It’s a well known result that if (G) is a connected and compact Lie group, then it’s de Rham cohomology is isomorphic to the cohomology of its Lie algebra. The compactness assumption is needed to turn arbitrary differential forms into left-invariant ones by integrating over (G). Why do we need to assume (G) is also connected?
@Diffgeometer1 - the Lie algebra of a Lie group is isomorphic to the Lie algebra of its identity component, so there's no way the Lie algebra can know what's going on in other components.
Consider the Lie groups U(1) and U((1)×ℤ/2. They have isomorphic Lie algebras but they have different zeroth and first cohomology groups!
@johncarlosbaez Thanks! That makes sense. As you pointed out, if (G) has more than one component then, for example, the degree (0) de Rham cohomology will be isomorphic to (k)-copies of (\mathbb{R}) where (k) is the number of components of (G). On the other hand, the degree (0) cohomology of its Lie algebra is always isomorphic to (\mathbb{R}). So there's no way (H^\bullet_{dR}(G)\simeq H^\bullet(\mathfrak{g})) could be true without the assumption that (G) is connected.
@Diffgeometer1 - More generally if any space X is the disjoint union of n copies of some space Y, the de Rham cohomology of X in any degree will be the direct sum of n copies of the de Rham cohomology of Y. This lets you easily work out the de Rham cohomology of a Lie group with multiple connected components if you know the cohomology of any one component (since the components are all diffeomorphic).
@johncarlosbaez I didn’t consider that! All the components of a Lie group are diffeomorphic. I forgot! So if (G) is compact and not necessarily connected, then its entire de Rham cohomology can be computed from its Lie algebra cohomology. As you said, if (G) has (n) components, then it’s de Rham cohomology is just (n) copies of (H^\bullet(\mathfrak{g}))! So the whole thing is known! Thanks!
@johncarlosbaez i looked at the proof which equates Lie algebra cohomology with de Rham cohomology for (G) a connected compact Lie group to see where the connectedness of (G) is being used exactly. As far as I can tell, this condition is needed so that there exists a curve joining every element (g\in G) to the identity element (e\in G). This implies that can (L_g) (left translation by (g\in G)) is homotopic to (L_e = id). From this one can construct a chain homotopy to show that the inclusion (\iota: \Omega^\bullet_L(G)\hookrightarrow\Omega^\bullet(G)) induces an isomorphism between cohomology of left-invariant forms and de Rham cohomology of all forms. Compactness is needed not only for integration, but also to have a finite partition of unity. The chain homotopy is then a finite sum of linear maps (\Omega^k(G) \rightarrow \Omega^{k-1}(G)) using the finite partition of unity. The proof is not hard hard but also not trivial. I’ve been curious about this result for awhile so it’s interesting to go through the proof and learn new techniques along the way.
Also the same proof can be used to show that if a compact connected Lie group (G) acts on a manifold (M) (from the left), then [H^\bullet_{dR}(M)\simeq H^\bullet_G(M)] where (H^\bullet_G(M)) is the cohomology of (G)-invariant forms on (M).
@Diffgeometer1 - the generalization you mention is new to me and interesting. It becomes more useful when there's a finite-dimensional space of G-invariant forms on M. At teh other extreme, when the action of G is trivial, the theorem is true but useless!
@johncarlosbaez Exactly! One place where we see a finite dimensional space of (G)-invariant forms is on reductive homogeneous manifolds. If (M=G/H) is a reductive homogeneous manifold, then one has a decomposition (\mathfrak{g}=\mathfrak{h}\oplus \mathfrak{b}) where (\mathfrak{b}) is merely a subspace of (\mathfrak{g}) which is also invariant under the adjoint action of (H). The space of (G)-invariant (k)-forms can then be identified with a subspace of (\wedge^k\mathfrak{b}^\ast) where (\mathfrak{b}^\ast) is the dual of (\mathfrak{b}). So this is more general than Lie groups and everything can still be reduced to algebra which is very nice.
@Diffgeometer1 - nice! It would be fun to try to use this to compute the de Rham cohomology of the Stiefel manifold Vₖ(ℝⁿ), which is the space of all lists of orthonormal vectors v₁, v₂, ..., vₖ in ℝⁿ. These lists are called "orthonormal k-frames". They're just (ordered) orthonormal bases when k = n, but they're interesting more generally.
The Stiefel manifold Vₖ(ℝⁿ) is usually described as G/H where
G = O(n)
H = O(n-k)
since rotations act transitively on orthonormal k-frames and the stabilizer is O(n-k). But I believe when k < n we can also describe this Stiefel manifold as G/H where
G = SO(n)
H = SO(n-k)
so we get G connected. For k = n we don't have O(n)/O(n-k) ≅ SO(n)/SO(n-k) because the former is O(n), the space of orthonormal bases, while the latter is SO(n), the space of oriented orthonormal bases. But there's no concept of an oriented k-frame for k < n so I think we get
O(n)/O(n-k) ≅ SO(n)/SO(n-k)
for k < n. In any event, I think the theorem you mentioned may let us compute the de Rham cohomology of SO(n)/SO(n-k). I'm aiming at an example that should be computable yet not trivial.
I’ve heard of the Stiefel manifold but never worked with it directly but I think I’ll start to.
I like that the (n)-dimensional sphere is just a special case of the Steifel manifold.[S^n=SO(n+1)/SO(n)=V_1(\mathbb{R}^{n+1})]The proof of the theorem which states[H^\bullet_G(M)\simeq H^\bullet_{dR}(M)]carries over virtually unchanged from the Lie group case if we assume that (G) acts transitively on (M). In other words, (M = G/H). If the (G)-action is not transitive, I’m not sure if the theorem is still true. In any case, the theorem remains true for the Steifel manifold so no problem with using the theorem to calculate the de Rham cohomology.
If you happen to find out if the theorem is still true for all actions (not just transitive ones), please let me know. I think it should remain true for all actions but the proof becomes more complicated.
Update: I’m pretty sure the result is true for all (G)-actions, not just transitive ones. The proof just requires a little modification. It’s not as difficult as I thought.
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