Tuesday, 28 April 2009

sg.symplectic geometry - How can I tell whether a Poisson structure is symplectic "algebraically"?

My primary motivation for asking this question comes from the discussion taking place in the comments to What is a symplectic form intuitively?.



Let $M$ be a smooth finite-dimensional manifold, and $A = cal C^infty(M)$ its algebra of smooth functions. A derivation on $A$ is a linear map ${}: A to A$ such that ${fg} = f{g} + {f}g$ (multiplication in $A$). Recall that all derivations factor through the de Rham differential, and so: Theorem: Derivations are the same as vector fields.



A biderivation is a linear map ${,}: Aotimes A to A$ such that ${f,-}$ and ${-,f}$ are derivations for each $fin A$. By the same argument as above, biderivations are the same as sections of the tensor square bundle ${rm T}^{otimes 2}M$. Antisymmetric biderivations are the same as sections of the exterior square bundle ${rm T}^{wedge 2}M$. A Poisson structure is an antisymmetric biderivation such that ${,}$ satisfies the Jacobi identity.



Recall that sections of ${rm T}^{otimes 2}M$ are the same as vector-bundle maps ${rm T}^*M to {rm T}M$. A symplectic structure on $M$ is a Poisson structure such that the corresponding bundle map is an isomorphism. Then its inverse map makes sense as an antisymmetric section of ${rm T^*}^{otimes 2}M$, i.e. a differential 2-form, and the Jacobi identity translates into this 2-form being closed. So this definition agrees with the one you may be used to of "closed nondegenerate 2-form".




Question: Is there a "purely algebraic" way to test whether a Poisson structure is symplectic? I.e. a way that refers only to the algebra $A$ and not the manifold $M$?




For example, it is necessary but not sufficient that ${f,-} = 0$ implies that $f$ be locally constant, where I guess "locally constant" means "in the kernel of every derivation". The easiest way that I know to see that it is necessary is to use Darboux theorem to make $f$ locally a coordinate wherever its derivative doesn't vanish; it is not sufficient because, for example, the rank of the Poisson structure can drop at points.



Please add tags as you see fit.

pr.probability - Hitting times for an N-dimensional random walk on a lattice with (strictly positive) random integer steps

Assuming you mean Leonid Kovalev's interpretation, the distribution of the hitting time in the $N = 1$ case is the same as the distribution of the number of cycles of a random permutation of $[n]$.



To be more specific, I'll change coordinates. Let $X_0 = (x_0^1, ldots, x_0^N) = (S, S, ldots, S)$. Let $X_1 = (x_1^1, ldots, x_1^N)$, where $x_1^j$ is chosen uniformly at random from $0, 1, ldots, x_0^j-1$. Define $X_2$ similarly in terms of $X_1$, and so on.



Then the sequence $(x_0^1, x_1^1, x_2^1, x_3^1, ldots)$ are as follows:



  • $x_0^1 = L$, of course.

  • $x_1^1$ is uniformly distributed on ${ 0, ldots, S-1 }$.

  • $x_2^1$ is uniformly distributed on ${ 0, ldots, x_1^1-1 }$.

and so on...
In particular the distribution of $x_1^1$ is the distribution of number of elements in a random permutation on $S$ elements which are {it not} in the cycle containing 1; In particular the distribution of $x_1^1$ is the distribution of number of elements in a random permutation on $S$ elements which are {it not} in any of the $k$ cycles with the smallest minima.



So the distribution of the smallest $k$ for which $x_k^1 = 0$ is exactly the distribution of the number of cycles of a random permutation of ${ 1, ldots, S }$; this is $1 + 1/2 + cdots + 1/S sim log S + gamma$, where $gamma = 0.577ldots$ is the Euler-Mascheroni constant.



In the higher-dimensional cases, the time to reach $(0, ldots, 0)$ is just the {it maximum} of the number of cycles of $N$ permutations chosen uniformly at random; this should still have expectation $log S$ plus a constant depending on $N$.

Sunday, 26 April 2009

na.numerical analysis - Finding the length of a cubic B-spline

You should first accept the fact that it's an elliptic integral, and therefore doesn't have an elementary expression without elliptic functions. If you had a numerical library with elliptic functions, then great. Otherwise, you need to either implement elliptic functions yourself, or implement numerical integration of your integral.



I recommend numerical integration, just because in context it is conceptually simple and reliable. Your integrand has a fairly tame form: It can't blow up, the integrand is continuous, and the integrand is also real analytic unless it touches zero. In this situation, Gaussian integration has excellent properties. I don't feel like doing a precise calculation, but I would expect that for any choice of the coefficients, Gaussian quadrature with just 5 evaluation points already has to be fairly close to the exact answer for any choices of the coefficients.




The above is part of an answer, but not a complete answer you really want 64 bits of accuracy. Assuming that the integrand is real analytic, Gaussian quadrature or Clenshaw-Curtis will converge exponentially. It seems reasonable enough to use Clenshaw-Curtis, which lets you recycle evaluation points and has a predictable formula for the numerical integration weights, with more and more points until the answer looks accurate.



The only problem is in the annoying case in which the integrand touches zero, or comes close to touching zero, which can be interpreted geometrically as a point on the spline with nearly zero velocity. (Typically it looks like a cusp.) Then the integrand is NOT real analytic and these numerical methods do not converge exponentially. Or, in the near-zero case, the integrand is analytic but the exponential rate of convergence is slow. I'm sure that there are tricks available that will handle this case properly: You could cut out an interval near the bad point and do something different, or you could subtract off a known integral to tame the bad point. But at the moment I do not have specific advice for an algorithm that is both reasonable fast and reasonably convenient. Clenshaw-Curtis is convenient and usually very fast for this problem, but not all that fast in bad cases if you push it to 64 bits of precision.



Also, these methods can be thought of as a more sophisticated version of chordal approximation. Chordal approximation faces the same issue, except worse: It never converges at an exponential rate for a non-trivial cubic spline. If you want 64 bits, you might need a million chords.




Meanwhile, the GNU Scientific Library does have elliptic function routines. If you have elliptic functions, then again, your integral is not all that simple, but it is elementary. I don't know whether GSL or equivalent is available for your software problem. If it is, then an elliptic function formula is probably by far the fastest (for the computer, not necessarily for you).




In a recent comment, bpowah says "All I wanted to know is whether or not it was faster to compute the given integral numerically or exactly." Here is a discussion. Computing an integral, or any transcendental quantity, "exactly" is an illusion. Transcendental functions are themselves computed by approximate numerical procedures of various kinds: Newton's method, power series, arithmetic-geometric means, etc. There is an art to coding these functions properly. A competitive implementation of a function even as simple as sin(x) is already non-trivial.



Even so, I'm sure that it's faster in principle to evaluate the integral in question in closed form using elliptic functions. It could be hard work to do this right, because the first step is to factor the quartic polynomial under the square root. That already requires either the quartic formula (unfortunately not listed in the GNU Scientific Library even though it has the cubic) or a general polynomial solver (which is in GSL but has unclear performance and reliability). The solution also requires elliptic functions with complex arguments, even though the answer is real. It could require careful handling of branch cuts of the elliptic functions, which are multivalued. With all of these caveats, it doesn't seem worth it to work out an explicit formula. The main fact is that there is one, if you have elliptic functions available but not otherwise.



The merit of a numerical integration algorithm such as Gaussian quadrature (or Clenshaw-Curtis, Gauss-Kronrod, etc.) is that it is vastly simpler to code. It won't be as fast, but it should be quite fast if it is coded properly. The only problem is that the integrand becomes singular if it reaches 0, and nearly singular if it is near 0. This makes convergence much slower, although still not as slow as approximation with chords. With special handling of the near-singular points, it should still be fine for high-performance numerical computation. For instance, a polished strategy for numerical integration might well be faster than a clumsy evaluation of the relevant elliptic functions.

Saturday, 25 April 2009

teaching - Curriculum reform success stories at an "average" research university

Here at University of Michigan, we have a completely separate introductory sequence for highly motivated freshmen. It's modeled after the ones at Harvard and Chicago, from what I've been told (although having not taken them, I can't honestly say for sure). We happened to use Spivak, but that's really the only book that is really 'phonebook-sized'.



