(x,y)$ with domain $mathbb{R}timesmathbb{R}$ (i. e., a subset of the domain defined by an explicit set-theoretic formula) along with a proof (in ZFC, say, or some natural extension) of the formal sentence "$P$ defines a well-ordering." Does there exist such a $P$, and does that answer relate to the independence result mentioned above?

More generally, we can consider an existential set-theoretic statement $exists P: F(P)$ where $F$ is some set-theoretic formula. Looking to the previous example, $F(P)$ could be the formal version of "$P$ defines a well-ordering on $mathbb{R}$." (We would probably begin by rewording that as something like "for all $zin P$, $z$ is an ordered pair of real numbers, and for all real numbers $x$ and $y$ with $xneq y$, $((x,y)in P vee (y,x)in P) wedge lnot ((x,y)in P wedge (y,x)in P)$, etc.) On the one hand, such a statement may be a theorem of ZF, or it may be independent of ZF but a theorem of ZFC. On the other hand, we can ask whether there is an explicit set-theoretic formula defining a set $P^*$ and a proof that $F(P^*)$ holds. How are these concepts related:

• the theoremhood of "$exists P: F(P)$" in ZF, or its independence from ZF and theoremhood in ZFC;




• the existence of an explicit $P^*$ (defined by a formula) with $F(P^*)$ being provable.

Are they related at all?

finite groups - Injective proof about sizes of conjugacy classes in S_n

For any cycle decomposition, we can uniquely order the cycles from smallest length to largest length, breaking ties between cycles of the same length in some fixed arbitrary way (say by maximal elements). Let us do this for concreteness.

Suppose there was an (injective) way to "join" and A-cycle and a B-cycle together to form an A+B cycle whenever |A| and |B| are > 1. Then, given any cycle decomposition as above, one can start bubbling the two largest cycles together to eventually form a single cycle of some length m.

If m = n - 1, we are done.

If m = n, write S = (1............) and then omit the last term.

If m = 1 the problem is very easy.

If n - 1 > m > 2, there is an injective map from m-cycles to elements whose cycle decomposition is a product of an m-cycle with a 2-cycle. For concreteness, one can add the 2-cycle with the two lowest missing entries. Now bubble again to form an m+2 cycle.

If m = 2, and n is at least 5, bubble with a 3-cycle.

If m = 2 and n = 4 (the last case), form whatever bijection you like between the two sets of 6 elements.

The key point is therefore to find a way to bubble an A-cycle and a B-cycle when |A|,|B| > 1.
We do this as follows.

Amongst the entries of A and B, there is a unique smallest integer, call it X.

Case 1. If X lies in A, Let Z denote the largest element of B.
Then one can (uniquely) write A = (X.....) and B = (.....Y,Z), where Y < Z. Then consider the A+B cycle obtained by concatenating A and B in this form, i.e. (X,......,Y,
Z).

Case 2. If X lies in B, let Z denote the smallest element of A.
Then one can (uniquely) write B = (X....) and A = (.....,Y,Z), where now Y > Z.
Then consider the A+B cycle (X,.....,Y,Z).

Given an A+B cycle, one can uniquely write it in the form (X,....,Y,Z), where X is the smallest entry. Then one can break it up into an A-cycle and B-cycle depending on whether Y < Z or Y > Z.

---

Since this was apparently a little confusing, suppose that the cycle lengths of S are
a_1 <= a_2 <= a_3 <= ..... <= a_r.
Here I omit the 1-cycle lengths, so a_1 > 1, and sum a_r = m for some m possibly less than n. Then the cycle lengths of the steps in the algorithm will have lengths:

(a_1, ...., a_(r-1),a_r),

(a_1, ...., a_(r-2),a_(r-1) + a_r),

(a_1, ...., a_(r-3),a_(r-2) + a_(r-1) + a_r),

....

(a_1 + a__2 + ... + a_r) = (m)

(2,m)

(m+2)

(2,m+2)

(m+4)

...

(n-1 or n, depending on m mod 2),

then n-1.

(if m = 2 and n is at least 5, then instead it should go

(2) --> (2,3) --> (5) --> (2,5) --> (7) --> (2,7) --> (9) ...,etc.

nt.number theory - Cycle Length of the Positive Powers of Two Mod Powers of Ten

And if you insist, let me write this out in detail. All you need is the following lemma.

