User alexander moll - MathOverflow

Evaluating an infinite product of q-exponentials

2013-03-19T03:11:42Z

For the $q$-exponential $$e_q(u) = \sum_{n=0}^{\infty} \frac{u^n}{[n]_q!}$$ with $[k]_q=\frac{1-q^k}{1-q}$ and $[n]_q! = [n]_q [n-1]_q \cdots [1]_q$, we don't have the property $e_q(u) e_q(v) = e_q (u+v)$. I learned this the hard way when trying to compute the infinite product $$\prod_{i=0}^{\infty} e_q (x q^i)$$ where $x$ is an indeterminate.

Does anyone see a way to get a closed form expression for this product?

Any help would be greatly appreciated!

Two appearances of the Jacobi Triple Product identity

2013-02-26T20:34:16Z

So far, I've seen two ways of deriving the Jacobi Triple Product (JTP) formula $$\sum_{m \in \mathbb{Z}} (-u)^m q^{\frac{m(m-1)}{2}}= \prod_{j=0}^{\infty} (1-uq^{j})(1-q^{j+1})(1-u^{-1} q^{j+1})$$ using infinite-dimensional representation theory. On the one hand, it is the Weyl-Kac denominator formula for $\widehat{\mathfrak{sl}_2}$. On the other hand, it is the character $$\text{Tr}(q^{L_0} u^{\alpha_0})$$ computed in two ways using the boson-fermion correspondence. Here $$\alpha_0 = \sum_{i \in \mathbb{Z}'} : \psi_i \psi_i^* :$$ is the zero mode of the Heisenberg algebra which computes the charge of the Dirac sea of free fermions, and $$L_0 = \sum_{i \in \mathbb{Z}'} i : \psi_i \psi_i^* :$$ is the energy operator for a particular representation of the Virasoro algebra on this semi-infinite wedge space.

Is there a structural relationship between these two appearances of JTP?

Bochner's Theorem and Total Positivity

2012-04-01T19:17:01Z

Bochner's Theorem for LCA groups applied to the case of $G = U(1)$ and $G^{\vee} = \mathbb{Z}$ tells us that through the Fourier transform, probability measures on the circle are in bijection with infinite positive semidefinite matrices with $1$s along the main diagonal. In the case of finite $N \times N$ matrices, we know from convex analysis that these are correlation matrices. Indeed, this corresponds to the case $G = \mathbb{Z} / N \mathbb{Z} \hookrightarrow U(1)$ of $N$th roots of unity with its dual group $\mathbb{Z} \twoheadrightarrow \mathbb{Z} / N \mathbb{Z} = G^{\vee}$.

Concretely, every principal minor of a positive semidefinite matrix has non-negative determinant. If a matrix satisfies the stronger condition that every minor has non-negative determinant, we call it a totally positive matrix.

Is there some nice condition on a positive semidefinite matrix which guarantees it is totally positive?
Which probability measures on the circle correspond to infinite totally positive matrices with $1$s on the main diagonal?

Recovering the Alexander Polynomial from Ocneanu's HOMFLYPT

2011-10-03T06:14:50Z

Let $H_q(n)$ denote the Hecke algebra associated to the symmetric group $S(n)$: this is the $\mathbb{Z}[q^{\pm 1/2}]$ algebra generated by $T_1, \ldots, T_{n-1}$ satisfying the braid relations along with $T_i^2=(q-1)T_i + q$. Ocneanu's trace $\tau_z (T_{\omega_{\mu}}) = z^{l(\omega_{\mu})}$ defined on fundamental elements is the unique normalized trace on our Hecke algebra that can be jiggled to yield invariants of oriented links: this gives the two-parameter HOMFLYPT polynomials $P_L(q,z)$. These polynomials satisfy the skein relation

$$ \Big ( \frac{z}{z-q+1} \Big )^{1/2} P_{L_+} (q,z) - \Big ( \frac{z}{z-q+1} \Big )^{-1/2} P_{L_-}(q,z) = (q^{1/2} - q^{-1/2}) P_{L_0} (q,z).$$

which motivates the common change of variables $x = \sqrt{\frac{z}{z-q+1}}$, $y = q^{1/2} - q^{-1/2}$. Note that the target ring of this trace has to be at least $\mathbb{Z}[q^{\pm 1/2}, z^{\pm 1/2}, (z-q+1)^{ \pm 1/2}]$ if we want to write down the associated invariant $P_L(q,z)$.