Essentially, within the first semester, we're fairly familiar with point-set topology (to the point that taking the introductory point-set topology course was a waste of time since I'd already seen it all), the basics of group theory, and a decent amount of analysis (last year, we finished the first semester with the proofs of taylor's theorem and the different error estimate formulas for series expansions.)



The second semester was finishing up covering some things like uniform and pointwise convergence that were left out of the first semester for time constraints, then linear algebra for pretty much the rest of the semester. We covered topics in a similar order to Hoffman and Kunze (i.e. abstract vector spaces before real/complex inner product spaces.). Then we returned at the end of the year to calculus on real and complex inner product spaces (frechet derivative, inverse function theorem, et cetera). During this semester, we were assigned mostly point-set topology and algebra problems unrelated to linear algebra. These problems usually led to the final step on a later homework being a proof of something like the fundamental theorem of Algebra or Sylow's first theorem (which are two I can remember explicitly.) (No textbook necessary for this course, although the official textbook was Hoffman and Kunze.)



The third semester was spent pretty much entirely spent on measure theory and integration, although a little bit of time was spent on complex analysis, but only enough to prove the statements about the fourier transform. (No textbook was assigned for this course. Anyone who actually purchased the course text indicated on the internet was urged to return the book.)



From what I've been told, next semester, we're doing differential geometry, but as of yet, I don't know how our professor intends to teach it (I asked him if he'd talk about the subject from a category-theoretic point of view (i.e., http://ncatlab.org/nlab/show/nPOV, but he laughed as though I were joking).



The upshot on this sequence is that by the end of the first three semesters, we also receive 'credit (toward completion of a math major, not actual credits)' for linear algeba and real analysis in addition to completion of the intro sequence.



Were I forced to take one of the other introductory sequences, I probably would not have become a math major. The rest of the math curriculum is that awful.

Friday, 24 April 2009

differential topology - When do two holonomy maps determine flat bundles that are isomorphic as just bundles (w/o regard to the flat connections)?

For an abelian structure group $A$, principal $A$-bundles over a smooth manifold $M$ can be described completely in terms of their holonomy functionals $f: LM to A$.



Here, $LM$ is the thin loop space whose elements are homotopy classes of loops, with the rank of the homotopies bounded by one. A map $f:LM to A$ is by definition smooth, if its pullback to the ordinary loop space is smooth (in the Fréchet sense).



Definition. A fusion map is a smooth map $f: LM to A$ such that for every triple $(gamma_1,gamma_2,gamma_3)$ of paths in $M$, with a common initial point and a common end point, and each path is smoothly composable with the inverse of each other, the map satisfies
$$
f(overline{gamma_2} star gamma_1) cdot f(overline{gamma_3} star gamma_2) = f(overline{gamma_3} star gamma_1).
$$



Theorem.



  1. The group of isomorphism classes of principal $A$-bundles with connection over $M$ is isomorphic to the group of fusion maps.


  2. The group of isomorphism classes of principal $A$-bundles with flat connection over $M$ is isomorphic to the group of locally constant fusion maps.


  3. The group of isomorphism classes of principal $A$-bundles over $M$ is isomorphic to the group of homotopy classes of fusion maps (where the homotopies are smooth and go through fusion maps).


Parts 1 and 2 are a theorem of J. Barrett, proved in "Holonomy and Path Structures in General Relativity and Yand-Mills Theory". Part 3 is in my paper "Transgression to Loop Spaces and its Inverse I".



Corollary.
Two principal $A$-bundles with connection (no matter if flat or not) are isomorphic as bundles, if and only if their holonomy functionals are homotopic in the sense of 3.



The non-abelian case should be similiar. In fact, Barrett's result is valid for non-abelian groups.

lie groups - Invariant Vector Fields for Homogenous Spaces

The short answer is that many (most? all?) homogeneous spaces do NOT have such a nice description. In particular, at any point $pin M$, the set of the $G$ or $H$ invariant vectors is a strict subset of $T_pM$. (Here, and below, I'm assuming $G$ and $H$ are compact - it's all I'm really familiar with)



Here's the quick counterexample when looking for $G$ invariance: Take any $Hsubseteq G$ with rank$(G)$ = rank$(H)$. Then the Euler characteristic of $G/H$ is greater than $0$. It follows that EVERY vector field has a 0, and hence, the only $G$-invariant vector field is trivial (since $G$ acts transitively).



For (much more) detail, including how it often fails even with rank$(H)<$rank$(G)$...



Suppose $M = G/H$ (and hence, $M$ is also compact), that is, I'm thinking of $M$ as the RIGHT cosets of $G$. For notation, set $ein G$ as the identity and let $mathfrak{g}$ and $mathfrak{h}$ denote the Lie algebras to $G$ and $H$.



Since $G$ is compact, it has a biinvariant metric. The biinvariant metric on $G$ is equivalent to an $Ad(G)$ invariant inner product on $mathfrak{g}$. This gives an orthogonal splitting (as vector spaces) of $mathfrak{g}$ as $mathfrak{h}oplus mathfrak{m}$. (The vector subspace $mathfrak{m}$ is not typically an algebra)



Now, the tangent space $T_{eH}M$ is naturally be identified with $mathfrak{m}$. In fact, this identification is $H$ invariant (where $H$ acts on $mathfrak{m}$ via $Ad(H)$ and $H$ acts on $M$ by left multiplication: $h * (gH) = (hg)H$).



Now, when you say "invariant vector fields", you must first specify $G$ invariant or $H$ invariant. Lets focus of $G$ first, then $H$.



Typically, there are not enough $G$-invariant vector fields to span the tangent space at any point. This is because not only does $G$ act transitively on $M$, but it acts very multiply transitive (an $H$s worth of elements move any element to any other given element). Thus, a $G$-invariant vector field on $M$ is actually equivalent to an $H$ invariant vector in $T_{eH}M$.



So, for example, viewing $S^{n} = SO(n+1)/SO(n)$, You'd find that $SO(n)$ acts transitively on the unit vectors in $mathfrak{m}$, and hence, there are no $SO(n+1)$ invariant vector fields on $S^{n}$.



Now, on to $H$ invariant vector fields. I haven't thought much about this, but even in this case, there are (often? always?) not enough. For example, viewing $S^2 = SO(3)/SO(2)$, we see that $H$ is a circle. In this case, viewing the north pole of $S^2$ as $eSO(2)$, the only $H$ invariant vector fields are the velocity vectors of rotation through the north-south axis - i.e., the flows are lines of lattitude.



Then we see that at generic points, the set of all vectors tangent to an $H$ invariant vector is a 1 dimensional vector space (and hence, doesn't span the whole tangent space), and the situation at the north and south pole is even worse: the only $H$ invariant vectors are trivial.



Even if you come up with a case where at some point, there were enough $H$ invariant vector fields, you can not translate them around all of $M$, since $H$ does NOT act transitively on $M$.

Thursday, 23 April 2009

Counting knots with fixed number of crossings

There are some known exponential bounds on the number. For example, if kn is the number of prime knots with n crossings, then Welsh proved in "On the number of knots and links" (MR1218230) that




2.68 ≤ lim inf (kn)1/n ≤ lim sup (kn)1/n ≤ 13.5.




The upper bound holds if you replace kn by the much larger number ln of prime n-crossing links.



Sundberg and Thistlethwaite ("The rate of growth of the number of prime alternating links and tangles," MR1609591) also found asymptotic bounds on the number an of prime alternating n-crossing links: lim (an)1/n exists and is equal to (101+√21001)/40.

Tuesday, 21 April 2009

gr.group theory - Is there any criteria for whether the automorphism group of G is homomorphic to G itself?

This answer is essentially a series of remarks, but ones which I hope will be helpful to you.



(1) There are two ways to interpret the condition that $G$ be isomorphic to its automorphism group: canonically and non-canonically.



a) Say that $G$ is complete if every automorphism of $G$ is inner (i.e., conjugation by some element of $G$) and $G$ has trivial center. In this case, there is a canonical isomorphism $G stackrel{sim}{rightarrow} operatorname{Aut}(G)$.



The linked wikipedia article gives some interesting information about complete groups. As above, by definition having trivial center is a necessary condition; all nonabelian simple groups satisfy this. On the other hand, an interesting sufficient condition is that for any nonabelian simple group $G$, its automorphism group $operatorname{Aut}(G)$ is complete, i.e., we have canonically $operatorname{Aut}(G) = operatorname{Aut}(operatorname{Aut}(G))$.



b) It is possible for a group to have nontrivial center and outer automorphisms and for these two defects to "cancel each other out" and make $G$ noncanonically isomorphic to $operatorname{Aut}(G)$. This happens for instance with the dihedral group of order $8$.



2) It seems extremely unlikely to me that there is a reasonable necessary and sufficient condition for a general finite group to be isomorphic to its automorphism group in either of the two sense above.



But a lot of specific examples are certainly known: see for instance



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



in which the order of the outer automorphism group of each of the finite simple groups is given. So, for instance, exactly $14$ of the $26$ sporadic simple groups have trivial outer automorphism group, hence satisfy $G cong operatorname{Aut}(G)$.



I wouldn't be surprised if the outer automorphism groups of all finite groups of Lie type were known (they are not all known to me, but I'm no expert in these matters).

Monday, 20 April 2009

dg.differential geometry - Flux homomorphism for manifolds with boundary

Here are some remarks about your definition.



1) $H_1(A,partial A)$ is just one-dimensional, it is generated by a path that joins two sides of $A$.



2) The definition that you gave works for the annulus, and for surfaces with a boundary as well. This will also work for manifolds with boundary $M^n$, in the case
when you consider fluxes of volume-preserving maps (i.e. you work with $Omega^n$ and $H_{n-1} (M^n, partial M^n$). I have not seen this definition before, but it is so natural, that it would be strange if no one considered it.



3) It does not look that this definition will work for higher-dimensional symplectic manfiolds, if you want to study fluxes of symplectomorphisms (and you work with $omega$ and $H_1(M^{2n},partial M^{2n})$), because the restriction of $omega$ to $partial$ will be non-zero.

gn.general topology - Two commuting mappings in the disk

I'm sure what I write has been thought of by many, but it's a starting point that I thought should be written down.



First by the Brouwer fixed point theorem $f$ has at least one fixed point, say $bar{x}=f(bar{x})$.



If that fixed point is unique (contraction mappings spring to mind for a bunch of examples of this) we're done since $g(bar{x})=g(f(bar{x}))=f(g(bar{x}))$ and we see that $g(bar{x})$ is "another" fixed point of $f$, since the fixed point was unique $g(bar{x})=bar{x}=f(bar{x})$.



For "less nice" $f$ we still have that $f(g(bar{x}))=g(bar{x})$... in fact for any $ninmathbb{N}$ we have $f(g^n(bar{x}))=g^n(bar{x})$. If (without resorting to sequences) $g^n(bar{x})to y$, we can again claim success since we'll have $g(y)=y$ and $f(y)=y$.



Unless there's another "obvious" easy case I missed it seems like the interesting cases will be when $g^n(bar{x})$ does not converge. Two sub-cases spring to mind: when $g^{n}(bar{x})$ has finitely many accumulation points (like when $g^n(bar{x})$ is a periodic point of $g$), or ... it has lots. Intuition (really thinking about rational and then irrational rotations about the origin as one way to generate those two cases) tells me that in either of these cases what we really need to do drop the $bar{x}$ as a "starting point".



It "would be nice" if we can show $g$ conjugate to a rotation in the above two cases. Any thought on if that is true or not? I suspect not else $g$ would have a unique fixed point and we'd be done (as above)... Maybe semi-conjugate... but would that help? New minds, any thoughts?

Sunday, 19 April 2009

homological algebra - About higher Ext in R-Mod

So, in $R-Mod$, we have the rather short sequence



  • $mathrm{Ext}^0(A,B)cong Hom_R(A,B) $


  • $mathrm{Ext}^1(A,B)cong mathrm{ShortExact}(A,B)mod equiv $, equivalence classes of "good" factorizations of $0in Hom_R(A,B)congmathrm{Ext}^0(A,B)$, with the Baer sum.


Question:



  • $mathrm{Ext}^{2+n}(A,B) cong ??? $

While I suppose one could pose a conjugate question in algebraic topology/geometry, where the answer might look "simpler", I'm asking for a more directly algebraic/diagramatic understanding of the higher $mathrm{Ext}$ functors. For instance, I'd expect $mathrm{Ext}^2(A,B)$ to involve diagrams extending the split exact sequence $Arightarrow Aoplus Brightarrow B$, but precisely what sort of extension? Or is that already completely wrong?

Saturday, 18 April 2009

homological algebra - Mittag-Leffler condition: what's the origin of its name?

The wording of your question suggests that you're familiar with the "classical" Mittag-Leffler theorem from complex analysis, which assures us that meromorphic functions can be constructed with prescribed poles (as long as the specified points don't accumulate in the region).



It turns out - or so I'm told, I must admit to never working through the details - that parts of the proof can be abstracted, and from this point of view a key ingredient (implicit or explicit in the proof, according to taste) is the vanishing of a certain $lim_1$ group -- as assured by the "abstract" ML-theorem that you mention.



I'm not sure where this was first recorded - I hesitate to say "folklore" since that's just another way of saying "don't really know am and not a historian". One place this is discussed is in Runde's book A taste of topology: see Google Books for the relevant part.



IIRC, Runde says that the use of the "abstract" Mittag-Leffler theorem to prove the "classical" one, and to prove things like the Baire category theorem, can be found in Bourbaki. Perhaps someone better versed in the mathematical literature (or at least better versed in the works of Bourbaki) can confirm or refute this?

nt.number theory - Generalization of primitive roots

The standard definition is that $ainmathbb{Z}$ is a primitive root modulo $p$ if the order of $a$ modulo $p$ is $p-1$.



Let me rephrase, to motivate my generalization: $ainmathbb{Z}$ is a primitive root modulo $p$ if the linear recurrence defined by $x_0=1$, $x_n=ax_{n-1}$ for $ngeq1$ has maximum possible period in $mathbb{Z}/pmathbb{Z}$.



So, we could define $(a_1,ldots,a_r)inmathbb{Z}^r$ to be an $r$-primitive root modulo $p$ when the order $r$ linear recurrence defined by $x_0=cdots=x_{r-2}=0$, $x_{r-1}=1$, and $x_n=a_1x_{n-1}+cdots+a_rx_{n-r}$ for $ngeq r$ has the maximum possible period (for such sequences) in $mathbb{Z}/pmathbb{Z}$.



Is anything known about the generalized version of primitive roots I've described? I chose my initial values $x_0=cdots=x_{r-2}=0$, $x_{r-1}=1$ in analogy with the Fibonacci numbers, but is there a standard / most general choice? The starting values affect the period a lot, so it's important to be working with the "right" ones, I guess. What is the maximum achievable period for a linear recurrence in $mathbb{Z}/pmathbb{Z}$? An obvious upper bound is $p^r-1$, but it's not clear to me that this is the correct number to replace $p-1$ in the standard definition of primitive root. Is anything known in the direction of Artin's conjecture for "$r$-primitive roots", as I've called them?



Other thoughts: because there is no "formula" for the order of $a$ modulo $p$ (this would amount to a formula for the discrete logarithm), there certainly isn't a formula for the period of the linear recurrence defined by $(a_1,ldots,a_r)$ modulo $p$. I tried to come up with one for a while before I realized this, so I just wanted to save anyone else the trouble.

Thursday, 16 April 2009

riemannian geometry - Injectivity radius and the cut locus

Consider a connected, complete and compact Riemannian manifold $M$. Is it correct that the following equality holds: $text{inj}(x)=text{dist}left(x,text{CuL}(x)right)$? Or in words that the injectivity radius of a point is the distance from the point to its cut locus.



Here is my explanation: As the manifold is compact and complete, then the cut locus $text{CuL}(x)$ is compact as well[1]. Thus, there exists a point $yin text{CuL}(x)$ such that $text{dist}left(x,text{CuL}(x)right)=text{dist}(x,y)$. Since $y$ is a cut point of $x$, there exists a tangent vector $xi_0in T_x M$ such that $y=exp_xleft(c(xi_0)xi_0right)$[2], where $c(xi_0)$ is the distance from $x$ to its cut point in the $xi_0$ direction. This in turn means that $text{dist}(x,y) = c(xi_0)$.



