Vtyjkyuk

I am working on a paper, which states that the dual of a finite dimensional cocyclic module is cyclic. I tried to write down a proof, but I failed and I do not know if this is true in full generality or of it is true just in my case.

Let $A$ be a complex Hopf algebra and $M$ a finite $mathbbC$-dimensional cocyclic $A$-module with cocyclic vector $m in Msetminus lbrace 0 rbrace$, i.e. every nontrivial submodule $U subset M$ contains $m$. Let $M^* = operatornameHom(M,mathbbC)$ be the dual module. The statement is, that $M^*$ is cyclic with cyclic vector $m'$ corresponding to $m$.

What I tried so far, is the following. Let $e_1, dots, e_n$ be a $mathbbC$-basis of $M$ and let $delta_1, dots, delta_n$ be the dual basis of $M^*$, i.e. $delta_i(e_j) = delta_ij$.

We can assume that $m = e_1$, in particular $m' = delta_1$. Now since $m = e_1$ is cocyclic, there exists for every $e_i$ an element $a_i in A$ such that $a_ie_i = e_1$. To show that $M^*$ is cyclic with cyclic vector $delta_1$, it suffices to find elements $tilde a_i$ such that $tilde a_i delta_1 = delta_i$ for every $1 le i le n$.

The first obvious idea is to try to act with $a_i$ on $delta_1$ and look what happens. It is clear that
beginalign*
(a_i delta_1)(e_i) = delta_1(a_i e_i) = delta_1(e_1) = 1,
endalign*
but if we write $a_i e_j = sum_k a_k^ij e_k$, we have
beginalign*
(a_i delta_1)(e_j) = delta_1(a_i e_j) = a_1^ij
endalign*
which a priori can be nonzero.

I tried to correct the elements $a_i$ to make $a_1^ij$ be $0$, but this only works if $n = 2$, and moreover I think, that there should be a more intrinsic reason why the dual of a cocyclic is cyclic.

In the paper I am working on, my algebra is the universal enveloping algebra $U(mathfrakg otimes mathbbC[t])$ of the current algebra, where $mathfrakg$ is a semisimple finite dimensional Lie algebra. My module $M$ is in fact graded, but I do not think, that this is important.

It's not true that cyclic or cocyclic modules have to be simple (e.g. Verma modules are cyclic but not always simple).

A module is cyclic if (not only if) it has unique maximal proper submodule, and a cocyclic module is one with a unique minimal nonzero submodule ($Am$, in your notation - let's call it $S$).

To prove $M^*$ is cyclic, we show it has a unique maximal submodule. Consider the proper submodule $S^perp=fin M^* : f(S)=0 leqslant M^*$ which I claim is unique maximal. To see this, let $T < M^*$. Then $0 neq T^o = x in M : forall t in T, t(x)=0 leqslant M$, so $S leqslant T^o$, so $T^o perp leqslant S^perp$, so $T leqslant S^perp$ as $T leqslant T^o perp$. The only reason you need Hopf algebras is so that all of these things are submodules.