3
$\begingroup$

Consider: a vector field $X=\nabla \phi$ on a compact, semi-simple, connected matrix Lie group $G$ where $\phi$ as a smooth scalar field on $G$ possessing only a single maxima which topologically is a point, and $\nabla$ is the gradient w.r.t to the bi-invariant metric (unique up to a constant factor which makes no difference here).

Is the following true: the set $S = \left\{ X|_g g^{-1} \ \big| \ g \in G \right\}$ must span $\mathfrak{g}(n)$ (the lie algebra of $G$)?

here $X|_g g^{-1}$ is the right translation of $X|_g$ to the identity of $G$, as we have a matrix Lie group, this composition of a tangent vector and a group element is just matrix multiplication.

$\endgroup$
8
$\begingroup$

Sure. The metric is a red herring here: this is the same as asking if the translates of the differential $d\phi$ span $\mathfrak{g}^*$. Phrased this way, the answer is easy to see. If they didn't span, then there would a non-zero element of the perpendicular to the smaller subspace they span, that is, a non-zero vector $X\in \mathfrak{g}$ such that $d\phi$ is perpendicular to the right invariant vector field $R_X$. If this were the case, then $\phi$ would be constant on trajectories of the form $e^{tX}g$ (since $R_X$ is tangent to these trajectories), so it could not have an isolated maximum.

Note that I didn't use compact, semi-simple or connected here, or for that matter that there is a single maximum; I only need that an isolated maximum or minimum (or an isolated critical point) exists. Actually, this argument shows that only considering the values on any neighborhood of the maximum suffice for a spanning set.

$\endgroup$
1
  • $\begingroup$ I should have said: at least one isolated maximum. $\endgroup$
    – Benjamin
    Jul 31 '15 at 17:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.