7
$\begingroup$

Tischler theorem states that the existence of a nowhere vanishing closed $1$-form in a compact manifold $M$ implies that the manifold fibers over $S^1$. Do you know any other differential topology results of this kind? By this kind I mean $$ \text{Existence of some differential form/s} \implies \text{Topological consequences on $M$}$$

$\endgroup$

7 Answers 7

6
$\begingroup$

If a compact smooth manifold has a closed, non-degenerate $2$-form, i.e. it is a symplectic manifold, then it must be even dimensional and all the even Betti numbers $b_{2k}(M) = \dim(H_{2k}(M,\mathbb{R}))$ are non-zero.

To prove it is even dimensional is just linear algebra, a vector space with a non-degenerate bilinear form must be even dimensional. To prove the fact about the Betti numbers, note that if a manifold $M$ of dimension $2n$ has a closed non-degenerate $2$-form $\omega$ then $[\omega]^{k} \neq 0 \in H^{2k}(M,\mathbb{R})$ (for $1 \leq k \leq n$), since $[\omega]^{n}$ is the class of a volume form on $M$.

This is not really considered a theorem, but I thought it was worth mentioning since the hypothesis are very similar to Tischler theorem only for forms of one dimension greater.

$\endgroup$
5
$\begingroup$

On a $2k+1$-dimensional manifold, a 1-form $\alpha$ such that $\alpha\wedge (d\alpha)^k \neq 0$ at each point gives a contact structure. I believe that it is still open which $2k+1$-manifolds for $k>1$ admit a contact structure ( all 3-manifolds have a contact structure).

On a 3-manifold, a nowhere vanishing $1-$form $\alpha$ such that $\alpha\wedge d\alpha=0$ is equivalent to having a foliation ($ker\alpha$ gives an integrable plane field by the Frobenius theorem). This does not give much topological information, since every 3-manifold admits a (2-dimensional smooth orientable) foliation. However, if there is also a closed 2-form $\omega$ such that $\omega \wedge \alpha >0$ (i.e. nowhere vanishing), then the foliation is taut. This is a non-trivial topological condition, as many 3-manifolds do not admit a taut foliation, and holds for fibered 3-manifolds, so strictly generalizes Tischler's theorem in the 3-dimensional case in some sense.

$\endgroup$
3
$\begingroup$

Maybe the cheapest possible answer: if the dimension of $M$ is $m$, and there exists a nowhere-vanishing $m$-form, then $M$ is orientable.

$\endgroup$
2
$\begingroup$

There has been a couple of works from Stefan Schreieder and coauthors, around Kotschick's conjecture, that states that a compact Kähler manifold X fibres smoothly over the circle if and only if it admits a holomorphic 1-form without zeros.

$\endgroup$
2
$\begingroup$

If a Riemann surface admits a nowhere vanishing holomorphic one form, then it must have genus one (which is a special case of Tischler theorem).

A theorem of Popa and Schnell generalizes this to all complex projective manifolds:

If a smooth complex projective variety admits a nowhere vanishing holomorphic 1-form, then it is not of general type.

Of course, this is not a genuine topological implication as in the case of Riemannian surfaces. However, some topological information can be extracted, at least in the case of complex dimension 2.

$\endgroup$
1
$\begingroup$

There is a similar situation happening in graph theory. Specifically, one can interpret "nowhere vanishing 1-forms" on graphs as nowhere-zero flows on graphs. Then, it's an easy theorem that graphs with nowhere-zero flows are precisely the bridgeless graphs (which is a topological property in some sense).

$\endgroup$
1
$\begingroup$

If a smooth compact oriented n-manifold M has a closed k-form that is not a coboundary, it then defines a non-trivial k-dimensional cohomology class, which is a kind of k-dimensional hole in M.

This is most definitely topological information, since by the de Rham theorem combined with Poincaré duality it corresponds to a class of singular k-dimensional homology (over the reals) — an almost visceral "hole" in M.

This is so basic that it's a bit strange to post it, but it should not be overlooked.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .