# Realizing a homology by a smooth immersion

An alternative title is: When can I homotope a continuous map to a smooth immersion?

I have a simple topology problem but it's outside my area of expertise and I worry may be rather subtle. Any help would be appreciated.

The set-up is the following: Let $M$ be some (closed say) $n$ dimensional manifold and suppose that $\Sigma_1$ and $\Sigma_2$ are two closed submanifolds of $M$ of dimension $k$. Note that $\Sigma_1$ and $\Sigma_2$ are allowed to intersect (in my situation they are also embedded but I don't believe this is effects anything). Suppose in addition that $\Sigma_1$ and $\Sigma_2$ are homologous. If $k\leq n-2$ I would like a compact manifold with boundary $\Gamma$ with $\partial \Gamma=\gamma_1\cup \gamma_2$ and a smooth immersion $F:\Gamma\to M$ so that $F(\gamma_i)=\Sigma_i$. In other, words the homology between $\Sigma_1$ and $\Sigma_2$ can be realized by a smooth immersion.

I believe by approximation arguments one can always get a smooth such $F$ without restriction on $k$ but it need not be an immersion (especially if $k=n-1$). My gut is that when you have $k\leq n-2$ since the dimension of the image of $F$ is codimension one you have enough room to perturb it to be an immersion. That is that $F$ is homotopic rel boundary to our desired immersion.

Unfortunately, I don't know enough to formalize this and all my intuition comes from considering curves and domains in $\mathbb{R}^3$ so I'm afraid there may be obstructions in general.

References would be greatly appreciated.

Thanks!

Edit:

As suspected, the question is somewhat subtle . To make it tractable lets assume that $M$ is a $C^\infty$ domain in $\mathbb{R}^3$ (so is a fairly simple three-manifold with boundary) and that the $\Sigma_i$ are curves. This is where my intuition says that there should be such a smooth immersion.

• Modulo 2, you can always realize by a smooth map from a manifold. You can assume that map is an immersion provided $k \leq n/2$ by Whitney's approximation theorem (these results are due to Rene Thom). To get the realization with integer coefficients is more difficult -- sometimes you can, sometimes you can't. Are you working in a fairly general context, or is it quite specific? – Ryan Budney Oct 27 '10 at 1:02
• You can get any mod 2 homology class of $M$ by a map of a manifold into $M$ (homology group surjects onto bordism group), while this fails in general for integral homology and oriented manifolds. But even in the mod 2 case homology does not inject to bordism; the first obstruction in the integral case is signature of a 4-manifold (see my answer), and the first obstruction in the mod 2 case is mod 2 Euler characteristic of a 2-manifold. – Tom Goodwillie Oct 27 '10 at 1:19
• My ignorance is revealed! I want integer homology. My $\Sigma_i$ are actually curves and my $M$ is a $C^\infty$ domain in $\Real^3$ (I.e is a smooth 3-manifold with boundary). – Rbega Oct 27 '10 at 2:25
• For 3-manifolds you can realize all 1 and 2-dimensional integral homology classes by embedded submanifolds. For H1 it comes from the Hurewicz theorem + general position. For H2 it comes from Serre's theorem and Poincare duality $H_2(M) \simeq H^1(M) \simeq [M,S^1]$. – Ryan Budney Oct 27 '10 at 3:30
• Bredon's text "Geometry and Topology" would be a good reference for these details. – Ryan Budney Oct 27 '10 at 14:43

## 3 Answers

There is a general strategy for these kind of problems, which sometimes helps (the ''h-principle''): separate the homotopical and smooth aspects of the problem. Setup: $f:N \to M$ a map of smooth manifolds, $dim (N) < dim (M)$, $f|_{\partial N}$ is an immersion.

Step 1: if your $f$ is going to be homotopic to an immersion $g$, then there should exist a bundle monomorphism $\phi:TN \to f^* TM$, extending the derivative of $f|_{\partial}$ to all of $N$. Otherwise, the hypothetical immersion won't exist, because there is no candidate for its derivative. Constructing this monomorphism is a purely bundle theoretic problem. Whether this can be successfully attacked depends very much on the concrete situation -- maybe it is trivial, maybe it is very hard. In any case -- if there is no candidate for an immersion in sight -- it is be easier to construct a bundle map than an immersion.

Step 2: if you succeed in solving the homotopy problem, you are done! This is the content of the Smale-Hirsch-Theorem: once $\phi$ exists, $f$ will be homotopic to an immersion $g$ and the derivative is homotopic (as a bundle mono) to $\phi$. There is a relative statement as well; you have to extend the immersion $f_{\partial M}$ to an immersion of a collar of $\partial M$ in $M$. This is not trivial since the collar has one dimension more. If $dim(N)=dim(M)$, then the result still holds (in your situation, where $N$ is a connected cobordism with nonempty boundary), but you cannot fix $f$ on the boundary.

Your intuition in the low-dimensional case seems to be true: if $N$ is a surface and $M$ a domain in $R^3$, then both are parallelizable.

References: Eliashberg-Michachev: Introduction to the h-principle, Haefliger: Lectures on the theorem of Gromov, Ponaru: Homotopy theory and differentiable singularities. Adachi: Immersions and embeddings.

• This seems to be more or less what I was looking for. Thanks. – Rbega Oct 28 '10 at 18:32
• Does it help with your example? Great! – Johannes Ebert Oct 28 '10 at 19:14

Reading between the lines, I suppose you have oriented manifolds $\Sigma_i$, so that each of them determines an element of the integral homology group $H_k(M)$. In fact the assumption that these two classes are equal does not even imply that there is a compact oriented $(k+1)$-manifold $\Gamma$ with boundary such that the boundary is $\Sigma_1\coprod \Sigma_2$. For example, $\Sigma_1$ could be $\mathbb CP^2$, $\Sigma_2$ could be $S^4$, and they could both be embedded in $S^n$ for $n>>0$. The boundary of a compact oriented $5$-manifold always has signature zero.

Maybe modify the question, so that insteading of assuming homologous you assume cobordant, meaning that you assume there is a compact oriented $\Gamma_0$ mapped into $M$ in such a way that $\partial \Gamma_0\to M$ is an immersion, and you ask whether there exists another thing $\Gamma\to M$ with $\partial\Gamma\to M$ isomorphic to $\partial \Gamma_0\to M$ but this time with $\Gamma\to M$ an immersion.

• Thanks! Given this I made the question (much) more specific. In particular, failure to be cobordant is no longer an issue. – Rbega Oct 27 '10 at 2:32

The answer is known in the special case you mentioned (curves in a 3-manifold) Actually it is true more generally: Namely if k = n-2 and your k-dimensional oriented submanifolds $\Sigma_1$ and $\Sigma_2$ represent the same INTEGER homology class in the ambient ORIENTED n-manifold M, and they are disjoint and embedded, then there is an embedded k+1 dimensional submanifold in the cylinder $M \times I$ with boundary $\Sigma_1$ in $M\times {0}$ and $\Sigma_2$ in $M \times {1}.$

This a trivial consequence of the fact, that $K(Z,2) = MSO(2).$ The homotopy classes [M, K(Z,2)] give the 2-dimensional cohomology group, hence also the k-th homology group of M. The homotopy classes [M, MSO(2)] give the codimension 2 embedded submanifolds up to an embedded cobordism (by the Thom construction). QED.

In general the answer is negative, as it was mentioned by others.