Many people take the HOMFLYPT polynomials to be those obtained after the specialization $z=q^{N}/[N]$, which seems to be equivalent to $x=\sqrt{\frac{z}{z-q+1}} = q^{N/2}$. Setting $N=2$ recovers the Jones polynomial, and setting $N=0$ is supposed to recover the Alexander polynomial.

How am I supposed to correctly obtain the Alexander polynomial in terms of the original $q$ and $z$?

It seems that $x=q^{0/2}=1$ gives the correct specialization. Still, doesn't $\sqrt{\frac{z}{z-q+1}}=1$ force $q=1$, leaving $z$ free? How does the right-hand side of the skein relation survive then?

Two curious asymptotic results for dimensions of type A objects

2012-03-31T06:35:33Z

Let $V_{\lambda}$ and $W_{\lambda}$ be the irreducible representations of $S(n)$ and $\mathfrak{su}(N,\mathbb{C})$ associated to the partition $\lambda \in \mathbb{Y}$ of size $| \lambda |=n$ and length $l(\lambda) \leq N$. The following limit $$\frac{\dim V_{\lambda}}{n!} = \lim_{N \rightarrow \infty} \frac{\dim W_{\lambda}}{N^{n}}$$ follows immediately from the well known hook (content) formulas $$\dim V_{\lambda} = \prod_{\square \in \lambda} \frac{n!}{h(\square)} \ \ \ \ \dim W_{\lambda} = \prod_{\square \in \lambda} \frac{N + c(\square)}{h(\square)}$$ which can be found in Macdonald's book. Notice that $n! = \dim_{\mathbb{C}} \mathbb{C}[S(n)]$ and $N^n = \dim_{\mathbb{C}} (\mathbb{C}^N)^{\otimes n}$, so what we're seeing is that as $N \rightarrow \infty$, the relative multiplicity of $V_{\lambda}$ in Schur-Weyl duality approaches the relative multiplicity of $V_{\lambda}$ in the regular representation.

Does anyone have a good feeling for why this is true?

Also, let us not forget the Peter-Weyl theorem! If for a compact group $G$ we write $G^{\vee}$ for its set of finite dimensional irreducible representations over $\mathbb{C}$, we have $$L^2(SU(N)) = \widehat{\bigoplus_{\lambda \in SU(N)^{\vee}}} W_{\lambda} \boxtimes W_{\lambda} $$ $$(\mathbb{C}^N)^{\otimes n}=\bigoplus_{\lambda \in SU(N)^{\vee} \cap S(n)^{\vee}} V_{\lambda} \boxtimes W_{\lambda} $$ $$\mathbb{C}[S(n)] = \bigoplus_{\lambda \in S(n)^{\vee}} V_{\lambda} \boxtimes V_{\lambda}$$

The limit we discussed above relating the second to the third line here actually also happens when we pass from the first to the second line: the ``relative multiplicity'' of $W_{\lambda}$ in its regular representation approaches the relative multiplicity of $W_{\lambda}$ in Schur-Weyl duality.

Can anyone give me some intuition for what's going on here + why I might expect such a result?

Whitehead products and a realization problem for graded Lie algebras

2012-02-16T16:15:09Z

Many $\mathbb{Z}$-graded Lie algebras $\mathfrak{g}$ over $\mathbb{C}$ we would like to study are non-degenerate in the sense that

$\dim_{\mathbb{C}} \mathfrak{g}_n < \infty \ \forall n \in \mathbb{Z}$
$\mathfrak{g}_0$ is abelian
For generic $\lambda \in \mathfrak{g}_0^*$, $\forall n \in \mathbb{N}$ the bilinear form $\mathfrak{g_n} \otimes \mathfrak{g}_{-n} \rightarrow \mathbb{C}$ given by $$a \otimes b \mapsto \lambda ([a,b])$$ is non-degenerate.

This includes simple Lie algebras, Loop algebras, affine Kac-Moody algebras, the Heisenberg algebra, the Witt algebra, and the Virasoro algebra.

For any of these examples, can we find some topological space $X$ such that $\pi_{*+1}(X) \cong \mathfrak{g}_{\geq 0}$ under the Whitehead bracket?

Coincidences amongst classifying spaces and Eilenberg Mac-Lane spaces

2011-11-16T04:17:50Z

Given that $$\mathbb{R}P^{\infty} = B O(1) = K(\widehat{O(1)}, 1)$$ $$\mathbb{C} P^{\infty} = B U(1) = K( \widehat{U(1)}, 2)$$ is there any way to make sense of $$\mathbb{H}P^{\infty} = B Sp(1)$$ in a similar manner using the representation theory of the non-abelian group $Sp(1) \cong Spin(3) \cong SU(2)$?

