How to calculate $V^i \frac{\partial}{\partial u^i}\frac{2R^2u^j}{|u|^2+R^2}$

Question

This is part of the calculation from page 37 of Lee's Riemannian Manifolds where we are trying to prove that Stereographic projection is a conformal equivalence.

The book says:

$V^i \frac{\partial}{\partial u^i}\frac{2R^2u^j}{|u|^2+R^2}=\frac{2R^2V_j}{|u|^2+R^2} - \frac{4R^2u^j<V,u>}{(|u|^2+R^2)^2}$

Where $u=(u_1,....u_n,0) \in \mathbb{R}^{n+1}$ and $V(|u|^2)=2 \Sigma_k V^ku^k=2<V,u>$

I'm having problems getting from one side of the equality to the others... This is ultimately just the quotient rule, right? Can somebody help me with this? Thanks

Answer 1

I don't know what any of these terms mean, but it does appear to be a simple product rule type equation. The key thing that is absorbed into the notation is that $|u|^2$ is a function of $u_1,u_2,...,u_n$ and must be considered so for the derivatives.

Assuming $R$ is a constant, consider a lower dimensional example $$ \frac{\partial}{\partial a} \frac{1}{a^2+b^2+R^2} = -\frac{2 a}{(a^2+b^2+R^2)^2} $$ so in the case where $i \ne j$, if $|u|^2=\vec{u}\cdot\vec{u}=\langle u,u\rangle$ $$ \frac{\partial}{\partial u^i} \frac{u^j}{|u|^2+R^2} = u^j\frac{\partial}{\partial u^i} \frac{1}{|u|^2+R^2} = -\frac{2 u^i u^j}{(|u|^2+R^2)^2} $$ and $$ V^i\frac{\partial}{\partial u^i} \frac{u^j}{|u|^2+R^2} = -\frac{2 V^i u^i u^j}{(|u|^2+R^2)^2} $$ with the implied sum over $i$ (Einstein notation) this gives the inner product term $$ V^i\frac{\partial}{\partial u^i} \frac{u^j}{|u|^2+R^2} = -\frac{2 \langle V,u\rangle u^j}{(|u|^2+R^2)^2} $$ the origin of the other term is going to be solely from the case that $i=j$, which will invoke a product (or quotient, same thing) like rule. Consider:

$$ \frac{\partial}{\partial u^j} \frac{u^j}{|u|^2+R^2} = \frac{1}{|u|^2+R^2}\frac{\partial}{\partial u^j}u^j + u^j\frac{\partial}{\partial u^j}\frac{1}{|u|^2+R^2} $$ the second term we know how to deal with as above, and is needed to contribute to the terms as before, in the first term here $$ \frac{\partial}{\partial u^j}u^j = 1 $$ but now the $V^i$ is definately a $V^j$ because it's the case where $i=j$. I don't know why the index dropped, it might be a typo, or perhaps this is some kind of contra-co-variant magic that I don't have the background to fully understand. Otherwise the $2R^2$ looks like a constant which will factor in.

Hope this helps!

Edit:

I think I see the raising and lowering of the index must be that $V^i$ is a row vector, and turns into a column vector $V_j$ after the product with the matrix of values.