Lemma: Let f(n) be periodic with period p and let g be injective. Then g(f(n)) is periodic with period p.

Proof. Clearly g(f(n+p)) = g(f(n), so g(f(n)) has some period q dividing p. On the other hand, g(f(n+q)) = g(f(n)) for all n if and only if f(n+q) = f(n) for all n by injectivity, so q = p.

As I remarked above we have b_n = b for all but finitely many n and x -> CRT(x, b) is an injection. The result follows.

ag.algebraic geometry - Counting branched covers of the projective line and Spec Z

I've asked a question like this before, but now I'm more interested in counting the number of covers.

We suppose given the following data.

1. A positive integer $d$




2. A finite set of closed points $B= ( b_1,ldots,b_n )$ in $mathbf{P}^1_mathbf{C}$




3. Branch types $T_1,ldots, T_n$.

Question. How many branched covers of $mathbf{P}^1_mathbf{C}$ exist which are branched only over $b_i$ (with branch type $T_i$ over each $b_i$)?

The answer lies within the Hurwitz number for $(T_1,ldots,T_n)$. This translates the problem to combinatorial group theory.

Now, for my main question:

Q1. Can one ``count'' covers of $textrm{Spec} mathbf{Z}$ as above? That is, can one count the number of finite field extensions $$mathbf{Q}subset K$$ of given degree $d=[K:mathbf{Q}]$ which are unramified outside a given set of prime numbers $p_1,ldots,p_n$ with ramification types $T_1,ldots,T_n$?

I know that one can use Minkowski's Geometry of Numbers to give some nontrivial bounds on the discriminant. Is this the best we can do?

ct.category theory - Is there a natural way to give a bisimplicial structure on a 2-category?

I mean by the nerve functor.

Given a 2-category $mathcal{C}$, if we forget the 2-category structure (just view $mathcal{C}$ as a category), the nerve functor will give us a simplicial set $Nmathcal{C}$. However, $mathcal{C}$ is a 2-category, thus for any two objects $x,yinmathcal{C}$, $Hom_{mathcal{C}}(x,y)$ is a category, applying the nerve functor gives us a simplicial set $N(Hom(x,y))$.

My question is, can these two simplicial set structure compatible in some way, gives us a bisimplicial set $N_{p,q}(mathcal{C})$, say? Or is there another way to give a bisimplicial structure on a 2-category?

computational complexity - Why is proving P != NP so hard?

On the contrary, there are two major results in complexity theory that rule out a wide class of methods to show that $P ne NP$. The first is the theorem of Baker, Gill, and Solovay, that a proof that $P ne NP$ (or a proof that they are equal) cannot relativize. In other words, they showed that there exists an oracle relative to which they are equal, and an oracle relative to which they are different. The second result is the theorem of Razborov and Rudich, that if a widely accepted refinement of the $P ne NP$ conjecture is true, then there does not exist a "natural proof" that they are different. By a natural proof, they mean a proof from a large class of combinatorial constructions. In light of those two theorems, there actually aren't very many known promising techniques left, even though there is a lot of evidence by example that the conjecture seems to be true. As Razborov and Rudich explain, these two results rule out candidate approaches to P vs NP for sort-of opposite reasons.

There is a CS professor named Ketan Mulmuley who has expressed some optimism that P vs NP can be solved with "geometric complexity theory". I can believe that Mulmuley is doing interesting work of some kind (which seems to involve quantum algebra and representation theory), but I haven't heard of many complexity theorists who are optimistic along with him that he can really solve P vs NP. (But hey, Perelman surprised everyone with a proof of the Poincare conjecture, so who knows.)

---

Some additional remarks. First, there are plenty of conjectures have ample evidence yet are difficult for no obvious reason. The P vs NP problem has an unusual status in that people have thought of rigorous reasons that it's hard.

Second, when people prove a "barrier result" (meaning, a negative result about how not to prove a conjecture), obviously the community will take it as a challenge to find new ideas that circumvent the barrier. As mentioned in the comments, there was even a conference last year on doing exactly that! Baker, Gill, and Solovay was published in 1975, and it took about 15 years to find convincing exceptions to their point about relativization. (Unconvincing exceptions that can be explained as unfairly restricted oracle access came more quickly.) Nonetheless, when I did a computer-assisted survey of binary relations between complexity classes a few years ago, it was clear that the vast majority of these proven relations still relativize. It is true that by now the research focus is on non-relativizing results, with the exception of quantum complexity classes, where relativizing results are still popular.

