# Flowing a vector along a vector field $X$ using the pushforward of the flow of $X$

On the three-sphere $S^3$, I'm given three vector fields $X$, $Y$ and $Z$, such that at each point $p\in S^3$, the tangent vectors $X_p$, $Y_p$ and $Z_p$ form an orthogonal basis of the tangent space $T_pS^3$.

I denote by $X^t:S^3\rightarrow S^3$ the flow of the vector field $X$, so $X^0(p)=p$ and $\frac{d}{dt}X^t(p)=X_{X^t(p)}$.

Its pushforward at the point $p$ is denoted $(X^t)_{*,p}$, so this is a map from $T_pS^3$ to $T_{X^t(p)}S^3$.

Now let $v$ be an arbitrary tangent vector in $T_pS^3$. We can write $v=v_xX_p+v_yY_p+v_zZ_p$. I want to prove the following:

$(X^t)_{*,p}(v)=v_x X_{X^t(p)}+...$.

Here the dots denote other terms in the span of $Y_{X^t(p)}$ and $Z_{X^t(p)}$, but it's only the $X_{X^t(p)}$ component I care about (plus I don't know the other components).

Is this maybe immediate from the definition of the flow? I tried using the defining equation $\frac{d}{dt}X^t(p)=X_{X^t(p)}$ of the flow of $X$, but I don't know how I can relate the time derivative with the pushforward...

Edit: so I think I can visualize my question as follows. Starting with a tangent vector $v$ at $p$, flowing this vector along $X$ (using the pushforward of the flow) does not change the component of $v$ along $X$. Sounds plausible...

Just to set notation, let $$X$$ a vector field on $$M$$ and suppose the flow $$\phi_t$$ of $$X$$ exists for all $$t$$ (which is the case if $$M$$ is compact, like $$S^3$$ is). So, we are going to prove that $$D_p\phi_s(X_p)=X_{\phi_s(p)}.$$

We just need to use the "correct" definition of the pushforward: In general, for a map $$f:M\rightarrow N$$ we can compute $$D_pf(u)$$ by the following method. Let a curve $$c:(-\epsilon,\epsilon)\rightarrow M$$ such that $$c(0)=p$$ and $$c'(0)=u$$. Then $$D_pf(u)=(f\circ c)'(0).$$

In words, just pick a curve with velocity $$u$$, push forward the curve and compute the new velocity.

In our case now: To show our formula we have first to find a curve $$c$$ such that $$c(0)=p$$ and $$c'(0)=X_p$$. By definition, we can pick $$c(t)=\phi_t(p)$$! Therefore, to compute the right hand side of our formula all we have to compute is the velocity of $$\sigma(t)=\phi_s(\phi_t(p))$$. But, we have that $$\phi_s(\phi_t(p))=\phi_t(\phi_s(p))$$ therefore $$D_p\phi_s(X_p)=\sigma'(t)=\frac{d}{dt}\phi_t(\phi_s(p))=X_{\phi_s(p)}.$$

Let me just mention that this property holds because in fact $$t\rightarrow \phi_t$$ is a homomorphism from $$\mathbb{R}\rightarrow Diff(M)$$. If we have a $$\textit{time-dependent}$$ vector field $$X_t$$ then tthis property need not hold.

It may be easier to see at first with derivations. The flow is a smooth $$\theta:\mathbb R\times S^3\to S^3$$ that satisfies $$\theta(0,p)=p$$ and $$X_{\theta(0,p)}=\frac{d\theta(t,p)}{dt}|_{t=0}.\$$ Fixing $$p\in S^3$$ and $$t_0\in \mathbb R$$, we flow from time $$t=0$$ to time $$t=t_0$$. This means that we move from $$p$$ to $$\theta(t_0,p).$$ We want to prove that $$X_{\theta(t_0,p)}=\frac{d\theta(t,p)}{dt}|_{t=t_0}.\$$

There is an induced map between $$T_pS^3$$ and $$T_{\theta(t_0,p)}S^3$$ that takes the derivation $$X_p$$ to a derivation $$X_{\theta(t_0,p)}$$ given by $$X_{\theta(t_0,p)}(f)=X_p(f\circ \theta(t_0,-)).$$ This is the pushforward.

Now, $$\theta(-,\theta(t_0,p))$$ is a curve from some interval $$(-\epsilon,\epsilon)$$ into $$S^3$$ such that $$\theta(0,\theta(t_0,p))=\theta(t_0,p)$$ so $$X_{\theta(t_0,p)}(f)=\frac{d f\circ \theta(t,\theta(t_0,p))}{dt}|_{t=0}.$$

But since $$\theta$$ is a flow, $$\theta(t,\theta(t_0,p))=\theta(t+t_0, p)$$ so in fact $$X_{\theta(t_0,p)}(f)=\frac{d f\circ \theta(t+t_0, p)}{dt}|_{t=0}=\frac{d f\circ \theta(t, p)}{dt}|_{t=t_0}.$$

We conclude that $$X_{\theta(t_0, p)}=\frac{d\theta(t, p)}{dt}|_{t=t_0}$$ as expected.