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How does one show that the surface element of $S^{3} = \{x=(x_{1}, \dots, x_{4}) \in \mathbb{R}^{4} \mid |x|^2=1\}$ is given by the following $3$-form:

$$\omega=x_{1}dx_{2}\wedge dx_{3}\wedge dx_{4}-x_{2}dx_{1}\wedge dx_{3}\wedge dx_{4}+x_{3}dx_{1}\wedge dx_{2}\wedge dx_{4}-x_{4}dx_{1}\wedge dx_{2}\wedge dx_{3}?$$

Is there a standard way of calculating this? How to begin with it?

Thanks in advance.

Michael Albanese
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user124471
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    you take the unit normal vector - in this case $\sum_{i=1}^4 x_i,\partial/\partial x_i$, and contract it with the volume form $dx^1\wedge\dots\wedge dx^4$ on the ambient space $\mathbb R^4$ – user8268 May 25 '14 at 19:34
  • @user8268 Would you recommend a reference textbook for this definition? – rych May 27 '14 at 08:29
  • You really shouldn't call it a surface element as $S^3$ is not a surface. – Michael Albanese Jan 05 '15 at 19:25

1 Answers1

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I addressed how to calculate an orientation form on $S^n$ in this answer. It is given by

$$\alpha = \left.\sum_{j=1}^{n+1}(-1)^{j-1}x^jdx^1\wedge\dots\wedge dx^{j-1}\wedge dx^{j+1}\wedge\dots\wedge dx^{n+1}\right|_{S^n}.$$

In your case, $n = 3$ and instead of denoting the orientation form by $\alpha$, you are using $\omega$. So we have

\begin{align*} \omega =&\ (-1)^{1-1}x^1dx^2\wedge dx^3\wedge dx^4 + (-1)^{2-1}x^2 dx^1\wedge dx^3\wedge dx^4 + (-1)^{3-1}x^3 dx^1\wedge dx^2\wedge dx^4\\ &+ (-1)^{4-1}x^4 dx^1\wedge dx^2\wedge dx^3\\ =&\ x^1dx^2\wedge dx^3\wedge dx^4 - x^2 dx^1\wedge dx^3\wedge dx^4 + x^3 dx^1\wedge dx^2\wedge dx^4 - x^4 dx^1\wedge dx^2\wedge dx^3 \end{align*}

where of course $\omega$ is restricted to $S^3$ (it is actually a well-defined form on $\mathbb{R}^4$).

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Michael Albanese
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