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Let $G$ be a compact connected $n$-dimensional Lie group, and let $1 \le k < n$. Do there exist a non-zero bi-invariant differential $k$-form on $G$?

I know that the answer is positive for $k=n$ (top forms)-but the proof I know does not adapt to the case $k<n$. (That proof shows that every left-invariant $n$-form is in fact bi-invariant).

Edit:

I guess that we can always take a bi-invariant top-form $\text{Vol} \in \Omega^n(G)$ and given some right-invariant $k$-form $\omega$ on $G$, define

$$ \tilde \omega=\int_G (L_g)^*\omega \cdot \text{Vol},$$

or more explicitly

$$ \tilde \omega_x=\int_G ((L_g)^*\omega)_x \cdot \text{Vol}\text{ for every } x\in G,$$

where the last integral is an integral of a $\bigwedge^k (T_xG)^*$- valued function, hence makes sense. (as we think of the point $x$ as a constant).

Then if I am not mistaken, $\tilde \omega$ should be bi-invariant; this is similar to how one constructs a bi-invariant Riemannian metric on a compact group, but I guess that in the case of a form, we cannot guarantee that the averaging process won't give zero as a result. (This is in contrast to the metric case, where we have positivity, which is preserved by averaging).

It would be nice to see this in a concrete example; the argument given by Nate Eldredge shows that any bi-invariant $1$-form on a simple Lie group is zero. So the averaging process described above should give zero on any such group, e.g. $\text{SO}(5)$.

I wonder if there is a way to "compute directly" the average, thus verifying that it is zero.

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Take $k=1$ for instance. If $\eta$ is a bi-invariant one-form, then it seems to me we get $\eta(\operatorname{Ad}_g X) = \eta(X)$ for every $g \in G$ and every $X \in \mathfrak{g}$. Taking $g = \exp(tY)$ and differentiating at $t=0$, we get $\eta([Y,X]) = 0$. So if $[\mathfrak{g}, \mathfrak{g}]=\mathfrak{g}$, as for instance when $\mathfrak{g}$ is simple, this forces $\eta = 0$.

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  • $\begingroup$ Thanks, that's a very nice answer. By the way, I guess that one can start with a right (or left) invariant $1$-form, and average on its translates, similarly to how one constructs a bi-invariant Riemannian metric on a compact group; I guess that the difference is that in the case of a form, we cannot guarantee that the averaging process won't produce the zero form. (This is in contrast to the metric case, where we have positivity, which is preserved by the averaging). Does this reasoning sound reasonable to you? (I have edited the question to make the averaging suggestion more explicit). $\endgroup$ – Asaf Shachar Jul 9 at 15:52
  • $\begingroup$ By the way, do you have a guess regarding the situation for $k$-forms where $k >1$? I guess that at least for $k=n-1$, we can conclude by duality that there is no $n-1$ bi-invariant form (since we can use the Hodge star operator w.r.t a bi-invariant metric). $\endgroup$ – Asaf Shachar Jul 9 at 15:58
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There is a classical theorem by E. Cartan that the space of bi-invariant differential forms on a compact Lie group is isomorphic to its de Rham cohomology. (Basically via an averaging argument one shows that the inclusion of the subcomplex formed by bi-invariant forms into the de-Rham complex of a compact Lie group induces an isomorphism in cohomology. Then one shows that the restriction of the exterior derivative to bi-invariant forms is identically zero.) Hence vanishing of $H^k(G,\mathbb R)$ obstructs existence of a non-zero invariant $k$-form.

I doubt that there will be a simple direct argument for vanishing of the form obtained from averaging a left-invariant $k$-form. Compare to the proof of the Schur orthogonality relations for compact groups, for which the vanishing of the average is a rather deep result.

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  • $\begingroup$ Thank you, this is very interesting. Does Cartan's theorem imply that every harmonic form w.r.t a bi-invariant metric on a compact Lie group is bi-invariant? (The converse is true: every bi-invariant form is harmonic, since it's closed, and the Hodge dual preserves bi-invariance). $\endgroup$ – Asaf Shachar Jul 10 at 10:30
  • $\begingroup$ I think it does, and basically you have given the argument: Bi-invariant forms are a subspace of harmonic forms, but since the dimensions of both spaces coincide with the dimension of de Rham cohomology, they have to agree. $\endgroup$ – Andreas Cap Jul 10 at 11:11

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