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.
Best Answer
Edit: I misread the question.
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.