User blake - MathOverflow

Second homotopy of a torus complement in the 4-sphere

2013-02-14T03:58:17Z

Let $T$ be the boundary of a solid torus in $S^4$. Are there any theorems or methods which would help one to compute $\pi_2(S^4 -T)$? Or to say if, e.g., it had finite rank and no torsion?
More generally, suppose we have a closed manifold $X$ and we remove a submanifold $Y$ of at least codimension 2. Is there a general method for computing the homotopy groups of $X-Y$? The above case was the original example that got me to think about this but a more general result would also perhaps be helpful.

Line bundles, connections, and covariantly holomorphic sections

2012-09-11T03:01:49Z

I have a confusion regarding the line bundles arising in Kahler quantization for the torus. I know of course that the space of holomorphic sections should be isomorphic to a space of theta functions. However, there is a gap in my understanding of these bundles. Although a bit pedantic let me first specify the construction of these bundles which I am thinking of.

We may regard the torus as the quotient $\mathbb{C}/\Lambda$, where $\Lambda$ is a lattice generated by two complex numbers, and $\mathbb{C}$ has the standard symplectic form $dx\wedge dy$. Let us for concreteness take $\Lambda$ to be generated by 1 and $i$. Any bundle over the torus can be regarded as a quotient of $\mathbb{C}\times\mathbb{C}$ by an action of $\Lambda$, with $\lambda(z,s)=(z+\lambda,g(\lambda, z)s)$ where $g$ is a cocycle.

The line bundles over the torus are classified by their Chern class, and hence are classified by a level $k$. To obtain the bundle $L_k$ for level $k$ we take $g(\lambda)=e^{\pi ik\omega(\lambda, z)}$, where the hermitian form $h$ on $\mathbb{C}\times\mathbb{C}$ is given by $h((z,s),(z,s'))=s\overline{s'}$. Clearly the cocycle $g$ preserves the hermitian form.

A connection $D$ on $\mathbb{C}\times\mathbb{C}$ with curvature $-2\pi ik\omega$ may be taken to be of the form $D=d-2\pi ikzd\overline{z}$. Now the covariantly constant sections are represented by functions $f$ obeying $\partial_\overline{z}f=2\pi ikzf$. This has solutions $f(z,\overline{z})=\phi(z)e^{2\pi i kz\overline{z}}$ where $\phi$ is holomorphic.

However, for $|f|$ to be invariant under $\Lambda$, we must have $\phi$ invariant under $\Lambda$, since $|f|=|\phi|$. This is impossible unless $\phi$ is constant, as may be argued using the maximum modulus principle, for example. Therefore the space of covariantly holomorphic functions which descend to sections on $L_k$ appears to be 1-dimensional.

Obviously there must be a mistake somewhere - one should get something related to theta functions. But it is not clear to me why on these bundles $L_k$ constructed as above the space of covariantly holomorphic sections is one dimensional? What is the problem with this approach, and what is the correct way to approach this?

pairing theta functions for different complex structures

2012-08-31T15:10:15Z

I apologize for my previous attempt to ask this, which was very badly written.

Let us start with $\mathbb{C}\times\mathbb{C}$. To form an Hermitian line bundle over a complex torus with complex structure given by $\tau$ we may quotient $\mathbb{C}\times\mathbb{C}$ by the lattice generated by 1 and $\tau$, using an appropriate cocycle to define the action on the line bundle. Holomorphic sections of this line bundle will lift to linear combinations of theta functions on $\mathbb{C}$ (in an appropriate trivialization$. We can also define an inner product on the space of holomorphic sections in the usual way, by integrating the sesquilinear product of the two sections.

However, what if we want to integrate the sesquilinear product of two theta functions which come from different $\tau$? So, for example, if $\theta_1$ and $\theta_2$ are lifts of holomorphic sections in the $\tau_1$ and $\tau_2$ complex structures, is there a standard formula for the integral of $\theta_1 \overline{\theta_2}$ over the torus? What complicates this seems to be that the bundles (and the tori) are constructed by taking two different quotients, thus one must first find a unitary bundle diffeomorphism between them.

