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The volume form on the unit sphere $S^{n}$ in $\mathbb{R}^{n+1}$ is given by

$$i_{\bf r}\ dx^1 \wedge \cdots \wedge dx^{n+1}=\sum (-1)^{i-1} x^i \, dx^1 \wedge\cdots \wedge \widehat{dx^i} \wedge \cdots \wedge dx^{n+1}$$

Why must the volume form $dx^1 \wedge \dots \wedge dx^{n+1}$ act on the vector ${\bf r}$ to give the volume form on the unit sphere?

Also, how do I get the form of the volume form on the right-hand side of the equation?

Michael Hardy
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nightmarish
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1 Answers1

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The vector $r=\sum_ix^i\frac{\partial}{\partial x^i}$ is the extension of the unit normal of $S^n$ to a vector field on $\mathbb R^{n+1}$. (It is half of the gradient of the function $r=\sum_i(x^i)^2$ for which the sphere is a level set, and along the unit sphere it has length one.) So what you do is just inserting the unit normal into the volume form on $\mathbb R^{n+1}$ to get a volume form on the hypersurface $S^n$. The explicit form readily follows from the fact that $dx^i(r)=x^i$ which is obvious from the definition of $r$ and the compatibility of an insertion operator with a wedge product.

Andreas Cap
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  • I don't understand why $\displaystyle{{\bf r} = \sum_{i=1}^{n+1} x^{i} \frac{\partial}{\partial x^{i}}}$.

    My hunch is that $\displaystyle{{\bf r} = \sum_{i=1}^{n+1} \frac{\partial}{\partial x^{i}}}$ since $\bf r$ is a unit vector and must have components which are unity. But, I don't see why my hunch is wrong.

     – nightmarish Nov 30 '16 at 00:28
  • The normalized vector $\mathbf{r}$, interpreted as a vector field, is $\mathbf{r} = \frac{1}{\sqrt{\sum_{i=1}^{n+1} (x^i)^2}} \sum_{i=1}^{n+1} x^i \frac{\partial}{\partial x^i}$ (this is identified with the vector field $\frac{1}{\sqrt{\sum_{i=1}^{n+1} (x^i)^2}} (x^1, \dots, x^{n+1})$ by replacing $\frac{\partial}{\partial x^i}$ with $e_i$) and then each component is of unit length. It is not $\sum_{i=1}^{n+1} \frac{\partial}{\partial x^i}$ which is identified with the constant vector field $(1,\dots,1)$. – levap Nov 30 '16 at 01:36
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    Since you did not write in the question what $r$ is, I made a reasonable assumption. Anyway, the details are not important as long as $r$ restricts to the unit normal along $S^n$. This is the case both for my suggestion and the one in the comment by @levap . – Andreas Cap Nov 30 '16 at 09:05