Graduate What's the Proper Way to Push Forward a Vector Field in Differential Geometry?

[kiuhnm]

I'm learning Differential Geometry on my own for my research in ML/AI. I'm reading the book "Gauge fields, knots and gravity" by Baez and Muniain.

An exercise asks to show that "if \phi:M\to N we can push forward a vector field v on M to obtain a vector field (\phi_*v)_q = \phi_*(v_p) whenever \phi(p)=q."

I get a q in both places:
(\phi_*v)_q(f) = (\phi_*v)(f)(q) = (v(f\circ \phi))(q) = v_q(f\circ \phi)=(\phi_* v_q)(f)

What's wrong with my steps?

 

[andrewkirk]

The second step (second '=' sign) is incorrect. The RHS, which is $(v(f\circ\phi))(q)$, is undefined because $(v(f\circ\phi))$ is a vector field on $M$ (ie a map from $M$ to $TM$), and so cannot take $q\in N$ as an argument.

 

[kiuhnm]

I don't understand if I'm asked to find a proper definition for the pushforward of vector fields or if I can prove the equality by knowing the pushforward of a vector field at a point. I only know that (\phi_*v)(f) = v(\phi^* f) where v\in T_p M and \phi:M\to N.

 

[fresh_42]

There is barely more to say than what @andrewkirk already has said. It might help to write $q=\phi(p)$ from the second step on. The overall situation is as follows:
\begin{equation*}
\begin{matrix}
TM & \overset{\phi_*=D\phi}{\longrightarrow} &TN \\
\pi_M \downarrow & & \;\; \downarrow \pi_N \\
M & \overset{\phi}{\longrightarrow} &N \\
\end{matrix}
\end{equation*}
and the task is to make this diagram commutative. The tangent vectors can be seen as directional derivatives. The maps are given by $\phi_* : TM \rightarrow TN$
\begin{equation*}
\phi_* : T_{p}M \rightarrow T_{\phi(p)} N \\
(\phi_*(v))(f)=v(f \circ \phi)
\end{equation*}
with $p \in M, q \in N, v \in T_pM, f \in C^\infty(N)$. So at some stage you have to switch from $q$ to $p$, because the derivatives at $p$ are what we have, and the derivatives at $q$ are what we want to get. The very last $\phi$ in the above is there to do exactly this: reducing what we want at $q$ to what we have at $p$ because $\phi(p)=q$.

 

[kiuhnm]

I know that (\phi_* v)_q(f) = v_p(f\circ \phi) makes perfect sense type-wise, but I can't show that algebraically because I seem to miss some key property. I have no way of manipulating \phi_*v because I can only do it when v is in T_p M for some p\in M.
I have no idea what \phi_*=D\phi means and I guess it's important.
edit: are you saying that \phi_* is a tangent vector?

 

[fresh_42]

I know that (\phi_* v)_q(f) = v_p(f\circ \phi) makes perfect sense type-wise, but I can't show that algebraically because I seem to miss some key property.

That is the key property. We don't know $\phi_*$ but we know the tangent spaces $T_pM\; , \; T_qN$ and the differentiable function $\phi$. So this line defines what we are looking for.

I have no way of manipulating \phi_*v because I can only do it when v is in T_p M for some p\in M.
I have no idea what \phi_*=D\phi means and I guess it's important.

This is the other way to look at it. It means that $\phi_*$ is the derivative $D$ of $\phi$, a linear transformation between the tangent spaces.

edit: are you saying that \phi_* is a tangent vector?

No. $v$ is a tangent vector - at $p$. $\phi_*$ maps it to a tangent vector at $q=\phi(p)$.