Recall that $text{inj}(x)=inf_{xiin T_x M}(c(xi))$. This means that $text{inj}(x) leq c(xi_0) = text{dist}(x,y)=text{dist}left(x,text{CuL}(x)right)$. If $text{inj}(x)< c(xi_0)$, then since $M$ is compact, it means that there exists some other tangent vector $xiin T_x M$ with $c(xi) < c(xi_0)$. But this means that $exp_x(c(xi)xi)$ is a cut point of $x$ closer to it then $y$, and this is a contradiction.




[1] See Contributions to Riemannian Geometry in the Large by W. Klingenberg



[2] Here I'm using the notation of I. Chavel in his book Riemannian Geometry - Modern Introduction.




Update(@dror)
Today I finally found a copy of the book *Riemannian Geometry" by Takashi Sakai, and there the above is stated as proposition 4.13 in chapter 3.
Thanks anyway.

The topology of periodic permutohedral boundary conditions

$mathbb{R}^n$ admits a tessellation by permutohedra. The corresponding identification of facets of a permutohedron therefore gives a well-defined space: call it $X_n$. For example, $X_2$, the hexagon with opposite sides identified, can be shown to be a 2-torus (see figure 2 in http://arxiv.org/pdf/cond-mat/0703326v2). Is $X_n cong (S^1)^n$? Why (or why not)?

Wednesday, 15 April 2009

ag.algebraic geometry - Crystalline cohomology of abelian varieties

I am trying to learn a little bit about crystalline cohomology (I am interested in applications to ordinariness). Whenever I try to read anything about it, I quickly encounter divided power structures, period rings and the de Rham-Witt complex. Before looking into these things, it would be nice to have an idea of what the cohomology that you construct at the end looks like.



The l-adic cohomology of abelian varieties has a simple description in terms of the Tate module. My question is: is there something similar for crystalline cohomology of abelian varieties?



More precisely, let $X$ be an abelian scheme over $mathbb{Z}_p$. Is there a concrete description of $H^1(X_0/mathbb{Z}_p)$? (or just $H^1(X_0/mathbb{Z}_p) otimes_{mathbb{Z}_p} mathbb{Q}_p$?) I think that this should consist of three things: a $mathbb{Z}_p$-module $M$, a filtration on $M otimes_{mathbb{Z}_p} mathbb{Q}_p$ (which in the case of an abelian variety has only one term which is neither 0 nor everything) and a Frobenius-linear morphism $M to M$.



I believe that the answer has something to do with Dieudonné modules, but I don't know what they are either.

Tuesday, 14 April 2009

Trivial valuation

Maybe you meant for the extension $L/K$ to be algebraic, in which case it is true that any extension of the trivial valuation on $K$ to $L$ is trivial. This clearly reduces to
the case of a finite extension, and then -- since a trivially valued field is complete --
this follows from the uniqueness of the extended valuation in a finite extension of a complete field. Maybe you view this as part of the sledgehammer, but it's not really the heavy part: see e.g. p. 16 of



http://math.uga.edu/~pete/8410Chapter2.pdf



for the proof. (These notes then spend several more pages establishing the existence part of the result.)



Addendum: Conversely, if $L/K$ is transcendental, then there exists a nontrivial extension on $L$ which is trivial on $K$. Indeed, let $t$ be an element of $L$ which is
transcendental over $K$, and extend the trivial valuation to $K(t)$ by taking $v_{infty}(P/Q) = deg(Q) - deg(P)$. (The completion of $K$ with respect to $v_{infty}$ is the Laurent series field $K((t))$, so this is really the same construction as in Cam's answer.) Then I prove* in the same set of notes that any non-Archimedean valuation can be extended to an arbitrary field extension, so $v$ extends all the way to $L$ and is certainly nontrivial there, being already nontrivial on $K(t)$.



*: not for the first time, of course, though I had a hard time finding exactly this result in the texts I was using to prepare my course. (This does use the sledgehammer.)

linear algebra - Is this an if-and-only-if definition of affine?

Usually, when we say "affine transformation", we mean an invertible one. Either way, any affine transformation is indeed of the form $x mapsto Ax + b$, where $A$ is a (invertible) linear transformation and $b$ is a fixed vector.



If by "scaling", you mean a scalar multiple of the identity matrix and by a shear, you mean an upper triangular matrix with 1's along the diagonal, then rotations, scalings, shears, and translations do not generate all possible affine transformations, because rotations, scalings, and shears do not generate all possible linear transformations.



On the other hand, any linear transformation can always be written as $A = RDS$, where $R$ and $S$ are orthogonal transformations (i.e., rotations) and $D$ is diagonal. The action of a diagonal matrix can be viewed as rescaling by different factors in different amounts in each co-ordinate direction. So compositions of translations, rotations, and co-ordinate scalings generate all affine transformations.



Any linear transformation can also be written as $A = RDU$, where $R$ is orthogonal, $D$ is diagonal, and $U$ is upper triangular with all $1$'s along the diagonal. In this sense any invertible affine transformation can be written as a composition of a translation, a rotation, co-ordinate scalings, and a composition of shears.

Dwork's use of p-adic analysis in algebraic geometry

The premise of the question is not correct. Dwork's methods (and modern descendants of them) are a major part of modern arithmetic geometry over $p$-adic fields, and of $p$-adic analysis.
You could look at the two volume collection Geometric aspects of Dwork theory to get
some sense of these developments.



Just to say a little bit here: Dwork's approach led to Monsky--Washnitzer cohomology,
which in turn was combined with the theory of crystalline cohomology by Bertheolt to develop the modern theory of rigid cohomology. The $p$-adic analysis of Frobenius actions is also
at the heart of the $p$-adic theory of modular and automorphic forms, and of much of the machinery underlying $p$-adic Hodge theory. The theory of $F$-isocrystals (the $p$-adic analogue of variations of Hodge structure) also grew (at least in part) out of Dworks
ideas.



To get a sense for some of Dwork's techniques, you can look at the Bourbaki report Travaux de Dwork, by Nick Katz, and also at Dwork's difficult masterpiece $p$-adic cycles, which has been a source of insipiration for many arithmetic geometers.



In some sense the $p$-adic theory is more difficult than the $ell$-adic theory, which is
why it took longer to develop. (It is like Hodge theory as compared to singular cohomology.
The latter is already a magnificent theory, but the former is more difficult in the sense that it has more elaborate foundations and a more elaborate formalism, and (related to this) has ties to analysis (elliptic operators, differential equations) that are not present
in the same way in the latter.) [For the experts: I am alluding to $p$-adic Hodge theory,
syntomic cohomology, $p$-adic regulators, Serre--Tate theory, and the like.]

Monday, 13 April 2009

ds.dynamical systems - Partitions and Expansiveness

Why if one have an $varepsilon$-expansive homeomorphism $T:X rightarrow X$ ($X$ a compact metric space) and a given partition $D$ of $X$ which has diameter smaller than $varepsilon$ the sequence of refined partitions $D_n = bigvee_{i = -n}^n T^{-i} D$ has diameter converging to zero ?



Recall that a $varepsilon$-expansive homeomorphism $T$ is such that given any two distinct points $x$ and $y$ there exist $n in mathbf{Z}$ such that $d(T^nx, T^ny) > varepsilon$



I can see intituively why this is true, somehow the refined partitions have less an less points in its members precisely because they have diameter less than epsilon but $T$ keeps separating points (and i fact open sets of points) at distance greater than $varepsilon$.



Thanks in advance!

Sunday, 12 April 2009

qa.quantum algebra - Categorifications of the Fibonacci Fusion Ring arising from Conformal Field Theory

The Virasoro minimal model $mathcal{M}(2,5)$ (or in some conventions also $mathcal{M}(5,2)$ is the conformal field theory which describes the critical behaviour of the Lee-Yang edge singularity. It is described, for example, in Conformal Field Theory, by di Francesco, Mathieu and Sénéchal; albeit the description of the Lee-Yang singularity itself is perhaps a little too physicsy. Still their treatment of minimal models should be amenable to mathematicians without prior exposure to physics.



At any rate, googling Lee-Yang edge singularity might reveal other sources easier to digest. In general it is the Verlinde formula which relates the fusion ring and the Virasoro characters, and at least for the case of the Lee-Yang singularity, these can be related in turn to Temperley-Lieb algebras and Ocneanu path algebras on a suitable graph. Some details appear in this paper.



