Ngô Quốc Anh

April 10, 2011

A property of the Weingarten map (the sharp operator)

Filed under: Riemannian geometry — Ngô Quốc Anh @ 23:29

Let $M$ be a surface, $p \in M$ and $\eta$ the unit normal vector (field) of $M$ defined along a neighborhood $W$ of $p$.

Definition (the Weingarten map). The Weingarten map or the sharp operator $L: T_pM \to T_pM$ is defined to be $\displaystyle L(X)=\nabla_X\eta$

for any $X\in T_pM$.

To see this definition is well-defined, in this note, we shall prove that indeed $L(X) \in T_pM$ for any tangent vector $X\in T_pM$. To this purpose, we have to show that $\langle \nabla_X\eta,\eta \rangle =0$.

Be means of the normality, it holds $\langle \eta,\eta \rangle =1$ which implies $\nabla_X\langle\eta,\eta \rangle =0$.

By the property saying that the connection $\nabla$ is compatible with the metric, it holds $\nabla_X\langle\eta,\eta \rangle =\langle\nabla_X\eta,\eta \rangle +\langle\eta,\nabla_X\eta \rangle$.

Thus $2\langle\nabla_X\eta,\eta \rangle=0$.

The proof is complete.

The role of the Weingarten map is important since it is a tool to introduce the second fundamental form $II:T_pM \times T_pM \to \mathbb R$ given by $II(X,Y)=\langle L(X),Y\rangle .$

Apparently, for a plan, the second fundamental form is identically zero and for a sphere $S$ of radius $r$, its second fundamental form is nothing but $\frac{1}{r} {\rm id}$.

1. Notice that the Weingarten map is known as the shape operator. This has nothing to do with sharp!

Comment by OrbiculaR — April 12, 2011 @ 4:27

• Thank you, this is a typo :(.

Comment by Ngô Quốc Anh — April 12, 2011 @ 4:29

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