Can we make homology from interior derivative (interior product)?

Question

I have a question that I have been curious about for years.

In differential geometry, since the exterior derivative satisfies property $d^2=0$, we can make a de Rham cohomology from it.

Then if we write $\iota_X:\Omega^n\rightarrow\Omega^{n-1}$ as the interior derivative(also called as interior product) for a vector field $X$, then $\iota_X^2=0$ holds. Can we make a homology for a suitable vector field $X$ from this?

And if you can create such a homology, are there any useful properties about it? Like the de Rham theorem.

I would really appreciate it if you could let me know.

Answer 1

This is more of a comment that is far too long for a comment. But I have also wondered about this, so I thought I would get started with the simplest possible examples. If anyone has a good interpretation for what these cohomology groups are, I would also love to know it.

For the simplest example, let's take $S^1$ as our manifold. The (co)tangent bundles are both trivial. Write their generators over $C^\infty(S^1)$ and $\partial_\theta$ and $d\theta$. Take a vector field $v=f\partial_\theta$ for some $f\in C^\infty(S^1)$. We consider the chain complex $$0\to\Omega^1(S^1)\xrightarrow{\iota_v}C^\infty(S^1)\to 0$$ Write $\eta\in\Omega^1$ as $\eta=fd\theta$. Then $\iota_v(\eta)=fg$. Therefore, $$\ker\iota_v=\{\eta=fd\theta\in \Omega^1(S^1)|fg=0\}=\text{Ann}_{C^\infty(S^1)}\langle g\rangle$$ That is, $H^1(S^1,g\partial_\theta)$ is the annihilator of the ideal generated by $g$, as a module over $C^\infty(S^1)$. By definition, interior multiplication acts trivially on functions. So to find $H^0(S^1,g\partial_\theta)$, we only need to know the image of $\iota_v$. Clearly, we have $$\text{im }\iota_v=\{fg\mid f\in C^\infty(S^1)\}$$ since $fd\theta$ defines a $1$-form on $S^1$. As such, we get $$H^0(S^1,g\partial_\theta)=C^\infty(S^1)/\langle g\rangle$$ If this were algebraic geometry, we could think about $H^1(S^1,g\partial_\theta)$ as the annihilator of an ideal sheaf, and of $H^0(S^1,g\partial_\theta)$ as the coordinate ring of the subscheme defined by said ideal sheaf.

At any rate, these vector spaces are not finite dimensional, so I would say this is not a particularly "nice" cohomology. It reminds me of Poisson cohomology, which is likewise ill-understood even for simple manifolds.

Another case to look at, is a manifold $X$ with an action of a Lie group $G$. Then we can look at the fundamental vector fields $v_\xi$ for $\xi\in\mathfrak{g}$ and ask ourselves what these cohomology groups might represent. The easiest case to consider, then, would again be a torus, say $T^2=S^1\times S^1$, viewed as a principal $S^1$-bundle over the circle. Then we can look at the vector field $v=\partial_{\theta_2}$, where I denote the coordinates by $(\theta_1,\theta_2)$. This is the fundamental vector field which corresponds to $1\in\mathfrak{u}(1)\cong\mathbb{R}$. Once again, $\Omega^2(T^2)$ and $\Omega^1(T^2)$ are free modules over $C^\infty(T^2)$, so computations become very easy (and maybe this is why nothing interesting is happening). Similar reasoning to the above shows that, this time, we get $H^2(T^2,v)=0$, $H^1(T^2,v)= C^\infty(T^2)\cdot d\theta_1/(-C^\infty(T^2)\cdot d\theta_1)=0$ and $H^0(T^2,v)=C^\infty(T^2)/C^\infty(T^2)=0$.

Perhaps if we took the Hopf fibration $S^3\to S^2$ as a principal $S^1$-bundle, it would yield something more interesting. This is a little bit more involved.

EDIT: The response by Mariano Suárez-Álvarez is of course correct, and it tells us that this (co)ohomology will be trivial whenever the vector field is nowhere vanishing. As such, it is not particularly interesting to look at, I believe. This is largely due to the fact that the operation of interior multiplication is really a "pointwise" operation, not a local operation like differentiation. Hence it acts trivially on $C^\infty(X)$, which is why it does not give "interesting" local information about the manifold $X$.

It also answers what happens on the Hopf bundle: the cohomology will again be trivial, because the group action is free. Rather, when we have an $S^1$-action (or an $\mathbb{R}$-action) on a manifold $X$, the cohomology that you obtain from this complex, using infinitesimal generators for the action, is going to tell you something about the fixed point locus of the action. I suppose that's about as satisfactory of an answer that one could hope for, in this case.

Answer 2

You can do this calculation purely algebraically. I'd suggest doing first the following variation.

Consider a vector space $V$, a vector $v$ in $V$, the exterior algebra $\Lambda^\bullet V$, and the map $w\mapsto v\wedge w$ from the exterior algebra to itself of degree $+1$. One can easily check that this complex is exact if and only if the vector v is non-zero.

If you do that, which is easier than your problem because of less notation, you can look at the next thing:

Let now $A$ be the graded vector space $\Lambda^\bullet V^*$, the exterior algebra on the dual space of $V$, let $v$ be a vector in $V$, and now let $d$ be the map of degree $-1$ on $A$ that is given by contraction with $v$. Can you compute the homology of $A$?