The relation between the Virasoro minimal models and the representations of $widehat{sl}_2$ goes by the name of the coset construction in the Physics conformal field theory literature or also Drinfeld-Sokolov reduction. This procedure gives a cohomology theory (a version of semi-infinite cohomology for a nilpotent subalgebra) which produces Virasoro modules from $widehat{sl}_2$ modules. Relevant words to google are W- algebras, Casimir algebras,... Of course here we are dealing with the simplest case of $widehat{sl}_2$ and Virasoro, which is the tip of a very large iceberg. The case of the Lee-Yang edge singularity is simple enough that it appears in many papers as an example from which to understand more general constructions.



I know less about the quantum group story, but this paper of Gaberdiel might be a good starting point.

Friday, 10 April 2009

gr.group theory - Is $varphi(n)/n$ the maximal portion of $n$-cycles in a degree $n$ group?

It is true for all primitive groups: The primitive groups of degree n containing an n-cycle were independently classified in



Li, Cai Heng The finite primitive permutation groups containing an abelian regular subgroup.
Proc. London Math. Soc. (3) 87 (2003), no. 3, 725--747. )



and



Jones, Gareth A.
Cyclic regular subgroups of primitive permutation groups.
J. Group Theory 5 (2002), no. 4, 403--407.



They are the groups G such that



-$C_pleqslant Gleqslant AGL(1,p)$ for p a prime



-$A_n$ for n odd, or $S_n$



-$PGL(d,q)leqslant G leqslant PGamma L(d,q)$: here there is a unique class of cyclic subgroups generated by an n-cycle except for $G=PGamma L(2,8)$ in which case there are two.



-$(G,n)=(PSL(2,11),11), (M_{11},11), (M_{23},23)$



All these groups satisfy the bound.



Gordon Royle has pointed out to me that the bound does not hold for elements of order n. The smallest examples which do not meet the bound are of degree 12 and are the groups numbered 263 and 298 in the catalogue of groups of degree 12 in Magma.

Thursday, 9 April 2009

measure theory - A question about sigma-ring

How about choosing $B = B_1 - B_2 $ and $C = C_1 -C_2$, we have $ B in S(E cap A)$ by property of $ S(E cap A) $, and $ C in S(E) $, by property of $S(E)$.



The fact that this works can be seen by observing that both $B1 cup (C1−A)$, and $B2cup (C2−A)$ are in fact both disjoint unions. $B_1,B_2$ has everything to do with $A$ and $C_1 -A, C_2 -A$ has nothing to do with $A$ at all.



ps: I don't know why you use the and and or symbol here instead of intersection and union symbol.

Tuesday, 7 April 2009

matrices - How does eigenvalues and eigenvectors change if the original matrix changes slightly?

As Matthew Daws pointed out with his link above, Terry Tao answers your specific question. However, you might be interested in the larger framework of random matrix theory, which is includes the study of eigenvalue distributions of large random matrices. This subject makes precise Douglas Zare's comment:

I would guess that there will be some properties of the eigenvalues which hold for at least 98% of the alterations of a large matrices.


By putting special probability distributions on the space of $N times N$ matrices, one can do everything explicitly using orthogonal polynomials, Riemann-Hilbert theory, and a whole slew of other exact tools. For example, the Gaussian Unitary Ensemble (GUE) is the class of complex $N times N$ matrices with i.i.d. Gaussian entries constrained so that the matrix is Hermitian. Here, the distribution of largest eigenvalue follows the Tracy-Widom law, the cumulative distribution of eigenvalues is Wigner's semicircle law, and one can calculate explicitly pretty much any other quantity of interest.



Just as the central limit theorem holds for a much wider class of i.i.d. random variables than just Gaussians, much current work in the analysis of random matrix theory is to show that these properties are universal, and don't rely on the specific Gaussian structure.

Friday, 3 April 2009

convex polytopes - Finding a bounding volume (line segments) from a kDop definition.

OK, it seems like your $N=52$ then and the normals to the planes are fixed. Then you are right that $N^3$ method takes 1ms time. Indeed, to find the parametric coefficients of the intersection of any 2 planes once you prestore the line directions and the inverse matrices takes just 6 multiplications and 3 additions, to find the cut takes 52 cycles of 6 multiplications, 5 additions, one division, and two comparisons each (keeping the max and the min), and doing all 1326 pairs that requires 1326*(9+52*14)=977262 elementary operations, which puts your processor speed at 10^9 multiplications per second, which is about right. There is an $O(N^2)$ algorithm if you know a point inside the polytope, which, once you are talking about kDop, you, probably do, but we need to count operations very carefully to see if it beats the $N^3$ one.



The key problem is that in the worst case when every plane matters, you may have about 3*52=156 edges which is just one tenth of the maximal possible amount and if you spend the same 9+52*14 operations on each edge, you will be just on the edge of what you need. Still, it is worth trying. It works under the assumption that no 4 planes intersect at one point. To make sure that it is the case, just perturb your Mins and Maxes by some random numbers that are much bigger than the precision with which you can compute the points but much less than what is visible to the eye (say, $10^{-4}$ times the object size, also make sure that you decrease Min's and increase Max's) but do all comparisons with much higher precision (like $10^{-12}$ times the object size) where the object size is just the maximal span in your directions, which you can find in linear time.



Start with finding one edge. To this end, just take any point inside the polytope, compute the vertices of the dual polytope with respect to that point, choose any direction, and take the furthest vertex of the dual polytope from the origin. Now, you know that this plane certainly matters. It will take you just linear time, so I'll not even count that. Once you figured out one plane, you can find an edge on that plane in the time 50*(9+52*14)=36850 elementary operations just as in the $N^3$ method. Keep track on which planes give you the boundary points.



Now, at each moment, keep track of edges you already have, and the vertices you have with their planes. For each vertex also keep the track of whether you know each of the adjacent edges (according to the pair of planes) completely. In the beginning, you have the list of 2 vertices with one known and two unknown edges.



Take the first vertex in the list with at least one unknown adjacent edge. You know the pair of planes for this edge, so you can find the new edge in time (9+26*15)=737, where you have 26 instead of 52 because you can find whether you'll need Max or Min in each direction in one comparison. Once you have found the edge, see if its other end is one of the already found vertices and mark the corresponding edge at that vertex as known (about 156*2=312 operations). If not, add its other vertex to the end of the list. This gives you 1049 operations per edge.



Thus, your total is 36850+156*1049=200494 operations, which is 5 times better than my count for the $N^3$ algorithm but still a bit slower than you need.



There is a room for improvement. The most obvious one is that you may want to keep the list of already found vertices for each pair of planes (the update costs 3 operations and reduces 312 to just 6. Thus, you are down to 36850+156*153=154318, which is within the limits, but just barely. Another thing is that you can try to reduce the time you need to find the first edge. but I do not see how to do it at the moment.




All right, first thing first. The main part of the cure is to construct the admissibility table that, for each pair of directions, lists all directions that may form a vertex with the given two. Surprisingly, it is very short. I'll keep the directions in the 0,$pm 1$ format like you (not normalized). I assume that you keep an array of these directions p of length 26. The only thing I want you to make sure about the order is that p[2k] and p[2k+1] are opposite for each k=0,1,...,12. This will spare a few operations in the edge tracing algorithm.



First, let us exclude all degenerate pairs. That is pretty easy. The pair (A,B) is degenerate iff A+B is a multiple of some other direction in the list. In practical terms, it means that all non-zero entries of $A+B$ are equal in absolute value. The first thing
is to create the table of pairs of directions and precompute some information for them.
To make it working for both algorithms, I suggest that for each pair (i,j) of indices, you create the structure Q[i,j] with the following fields



Boolean nondegeneracy; true if the pair is non-degenerate, false if it is degenerate.
You should get 288 non-degenerate pairs if the order of i,j matters (144 with i

vector normal; normal=$p[i]times p[j]$ is the direction orthogonal to both p[i] and p[j]



iprojector, jprojector; those are just found as $v=p[j]timesoperatorname{normal}$; iprojector$=v/(vcdot p[i])$ and similarly for jprojector. This will allow you to find a point X with given scalar products $a=p[i]cdot X$ and $b=p[j]cdot X$ as a times iprojector plus b times jprojector (6 multiplications and 3 additions)



an array goodindices of structures. Each structure element goodindices[m] will have 3 fields: integer k such that p[i],p[j],p[k] may make a vertex, and 3 scalar products iscal=iprojector$cdot$p[k],jscal=jprojector$cdot$p[k],and nscal=normal$cdot$p[k]. To construct it, you take your pair a=p[i],b=p[j] of directions and include the structure with the first field k (once you know k, you can compute the other three fields) into goodindices if the following two conditions hold:



1)the determinant det=det(a,b,c) is not 0



2) for every index m different from i,j,k at least one of the determinants det(a,b,d), det(a,d,c), and det(d,b,c) where d=p[m] has sign opposite to the sine of det, i.e., if, say, det > 0, one of them is strictly less than 0 (since all coordinates are integer, compare to -0.5). This checks that there is no direction in the cone formed by a,b,c.



