If I think to the surface genus $g$-torus as the boundary of a solid genus $g$-torus in $\mathbb{R}^{3}$, is it possible to have a nonvanishing smooth vector field such that in every point is pointing inward ?
I think the answer is no if $g \ne 1$ : this should follows from the fact that for the Hopf-Lemma given such field than the sum of the index, i.e. $0$ for the sake of contradiction should be equal the gauss map on the boundary, which means the genus $g$-torus, which is $1-g$ for a known fact.
So $g=1$ is the only problem to deal with, and I think that even for $g = 1$ the statement can't hold but I couldn't find any contradiction in that,
Any help would be appreciated.
Such a vector field exists: The solid torus is the surface of revolution of the closed disc, so it can be described with three coordinates $(r,\theta,\varphi)\in[0,1]\times[0,2\pi)^2$, e.g. with the embedding $$ \begin{pmatrix}r\\\theta\\\varphi\end{pmatrix}\mapsto\begin{pmatrix}(2+r\cos\varphi)\cos\theta\\(2+r\cos\varphi)\sin\theta\\r\sin\varphi\end{pmatrix}\in\mathbb{R}^3 $$ Using $\partial_r,\partial_\theta,\partial_\varphi$ as the corresponding tangent frame, I claim the vector field $V=\partial_\theta-r\partial_r$ is a smooth nonvanishing vector field that is inward pointing on the boundary. One can check smoothness easily enough by writing $V$ in $\mathbb{R}^3$.