soft question - Which math paper maximizes the ratio (importance)/(length)?

Any of three papers dealing with primality and factoring that are between 7 and 13 pages:

First place: Rivest, R.; A. Shamir; L. Adleman (1978). "A Method for Obtaining Digital Signatures and Public-Key Cryptosystems". Communications of the ACM 21 (2): 120–126.

Runner-up: P. W. Shor, Algorithms for quantum computation: Discrete logarithms and factoring, Proc. 35nd Annual Symposium on Foundations of Computer Science (Shafi Goldwasser, ed.), IEEE Computer Society Press (1994), 124-134.

Honorable mention: Manindra Agrawal, Neeraj Kayal, Nitin Saxena, "PRIMES is in P", Annals of Mathematics 160 (2004), no. 2, pp. 781–793.

ag.algebraic geometry - Are there "motivic" proofs of Weil conjectures in special cases?

Of course, there's Serre's Analogues kählériens de certaines conjectures de Weil, Annals 1960, where he deduces an analogue of the Weil-Riemann hypothesis over $mathbb{C}$ using standard facts from Hodge theory.
This is technically not an answer at all, but I thought I'd mention it since I had the
(perhaps mistaken) impression that this was partly the inspiration for the standard conjectures.

The other more relevant comment is that one can give an elementary proof of the Weil conjecture for any smooth variety whose Grothendieck motive lies in the tensor
category generated by curves. I should explain,
especially in light of Minhyong's comments, that this could be understood as shorthand
for saying the variety can built up from curves by taking products, taking images, blow ups along centres which of the same type, and so on. Actually, for such varieties, the
Frobenius can be seen to act semisimply. I think this open in general. So perhaps there's
some value in this.

set theory - Some consequences of internally approachable structures

I just read for the first time the definition of an internally approachable set, which says:

A set $N$ is internally approachable (I.A.) of length $mu$ iff there is a sequence $(N_{alpha} : alpha < mu)$ for which the following holds: $N=bigcup_{alpha< mu} N_{alpha}$ and for all $beta < mu$ $( N_{alpha} : alpha < beta ) in N$.

Now if $N prec (H(theta), in, < )$ is I.A. of length $mu$. Is it true that

(a) If $alpha < mu$ then $alpha in N$

(b) If $alpha < mu$ then $N_{alpha} in N$ ?

This is trivial if $N$ is transitive, and I'm quite sure that both (a) and (b) hold but I need a good argument.

ac.commutative algebra - Nilradicals without Zorn's lemma

I assume the argument you have in mind is the folowing: suppose $f in cap mathfrak{p}$; then to show that $f$ is nilpotent, it suffices to show that the localized ring $A_f$ is zero. And indeed, if $f$ is in every prime ideal of $A$, then $A_f$ has no prime ideals at all; since every nonzero ring has a maximal ideal by Zorn's Lemma, we must have $A_f = 0$.

This line of reasoning easily adapts to show that in fact, the statement that $operatorname{Nil}(A) = cap mathfrak{p}$ implies that every nonzero ring has a prime ideal. Indeed, suppose that $A$ were nonzero with no prime ideals; then $cap mathfrak{p} = A$, so every element of $A$ is nilpotent. In particular, $1 = 1^n = 0$, so $A = 0$.

Following Eric Rowell's answer, this is very close to being equivalent to the axiom of choice (however, it does not obviously imply the existence of maximal ideals).

differential topology - Is a conceptual explanation possible for why the space of 1-forms on a manifold captures all its geometry?

Let $M$-be a differentiable manifold. Then, suppose to capture the underlying geometry we apply the singular homology theory. In the singular co-chain, there is geometry in every dimension. We look at the maps from simplexes, look at the cycles and go modulo the boundaries. This has a satisfying geometric feel, though I need to internalize it a bit more(which matter I tried to address in other questions).

Now on the other hand, let $Omega^1$ be the space of $1$-forms on the space. The rest of the de Rham complex comes out of this object, wholly through algebraic processes, ie by taking the exterior powers and also the exterior derivative. After getting this object in hand, the journey upto getting the de Rham cohomology ring is entirely algebraic.

And by the de Rham theorem, this second method is equally as good as the more geometric first method. In the second method no geometry is explicitly involved anywhere in any terms after the first term. So the module of $1$-forms somehow magically capture all the geometry of the space $M$ without need of any explicit geometry. This is amazing from an algebraic point of view since we have less geometric stuff to understand.

This makes me wonder for the conceptual reason why this is true. I know that one should not look a gift horse in the mouth. But there is the need to understand why there is such a marked difference in the two approaches to capturing the geometry in a manifold, viz, through de Rham cohomology in differential topology, and through singular homology in algebraic topology. I would be grateful for any explanations why merely looking at all the $1$-forms is so informative.

Edited in response to comments: I meant, the de Rham cohomology is as good as singular homology for differentiable manifolds. What is "geometry of rational homotopy type"? And what is "geometry of real homotopy type"?

Are there other nice math books close to the style of Tristan Needham?

If I may add my two cents, I would add two more books that are an integral part of my library, and which I have presently lent to a gifted middle school student. One is the 'shape of space' by Jeff Weeks, and the other is 'Symmetry of things' by John Conway

Jeff Week's book is an incredibly enjoyable account of the topology of 3-manifolds. I came across someone mentioning the late Bill Thurston's book in this post. While Thurston's book is definitely more rigorous, I would say that Week's book is an overlooked classic. His invitation to experiment with intuition to extrapolate to the abstract, and tying in a theoretician's mental forays with cosmological measurements is quite an eye-opener.

John Conway's book, on the other hand, while it showcases some ideas of symmetry through the work of some artists like Bathsheba Grossman, is largely about abstraction. It is a major work, the latter part technical enough to challenge and inspire mathematicians on the forefront of their field (in his words, not mine!).

na.numerical analysis - Numerical instability using only Heun's method on a simple PDE.

I'm trying to simulate the evolution of the Wigner function (a pseudo probability distribution over phase space) for a point particle moving in a chaotic potential. I'll provide background first, as well as some ignorant speculation at the end, but I've isolated my problem to a very simple case, so you should probably just skip to the "Reduced Problem".

Background

The PDE governing the Wigner function $W(x,p,t)$ can be approximated as

$partial_t W = -frac{p}{m} partial_x W + V^prime (x) partial_p W - frac{hbar^2}{24} V^{prime prime prime} (x) partial_p^3 W$

or, in the dimensionless PDE notation,

$W_t = -p W_x + f(x) W_p - f^{prime prime}(x) W_{ppp} .$

However, the problem I'm having occurs for zero potential. In this case, the $p$-dependance of the Wigner function decouples from it's evolutions, so we can just fix $p$.

Reduced Problem

Consider merely the equation

$partial_t W = - c partial_x W$

for $W(x,t)$. For initial conditions $W(x,t=0) = g(x)$, it is solved by $W(x,t) = g(x-ct)$, which is just the original wave sliding at speed $c$.

