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In my lecture notes it says:

Let H be a $k$-Hopf algebra. Let M be a left H-module. The invariants of H on M are defined as the $k$-vector subspace
$M^{H}$:= {m $\in$ M | h.m=$\epsilon$(h)m for all h $\in$ H} of M.

Why does the counit $\epsilon$ appear?
I am aware that this is not a particularly precise question but maybe answers will be insightful still.

Max Demirdilek
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  • Ups, sorry, that is right: $0_M$=$0_H$.m=m. Why the counit is still not clear to me. I'll edit the question. – Max Demirdilek Feb 25 '20 at 13:56
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    Well, how else would you make a sensible definition? You probably have the group case in mind, but note that even in a trivial module $hm=m$ only for certain special elements $h$ of the group algebra. The point of invariants is that they are things which behave like the trivial module, and the trivial module is defined using the count. – Matthew Towers Feb 25 '20 at 14:02
  • I see: Having the group case in mind (which I indeed had), one would want to define $M^{H}$:={m $\in$ M| h.m = c$\cdot$m for all h $\in$ H} with $c$ a scalar that is zero if $h$ is zero. So one somehow looks for a map from H to K that vanishes if $h$ is zero. The linear map $\epsilon$ seems a natural candidate. – Max Demirdilek Feb 26 '20 at 08:08
  • The definition seems more approachable know: It helped a bit to just rewrite it and to notice that the space of invariants is simply an intersection of certain eigenspaces of various linear maps to the eigenvalue of $\epsilon(h)$, i.e.

    $M^{H} = $$\bigcap_{h\in H} {m\in M |h.m=\epsilon (h)\cdot m} $

     – Max Demirdilek Feb 26 '20 at 08:25
  • Regarding ‘trivial module’: How do the notions in the Hopf algebra and in the ring case align? Consider an R-module with R a ring acting on an abelian group. The trivial module is usually said to be the abelian group where action is given by the zero homomorphism. In our case, (Hopf) algebra acting on vector space, the trivial module is a vector space where the action is given by $h.m=\epsilon (h) \cdot m$. Hence, in the special case where the counit is the zero homomorphism one recovers (in a sense) the notion in the abelian group case, right? – Max Demirdilek Feb 26 '20 at 08:41
  • I don't think there is a "trivial module" defined for an arbitrary ring. You normally require the unit in the ring to act as the identity which it wouldn't with your suggestion. For hopf algebras, the counit can't ever be the zero homomorphism. – Matthew Towers Feb 26 '20 at 11:58
  • @Matthew: You are right. What I wrote was (mostly) nonsense. That the counit cannot be zero follows immediately from its definition. Besides, I should have know better as the augmentation ideal has codimension 1. Regarding ‘trivial module’ over an arbitrary ring: I was referring to Berci’s definition: https://math.stackexchange.com/questions/1316867/what-exactly-is-a-trivial-module. Thanks anyway for your help and patience. – Max Demirdilek Feb 26 '20 at 22:13
  • ... refers to what? – Max Demirdilek Feb 27 '20 at 13:14
  • sorry, accidental post – Matthew Towers Feb 27 '20 at 13:31
  • the point is that, as Berci says, that definition refers only to the zero module. I would call that the zero module and not the trivial module, because the trivial module - when one is defined - always refers to something else. – Matthew Towers Feb 27 '20 at 13:50

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