Taylor expansion of a q-analog of the negative binomial distribution

2011-11-16T00:40:17Z

Given $A,B \in \mathbb{Z}_+$ and $ 0 < t, q< 1$, I'd like to compute the coefficients $c_n(q,A,B)$ in the expansion of the product $$\prod_{i=0}^{A-1} \prod_{j=0}^{B-1} \frac{1}{1-t q^{i+j}} = \sum_{n=0}^{\infty} c_n t^n.$$ As $q \rightarrow 1$, this returns the well known formula $$\frac{1}{(1-t)^{AB}} = \sum_{n=0}^{\infty} \binom{AB+n-1}{n} t^n$$ which has a quick enumerative proof.

So far I've determined that the highest power of $q$ in $c_n(q,A,B)$ is $(A-1)(B-1)n$ less than the highest power of $q$ in the $q$-binomial coefficient $\binom{AB+n-1}{n}_q$. Can anyone see what these $c_n$ count in the expansion of the product? Any help would be greatly appreciated!

Does there exist a complex Lie group G such that ...

2011-10-12T01:50:50Z

... every Riemann surface of genus $1$ appears as a complex one-parameter subgroup of $G$?

Analytic Characterization of Parallel Transport of Fundamental Groups

2011-09-28T03:43:35Z

(Note that I've edited the main body of the question to make it clear for other readers.)

Fix a principal $G$-bundle $\rho: P \rightarrow X$ and fix a point $p \in P_x$ in the fiber above $x \in X$. If $\rho$ is equipped with a flat connection $\omega$, we get a surjective homomorphism $\pi_1(X,x) \rightarrow \text{Hol}_p(\omega)$.

When is the map $\pi_1(X, x) \rightarrow \text{Hol}_p(\omega) \hookrightarrow G \rightarrow \text{Inn}(G)$ surjective?

The motivation for this question comes from the fact that such a composition is surjective in the case of ``parallel transport'' of fundamental groups in $X$ along curves changing basepoints.

Two ways of generalizing factorials via symmetric groups

2011-09-20T21:28:26Z

By the Bruhat decomposition of $GL(n, \mathbb{F}_q) / B_n$ we know that $$[n]! = \sum_{ \sigma \in S(n)} q^{l(\sigma)}$$ where $[n]! = \prod_{j=1}^n (1+q + \cdots + q^{j-1})$ and $l(\sigma)$ is the length of the permutation $\sigma \in S(n)$ (also known as the number of involutions of $\sigma$).

We also know that $$\theta^{(n)} = \sum_{\sigma \in S(n)} \theta^{[\sigma]}$$ where $[\sigma]$ is the number of cycles of the permutation $\sigma \in S(n)$ and $\theta^{(n)} = \theta(\theta+1) \cdots (\theta+n-1)$. Notice that $$\lim_{q \rightarrow 1} \ [n]! = n! = \lim_{\theta \rightarrow 1} \ \theta^{(n)}.$$ Is there a way to write $$\sum_{\sigma \in S(n)} q^{l (\sigma)} \theta^{[\sigma]}$$ explicitly as a function of $q$ and $\theta$?

Infinitely Divisible Distributions and Maximal Entropy

2011-06-18T06:54:22Z

The normal distribution on $\mathbb{R}$, the exponential distribution on $\mathbb{R}_{\geq 0}$, and the geometric distribution on $\mathbb{N}$ are examples of distributions that are both infinitely divisible and entropy maximizers. On the other hand, the Poisson distribution is an infinitely divisible distribution on $\mathbb{N}$ without maximizing entropy, while the uniform distribution on the interval $[a,b]$ maximizes entropy but is not infinitely divisible.

Can anything be said about the relationship between these two classes of distributions?

Sums of Partitions and Stirling's formula

2011-07-18T01:08:20Z

Stirling's formula $$N! \sim \sqrt{2 \pi}\ N^{N+ \frac{1}{2}} e^{-N}$$ follows easily from Laplace's method in light of the famous integral representation $$N! = \int_0^{\infty} e^{-z} z^N dz.$$ Basic representation theory of the symmetric group $S(N)$ gives the remarkable finite sum $$N! = \sum_{\lambda \in \mathbb{Y}_N} (\dim \lambda)^2$$ over all partitions of $N$, where $\dim \lambda$ counts the number of paths from $\emptyset$ to $\lambda$ in Young's lattice $\mathbb{Y}$.

