$G$ be a subgroup of $GL(n,\mathbb R)$ such that $G$ is also a surface , does there exist a no where vanishing tangent vector field on $G$?

Question

Let $G$ be a subgroup of $GL(n,\mathbb R)$ which is also a $m$-surface for some $m$, how to find a no where vanishing tangent vector field for $G$ ?

Answer 1

Recall that for any Lie group $G$ that left multiplication $L_g: G \to G$ given by $L_g(h) = gh$ is a diffeomorphism (the inverse given by $L_{g^{-1}}$), which furthermore shows that $d(L_g)_h: T_hG \to T_{gh}G$ is of full rank for all $h \in G$ (meaning that $\ker d(L_g)_h = \{0\} \subset T_hG$). Notice that for any Lie subgroup $H \subseteq G$ there exists some nonzero tangent vector $X_e \in T_eH$, hence the corresponding tangent vector $X_h \equiv d(L_h)_e(X_e)$ is nonzero since the Jacobian of left-translation is non-singular. By defining the vector field $X$ pointwise by $X(p) = X_p$, we see that this is smooth since if $\gamma:(-\epsilon, \epsilon) \to H$ is a curve with $\gamma(0) = e$ and $\gamma'(0) = X_e$ we find that for $f \in C^\infty(H)$

$$ (Xf)(p) \;\; =\;\; d(L_p)_e(X_e)f(p) \;\; =\;\; X_e(f\circ L_p)(e) \;\; =\;\; \left . \frac{d}{dt} (f\circ L_g \circ \gamma) \right |_{t=0} $$

which is a smooth map. This proves moreover that $X$ is a smooth vector field that doesn't vanish anywhere.

We can see that we can simply apply the above argument to $GL(n,\mathbb{R})$ and to any appropriate Lie subgroup.