# Invariant differential forms on commutative group schemes are closed!?

Please assume that $G/S$ is an abelian scheme if it helps.

Let $G/S$ be a commutative group scheme and let $(\Omega^\bullet_{G/S},d)$ be the algebraic de Rham complex. Let $\omega\in \Omega^r_{G/S}(G)$ be an invariant r-form. Is it true in this generality that $d\omega=0$? Is there a reference for such a fact?

By ulrich's answer below the statement is true whenever 2 is invertible on $G$. But what if $G$ has points of characteristic 2?

I'm trying to understand the proof of the degeneracy of the Hodge-de Rham spectral sequence for abelian schemes from Berthelot-Breen-Messing, Théorie de Dieudonné cristalline II, specifically Lemma 2.5.3. In its proof they say "Comme les différentielles invariantes sont fermées[...]".

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I suppose it depends on what it means to be "invariant". Neither the left- or right-invariant differential forms on a Lie group are necessarily closed, but the bi-invariant forms are. –  José Figueroa-O'Farrill May 21 '11 at 16:06
Thanks, I've simply added the assumption that the group scheme should be commutative. –  Philipp Hartwig May 23 '11 at 15:44
For abelian varieties, the claim (in any characteristic) follows from the fact that they can be lifted to characteristic 0, a result of Norman and Oort (and Grothendieck and Mumford). However, I believe that there should be a more elementary proof that also works in characteristic 2. –  ulrich May 24 '11 at 9:22
I'm afraid I wouldn't be comfortable using that. My goal is to really understand some of the proofs in BBM and this would not be achieved by quoting far more difficult results. –  Philipp Hartwig May 27 '11 at 9:34
Philipp, it seems the usual Maurer-Cartan equations are all you need. See, for example, Proposition 51 in Chapter 3 Section 14 of Bourbaki's Lie Groups and Lie Algebras, which is stated in terms of Lie groups and Lie algebras, of course---possibly over a field of characteristic > 0---but which also applies to smooth algebraic groups. (Bourbaki's formulation avoids the "1/2" that may have worried you in other treatments.) The starting point is the familiar formula d\omega(X,Y) = X\omega(Y) - Y\omega(X) - \omega([X,Y]): if X,Y, and \omega are left invariant, then the first two terms vanish... –  user2490 May 30 '11 at 4:13

It seems to me that in the proof they only use the fact for abelian varieties: For this you can use the fact that $[n]^*$ acts on invariant $1$-forms by multiplication by $n$ -- this follows from the fact that $[n]_*$ acts by multiplication by $n$ on the tangent space at $0$ -- hence by $n^r$ on $r$-forms, and the compatibility of $d$ with pullback to prove the claim (if the characteristic is not $2$).

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The fact that you quote should be true for invariant forms on any commutative group scheme. After all, multiplication by $n$ is just composition of $n$-fold multiplication with the $n^{\text{th}}$ diagonal map. You just have to pick an $n$ that's invertible, and your argument should go through. –  Keerthi Madapusi Pera May 21 '11 at 18:37
Thanks; I realized this later. It's annoying though that the argument doesn't work in characteristic 2. –  ulrich May 22 '11 at 5:55
Thanks a lot, this is a very neat argument. So we get the result whenever 2 is invertible on G. I'd still be very interested in an argument that also works in characteristic 2 though. –  Philipp Hartwig May 23 '11 at 15:33

I would be a little bit nervous about things when the group scheme is not smooth (there may not be any problems though) but you are interested in a smooth case anyway. To me it seems that the most natural way of proving this is to look at the relation between commutators of vector fields and the differential of forms. Thus if $X$ and $Y$ are vector fields and $\omega$ is a $1$-form we have $d\omega(X,Y)=X(\omega(Y))-Y(\omega(X))+\omega([X,Y])$. Applying this to the case when $X$, $Y$ and $\omega$ are (left say) translation invariant gives $d\omega(X,Y)=\omega([X,Y])$ as $\omega(X)$ and $\omega(Y)$ are constant functions. This shows that all translation invariant $1$-forms are closed precisely when the Lie algebra of the group is commutative. Of course the Lie algebra is commutative if the group is (I guess the converse does not hold in positive characteristics though I cannot offhand come up with an example).

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For abelian varieties this is due to Igusa, I believe (but I have not verified this reference). The argument goes like this, IIRC.

Suppose first that $C$ is a curve over an algebraically closed field $k$. Then for every $d>0$ there are canonical identifications $$H^0(Pic^dC,\Omega^1)=H^0(C^{(d)},\Omega^1)=H^0(C^d,\Omega^1)^{{\frak S}_d}=H^0(C,\Omega^1)$$ where $C^d$ is the Cartesian product, $C^{(d)}$ the symmetric product and the superscript ${\frak S}_d$ indicates the subspace of invariants under the symmetric group ${\frak S}_d$. (This is proved in Milne's article on Jacobians in Cornell-Silverman.) Take $d=g$, so that $C^g\to C^{(g)}$ is separable and $C^{(g)}\to Pic^gC$ is birational. Take a $1$-form $\omega''$ on $Pic^gC$ and pull it back to $\omega'$ on $C^g$; then $\omega'=\sum_1^g pr_i^*\omega$ where $\omega$ is a $1$-form on $C$ and $\omega''=\omega'=\omega$ under the identifications in question. Now $d\omega=0$ for reasons of dimension, so $d\omega''=0$ and we are done for Jacobians.

Now suppose that $A=G$ is any abelian variety and take a generic curve $C$ on $A$ (a complete intersection of very ample divisors). Identify $Pic^1C$ with $Alb\ C$. Then the natural map $Alb\ C\to A$ induces an injection on global $1$-forms ($C$ is in general position on $A$) and so on $2$-forms. Then for a $1$-form on $A$ we have $d\omega=0$ on $Alb\ C$ and then $d\omega=0$ on $A$.

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