Third, the newer Razborov-Rudich theorem made people start all over again to look for barrier loopholes. Moreover the Baker-Gill-Solovay theorem, as an obstruction to P vs NP, was sharpened somewhat by Aaronson and Wigderson in their paper on "algebrization" of complexity class relations. My third point is that I can't speak with any authority on efforts to overcome the current set of barriers.

nt.number theory - Class number of non-maximal order in imaginary quadratic function field?

There is a formula that works in all degrees, not just imaginary quadratic. In a global field $K$, let $O$ be integral over ${mathbf Z}$ or ${mathbf F}[t]$ (${mathbf F}$ a finite field) and be "big", i.e., it has fraction field $K$. Let $mathfrak c$ be the conductor ideal of $O$ in its integral closure $R$. Then
$$h(O) = frac{h(R)}{[R^times:O^times]}frac{varphi_{R}({mathfrak c})}{varphi_O(mathfrak c)},$$
where $varphi_O(mathfrak c)$ is the number of units in $O/mathfrak c$ and $varphi_R(mathfrak c)$ is the number of units in $R/mathfrak c$. This is derived in Neukirch's alg. number theory book in the number field case, but it goes through to any one-dimensional Noetherian domain with a finite residue rings and a finite class group.
In the imag. quadratic case the unit index $[R^times:O^times]$ is 1 most of the time so you don't notice it.

Both $varphi_R(mathfrak c)$ and $varphi_O(mathfrak c)$ can be written in the form ${text N}(mathfrak c)prod_{mathfrak p supset mathfrak c}(1 - 1/{text{N}(mathfrak p)})$, where the ideal norm $text N$ means the index in $R$ or $O$ and $mathfrak p$ runs over primes in $R$ or $O$ for the two cases.

ag.algebraic geometry - Differential of the Torelli morphism at the boundary

Let consider the Torelli morphism $T:mathcal{M}_g rightarrow mathcal{A}_g$, from the moduli space of curves of genus $g$ to the moduli space of principal polarized abelian varieties of dimension $g$, that maps a curve to its Jacobian. The differential of $T$ at a point $[C]$ is the natural map
$$H^1(C, T_C) rightarrow Sym^2H^1(C, mathcal{O}_C).$$
I know that $T$ can be extended to a map
$$T:bar{mathcal{M}_g} rightarrow bar{mathcal{A}_g}$$
from the Deligne-Mumford compactification of $mathcal{M}_g$ to some compactification of $mathcal{A}_g$.

I would like to know if there is a way to describe the differential of $T$ at a point representing a nodal curve. More specifically, how can we describe the deformations space of a semi-abelian variety and in particular of a generalized Jacobian variety?
Over $mathbb{C}$, by computing the period matrix, one can show that the differential of $T$ has maximal rank at each point representing a nodal curve with non-hyperelliptic normalization. I'm wondering if, perhaps, there is a more algebraic way to see it.

gn.general topology - Galois Groups vs. Fundamental Groups

I saw this question a while ago and felt something in the way of a (probably misguided) missionary zeal to make at least a few elementary remarks. But upon reflection,
it became clear that even that would end up rather long, so it was difficult to find the time until now.

The point to be made is a correction: fundamental groups in arithmetic geometry are not the same as Galois groups, per se. Of course there is a long tradition of
parallels between Galois theory and the theory of covering spaces, as when Takagi writes of being misled by
Hilbert in the formulation of class field theory essentially on account of
the inspiration from Riemann surface theory. And then, Weil was fully aware that
homology and class groups are somehow the same, while speculating that a sort of
non-abelian number theory informed by the full theory of the 'Poincare
group' would become an ingredient of many serious arithmetic investigations.

A key innovation of Grothendieck, however, was the formalism for refocusing attention on the
base-point. In this framework, which I will review briefly below, when one says
$$pi_1(Spec(F), b)simeq Gal(bar{F}/F),$$
the base-point in the notation is the choice of separable closure
$$b:Spec(bar{F})rightarrow Spec(F).$$
That is,

Galois groups are fundamental groups with generic base-points.

