# Do the values of the differential of a function on a Lie group with a single maximum span the Lie algebra?

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.

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.