q-deformed group characters - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T19:23:11Z http://mathoverflow.net/feeds/question/110556 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110556/q-deformed-group-characters q-deformed group characters John Mangual 2012-10-24T16:11:55Z 2012-10-24T20:38:29Z <p>In a paper by <a href="http://front.math.ucdavis.edu/1207.3497" rel="nofollow">Yuji Tachikawa</a>, I found a q-deformed "2d Yang-Mills paritition function for a cylinder". Here it is (adapted):</p> <p>$$Z(q, x_L, x_R) = \mu(q, x_L)^{-1/2} \langle x_L | \bigg[ \sum_{R \in \mathrm{Irr}(G)} | R \rangle e^{- aC_2(R) } \langle R | \bigg] |x_R \rangle \mu(q, x_R)^{-1/2}$$</p> <p>Here's some stuff to help you interpret:</p> <ul> <li>$G$ is a compact lie group and the Irreducible representations should be indexed by the root lattice.</li> <li>conjugacy classes are indexed by elements of maximal torus $\vec{x} \in \mathbb{T}^n \subset G$</li> <li>$C_2(R)$ is the quadratic Casimir of the representation. </li> <li>In my notation, borrowed from quantum mechanics $\langle R|x \rangle = \chi_R(x), \langle x|R \rangle=\overline{\chi_R(x)}$.</li> <li>$\displaystyle \mu(q, X) = \exp \left[ \sum_{n=1}^\infty \frac{-2q^n}{1-q^n}\chi_{\mathrm{adj}}(x^n) \right]$</li> <li>The partition function depends on the area $a$ of the cylinder.</li> </ul> <p>In fact, let's turn this into a statement about the Laplacian: The $q$-dependence is hidden:</p> <p>$$e^{- a \Delta} =<br> \sum_{R \in \mathrm{Irr}(G)} | R \rangle e^{- aC_2(R) } \langle R |$$</p> <p>Let's set the area to $0$. From the last line, we should get the identity matrix. However,</p> <p>$$<br> \sum_{R \in \mathrm{Irr}(G)} \langle x_L | R \rangle \overline{ \langle x_R | R \rangle } = \mu(q, x_L) \delta(x_L = x_R)$$</p> <p>This really looks like orthogonality of characters for compact groups, except the right side should be the identity. <hr> <strong>What are these characters $\langle x | R \rangle$ ?</strong> </p> <p>Originally, I wanted to ask about an analogue for finite $G$, but I don't even have a point of reference.</p> http://mathoverflow.net/questions/110556/q-deformed-group-characters/110583#110583 Answer by John Mangual for q-deformed group characters John Mangual 2012-10-24T20:21:41Z 2012-10-24T20:38:29Z <p>They are proportional to the Schur polynomials. For representation $\lambda \in \mathrm{Irr}(G), \chi \in \mathbb{T}^n \subset G$:</p> <p>$$\langle \lambda | x \rangle = \mu(q,x) \chi^\lambda(x)$$</p> <p>This weight is is independent of the representation so we can do $\displaystyle \sum_\lambda$ no problem!</p> <p>Here the Schur polynomial is defined by $\displaystyle \chi^\lambda(a) = \frac{ \det a_i^{\lambda_j + k - j}}{\det a_i^{k-j}}$</p> <p>The number $\mu(q,x)$ is called <em>superconformal index</em> denoted $\mathcal{I}_q^V(a)$ in Section 6 of <a href="http://arxiv.org/abs/1110.3740" rel="nofollow">Gauge Theories and Macdonald Polynomials</a> by Gadde, Rastelli, Razamat &amp; Yan.</p> <hr> <p>This superconformal index is the trace over representations over a certain superalgebra. It can also be considered a matrix integral over Haar measure. In the case of the cylinder</p> <p>$$\mu(a,b) = \Delta(a) \mathcal{I}^V(a) \delta(a, b^{-1})$$</p> <p>at least, in the zero area limit.</p> <p>In special cases, they find functions $f^\alpha(a)$ (in our case $=\langle \lambda|x \rangle$) orthogonal with respect to "propagator measure":</p> <p>$$\oint [da] \Delta(a)\mathcal{I}^V(a) f^\alpha(a) f^\beta (a^{-1}) = \delta^{\alpha \beta}$$</p> <p>Then they can define a new metric and structure constants</p> <p>\begin{eqnarray} \mathcal{I}(a,b,c) &amp;=&amp; \sum_{\alpha, \beta, \gamma} C_{\alpha\beta\gamma}f^\alpha(a)f^\beta(b)f^{\gamma}(c)\ \mu^{\alpha\beta} &amp;=&amp; \oint [da]\oint [db] \mu(a,b) f^\alpha(a)f^\beta(b)\ \end{eqnarray}</p> <p>These generalize the orthogonality of characters relations (at least for compact Lie groups).</p> <hr> <p>This "superconformal index" for a surface with punctures has an expression in terms of these generalized group characters.</p> <p>$$\mathcal{I}_{g,s} (a_1, a_2, \dots, a_n) = \sum_\alpha (C_{\alpha\alpha\alpha})^{2g-2+s} \prod_{i=1}^s f^\alpha(a_i)$$</p>