The meaning of this is clearer in the Galois-theoretic interpretation of the fundamental group of
a smooth variety $X$. There as well, the choice of a separable closure
$k(X)hookrightarrow K$ of the function field $k(X)$ of $X$ can be viewed as a base-point
$$b:Spec(K)rightarrow X$$
of
$X$, and then
$$pi_1(X,b)simeq Gal(k(X)^{ur}/k(X)),$$
the Galois group of the maximal sub-extension $k(X)^{ur}$ of $K$ unramified over $X$.
However, it would be quite limiting to take this last object as the definition of the fundamental group.

We might recall that even in the case of a path-connected pointed topological space $(M,b)$ with universal covering space $$M'rightarrow M,$$
the isomorphism $$Aut(M'/M)simeq pi_1(M,b)$$ is not canonical. It comes rather
from the choice of a base-point lift $b'in M'_b$. Both $pi_1(M,b)$ and $Aut(M'/M)$
act on the fiber $M'_b$, determining bijections
$$pi_1(M,b)simeq M'_bsimeq Aut(M'/M)$$
via evaluation at $b'$. It is amusing to check that the isomorphism of groups obtained thereby is independent of
$b'$ if and only if the fundamental group is abelian. The situation here is an instance of the choice involved in the isomorphism
$$pi_1(M,b_1)simeq pi_1(M,b_2)$$
for different base-points $b_1 $ and $b_2$.
The practical consequence is that when fundamental groups are equipped with natural
extra structures coming from geometry, say Hodge structures or Galois actions, different base-points give rise to enriched groups that are
are often genuinely non-isomorphic.

A more abstract third group is rather important in the general discussion of base-points. This is
$$Aut(F_b),$$
the automorphism group of the functor
$$F_b:Cov(M)rightarrow Sets$$
that takes a covering $$Nrightarrow M$$ to its fiber $N_b$. So elements of $Aut(F_b)$ are
compatible collections $$(f_N)_N$$ indexed by coverings $N$ with each $f_N$ an automorphism of the set $N_b$.
Obviously, newcomers might wonder where to get such compatible collections, but
lifting loops to paths defines a natural map
$$pi_1(M,b)rightarrow Aut(F_b)$$
that turns out to be an isomorphism. To see this, one uses again the fiber
$M'_b$ of the universal covering space, on which both groups act compatibly.
The key point is that while $M'$ is not
actually universal in the category-theoretical sense, $(M',b')$ is universal
among pointed covers. This is enough to show that an element of $Aut(F_b)$ is completely determined by its action
on $b'in M'_b$, leading to another bijection $$Aut(F_b)simeq M'_b.$$
Note that the map $pi_1(M,b)rightarrow Aut(F_b)$ is entirely canonical,
even though we have used the fiber $M'_b$ again to prove bijectivity, whereas the identification with $Aut(M'/M)$
requires the use of $(M'_b,b')$ just for the definition.

Among these several isomorphic groups, it is $Aut(F_b)$ that ends up most relevant for the
definition of the etale fundamental group.

So for any base-point $b:Spec(K)rightarrow X$ of a connected scheme
$X$ (where $K$ is a separably closed field, a 'point' in the etale theory), Grothendieck defines
the 'homotopy classes of etale loops' as
$$pi^{et}_1(X,b):=Aut(F_b),$$
where $$F_b:Cov(X) rightarrow mbox{Finite Sets}$$ is the functor that sends a finite etale covering
$$Yrightarrow X$$ to the fiber $Y_b$. Compared to a construction like
$Gal(k(X)^{ur}/k(X))$, there are three significant advantages to this definition.

(1) One can easily consider small base-points, such as might come from
a rational point on a variety over $mathbb{Q}$.

(2) It becomes natural to study the variation of $pi^{et}_1(X,b)$ with $b$.

(3) There is an obvious extension to path spaces $$pi^{et}_1(X;b,c):=Isom(F_b,F_c),$$ making up a two-variable
variation.

