# Multiplication of extreme vector

This question might be elementary and standard.

Standard Notions: Let $g$ be a semisimple Lie algebra. Let $\pi=({\alpha_{1},....\alpha_{n}})$ be simple roots

$P^{+}(\pi)=\Sigma\mathbb{N}\alpha _{i}$

Suppose $\lambda,\nu\in P^{+}(\pi)$ and $e_{w\lambda}, e_{w\nu}$ are extreme vector. ($\lambda,\nu$ are highest weight and $w\in W$, Weyl group).

Claim 1: $e_{w\lambda}\otimes e_{w\nu}$ is an extreme vector of weight vector $e_{w(\lambda+\nu)}$.(This has been done, easily follows from the definition of tensor products of representations)

Claim 2: $e_{w\lambda}\otimes e_{w\nu}=A^{*}e_{w(\lambda+\nu)}$, where $A^{*}$ is multiplicative set of invertible elements

There is a paper by A.Joseph talking about this observation. He claimed that it follows from Weyl Character formula. I think he might talk about the decomposition of tensor product of irreducible representations. However, I can not find a proof.

Any comments are welcome.

Edit: The paper I talked about is "Faithfully flat embeddings for minimal primitive quotients of quantized enveloping algebras" by Anthony Joseph. There is another paper by A.Rosenberg and V.Lunts on "localization for quantum group" page 138

Section 2: Ore localization of rings $R_{A}$. They claimed

$e_{w\lambda}\otimes e_{w\nu}=A^{*}e_{w(\lambda+\nu)}$, where $A^{*}$ is multiplicative set of invertible elements by using Weyl character formula,but how?

• I don't understand what is $A$. Also, recall that highest weight vectors are determined only up to scalar multiple, as are any eigenvectors of any action. – Theo Johnson-Freyd Apr 20 '10 at 4:07
• just consider the set $A=k$,where $k$ is ground field of Lie algebra.(say complex semisimple Lie algebra, then $k=C$) – Shizhuo Zhang Apr 20 '10 at 4:17
• I believe this is just the observation that the $w(\lambda)+w(\nu)$ weight space in $V_\lambda \otimes V_\nu$ is 1-dimensional. Since both $e_{w\lambda}\otimes e_{w\nu}$ and $e_{w(\lambda +\nu)}$ (however you want to precisely define that) both lie in this weight space, one is a scalar multiple of the other. To see that this weight space is 1-dimensional, it suffices to notice that the $\lambda+\nu$ weight space is 1-dimensional and to recall that weight-multiplicities are invariant under weyl group reflections. – Peter Tingley Apr 21 '10 at 17:18

## 2 Answers

If $e_\lambda$ and $e_\nu$ are highest weight vectors, of their respective representations, then $e_\lambda \otimes e_\nu$ is a highest weight vector in the tensor product: being a highest weight vector just means that it's an eigenvector for our fixed Borel subalgebra, which remains true using the action of a Lie algebra on a tensor product of representations. Applying $w$, we see that $e_{w\lambda} \otimes e_{w\nu}$ is an extreme vector.

• yes, this just follows from definition of tensor products of representations. But what is the relation of $e_{w\lambda}\otimes e_{w\nu}$ and $e_{w(\lambda+\nu)}$ – Shizhuo Zhang Apr 19 '10 at 21:11
• Is there a equality(with coefficient)describing the relationship? – Shizhuo Zhang Apr 19 '10 at 21:12
• What does $e_{w(\lambda + \nu)}$ mean? $e_{w\lambda} \otimes e_{w\nu}$ is a weight vector of $V_\lambda \otimes V_\nu$ of weight $w(\lambda + \nu)$... is that what you want? – Steven Sam Apr 19 '10 at 21:21
• @Sam, I will formulate the problem again. Sorry for mislead – Shizhuo Zhang Apr 19 '10 at 22:00
• As Steven implies, the language is too fuzzy at times. Some features of the tensor product of modules over $\mathbb{C}$ are easy to describe in terms of weights, but detailed module structure gets very complicated. The solution by Shrawan Kumar of the old PRV Conjecture (Parthasarathy, Range Rao, Varadarjan) in Invent. Math. 93 (1988) is a sample of this. "Extremal" weights in the irreducible case are just the Weyl group conjugates of the highest weight (all have multiplicity 1), but in a tensor product what is extreme/extremal? (And which Joseph paper do you refer to?) – Jim Humphreys Apr 19 '10 at 22:07

Maybe I understand what the basic question is, after looking at the two papers mentioned:

MR1261902 (94m:17013) 17B37 (17B35) Joseph, Anthony (F-PARIS6-F), Faithfully flat embeddings for minimal primitive quotients of quantized enveloping algebras. Quantum deformations of algebras and their representations (Ramat-Gan, 1991/1992; Rehovot, 1991/1992), 79–106, Israel Math. Conf. Proc., 7, Bar-Ilan Univ., Ramat Gan, 1993.

MR1694897 (2001f:17028) 17B37 (16S32 22E47) Lunts,V. A. (1-IN); Rosenberg, A. L. [Rosenberg, Alexander L.] (D-MPI), Localization for quantum groups. Selecta Math. (N.S.) 5 (1999), no. 1, 123–159.

Fortunately I have the conference volume containing Joseph's paper and can get the other paper online via the library Springer subscription.

Both papers discuss how to pass from classical results on localization and such to quantized enveloping algebras (not easy to do), but I think the immediate issue just concerns the classical theory of finite dimnsional representations of a semisimple Lie algebra. Here each simple module $E(\lambda)$ has a unique highest weight vector $e_\lambda$ up to scalars (with French espace $E$ instead of English vector space $V$). When tensoring two such modules $E(\lambda) \otimes E(\mu)$, a summand $E(\lambda +\mu)$ occurs uniquely and involves the highest weight of the tensor product. By taking the direct sum of all simple modules, one gets a transparent description of the representation ring due to complete reducibility. To get a unique choice of highest weight vector $e_\lambda$ for each dominant integral weight, it is enough here to specify arbitrary choices for the fundamental weights and then tensor systematically.

In Joseph's 2.2 the wording may be misleading. He comments parenthetically that the analogous simple modules for the quantized enveloping algebra also satisfy the Weyl character formula. But this is unrelated to the following line "Consequently ..." (Joseph's papers are interesting but not always easy to read in detail.)

• Jim, actually, I have figured out how to prove this not based on this answer you gave but based on your comments. – Shizhuo Zhang Apr 22 '10 at 2:54