As is well known, the quantum groups $SU_q(n)$, amongst others, arise from $R$-matrix solutions of the Yang-Baxter Equation. My question is: For any subalgebra of $GL(n)$, does there exist an $R$-matrix Yang-Baxter solution that q-deforms it?

  • $\begingroup$ What do you mean by "arise from R-matrix solutions of the Yang-Baxter Equation"? $\endgroup$ – Mariano Suárez-Álvarez Jan 18 '10 at 18:25
  • $\begingroup$ The entries of the $R$-matrix encode in a direct way the commutation relations of the algebra deforming the coordinate algebra in question. $\endgroup$ – John McCarthy Jan 18 '10 at 18:44
  • $\begingroup$ Are you referring to the- FRT construction? $\endgroup$ – Sebastian Burciu Jan 18 '10 at 19:04
  • $\begingroup$ Yes, that's exactly what I'm talking about. $\endgroup$ – John McCarthy Jan 18 '10 at 19:20
  • $\begingroup$ In Kassel's book, VIII.7, FRT construction is illustrated for GLq(2) and SLq( 2) . I don't know a reference for SLq(n). One has to start with a given solution for QYBE and, as far as I know, there are very few that can be written down explicitly. Do you have a reference for SUq(n)? $\endgroup$ – Sebastian Burciu Jan 18 '10 at 20:05

There are certainly FRT-type constructions for all the classical simple Lie groups, types A,B,C,D (and a modification of type A to encompass $GL_n$ instead of $SL_n$). See e.g. Klymik and Schmudgen, Quantum Groups and Their Representations (I think that's the title). Using the theory of roots, etc. it is also possible to construct subalgebras of $U_q(\mathfrak{g})$ corresponding to Borel subgroups, parabolic subgroups, and subgroups corrresponding to symmetric pairs (see papers by Dijkhuizen, Joseph, Kolb, Letzter, Noumi, Stokman, etc. with the words "quantum symmetric pairs" in the titles).

More generallly, there seems to be a philosophy that any construction involving simple algebraic groups of classical type has a quantum group generalization, but there are often interesting extra degrees of freedom when you quantize and unexpected subtleties. If you elaborate which subgroups you're interested in, I can try to elaborate where to find sources (it would comprise a whole compendium to list sources for all such constructions =]).

One thing to clarify as regards your question: In the cases I mentioned above, one doesn't seek a new R-matrix to describe the subgroups of $GL_n$. Remember that the R-matrix is used to define the commutativity relations of the "coordinates" $a^i_j$. This defines the "quantized coordinate algebra" O_q(G), which plays the role of O(G) in algebraic groups. finding subgroups of G in terms of the coordinate algebra O(G) means finding quotient Hopf algebras. There is a similar picture for O_q(G), although depending on the situation, there can be extra difficulties. I believe that the Borel subgroup is quantized in a fairly straightforward way (dually, there is a Hopf subalgebra $U_q(\mathfrak b) \subset U_q(\mathfrak g)$). I suspect it's slightly more tricky when you consider $U_q(\mathfrak p)$ for parabolics, but I believe it is understood by now (perhaps someone more knowledgeable can post another answer with references?). For symmetric pairs, one considers one-sided co-ideal subalgebras of $U_q(\mathfrak g)$ instead of Hopf subalgebras. The details in each example are technical, but the point I mean to clarify is that one doesn't find a new R-matrix for each sub-algebra, but rather uses the same R-matrix to define commutation relations and then studies quotients (dually, subalgebras) of the original (or dual) algebra.

By the way, I believe that the way one handles real forms, for instance $O_q(SU_n)$, is to observe that $O_q(SL_n)$ has an involution (discussed in Klymik and Schmudgen, for instance), and one studies $O_q(SL_n)$ representations which are compatible with the involution (meaning they have an inner product with a condition relating inner product to convolution). This comes from thinking about $SU_n$ as a subgroup of $SL_n$ fixed by $X\mapsto (X^\dagger)^{-1}$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.