§ Gauss, normals, fundamental forms [TODO ]

• consider a parametrization $r: u, v \to \mathbb R^3$
• at a point $p = r(u, v)$ on the surface, the tangent vectors are $r_u \equiv \partial_u r$ and similarly $r_v \equiv \partial_v r$.
• Let $k = xr_u + y r_v$. Then $k \cdot k$ is the first fundamental form . Computed as $k= (xr_u + y r+v) \cdot (x r_u + y r_v)$. Write this as $E x^2 + 2F x y + G y^2$. These numbers depend on the point $(u, v)$, or equally, depend on the point $p = r(u, v)$.
• Further, we also have a normal vector to the tangent plane. $N(p)$ is the unit normal pointing outwards. We can describe it in terms of a parametrization as $n \equiv r_u \times r_v / ||r_u \times r_v||$.
• Gauss map / Gauss Rodrigues map ( $N$): map from the surface to $S^2$. $N$ sends a point $p$ to the unit normal at $p$.
• The tangent plane to $N(p)$ on the sphere is parallel to the tanent plane on the surface at $p$, since the normals are the same, as that is the action of $N$ which sends the normal at the surface $p \in S$ to a point of the sphere / normal to the sphere.
• Thus, the the derivative intuitively "preserves" tangent planes! [as normal directions are determined ].
• If we now think of $dN$, it's a map from $T_p S$ to $T N(p) S^2 = T_p S$. Thus it is a map to the tangent space to itself .
• In terms of this, gauss realized that gaussian curvature $K_2 = K = k_1 k_2$ is the determinant of the map $dN_p$ [ie, the jacobian ]. Curvature is the distortion of areas by the normal. So we can think of it as the ratio of areas area of image/area of preimage.

https://www.youtube.com/watch?v=drOldszOT7I&list=PLIljB45xT85DWUiFYYGqJVtfnkUFWkKtP&index=34