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Question Is there any relation between Weyl groups for orthogonal and symplectic groups vs. Brauer algebra ?

Motivation For GL the symmetric group $S_n$ plays two roles: 1) Weyl group 2) appears in Schur–Weyl duality.

Brauer algebra plays the role of S_n in Schur-Weyl duality for SO, SP (for both groups with the only difference that wheel = N or -N). And it is NOT (as far as I understand) the group algebra of corresponding Weyl group. However since in GL case it is so, may be there is some relation ?

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I think the answer to your basic question is negative, though I don't have all the relevant literature at my fingertips. While $S_n$ does happen to play simultaneous roles for Lie type $A_{n-1}$ as the Weyl group and as a key player in Schur-Weyl duality, this seems to be an accident. For example, types $B,C$ (odd orthogonal and symplectic) have the same Weyl group but not the same Brauer algebras, while type $D$ (even orthogonal) has a different Weyl group.

In terms of highest weight theory, the Weyl group always plays an essential theoretical role but this is not usually related directly to the role of a symmetric group permuting factors in a tensor power of some "natural" representation as in Schur-Weyl duality. As far as I know, the analogy you are looking for isn't helpful.

ADDED: I should emphasize that for fixed $n$ and $\mathrm{GL}_n$ (originally over the complex field), each symmetric group $S_d$ plays a role in Schur-Weyl duality by permuting factors in the tensor product of $d$ copies of the natural $n$-dimensional module. Only when $d=n$ does this happen to involve the Weyl group of $\mathrm{GL}_n$ (or $\mathrm{SL}_n$).

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    $\begingroup$ @Jim Thank you very much for yours answer. Let me mention what I kept in mind asking. By virtue of SW-duality we can construct with the help of Brauer algebra irreps of G, character of irreps are W-invariant, so it may naively and informally seem that Brauer algebra "knows" about W-invariants - and hence "knows" about W itself... Well, of course, this is so "informal" that probably does not make sense... Selfcritics: to construct irreps we need to start with V^{\otimes N} and act on it by "G" itself, "G" is igredient of construction not only Brauer, so may be not Brauer, but "G" knows about W $\endgroup$
    – Alexander Chervov
    Commented Jun 6, 2012 at 5:56

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