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 $q$?

2$\begingroup$ What do you mean by $\operatorname{Sym}$ in the quantum case? Homogeneous components of the "symmetric" algebra constructed using the braiding, a.k.a. Nichol's algebra? $\endgroup$ – Mariano SuárezÁlvarez Jul 28 '10 at 17:17

$\begingroup$ The inner of sym is clear it is the corresponding irreducible representation of the quantum group. But indeed the outer sym is not clear at least to me. In fact my question can be rephrased as: is there a good definition of the outer sym for which one has a Hermite isomorphism? $\endgroup$ – Abdelmalek Abdesselam Jul 28 '10 at 17:48
There is in fact a reasonable way to define quantum analogues of symmetric and exterior powers of a finitedimensional 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 Rmatrix.
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 welldefined 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 2dimensional 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 $n1$ 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 nottoobad 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 qsymmetric powers (or equivalently, just the qsymmetric cube) are the right size.
The flat simple modules $V_\lambda$ have been classified by Sebastian Zwicknagl in his paper RMatrix 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 2dimensional 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.

$\begingroup$ @MTS: that seems related to the comment by Sasha below, but with much more details. Thanks, that's very helpful. $\endgroup$ – Abdelmalek Abdesselam Aug 17 '12 at 22:09
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 noone has come up with a way of defining $Symm^n$ even for $n=2$ for crystal graphs.

$\begingroup$ How about this for Sym^n(Sym^m): one could try to build the corresponding projector in V tensored with itself nm times seen as n groups of m strands. For each group use the formula for the mth JonesWentzl projector. There is a formula for that with a sum over permutations. I am wondering if there is a similar formula which can be found at the next level when permuting the groups. I guess crossings would be implemented by Rmatrices. Couldnt' something like this work? i.e. give an equivariant projection which has the same decomposition into irreducibles as in the classical case? $\endgroup$ – Abdelmalek Abdesselam Jul 29 '10 at 2:05

$\begingroup$ I don't see a way to do this; hence my pessimistic answer. Maybe someone else will come up with something? $\endgroup$ – Bruce Westbury Jul 29 '10 at 4:43

2$\begingroup$ One can define $Sym^n(V_\lambda)$ as the subspace
$\{v: R_iv=c_\lambda v\}$
, where $c_\lambda$ is the (easily computable) eigenvalue of $R$ on $v_\lambda\otimes v_\lambda\in V_\lambda\otimes V_\lambda$. The difficulty of generalizing it to reducible representations is that $c_\lambda$ depends on $\lambda$.... $\endgroup$ – Sasha Kirillov Nov 18 '10 at 11:55 
$\begingroup$ @ Sasha: what do you mean by $v_\lambda$? Is it a highest weight vector? Somebody else could be interested in reducible representations but I am not. So what you say works in the case when $V_\lambda$ is the symmetric power of the fundamental representation i.e. $\lambda$ has one part, right? Is it obvious that the iterated symmetric power defined in this way is an sl_2,q module? If this is the case, do you think Hermite reciprocity holds? $\endgroup$ – Abdelmalek Abdesselam Nov 19 '10 at 15:37