Is there a quantum Hermite reciprocity?

Question

It is well known that there is an isomorphism of $SL_2=SL(V)$ representations $$ Sym^n(Sym^m(V))\simeq Sym^m(Sym^n(V)) $$ called Hermite reciprocity (discovered in 1854). My question is: Is there anything like this isomorphism for $U_q(sl_2)$, at least for generic $q$?

Answer 1

There is in fact a reasonable way to define quantum analogues of symmetric and exterior powers of a finite-dimensional representation of $U_q(\mathfrak{g})$. Let $V$ be such a representation, and let $\hat{R} : V \otimes V \to V \otimes V$ be the braiding of $V$ coming from the universal R-matrix.

It is a fact (see, for instance, Proposition 22 and Corollary 23 in Chapter 8 of the book Quantum Groups and Their Representations, by Klimyk and Schmudgen) that the eigenvalues of $\hat{R}$ are all of the form $\pm q^{t_i}$, where $t_i \in \mathbb{Q}$. Call the eigenvalues of the form $+q^{t_i}$ positive, and those of the form $-q^{t_i}$ negative (this notion is well-defined if $q$ is not a root of unity). Then call eigenvectors for $\hat{R}$ positive or negative, respectively, if their eigenvalues are positive or negative. The idea is that positive eigenvectors are $q$-symmetric, while negative eigenvectors are $q$-antisymmetric.

Then define $$ S_q^2 V = \mathrm{span} \{ \text{positive eigenvectors} \} $$ and $$ \Lambda_q^2 V = \mathrm{span} \{ \text{negative eigenvectors} \}. $$ Since $\hat{R}$ is diagonalizable, $V \otimes V = S_q^2V \oplus \Lambda^2_q V$. For example, when $V$ is the 2-dimensional representation of $U_q(\mathfrak{sl}_2)$ with weight basis $x,y$, where $x$ is the highest weight vector, we have $$ S_q^2 V = \mathrm{span} \{ x \otimes x, y \otimes x - q x \otimes y, y \otimes y \}, $$ and $$ \Lambda_q^2 V = \mathrm{span} \{ y \otimes x + q^{-1} x \otimes y \}. $$

Finally, you can define higher quantum symmetric powers $S^n_qV$ to be the submodules of $V^{\otimes n}$ created by intersecting the submodules of tensors that are $q$-symmetric in all $n-1$ consecutive pairs of entries: $$ S^n_q V = (S^2_q V \otimes V^{\otimes n -2}) \cap \dots \cap (V^{\otimes n -2} \otimes S_q^2 V). $$ There is also a closely related notion of quantum symmetric algebra, which is a graded $U_q(\mathfrak{g})$-module algebra whose homogeneous components are isomorphic to the quantum symmetric algebra defined above.

Anyway, that's the good news; there is a not-too-bad definition of quantum symmetric powers. The bad news is that it doesn't always give you the classical result. The quantum symmetric powers of a module are no larger than their classical counterparts, and the module is called flat (in a different sense than the usual homological one) if all of its q-symmetric powers (or equivalently, just the q-symmetric cube) are the right size.

The flat simple modules $V_\lambda$ have been classified by Sebastian Zwicknagl in his paper R-Matrix Poisson Algebras and their Deformations. For each semisimple Lie algebra there are only finitely many flat simple modules.

In the paper Braided Symmetric Algebras of $U_q(\mathfrak{sl_2})$-modules and Their Geometry, he computes all of the quantum symmetric powers of simple $U_q(\mathfrak{sl_2})$-modules. It turns out that if $V$ is the 2-dimensional simple module, then its symmetric powers are the right size, i.e. $$ S^n_q V \cong V_n, $$ where $V_n$ is the $(n+1)$-dimensional simple module. So your question about Hermite Reciprocity boils down to the question: are $S^m_q V_n$ and $S^n_q V_m$ isomorphic for any $m$ and $n$?

The answer is that they are not. The first example is $m=3$ and $n = 4$, which follows from the computation in Theorem 3.1 of the second paper I referenced. The decompositions into simple modules are: $$ S^4_q(V_3) \cong V_{12} \oplus V_8, $$ while $$ S^3_q(V_4) \cong V_{12} \oplus V_{8} \oplus V_{4} \oplus V_{0}. $$ Of course, this doesn't rule out the possibility of a better definition which does satisfy Hermite Reciprocity, but nobody has come up with one yet. And if you want everything to be $U_q(\mathfrak{g})$-equivariant, then your choices are pretty rigid. But perhaps if you let go of that requirement then something more is possible.

Answer 2

Take the question to be: Can we define a $q$-analogue of $Symm^n$? Then we can cheat and declare it has the same character as for $q=1$. I suspect there is not a principled way of doing this. My circumstantial evidence is that no-one has come up with a way of defining $Symm^n$ even for $n=2$ for crystal graphs.