In particular, this integral defines a pairing between the holomorphic sections from the $\tau_1$ structure and the holomorphic sections coming from $\tau_2$. Does this pairing induce a linear map between the spaces of functions which is unitary, or projectively unitary? And, is this related to what I have seen called the 'Hermite-Jacobi action?'

Theta functions and Fourier transforms

2012-08-29T14:01:07Z

Let $T_\tau$ be the 2-dimensional torus, with the complex structure induced by the lattice generated by $1$ and $\tau$. Then for a line bundle $L_k$ over $T$ with level $k$, there is an orthonormal basis whose sections lift to theta functions on the complex plane.

My question is, what if we try to compare the Hilbert spaces obtained for different $\tau$? One method would be to look at the Hitchin connection. However, if we instead tried to use the Blattner-Kostant-Souriau (BKS) pairing between polarizations, would this just amount to doing a Fourier transform of sorts? In particular will this just give us something proportional to the Hermite-Jacobi action?

High-dimensional ribbon knots

2012-06-27T02:41:33Z

Let us suppose that we have a ribbon embedding $S^n \rightarrow S^{n+2}$ for $n\geq 3$. Call this knot $K$. By a theorem of Levine (and Trotter for $n=3$ I believe) we know $K$ is unknotted if the complement is homotopy equivalent to $S^1$. The question which I wonder is, does it suffice to show that the fundamental group is $\mathbb{Z}$, given the fact that $K$ is ribbon? For $n=2$ this would be true, as Freedman mentions (that having infinite cyclic fundamental group implies being homotopy equivalent to a circle) but I am not terribly familiar with the high-dimensional case.

Answer by Blake for When has pure mathematics been influenced by the social context of mathematicians?

2012-07-01T20:35:25Z

I think there's two related issues here. I've heard some sociological theories suggest that the answers we get in math are socially influenced. This is clearly mistaken, because math is a purely logical discipline: once you pick axioms (and a system of logic, to be pedantic), your answers are fixed. Similarly in the physical sciences, once you pick your experiment, you have no control over the outcome.
The exception to this of course is when people disregard the strict rules of logic. Example: philosophers used to believe that Euclidean geometry was somehow 'automatic' (see e.g. Kant). There was a mathematician whose name escapes me who came extremely close to developing non-Euclidean geometry, but, after proving a number of theorems about it, concluded that there was a contradiction because the system was 'absurd' or something like that. But this is an example of somebody doing math 'wrongly.' Math done 'right' has consequences that are not socially influenced.

On the other hand, there's the question of what we choose to study. That is obviously going to be influenced by social and personal factors. For example I recall some stuff about how the ancient Greeks liked to think geometrically, whereas the ancient Chinese liked to think algebraically (and I think the ancient Arabic mathematicians as well). You can see this in the types of discoveries that they made. There are studies out there on what sorts of differences made this happen; I think there's even someone who conjectures that Westerners still tend to think more geometrically, and Easterners more algebraically, or something like that.

Homology and homotopy type for knot complements

2012-06-30T07:00:12Z

I'm trying to understand a paper by Kawauchi and Matumoto, 'An estimate of infinite cyclic coverings and knot theory.' In one of their proofs they have a manifold $E$ which is the complement of a codimension-2 $n$-knot, with its infinite cyclic cover $\widetilde{E}$. They show that $\widetilde{H}_{*}(\widetilde{E};\mathbb{Z})=0$. Then they claim that because of this we must have that $E$ is homotopy equivalent to $S^{1}$.

It is clear that this shows $E$ is homologically equivalent to $S^1$, but why do we also get the result that it is homotopy equivalent?

Answer by Blake for High-dimensional ribbon knots

2012-06-27T16:45:12Z

I found a paper which discusses this issue: Ribbon knots and ribbon disks from Asano, Marumoto, and Yanagawa. They establish that for $n\geq 3$ a ribbon knot with infinite cyclic fundamental group is trivial.

http://ir.library.osaka-u.ac.jp/metadb/up/LIBOJMK01/ojm18_01_12.pdf

Symplectic structures from Lagrangians?

