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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?

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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

1 Answer 1

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.

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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

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