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I'm reading reading about differential forms and I keep finding alternative notions and I'm quite fed up with the inconsistency. The way I first learned about the is that a $k$-form $\omega$ is a smooth section of the $k$th exterior power of the cotangent bundle i.e. $\omega \in \Gamma \left(\bigwedge^k(T^\ast M)\right)$. This makes sense since $\omega : M \to \bigwedge^k(T^\ast M)$ gives us $\omega_p \in \bigwedge^k(T^\ast M)=\bigcup_{p \in M} \bigwedge^k(T^\ast_pM)$ which further means that $\omega_p$ is a multilinear map $T_pM \times \dots \times T_pM \to \mathbb{R}$. And since $\bigwedge^k(T^\ast M)$ is a vector space with sufficiently nice basis we can express $\omega$ as $$\omega = \sum_{I} a_I dx_I.$$

I'm now reading an example where they consider $\mathbb{R}^n -\{0\}$ and $\omega:\mathbb{R}^n-\{0\} \to \bigwedge^k(\mathbb{R}^n-\{0\})$ as $$\omega_p(v_1,\dots,v_{n-1})=dx_1 \wedge \dots \wedge dx_n \left(\frac{p}{|p|},v_1,\dots, v_{n-1} \right)$$ and this is supposed to be an $(n-1)$-form.

The issue here is that no general formula of $\omega$ is being given. It's only given that the evaluation at $p$ has that formula. How can I recover $\omega$ from this if I only now the evaluation at $p$? I have no information about $a_I(p)$ except that it's apparently $1$ all the time so is it safe to assume that $a_I =1$ since it evaluates to $1$ for all inputs?

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  • $\begingroup$ $\newcommand\R{\mathbb{R}}$ $$\omega: \R^n-\{0\} \rightarrow \bigwedge^k(\R^n-\{0\})$$ should be $$\omega: \R^n-\{0\} \rightarrow \bigwedge^k\R^n$$ because $T_p\R^n = \R^n$. $\endgroup$
    – Deane
    Feb 1, 2023 at 15:30
  • $\begingroup$ Here, $p$ represents both a point in $\mathbb{R}^n-\{0\}$ and a vector field $p: \mathbb{R}^n \rightarrow \mathbb{R}^n$. In the formula, $$V(p) = \frac{p}{|p|}$$ is a vector field on $\mathbb{R}^n-\{0\}$. So for each $p \in \mathbb{R}^n-\{0\}$, So the right side of the equation means $$(dx_1\wedge\cdots\wedge dx_n)_p(V(p), v_1, \dots, v_{n-1}),$$ where $V(p), v_1, \dots, v_{n-1} \in T_p( \mathbb{R}^n-\{0\}) = \mathbb{R}^n$. If you write each $v_k = (v_k^1, \dots, v_k^n)$ and $p = (p^1, \dots, p^n)$, this can be evaluated explicitly. $\endgroup$
    – Deane
    Feb 1, 2023 at 17:19

2 Answers 2

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What is meant is that you take the constant form $dx_1\wedge\dots\wedge dx_n$, then plug in $\tfrac{p}{|p|}$ as the first entry and leave the other $n-1$ entries free to obtain an $(n-1)$-Form. You can easily express this as a linear combination of the possible wedge products of $(n-1)$ of the $dx_i$ by plugging into the definition of the wedge product. The result should essentially be $$ \omega_x=\sum_{i=1}^n(-1)^{i-1}\frac{x_i}{|x|}dx_1\wedge\dots\wedge dx_{i-1}\wedge dx_{i+1}\wedge\dots\wedge dx_n. $$

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  • $\begingroup$ This is $\omega$, it is a function depending on $x$. $\endgroup$ Feb 1, 2023 at 14:09
  • $\begingroup$ @Laurens What? Of course a differential form depends on the point $x$, since it's an algebra over functions which depend on $x$. As the simplest example, a $1$ form over $\mathbb R$ is of the general form $\omega = f(x) dx$ for some smooth $f \colon \mathbb R \to \mathbb R$, which obviously depends on $x$ and whose exterior derivative is trivial to calculate.... If differential forms didn't depend on the point, you couldn't do calculus $\endgroup$ Feb 1, 2023 at 14:23
  • $\begingroup$ In the expansion you have in your question, the $a_I$ are smooth functions that depend on $x$. So in this concrete case $a_{1,\dots ,i-1,i+1,\dots,n}=\frac{x_i}{|x|}$. Computing the exterior derivative amounts to computing partial derivatives of the $a_I$. $\endgroup$ Feb 1, 2023 at 14:41
  • $\begingroup$ I find this a bit confusing. The exterior derivative is defined such that in this case $$d\omega = \sum_{1 \le i_1 < \dots < i_{n-1}\le n} da_{i_1, \dots, i_{n-1}} dx_{i_1} \wedge \dots \wedge dx_{i_{n-1}}$$ but I'm still not seeing how ur getting that these $a$'s all equal $\frac{x_i}{|x|}$. @AndreasCap $\endgroup$
    – Laurens
    Feb 1, 2023 at 18:21
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    $\begingroup$ If you look at the definition of the wedge product and use $dx_j(e_i)=\delta_{i,j}$ then you see that $dx_1\wedge\dots\wedge dx_n(e_i,v_1,\dots,v_{n-1})=(-1)^{i-1}dx_1\wedge\dots\wedge dx_{i-1}\wedge dx_{i+1}\wedge\dots\wedge dx_n(v_1,\dots,v_{n-1})$. $\endgroup$ Feb 3, 2023 at 9:54
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If $\alpha\in \Omega^{n}(M)$ is a differential form and $V\in \mathcal X(M)$ is any vector field, you can create a differential form $\iota_V\alpha\in \Omega^{n-1}(M)$ defined via $$\iota_V\alpha(X_1, \dotsc, X_{n-1}) = \alpha(V, X_1, \dotsc, X_{n-1}).$$ This is called the interior product.

In your case $\alpha = dx_1\wedge \dotsc\wedge dx_n$ is the standard volume form on $M=\mathbb R^n\setminus \{0\}$, $V$ is the radial vector field $V(x) = x/||x||$, and $\omega=\iota_V\alpha$.

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