# The Unapologetic Mathematician

## The Lie Derivative on Cohomology

With Cartan’s formula in hand we can show that the Lie derivative is a chain map $L_X:\Omega(M)\to\Omega(M)$. That is, it commutes with the exterior derivative. And indeed, it’s easy to calculate

\displaystyle\begin{aligned}L_X\circ d=(d\circ\iota_X+\iota_X\circ d)\circ d&=d\circ\iota_X\circ d\\d\circ L_X=d\circ(d\circ\iota_X+\iota_X\circ d)&=d\circ\iota_X\circ d\end{aligned}

And so, like any chain map, the Lie derivative defines homomorphisms on cohomology: $L_X:H^k(M)\to H^k(M)$. But which homomorphism does it define?

Well, it turns out that Cartan’s formula comes in handy here as well, for it’s exactly what we need to say that the Lie derivative is null-homotopic. And like any null-homotopic map, it defines the zero map on cohomology. That is, if we take some closed $k$-form $\omega\in Z^k(M)$, which defines a cohomology class in $H^k(M)$ — any cohomology class has such a representative $k$-form — and hit it with $L_X$, the result is an exact $k$-form.

Actually, this shouldn’t be very surprising, considering Cartan’s formula. Indeed, we can calculate directly

$\displaystyle L_X\omega=d(\iota_X\omega)+\iota_X(d\omega)=d(\iota_X\omega)$

since by assumption $\omega$ is closed, which means that $d\omega=0$.

July 28, 2011 - Posted by | Differential Topology, Topology