Show that $d\beta=0 \iff p=n/2$

Question

Let $\beta$ be the $(n-1)$-form on $\mathbb{R}^n \setminus \{0\}$ given by

$\displaystyle \beta = \sum_{i=1}^{n}(-1)^{i-1}\frac{x^i dx^1 \wedge dx^2 \wedge \dots \wedge \hat{dx^i} \wedge \dots \wedge dx^n}{((x^1)^2+\dots+(x^n)^2)^p}$

where $\hat{dx^i}$ means that $dx^i$ is to be omitted. Show that $d\beta=0 \iff p=n/2$.

Let $r=((x^1)^2+\dots+(x^n)^2$. So

$\displaystyle \begin{align} d\beta &= d(\sum_{i=1}^{n}(-1)^{i-1}\frac{x^i dx^1 \wedge dx^2 \wedge \dots \wedge \hat{dx^i} \wedge \dots \wedge dx^n}{((x^1)^2+\dots+(x^n)^2)^p}) \\ &= \sum_{i=1}^{n}(-1)^{i-1} d(\frac{x^i dx^1 \wedge dx^2 \wedge \dots \wedge \hat{dx^i} \wedge \dots \wedge dx^n}{((x^1)^2+\dots+(x^n)^2)^p}) \\ &= \sum_{i=1}^{n}(-1)^{i-1} d(\frac{x^i}{r^p} dx^1 \wedge dx^2 \wedge \dots \wedge \hat{dx^i} \wedge \dots \wedge dx^n) \\ &= \sum_{i=1}^{n}(-1)^{i-1} d(\frac{x^i}{r^p}) \wedge dx^1 \wedge dx^2 \wedge \dots \wedge \hat{dx^i} \wedge \dots \wedge dx^n \\ &= \sum_{i=1}^{n}(-1)^{i-1} [\frac{1}{r^p}dx^i - \frac{p}{r^{p+1}}dr] \wedge dx^1 \wedge dx^2 \wedge \dots \wedge \hat{dx^i} \wedge \dots \wedge dx^n \\ &= \sum_{i=1}^{n}(-1)^{i-1} [\frac{1}{r^p}dx^i] \wedge dx^1 \wedge dx^2 \wedge \dots \wedge \hat{dx^i} \wedge \dots \wedge dx^n - \sum_{i=1}^{n}(-1)^{i-1} [\frac{p}{r^{p+1}}dr] \wedge dx^1 \wedge dx^2 \wedge \dots \wedge \hat{dx^i} \wedge \dots \wedge dx^n \\ &= \sum_{i=1}^{n}(-1)^{i-1}(-1)^i \frac{1}{r^p} \wedge dx^1 \wedge \dots \wedge dx^i \wedge \dots \wedge dx^n - \sum_{i=1}^{n}(-1)^{i-1} [\frac{p}{r^{p+1}}\frac{\partial r}{\partial x^i}dx^i] \wedge dx^1 \wedge dx^2 \wedge \dots \wedge \hat{dx^i} \wedge \dots \wedge dx^n \\ &= \sum_{i=1}^{n} \frac{1}{r^p} \wedge dx^1 \wedge \dots \wedge dx^i \wedge \dots \wedge dx^n - \sum_{i=1}^{n} [\frac{p}{r^{p+1}}2x^idx^i] \wedge dx^1 \wedge dx^2 \wedge \dots \wedge \hat{dx^i} \wedge \dots \wedge dx^n \\ &= \sum_{i=1}^{n} \frac{1}{r^p} \wedge dx^1 \wedge \dots \wedge dx^i \wedge \dots \wedge dx^n - \sum_{i=1}^{n} [\frac{2p}{r^{p+1}}x^i] \wedge dx^1 \wedge dx^2 \wedge \dots \wedge x^i \wedge \dots \wedge dx^n \end{align}$

I do not see what I have done wrong or how this could imply that $p=n/2$.

Answer 1

I don't know where your error is, but using the exterior derivative is simpler: $$ \begin{align} d\beta &= d\left(\sum_{i=1}^{n}(-1)^{i-1}\frac{x^i dx^1 \wedge dx^2 \wedge \dots \wedge \hat{dx^i} \wedge \dots \wedge dx^n}{((x^1)^2+\dots+(x^n)^2)^p}\right) \\ % &= \sum_{i=1}^{n}(-1)^{i-1} \sum_{k=1}^n\frac{\partial}{\partial x^k}\frac{x^i }{((x^1)^2+\dots+(x^n)^2)^{p}} dx^k \wedge dx^1 \wedge dx^2 \wedge \dots \wedge \hat{dx^i} \wedge \dots \wedge dx^n \\ % &= \sum_{i=1}^{n}(-1)^{i-1}\frac{\partial}{\partial x^i}\frac{x^i }{((x^1)^2+\dots+(x^n)^2)^{p}} dx^i \wedge dx^1 \wedge dx^2 \wedge \dots \wedge \hat{dx^i} \wedge \dots \wedge dx^n \\ % &= \sum_{i=1}^{n}(-1)^{i-1}\frac{\partial}{\partial x^i}\frac{x^i }{((x^1)^2+\dots+(x^n)^2)^{p}} (-1)^{i-1} dx^1 \wedge dx^2 \wedge \dots \wedge dx^i \wedge \dots \wedge dx^n \\ % &= \left(\sum_{i=1}^{n} \frac{\partial}{\partial x^i}\frac{x^i }{((x^1)^2+\dots+(x^n)^2)^{p}}\right) dx^1 \wedge dx^2 \wedge \dots \wedge dx^i \wedge \dots \wedge dx^n \end{align} $$ Now, each term of the sum is: $$ \frac{\partial}{\partial x^i}\frac{x^i }{((x^1)^2+\dots+(x^n)^2)^{p}} = \frac{r^p - 2pr^{p-1}(x^i)^2}{r^{2p}} = \frac{r - 2p(x^i)^2}{r^{p+1}}, $$ so putting everything together: $$ d\beta = \frac{nr-2pr}{r^{p+1}} dx^1 \wedge dx^2 \wedge \dots \wedge dx^i \wedge \dots \wedge dx^n. $$ EDIT: You're missing the square in the $(x^i)$ term! Everything else seems right.