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...


2 Answers 2


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .