Maps Intertwining Vector Fields
Let 
be a smooth map between manifolds, with derivative 
, and let 
and 
be smooth vector fields. We can compose them as 
and 
, and it makes sense to ask if these are the same map.
To put it another way, 
is the vector 
specifies for the point 
. On the other hand, 
is the image of the vector 
specifies for the point 
. If these two vectors are the same for every 
, then we say that 
“intertwines” the two vector fields, or that and are “-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 is a diffeomorphism we can actually use this to transfer vector fields from one manifold to the other. Given a point 
, which point should it be compared to? the inverse image 
, of course. This point gets the vector 
, which then gets sent to 
. That is, if we define 
, then is guaranteed to intertwine and .
Since 
is a linear map on each stalk it’s clear that if intertwines 
and 
, as well as 
and 
, then intertwines 
and 
. 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]](https://s0.wp.com/latex.php?latex=%5BX_1%2CX_2%5D&bg=e6e6e6&fg=333333&s=0&c=20201002)
and ![[Y_1,Y_2]](https://s0.wp.com/latex.php?latex=%5BY_1%2CY_2%5D&bg=e6e6e6&fg=333333&s=0&c=20201002)
. The catch is that we don’t just compute these point-by-point.
Let’s pick a text function 
and a point . We first check that
&=(Y_i\phi)(f(q))\\&=Y_{i,f(q)}\phi\\&=[f_*X_{i,q}]\phi\\&=X_{i,q}(\phi\circ f)\end{aligned}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cbegin%7Baligned%7D%7B%7D%5B%28Y_i%5Cphi%29%5Ccirc+f%5D%28q%29%26%3D%28Y_i%5Cphi%29%28f%28q%29%29%5C%5C%26%3DY_%7Bi%2Cf%28q%29%7D%5Cphi%5C%5C%26%3D%5Bf_%2AX_%7Bi%2Cq%7D%5D%5Cphi%5C%5C%26%3DX_%7Bi%2Cq%7D%28%5Cphi%5Ccirc+f%29%5Cend%7Baligned%7D&bg=e6e6e6&fg=333333&s=0&c=20201002)
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}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cbegin%7Baligned%7D%7B%7D%5BY_1%2CY_2%5D_%7Bf%28p%29%7D%5Cphi%26%3DY_%7B1%2Cf%28p%29%7D%28Y_2%5Cphi%29-Y_%7B2%2Cf%28p%29%7D%28Y_1%5Cphi%29%5C%5C%26%3Df_%2AX_%7B1%2Cp%7D%28Y_2%5Cphi%29-f_%2AX_%7B2%2Cp%7D%28Y_1%5Cphi%29%5C%5C%26%3DX_%7B1%2Cp%7D%28%28Y_2%5Cphi%29%5Ccirc+f%29-X_%7B2%2Cp%7D%28%28Y_1%5Cphi%29%5Ccirc+f%29%5C%5C%26%3DX_%7B1%2Cp%7D%28X_2%28%5Cphi%5Ccirc+f%29%29-X_%7B2%2Cp%7D%28X_1%28%5Cphi%5Ccirc+f%29%29%5C%5C%26%3D%5BX_1%2CX_2%5D_p%28%5Cphi%5Ccirc+f%29%5C%5C%26%3D%28f_%2A%5BX_1%2CX_2%5D_p%29%5Cphi%5Cend%7Baligned%7D&bg=e6e6e6&fg=333333&s=0&c=20201002)
So intertwines and , as we asserted. In the case where is a diffeomorphism, this means that the construction above gives us a homomorphism of Lie algebras from 
to 
.
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Posted by John Armstrong |
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