This last, in particular, has no analogue at all in the Galois group approach to
fundamental groups. When $X$ is a variety over $mathbb{Q}$, it becomes possible, for example, to study $pi^{et}_1(X,b)$ and
$pi^{et}_1(X;b,c)$ as sheaves on $Spec(mathbb{Q})$, which encode rich information about
rational points. This is a long story, which would be rather tiresome to expound upon here
(cf. lecture at the INI ).
However, even a brief contemplation of it might help you to appreciate the arithmetic perspective that general $pi^{et}_1$'s
are substantially more powerful than Galois groups. Having read thus far, it shouldn't surprise you that
I don't quite agree with
the idea explained, for example, in this post
that a Galois group is only
a 'group up to conjugacy'. To repeat yet again, the usual Galois groups are just fundamental groups with specific large base-points.
The dependence on these base-points as well as a generalization to small base-points
is of critical interest.

Even though the base-point is very prominent in Grothendieck's definition, a curious fact is that it took quite a long time for even the experts to fully metabolize its significance.
One saw people focusing mostly on base-point independent constructions
such as traces or characteristic polynomials associated to representations. My impression is that the initiative for allowing the base-points a truly active role
came from Hodge-theorists like Hain, which then was taken up by arithmeticians like Ihara and Deligne.
Nowadays, it's possible to give entire lectures just about base-points, as Deligne has actually done on several occasions.

Here is a puzzle that I gave to my students a while ago: It has been pointed out that
$Gal(bar{F}/F)$ already refers to a base-point in the Grothendieck definition. That is,
the choice of $Fhookrightarrow bar{F}$ gives at once a universal covering space and a base-point.
Now, when we turn to the manifold situation $M'rightarrow M$, a careful reader may have noticed a hint above that there is
a base-point implicit in $Aut(M'/M)$ as well.
That is, we would like to write $$Aut(M'/M)simeq pi_1(M,B)$$ canonically for some base-point $B$. What is $B$?

Added:

-In addition to the contribution of Hodge-theorists, I should say that Grothendieck himself urges attention to many base-points in his writings from the 80's, like 'Esquisse d'un programme.'

-I also wanted to remark that I don't really disagree with the point of view in JSE's answer either.

Added again:

This question reminds me to add another very basic reason to avoid the Galois group as a definition of $pi_1$. It's rather tricky to work out the functoriality that way, again because the base-point is de-emphasized. In the $Aut(F_b)$ approach, functoriality is essentially trivial.

Added, 27 May:

I realized I should fix one possible source of confusion. If you work it out, you find that the bijection $$pi_1(M,b)simeq M'_bsimeq Aut(M'/M)$$ described above is actually an anti-isomorphism. That is, the order of composition is reversed. Consequently, in the puzzle at the end, the canonical bijection $$Aut(M'/M)simeq pi_1(M,B)$$ is an anti-isomorphism as well. However, another simple but amusing exercise is to note that the various bijections with Galois groups, like $$pi_1(Spec(F), b)simeq Gal(bar{F}/F),$$ are actually isomorphisms.

Added, 5 October:

I was asked by a student to give away the answer to the puzzle. The crux of the matter is that
any continuous map $$B:Srightarrow M$$ from a simply connected set $S$ can be used as a base-point for the
fundamental group. One way to do this to use $B$ to get a fiber functor
$F_B$ that associates to a covering $$Nrightarrow M$$the set of splittings of the covering $$N_B:=Stimes_M Nrightarrow S$$ of $S$.
If we choose
a point $b'in S$, any splitting is determined by its value at $b'$, giving
a bijection of functors
$F_B=F_{b'}=F_b$ where $b=B(b')in M$. Now, when $$B:M'rightarrow M$$ is the universal
covering space, I will really leave it as a (tautological) exercise
to exhibit a canonical anti-isomorphism
$$Aut(F_B)simeq Aut(M'/M).$$ The 'point' is that $$F_B(M')$$ has a canonical
base-point that can be used for this bijection.

ag.algebraic geometry - Why is the fibered coproduct of affine schemes not affine?

The short answer is that the category of affine schemes does have pushouts, but these are not the same as pushouts of affine schemes calculated in the category of all schemes.

