Relationship between negative curvature and positive definite second fundamental form on a manifold with smooth boundary

Let us consider a smooth manifold with a smooth boundary, and let $$\gamma$$ be an oriented unit-speed curve that parameterizes the boundary. The signed curvature of the boundary is given by $$\kappa := \langle D_t \gamma'(t), N \rangle_g$$, where $$N$$ is the unit normal vector field along the boundary and $$D_t$$ is the covariant derivative with respect to the parameter $$t$$.

We define second fundamental form on boudary as $$\mathbb{I}_x(v, w):=-\left\langle\nabla_v \nu, w\right\rangle_g.$$If the second fundamental form of the boundary, denoted by $$\mathbb{I}_x$$, is positive definite, then the manifold is strictly convex. However, it may seem counterintuitive that a negative curvature can arise from a positive definite second fundamental form, and it is not immediately clear from the formula how this relationship works.

Could anyone please shed some light on this topic and explain how negative curvature can be related to a positive definite second fundamental form? Thank you in advance for any insights or help.

• To clarify: You're talking about a $2$-manifold with boundary in the first paragraph. When you're talking about convexity in the second paragraph, it doesn't make sense to me; are you now talking about the second fundamental form of a surface or still of a curve? Please edit your post to give a concrete example illustrating your topic/question. Apr 27, 2023 at 18:48

Let $$N$$ be a unit normal for $$\partial M$$. Since $$\gamma$$ lies in $$\partial M$$, $$\gamma'$$ lies in $$T\partial M$$ and is therefore everywhere orthogonal to $$N$$. It follows that the function $$\langle \gamma',N\rangle$$ is everywhere zero. Deriving this equality with respect to $$t$$ yields $$\langle D_t\gamma',N\rangle + \langle \gamma',\nabla_{\gamma'}N\rangle = 0.$$ Hence, $$\kappa = -\langle \gamma',\nabla_{\gamma'}N\rangle.$$ Now, everything boils down to either $$N = \nu$$ or $$N = -\nu$$ (the normal can be either inward pointing or outward pointing). This will give you a sign for $$\kappa$$ depending on your assumption on the second fundamental form.
• @Curiousstudent I might not have been clear. Convexity does not implies negative curvature, only that the curvature is bigger than that of the ambient space. For the formula I gave, it is just usual Gauss formula between the two Riemann tensors, applied to $(u,v,v,u)$ and divided by $g(u,u)g(v,v)-g(u,v)^2$ to obtain sectional curvatures. As I don't have Lee's book at hand right now, I don't know what formula he gives: if it involves the vector valued second fundamental form $\alpha$, this just comes from $\alpha(u,v) = I\! I_x(u,v)N$ with $N$ the normal vector field. Apr 27, 2023 at 10:15