Suppose my initial condition is a normal distribution: $g(x) propto e^{-x^2}$. If I were to simulate this with just Euler's method,

$W (x_i,t_{j+1}) = W (x_i,t_j)+ Delta W (x_i,t_{j+1}), $

$Delta W (x_i,t_j+1) = -c [W(x_{i+1})-W(x_{i-1})]/(2 Delta x),$

I would quickly run into a serious instability: On the leading and trailing edges of the traveling Gaussian (where the function is concave up) Euler's method will under-estimate the time derivative, yielding overly steep leading and trailing edges. Here's an animated gif of what it looks like. Eventually, the trailing edge dips below zero and all hell breaks loose.

Furthermore, this instability scales poorly: $t_{breakdown} propto Delta t$, so if I want to double the length of the simulation it requires 4 times as many total time steps.

So, just use Heun's method or the 4th order Runge-Kutta method, right? Well, here's the thing. Switching to Heun's method does significantly improves the simulation; $t_{breakdown}$ increases by about a factor of 10. But, once the method is fixed (Heun, RK4, whatever), the scaling problem remains. This puts a pretty hard ceiling on the length of my simulation.

Now, when I use Heun's method, the breakdown is slightly different. Basically, the wave dips below zero on the distantly trailing tails of the Gaussian where the function is very small (like $10^{-40}$). But as soon as it does dip below zero, this error grows exponentially and swamps the simulation. Here is an animated gif zoomed in on the trailing edge on a logarithmic scale$^dagger$.

Question

Is there a way to fix this instability, or am I doomed to the same time scaling? Do I need to just keep increase the order of my Runge-Kutta method (RK5, RK6, etc.) in order to simulate for longer times?

(By the way, RK4 does not seem to improve much over Heun's method and take significantly longer to run.)

Ignorant Speculation

(Don't read...)

I think this has something to do with the nature of Gaussians. Euler's method breaks down because it calculates using the derivative at the beginning of the interval (which basically assumes the derivative is constant along the interval) so it performs badly for appreciable 2nd derivatives. Heun's method is better because it averages derivative at the beginning and end of the interval; it is vulnerable to strong third derivatives. I figure 4rth order Runge-Kutta is vulnerable to strong forth derivatives, although it's probably more complicated than that.

The problem with the Gaussian is that all derivatives (when normalized to the magnitude of the function) become large away from the center, so the evolution of the tails always breaks:

$g = e^{-x^2}$

$g^prime/g = -2x$

$g^{prime prime}/g = 4x^2-2$

$g^{prime prime prime}/g = -8x^3+12x$

But then again, it's not like Gaussians are unusal functions; if this were a problem, people would know about it and have a solution.

Also, the extreme tails of the Gaussian really shouldn't break the simulations. You think that if there were problems with derivatives for such tiny values then numerical errors introduced by machine precision would be worse.

---

$^dagger$Actually, it's a crudely modified logarithmic scale which cuts out the range $[-10^{-50},10^{-50}]$ in order to displays negative values. I am about 50% sure that the apparent discontinuity of the "jump" as it falls below zero is due to crudeness of the scale. Here is the absolute value on a normal log scale. There does seem to be a weird "bounce" when it goes below zero.

---

Edit

Seeing as that I am getting errors to develop on the far tails of the Gaussian, I don't think adaptively adjusting will help. It doesn't seem like a special case/situation is causing the problem.

Also, I can't really check easily for trouble spots because I need to prevent this error even when I'm not in such a simple situation. Plus, the Wigner function is supposed to go negative occasionally, just not when it starts out positive and the equations of motion are so simple.

Edit2

I don't think I can use an implicit method because, for non-trivial p-dependence, solving the implicit equation

$W(x_i,p_j,t_{k+1}) - W(x_i,p_j,t_k) -Delta t f[W(x_1,p_j,t_{k+1})]=0$

for $W(x_i,p_j,t_{k+1})$ isn't computationally feasible because $W(x_i,p_j,t=t_0)$ is a huge 2 dimensional matrix. The Scholarpedia article by Hamdi et al (suggested by Jitse Niesen below) supports this.

Associated graded and flatness

Let me suppose, as in your examples, that we have a base field $k$.

It is well known that to check that a right $A$-module $M$ is flat it is enough to show that whenever $Ileq_ell A$ is a left ideal, the map $Motimes_AIto Motimes_A A$ induced by the inclusion $Ito A$ is injective. This condition can be rewritten: $M$ is flat iff for each left ideal $Ileq_ell A$ we have $mathrm{Tor}^A_1(M,A/I)=0$.

So now suppose $A$ and $M$ are (exhaustively, separatedly, increasingly from zero) filtered in such a way that $mathrm{gr}M$ is a flat $mathrm{gr}A$-module.

Pick a left ideal $Ileq_ell A$; notice that the filtration on $A$ induces a filtration on the quotient $A/I$. We can compute $mathrm{Tor}^A_bullet(M,A/I)$ as the homology of the homologically graded complex $$cdotsto Motimes_kA^{otimes_kp}otimes_kA/Ito Motimes_kA^{otimes_k(p-1)}otimes_kA/Itocdots$$ with certain differentials whose formula does not fit in this margin, coming from the bar resolution. Now the filtrations on $M$, on $A$ and on $A/I$ all collaborate to provide a filtration of our complex. We've gotten ourselves a positively homologicaly graded with a canonically bounded below, increasing, exhaustive and separated filtration. The corresponding spectral sequence then converges, and its limit is $mathrm{Tor}^A_bullet(M,A/I)$. Its $E^0$ term is the complex
$$cdotstomathrm{gr}Motimes_kmathrm{gr}A^{otimes_kp}otimes_kmathrm{gr}(A/I)to mathrm{gr}Motimes_kmathrm{gr}A^{otimes_k(p-1)}otimes_kmathrm{gr}(A/I)tocdots$$
with, again, the bar differential, and its homology, which is the $E^1$ page of the spectral sequence, is then precisely $mathrm{Tor}^{mathrm{gr}A}_bullet(mathrm{gr}M,mathrm{gr}(A/I))$. Since we are assuming that $mathrm{gr}M$ is $mathrm{gr}A$-flat, this last $mathrm{Tor}$ vanishes in positive degrees, so the limit of the spectral sequence also vanishes in positive degrees. In particular, $mathrm{Tor}^A_1(M,A/I)=0$.

NB: As Victor observed above in a comment, Bjork's Rings of differential operators proves in its Proposition 3.12 that $mathrm{w.dim}_AMleqmathrm{w.dim}_{mathrm{gr}A}mathrm{gr}M$ (here $mathrm{w.dim}$ is the flat dimension) from which it follows at once that $M$ is flat as soon as $mathrm{gr}M$ is; the argument given is essentialy the same one as mine. I am very suprised about not having found this result in McConnell and Robson's!

at.algebraic topology - Algebras over the little disks operad

This was supposed to be a comment on Neil's answer but it was too long.

Here is a more general class of tree that has a monoid structure. Let $X$ be a tree. Consider the tree obtained by considering only internal vertices (vertices with valence greater than 1) and the edges between them. Assume that this subtree has no vertices with valence greater than 2. Then there is a monoid structure on $X$.