2012-06-08T00:44:15Z

In Witten's paper 'Quantization of Chern-Simons Gauge Theory with Complex Gauge Group,' he makes the statement (p. 35) that the symplectic form on the moduli space of flat $G_\mathbb{C}$ connections on a surface can be derived from the Chern-Simons Lagrangian and depends upon the coupling chosen.
Now I am familiar with the bracket on the moduli space of connections obtained via Goldman's construction, but what is this method for getting a symplectic form from a Lagrangian and where could I learn more about it? It sounds like this is a general construction, too, that could apply any time a phase space is derived from a configuration space, not just something that applies to this particular case?

Symplectic forms and 1-forms

2012-06-08T08:43:47Z

Suppose we have a real symplectic manifold $(M,\omega)$. Under what conditions can we find a global 1-form $\alpha$ such that $\omega = \alpha \wedge\alpha$?
Obviously there are some simple obstructions, for example, the cotangent bundle must admit a non-vanishing section (thus, surfaces of genus $>1$ are out). However it does not seem easy to come up with sufficient conditions.

Connections on line bundles over the torus

2012-04-19T02:27:58Z

If I understand correctly, every line bundle $L$ over the (2-dim) torus can be obtained from a quotient of $\mathbb{R}^2 \times \mathbb{C}$ by a $\mathbb{Z}^2$ lattice action. Different line bundles are obtained depending on how the lattice 'twists' the fibers. The connection $D = d + (1/2)(pdq-qdp)$ descends to a connection on $L$, as does the canonical Hermitian pairing.
My question is this: Suppose we wanted to geometrically quantize the torus by a Kähler polarization. Then we would want to find sections $s$ of $L$ such that $D_v s = 0$ for antiholomorphic $v$. How can we write down the explicit sections satisfying this equation in the case of the torus with a Hermitian bundle $L$ as above? And how can we prove that for a general compact manifold the dimension of the space of solutions is finite? (I know how to solve the differential equation for the case of $\mathbb{R}^2 \times \mathbb{C}$ and get the Bargmann-Fock space, but it is unclear how one can solve the analogous problem for compact manifolds - at least unclear to me)

BKS pairing in the SU(2) Chern-Simons theory

2012-04-25T06:13:50Z

I know that usually, the way to compare the Hilbert spaces arising from $SU(2)$ Chern-Simons theory with different Kähler polarizations is via the Hitchin connection. However, it should be possible, I would think, to try to use a BKS pairing between them. Are any results known on whether this pairing is unitary? (My guess would be that the answer is no, based upon the results of Kirwin et al for toric varieties, but then the character varieties that are quantized in Chern-Simons theory aren't really toric varieties).
Likewise for real polarizations, I know Jeffrey and Weitsman did some work on BKS pairings but I don't recall seeing any theorems in their papers about whether the pairing was unitary for real polarizations in general. Has there been further work done on the pairings for real polarizations?

SL(2,C) Chern-Simons theory in genus 1

2011-11-02T14:55:03Z

In http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.cmp/1104202513, Witten claims (p. 54) that to quantize the moduli space of flat $SL(2,\mathbb{C})$ connections on a torus, one can simply quantize the cotangent bundle of a real torus and take the part invariant by the Weyl group $W$ .
I agree that $T^*(T\times T)/W$ is the moduli space of flat connections on the torus. However, it seems to me that the correct symplectic form should be the following: if we parametrize $T\times T$ with coordinates $(r_1, \theta _1, r_2, \theta _2)$ (quotienting the $\theta$ coordinates appropriately) then the symplectic form $\omega$ should be induced by $dr_1\wedge dr_2+d\theta _1 \wedge d\theta _2$. This restricts to the correct symplectic form on the real torus $T\times T$ obtained by taking the $r_1 = r_2 = 1$ subspace (which appears to be the $SU(2)$ character variety with its correct symplectic structure). But this is not the symplectic form induced by the cotangent structure, which would be $\omega'=d\theta _1 \wedge dr_1 + d\theta _2 \wedge dr_2$. On the other hand $\omega'$ vanishes on the real torus corresponding to $SU(2)$ representations, and does not seem to be the form induced by the character variety, which comes equipped with a natural symplectic form.
This makes a substantial difference in quantization because $\omega'$ is exact and hence we can take the trivial line bundle for prequantization, and then in a real polarization obtain a Hilbert space $L^2(T\times T)$ (which is what Witten claims to be the quantization of the moduli space). However, if we take $\omega$ as the symplectic form, $\omega$ is non-zero in cohomology, and we will end up with a more complicated quantization.

