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Recall that for k a field, a finite dimensional k-algebra A is called symmetric if it is isomorphic to its dual as a bimodule of itself. Which is to say, there's a trace map t:A -> k such that t(ab)=t(ba) and for any nonzero a, there is a b such that t(ab) is not 0.

The most popular examples of symmetric algebras are the cohomology rings of compact manifolds. In this case, the trace is integration of top forms.

Various algebras arising in representation theory are symmetric and this has some nice consequences for their categories of representations (for example, the Serre functor is trivial).

Khovanov and Lauda have defined some algebras by explicit generators and relations (see, for example 0909.1810, the cyclotomic KLR algebras, which I strongly believe are symmetric, but after quite a bit of trying, have had no luck with. Anyone else care to try their hand?

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I wish the term "symmetric algebra" weren't overloaded like this, especially since trace is only invariant under cyclic shifts. Maybe something like "E[1]-self-dual"? (btw, I'm not asking you to wage war on the responsible parties) –  S. Carnahan Oct 16 '09 at 4:31
    
Better would be to call them "Serre trivial algebras" since that reminds you they have trivial Serre functor (actually an equivalent condition for finite dimensional algebras). –  Ben Webster Oct 16 '09 at 13:23

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up vote 2 down vote accepted

This is now proven both as Theorem 1.7 in my paper "Knot invariants and higher representation theory I" and by Kang and Kashiwara in "Categorification of Highest Weight Modules via Khovanov-Lauda-Rouquier Algebras."

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Hi Ben, just came across this. I didn't know that this was also in Kang-Kashiwara. In type A only, Jun Hu and I proved this is in our paper “Graded cellular bases for the cyclotomic Khovanov-Lauda-Rouquier algebras of type A”, Adv. Math., 225 (2010), 598-642. arXiv:0907.2985 –  Andrew Dec 22 '13 at 20:38
    
@Andrew It's not explicitly proven in Kang and Kashiwara, but the existence of a Frobenius structure, at least, follows immediately from their results by work of Rouquier (5.16 in 0812.5023). I think you can assure that this Frobenius structure is symmetric on general principle (this is a slightly tricky point; even in the more explicit proof in my paper, the "most natural" Frobenius structure is only symmetric up to scalars, which you have to go back in and fix by hand). –  Ben Webster Dec 23 '13 at 2:41
    
Thanks for this, I haven't had a chance to look at [KK] again yet. Of course, we got the symmetry for free but I wouldn't claim that our construction of the form in type A is in any way natural... –  Andrew Dec 23 '13 at 22:37

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