Can projective plane be a minimal surface?

Question

Let $(M,g)$ be a Riemannian 3 manifold and $\Sigma\subset M$ be a closed minimal surface. Can $\Sigma$ be diffeomorphic to a projective plane or can we notice certain necessary properties for a minimal surface that are invariant for diffeomorphism such that projective plane doesn't have?

Answer 1

Let $\Sigma$ be a submanifold of a Riemannian manifold $(M, g)$. The second fundamental form of $\Sigma$ is $\mathrm{II}(X, Y) = (\nabla_XY)^{\perp}$ and the mean curvature is $H = \operatorname{tr}_g(\mathrm{II})$. The submanifold $\Sigma$ is totally geodesic if $\mathrm{II} \equiv 0$ and minimal if $H \equiv 0$. Note that a totally geodesic submanifold is minimal, which is why Ted Shifrin's comment gives an answer to the question.

More generally, any manifold can be realised as a totally geodesic submanifold and hence as a minimal submanifold. One way to do this is to use the fact that if $(M_1, g_1)$ and $(M_2, g_2)$ are Riemannian manifolds, then $M_1$ and $M_2$ are totally geodesic submanifolds of $(M_1\times M_2, g_1 + g_2)$. In particular, $M_1$ is a minimal hypersurface of $(M_1\times S^1, g_1 + dt^2)$.

Even more generally, if $\Sigma$ is a submanifold of $M$, then $M$ admits a Riemannian metric for which $\Sigma$ is totally geodesic and hence minimal, see this MathOverflow question.