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I ran across the following statement in a paper, and it seems fishy to me:

Lemma: If $A$ is any Hopf algebra, and if $U$ is an $\mathbb{N}_0$-filtered $A$-module algebra, then $U$ and $\mathrm{gr} (U)$ are isomorphic as $A$-modules.

There is no proof of the lemma, it just states that it is a well-known fact.

Such an isomorphism cannot be canonical: consider just the case that $A = k$ is a field and $U$ is any filtered $k$-algebra. In this situation there are plenty of vector space isomorphisms of $U$ with $\mathrm{gr}(U)$, just by pulling back a basis of each $U_{n} / U_{n-1}$, but these are hardly canonical.

So if the lemma is true, it is saying that there is some way to choose one of these maps so that it is an isomorphism of $A$-modules.

Question

Is the lemma true? If no, what is a counterexample? If yes, could you please provide a proof or a reference?

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1 Answer 1

up vote 2 down vote accepted

The statement is definitely false. For example, let $A = \mathbb k[x]$ be the group algebra of $\mathbb Z$. Let $U$ be the two-dimensional module in which $x$ acts by $\bigl( \begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix} \bigr)$. The span of $\bigl( \begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\bigr)$ is a submodule on which $x$ acts by $1$, and the quotient is also one-dimensional with $x=1$. So we give $U$ a filtration $0 \subset \mathbb k \bigl( \begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\bigr) \subset U$. Then the associated graded $\operatorname{gr}(U)$ is the two-dimensional $A$-module on which $x$ acts by $\bigl( \begin{smallmatrix} 1 & 0 \\ 0 & 1 \end{smallmatrix} \bigr)$. This is not isomorphic to $U$ as and $A$-module, since in $U$ the $x$-action is not diagonalizable.

In general, you should only expect an isomorphism $\operatorname{gr}(U) \cong U$ if you have some semisimplicity. For example, when $A$ is the universal enveloping algebra of a semisimple Lie algebra, and $U$ is locally finite-dimensional (each filtered piece is finite-dimensional), or the group algebra of a finite group in characteristic $0$ (or prime to the order of the group) and $U$ is arbitrary, then you can build a noncanonical $A$-module isomorphism $\operatorname{gr}(U) \cong U$.

You should write to the author of the paper.

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Nice answer Theo, thank you. –  MTS Apr 21 '11 at 15:51
    
Do you know any examples where the algebra structure on $U$ is nontrivial? In particular where $U$ is unital? –  MTS Apr 22 '11 at 19:42
    
@MTS: I don't understand your question. The module $U$ in this case is not an algebra, but an $A$-module; I thought that was what you were asking about. Are you now asking that $U$ be a unital $A$-algebra? If so, then the statement is false even over a field $A = \mathbb k$: there are plenty of filtered algebras that are not isomorphic as algebras to their associated graded algebras, e.g. universal enveloping algebras. But if you want an $A$-algebra which is not iso to its as.gr as even an $A$-module, take the free (filtered) algebra on the (filtered) module $U$ above. –  Theo Johnson-Freyd Apr 23 '11 at 0:03
    
It's possible that I misread your question the first time --- I now see that you do ask for $U$ to be an $A$-module. Anyway, I think you can take the free algebra on the example I gave; even better, use the example I gave as a square-zero extension of $A$, so that your $A$-algebra is $A\oplus U$ as an $A$-module, with multiplication = 0 on the $U\otimes U$ part, and given by the $A$-action on the rest. (This is well-defined since $A$ is commutative.) –  Theo Johnson-Freyd Apr 23 '11 at 0:06

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