Specifically, this property means that there is a sequence of vertices $v_1,ldots, v_n$ so that every vertex is either one of the $v_i$ or has valence one and shares its unique edge with one of the $v_i$. Further, $v_i$ shares an edge with $v_{i+1}$. For $i lt n$, let $r_i$ be one less than the valence of $v_i$. For $i=n$, let $r_n$ be the valence of $v_n$. This number is at least $1$. Let $A_i$ be Neil's monoid ${zin mathbb{C}: z^{r_i}in [0,1]}$. Consider the monoid $A_1timescdots times A_n$, and consider the subsets $M_i={0}timescdotstimes{0}times A_itimes{1}times cdots times{1}$. $M_1$ contains the unit of the product; $M_i$ is closed under the monoidal product, and if $ilt j$, then $M_itimes M_jsubset M_j$.

Now let $M$ be the union of the $M_i$. topologically, $M_i$ is $A_i$, that is, a corolla with $r_i$ edges. The corolla $M_i$ is glued to $M_{i+1}$ along the point $(underbrace{0,ldots, 0}_i, 1,ldots, 1)$, which is the central vertex of $M_i$ and an extremal vertex of $M_{i+1}$. This yields $X$.

Note that not every boundary point of $X$ arises as the unit of a monoid structure under this construction.

Edit:

My answer to mathoverflow.net/questions/91327 shows that the universal cover of the theta graph (and many similar trees) cannot carry a monoid structure.

pr.probability - Surfaces that are 'everywhere accessible' to a randomly positioned Newtonian particle with an arbitrary velocity vector

Consider an idealized classical particle confined to a two-dimensional surface that is frictionless. The particle's initial position on the surface is randomly selected, a nonzero velocity vector is randomly assigned to it, and the direction of the particle's movement changes only at the surface boundaries where perfectly elastic collisions occur (i.e. there is no information loss over time).

My question is - Does there exist such a bounded surface where the probability of the particle visiting any given position at some time 't', P(x,y,t), becomes equal to unity at infinite time? In other words, no matter where we initialize the particle, and no matter the velocity vector assigned to it, are there surfaces that will always be 'everywhere accessible'?

(Once again, I welcome any help asking this question in a more appropriate manner...)

reference request - Vector-valued valuations on lattices

There's a fair amount of work on valuations on (modular) lattices, by which I mean functions $v : mathcal{L} rightarrow R$ that satisfy the modular expression
$$v(x) + v(y) = v(x wedge y) + v(x vee y) $$

I'm wondering if there's been any work on vector-valued valuations (where the range of v is $R^k$ and the same relation holds) ?

In addition, I'm also interested in lower valuations (I'm not sure if this name is standard) that satisfy the submodular inequality
$$v(x) + v(y) ge v(x wedge y) + v(x vee y) $$
and possibly the generalization to $R^k$ where we replace the above by
$$v(x) + v(y) succeq v(x wedge y) + v(x vee y) $$
($succeq$ being the coordinate-wise partial order)

This is a reference request, for the most part.

singularity theory - Is the desingularization of a normal variety with only quotient singularities projective

Here is an explicit example that this does not work for arbitrary resolutions. The assumptions below are only needed to get the flat morphism over the curve and they are not really necessary but make life a little easier. I think the example Karl is referring to is Hartshorne, GTM 52, Appendix B, Example 3.4.1. Anyway, here we go.

Let $Y=Ctimes S$ where $C$ is a smooth projective curve and $S$ is a smooth projective surface. The projection $pi:Yto C$ is not only flat, but smooth and projective. Assume further that there exists a morphism $alpha: Cto S$ such that it is an embedding everywhere except at two points, whose images are the same point $Pin S$. Consider the following curves: $C_1=Ctimes {P}$,$C_2={(x,alpha(x)mid xin C }$. By the previous assumptions, $C_1$ and $C_2$ meet in exactly $2$ points. Now perform Hironaka's trick to produce a non-projective smooth $3$-fold: blow-up $C_1$ and $C_2$ in the two possible orders, but in different ways near the two intersection points (need to do it locally and then glue). See Hartshorne, GTM 52, Appendix B, Example 3.4.1 for details on this construction. Let $sigma: Xto Y$ be the $3$-fold obtained this way. It is easy to see that this is not projective using intersection numbers. See Hartshorne for the computation.

Perhaps the more interesting point about this example is that it shows that a morphism being projective is not local on the target. (Like being proper is, [Hartshorne, II.4.8(f)]).

Obviously $picircsigma$ is projective locally on $C$, since it is a combination of the original projective morphism $pi$ and two (projective) blow-ups, but $picircsigma$ cannot be projective, since then so would be $X$ (over the base field, since $C$ is).

Finally, to answer your question in the comments, whether there is "a" resolution that
's projective, the answer is certainly "yes". If you pick one that's a sequence of blow-ups, then the morphism is projective and hence so is $X$. In fact, using Chow's Lemma, you can "correct" any resolution that may accidentally be non-projective: by Chow's Lemma, you can always dominate birationally your non-projective (but proper) variety by a projective one and then applying a projective resolution of that will save the day.

dg.differential geometry - A local transitivity property of the automorphism group of a foliated manifold

Let $(M,mathcal F)$ be a smooth foliated manifold. An automorphism of $(M,mathcal F)$ is a diffeomorphism of $M$ that takes leaves of $mathcal F$ onto leaves. Let now $L$ be a leaf of $mathcal F$. It may happen that $L$ has an open neighborhood $U$ which is a sum of leaves and such that for every leaf $L'subset U$ there exists an automorphism $phi$ taking $L$ onto $L'$.

My questions are:

Does every $(M,mathcal F)$ have a leaf $L$ with the described property?

If the answer is no, how much can we hope for instead? And is there a simple counterexample? Are there some natural classes of foliations which still have this property?

Context:
Ideally, I would like to consider a smooth foliated manifold $(M,mathcal F)$ such that in some flat chart $psicolon Uto mathbb R^n$ there exists an open set $Vsubset U$ and a family $tau_x$ of automorphisms of $(M,mathcal F)$ indexed by a neighborhood of 0 in $mathbb R^n$, where $tau_x$ acts on $V$ like a translation by $x$ in the chart $psi$. This condition is satisfied for instance when there is a neighborhood $U$ of a leaf $L$ and a foliation-preserving diffeomorphism $Uto Ltimes W$, where $Ltimes W$ has the corresponding product foliation.

Structure Theorem for finitely generated commutative cancellative monoids?

Is there a Structure Theorem for finitely generated commutative cancellative monoids?

Of course they can be densely embedded into a finitely generated abelian group, whose structure is known. Also, in the book of J. C. Rosales and P. A. García-Sánchez there are some special embedding theorems: If the monoid is torsionfree, it even embeds to some free abelian group, and if the monoid is also reduced, it embeds in some free commutative monoid.

But I want to know if it is possible to give a complete classification (for example, in terms of generators and relations, as in the case of groups).

ct.category theory - What is the underlying graphical calculus of the Interactions-Round-a-Face lattice model?

