Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In the paper Seminormal representations of Weyl groups and Iwahori-Hecke algebras, Arun Ram defines a seminormal basis as follows: given a chain of split semisimple $K$-algebras $K\cong H_0 \subseteq H_1 \subseteq \dots \subseteq H_r$ and an $H_r$-irreducible $N_\lambda$, a seminormal basis of $N_\lambda$ is a $K$-basis $B$ of $N_\lambda$ compatible with the restrictions in the following sense: there is a partition $B = B_{\mu_1} \sqcup \dots \sqcup B_{\mu_k}$ such that if $N_{\mu_i} = K B_{\mu_i}$ then $N_\lambda = N_{\mu_1} \oplus \dots \oplus N_{\mu_k}$ as $H_{r-1}$-modules. Further, there is a partition of each $B_{\mu_i}$ that gives rise to a decomposition of $N_{\mu_i}$ into $H_{r-2}$-irreducibles, and so on, all the way down to $H_0$.

Note that if the restriction of an $H_i$-irreducible to $H_{i-1}$ is multiplicity-free, then a seminormal basis is unique up to a diagonal transformation.

Now let $H_r$ be the type $A$ Hecke algebra defined over $K = \mathbb{C}[q^{1/2},q^{-1/2}]$ and $e_i$ the idempotent corresponding to the sign representation in the copy of $H_2 \subseteq H_r$ generated by $T_i$. In the paper Hecke algebras of type $A_n$ and subfactors, Wenzl defines a version of Young's orthogonal basis for each irreducible representation $M_\lambda$ of $H_r$. For each $\lambda$, this is a seminormal basis with respect to the chain $H_1 \subseteq \dots \subseteq H_r$, where $H_i$ is the subalgebra generated by $T_1,\ldots,T_{i-1}$. Also, in this basis the matrix corresponding to left multiplication by $e_i$ is equal to its transpose.

I am looking for another reference aside from Ram and Wenzl's papers that discusses this basis. In particular, I would like a reference that discusses how this basis is the unique seminormal basis satisfying this self-adjointness property. What is the correct definition of self-adjoint in this setting? Is there a good general way of defining the inner product with respect to which the $e_i$'s are to be self-adjoint? Why choose the $e_i$'s to be self-adjoint and not some other generators for the Hecke algebra?

share|improve this question
Have you looked at Vershik-Okounkov? –  Ben Webster Jan 16 '11 at 4:33
No, I'll take a look at it. –  Jonah Blasiak Jan 16 '11 at 7:50
add comment

1 Answer

I don't quite see this as something you would expect to find explained in a paper. As I see it there are three ways to define the orthogonal basis: one, as you describe where when you restrict from $H_i$ to $H_{i-1}$ you get a direct sum decomposition (without a change of basis); secondly, as a basis in which the Gelfand-Tsetlin subalgebra is represented by diagonal matrices; thirdly, the Gram matrix of an inner product is diagonal.

I don't think I can add much to the equivalence of the first two definitions. This is the Vershik-Okounkov approach that Ben has mentioned.

For the third approach you need an anti-involution $*$ on the algebra. This constructs the dual of a representation. An inner product on a representation is an isomorphism with the dual. If the basis is orthonormal then self-dual elements of the algebra will be represented by symmetric matrices and $e_i$ is self-dual.

The equivalence of the third approach with the first two is just the observation that the chain of subalgebras is invariant under the anti-involution. Alternatively the Gelfand-Testlin subalgebra is self-dual.

share|improve this answer
I agree that I might not find this in a paper. If there is a good discussion of this in a book somewhere, I would be happy with that too. I'm somewhat confused by your response. A seminormal basis is not necessarily equal to the orthogonal basis defined by Wenzl because a seminormal basis is only unique up to diagonal transformation. I think that additionally requiring that the $e_i$ are self-dual then forces the basis to be equal to this orthogonal basis (up to global scale). I don't understand how these second and third items are ways to define an orthogonal basis. –  Jonah Blasiak Jan 18 '11 at 0:24
What anti-involution should we use in this context? I would guess the most natural one is the one sending $T_w$ to $T_{w^{-1}}$. However, this anti-involution fixes the chain $H_1 \subseteq H_2 \dots H_r$ above as well as $K \subseteq <T_{n-1}> \subseteq <T_{n-1}, T_{n-2}> \subseteq \dots H_r$. But Wenzl's orthogonal basis is not seminormal with respect to this chain. –  Jonah Blasiak Jan 18 '11 at 0:29
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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