complex vector bundles and curvature

2012-02-24T05:07:08Z

Let us suppose that $X$, with a 2-form $\omega$. Suppose $J$ is an element of $su(2)$ such that $J^2=-e$ for $e$ the identity. Is there a necessary and sufficient condition on $\omega$ which will give the existence of $E\rightarrow X$ a $SU(2)$ bundle with a connection whose curvature is $J\omega$?

BKS pairing for distributional sections

2012-02-01T03:14:38Z

I am trying to understand the Blattner-Kostant-Sternberg pairing as it applies to geometric quantization in real polarizations whose integral manifolds are, for simplicity, compact. I have been trying to follow Sniatycki's account (Geometric Quantization and Quantum Mechanics, pp 73-75). Thus, we have a symplectic manifold $(X,\omega)$, of dim $2n$ a Hermitian bundle $L$ with connection $\nabla$ whose curvature is $\omega$.
We choose a polarization D by Lagrangian submanifolds, and a bundle of half-forms. We obtain the Bohr-Sommerfeld variety S consisting of points lying in integral submanifolds of D with trivial holonomy. This will be a finite collection of tori $S_i$ if $X$ is compact and $D$ is sufficiently nice. Now we wish to define a pairing on them.
In Sniatycki's account we project each $S_i$ to a $Q_i$ in $X/D$. We wish to obtain a density on each $Q_i$. He does this by lifting a basis for $T_x(Q_i)$ and using $\omega$ to obtain a basis for $S_i$ from this, on which the half-form has a well-defined value.
This is fine so long as $Q_i$ is of dimension $n$. But it seems to me that in many cases, e.g. a torus, the dimension will actually be 0, in which case the prescription breaks down. Obviously I am missing something. Is there a source that discusses this issue in more depth?

Line bundles with complex connection

2012-01-22T05:25:05Z

Suppose that we have a complex manifold $X$, and a line bundle $L$ over $X$. It is known that the line bundles over $X$ are parametrized by their Chern class, the Chern class being the cohomology class of the curvature of a connection on $L$, which must be integral. Thus $L$ admits a connection with curvature $\omega$ iff $[\omega]$ is an integral cohomology class.
My question is this: what if we are interested instead in complex connections on $L$ (connections that are $\mathbb{C}$-linear instead of $\mathbb{R}$-linear)? Given a complex (2,0)-form $\omega$ on $X$, under what circumstances will $L$ admit a complex connection with curvature $\omega$?

2-tangles and quantum groups and 2-groups

2011-11-19T09:39:01Z

Turaev developed the notion of a quantum group by considering the category of tangles (thought of with objects as collections of 2$n$ points and with morphisms being braids between them with cups and caps) and considering the algebraic version of this.
I see to recall that someone has developed a notion of a quantum 2-group, or something like that, by considering the 2-category defined by 2-link movies. However, I have no recollection of where I saw this (I think it was a dissertation or something). It also seems that the 2-knot category, considered as a standard category, should define some family of algebraic objects - has this been studied, and if so where should I look for information on this?

Deriving the Hilbert spaces for Chern-Simons TQFTs with complex gauge group

2011-10-31T23:58:37Z

One method for finding the Hilbert spaces corresponding to surfaces in Chern-Simons TQFT is by geometrically quantizing the phase space, which is just the character variety of the surface. I know that this has been done by Anderson and others in the holomorphic polarization and by Jeffrey and Weitsman in the real polarization when the gauge group is $SU(2)$.
Is there a derivation of the state spaces for Chern-Simons TQFTs with gauge group $SL(n,\mathbb{C})$ that uses geometric quantization? Equivalently, is there a geometric quantization of the $SL(n,\mathbb{C})$ character varieties of surfaces? Or at least for $n=2$?

