# Gradient of norm of embedding

Let $\varphi:(M,g,\nabla)\to\mathbb{R}^n$ be a smooth embedding of a convex hypersurface. I want to explicitly calculate $$\langle \varphi,\varphi_{\ast}(\nabla\|\varphi\|^2)\rangle.$$ In particular, I want to know if $\langle \varphi,\varphi_{\ast}(\nabla\|\varphi\|^2)\rangle\geq0.$ Here $\langle,\rangle$ denotes the Euclidean inner product.

Remark: A version of this question was posted by me earlier, however, since it was misrepresented, I decided to erase.

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I assume that ($g, \nabla$) are induced metric and connection from the embedding $\varphi$. Note that $\nabla \| \varphi\|^2 = 2 \varphi ^\top$, the tangential part of $\varphi$. Thus

$$\langle \varphi , \nabla \|\varphi\|^2 \rangle = 2\|\varphi^\top\|^2 \geq 0.$$

(It's true for any immersion $\varphi$)

Remark: To show that $\nabla \| \varphi\|^2 = 2 \varphi ^\top$, let $x\in M$ and $e_1, \cdots e_{n-1}$ be an orthonormal bases of $T_xM$. Then in general for any smooth function $f:M \to \mathbb R$,

$$\nabla f = \sum_{i=1}^{n-1} (\nabla_{e_i} f) \ e_i.$$

Hence

$$\nabla \|\varphi\|^2 = \sum (\nabla_{e_i} \|\varphi\|^2) \ e_i = 2 \sum \langle \varphi, e_i\rangle \ e_i = 2\varphi^\top.$$

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How does one demonstrate the formula $\nabla \Vert \varphi \Vert^2 = 2\varphi^\top$? – Robert Lewis Jul 19 '14 at 16:30
@RobertLewis: I have edited the answer. Please have a look. – Arctic Char Jul 19 '14 at 17:09
I think your answer is correct, I calculated in the geodesic polare coordinate that $\langle \varphi,\nabla\|\varphi\|^2\rangle=2r^2(1-\langle \nu,\partial_r\rangle^2),$ where $\nu$ is the unit normal vector of the convex hypersurface. Since $|\partial_r|=1$, positivity also follows. However, I like your answer. – timhortons Jul 19 '14 at 17:15
@John: thanks. Am meditating on your derivation . . . Om!!! – Robert Lewis Jul 19 '14 at 17:28