Cohomology defined by a different map is isomorphic to de Rham cohomology

Question

Let $d$ be the derivative map for de Rham cohomology. If we define $d_f=m_{-f}\circ d\circ m_f$ where $m_f(\omega)=e^f\omega$ for a smooth function $f:M\to\mathbb{R}$. Notice that $d_f^2=0$. How do we show that the cohomology defined by $d_f$ is isomorphic to the de Rham cohomology? I saw another question Is this cohomology isomorphic to De Rham Cohomology? that talks about a similar problem so it would be great if anyone could elaborate on the comment. Thanks!

Answer 1

To show that the cohomology defined by $d_f$ is isomorphic to the de Rham cohomology, let's elaborate on the structure of $d_f$ and the isomorphism between these cohomologies. The operation you've defined, $d_f=m_{-f} \circ d \circ m_f$, is a twisted derivative, where $m_f(\omega)=e^f \omega$ for a smooth function $f: M \rightarrow \mathbb{R}$, and $d$ is the standard exterior derivative of de Rham cohomology.

Given $d_f^2=0$, we can consider the cohomology defined by $d_f$, which consists of equivalence classes of forms $\omega$ such that $d_f \omega=0$, with two forms considered equivalent if they differ by $d_f \eta$ for some form $\eta$.

Step 1: Show $d_f^2=0$

First, let's confirm that $d_f^2=0$. We have $d_f=m_{-f} \circ d \circ m_f$. Applying $d_f$ twice gives: $$ d_f^2=m_{-f} \circ d \circ m_f \circ m_{-f} \circ d \circ m_f $$

Given $m_{-f} \circ m_f=\mathrm{id}$ and $d^2=0$, it follows that $d_f^2=0$.

Step 2: Construct an Isomorphism

To show that the cohomology defined by $d_f$ is isomorphic to the standard de Rham cohomology, we look for a map that relates $d$-closed forms to $d_f$-closed forms and preserves the cohomology classes. The map $m_f$ effectively provides this connection.

$\textbf{Isomorphism}$: Consider the map induced by $m_f$ on cohomology. Specifically, $m_f$ maps $d$-closed forms to $d_f$-closed forms, because if $d \omega=0$, then $d_f\left(m_f(\omega)\right)=m_{-f} \circ d \circ m_f\left(m_f(\omega)\right)=$ $m_{-f} \circ d\left(e^f \omega\right)=0$ since applying $d$ on $e^f \omega$ yields a form that is still in the kernel of $m_{-f}$ after application due to the linearity and commutation properties of $d$ and multiplication by functions.

$\textbf{Injectivity and Surjectivity}$: This map is injective and surjective. Injectivity follows because if $m_f(\omega)$ is $d_f$-exact, then $\omega$ is $d$-exact, given the invertibility of $m_f$. Surjectivity comes from the ability to take any $d_f$-closed form, apply $m_{-f}$ to it, and get a $d$-closed form, which can then be mapped back by $m_f$.

$\textbf{Preservation of Cohomology Classes}$: Two forms that are cohomologous in the de Rham sense are also cohomologous in the $d_f$-cohomology, and vice versa, due to the bijectivity of $m_f$ and its inverse $m_{-f}$ on the space of forms.

Therefore, the cohomology defined by $d_f$ is isomorphic to the de Rham cohomology. This isomorphism is essentially due to the fact that $m_f$ and $m_{-f}$ act as conjugations that translate between the standard exterior derivative and the twisted derivative $d_f$, while preserving the structure of cohomology classes.