Does every compact Lie group admit a bi-invariant differential $k$-form? 
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.

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 $\eta = 0$.
