# The Unapologetic Mathematician

## Maps Intertwining Vector Fields

Let $f:M\to N$ be a smooth map between manifolds, with derivative $f_*:\mathcal{T}M\to\mathcal{T}N$, and let $X:M\to\mathcal{T}M$ and $Y:N\to\mathcal{T}N$ be smooth vector fields. We can compose them as $f_*\circ X:M\to\mathcal{T}N$ and $Y\circ f:M\to\mathcal{T}N$, and it makes sense to ask if these are the same map.

To put it another way, $Y\circ f:p\mapsto Y_{f(p)}$ is the vector $Y$ specifies for the point $f(p)$. On the other hand, $f_*\circ X:p\mapsto f_*(X_p)$ is the image of the vector $X$ specifies for the point $p$. If these two vectors are the same for every $p\in M$, then we say that $f$ “intertwines” the two vector fields, or that $X$ and $Y$ are “ $f$-related”. The latter term is a bit awkward, which is why I prefer the former, especially since it does have that same commutative-diagram feel as intertwinors between representations.

Anyway, in the case that $f$ is a diffeomorphism we can actually use this to transfer vector fields from one manifold to the other. Given a point $q\in N$, which point should it be compared to? the inverse image $f^{-1}(q)$, of course. This point gets the vector $X\left(f^{-1}(q)\right)$, which then gets sent to $f_*\left(X\left(f^{-1}(q)\right)\right)$. That is, if we define $Y=f_*\circ X\circ f^{-1}$, then $f$ is guaranteed to intertwine $X$ and $Y$.

Since $f_*$ is a linear map on each stalk it’s clear that if $f$ intertwines $X_1$ and $Y_1$, as well as $X_2$ and $Y_2$, then $f$ intertwines $c_1X_1+c_2X_2$ and $c_1Y_1+c_2Y_2$. But we’ve just seen that vector fields form a Lie algebra, and it would be nice if we could say the same for $[X_1,X_2]$ and $[Y_1,Y_2]$. The catch is that we don’t just compute these point-by-point.

Let’s pick a text function $\phi\in\mathcal{O}N$ and a point $p\in M$. We first check that \displaystyle\begin{aligned}{}[(Y_i\phi)\circ f](q)&=(Y_i\phi)(f(q))\\&=Y_{i,f(q)}\phi\\&=[f_*X_{i,q}]\phi\\&=X_{i,q}(\phi\circ f)\end{aligned}

Now we can calculate \displaystyle\begin{aligned}{}[Y_1,Y_2]_{f(p)}\phi&=Y_{1,f(p)}(Y_2\phi)-Y_{2,f(p)}(Y_1\phi)\\&=f_*X_{1,p}(Y_2\phi)-f_*X_{2,p}(Y_1\phi)\\&=X_{1,p}((Y_2\phi)\circ f)-X_{2,p}((Y_1\phi)\circ f)\\&=X_{1,p}(X_2(\phi\circ f))-X_{2,p}(X_1(\phi\circ f))\\&=[X_1,X_2]_p(\phi\circ f)\\&=(f_*[X_1,X_2]_p)\phi\end{aligned}

So $f$ intertwines $[X_1,X_2]$ and $[Y_1,Y_2]$, as we asserted. In the case where $f$ is a diffeomorphism, this means that the construction above gives us a homomorphism of Lie algebras from $\mathcal{T}M$ to $\mathcal{T}N$.