I'm trying to demonstrate the Liouville-Gibbs theorem : Given $X$ a $C^1$ vector field and its flow $F_t(x) = F(t, x)$, we want to find a condition such that all the $F_t$ preserve the volume. It is equivalent to show that the Jacobian $det(d(F_t)_x)$ is 1 for all $x$ and $t$ (by integration).

Then we derive it regarding to t : $$det(d(F_t)_x)Tr(d(F_t)_x^{-1} \frac{d}{dt}d(F_t)_x) = 0$$ We can simplify and have : $$Tr(d(F_t)_x^{-1} \frac{d}{dt}d(F_t)_x) = 0 \ (1)$$

I would like to show that $Tr(dX_x) = 0$ (i.e. the divergence is zero), but I have trouble with this expression.

I even showed (falsely) that $(1)$ is always true by applying the product rule and the properties of the trace…


1 Answer 1


Sorry, I did not see the simple chain rule : $$\frac{d}{dt}d(F_t)_x = d(\frac{dF_t}{dt})_x = d(X \circ F_t)_x$$ by Schwarz and the differential equation. Thus we apply the chain rule and : $$d(X \circ F_t)_x = d(X)_{F_t(x)} \circ d(F_t)_x $$ and we have $Tr(dX_{F_t(x)}) = 0$

Then just evaluate at t=0 and we have : $$Tr(dX_x) = 0$$


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