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This is bound to have some easy explanation in terms of aligning the differential of the normal vector to the surface $S$ with the direction of the vector taken into the quadratic $\vec v$, but I am not sure why. Since it is a definition, the temptation is to just take it at face value, but it has to have an explanation.

$$II_p=\color{red}-\langle dNp(\vec v), v \rangle$$


I found a good explanation on Reddit, and I will keep the question open to see if I get better answers, although this is really good:

The second fundamental form measures the extrinsic curvature. The actual definition is a bit cumbersome, but interpretation is very geometric. The value on a unit vector (unit with respect to the first fundamental form) is how much the surface bends in that direction. Larger number means more bending. The sign tells whether the surface bends upwards or downwards - for this you need to choose which side of the surface is a "top side" and "bottom side", mathematically this is a choice of the unit normal vector field. Positive signs says that standing on the top side it looks like $-x^2$ (a hill), negative signs says that standing on the top side it looks like $x^2$ (a valley).

If the second fundamental form is positive definite or negative definite, the surface looks locally like a portion of a sphere (and the definiteness just tells you whether the unit normal points inside or outside). If there exist both positive and negative directions, then the surface locally looks like a saddle surface.

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There's no intrinsic geometric way to distinguish the sign, as you can change the direction of the unit normal field $N$. But positive normal curvature means that the the curve is bending toward $N$ rather than away from $N$.

The main reason for the negative sign is to remove a negative sign from the second fundamental form when you compute it with second-order partial derivatives using a parametrization. That is, if you have a parametrization $\mathbf x(u,v)$, then, for example, $$\text{II}_p(\mathbf x_u,\mathbf x_u) = \mathbf x_{uu}\cdot N.$$ This is how one usually computes the matrix $\text{II}$. The negative sign arises because when we differentiate $\mathbf x_u\cdot N = 0$, we get $$\mathbf x_{uu}\cdot N = -N_u\cdot \mathbf x_u = -dN_p(\mathbf x_u)\cdot \mathbf x_u.$$

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  • $\begingroup$ Thank you! Right... differentiating $\langle \vec x_u, N\rangle=0$ we get $\langle \vec x_{uu}, N\rangle + \langle \vec x_u, N_u\rangle=0,$ which means that $\langle x_{uu},N\rangle = -\langle dN_p(\vec x_u), \vec x_u\rangle..$ $\endgroup$ Apr 18, 2020 at 19:07

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