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First, we make the following observation: let $X: M \rightarrow TM $ be a vector field on a smooth manifold. Taking the contraction with respect to $X$ twice gives zero, i.e. $$ i_X \circ i_{X} =0.$$ Is there any "name" for the corresponding "homology" group that one can define (Kernel mod image)? Has this "homology" group been studied by others (there are plenty of questions that one can ask........is it isomorphic to anything more familiar etc etc).

Similarly, a dual observation is as follows: Let $\alpha$ be a one form; taking the wedge product with $\alpha$ twice gives us zero. One can again define kernel mod image. Does that give anything "interesting"?

If people have investigated these questions, I would like to know a few references.

My purpose for asking the "name" of the (co)homology group is so that I can make a google search using the name. I was unable to do that, since I do not know of any key words under this topic (or if at all it is a topic).

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  • $\begingroup$ I haven't read the answers (yet) but it reminds me of the Koszul complex. $\endgroup$ – Qfwfq Nov 19 '14 at 14:12
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If the vector field $X$ never vanishes, the homology corresponding to $i_X$ is trivial. Suppose $\alpha\in\Omega^p(M)$ is in the kernel of $i_X$. If we take $\beta=X^\flat\wedge\alpha$, we have $i_X\beta=i_X(X^\flat)\wedge\alpha=|X|^2\alpha$. Thus if $X$ does not vanish, we have $\alpha=i_X(|X|^{-2}X^\flat\wedge\alpha)$ so $\alpha$ is in the image of $i_X$.

Consider then the homology corresponding to a one-form $\alpha$. If $\alpha$ never vanishes, the corresponding cohomology is trivial. Let $w_\alpha(\beta)=\alpha\wedge\beta$ for any differential form $\beta$. Suppose then that $\beta\in\Omega^p(M)$ is in the kernel of $w_\alpha$. If $\gamma=i_{\alpha^\sharp}\beta$, then $w_\alpha\gamma=i_{\alpha^\sharp}(\alpha)\wedge\beta=|\alpha|^2\beta$. Thus $\beta=w_\alpha(|\alpha|^{-2}i_{\alpha^\sharp}\beta)$.

The results above were heavily based on the fact that the vector field and the one-form do not vanish. Things get more interesting if they have zeroes. In the following example the first homology group is nontrivial but it is infinite dimensional which makes the theory less nice than classical homology theory. I don't know if the homology groups corresponding to vector fields can be finite dimensional but nontrivial. Perhaps this is possible if the zeroes are isolated.

Consider the case when $X$ is a vector field in the plane and let $\beta$ be a one-form. Now $i_X\beta=0$ means that $X$ and $\beta^\sharp$ are orthogonal. Identifying two-forms with scalars, $\iota_Xf=fX^\perp$ for any scalar $f$, where $X^\perp$ is $X$ rotated by 90 degrees. If the first homology group is trivial, then every vector field $Y$ ($=\beta^\sharp$) orthogonal to $X$ is of the form $fX^\perp$ for some scalar $f$. In particular $Y$ must vanish where $X$ vanishes, which is clearly false. Consider for example the vector fields $X(x,y)=(0,x)$ and $Y(x,y)=(1,0)$. Note that for this $X$ the corresponding first homology group is infinite dimensional.

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  • $\begingroup$ This reminds of the hairy ball theorem. On the plane the "vector field homology" is boring, but so is the plane in general. But on manifolds where vector fields necessarily vanish, interesting things may happen. $\endgroup$ – shuhalo Oct 17 '14 at 22:23
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You can get something "interesting" if you couple the $1$-form version with the de Rham differential. Namely, if $\alpha \in \Omega^1(X)$ is a closed $1$-form on a smooth manifold $X$, then the de Rham complex can be equipped with a twisted de Rham differential $d + \alpha$, by which I mean

$$(d + \alpha) \beta = d \beta + \alpha \wedge \beta.$$

This is a differential since

$$(d + \alpha)^2 \beta = d (\alpha \wedge \beta) + \alpha \wedge (d \beta) = 0$$

(since we assumed $d \alpha = 0$), and its cohomology is a twisted version of de Rham cohomology; in fact it's precisely de Rham cohomology with local coefficients given by the local system associated to the (exponential of the?) image of $\alpha$ in $H^1(X, \mathbb{R}) \cong \text{Hom}(\pi_1(X), \mathbb{R})$. One way to think about this construction is to think of $d + \alpha$ as a flat connection on the trivial line bundle over $X$.

