Two curious asymptotic results for dimensions of type A objects - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T04:58:44Z http://mathoverflow.net/feeds/question/92736 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/92736/two-curious-asymptotic-results-for-dimensions-of-type-a-objects Two curious asymptotic results for dimensions of type A objects Alexander Moll 2012-03-31T06:35:33Z 2012-04-03T14:42:48Z <p>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. </p> <p>Does anyone have a good feeling for why this is true?</p> <p>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 $$L^2(SU(N)) = \widehat{\bigoplus_{\lambda \in SU(N)^{\vee}}} W_{\lambda} \boxtimes W_{\lambda}$$ $$(\mathbb{C}^N)^{\otimes n}=\bigoplus_{\lambda \in SU(N)^{\vee} \cap S(n)^{\vee}} V_{\lambda} \boxtimes W_{\lambda}$$ $$\mathbb{C}[S(n)] = \bigoplus_{\lambda \in S(n)^{\vee}} V_{\lambda} \boxtimes V_{\lambda}$$</p> <p>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.</p> <p>Can anyone give me some intuition for what's going on here + why I might expect such a result?</p> http://mathoverflow.net/questions/92736/two-curious-asymptotic-results-for-dimensions-of-type-a-objects/92772#92772 Answer by Bruce Westbury for Two curious asymptotic results for dimensions of type A objects Bruce Westbury 2012-03-31T20:22:29Z 2012-03-31T20:22:29Z <p>The $q$-analogue of $\dim W_\lambda$ is the specialisation $s_\lambda(q,q^2,\ldots ,q^n)$ and the $q$-analogue of $\dim V_\lambda$ is the principal specialisation $s_\lambda(q,q^2,\ldots )$ which is manifestly given by taking $n\rightarrow\infty$. In fact in Enumerative Combinatorics II by Stanley this is how the hook length formula for $\dim V_\lambda$ is derived.</p> http://mathoverflow.net/questions/92736/two-curious-asymptotic-results-for-dimensions-of-type-a-objects/93005#93005 Answer by Hugh Thomas for Two curious asymptotic results for dimensions of type A objects Hugh Thomas 2012-04-03T14:42:48Z 2012-04-03T14:42:48Z <p>This is an answer to Alexander's combinatorial reformulation of the question in comments to Bruce's answer. </p> <p>dim $V_\lambda$/$n$! is the chance that you will get a standard Young tableau if you assign the values 1 to $n$ to the boxes of a tableau of shape $\lambda$ according to a random permutation.</p> <p>dim $W_\lambda/N^n$ is the chance that you will get a semi-standard Young tableau if you assign a value in $[1,N]$ to each box in a tableau of shape $\lambda$. </p> <p>Think of the second procedure in the following way: first choose a set of $n$ numbers from 1 to $N$ to serve as entries, and then assign them to boxes. If the entries are all different, then the chances that what you get is a semistandard tableau is the same as the chance that you get a standard tableau starting with 1..$n$. </p> <p>As $N$ tends to infiniity, the chance that you will choose two entries the same becomes vanishingly small, so in the limit, dim $W_\lambda/N^n$ tends to dim $V_\lambda$/$n$!. </p>