Does there exist bi-invariant inner products on Sp(2N,R)? I'm a little confused: I know that Sp(2N,R) is non-compact and its Killing form is certainly not positive definite (as required for a proper inner product). I recall that a Lie Group has bi-invariant inner products if it is product of a compact Lie group and $\mathbb{R}^n$, but I don't have this intuition for Sp(2N,R).
The reason I'm asking is the following:
- Given an arbitrary point $g\in Sp(2N,R)$ and some left-invariant inner product on the tangent spaces of the group, we can find the shortest geodesics connecting the identity $e$ and the point $g$. Let's call this geodesics $g(s)$.
- Left-invariance ensures that the shortest path from $h$ to $h\cdot g$ is given by $h\cdot g(s)$.
- I'm now looking for the shortest path from $e$ to $h\cdot g\cdot h^{-1}$. In this case, the path $h\cdot g(t)\cdot h^{-1}$ is in general not a geodesic. However, provided that the metric is invariant under the adjoint action of $h$, the map $G\to G: g\mapsto h\cdot g\cdot h^{-1}$ would be an isometry and therefore the if $g(s)$ was the shortest path between $e$ and $g$, $h\cdot g(s)\cdot h^{-1}$ is the shortest path between $e$ and $h\cdot g\cdot h^{-1}$. Being invariant under the adjoint action is equivalent to bi-invariance.
Is this correct? Are there any bi-invariant inner products on Sp(2N,R)? If yes, is there just one or several?
No, $SP(2n)$ has no bi-invariant Riemannian metric. As you say earlier, you can see this from the Lie algebra. An easier argument is that a bi-invariant Riemannian metric would restrict to a bi-invariant inner product on one of the embedded $SL(2)$'s, for example, the group $$\begin{pmatrix} * & 0 & * & 0 \\ 0 & I & 0 & 0 \\ * & 0 & * & 0 \\ 0 & 0 & 0 & I \end{pmatrix},$$ and $SL(2)$ does not admit bi-invariant Riemannian metrics.