📂Geometry

The Relationship between the Second Normal Form and the Vingarten Map

Theorem¹

Let’s call the tangent vector for a point $p$ on a surface $M$ as $\mathbf{X}, \mathbf{Y} \in T_{p}M$. Then the following holds.

$$II(\mathbf{X}, \mathbf{Y}) = \left\langle L(\mathbf{X}), \mathbf{Y} \right\rangle = \left\langle \mathbf{X}, L(\mathbf{Y}) \right\rangle$$

Here, $L$ is the Weingarten map.

Description

In other words, the Weingarten map $L$ is a self-adjoint linear transformation.

Proof

Properties of the Weingarten Map

If we define ${L^{l}}_{k} = \sum \limits_{i} L_{ik}g^{il}$, then the following holds.

$$L(\mathbf{x}_{k}) = \sum_{l} {L^{l}}_{k}\mathbf{x}_{l}$$

Here, $L_{ij}$ is the coefficient of the second fundamental form, $[g^{il}]$ is the inverse matrix of the first fundamental form coefficient matrix.

Let’s say $\mathbf{X} = X^{i}\mathbf{x}_{i}, \mathbf{Y} = Y^{j}\mathbf{x}_{j}$. Then, by the properties of the Weingarten map, the following holds. Using Einstein notation,

$$\begin{align*} \left\langle L(\mathbf{X}) , \mathbf{Y} \right\rangle =&\ \left\langle X^{i}{L^{l}}_{i}\mathbf{x}_{l}, Y^{j}\mathbf{x}_{j} \right\rangle \\ =&\ X^{i}Y^{j}{L^{l}}_{i} \left\langle \mathbf{x}_{l}, \mathbf{x}_{j} \right\rangle \\ =&\ X^{i}Y^{j}{L^{l}}_{i} g_{lj} \\ =&\ X^{i}Y^{j}L_{ki}g^{kl} g_{lj} \\ =&\ X^{i}Y^{j}L_{ki}\delta_{j}^{k} \\ =&\ X^{i}Y^{j}L_{ji} \\ =&\ II(\mathbf{X}, \mathbf{Y}) \end{align*}$$

The same result is obtained by the same method for $\left\langle \mathbf{X}, L(\mathbf{Y}) \right\rangle$ as well.

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