Funnily enough, the total number of indices in goodindices is never greater than 6.



Also, let Max[i] be the maximal scalar product of your object points with p[i]
(I prefer to keep all 26 directions and Max array to keeping 13 and Min and Max arrays just to avoid any casework anywhere)



Now, when the table is ready, it is very easy to run the N^3 algorithm. For each pair i,j with i < j do the following.



If Q[i,j].nondegeneracy=false, just skip the pair altogether



Determine tmin as the maximum over all structures S in Q[i,j].goodindices such that S.nscal<0 of the ratios (M[S.k]-M[i]S.iscal-M[j]S.jscal)/S.nscal .



Determine tmax as the minimum over all structures S in Q[i,j].goodindices such that S.nscal>0 of the ratios (M[S.k]-M[i]S.iscal-M[j]S.jscal)/S.nscal.



If tmin < tmax, put X=M[i]Q[i,j].iprojector+M[j]Q[i,j].jprojector.; compute the points X+tmin Q[i,j].normal and X+tmax Q[i,j].normal and draw the edge between them (or store it for later drawing).



You should optimize here updating tmin and tmax when looking at each k and stopping immediately if tmax gets less than or equal to tmin. This cycle over k is the most time-consuming part when defining the edge, so how you write it determines the real speed of this algorithm. Optimize as much as possible here to make sure that you do not do any extraction from array twice, etc. Each little extra operation counts and memory is not a problem. That's why I suggested to prestore the scalar products as well, though it might seem a bit silly when we are talking of cycle of length 6 within a cycle of length 144



That's all. It should run fast enough to be acceptable already and it is robust in the sense that you do not need to perturb your maxes, check for paralellness anywhere, etc., etc.



The other algorithm is still about 4 times faster but I should check that it has no stability issues before posting it. Try this first and tell me the running time.

co.combinatorics - Canonical bases for modules over the ring of symmetric polynomials

One profitable thing to look at might be geometric Satake:



Roughly, one can categorify the symmetric polynomials acting on all polynomials as perverse sheaves on $GL(n,mathbb{C}[[t]])setminus GL(n,mathbb{C}((t)))/GL(n,mathbb{C}[[t]])$ acting on perverse sheaves on $GL(n,mathbb{C}[[t]])setminus GL(n,mathbb{C}((t)))/I$ where $I$ is the Iwahori (matrices in $GL(n,mathbb{C}[[t]])$ which are upper-triangular mod $t$).



The maps to polynomials are take a sheaf and send it to the sum over sequences $mathbf{a}$ of $n$ integers of the Euler characteristic of its stalk at the diagonal matrix $t^{mathbf{a}}$ times the monomial $x^{mathbf{a}}$.



One nice thing that happens in this picture is the filtration of polynomials by the invariants of Young subgroups appears as a filtration of categories. Thus, one can take quotient categories and get a nice basis, with lots of good positivity, for the isotypic components.



For the multiplicity space, one might be able to do some trick using cells. It's not immediately clear to me how.

rt.representation theory - Is it useful to consider cohomology of group representations?

In group representation theory, one attempts to explain and classify (some of) the modules over the group ring $k[G]$, for some field $k$. In group cohomology, one develops the machinery of the cohomology of groups out of the category of modules over the ring $mathbb{Z}[G]$. In both cases, one has tensor products, restriction maps, induced modules, etc, and all these constructions look very similar in both cases, with $k$ replaced by $mathbb{Z}$ in group cohomology.



My question is, might one develop a theory of cohomology of $G$-representations over a field $k$ analogous to the theory of cohomology of $G$-modules? Essentially, one would take a representation $M$, take an injective resolution of $M$ in the category of $k[G]$-modules, then find the fixed elements to get the cohomology. I believe this would be $mathrm{Ext}_{k[G]}(k,M)$, where $k$ is given the trivial $G$-action.



Given that we know a lot about the structure of finite-dimensional representations of finite groups (character tables), what would the character table tell us about the cohomology? What might the cohomology tell us about the character table? What happens if we consider representations of infinite groups, such as Lie groups, algebraic groups, or Galois groups?

nt.number theory - Q-lattices and commensurability, any insight into the definition and intuition?

I've been coming across $mathbb{Q}$-lattices in $mathbb{R}^n$ in my reading, and I'm having trouble understanding the definitions. Connes and Marcolli define it as a lattice $Lambda in mathbb{R}^n$ together with a homomorphism $phi : mathbb{Q}^n / mathbb{Z}^n to mathbb{Q} Lambda / Lambda$. Moreover, two $mathbb{Q}$-lattices $Lambda_1$ and $Lambda_2$ are commensurable iff 1) $mathbb{Q} Lambda_1 = mathbb{Q}Lambda_2$ and 2) $phi_1 = phi_2$ mod $Lambda_1 + Lambda_2$.



I think I understand condition 1): the lattices must be rational multiples of each other to be commensurable. I don't even understand the notation for condition 2). The best I can gather is that the homomorphism $phi$ labels which positions in $mathbb{Q} Lambda / Lambda$ come from your more normal "discrete hyper-torus" $mathbb{Q}^n / mathbb{Z}^n$. Condition 2) then says that the same points are labelled. Is this anywhere near the right ballpark? Can anyone recommend any literature on the subject?



I'm a pretty young mathematician (not even in a PhD program...yet) so please forgive me if this question seems basic.



Thanks.

Thursday, 2 April 2009

cv.complex variables - conformally embedding complex tori into R^3

Let $L$ be a lattice in $mathbb{C}$ with two fundamental periods, so that $mathbb{C}/L$ is topologically a torus. Let $p:mathbb{C}/L mapsto mathbb{R}^3$ be an embedding ($C^1$, say). Call $p$ conformal if pulling back the standard metric on $mathbb{R}^3$ along $p$ yields a metric in the equivalence class of metrics on $mathbb{C}/L$ (i.e. a multiple of the identity matrix).




Is there an explicit formula for such a p in the case of L an oblique lattice?





Backgorund



The existence of such $C^1$ embeddings is implied by the nash embedding theorem (fix a metric on $mathbb{C}/L$, pick any short embedding, apply nash iteration to make it isometric and hence conformal).



For orthogonal lattices, the solution is simple: Parametrise the standard torus of radii $r_1$, $r_2$ in the usual way. Make the ansatz $pi(theta, phi) = (f(theta), h(phi))$, pull back the standard metric on $mathbb{C}/L$ and solve the resulting system of ODEs. This relates $r_1/r_2$ to the ratio of the magnitudes of the periods. This shows that no standard torus can be the image of $p$ in the original question (oblique lattice), although that is geometrically clear anyway.

fa.functional analysis - Generalize Fourier transform to other basis than trigonometric function

You need the orthogonality condition to get such an integral representation for the coefficients; otherwise it would probably be more complicated.



The Fourier series of any $L^2$ function converges not only in the norm (which follows from the fact that ${e^{inx}}$ is an orthonormal basis) but also almost everywhere (the Carleson-Hunt theorem). Both these assertions are also true in any $L^p,p>1$ but at least the first one requires different methods than Hilbert space ones. In $L^1$, by contrast, a function's Fourier series may diverge everywhere.



