What_is_pullback_of_a_vector_field@20250813_181339

user: What is pullback of a vector field? [Timestamp: 2025/08/13 18:13:39]

ai: In differential geometry one usually defines the pushforward of a vector field—not its pullback. Given a smooth map F: M → N and a vector field X on M, one can push it forward to a vector field F₍*₎X on N (provided F is smooth), but “pulling back” a vector field on N to M is not generally defined unless F is a diffeomorphism (or at least locally invertible). In the case where F is a diffeomorphism, one can define the pullback of a vector field Y on N by

  FY = (F⁻¹)₍₎Y

so that F*Y is a vector field on M. This definition is possible because F⁻¹ exists and allows one to take the pushforward in the reverse direction.

In summary, while pulling back differential forms is always well-defined via the pullback F*ω, pulling back vector fields is not done in general. Instead, one typically works with the pushforward, and a pullback of a vector field is only available when F is a diffeomorphism (or under certain additional conditions enabling the definition of an inverse mapping on the tangent spaces).

Sources:

• Differential Forms [Timestamp: 2025/08/13 18:18:08]

user: Is it true that $Ph{i}_s^{\ast }X=X$ if ${\Phi }_s$ is a flow generated by $X$? [Timestamp: 2025/08/13 18:20:05]

ai: Yes, that’s correct, with the proper interpretation of the pullback for vector fields. For a flow Φₛ generated by a vector field X, if we define the pullback of a vector field Y (when the map is a diffeomorphism) as

  Φₛ* Y = (Φₛ⁻¹)₍*₎ Y,

then in particular we have

  Φₛ* X = X.

To explain briefly: Since X generates the flow Φₛ, the pushforward of X by its own flow satisfies

  (Φₛ)₍*₎ X = X.

Taking the pullback (which is defined via the inverse of Φₛ), we recover

  Φₛ* X = (Φₛ⁻¹)₍*₎ X = X.

Another way to see this is by recalling that the Lie derivative of X in the direction of itself is zero, i.e. 𝓛ₓX = 0. Since the Lie derivative is the infinitesimal generator of the flow, this means that the flow leaves the vector field invariant.

Thus, if Φₛ is the (local) flow of X, then indeed Φₛ* X = X.

Sources:

• Integral Curves and Flows
• Symplectic Structures [Timestamp: 2025/08/13 18:20:14]

user: Can you show this in detail? [Timestamp: 2025/08/13 18:23:33]

ai: Let’s work through a detailed explanation. Assume ${\Phi }_s$ is the local flow of $X$, so that for each point $p$ in the manifold, the curve

$$\begin{gathered} s \\ mapsto \\ Ph{i}_s(p) \end{gathered}$$

satisfies the initial value problem

$$\begin{gathered} fracdds \\ Ph{i}_s(p)=X( \\ Ph{i}_s(p)) \\ quad \\ textand \\ quad \\ Ph{i}_0(p)=p. \end{gathered}$$

Because $X$ generates the flow, its Lie derivative along itself must vanish; that is,

$$mathcal{L}_{X}X=0.$$

This fact underlies the invariance of $X$ under the flow. To see this more directly, let’s recall the definition of the pullback of a vector field under a diffeomorphism. For any diffeomorphism $varphi$, the pullback of a vector field $Y$ is given by

$$\begin{gathered} varph{i}^{\ast }Y=( \\ varph{i}^{-1}{)}_{\ast }Y, \end{gathered}$$

which means that for any point $p$, we have

$$\begin{gathered} ( \\ varph{i}^{\ast }Y)(p)=( \\ varph{i}^{-1}{)}_{\ast } \\ Big(Y( \\ varphi(p)) \\ Big). \end{gathered}$$

Now, let’s plug in $$\begin{gathered} varphi= \\ Ph{i}_s \end{gathered}$$ and set $Y=X$. Then

$$\begin{gathered} ( \\ Ph{i}_s^{\ast }X)(p)=( \\ Ph{i}_s^{-1}{)}_{\ast } \\ Big(X( \\ Ph{i}_s(p)) \\ Big). \end{gathered}$$

Since $Ph{i}_s$ is generated by $X$, by the definition of a flow we have

$$\begin{gathered} fracdds \\ Ph{i}_s(p)=X( \\ Ph{i}_s(p)). \end{gathered}$$

The key observation is that the flow “moves forward” in such a way that pushing forward the vector field $X$ along its own flow reproduces itself. More specifically, the pushforward of $X$ by $Ph{i}_s$ satisfies

$$\begin{gathered} ( \\ Ph{i}_s{)}_{\ast }X(p)=X( \\ Ph{i}_s(p)). \end{gathered}$$

Now, applying the definition of the pullback using the inverse flow, we get:

$$\begin{gathered} ( \\ Ph{i}_s^{\ast }X)(p)=( \\ Ph{i}_s^{-1}{)}_{\ast } \\ Big(( \\ Ph{i}_s{)}_{\ast }X(p) \\ Big). \end{gathered}$$

Since pushing forward and then pulling back by the inverse cancels out, we have

$$\begin{gathered} ( \\ Ph{i}_s^{-1}{)}_{\ast } \\ Big(( \\ Ph{i}_s{)}_{\ast }X(p) \\ Big)=X(p). \end{gathered}$$

This shows that

$$\begin{gathered} ( \\ Ph{i}_s^{\ast }X)(p)=X(p) \\ quad \\ textforeverypointp. \end{gathered}$$

Thus, we conclude:

$$Ph{i}_s^{\ast }X=X.$$

To summarize in simple terms:

1. The flow $Ph{i}_s$ generated by $X$ moves points along the integral curves of $X$.
2. Pushing forward $X$ by the flow gives back the value of $X$ at the new point.
3. Pulling back with the inverse of the flow reverses this operation, returning to the original vector $X(p)$.
4. Therefore, pulling back $X$ by $Ph{i}_s$ leaves it unchanged.

This completes the detailed demonstration.

Sources:

• Explain like I am 5
• Microarchitecture
• Simplify
• Make longer
• Algorithms [Timestamp: 2025/08/13 18:23:50]