0
$\begingroup$

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!

$\endgroup$
2
  • 1
    $\begingroup$ What article? ${}$ $\endgroup$ Sep 4 '11 at 6:48
  • 4
    $\begingroup$ 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. $\endgroup$ Sep 4 '11 at 7:50
5
$\begingroup$

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

$\endgroup$
2
  • 1
    $\begingroup$ 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 $dim End(X_k)$. $\endgroup$ Sep 4 '11 at 15:05
  • $\begingroup$ Yes, of course over algebraically closed field should have been in my post. Thanks! $\endgroup$ Sep 4 '11 at 21:58
3
$\begingroup$

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).

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.