Background

Let $mathcal{L}$ be an $m times n$ square lattice on a torus, and let $Sigma$ be a finite set. We think of $Sigma$ as the possible spin values that can be assigned to the points of the lattice; a state will therefore be a map $mathcal{L} to Sigma$. In the Interactions-Round-a-Face (IRF) model, there is a weight function $W: Sigma^4 to mathbb{R}$, and the total Boltzmann weight of a state $s: mathcal{L} to Sigma$ will be $overline{W}(s) = prod_{text{faces } (i, j, k, l)} W(s_i, s_j, s_k, s_l)$. (The vertices of a face are labeled starting from the upper left and going clockwise, say.) The partition function is defined to be $sum_{text{states } s} overline{W}(s)$. This model has many exactly solvable 2-dimensional lattice models as special cases.

The standard analysis of this partition function begins by considering the row transfer matrix, defined as follows. Given a pair of adjacent rows and spin vectors $phi = (s_1, ldots, s_n)$, $phi' = (s_1', ldots, s_n')$, we define $T_{phi, phi'} = prod_{i = 1}^n W(s_i, s_{i + 1}, s_{i + 1}', s_i')$. We think of $T$ as defining a $lvert Sigma rvert^n times lvert Sigma rvert^n$ matrix. One then observes that the partition function is simply the trace of $T^m$.

More generally, one could take each face to have a different weight function $W$; we would then get a different transfer matrix for each row, and the partition function would be the trace of the product of all these transfer matrices.

Question

Computations in the IRF model can be viewed as a sort of graphical calculus, where we think of the row transfer matrix as a map from $V = mathbb{C}^{Sigma^n}$ to itself, where the input is given by the assignment of spins to the top row, and the output is given by the assignment of spins to the bottom row. More explicitly, $T$ should be thought of as a multilinear form on $V otimes V$, which we then convert into a map $V to V$ using the given basis to identify $V$ with $V^ast$. Stacking rows corresponds to composing $T$ with itself, and identifying the $(m + 1)$-st row with the first row corresponds to taking the trace. This interpretation also works when $T$ varies from row to row.

What's not clear to me is how to extend this graphical calculus to horizontal composition. One would like to start not from the row transfer matrix $T$, whose definition seems somewhat artificial, but the face weight $W$, which can be thought of as a 4-linear form on $V' = mathbb{C}^{Sigma}$, or as a map $W: V'^{otimes 2} to V'^{otimes 2}$ by identifying the "bottom" two copies of $V'$ with $V'^{ast}$ as above. Vertical composition is again given by powering $W$, but the horizontal composition is strange. It sends the $(2, 2)$-tensors $W_{s_1's_2'}^{s_1s_2}$, $W_{s_2's_3'}^{s_2s_3}$ to the "$(3, 3)$-tensor" $X_{s_1's_2's_3'}^{s_1s_2s_3} = W_{s_1's_2'}^{s_1s_2} W_{s_2's_3'}^{s_2s_3}$, which doesn't seem to be a "categorical" construction. The problem is that we need to use the middle input (and output) twice, so the resulting map will be quadratic, rather than linear, in these variables.

What is the correct categorical framework for the IRF model?

dg.differential geometry - Equivalent definitions of Gaussian curvature

It seems to me that the most difficult step is learning and understanding the intrinsic definition of Gauss curvature. The wikipedia page, http://en.wikipedia.org/wiki/Gaussian_curvature, provides a few different versions.

My favorite way to describe the extrinsic definition is the following: To calculate the Gauss curvature at a point $p$ on the surface: Move the surface so that $p$ is at the origin. Rotate the surface so that the $xy$-plane is tangent to the surface at the origin. The surface near the origin is now given as a graph $z = f(x,y)$, where the partials $ partial_x f $ and $partial_y f$ vanish at the origin. The second fundamental form of the surface at $p$ is given by the Hessian $partial^2 f(0)$, and the Gauss curvature at $p$ is the determinant of the Hessian.

Proving the theorem egregium is now pretty straightforward, especially if you remember that you don't need exact formulas for anything but only second order approximations at the origin.

Why is continuity so crucial in arithmetic/algebraic geometry?

I'm interested in the arithmetic/diophantine equation applications of arithmetic/algebraic geometry. From what I understand, many of the difficult/technical aspects of the latter theories (sheaves, cohomologies, schemes, ...) are due to the desire to access continuity. The benefits of this continuity must be great due to the substantial difficulties in setting it up, but what are they? For example, I have read that Grothendieck's cohomology was crucial in the solution of the Riemann hypothesis for varieties over finite fields, but why is continuity so important for this result?

ag.algebraic geometry - Stacks in the Zariski topology?

1.) It's possible to define stacks on ANY category equipped with a Grothendieck topology (such a category with a topology is called a site). In particular, this holds true for the Zariski site. Moreover, there is always a way to define an "Artin stack"- these are those stacks which arise as torsors for a groupoid object in your site. Outside of algebraic geometry, these give rise to notions of topological and differentiable stacks, for instance.

EDIT: As long as groupoid objects exist in your category.

2.) As in Harry's post, any stack which is a stack in a site which is finer than the Zariski topology is also a stack in the Zariski topology.

To address your general question as to "why a new notion of open cover is necessary if all I am interested in is remembering stabilizers", you should learn a bit about Grothendieck topologies. I'll make a couple remarks:

i) If all you cared about were stabilizers, then you wouldn't need to use any covers at all; ordinary fibered categories would do the trick!

Indeed, take a group object in your site acting an object, and take the action groupoid- it is a groupoid object. Look at the pseudo-functor which assigns each object of your site the groupoid of maps into this groupoid object (considering the object as a groupoid with all identity arrows). This remembers the stabilizers for this action.

ii) (subcanonical) Grothendieck topologies are a choice of a type of cover for your objects, in such a way that this object is the colimit of these covers, AND "this is important to remember". This is a little imprecise, so, allow me to elaborate via an example from topology:

Let $U_i$, $iin I$ be an open cover of a space X. Then, continuous maps from X to another space Y are in bijection with with continuous maps $f_i:U_i to Y$ which agree on their intersection. This is just saying that X is the colimit of this open cover. Instead, we can view this a property of the presheaf $Hom(blank,Y)$ represented by $Y$ on the category of topological spaces (for you set theorists, choose a Grothendieck universe).

For any $X$ and any open cover $U_i$, $iin I$ of $X$, (let $Hom(blank,Y)=F$)

the natural map $F(X) to varprojlim left[{prod{F(U_i)}} rightrightarrows {prod{F(U_{ij})}}right]$

is a bijection.

If $F$ is any presheaf, this is just saying $F$ is a sheaf. Since this is NOT true for an ARBITRARY presheaf $F$, X is no longer the colimit of its open covers in the full category of all presheaves. The same argument holds for all fibred categories- it's only true if we restrict to STACKS (and $X$ then becomes the weak colimit of this cover, but, never mind).

The reason you add the condition for descent for covers, is so that "all maps into your stack from a space are continuous". More precisely, and more generally, it's so that maps from a space, scheme, whatever you site is, into a stack can be determined by mapping out of elements of some covering of your object in a way that glues (for stacks, rather than sheaves, they don't need to AGREE on the intersection, but, agree up to an invertible 2-cell, plus some coherency conditions).

