Dimension of a Hopf algebra == sum of squares of its simple modules? - MathOverflow [closed]

Dimension of a Hopf algebra == sum of squares of its simple modules?

2011-09-04T06:32:00Z

when I read an article,I find it seems there is a conclusion like the followings.

$H$ is an Hopf algebra(or an abstract group). Then $dimH=\sum_{V:simple ~module ~of ~H}(dimV)^2$.

who can tell me where I can find this content?Thank you very much.

oh...It seems $H$ need to be semisimple!

Oh...I think I already understand this.~When $H$ is a semisimple algebra.The above conclusion is right.For any Hopf algebra,maybe it's wrong. So please vote to close.Thanks everyone!

Answer by Ralph for Dimension of a Hopf algebra == sum of squares of its simple modules?

2011-09-04T10:19:24Z

In general, the statement is false: Let $G$ be a non-trivial $p$-group and $k$ a field of characteristic $p$. Then the group ring $kG$ is a Hopf algebra of dimension $|G| > 1$, while the only simple $kG$-module is $k$ with trivial $G$-operation (see: Benson: Representations and Cohomology I, Lemma 3.14.1).

Answer by David Jordan for Dimension of a Hopf algebra == sum of squares of its simple modules?

2011-09-04T13:34:49Z

As Torsten Ekedahl observers, this is a fact about finite dimensional algebras, and doesn't concern the coproduct on $H$ at all. And as you have noted, it's not true as stated for non-semisimple algebras.

However, there is a natural modification, which is true for all finite dimensional algebras $A$. Let $X_1,\ldots, X_k$ denote the isomorphism classes of simple objects of $Rep(A)$, and let $P_1,\ldots P_k$ denote their projective covers. Then we have:

$dim(A) = \sum_k (dim X_k) (dim P_k)$.

Of course if $H$ is semi-simple then this recovers the well-known result you mentioned, since $P_k=X_k$ then.

See, for instance, the comprehensive lecture notes:

http://ocw.mit.edu/courses/mathematics/18-712-introduction-to-representation-theory-fall-2010/lecture-notes/MIT18_712F10_ch7.pdf