Are there metrics of nonnegative Gaussian curvature on these surfaces?

Question

Let $\Sigma$ be a compact surface of genus $g \geq 1$ and having $r \geq 1$ boundary components. Are there metrics of nonnegative Gaussian curvature on $\Sigma$? If $\Sigma$ were closed, then the Gauss Bonnet theorem implies that no such metrics exist, except for the flat torus ($g = 1$).

Answer 1

There do indeed exist such metrics. The idea is that any compact surface with nonempty boundary can be smoothly immersed in the plane, and hence in $S^2$, and so you can get a metric of constant Gaussian curvature $+1$ by pulling back the Riemannian metric on $S^2$ under the immersion. What I say is true as long as $\Sigma$ is orientable, an assumption that I believe you are using too, so I will proceed with that assumption. (One could do the non-orientable case too by immersion into the projective plane).

Your surface $\Sigma$ can be obtained from a regular polygon $P \subset \mathbb R^2$ with $4k$ sides, $k \ge 1$, by indexing the sides in some order, choosing a pairing of the even-indexed sides, and then, for any pair of even-indexed sides $E,E'$, choosing a gluing diffeomorphism $E \leftrightarrow E'$ that reverses counter-clockwise orientation.

For instance, think of the torus as obtained by identifying opposite sides in the usual fashion, but before doing the identifications first truncate the corners to get an octagon; the resulting identifications of the even-indexed opposite side pairs of that octagon gives a one-holed torus, $g=1$, $r=1$.

So then, instead of actually identifying a certain side pair $E,E'$ of $P$, instead use a smooth embedding $[0,1] \times [0,1] \hookrightarrow \mathbb R^2$ so that the image of this embedding intersects $P$ along $[0,1] \times 0 \approx E$ and $[0,1] \times 1 \approx E'$ (the existence of this smooth embedding uses the assumption that the gluing diffeomorphism $E \leftrightarrow E'$ reverses counterclockwise orientation). This gives a topological immerison of $\Sigma$ which is smooth except at each corner of $P$, but a bit of smoothing will iron out the corners.

Answer 2

Here's a different answer using conformal deformations.

First observe that the Gauss--Bonnet theorem for a compact surface $(M^2,g)$ with boundary states that $$ 2\pi\chi(M) = \int_M K\,d\sigma + \oint_{\partial M} k\,ds , $$ where $k$ is the geodesic curvature of the boundary. In particular, there is no reason to expect that the Euler characteristic obstructs the existence of a metric of nonnegative Gaussian curvature unless you also require that the boundary is (weakly) convex.

Next observe that if $\hat g = e^{2u}g$, then \begin{align} e^{2u}K_{\hat g} & = K_g - \Delta_g u, \\ e^u k_{\hat g} & = k_g + \partial_n u, \end{align} where $\partial_n$ is the outward-pointing unit normal (with respect to $g$). In particular, if $u$ is the solution of the Dirichlet problem $$ \begin{cases} \Delta_gu = K_g, & \text{in $M$}, \\ u = 0, & \text{on $\partial M$} , \end{cases} $$ then the metric $e^{2u}g$ will be flat (the boundary condition is irrelevant here; I just give an example where you know there exists a unique smooth solution, provided $\partial M\not=\emptyset$). In other words, you can conformally deform $any$ compact surface with boundary to have Gaussian curvature identically zero.

Most interestingly, Osgood, Philips and Sarnak proved [Extremals of determinants of Laplacians, Theorem 1(c)] that given a compact surface $(M^2,g)$ with boundary, there is a unique (up to homothety) conformally related metric $e^{2u}g$ with Gaussian curvature identically zero and boundary of constant geodesic curvature.