# Parallel forms and cohomology of symmetric spaces

Let $G/H$ be a compact symmetric space. Then I believe the following is true: if $\alpha \in \Omega^k(G/H)$ and $\nabla$ the Levi-Civita connection, then $$(\alpha \text{ is induced by an \operatorname{Ad} H element of \Lambda^* \mathfrak h^\perp}) \Leftrightarrow \alpha \text{ is G-invariant} \Leftrightarrow \nabla \alpha = 0 \Leftrightarrow \alpha \text{ is harmonic}.$$ This means the real cohomology of $G/H$ is isomorphic to the space of $\operatorname{Ad} H$ invariant elements of $\Lambda^* \mathfrak h^\perp$, which seems like a reasonable thing to be able to compute.

For what compact symmetric spaces have these been computed explicitly, and where can this sort of thing be found in the literature?

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This is more or less the content of the paper by Chevalley and Eilenberg where they defined Lie algebra cohomology, no? –  Mariano Suárez-Alvarez Apr 8 '13 at 22:51
See Section 7.4.2 of these lectures www3.nd.edu/~lnicolae/Lectures.pdf –  Liviu Nicolaescu Dec 2 '14 at 13:13

As mentioned above, an invariant form on G/H is harmonic. The converse is proved in Helgason's book (1978), pages 227 and 564. The case of G itself is a theorem of Hodge.

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I think there's some confusion in the question. For example by "Levi-Civita connection" you must really mean some kind of Laplacian. Anyway, your end result about the cohomology of $G/H$ is essentially correct. To help clarify the situation, let's bring in some relative Lie algebra cohomology.

The first basic fact is that a Lie group $G$ acts naturally on the space $\Omega^\ast(G/H)$ of differential forms on $G/H$, where $H$ is a closed subgroup of $G$. If $G$ and $H$ are connected, then evaluation at the identity gives an isomorphism from the space $\Omega^\ast(G/H)^G$ of $G$-invariant forms onto the space $C^\ast(\mathfrak g, \mathfrak h)$ of relative Lie algebra cohomology cocycles. This isomorphism descends to an isomorphism between the cohomology of the complex of $G$-invariant forms on $G/H$ and relative Lie algebra cohomology $H^\ast(\mathfrak g, \mathfrak h)$.

Now assume that $G$ is reductive and $H$ maximal compact in $G$ with corresponding Cartan decomposition $\mathfrak g = \mathfrak h \oplus \mathfrak p$. Then there are natural isomorphisms $$C^q(\mathfrak g, \mathfrak h) = \mathrm{Hom}_{\mathfrak h} (\wedge^q \mathfrak p, \mathbb R) = (\wedge^q \mathfrak p)^{\mathfrak h},$$ where $\mathfrak h$ is acting via the adjoint representation. If you want, this can be rephrased in terms of the action of $H$, and moreover it's possible to introduce a bilinear form so that the above isomorphisms yield $$C^q(\mathfrak g, \mathfrak h) = (\wedge^q \mathfrak h^\perp)^H.$$ The next key result is that the differential of this complex is zero, whence $$H^q(\mathfrak g, \mathfrak h) = (\wedge^q \mathfrak h^\perp)^H.$$ In some precise sense this is a manifestation of a theorem of E. Cartan which states that an invariant form on $G/K$ is automatically closed (and harmonic). Indeed, we can combine this last isomorphism with the isomorphisms mentioned in the second paragraph to conclude that $$\Omega^q(G/H)^G = (\wedge^q \mathfrak h^\perp)^H$$ and both are isomorphic to the cohomology of the complex of $G$-invariant forms on $G/H$.

In general we can't deduce from this that the cohomology of $G/H$ is isomorphic to $(\wedge^\ast \mathfrak h^\perp)^H$, because not every closed form will necessarily have an invariant form in its cohomology class. However, if $G$ is compact, then this is not a problem, because we can use the usual averaging trick.

Anyway, the point of introducing Lie algebra cohomology into the picture is that it makes things very computable. It will also be an important key phrase for any literature search.

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Thanks for the answer. I think I do mean the Levi-Civita connection and not a Laplacian. At least for $G$ compact, isn't any harmonic form necessarily $G$ invariant and for a symmetric space any invariant form is parallel. I will look into using that Maple package, though I'm still wondering if there's a list out describing $\Lambda^* \mathfrak h^\perp)^H$ for a large class of examples. –  Eric O. Korman Apr 9 '13 at 0:03
Each cohomology class has a unique harmonic representative by deRham. But under the action of the group $G$, since $G$ is connected, the cohomology class of a harmonic form is unchanged by $G$ action. Therefore the harmonic representative is $G$-invariant. I can't remember if that ensures that it is parallel. –  Ben McKay Apr 11 '13 at 12:46