For a longer answer, consider an example that's small enough to compute: The projective line (over the complex numbers, say) is not affine but it is gotten by gluing two copies of the affine line along the punctured affine line, so it is the pushout (in the category of schemes) of a diagram of affine schemes. Now what's the pushout of that same diagram in the category of affine schemes? Well, the rings involved are two copies of $C[x]$ and a copy of $C[x,x^{-1}]$. The two maps are the two injections of $C[x]$ into $C[x,x^{-1}]$, one sending $x$ to $x$, and the other sending $x$ to $x^{-1}$. The pullback of these, in the category of commutative rings, is just $C$, because the only way a polynomial in $x$ can equal a polynomial in $x^{-1}$ is for both of them to be constant. Therefore, the affine-scheme pushout is not the projective line but a point.

Intuitively, if you glue together two copies of the line along the punctured line "gently," allowing the result to be non-affine, you get the projective line, but if you demand that the result be affine then your projective line is forced to collapse to a point.

ag.algebraic geometry - etale fundamental group and etale cohomology of curves

Given a curve $C$. Is there any relation between the etale fundamental group $pi_1(C)$ and the first etale cohomology of the constant sheaf , say $Z/nZ$, on $C$ ?

For example, if $C$ is a complex curve, then the singular cohomology $H^1(C,Z)$ is the dual of the topological fundamental group divided by the commutators ( which is the same as Hom$(pi_1(C),Z) )$.

So it seems that there should be some relation between Hom$(pi_1(C),Z/nZ)$ and $H^1(C,Z/nZ)$ in the etale case, but how?

big list - Older editions of which books were better than the new ones?

Hausdorff's book Mengenlehre in the first edition had an appendix, omitted in subsequent editions, on the Banach paradox. (Later made into the Banach-Tarski paradox by Tarski...) Someone once told me this was the best, most elementary, presentation of it -- I haven't compared different versions of the proof myself.

nt.number theory - A limit involving the totient function

P. Erdős and Leon Alaoglu proved in [1] that for every $epsilon > 0$ the inequality $phi(sigma(n)) < epsilon cdot n$ holds for every $n in mathbb{N}$, except for a set of density $0$.

C. L. mentioned in [2] that as a consequence of the previous result one can ascertain that $displaystyle lim_{n to infty} frac{phi(sigma(n)) }{n} = 0.$

Does anybody know how it is that C. L. proceeded in order to arrive at such a conclusion?

Clearly enough, the fact that an inequality of the type $a_{n} < epsilon cdot n$ holds for every $epsilon > 0$ and a subset of $mathbb{N}$ of density $1$ does not imply, in general, that the sequence $displaystyle frac{a_{n}}{n}$ goes to $0$ as $n to infty$.

Hope you guys can shed some light on this inquiry of mine. Let me thank you in advance for your continued support.

References

[1] L. Alaoglu, P. Erdős: A conjecture in elementary number theory, Bull. Amer. Math. Soc. 50 (1944), 881-882.

[2] Mathematical Reflections, Solutions Dept, Issue #3, 2009, page 23.

lie algebras - Under what conditions will a unitary matrix fix a subspace which does not diagonalize the generating Hamiltonian?

OK, I think I have a comprehensive answer. Since this is physics-inspired, I'll use bra-ket notation.

First, in my comment above, you can see that if $U_{t_0}$ fixes a subspace $V_n subset V$, then $U_{t_0}$ is block diagonal:

If we write $U$ in the basis ${|v_1rangle,...,|v_Nrangle}$ and see how it acts on the basis of $V_n$, we must have the lower left block of $U$ be zero. Then, since the columns of unitary matrices form an orthonormal basis of the whole space, the columns of the upper left block of $U$ must be an orthonormal basis of $V_n$ . So, the columns of the right side of $U$ must span ${V_n}^{perp}$, so that the upper right hand block is zero.

Next, let $W:=U_{t_0}|_{V_n}$ be the upper block of $U_{t_0}$ which acts on $V_n$:

$quad U_{t_0} = \left[ begin{matrix} W & 0 \ 0 & W' end{matrix} right]$

$W$ is unitary $n times n$ matrix, and therefore can be diagonalized. Without loss of generality, assume the orthonormal vectors ${|v_1rangle,...,|v_nrangle}$ diagonalize $W$ with respective eigenvalues ${exp(i w_i t_0)}$. Also, let ${|h_1rangle,...,|h_Nrangle}$ be the orthonormal basis (in general, completely different than the $|v_irangle$) which diagonalizes $H$ with eigenvalues $h_j$.