But actually the argument above proceeds just fine if we generalize $\alpha$ to an odd form $\alpha \in \Omega^{2k+1}(X)$, although this collapses the $\mathbb{Z}$-grading on de Rham cohomology to a $\mathbb{Z}_2$-grading. When $\alpha \in \Omega^3(X)$ the corresponding twisted de Rham cohomology groups are the recipients of twisted Chern characters for twisted K-theory; see, for example, Atiyah and Segal's Twisted K-theory and cohomology.

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  • $\begingroup$ Do you know what happens if $d\alpha \neq 0$ holds? I would be highly interested in a reference that develops this case for people with differential geometric background. $\endgroup$ – shuhalo Jan 12 '15 at 13:54
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I believe that the answer is something in between "not really" and "kind of" and was indicated by Qiaochu Yan.

In Wittens famous paper "Morse Theory and Supersymmetry", he considers operators of the form $\sigma d + \alpha$ for a one-form $\alpha$ and a coefficient $\sigma$ with only non-degenerate critical points and shows that for $\sigma \rightarrow 0$, one gets the Morse Homology of a manifold: The Homology of the Complex at level $k$ is $n_k$-dimensional, where $n_k$ is the number of critical points of index $k$. It might be a reasonable assumption that you get the same kind of information with your approach.

However, you cannot set $\sigma = 0$ right away, I believe. I made a few example calculations, and I think that if you assume that your form $\alpha$ has only non-degenerate critical points, you always obtain a zero homology (if you drop the assumption of non-degeneracy, you will always end up with something infinite-dimensional, as pointed out above). At least this is easy to see in two dimensions, and in higher dimensions, it should not be difficult to show, only a little annoying probably ;-).

The physical intuition, is the following: The particles move along the flow lines of the vector field $\alpha^\sharp$, and if you let the diffusion coefficient $\sigma \rightarrow 0$, they concentrate in the minima (where "minima for $k$-forms" turn out to be critical points of higher index somehow). But: If you set the diffusion $\sigma$ to zero right away, the particles will not move; they are just pinned to where you are, so you don't notice anything of your potential term.

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These constructions certainly are very recognisable and meaningful in the algebraic context, that is if you think of algebraic differential forms on an affine algebraic variety. Namely, the corresponding complexes both implement versions of the Koszul complex for a sequence of elements $f_1,\ldots,f_n$ in $R$, where $R$ is the ring of functions on $M$, and $n=\dim M$. Indeed, the algebra of differential forms $\Omega^\bullet(M)$ is the exterior algebra $\Lambda_R^\bullet(R\,dx_1\oplus\ldots\oplus R\,dx_n)$, and for the given vector field $X=f_1\partial_1+\cdots+f_n\partial_n$, the differential $i_X$ on $\Omega^\bullet(M)$ is the $R$-algebra (super)derivation taking $dx_i$ to $f_i$, which essentially is a commonly used definition of the Koszul complex. For the $1$-form $\omega=f_1\,dx_1+\cdots+f_n\,dx_n$, the differential $\omega\wedge$ on $\Omega^\bullet(M)$, which as we remember, is $\Lambda_R^\bullet(R\,dx_1\oplus\ldots\oplus R\,dx_n)$, is dual to the differential corresponding to $i_X$ under the isomorphism of $R$-modules $$\mathop{\mathrm{Hom}}\nolimits_R(\Lambda_R^\bullet(R\,dx_1\oplus\ldots\oplus R\,dx_n),R)\simeq \Lambda_R^\bullet(R\,dx_1\oplus\ldots\oplus R\,dx_n).$$

In any case, the Koszul complex is acyclic if and only if $f_1,\ldots,f_n$ define a complete intersection (this I think also holds in the analytic case, I have no intuition about the smooth one). There are many references, e.g Principles of Algebraic Geometry by Griffiths & Harris, or Commutative Algebra by Bourbaki.

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Actually, I have seen something like this used before in the paper Holomorphic vector fields and Kaehler manifolds, Invent 1973, by Carrel and Lieberman. They prove that if a compact Kähler manifold admits a holomorphic vector field $v$ with nonempty zero locus $Z(v)$, then the Hodge numbers vanish for $|p-q|> \dim Z(v)$. The key idea is to analyze the complex of sheaves $$\ldots \Omega^p\stackrel{i_v}{\to}\Omega^{p-1}\stackrel{i_v}{\to}\ldots $$ which is really a Koszul complex. The complex given by wedging by a $1$-form is also interesting, and it has been studied on the same class of manifolds by Green and Lazarsfeld, Deformation theory, generic vanishing theorems…, Invent (1987).

(I deleted this soon after I wrote it, but perhaps it does give a sort of answer.)

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