Is it possible to derive Stirling's formula directly from this finite sum?

[Edit] I'd like to thank everyone for the very interesting comments below. For those looking at this thread for the first time, I was hoping that an answer to the question above might help us calculate the asymptotics of $$f(N,a) = \sum_{\lambda \in \mathbb{Y}_N} (\dim \lambda)^{a}.$$ Notice that the case $a=0$ is the partition function $|\mathbb{Y}_N|$.

Continuous Measurement in Quantum Mechanics

2011-07-19T05:12:40Z

Let $\mathcal{P}(S^{\infty})$ denote the set of probability measures on the unit sphere $S^{\infty} \subset \mathcal{H}$ in the Hilbert space of states of a quantum mechanical system. Measurement of an observable $\Omega$ corresponds to orthogonal projection, sending $\delta_{|\psi \rangle}$ to a particular measure supported on the eigenvectors of $\Omega$, thus inducing a map $T_{\Omega}: \mathcal{P}(S^{\infty}) \rightarrow \mathcal{P}(S^{\infty})$. If we think of $T_{\Omega}$ as the transition matrix of a Markov chain, we can say that a continuum $\lbrace \Omega_t \rbrace_{t \in \mathbb{R} \geq 0}$ of observables induces a stochastic process on $S^{\infty}$.

If we let $\Omega_t = H(t)$, the time-dependent Hamiltonian of our system, is the associated stochastic process the deterministic one described by the Schrödinger equation?
What does this construction have to do with the time-energy uncertainty relation?

Probabilistic Solution of the Porous Medium Equation

2011-07-03T04:17:58Z

It is well known that the transition density for standard Brownian motion $B_t$ in $\mathbb{R}^d$ yields a solution to the global Cauchy problem for the heat equation $$u_t = \Delta u$$ with initial condition given by the Dirac distribution $\delta_0$.

Unlike the heat equation, the porous medium equation $$u_t = \Delta(u^m)$$ with exponent $m>1$ has finite speed of propagation.

When is the infinitesimal generator of a stochastic process linear?
Is there a probabilistic solution of this non-linear diffusion equation?

Least-squares regression and differential geometry

2011-06-05T23:56:30Z

For $k, n \in \mathbb{N}$, let $\mathcal{C}_n \mathbb{R}^k$ denote the configuration space of $n$ distinct points in $\mathbb{R}^k$.

(1) Is there a description of the tangent space $T_C \mathcal{C}_n \mathbb{R}^k$ in terms of the configuration $C$?

Equipping $\mathbb{R}^k$ with the usual metric, a regression line $l_C$ of a collection $C \in \mathcal{C}_n \mathbb{R}^k$ is a line minimizing the quantity $E_C = \sum\limits_{p \in C} d(p, l_C)^2.$ We can see this as variational problem $E_C: M^k \rightarrow \mathbb{R}$ where $M^k$ is the parameter space of all lines in $\mathbb{R}^k$.

(2) Is there an explicit parametrization of $M^k$?

Without this knowledge, I'm not sure how to proceed to check whether $E_C$ is a Morse function.

[Note for $k=2$: given $C \in \mathcal{C}_n \mathbb{R}^k$, since $n<\infty$ we can always find an angle $\theta$ such that a rotation of our axes by $\theta$ yields coordinates $(x,y)$ on $\mathbb{R}^2$ for which the $x$-values of the $p \in C$ are all distinct, bringing us back to function-fitting and the usual least-squares regression which minimizes only the distances in the $y$ direction.]

On the decomposition of two representations of the Iwahori-Hecke algebra of type A_n

2011-03-24T16:00:47Z

Consider the Iwahori Hecke algebra $H_q(n)$ of the symmetric group $S(n)$, the $\Bbb{Z}[q^{1/2},q^{-1/2}]$ algebra with generators $T_i$, $1 \leq i \leq n-1$, subject to the braid relations and the quadratic relation $T_i^2 = (q-1)T_i + q$.

(i) Let $V_k$ be the defining representation of the quantum super-group $U_{q,k,m}:=U_q(\mathfrak{gl}(k+m|m))$. The commutant of the diagonal action of $U_{q,k,m}$ on $V_k^{\otimes n}$ is generated by the representation of $H_q(n)$ on this space $\forall k, m \in \Bbb{N}$.

(ii) Consider the left action of the group of invertible $n \times n$ matrices $GL(n, q)$ with coefficients in $\mathbb{F}_q$ on the full flag variety $X(n) := GL(n,q) / B$ and take the corresponding representation in the vector space $L[X(n)]$ of functions (or measures) on $X(n)$. The space of intertwiners of this representation of $GL(n,q)$ is isomorphic to $H_q(n)$.

