This is a follow-up question to my previous one where I was trying to understand the classes of Legendrian immersions of circles into contact manifolds.

I'm interested in classifying isotropic immersions (of spheres, in my particular application, if that makes a difference) into contact manifolds up to regular homotopy through isotropic immersions. Let $(M,\xi)$ be a contact (2n+1)-manifold; here $\xi$ is a distribution of $2n$-planes in $TM$. An isotropic immersion of a manifold N (with dimension $k\leq n$, for $k=n$ this is a Legendrian immersion) is an immersion $f:N\rightarrow M$ where the image of $df_x:T_x N\rightarrow T_{f(x)}M$ lies in the plane $\xi_{f(x)}$ for all $x$.

Mike Usher's answer to my previous question directed me to a paper by Ekholm, Etnyre and Sullivan, in particular section 3.3 (p.19), which defines the rotation class of a Legendrian $f$ as follows.

If $\alpha$ is a contact 1-form on M, that is Ker $\alpha=\xi$, then $d\alpha_p|\xi_p$ is a symplectic form on $\xi_p$. If one chooses any complex structure $J$ for $\xi$ that is compatible with the symplectic structure, then complexifying $df$ to be $df_{\mathbb C}:TL\otimes \mathbb{C}\rightarrow\xi$ is a fiberwise bundle isomorphism.

The rotation class is the homotopy class of $(f,df_{\mathbb C})$ in the space of complex fiberwise isomorphisms $TL\otimes\mathbb{C}\rightarrow \xi$ and is denoted $r(f)$. An h-principle for Legendrian immersions implies that $r(f)$ is a complete invariant for $f$ up to regular homotopy through Legendrian immersions.

First, I can see how the above definition generalizes to isotropic immersions (namely, instead of bundle isomorphisms, I'll get bundle monomorphisms), but how do I figure out what the possible rotation classes in that setting are? Actually, I'm not exactly sure how to do it in the Legendrian case either. It seems one has to understand certain homotopy classes of bundle maps, which I don't know anything about. (It's OK if the answer is just a reference to a section of a book on bundle theory or something)

Second, what happens when $\xi$ is not coorientable, that is, there is no global $\alpha$? In this case, I don't even know how to define a complex structure compatible with $\xi$ which seems to be what the rotation class relies on.


1 Answer 1


jc, you'll have fun working out answers to examples of your first question. I'm only going to address the Legendrian case. If I didn't make mistakes, I'll conclude that Legendrian immersions of $S^n$ into $S^{2n+1}$ are all regular homotopic if $n$ is even, while if $n$ is odd they are classified by a rotation number. For the second question, I'll just suggest that you think about 1-forms valued in the orientation line bundle.

The h-principle you quoted (cf. Eliashberg-Mishachev's book) tells us that we need to understand homotopy classes $[f,l]$, where $f$ is a map $L \to M$ and $l \colon TL\otimes \mathbb{C}\to f^* \xi$ an injective bundle map. Understanding the set $[L,M]$ of homotopy classes of unbased maps is an obstruction theory problem, and in general pretty intractable, but $[S^k,M]$ is the set of $\pi_1(M)$-orbits in $\pi_k(M)$. The fibre of the forgetful map $[f,l]\mapsto [f]$ is exactly the set of injective bundle maps $l \colon TL\otimes \mathbb{C}\to f^\ast \xi$ (this is an application of the covering homotopy property of bundles, bearing in mind that the injective bundle maps themselves form a fibre bundle).

Now fix $f$. In the Legendrian case, what we're looking is $\pi_0$ of the space $I(f)$ of isomorphisms $TL\otimes \mathbb{C}\to f^\ast \xi$. In terms of classifying maps, $\pi_0 I(f)$ is the set of homotopy classes of homotopies from $t_{\mathbb{C}}$, the composite of the tangent map $t\colon L \to BO(n)$ with $BO(n)\to BU(n)$, to $\tilde{\xi}f$. Necessary conditions for $I(f)\neq \emptyset$ are that, for all $j$, $p_j(TL) = \pm f^\ast c_{2j}(\xi)$ (I forget what the sign should be) and $2 f^\ast c_{2j+1}(\xi)=0$. When $L=S^n$, you have to compare two elements of $\pi_n BU(n)=\pi_{n-1}U(n) = \pi_{n-1}U(\infty)$ which is $\mathbb{Z}$ or $0$ according to whether $n$ is even or odd. Since the Pontryagin classes of $S^n$ are zero by the signature theorem, I think what you actually have to check is whether $f^* c_{n/2}(\xi)=0$.

If one isomorphism exists, there will be more: we can compose with any complex automorphism of $T^\ast L\otimes \mathbb{C}$. Up to homotopy, automorphisms correspond to isomorphism classes of vector bundles over $S^1\times L$ extending $T^\ast L\otimes \mathbb{C}\to L$, and so one has to classify such extensions. This is another obstruction theory problem. When $L=S^n$, any two extensions agree over $S^1 \vee L$, and over the last $(n+1)$-cell they differ by a map $S^{n+1} \to BU(n)$.

So, when $L=S^n$, there's a freely transitive action of $\pi_{n+1}BU(n)=\pi_n U(n) = \pi_n U(\infty)$ on $I(f)$.

Examples: say $M=S^{2n+1}$, $\xi$ any contact structure, $L=S^n$, $f$ any map (necessarily homotopically trivial). Then $f^\ast \xi$ is trivial, so $I(f)$ is non-empty. It is a singleton if $n$ is even, and is a freely transitive $\mathbb{Z}$-set if $n$ is odd. When $n=1$, the $\mathbb{Z}$ should be the classical rotation number.

  • $\begingroup$ Can you recommend some good obstruction theory references so that I can learn to work out the calculations you describe? I am still slowly working my way through Hatcher's algebraic topology book; maybe what you mention is in there in a form I can't quite grasp yet? $\endgroup$
    – j.c.
    Apr 24, 2010 at 18:53
  • 1
    $\begingroup$ There are two ways of setting up obstruction theory - an abstract approach using Postnikov towers, and a more concrete one using cell complexes. The second method is the one that I find easier to remember and apply to manifolds. Hatcher covers the first approach nicely; for the second, I think I read Milnor-Stasheff and Steenrod's classic on fibre bundles. (Other MO-ers may have better suggestions...) $\endgroup$
    – Tim Perutz
    Apr 24, 2010 at 21:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.