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

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!

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What article? ${}$ – Mariano Suárez-Alvarez Sep 4 2011 at 6:48
This has nothing to do with $H$ being a Hopf algebra but is true for any semi-simple algebra (over say an algebraically closed field so there are no division algebras involved) and follows directly from the Wedderburn classification of semi-simple algebras. – Torsten Ekedahl Sep 4 2011 at 7:50

## closed as no longer relevant by Todd Trimble, Dan Petersen, Qiaochu Yuan, Scott Morrison♦Sep 4 2011 at 18:15

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).

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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

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This result is not true for all algebras. The proof needs that the field is a splitting field for $A$. So it works for algebraically closed field and the like but not in general. The general version should include an additional factor like $dim End(X_k)$. – Johannes Hahn Sep 4 2011 at 15:05
Yes, of course over algebraically closed field should have been in my post. Thanks! – David Jordan Sep 4 2011 at 21:58