Q1) How does the representation in (ii) decompose into irreducible representations of $GL(n,q)$?

Q2) Is there any relationship between $GL(n,q)$ and $U_{q,k,m}$?

On the cohomology ring of the Grassmannian

2010-07-03T16:38:44Z

The basis of Schubert classes for the cohomology ring $H^*(\text{Gr}(m,N))$ of the Grassmannian of $m$-dimensional subspaces of $\mathbb{C}^N$ is indexed by $L(m,N-m)$, the poset of all partitions fitting inside a $m \times (N-m)$ box. This is the quotient of the powerset $2^{m(N-m)}$ by the action of the wreath product $S(m) \wr S(N-m)$. How does this come from the fact that $\text{Gr}(m,N) \cong U(N) / \big (U(m) \times U(N-m) \big )$? Can this be extended to other homogeneous spaces?

Comment by Alexander Moll

2013-03-19T15:28:06Z

Hey Ben - use the last comment in en.wikipedia.org/wiki/Q-exponential

Comment by Alexander Moll

2013-03-19T03:32:46Z

Hopefully something else - but you're right that for a certain value of $x$ as a function of $q$ this will return the MacMahon generating function for plane partitions.

Comment by Alexander Moll

2013-03-15T04:12:26Z

Oh good point! It seems I didn't scroll down far enough when I searched "plane partitions" within MO.

Comment by Alexander Moll

2013-02-26T23:05:23Z

Also, I'm hoping to use this observation as an opportunity to start investigating such papers - I'm just not sure where to begin!

Comment by Alexander Moll

2013-02-26T22:57:16Z

The second approach is essentially the proof given in en.wikipedia.org/wiki/Jacobi_triple_product

Comment by Alexander Moll

2012-04-03T22:27:03Z

This is a very nice argument Hugh - thank you!

Comment by Alexander Moll

2012-04-01T00:16:51Z

i.e. is there a ``bijective'' proof of the limit above?

Comment by Alexander Moll

2012-04-01T00:16:16Z

Sorry - I'm familiar with this approach to calculating the limit, and should have included it in my original post, where I was aiming for more of a representation theoretic "Why?". Maybe we can ask this question combinatorially: is there a reason why the number of semi standard Young tableaux of shape $\lambda$ with entries from $\{1, \ldots, N\}$ after dividing by $N^n$ is asymptotic to the number of standard Young tableaux with shape $\lambda$ if we divide by $n!$?

Comment by Alexander Moll

2012-03-31T07:09:59Z

Also, something I really want is some representation theoretic place where I might find a direct sum of $W_{\lambda, N} \boxtimes W_{\lambda, M}$ over all $\lambda \in \mathbb{Y}$ with length $l(\lambda) \leq \min (N,M)$ - a sort of ``unbalanced'' Peter-Weyl theorem for $SU(N)$ on the left and $SU(M)$ on the right. Any thoughts?

Comment by Alexander Moll

2011-11-18T05:42:45Z

I like Prof. Teschl's notes mat.univie.ac.at/~gerald/ftp/book-ode/index.html

Comment by Alexander Moll

2011-11-16T15:35:54Z

Thanks for the reference, Mark - that book looks really cool!

Comment by Alexander Moll

2011-11-16T15:21:27Z

My first reaction was Tannaka-Krein dualtiy, but I feel like there should be something more concrete for SU(2), probably its Langlands dual group since it's reductive. I'd be surprised if there wasn't any pre-existing work on/near this topic...

Comment by Alexander Moll

2011-11-16T14:18:43Z

Thanks for the response - I'm certainly glad to see a 4. I'm not familiar with rational homotopy theory, but I'd like to ask: is there some way of understanding the rationalization of BG in terms of the representation theory of G? How can one explain this $\mathbb{Z}$ in the result for $BSp(1)$?

Comment by Alexander Moll

2011-11-16T07:24:49Z

Hi - though I'm no expert, the paper of Carqueville and Murfet arxiv.org/abs/1108.1081 seems relevant.

Comment by Alexander Moll

2011-11-16T06:32:24Z

$\widehat{G}$ denotes the dual group of a locally compact abelian group. In particular, $\widehat{G} \cong G$ for finite groups (but not canonically), and you may recall $\widehat{U(1)} \cong \mathbb{Z}$ from Fourier series. For a non-abelian group, the machinery of K(G,n)s cannot possibly work, which is the point of the question.