# Pushforward of Lie bracket

1. Dec 8, 2003

### lethe

one elementary result that you see when you first learn differential geometry is that the pushforward of the Lie bracket of two vector fields is the Lie bracket of the pushforward of the two vector fields, i.e.

let $\phi$ be a diffeomorphism from manifold M to N, and let v, w be two vector fields on M. then

$$\phi_*[\mathbf{v},\mathbf{w}]=[\phi_*\mathbf{v},\phi_*\mathbf{w}]$$

where the Lie bracket is defined by its action on $C^\infty(M)$:
$$[\mathbf{v},\mathbf{w}](f)=\mathbf{v}(\mathbf{w}(f))-\mathbf{w}(\mathbf{v}(f))$$

this amounts to saying that the Lie bracket is a natural object. I assume that the proof of this statement is elementary, every textbook i have leaves this proof as an exercise (actually, some books do prove the corresponding more general statement in the case that $\phi$ is not a bijection, in which case there is no well defined pushforward operation)

but for some reason, i have never been able to prove this statement. it should be just a straightforward application of the definitions, so my inability to do it bothers me. let me show you what happens when i try to prove it, in the hopes that someone can spot my mistake. or if someone just knows the answer, feel free to pipe up.

first the left-hand side:

$$\phi_*[\mathbf{v},\mathbf{w}](g)=[\mathbf{v},\mathbf{w}](\phi^*g)=\mathbf{v}(\mathbf{w}(\phi^*g))-\mathbf{w}(\mathbf{v}(\phi^*g))$$
where $g\in C^\infty(N)$

now the right-hand side

$$[\phi_*\mathbf{v},\phi_*\mathbf{w}](g)=\phi_*\mathbf{v}(\phi_*\mathbf{w}(g))-\phi_*\mathbf{w}(\phi_*\mathbf{v}(g))$$

and i can t go much past this point. when i try to move the $\phi$ around using the identities of the pullback and the pushforward:

$$\begin{gather*} \phi^*g=g\circ\phi\\ \phi_*\mathbf{v}(g)=\mathbf{v}(\phi^*g) \end{gather*}$$

i usually end up with a nonsensical expression. for example, continuing with the last line above:

$$\phi_*\mathbf{w}(\phi_*\mathbf{v}(g))=\phi_*\mathbf{w}(\mathbf{v}(\phi^*g))$$

this last expression, that i obtained by na&iuml;vely applying the definition of the pushforward (presumably incorrectly), is meaningless, since it has a vector in N trying to act on a function in $C^\infty(M)$, which is, of course, not the proper domain of definition for a vector on N.

i believe that the error i made above has something to do with keeping track of where on the manifold the various objects are to be evaluated. many textbooks choose to make this explicit by carrying a subscript for the point of evaluation, a practice that seemed pointless to me, but i think here one sees the wisdom in it.

anyway, with or without that error, i have been unable to make the two expressions equal, so, anyone know how this one goes?

2. Dec 8, 2003

### lethe

although i would much prefer to do this proof in a coordinate independent form, i thought perhaps trying it in local coordinates would shed some light on the problem:

$$\mathbf{v}=v^i\partial_i$$

so

$$(\phi_*\mathbf{v})^i=v^j\partial_j\phi^i$$

and in local coordinates, the Lie bracket looks like:

$$([\mathbf{v},\mathbf{w}])^i=v^j\partial_jw^i-w^j\partial_jv^i$$

so

$$(\phi_*[\mathbf{v},\mathbf{w}])^i=(v^j\partial_jw^k-w^j\partial_jv^k)\partial_k\phi^i$$

on the other hand:

$$\begin{multline*} ([\phi_*\mathbf{v},\phi_*\mathbf{w}])^i=(\partial_k\phi^j)v^k\partial_j(w^\ell\partial_\ell\phi^i)-(\partial_k\phi^j)w^k\partial_j(v^\ell\partial_\ell\phi^i)\\ =v^kw^\ell\partial_k\phi^j(\partial_\ell\partial_j\phi^i)+v^k(\partial_k\phi^j)\partial_\ell\phi^i\partial_jw^\ell\\ -w^kv^\ell\partial_k\phi^j\partial_j\partial_\ell\phi^i-w^k\partial_k\phi^j\partial_\ell\phi^i\partial_jv^\ell\\ =(v^kw^\ell-v^\ell w^k)\partial_k\phi^j(\partial_j\partial_\ell\phi^i)+(v^k\partial_jw^\ell-w^k\partial_jv^\ell)\partial_k\phi^j\partial_\ell\phi^i \end{multline*}$$

at first it looks like some nice cancellation might happen, because of the mixed partial derivatives from the product rule, and the subtraction. but alas, that didn t seem to work out....

and needless to say, the two expressions do not match, although the second term of the last calculation looks pretty close to the term from the left-hand side.

Last edited: Dec 8, 2003
3. Dec 12, 2003

### lethe

i think i have resolved this correctly.

4. Jun 16, 2011

### Yannick30

if not it is almost obvius when insead of taking the algebraic definition of the lie bracket you take the geometric one i.e. let
$$\varphi^t$$

be the flow corresponding to v and
$$\psi^t$$
the one of w then
$$[v,w](p) =\frac{\partial}{\partial t} \varphi^t\circ \psi^t\circ \varphi^{-t}\circ\psi^{-t}(p)$$

Last edited: Jun 16, 2011