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According to Schur--Weyl duality, the centralizer of $\mathrm{GL}(V)$ acting diagonally on $V^{\otimes N}$ is the group algebra of the symmetric group $\mathbb S_N$. An equivalent formulation is the existence of a decomposition $$ V^{\otimes N} \cong \bigoplus_\lambda s_\lambda \otimes V_\lambda $$ where $s_\lambda$ (resp. $V_\lambda$) denote irreducible representations of $\mathbb S_N$ (resp. $\mathrm{GL}(V)$), and $\lambda$ ranges over partitions of $N$.

If we suppose that $V$ has a bilinear form and replace $\mathrm {GL}(V)$ by $\mathrm O(V)$, then the centralizer is a quotient of the Brauer algebra $B_N$. The analogous decomposition reads $$ V^{\otimes N} \cong \bigoplus_\lambda \pi_\lambda \otimes V_{\langle \lambda \rangle} $$ where $\pi_\lambda$ (resp. $V_{\langle \lambda \rangle}$) denote irreducible representations of $B_N$ (resp. $\mathrm{O}(V)$), and $\lambda$ ranges over partitions of $N - 2k$, $k \geq 0$.

In the group algebra of $\mathbb S_N$, there are explicit idempotents whose images are the irreducibles $s_\lambda$, called Young symmetrizers. What's the analogue in $B_N$? Is there a formula for an idempotent in $B_N$ which projects onto $\pi_\lambda$?

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    $\begingroup$ A few years ago I have asked Arun Ram a similar question (I was looking for a lift to the partition algebra, not the Brauer algebra), and he referred me to one of his oldest papers (this is what he said). It might be Arun Ram and Hans Wenzl, Matrix Units for Centralizer Algebras ( sciencedirect.com/science/article/pii/002186939290109Y ). I did not, unfortunately, dig into this; I am not sure if the paper actually settles the question (it seems to be more about generalizing seminormal units). $\endgroup$ Mar 12, 2015 at 12:07
  • $\begingroup$ Does arxiv.org/abs/1408.3592, Definition 7.4.3. help? $\endgroup$ Mar 12, 2015 at 13:34
  • $\begingroup$ Dear Darij, dear Martin: thanks for the pointers! It'll take me a few days to say whether they answer my question. $\endgroup$ Mar 12, 2015 at 20:06
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    $\begingroup$ There is an explicit formula in Nazarov's 2002 ICM address arxiv.org/abs/math/0209129 or his previous paper with a similar title. His setting is slightly more general but it's not hard to specialize to the Brauer algebra. $\endgroup$ Mar 12, 2015 at 23:14
  • $\begingroup$ @GjergjiZaimi I just looked at this question again and realized I didn't thank you -- Nazarov's formulas are exactly what I was looking for. It's amazing that this wasn't known until so recently!! $\endgroup$ Mar 27, 2015 at 10:47

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