13
$\begingroup$

In a Hopf algebra $H $ (over some field $ k $), there is the notion of a Haar element $ h \in H$. This is an element of the algebra which has the property that if $ V $ is a representation of $ H $, then the action of $ h $ projects onto the isotypic component of the trivial subrepresentation of $ V $.

If $ A $ is just a $k$-algebra, then we have no notion of isotypic component of the trivial representation. However, there is one interesting exception. Suppose that $ V, W $ are two $A$-modules, then $Hom_k(V,W) $ is an $ A^{op} \otimes A $ module and it contains the space of homomorphisms $ Hom_A(V, W) $. So it is natural to look for some element $ T \in A^{op} \otimes A $ which when acting on $ Hom_k(V,W) $ projects onto $ Hom_A(V, W) $.

More generally, if $ M $ is an $ A $-bimodule, then we would want $ T $ to project onto the subspace $$ \{ m\in M : am = ma \, \text{ for all } \, a \in A \}$$

There are a number of conditions that such an element $ T $ should satisfy.

  1. $ T (a \otimes 1) = (1 \otimes a) T $ for all $ a \in A. $

  2. $ m(T) = 1$, where $m : A \otimes A \rightarrow A $ is the multiplication map

  3. $T^2 = T$ (computed in the algebra $ A^{op} \otimes A $.)

Is there a name for such an element $ T$? Are there any existence/uniqueness results regarding such elements?

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ If M is any A-bimodule you always have the subspace Z(M) of m with am=ma for all a. Would you want T to work for all of these? $\endgroup$
    – Benjamin Steinberg
    Sep 7, 2017 at 11:01
  • $\begingroup$ Yes, that is a good point. I have added this and some conditions on $ T $ which ensure this. $\endgroup$
    – Joel Kamnitzer
    Sep 7, 2017 at 13:46
  • 7
    $\begingroup$ The three conditions 1, 2, 3 you found define what is called a separability idempotent for $A$; see en.wikipedia.org/wiki/Separable_algebra . $\endgroup$
    – darij grinberg
    Sep 7, 2017 at 14:27
  • $\begingroup$ Great, that seems like a good answer to my question. In particular, if $ k $ is algebraically closed, the existence of the separability idempotent is equivalent to $ A $ being a product of matrix algebras. $\endgroup$
    – Joel Kamnitzer
    Sep 7, 2017 at 14:58
  • 3
    $\begingroup$ It means $A$ is projective as an $A$-bimodule. In general it is equivalent to $A$ being semisimple under all base extension or equivalently a product of matrix algebras over separable extensions of the ground field. $\endgroup$
    – Benjamin Steinberg
    Sep 7, 2017 at 15:57

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.