There are many conditions that describe when a function's Fourier series converges to the appropriate value at a given point (e.g. having a derivative at that point should be sufficient). Simple continuity is insufficient; one can construct continuous functions whose Fourier series diverge at a dense $G_{delta}$. The problem arises because the Dirichlet kernels that one convolves with the given function to get the Fourier partial sums at each point are not bounded in $L^1$ (while by contrast, the Fejer kernels or Abel kernels related respectively to Cesaro and Abel summation are, and consequently it is much easier to show that the Fourier series of an $L^1$ function can be summed to the appropriate value using either of those methods). Zygmund's book Trigonometric Series contains plenty of such results.



There is a version of the Carleson-Hunt theorem for the Fourier transform as well.

quantum mechanics - Which functions are Wiener-integrable?

I'm looking for either a few precise mathematical statements about Wiener integrals, or a reference where I can find them.



Background



The Wiener integral is an analytic tool to define certain "integrals" that one would like to evaluate in quantum and statistical mechanics. (Hrm, that's two different mechanics-es....) More precisely, one often wants to define/compute integrals over all paths in your configuration or phase space satisfying certain boundary conditions. For example, you might want to integrate over all paths in a manifold with fixed endpoints. It's conventional to write the integrand as a pure exponential $exp(f)$. In statistical mechanics, the function $f$ in the exponent is generally real and decays to $-infty$ (fast enough) in all directions. If the path space were finite-dimensional, you would expect such integrals to converge absolutely in the Riemann sense. In quantum mechanics, $f$ is usually pure-imaginary, so that $exp(f)$ is a phase, and the integral should not be absolutely convergent, but in finite-dimensional integrals may be conditionally convergent in the Riemann sense. Typically, $f$ is a local function, so that $f(gamma) = int L(gamma(t))dt$, where $gamma$ is a path and $L(gamma(t))$ depends only on the $infty$-jet of $gamma$ at $t$. In fact, typically $L(gamma(t))$ depends only on the $1$-jet of $gamma$, so that $f(gamma)$ is defined on, for example, continuous piece-wise smooth paths $gamma$.



For example, one might have a Riemannian manifold $mathcal N$, and want to define:
$$U(q_0,q_1) = intlimits_{substack{gamma: [0,1] to mathcal N \ {rm s.t.}, gamma(0) = q_0,, gamma(1)=q_1}} expleft( - hbar^{-1} int_0^1 frac12 left| frac{partial gamma}{partial t}right|^2dt right)$$
where $hbar$ is a positive real number (statistical mechanics) or non-zero pure imaginary (quantum mechanics). The "measure" of integration should depend on the canonical measure on $N$ coming from the Riemannian metric, and the Wiener measure makes it precise.



On $mathcal N = mathbb R^n$, I believe I know how to define the Wiener integral. The intuition is to approximate paths by piecewise linears. Thus, for each $m$, consider an integral of the form:
$$I_m(f) = prod_{j=1}^{m-1} left( int_{gamma_j in mathbb R^n} dgamma_j right) exp(f(bargamma)) $$
where $bargamma$ is the piecewise-linear path that has corners only at $t = j/m$ for $j=0,dots,m$, where the values are $bargamma(j/m) = gamma_j$ (and $gamma_0 = q_0$, $gamma_m = q_1$). Then the limit as $mto infty$ of this piecewise integral probably does not exist for a fixed integrand $f$, but there might be some number $a_m$ depending weakly on $f$ so that $lim_{mto infty} I_m(f)/a_m$ exists and is finite. I think this is how the Wiener integral is defined on $mathbb R^n$.



On a Riemannian manifold, the definition above does not make sense: there are generally many geodesics connecting a given pair of points. But a theorem of Whitehead says that any Riemannian manifold can be covered by "convex neighborhoods": a neighborhood is convex if any two points in it are connected by a unique geodesic that stays in the neighborhood. Then we could make the following definition. Pick a covering of $mathcal N$ by convex neighborhoods, and try to implement the definition above, but declare that the integral is zero on tuples $gamma_{vecjmath}$ for which $gamma_j$ and $gamma_{j+1}$ are not contained within the same convex neighborhood. This would be justified if "long, fast" paths are exponentially suppressed by $exp(f)$. So hope that this truncated integral makes sense, and then hope that it does not depend on the choice of convex-neighborhood cover.



Of course, path integrals should also exists on manifolds with, say, indefinite "semi-"Riemannian metric. But then it's not totally clear to me that the justification in the previous paragraph is founded. Moreover, really the path integral should depend only on a choice of volume form on a manifold $mathcal N$, not on a choice of metric. Then one would want to choose a metric compatible with the volume form (this can always be done, as I learned in this question), play the above game, and hope that the final answer is independent of the choice. A typical example: any symplectic manifold, e.g. a cotangent bundle, has a canonical volume form.



One other modification is also worth mentioning: above I was imagining imposing Dirichlet boundary conditions on the paths I wanted to integrate over, but of course you might want to impose other conditions, e.g. Neumann or some mix.



Questions



Question 0: Is my rough definition of the Wiener integral essentially correct?



Question 1: On $mathcal N = mathbb R^n$, for functions $f$ of the form $f(gamma) = -hbar^{-1}int_0^1 L(gamma(t))dt$, for "Lagrangians" $L$ that depend only on the $1$-jet $(dotgamma(t),gamma(t))$ of $gamma$ at $t$, when does the Wiener integral make sense? I.e.: for which Lagrangians $L$ on $mathbb R^n$, and for which non-zero complex numbers $hbar$, is the Wiener integral defined?



Question 2: In general, what are some large classes of functions $f$ on the path space for which the Wiener integral is defined?



By googling, the best I've found are physics papers from the 70s and 80s that try to answer Question 1 in the affirmative for, e.g., $L$ a polynomial in $dotgamma,gamma$, or $L$ quadratic in $dotgamma,gamma$ plus bounded, or... Of course, most physics papers only treat $L$ of the form $frac12 |dotgamma|^2 + V(gamma)$.

Wednesday, 1 April 2009

Realizations and pinnings (épinglages) of reductive groups

OK, here's the deal.



I. First, the setup for the benefit of those who don't have books lying at their side. Let $(G,T)$ be a split connected reductive group over a field $k$, and choose $a in Phi(G,T)$ (e.g., a simple positive root relative to a choice of positive system of roots). Let $G_a$ be the $k$-subgroup generated by the root groups $U_a$ and $U_{-a}$. (Recall that $U_a$ is uniquely characterized as being a nontrivial smooth connected unipotent $k$-subgroup normalized by $T$ and on which $T$ acts through the character $a$.) This is abstractly $k$-isomorphic to ${rm{SL}}_ 2$ or ${rm{PGL}}_ 2$, and $G_a cap T$ is a split maximal torus.
So there is a central isogeny $phi:{rm{SL}}_ 2 rightarrow G_a$ (either isomorphism or with kernel $mu_2$), and since ${rm{PGL}}_ 2(k)$ is the automorphism group of ${rm{SL}}_ 2$ and of ${rm{PGL}}_ 2$ over $k$ there is precisely this ambiguity in $phi$ (via precomposition with its action on ${rm{SL}}_ 2$). The burning question is: to what extent can we use $T$ and $a$ to nail down $phi$ uniquely?



The action of $G_a cap T$ on $U_a$ is via the nontrivial character $a$, and among the two $k$-isomorphisms $mathbf{G}_ m simeq G_a cap T$ the composition with this character is $t mapsto t^{pm 2}$ in the ${rm{SL}}_ 2$-case and $t mapsto t^{pm 1}$ in the ${rm{PGL}}_ 2$-case. Fix the unique such isomorphism making the exponent positive.



Now back to the central isogeny $phi:{rm{SL}}_ 2 rightarrow G_a$. By conjugacy of split maximal tori (when they exist!) we can compose with a $G_a(k)$-conjugation if necessary so that $phi$ carries the diagonal torus $D$ onto $G_a cap T$. Recall that we used $a$ to make a preferred isomorphism of $G_a cap T$ with $mathbf{G}_ m$. The diagonal torus $D$ also has a preferred identification with $mathbf{G}_ m$, namely $t mapsto {rm{diag}}(t, 1/t)$. Thus, $phi: D rightarrow G_a cap T$ is an endomorphism of $mathbf{G}_ m$ with degree 1 or 2, so it is $t mapsto t^{pm 2}$ (${rm{PGL}}_ 2$-case) or $t mapsto t^{pm 1}$ (${rm{SL}}_ 2$-case). Since the standard Weyl element of ${rm{SL}}_ 2$ induces inversion on the diagonal torus, by composing with it if necessary we can arrange that $phi$ between these tori uses the positive exponent. That is exactly the condition that $phi$ carries the standard upper triangular unipotent subgroup $U^+$ onto $U_a$ (rather than onto $U_{-a}$).