A Follow-up About Connection Forms on Principal Bundles

2011-09-21T23:52:33Z

In this question I asked about proving that a connection form $\alpha$ on a $\mathbb{C}^*$ bundle had to have $2\pi i(\alpha - \overline{\alpha})$ be exact. From the answer to that question I understand why on any local trivialization, this quantity will be exact. However, this would ordinarily only imply global exactness if the 1st cohomology is trivial. I am probably being very dense, but how do we show that the global condition follows?

Principal bundle connections

2011-09-20T16:44:54Z

It is my understanding that a connection form on a principal G-bundle over a manifold X is defined to be a Lie algebra-valued 1-form $\alpha$ which reproduces the Lie algebra generators of the fundamental vector fields at every point. That is, if at $p$ a vector field $v$ takes the value $\frac{d}{dt}(exp(tg)p) |_{t=0}$ , then $\alpha(v)|_p = g$ .
I have noticed, however, that the fundamental vector fields defined for $\mathbb{C}^*$ bundles in Sniatycki's book on geometric quantization are not defined as $\frac{d}{dt}(exp(tg)p) |_{t=0}$ but rather as $\frac{d}{dt}(exp(2\pi itg)p) |_{t=0}$ , where $g$ is any complex number.
QUESTIONS:
1. Is my initial definition of a connection form correct?
2. Assuming the answer is yes, what will the relationship be between connection forms defined in the two different conventions?
3. Apart from the fact that Sniatycki's convention makes the real line generate the fundamental vector fields corresponding to the flow of the $U(1)$ action (if I am not in error here), is there some additional benefit to this convention? It appears to make $2\pi i$'s show up everywhere.

Connections with compatible Hermitian products on complex line bundles

2011-09-19T13:39:39Z

Let $X$ be a manifold, $L$ be a complex line bundle over $X$, and $L^{*}$ be the associated principal bundle. Suppose $\alpha$ is a connection form on $L^{*}$ , with associated connection $D$ on $L$. Suppose we wish to define a Hermitian product $\langle,\rangle$ on $L$ compatible with $D$ in the sense that for every pair of sections $s, s'$ and vector field $v$, we have $v\langle s,s'\rangle=\langle D_{v}s,s'\rangle+\langle s,D_{v}s'\rangle$ . According to Sniatycki's book 'Geometric Quantization,' such a pairing exists iff $2\pi i(\alpha - \overline{\alpha})$ is exact. However, I cannot figure out a proof of this, and he does not provide one.

Does anyone know where I can find a proof of this, or could anyone provide the proof here?

Comment by Blake

2012-07-03T16:00:42Z

Interpretations of physical theories are really debates about which set of axioms to use. There are further problems with, e.g., forgetting to account for an experimental variable. As an undergraduate I spent a great deal of time studying the experimental mistakes people made trying to measure orthopositronium's halflife... Anyway sure, claiming there's a national/cultural difference in mathematical ability is nonsense, but I think there's probably something to the claim that historically the development of mathematics proceeded along those sorts of lines, until the 'merging' of the two.

Comment by Blake

2012-07-02T11:23:28Z

You're probably right about the anachronisms. I read a description of some guy's research on this, not the research itself, so take it with a grain of salt. In my interactions I've known plenty of Asian geometers/topologists and plenty of Caucasian algebraists. As to the other point, that's what I meant by my second paragraph. Once the assumptions are chosen, your answers are fixed, but culture can play a huge role in which axioms you choose to take (in other words, what math you choose to study). So I stand by my confident assessment:)

Comment by Blake

2012-07-01T01:46:44Z

Oh, of course! I forgot the Hurewicz theorem applies to higher homotopy groups, and was just thinking about it in its application to the abelianization of the first homotopy group. Thanks!

Comment by Blake

2012-06-30T21:50:10Z

Ok - the point I am missing is: why do we get that $\pi_{2}(\widetilde{E})=0$ just from the triviality of $\pi_1$ and $\widetilde{H}(\widetilde{E};\mathbb{Z})$? Does this come from a long exact sequence of some sort? Because otherwise the second homology could be zero, without the second homotopy group vanishing. Then likewise once we know the first $n$ homotopy groups vanish, how does the triviality of the homology groups imply the next homology group vanishes?

Comment by Blake

2012-06-30T16:26:25Z

I was very careless and missed the crucial assumption that Aru Ray has filled in... I guess staring at the paper for hours had me taking it for granted.

Comment by Blake

2012-06-30T12:50:50Z

I'm not terribly familiar with all that has been done with large cardinals in logic, but I would guess the answer probably involves those set theories that deal with proper classes. The idea would be that V is just a class (not a set) and you can now start doing things with classes that you would with sets. You should run into exactly analogous hierarchies if you do, to the best of my knowledge. But informally speaking these 'classes' just behave like a further level of sets. However I might be missing some important details here.

Comment by Blake

2012-06-28T12:33:21Z

On the other hand Suciu claims that their result is based on an unproved result erroneously attributed to Lomonaco. maths.ed.ac.uk/~aar/papers/suciu1.pdf

Comment by Blake

2012-06-27T02:34:25Z

This question was incredibly badly phrased, for which I apologize. I meant to ask in the case where we tensor the exterior algebra with something nonabelian. But even then as Gal pointed out the answer is easy.

Comment by Blake

2012-04-25T19:31:37Z

Yes, sorry, BKS is Blattner-Kostant-Sternberg. They two constructions cannot be exactly equal. Hitchin depends on a path through Teichmuller space (although it is projectively path-independent). BKS doesn't, so they are at most projectively equivalent.

Comment by Blake

2012-04-20T16:01:47Z

I meant $(z,t)$ is identified with $(z+a+bi, T(a,b)t)$ of course, and in fact I think $T$ might depend on $z$ as well. But regardless it needs to be unitary to preserve the Hermitian form.

Comment by Blake

2012-04-20T15:59:55Z

Thank you. I'm stuck on one point: suppose I choose $A_1 = A_2 = 1/2$. Then on the complex plane a holomorphic section has the form $f(z)=h(z)exp((1/2)|z|^2)$, where $h$ is holomorphic in the usual sense. In order to define a section of the torus, however, we need to have $f(z+a+bi)=T(a,b)f(z)$ where T is always unitary (In particular, the bundle on the torus is obtained as $\mathbb{C}\times\mathbb{C}/(z,t)~(z+a+bi, T(a,b)t)$, and T must be unitary to preserve the Hermitian pairing). I don't see how we can choose h(z) to obtain such a condition - I also don't see why theta functions help?

Comment by Blake

2012-03-03T13:22:48Z

I see your point, however, in the problem I am considering I start with a real-valued $\omega$ on $X$ and wish to find a bundle with curvature $J\omega$ where $J$ is a section of the endomorphism bundle which is, at each point at least, a primitive 4th root of unity; if I'm not mistaken, this does imply the endomorphism bundle is trivial. So for my specific problem the assumption is natural.

Comment by Blake

2012-02-28T08:14:00Z

Yes, but I meant to ask for the case where the holomorphic structure is inherent to the manifold, and is not defined by the connection.

Comment by Blake

2012-02-28T08:12:43Z

That should read $d\alpha + \alpha \wedge \alpha = 2J\omega$ so we just need to normalize $\alpha$, but $d\alpha = 0$.

Comment by Blake

2012-02-28T08:11:33Z

I don't think $\omega^2\neq 0$ is enough for this. Let $\omega = \sum dp_i \wedge dq_i$ on $\mathbb(R)^n$; then $\omega ^2\neq 0$ but we can take the connection on a trivial bundle given by the potential $\alpha = Kdp_i + Idq_i$ where $I,J,K$ are the matrices corr. to the $i,j,k$ quaternions. Then $d\alpha + \alpha \wedge \alpha$ = 2J\omega$ so we just need to normalize it to get $J\omega$, but $d\alpha = 0$.