Then, for all $i$ with $1 le i le n$,

$quad exp(i w_i t_0) = langle v_i| U_{t_0}|v_irangle = langle v_i|exp(i H t_0) |v_irangle = sum_{j=1}^N exp (i h_j t_0) |langle v_i | h_j rangle|^2 $.

After multiplying both sides by $exp(-i w_i t_0) $ we get

$quad 1 = sum_{j=1}^N exp [i (h_j-w_i) t_0] |langle v_i | h_j rangle|^2$.

Now, since ${|h_jrangle}$ is an orthonormal basis, the values $|langle v_i | h_j rangle|^2$ sum to unity (for fixed, normalized $|v_irangle$). That means all the exponentials on the rhs for which $|langle v_i | h_j rangle|^2 neq 0$ must be equal to unity. So,

$quad (h_j-w_i) frac{t_0}{2 pi} in mathbb{Z}$

and

$quad (h_j-h_{j'}) frac{t_0}{2 pi} in mathbb{Z} quad forall j,j'$ such that $langle v_i | h_j rangle $ and $ langle v_i | h_{j'} rangle neq 0$.

In other words, for all $|h_jrangle$ that have non-vanishing inner product with $|v_irangle$, the corresponding $h_j$ are separated by integer multiples of $frac{2 pi}{t_0}$.

It's easy to show that this is also a sufficient condition. Given $H$ with eigenvalues $h_j$ and any subset $mathbf{H}_{t_0} subset {h_j}$ such that for all $h_j,h_{j'} in mathbf{H}_{t_0}$,

$quad (h_j-h_{j'}) frac{t_0}{2 pi} in mathbb{Z}$

then any vector $|vrangle$ in the span of eigenvectors corresponding to $mathbf{H}_{t_0}$ will be an eigenvector of $U_{t_0} = exp(i H t_0)$ with eigenvalue $exp(i h_j t_0)$ (which is the same for all $h_j in mathbf{H}_{t_0}$).

(Everything below is old)

Example (Old)

Still not an answer, but here's a simple but non-trivial example:

Let $U_t$ be the family of unitaries acting on $C^3$ which spatially rotate $R^3$ about the vector $(1,1,0)$ with period $T=2pi$. Let $t_0 = T/2=pi$. Then in the standard basis,

$quad H = frac{i}{sqrt{2}}left[ begin{matrix} 0 & 0 & 1 \ 0 & 0 & -1 \ -1 & 1 & 0 end{matrix} right] quad ,$

$quad U_{t_0/2} = frac{1}{2}left[ begin{matrix} 1 & 1 & -sqrt{2} \ 1 & 1 & sqrt{2} \ sqrt{2} & -sqrt{2} & 0 end{matrix} right]quad , quad U_{t_0} = left[ begin{matrix} 0 & 1 & 0 \ 1 & 0 & 0 \ 0 & 0 & 1 end{matrix} right] quad , quad U_{T} = I,$

so $U_{t_0}V_2 = V_2$, where $V_2 = mathrm{span}{(1,0,0),(0,1,0)}$. The eigenbasis is ${(1,-1,-isqrt{2}),(1,-1,isqrt{2}),(1,1,0)}$ with respective eigenvalues ${1,-1,0}$ for $H$.

Note that $H$ and $U_{t_0 /2}$ are not block diagonal in the standard basis, but $U_{t_0}$ is. In particular, no subset of the eigenvectors of $H$ form an orthonormal basis of $V_2$.

Commutation (Old)

I think the confusion over when you can infer that $H$ is block diagonal comes from the non-uniqueness of matrix roots. Positive matrices have unique positive roots, of course, but unitary matricies do not have unique unitary roots. For example, the $2 times 2$ identity matrix has itself and

$quad left[ begin{matrix} 0 & 1 \ 1 & 0 end{matrix} right]$

as square roots, but only the latter mixes the first and second rows. Likewise, I suspect that all non-trivial cases of unitaries fixing subspaces (where non-trivial is defined by Michael Underwood above) will have $t_0$ exactly such that non-degenerate eigenvectors of $H$ are degenerate for $U_{t_0}$.