Push vector field from Euclidean space to tangent space of manifold via charts

If ϕ:S2→R2\phi:S^2\to\mathbb{R}^2, (x,y,z)↦(x1−z,y1−z)(x,y,z)\mapsto (\frac{x}{1-z},\frac{y}{1-z}) (hence ϕ−1:(u,v)↦(2uu2+v2+1,2vu2+v2+1,u2+v2−1u2+v2+1)\phi^{-1}:(u,v)\mapsto (\frac{2u}{u^2+v^2+1},\frac{2v}{u^2+v^2+1},\frac{u^2+v^2-1}{u^2+v^2+1})), and I were to consider the vector field ∂∂u\frac{\partial}{\partial_u} on R2\mathbb{R}^2, im unsure as to how to push this into (x,y,z)(x,y,z) coordinates. Do I calculate the derivative matrix of ϕ−1\phi^{-1} and multiply it by the matrix (∂∂u0)
0 \end{pmatrix}
and then just use ϕ\phi to change all coordinates back from (u,v)(u,v) to (x,y,z)(x,y,z)? But then how do I change from ∂∂u\frac{\partial}{\partial_u} to ∂∂k\frac{\partial}{\partial_k} for k=x,y,zk=x,y,z?




What you are looking for is d(ϕ−1)(∂∂u)d(\phi^{-1})\left(\frac {\partial}{\partial_u}\right). So calculate d(ϕ−1)d(\phi^{-1}) and apply it to ∂∂u\frac {\partial}{\partial_u}.
– Paul Sinclair
2 days ago


1 Answer


You have a map ϕ:S2⊂R3→R2\phi: S^2 \subset \mathbb{R}^3 \to \mathbb{R}^2. Let u,vu,v be the coordinates function on S2S^2 and x,yx,y that on R2\mathbb{R}^2. You have ϕ−1:R2→S2⊂R3\phi^{-1}: \mathbb{R}^2 \to S^2 \subset \mathbb{R}^3 and you wish to sent a vector field on R2\mathbb{R}^2 to one on S2S^2. Then you want to use the push-forward ϕ−1∗:=ϕ−1∗,ϕ(p):Tϕ(p)R2→TpS2\phi^{-1}_*:=\phi^{-1}_{*,\phi(p)}: T_{\phi(p)} \mathbb{R}^2 \to T_pS^2 given by:

(ϕ−1∗Xϕ(p))(f)=Xϕ(p)(f∘ϕ−1)\left(\phi^{-1}_*X_{\phi(p)}\right) (f) = X_{\phi(p)} (f \circ \phi^{-1})

where ff is a germ at p∈U⊂S2p \in U \subset S^2 i.e ϕ−1∗:Xϕ(p)∈Tϕ(p)R2↦ϕ−1∗Xϕ(p)∈TpS2\phi^{-1}_*: X_{\phi(p)} \in T_{\phi(p)}\mathbb{R}^2\mapsto \phi^{-1}_*X_{\phi(p)} \in T_p S^2. Recall first that the bases for Tϕ(p)S2T_{\phi(p)}S^2 and TpR2T_p\mathbb{R}^2 are given by:

{∂∂u|ϕ(p),∂∂v|ϕ(p)}   {∂∂x|p,∂∂y|p}\left\{\frac{\partial}{\partial u}\Bigr|_{\phi(p)},\frac{\partial}{\partial v}\Bigr|_{\phi(p)}\right\} \ \ \ \left\{\frac{\partial}{\partial x}\Bigr|_{p},\frac{\partial}{\partial y}\Bigr|_{p}\right\}

Thus if you wish to push forward tangent vectors you have to compute:

ϕ−1∗(∂∂u|p)=a∂∂x|p+b∂∂y|p\phi^{-1}_* \left(\frac{\partial}{\partial u}\Bigr|_p\right) = a\frac{\partial}{\partial x}\Bigr|_{p}+ b \frac{\partial}{\partial y}\Bigr|_{p}

ϕ−1∗(∂∂v|p)=c∂∂x|p+d∂∂y|p\phi^{-1}_* \left(\frac{\partial}{\partial v}\Bigr|_p\right)=c\frac{\partial}{\partial x}\Bigr|_{p}+ d \frac{\partial}{\partial y}\Bigr|_{p}

The above follows from the fact that ϕ−1∗\phi^{-1}_* is linear i.e basis vectors map to basis vectors. You can recover the coefficients of the matrix:

A=[abcd]A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}

by evaluating both sides of the equations with x,yx,y. Remember x,yx,y are coordinate functions i.e x∘ϕ−1x \circ \phi^{-1} gives the first coordinate of the function ϕ−1\phi^{-1}. The given matrix above is the transition matrix between the two bases i.e A−1A^{-1} can take you the other way.