Combining these ideas, if you have a group acting on an object, the pseudo-functor produced by the action groupoid is rarely a stack with respect to your topology, but you can stackify it, and then it will become one and still remember all the stabilizers. I hope this helps!

at.algebraic topology - Splitting of the Universal Coefficients sequence

I would claim that the splitting (and indeed the whole universal coefficient
theorem) is not really a topological theorem. If we take the homological version
one really works with the chain complex $C_ast(X)$ in the derived category of
$mathbb Z$-complexes. We then have $C_ast(X,M)=C_ast(X)bigotimes M$ but as
$C_ast(X)$ is free this equals the derived tensor product
$C_*(X)bigotimes^{mathbb L} M$ and hence is a formula in the derived
category. One can then use the fact that in the derived category of $mathbb
Z$-modules every complex is isomorphic to the sum of its (shifted) homology:
$Ccong bigoplus_nH_n(C)[n]$ so that
$$
C_*(X)bigotimes^{mathbb L} M cong bigoplus_n(H_n(X)bigotimes^{mathbb L} M)[n]
$$
and as $Abigotimes^{mathbb L} Mcong Abigotimes Mbigoplus
mathrm{Tor}^1(A,M)[1]$ we get the universal coefficient formula including the
splitting.

This idea also demonstrates why the splitting is not canonical. We may for
instance consider a group $G$ acting on $X$. We then get at complex $C_*(X)$ in
the derived category of $G$-modules and a complex in that category is in general
not isomorphic to the sum of its homology.

On the other hand, this technique can be used (essentially) each time some
invariant of a topological space $X$ only depends on its chain complex in a way
that takes quasi-isomorphisms to isomorphisms. The conclusion is that it only
depends on the homology of $X$. A nice example is the homology of the $n$'th
symmetric product of $X$. It turns out to be the homology of a complex
constructed functorially from $C_*(X)$ and exactly in a way that preserves
quasi-isomorphisms. Hence it only depends on the homology of $X$ (and one can
also give explicit formulas).

However, the method that you declare a fondness for is also useful if one goes
beyond homology. It can be used to give a universal coefficient spectral
sequence (due to Adams I think) for the (co)homology with coefficients in module
spectrum over a ring spectrum. In general this spectral sequence does not
degenerate to short exact sequences so the problem of splitting is not even
(that) relevant. However, for for instance $K$-theory it does but I imagine
(though I don't know but others certainly do) that even there one can find
examples of non-splitting.

dg.differential geometry - Drawing of the eight Thurston geometries?

Here is a nice cyclic ordering of the eight geometries:

HxR, SxR, E^3, Sol, Nil, S^3, PSL, H^3

derived from staring at Peter Scott's table of Seifert fibered geometries. The table is organized by Euler characteristic of the base 2-orbifold and Euler class of the bundle. (See his BAMS article.) The cyclic ordering also has a bit of antipodal symmetry.

I didn't come up with geometric pictures of the eight geometrics but I have thought about "icons" to represent them. Here are my suggestions - I'm interested to hear what other people think/suggest.

• HxR -- triangular prism (where the triangle is slim ie ideal)


• SxR -- cylinder


• E^3 -- cube


• Sol -- tetrahedron with one pair of opposite edges truncated


• Nil -- annulus with a segment of a spiral (representing a Dehn twist)


• S^3 -- circle


• PSL -- trefoil knot


• H^3 -- figure eight knot (or possibly a slim tetrahedron)

I think it is also reasonable to ask for a "prototypical" three-manifold for each of the eight geometries. Here is an attempt:

• HxR -- punctured torus cross circle


• SxR -- two-sphere cross circle


• E^3 -- three-torus


• Sol -- mapping cylinder of [[2,1],[1,1]]


• Nil -- mapping cylinder of [[1,1],[0,1]]


• S^3 -- three-sphere


• PSL -- trefoil complement


• H^3 -- figure eight complement

Notice that all of the examples are either surface bundles over circles or circle bundles over surfaces, or both (ie products).

knot theory - A $k$-component link defines a map $T^{k}rightarrow mathrm{Conf}_{k} S^{3}$. Does the homotopy type capture Milnor's invariants?

It's been a while since I've thought about this but I think Koschorke answered much of your question back in 1997 "A generalization of Milnor's mu-invariants to higher-dimensional link maps" Topology 36 (1997), no 2. 301--324. Scanning through the paper I see he recovers many of the mu invariants but not all. He lists it as an open question (6.3) if the homotopy class of the map T^k --> C_k R^3 is a complete link homotopy invariant of the link.

Related:

Brian Munson put these Koschorke "linking maps" into the context of the Goodwillie calculus in a recent arXiv paper. I've wondered for a while if you could use these types of maps to create a direct construction of the Cohen-Wu correspondence between the homotopy groups of S^2 and their corresponding simplicial quotient object made from the brunnian braid groups.

dg.differential geometry - Cotangent bundle of a differentiable stack

If you ever wanted to construct the tangent bundle of a differentiable stack, it's relatively simple:

First, if $mathbf{X}$ is a stack coming from a Lie groupoid $mathcal{G}$, you could just say $mathbf{TX}$ is the stack associated to the tangent groupoid $mathcal{TG}$.

A more formal way is the following:

Consider the composite

$text{Mfd} stackrel{T}{rightarrow} text{VectorBundles} stackrel{text{forget}}{rightarrow} text{Mfd} stackrel{text{yoneda}}{rightarrow} text{St}(text{Mfd}).$

by restricting its weak left Kan extension to stacks we get a 2-functor (by abuse of notation)

$text{St}(text{Mfd}) stackrel{T}{rightarrow} text{St}(text{Mfd}).$

Now, suppose $mathbf{X}$ is a stack coming from a Lie groupoid $mathcal{G}$. Then $mathbf{X}$ is the weak colimit of the composition

$Delta^{rm op} stackrel{N(mathcal{G})}{rightarrow}Mfdstackrel{text{yoneda}}{rightarrow}text{St}(text{Mfd})$

(in fact even of its 2-truncation.)

Since $T$ is weak-colimit preserving, it follows that $T mathbf{X}$ is the weak colimit of

$Delta^{rm op} stackrel{N(mathcal{TG})}{rightarrow}text{Mfd}stackrel{text{yoneda}}{rightarrow}text{St}(text{Mfd})$

which is in turn just the stack associated to the tangent groupoid $mathcal{TG}$. So both definitions agree.

Now suppose you wanted to define the cotangent stack. The first line of attack, done naively, seems to fail. The cotangent functor $T^*$ is contravariant so it doesn't send groupoid objects to groupoid objects. However, less naively, in the literature (for instance here: http://arxiv.org/PS_cache/arxiv/pdf/0905/0905.4318v2.pdf), one can define the cotangent groupoid of $mathcal{G}$ to be a certain symplectic groupoid $mathcal{T^{*}G}_1 rightrightarrows text{Lie}(mathcal{G})^{*}$, where $text{Lie}(mathcal{G})^{*}$ is the dual Lie algebroid of $mathcal{G}$. (Is this invariant under Morita equivalence?)

