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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?

flag	asked Oct 15 2009 at 21:37 Ben Webster♦ 20.7k●1●44●114

	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 2009 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 2009 at 13:23

Ben Webster · Accepted Answer · 2011-06-14 22:28:55Z UTC

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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."

edited Jun 14 2011 at 22:28

answered Oct 23 2009 at 16:08

Ben Webster♦
20.7k●1●44●114

Are cyclotomic Khovanov-Lauda-Rouquier algebras symmetric?

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