# Submersions of closed manifolds

Let $f\colon\thinspace M\to N$ be a map of closed smooth manifolds, with $\dim M > \dim N$. Recall that a submersion is a smooth map whose differential is surjective at every point in the domain.

Can one give conditions which guarantee that $f$ is homotopic to an submersion?

These conditions would necessarily have to be homotopy invariants. I am thinking there may be something in terms of relations amongst characteristic classes.

I am aware of the theorem of Phillips

MR0208611 (34 #8420) Phillips, Anthony Submersions of open manifolds. Topology 6 1967 171–206.

which says roughly that, when $M$ is open, $f\colon\thinspace M\to N$ is homotopic to an immersion if and only if the differential $df\colon\thinspace TM\to TN$ is homotopic to a bundle epimorphism. I'm also aware of the subsequent work of Thomas

MR0225332 (37 #926) Thomas, Emery On the existence of immersions and submersions. Trans. Amer. Math. Soc. 132 1968 387–394.

giving applications of Phillips' theorem. However, I couldn't find any more modern references dealing with the case $M$ closed. Are there any, or is there a good heuristic reason why such conditions cannot be given?

Remark: By Ehresmann's Theorem, and since $M$ is compact, it is equivalent to ask whether $f$ is homotopic to the projection of a locally trivial fibration.

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Let $F$ be the homotopy fibre of $f$ (ie the space of pairs $(x,u)$, where $x\in M$ and $u$ is a path from $f(x)$ to a specified baspoint in $N$). If $f$ is homotopic to a submersion $f'$, then (using Ehresmann) $F$ will be homotopy equivalent to a closed manifold $(f')^{-1}\{\text{point}\}$, and in particular, it will be equivalent to a finite CW complex. I do not know whether the converse is also true, but I suspect that any counterexamples would be quite exotic. So the first thing to do is to try to estimate the size of $F$.

Let $k$ be a field, and take cohomology with coefficients in $k$. At least is $N$ is simply connected, there is an Eilenberg-Moore spectral sequence $\text{Tor}_{H^*N}(H^*M,k)\Rightarrow H^*F$, where $F$ is the homotopy fibre. If $H^*M$ is a free module over $H^*N$ then this collapses to an isomorphism $H^*F=H^*M\otimes_{H^*N}k$, and this will be a finite-dimensional $k$-algebra. I think it will also automatically have Poincare duality (provided that $M$ and $N$ are oriented). If $H^*M$ is not free over $H^*N$ then the $E_2$ page will typically be very large, and in some cases it may be possible to show that no possible pattern of differentials will leave a finite-dimensional $E_\infty$ page. If so, we can conclude that $f$ is not homotopic to a submersion.

Alternatively, if $f$ is homotopic to a submersion then the tangent bundle $\tau_M$ maps surjectively to $f^*(\tau_N)$, which implies that the Stiefel-Whitney polynomial $w(\tau_M,t)=\sum_iw_i(\tau_M)t^{\dim(M)-i}$ is divisible by $f^*w(\tau_N,t)$ in $H^*(M;\mathbb{Z}/2)[t]$. This should generally be easy to check. There is a similar criterion with Chern polynomials of the complexified tangent bundles in integral cohomology.

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@Neil: Thanks, the necessary condition involving SW-classes should be useful. –  Mark Grant Apr 13 '12 at 19:06
@Neil: "This should generally be easy to check": sorry to be dense, but could you offer me some more help here? I have my two polys $f,g\in R[t]$, where $R=H^\ast(M;\mathbb{Z}/2)$. But $R$ is not an integral domain. Are there any tips and tricks for deciding whether $f$ divides $g$ in this case? Or just brute force? –  Mark Grant Apr 17 '12 at 10:20
BTW, the manifold maps I am looking at are axial maps $\mathbb{R}P^n\times \mathbb{R}P^n\to \mathbb{R}P^r$ (so $n<r$ and the restriction to each factor is homotopic to the standard inclusion). Hence $R\cong\mathbb{Z}/2[a,b]/(a^{n+1},b^{n+1})$. –  Mark Grant Apr 17 '12 at 10:22
@Mark: the polynomials in question are monic, so you can just do the standard algorithm for long division and see if you are left with a remainder. This is straightforward in any particular case, but not necessarily for an infinite family of cases. In the case you mention we have $f(t)=(t+a+b)^{r+1}/t$ and $g(t)=(t+a)^{n+1}(t+b)^{n+1}/t^2$. The question of whether $f$ divides $g$ probably depends in a sensitive way on the binary expansions of $r+1$ and $n+1$. You could easily get Maple etc to calculate a few hundred cases, and a pattern would probably emerge. –  Neil Strickland Apr 17 '12 at 11:03
@Neil: Good idea, thanks! –  Mark Grant Apr 17 '12 at 11:53

Phillips theorem is plainly wrong in the compact case, and for fairly non subtle reasons.

Take a closed oriented $3$-manifold $M$. There are plenty of bundle epimorphisms $TM \to \mathbb{R}$ because $M$ is parallelizable; any form without zeroes gives you one. Now very often the kernel is a trivial vector bundle as well, so all possible characteristic classes I can think of are null. However, $f$ is never homotopic to a submersion since $\mathbb{R}$ is not compact.

The truth is that the fact that the immersion theorem holds for closed manifolds is an accident. Other cases of Gromov's h-principle (metric with signed curvature, symplectic structures) show that the h-principle fails completely in the closed case. This is not to say that there are no h-principles for closed manifolds, but definitely not for submersions.

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@Johannes: Thanks. Are there such simple counter-examples in the case when $N$ is also compact? For instance, replacing $\mathbb{R}$ by $S^1$, a map $M\to S^1$ is a class in $H^1(M;\mathbb{Z})$. Can we decide if this class contains a submersion? –  Mark Grant Apr 13 '12 at 9:06
@Mark: As I learnt from W. Steimle recently, it is a theorem of Browder--Levine, Farell, and Siebenmann that a map $f : M \to S^1$ is homotopic to a fibre bundle projection if and only if i) its homotopy fibre has the homotopy type of a finite CW complex, and ii) a certain torsion invariant $\tau \in Wh(\pi_1(M))$ vanishes. The torsion invariant is the Whitehead torsion of the $h$-cobordism one gets by splitting $M$ open over a regular value of $f$. –  Oscar Randal-Williams Apr 13 '12 at 9:33
And of course $M$ must have dimension $\geq 6$. –  Oscar Randal-Williams Apr 13 '12 at 9:33
@Oscar: Thanks for pointing out these references, in particular the work of Steimle and his collaborators seems to address my more general question. –  Mark Grant Apr 13 '12 at 19:05
The predecessor of the Browder-Levine theorem is Stallings Fibration Theorem which holds for a closed 3-manifold $M$ which is irreducible. The statement is that $f:M->S^1$ is homotopic to a fiber bundle projection if and only if $f_*:\pi_1(M)\to\pi_1(S^1)$ is surjective and has finitely generated kernel. –  Lee Mosher Apr 14 '12 at 13:57