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In my differential geometry notes they use a fact that I don't know where it comes from.

It says that any $k$-form can be written as:

$$\omega=\sum_{i=1}^k f_idx^{\mu_1}\wedge \cdots\wedge dx^{\mu_k}$$

That is, as a sum of terms like that, where the $f_i$ are functions.

Particularly, as $\mathbb R^n$ can be covered by a single chart, any $k$-form in $\mathbb R^n$ can be written as: $$\omega = \frac{1}{k!}\omega_{\mu_1\cdots \mu_k}dx^{\mu_1}\wedge \cdots\wedge dx^{\mu_k}$$

I don't understand any of those two statements.

MyUserIsThis
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    You don't understand what the statements are saying, or you don't understand how they can be proved? – Muphrid May 28 '14 at 21:24

1 Answers1

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If you have a differentiable manifold and a local chart $(U;x^1,\cdots, x^n)$ then $\{dx^{i_1}\wedge \cdots \wedge dx^{i_k}/1\leq i_1<i_2<\cdots < i_{k-1}<i_{k}\leq n\}$ is a basis of the vector space of $k$ forms at any point of $U.$ So, any $k$ form can be expressed in such a basis. That is, if $\omega$ is a $k$-form then there exists real numbers $\lambda_{i_1\cdots i_k},1\leq i_1<\cdots <i_k\leq n$ such that $$\omega_p=\sum_{1\leq i_1<\cdots <i_k\leq n} \lambda_{i_1\cdots i_k}(dx^{i_1}\wedge \cdots \wedge dx^{i_k})(p).$$

If we want to express the form in $U$ the numbers $\lambda_{i_1\cdots i_k},1\leq i_1<\cdots <i_k\leq n$ change with the point, that is, they are functions and, so

$$\omega=\sum_{1\leq i_1<\cdots <i_k\leq n} f_{i_1\cdots i_k}dx^{i_1}\wedge \cdots \wedge dx^{i_k}.$$

Finally, since $\mathbb{R}^n$ can be covered by a single chart the expression $$\omega=\sum_{1\leq i_1<\cdots <i_k\leq n} f_{i_1\cdots i_k}dx^{i_1}\wedge \cdots \wedge dx^{i_k}$$ holds on all $\mathbb{R}^n.$ Now if we don't consider the restriction $1\leq i_1<\cdots <i_k\leq n$ we can make $k!$ arrangements. That is the reason to have a $k!$ in

$$\omega = \frac{1}{k!}\omega_{\mu_1\cdots \mu_k}dx^{\mu_1}\wedge \cdots\wedge dx^{\mu_k}=\frac{1}{k!}\sum_{1\leq \mu_1,\cdots,\mu_k\leq n}\omega_{\mu_1\cdots \mu_k}dx^{\mu_1}\wedge \cdots\wedge dx^{\mu_k}.$$

mfl
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