II. So far, so good: we have used just $T$ and the choice of $a in Phi(G,T)$ to construct a central isogeny $phi_a:{rm{SL}}_ 2 rightarrow G_a$ carrying $U^+$ onto $U_a$ and $D$ onto $G_a cap T$, with the latter described uniquely in terms of canonically determined identifications with $mathbf{G}_ m$ (as $t mapsto t$ or $t mapsto t^2$). The remaining ambiguity is precomposition with the action of elements of ${rm{PGL}}_ 2(k)$ that restrict to the identity on $D$, which is to say the action of $k$-points of the diagonal torus $overline{D}$ of ${rm{PGL}}_ 2$. Such action restrict to one on $U^+$ that identifies $overline{D}(k)$ with the $k$-automorphism group of $U^+$ (namely, $k^{times}$ with its natural identification with ${rm{Aut}}_ k(mathbf{G}_ a)$).



Summary: to nail down $phi$ uniquely it is equivalent to specify an isomorphism of $U^+$ with $mathbf{G}_ a$. But $phi$ carries $U^+$ isomorphically onto $U_a$. So it is the same to choose an isomorphism of $U_a$ with $mathbf{G}_ a$. Finally, the Lie functor clearly defines a bijection $${rm{Isom}}_ k(mathbf{G}_ a, U_ a) simeq
{rm{Isom}}(k, {rm{Lie}}(U_ a)).$$
So we have shown that to specify a pinning in the sense of Definition A.4.12 of C-G-P is precisely the same as to specify a pinning in the sense of SGA3 Exp. XXIII, 1.1.



III. Can we improve a pinning to provide unambiguous $phi_c$'s for all roots $c$? No, there is a discrepancy of units which cannot be eliminated (or at least not without a tremendous amount of work, the value of which is unclear, especially over $mathbf{Z}$), and if we insist on working over $k$ and not $mathbf{Z}$ then there are further problems caused by degeneration of commutation relations among positive root groups in special cases in small nonzero characteristics (e.g., ${rm{G}}_ 2$ in characteristic 3 and ${rm{F}}_ 4$ in characteristic 2).
As we saw above, to nail down each $phi_c$ it is equivalent to do any of the following 3 things: fix an isomorphism $mathbf{G}_ a simeq U_c$, fix a basis of ${rm{Lie}}(U_ c)$, or fix a central isogeny ${rm{SL}}_ 2 rightarrow G_c$ carrying $D$ onto $G_c cap T$ via $t mapsto t$ or $t mapsto t^2$ according to the canonical isomorphisms of $mathbf{G}_ m$ with these two tori (the case of $G_c cap T$ being determined by $c$). This latter viewpoint provides $phi_{-c}$ for free once $phi_c$ has been defined (compose with conjugation by the standard Weyl element), so the problem is to really define $phi_c$ for $c in Phi^+$.



Consider the unipotent radical $U$ of the Borel corresponding to $Phi^+$, so $U$ is directly spanned (in any order) by the $U_c$'s for positive $c$. If we choose an enumeration of $Phi^+$ to get an isomorphism $prod_c U_c simeq U$ of varieties via multiplication, then for simple $c$ we have a preferred isomorphism of $U_c$ with $mathbf{G}_ a$ and one can ask if the isomorphism $mathbf{G}_ a simeq U_c$ can be determined for the other positive $c$ so that the group law on $U$ takes on an especially simple form. This amounts to working out the commutation relations among the $U_a$'s for $a in Delta$ (when projected in $U_c$ for various $c$), and by $T$-equivariance such relations will involve monomials in the coordinates of the $U_a$'s along with some coefficients in $k^{times}$ (and some coefficients of 0). These are the confusing "structure constants". Chevalley developed a procedure over $mathbf{Z}$ to make the choices so that the structure constants come out to be positive integers (when nonzero), but there remained some ambiguity of signs since



$${rm{Aut}}_ {mathbf{Z}}(mathbf{G}_ a) = mathbf{Z}^{times} = {pm 1}.$$



Working entirely over $k$, there are likewise $k^{times}$-scalings that cannot quite be removed. I am told that Tits developed a way to eliminate all sign confusion, but honestly I don't know a reason why it is worth the heavy effort to do that. For most purposes the pinning as above is entirely sufficient, and this is "justified" by the fact that all ambiguity from $(T/Z_G)(k)$-action is eliminated in the Isomorphism Theorem when improved to include pinnings (see Theorem A.4.13 in C-G-P).



IV. What about the realizations in the sense of Springer's book? All he's doing is making use of the concrete description of the group law on ${rm{SL}}_ 2$ to describe a central isogeny $phi_c$ for every positive root $c$. (His conditions relate $c$ with $-c$, which amounts to the link between $phi_c$ and $phi_{-c}$ in the preceding discussion.) As long as he restricts to $alpha in pm Delta$ then he's just defined a pinning in the above sense. But he goes further to basically do what is described in II but without saying so explicitly. He then has to confront the puzzle of the structure constants. (It is a real puzzle, since in the theory over $mathbf{Z}$, which logically lies beyond the scope of his book, the structure constants are not always $0$ and $pm 1$; in some rare cases one gets coefficients of $pm 2$ or $pm 3$, which implies that if one insists on working over fields and not over $mathbf{Z}$ then life in characteristic 2 and 3 will look a bit funny in some cases.) The only conceptual way I know of to overcome the puzzle of the structure constants is to work over $mathbf{Z}$ and to follow either SGA3 or Chevalley in this respect. For the former, one has to really do the whole theory of reductive groups over a base that is not necessarily a field. For the latter, perhaps the (unpublished?) Yale notes of Steinberg are the best reference.

ca.analysis and odes - Density of smooth functions under "Hölder metric"

This question came up when I was doing some reading into convolution squares of singular measures. Recall a function $f$ on the torus $T = [-1/2,1/2]$ is said to be $alpha$-Hölder (for $0 < alpha < 1$) if $sup_{t in mathbb{T}} sup_{h neq 0} |h|^{-alpha}|f(t+h)-f(t)| < infty$. In this case, define this value, $omega_alpha(f) = sup_{t in mathbb{T}} sup_{h neq 0} |h|^{-alpha}|f(t+h)-f(t)|$. This behaves much like a metric, except functions differing by a constant will differ in $omega_alpha$ value. My primary question is this:



1) Is it true that the smooth functions are "dense" in the space of continuous $alpha$-Hölder functions, i.e., for a given continuous $alpha$-Hölder $f$ and $varepsilon > 0$, does there exists a smooth function $g$ with $omega_alpha(f-g) < varepsilon$?



To be precise, where this came up was worded somewhat differently. Suppose $K_n$ are positive, smooth functions supported on $[-1/n,1/n]$ with $int K_n = 1$.



2) Given a fixed continuous function $f$ which is $alpha$-Hölder and $varepsilon > 0$, does there exist $N$ such that $n geq N$ ensures $omega_alpha(f-f*K_n) < varepsilon$?



This second formulation is stronger than the first, but is not needed for the final result, I believe.



To generalize, fix $0 < alpha < 1$ and suppose $psi$ is a function defined on $[0,1/2]$ that is strictly increasing, $psi(0) = 0$, and $psi(t) geq t^{alpha}$. Say that a function $f$ is $psi$-Hölder if $sup_{t in mathbb{T}} sup_{h neq 0} psi(|h|)^{-1}|f(t+h)-f(t)| < infty$. In this case, define this value, $omega_psi(f) = sup_{t in mathbb{T}} sup_{h neq 0} psi(|h|)^{-1}|f(t+h)-f(t)|$. Then we can ask 1) and 2) again with $alpha$ replaced by $psi$.



I suppose the motivation would be that the smooth functions are dense in the space of continuous functions under the usual metrics on function spaces, and this "Hölder metric" seems to be a natural way of defining a metric of the equivalence classes of functions (where $f$ and $g$ are equivalent if $f = g+c$ for a constant $c$). Any insight would be appreciated.