Simplifying $d(\frac{2y_1}{r^2+1})\wedge d(\frac{2y_2}{r^2+1})\wedge\cdots\wedge d(\frac{2y_n}{r^2+1})$, where $r^2=\sum y_i^2$

Question

I've been trying to find a nice, simple form for the following expression:

$$d\left(\frac{2y_1}{r^2+1} \right)\wedge d\left(\frac{2y_2}{r^2+1} \right)\wedge ... \wedge d\left(\frac{2y_n}{r^2+1} \right) $$ where $r^2 = y_1^2 + ... + y_n^2\ $, and I am regarding the $y_i$ as the standard coordinates for $\mathbb{R}^n$.

By trying the case $n=3$, I thought the general expression could be $$2^n \frac{1-r^2}{(r^2+1)^{n+1}}dy_1\wedge...\wedge dy_n$$ and I tried to show this by induction but got stuck. Any help is appreciated!

Edit: I came up with the formula by checking the case n=3. Then I tried showing by induction. Base case is saying $$d\left(\frac{2y_1}{y_1^2+1}\right) = \frac{2(1-y_1^2)}{(y_1^2+1)^2}dy_1$$ which is true. Then I tried applying the induction step as follows. Let $x_i = 2y_i/(r^2+1)$ $$dx_1\wedge ... \wedge dx_n = (-1)^{n-1}d(x_n dx_1\wedge ... \wedge dx_{n-1})$$ But I just realized this wouldn't work since I can't really apply the inductive hypothesis to $dx_1\wedge ...\wedge dx_{n-1}$ since I still have $n$ terms for $r^2$.

Answer 1

First, I would get rid of all the $2$'s. You can certainly put the $2^n$ in there at the end.

Write $u=1+r^2$. Note that $$d\left(\frac{y_i}{u}\right) = \dfrac{dy_i}u + y_i\,d\left(\frac1u\right)=\frac{dy_i}u - y_i\,\left(\frac{2r\,dr}{u^2}\right).$$ Then \begin{align*} \left(\frac{dy_1}u+y_1\,d\left(\frac1u\right)\right)&\wedge\dots\wedge\left(\frac{dy_n}u+y_n\,d\left(\frac1u\right)\right) \\ &= \frac{dy_1\wedge\dots\wedge dy_n}{u^n} + \sum\frac{dy_1}u\wedge\dots\wedge y_i\,d\left(\frac1u\right)\wedge\dots\wedge\frac{dy_n}u \\ &=\frac{dy_1\wedge\dots\wedge dy_n}{u^n} + \sum\frac{dy_1}u\wedge\dots\wedge y_i\frac{(-2y_i\,dy_i)}{u^2}\wedge\dots\wedge\frac{dy_n}u \\ &=\frac{dy_1\wedge\dots\wedge dy_n}{u^n} - \frac{2r^2}{u^{n+1}}dy_1\wedge\dots\wedge dy_n \\ &=\frac{(1+r^2)-2r^2}{u^{n+1}}dy_1\wedge\dots\wedge dy_n = \frac{1-r^2}{u^{n+1}}dy_1\wedge\dots\wedge dy_n, \end{align*} as desired. (Note that at most only one $d(1/u)$ factor can appear when you expand the product of sums, since $du\wedge du=0$.)