The second way would seem to be:

Consider the opposite of the composite

$text{Mfd}^{rm op} stackrel{T^*}{rightarrow} text{VectorBundles} stackrel{text{forget}}{rightarrow} text{Mfd} stackrel{text{yoneda}}{rightarrow} text{St}(text{Mfd}),$

$text{Mfd} to text{St}(text{Mfd})^{rm op}$

and take its weak left Kan extension. (So $mathbf{T^{*}X}$ is the weak limit of $T^{*} M$ over all $M to mathbf{X}$.)

The opposite of this 2-functor goes $text{St}(text{Mfd})^{rm op} to text{St}(text{Mfd})$.

How do these two notions of cotangent stack relate to each other? Is either reasonable? Is there a better notion than either of these?

I would like, for instance, if you have a Riemannian metric on an orbifold, to get an equivalence between its tangent stack and its cotangent stack.

Along these lines, the underlying space of the cotangent bundle of an orbifold is an orbifold (or is it a manifold, like its frame bundle?), so it should be represented by a orbifold groupoid. Is this the same as the cotangent groupoid?

Keep in mind, the correct definition of tangent stack should produce the correct notion of differentiable forms on a stack, and the cotangent stack should somehow have a symplectic structure.

st.statistics - "Main" diagonal of a matrix

One rigorous idea which is similar to your question is to look for the largest "transversal" of the matrix. A transversal is a collection of entries of the matrix with one entry from each row and column. The terms in the determinant of the matrix are multiplicative transversals, but for this question, you would be interested in the maximum additive transversal. There are polynomial time algorithms to find the largest transversal. You can view it as the optimum marriage problem, or as a min-cut, max-flow problem, but it ultimately comes from the fact that you are maximizing a linear functional (the value of the transversal) on the Birkhoff polytope, which is the convex hull of the permutation matrices. As such, it is a linear programming problem. The remaining question is how best to speed it up further, using the fact that it is a special case rather than general linear programming. There is a noted paper on this subject, On Algorithms for Obtaining a Maximum Transversal, by Duff.

Note that the question isn't an either-or between the diagonal and the anti-diagonal. There are many things that a transversal can do other than be close to the diagonal or anti-diagonal. Even so, you can look at how close the transversal is to one or the other extreme. For instance, you can look at its inversion number. The diagonal has the smallest inversion number, 0, while the antidiagonal has the unique largest inversion number, $n(n-1)/2$.

A slightly different idea is to look at correlations between the inversion number and the value of the transversal. I think some types of correlation can also be computed in polynomial time.

ct.category theory - Does the category of pro-sets have a generator?

Consider for simplicity the category of pointed sets $Set_*$. Let's call an object $G$ of a pointed category $C$ a generator if for every nonzero object $X in C$ there is a nonzero map $G to X$. In the category of pointed sets, for example, the two-point set is a generator (and a cogenerator as well).

Does the category $Pro-Set_*$ of cofiltered diagrams of pointed sets have a generator as well? If $X$ is a pro-pointed set with nontrivial limit then there is a nontrivial map $G to X$, where $G$ is a generator of pointed sets, considered as a one-object diagram. However, there are many pro-sets with trivial limit. For example, the pro-set $Xcolon mathbb N to Set$ given by $X(i) = mathbb N$ and $X(i+1 to i)(n) = n+1$ is a pro-set with empty limit, but which is nontrivial -- add a point in every degree to get an example in pointed sets.

Thus, my question is: Does the category of pro-pointed sets have a generator?

And, if the answer was yes, does more generally $Pro-C$ have a generator whenever $C$ has one?

nt.number theory - (Good) effective version of Kronecker's theorem?

Thm (Kronecker).- If all conjugates of an algebraic integer lie on the unit circle, then the integer is a root of unity.

Question: Can one provide a good effective version of this? That is: given that we have an algebraic integer alpha of degree <=d, can we show that alpha has a conjugate that is at least epsilon away from the unit circle, where epsilon depends only on d? It actually isn't hard to do this (from the standard proof of Kronecker, viz.: alpha, alpha^2, alpha^3... are all algebraic integers, and their minimal polynomials would eventually repeat (being bounded) if all conjugates of alpha lied on the unit circle) with
epsilon exponential on d, i.e., epsilon of the form epsilon = 1/C^d; what we actually want is an epsilon of the form 1/d^C, say.

(Question really due to B. Bukh.)

ag.algebraic geometry - Modules, Sheaves and Vector bundles

There are (at least) two details you are missing.

(1) This is not an equivalence of categories between finitely generated graded modules and coherent sheaves. If your module is $0$ in all sufficiently large degrees, then the corresponding sheaf will be zero. For example, let $S=k[x,y]$ and $M=S/langle x,y rangle$. The sheaf on $mathbb{P}^1$ corresponding to $M$ is the zero sheaf.

The category you want to work with is the one whose objects are finitely generated graded modules, and where we formally invert any map which is an isomorphism in sufficiently large degree. (Alternative formulation: we formally invert a map $f:M to N$ if, for any $s$ in the irrelevant ideal, $s^{-1} f: s^{-1} M to s^{-1} N$ is an isomorphism.)

(2) Not every coherent sheaf is a vector bundle. Correspondingly, not every finitely generated graded module will correspond to a vector bundle. If we were dealing with an affine variety, vector bundles would correspond to locally free modules. (Also called projective modules.) For projective varieties, things are a little trickier: the criterion is to be locally free away from the irrelevant ideal.

Sadly, I believe that there exist examples of modules which are locally free away from the irrelevant ideal, but are not isomorphic (in the above category) to any module which is locally free at the irrelevant ideal. This should be related to my question here. But you can go a long while without paying attention to this detail.

fa.functional analysis - In a Banach algebra, do ab and ba have almost the same exponential spectrum?

Let $A$ be a complex Banach algebra with identity 1. Define the exponential spectrum $e(x)$ of an element $xin A$ by $$e(x)= {lambdainmathbb{C}: x-lambda1 notin G_1(A)},$$ where $G_1(A)$ is the connected component of the group of invertibles $G(A)$ that contains the identity.

Is it true that $e(ab)cup{0} = e(ba)cup{0}$ for all $a,b in A$?

Equivalently, is it true that
$1-ab$ is in $G_1(A)$ if and only if $1-ba$ is in $G_1(A)$, for all $a,b in A$?

Note: The usual spectrum has this property.

Just an additional note:

We have $e(ab)cup{0} = e(ba)cup{0}$ for all $a,b in A$ if

1) The group of invertibles of $A$ is connected, because then the exponential spectrum of any element is just the usual spectrum of that element.

2) The set $Z(A)G(A) = {ab: a in Z(A), bin G(A)}$ is dense in $A$, where $Z(A)$ is the center of $A$. (One can prove this). In particular, we have $e(ab)cup{0} = e(ba)cup{0}$ for all $a,b in A$ if the invertibles are dense in $A$.

3) $A$ is commutative, clearly.

But what about other Banach algebras? Can someone provide a counterexample?

gr.group theory - When is a map given by a word surjective?

Let $w(x,y)$ be a word in $x$ and $y$.

Let $x$ and $y$ now vary in $SL_n(K)$, where $K$ is a field. (Assume, if you wish, that $K$ is an algebraically complete field of characteristic bigger than a constant.)

