2

I study the book "Geometry topology and physics" by Nakahara. And there is something I misunderstand at page 191.

Here, we compute by using Taylor series, the flow of a vector field $X$ :

$$ \sigma^\mu(t,x) = exp(t \frac{d}{ds}) \sigma^\mu(s,x)|_{s=0}$$

I totally agree with this formula.

But then he says :

The last expression can also be written as : $$ \sigma^\mu(t,x)=exp(tX)x^\mu $$

And then I don't understand...

I tried to see what happens when I derivate succesively $\sigma^\mu(t,x)$ but I get something like this :

$$ \frac{d^2}{dt^2}\sigma^\mu(t,x)|_{t=0}=\frac{d}{dt}X^\mu(\sigma(t,x))|_{t=0}=\frac{\partial X^\mu}{\partial \sigma^\nu}(\sigma(t,x))\frac{\partial \sigma^\nu}{\partial t}(t,x)|_{t=0}=\frac{\partial X^\mu}{\partial \sigma^\nu}(\sigma(t,x))X^\nu (\sigma(t,x))|_{t=0}$$

And I'm stuck.

I would like to understand why the expression of the book is true and also where is my mistake in my derivation.

A recall on definitions :

Given a vector field $X \in \mathbb{\chi}(M)$, we have :

$$ \frac{d}{dt}\sigma^\mu(t,x)=X^\mu(\sigma(t,x))$$ $$ \sigma^\mu(0,x)=x$$

And $\sigma : \mathbb{R} *M \rightarrow M $ is called the flow generated by $X$.

StarBucK
  • 689
  • 3
    You shouldn't take $\sigma^\mu(t,x) = \exp(tX)x^\mu$ to be a "theorem"! The expression $\exp (tX)$ is not even defined in the book! It's just a suggestive mnemonic to aid your memory, and it serves as shorthand for the correct equation, $\sigma^\mu(t,x) = \exp(t \tfrac d {ds} )\sigma^\mu(s,x)|_{s=0}$. You should forget about this! :-) – Kenny Wong Apr 05 '17 at 16:49
  • Ah ok. I thought it was some property that I didn't understand. But just to get why it should "help" : so the correct formula is $exp(t\frac{d}{ds})\sigma^\mu (s,x)|{s=0}$. The mnemotechnic is in the way to say "well I have $\sigma^\mu (0,x) =x^\mu$ so I can replace $\sigma^\mu (s,x)|{s=0}$ by $x^\mu$. And the derivation of the flow is equal to $X(\sigma^\mu(t,x))$ so i can consider that $\frac{d}{ds} <-> X$. Is it in this sense ? – StarBucK Apr 05 '17 at 17:01
  • 1
    Yeah, I think that's the idea! – Kenny Wong Apr 05 '17 at 17:03

0 Answers0