I would like to know for which words $w$ the map

$y rightarrow w(x,y)$

isn't surjective (or even dominant - that is, "almost surjective") for $x$ generic.

It is clear, for example, that the map is surjective for $w(x,y)=xy$, and that it isn't
surjective for $w(x,y)= y x y^{-1}$, or for $w(x,y) = y x^n y^{-1}$, $n$ an integer:
all elements of the image of $y rightarrow y x^n y^{-1}$ lie in the same conjugacy class.
A moment's thought (thanks, Philipp!) shows that $w(x,y) = x y x^n y^{-1}$
isn't surjective either: its image is just $x* im(yrightarrow y x^n y^{-1})$,
and, as we just said, $yrightarrow y x^n y^{-1}$ isn't surjective.

I would like to know if the only words $w$ for which the map
isn't surjective for $x$ generic are the $w$'s of the form
$w(x,y) = x^a v(x,y) x^b (v(x,y))^{-1} x^c$,
where $v$ is some word and $a,b,c$ are some integers. (This seems to me a sensible guess, though I would actually be quite glad if it weren't true.)

abstract algebra - Extension problem

The subtlety is in when two extensions are considered the same, in other words what are the morphisms of extension? There are different reasonable answers. For example a morphism could just be a morphism of short exact sequences (i.e. chain complexes).

However the statement that $Ext^1$ classifies extensions uses a very particular kind of morphism of extension: only those morphisms of short exact sequences which are the identity on A and B.

Let's look at extensions of $mathbb{Z}/3$ by $mathbb{Z}$, where the middle guy is a $mathbb{Z}$. Suppose that the map from $f:mathbb{Z} to mathbb{Z}$ is fixed and is multiplication by 3. Notice that there is a unique map $h: mathbb{Z} to mathbb{Z}$ such that $f = hf$,
namely $h=id$. This means that for any such extension, after you've identified the map $f$ as multiplication by 3, there is no choice for morphisms of extension. A map of these extensions has to be the identity on all three terms.

In short, the extensions $mathbb{Z} to mathbb{Z} to mathbb{Z}/3$, up to isomorphism fixing the first and last group, are in bijection with homomorphisms from $mathbb{Z} to mathbb{Z}/3$ which have kernel $3 cdot mathbb{Z}$. There are exactly two such homomorphisms, given by sending 1 to 1 or 1 to 2 in $mathbb{Z}/3$. So there are two distinct non-spilt extensions.

However you can also use $Ext^1(B;A)$ to get extensions up to the weaker notion of equivalence described above. Since Ext is functorial in both variables, you have an action by the automorphism groups of both A and B. Extensions up to this weaker notion are in bijection with equivalence classes in the quotient of $Ext^1$ by these actions.

You can check in this case that the action by automorphisms of $mathbb{Z}/3$ exchanges the two non-zero elements of $Ext^1(mathbb{Z}/3, mathbb{Z})$, and so there are exactly two extensions in this weaker sense: the spilt extension and the non-split one.

career - Advice on doing mathematical research

Terry Tao has some helpful ideas.

I can't say I've really mastered the practice of choosing problems. I guess if there's an area of math that you're really familiar with, questions will just start coming to you, but I can't tell how much of that ability comes from some osmotic process from the examples of peers.

When I'm stuck, I try to write down where I am in a lot of detail, then I take a break. If I'm lucky, I know what sort of knowledge might be able to crack my problem, and if I'm even luckier, I know who has that knowledge.

ac.commutative algebra - A local ring not a quotient of a regular local ring

A source available online is this paper "Examples of bad Noetherian rings" by Mariani (example 2.1).

The reason many of these types of construction work is because of the following vague and counter-intuitive phenomemon:

It is usually easier than we think for a complete local ring to be a completion of a Noetherian ring with certain properties.

For example, there is this amazing theorem by Heitmann that most complete local ring of depth at least $2$ is a completion of a UFD !

So back to Mariani's paper, the example is as follows: start with some local Artinian ring $(Q,m)$ such that $Q$ is not Gorenstein. Then $Q[[X]]$ is complete, and one can find a local domain $R$ such that $hat R=Q[[X]]$. Now if $R$ is a quotient of a regular local ring, then the comletion of $R$ is generically a complete intersection. But $Q[[X]]$ is not even generically Gorenstein, since $Q$ is not.

dg.differential geometry - Normal intersections of submanifolds

Let $M$ be a compact manifold, and let $M_1,ldots, M_k$ (k>2) be embedded submanifolds. Suppose that $pincap_{i=1}^k M_k$ and that for any subset $S$ of ${1,ldots, k}$ and any $jnotin S$ that $cap_{iin S}M_i$ intersects $M_j$ transversally at $p$.

I believe that in this case the fact that $cap_{i=1}^k M_k$ is nonempty is stable (still true after homotoping each $M_i$ a little bit). Does anyone have a reference for this fact?

at.algebraic topology - Describing the universal covering map for the twice punctured complex plane

As is well known, the universal covering space of the punctured complex plane is the complex plane itself, and the cover is given by the exponential map.

In a sense, this shows that the logarithm has the worst monodromy possible, given that it has only one singularity in the complex plane. Hence we can easily visualise the covering map as given by the Riemann surface corresponding to log (given by analytic continuation, say).

Seeing how fundamental the exponential and logarithm are, I was wondering how come I don't know of anything about the case when two points are removed from the complex plane.

My main question is as follows: how can I find a function whose monodromy corresponds to the universal cover of the twice punctured complex plane (say ℂ∖{0,1}), in the same way as the monodromy of log corresponds to the universal cover of the punctured plane.

For example, one might want to try f(*z*) = log(*z*) + log(*z*-1) but the corresponding Riemann surface is easily seen to have an abelian group of deck transformations, when it should be F₂.

The most help so far has been looking about the Riemann-Hilbert problem; it is possible to write down a linear ordinary differential equation of order 2 that has the required monodromy group.
Only trouble is that this does not show how to explicitly do it: I started with a faithful representation of the fundamental group (of the twice punctured complex plane) in GL(2,ℂ) (in fact corresponding matrices in SL(2,ℤ) are easy to produce), but the calculations quickly got out of hand.

My number one hope would be something involving the hypergeometric function ₂F₁ seeing as this solves in general second order linear differential equations with 3 regular singular points (for ₂F₁ the singular points are 0, 1 and ∞, but we can move this with Möbius transformations), but I was really hoping for something much more explicit, especially seeing as a lot of parameters seem to not produce the correct monodromy. Especially knowing that even though the differential equation has the correct monodromy, the solutions might not.

I'd be happy to hear about any information anyone has relating to analytical descriptions of this universal cover, I was quite surprised to see how little there is written about it.
Bonus points for anything that also works for more points removed, but seeing how complicated this seems to be for only two removed points, I'm not hoping much (knowing that starting with 3 singular points (+∞), many complicated phenomena appear).

abstract algebra - Canonical examples of algebraic structures

I often think of "universal examples". This is useful because then you can actually prove something in the general case - at least theoretically - just by looking at these examples.

Semigroup: $mathbb{N}$ with $+$ or $*$

Group: Automorphism groups of sets ($Sym(n)$) or of polyhedra (e.g. $D(n)$).

Virtual cyclic group: Semidirect products $mathbb{Z} rtimes mathbb{Z}/n$.

Abelian group: $mathbb{Z}^n$

Non-finitely generated group: $mathbb{Q}$

Divisible group: $mathbb{Q}/mathbb{Z}$

Ring: $mathbb{Z}[x_1,...,x_n]$

Graded ring: Singular cohomology of a space.

Ring without unit: $2mathbb{Z}$, $C_0(mathbb{N})$

Non-commutative ring: Endomorphisms of abelian groups, such as $M_n(mathbb{Z})$.

Non-noetherian ring: $mathbb{Z}[x_1,x_2,...]$.

Ring with zero divisors: $mathbb{Z}[x]/x^2$

Principal ideal domain which is not euclidean: $mathbb{Z}[(1+sqrt{-19})/2]$

Finite ring: $mathbb{F}_2^n$.

Local ring: Fields, and the $p$-adics $mathbb{Z}_p$

Non-smooth $k$-algebra: $k[x,y]/(x^2-y^3)$

Field: $mathbb{Q}, mathbb{F}_p$

Field extension: $mathbb{Q}(i) / mathbb{Q}, k(t)/k$

Module: sections of a vector bundle. Free <=> trivial. Point <=> vector space.

Flat / non-flat module: $mathbb{Q}$ and $mathbb{Z}/2$ over $mathbb{Z}$

Locally free, but not free module: $(2,1+sqrt{-5})$ over $mathbb{Z}[sqrt{-5}]$

... perhaps I should stop here, this is an infinite list.

Complexity of determining if two graphs have same cycle matroid?

I believe that they have the same complexity, but writing up the details has proved painful and it is possible I am missing something. Let me give the first part of my answer, and see whether you actually want more.

A graph is called 3-connected if it cannot be disconnected by removing two vertices. If G and H are 3-connected then they are isomorphic if and only if their matroids are isomorphic. (Follows immediately from Whitney's theorem, as there are no Whitney moves on 3-connected graphs.) So this subproblem of graph isomorphism and this subproblem of graphical matroid isomorphism are equivalent.

Now, it is easy to reduce graph isomorphism to 3-connected graph isomorphism. If G is any graph, let J_3(G) be the graph formed from G by adding 3 more vertices which are joined to each other and to every vertex of G. Then J_3(G) is isomorphic to J_3(H) if and only if G is isomorphic to H.

So [Graph isomorphism], [3-connected graph isomorphism] and [3-connected graphical matroid isomorphism] are equivalent, and [graphical matroid isomorphism] is at least as hard as these.

The part which is truly painful to write up is the reduction of graphical matroid isomorphism to 3-connected graphical matroid isomorpism. (And maybe I am missing something.) Here's how it should go. We can immediately reduce to the case of 2-connected graphs, because the matroid of a graph is determined by the matroids of its 2-connected components.

The next step is to break the graph into its representation as a 2-sum of 3-connected components. This is a less standard notion, so I'll sketch the idea. Some details are slightly wrong (mostly having to do with multiple edges); see the reference I link for full details. Let G be a 2-connected graph. If G is 3-connected, then it is its own 2-sum representation. If not, let G - {u,v} decompose as C_1 cup C_2 cup ... cup C_r. Take C_i and add two vertices u_i and v_i. There is an edge from u_i to each vertex of C_i which was joined to u, and similarly for v. Also, there is an edge from u_i to v_i. Call this modified graph G_i. If G_i is 3-connected, stop. If not, break it up in a similar manner. Keep proceeding in this way until we have a multiset of 3-connected graphs. It is not hard to see that, even implemented stupidly, this is a polynomial process.

A theorem of Cunningham and Edmonds states that this multiset is an isomorphism invariant of the matroid of G. Moreover, Cunningham and Edmonds also work out all the different ways to put these graphs together to obtain isomorphic graphical matroids. (One of their rules is the Whitney flip, which corresponds to gluing G_1 and G_2 together with u_2 and v_2 switched.) It shouldn't be too hard to see that their rule is effectively computable, but its giving me some trouble, so I'll stop here. ADDED: I'll say that I have worked with the Cunningham/Edmonds rule in hand computations and never had any trouble checking it; I'm just having trouble with formally showing it is efficiently checkable.

ag.algebraic geometry - Where does the splitting principle come from and does it generalize

We can think of the splitting principle as a condition on a "cohomology theory" (of some sort) $E^*$, coming about when working with Chern classes for instance, and then ask: When does $E^*$ satisfy this condition? First, let's make the condition more precise and reformulate it:

Condition 1: Given $X$ and a vector bundle $V$ on $X$, there exists $f: X' to X$ such that $f^* V'$ has a filtration with subquotients line bundles, and $f^*: E^*(X) to E^*(X')$ is injective.

But there is a universal choice for $X'$, namely the flag variety of $V$: $p: Fl(V) to X$. Any $f: X' to X$ with $f^* V'$ filtered with line bundle subquotients will factor through $p$, and so we're really just asking if $p^*: E^*(X) to E^*(Fl(V))$ is injective.

Condition 1': For all $X$ and $V$, $p^*: E^*(X) to E^*(Fl(V))$ is injective.

At this point there are two ways this answer can go, depending on ones tastes:

1. $Fl(V)$ is a very geometric object over $X$, so we might as well ask that we actually have a formula for $E^*(Fl(V))$ in terms of $E^*(X)$. If $E^*$ is "reasonable" (i.e., has Chern classes giving rise to a "projective bundle formula") then iteratively applying the projective bundle formula will give such a thing, and in fact show that $E^*(X)$ is a direct summand of $E^*(Fl(V))$.


2. (My favorite:) There's a nice way of strengthening Condition 1' that also holds in all reasonable cases, and that looks rather natural. You can ask that $Fl(V) to X$ behave like a "covering", i.e. that
   (Condition 2:)
   $$ E^*(X) to E^*(Fl(V)) to E^*left(Fl(V) times_X Fl(V)right) $$
   is an equalizer diagram. (So not only is pullback injective, but you can identify its image...) (In fact, in reasonable cases it'll be a split equalizer diagram, related to the direct summand thing above.)

If your question is one of proof + generalization (which I think it is), rather than vague motivation, then I haven't addressed it yet:

In topology. one can show that any complex-oriented cohomology theory (i.e., one with Chern classes for line bundles) $E^*$ has a projective bundle formula, satisfies all the conditions, etc.

In more-algebro-geometric contexts, you could deduce the Chow + K-theory (I don't know anything about the $gamma$-filtration) statements by either

1. Constructing $c_1$ + proving a projective bundle formula, and then feeding this into a general argument using these to prove the rest.


2. Going to the universal example of algebraic cobordism and then deducing the results for Chow + K-theory from the known relationships between them and algebraic cobordism. (Though this second approach is not so great, since those relationships hold under much more stringent hypotheses than are necessary to run the argument.)

One could also ask to generalize this in another direction, replacing vector bundles and $Fl(V)$ by more general $G$-bundles and their associated $G/B$-bundles. In general, that's a more complicated story...