If $M$ is a smooth, compact, orientable manifold, then any framed submanifold $N$ is the preimage $f^{-1}(y) $ for a smooth sphere-valued map $f$ transversal to $y$, with the framing of the normal bundle induced by $f$.

My question is: which framed submanifolds are induced by $\mathbb{R}^n$-valued maps? In other words, what is the condition on a framed submanifold $N$ of codimension $n$ to be the preimage $f^{-1}(0)$ for some $f: M\to\mathbb{R}^n$ transversal to $0$? "Framed-null-cobordand-ness" is probably necessary but not sufficient.

My motivation comes from this slightly more specific question that I'm trying to solve: if $M$ has a boundary and $g: \partial M\to S^n$ is given, then I want to find a common invariant of $f^{-1}(0)$ for all possible extensions $f: M\to D_n$ of $g$ ($D_n$ is the $n$-disc). At first, I though that a full invariant is a framed cobordism class in $M\setminus\partial M$ and that such cobordism classes are in 1-1 correspondence to homotopy classes $[\partial M, S^n]$. If I could prove that a framed cobordism $W$ between $N=f_1^{-1}(0)$ and $N'$ is the zero set of some $F: M\times [0,1]\to \mathbb{R}^n$ ($F$ nonzero on $\partial M)$, then I could use $F$ to define a homotopy between $F_0: (M,\partial M)\to (D_n, S^{n-1})$ and $F_1$ and then adjust it near the boundary so that $F$ is constant on $\partial M$. But I can't prove that $F$ exists.

I also tried to use a quotient $q: D_n/S^{n-1}\to S^n$, find a homotopy to $S^n$ and lift it: however, the homotopy lifting property works only for Serre fibration, which $q$ is not..

Any help will be much appreciated! (If necessary, I can also write more about the broader motivation behind these problems)

  • $\begingroup$ Should we assume $M$ to be closed? $\endgroup$ May 17, 2014 at 16:10
  • $\begingroup$ I rather didn't fix a too particular setting, any insight to the problem is welcome. I wonder how is it with the closed case, although, for the moment, I would be even happier to know whether a framed submanifold $W\subseteq M\times [0,1]$ is the zero set of a map, if $W$ doesn't intersect $\partial M\times [0,1]$ and $W\times \{0\}$ is the zero set of some map $f_0$ (transversal to $0$). $\endgroup$ May 17, 2014 at 16:31
  • $\begingroup$ In the closed case, the standard Pontryagin theorem seems to imply your "conjecture", that is: the zero set is always framed cobordant to the empty set. I'll think a bit about this other case, maybe. (Just to make sure, by $W\times\{0\}$ you rather mean $W\cap M\times\{0\}$?) $\endgroup$ May 17, 2014 at 16:50
  • $\begingroup$ Thanks a lot, I appreciate it! Yes, sorry for the mistake.. $\endgroup$ May 17, 2014 at 16:52
  • 2
    $\begingroup$ I think that nullcobordance is not sufficient: if we take 4 points on a circle with framing "+, +, -, -", than this set is framed nullcobordant but not preimage of zero under any $f: S^1\to \mathbb{R}$ transversal to $0$. $\endgroup$ May 17, 2014 at 18:50

3 Answers 3


edit: I've cleaned it up to make the dimension restrictions clearer, and to cover Peter's case of $M$ having boundary.

Let $M$ have dimension $m$, be a manifold possibly with boundary. $N \subset M$ a submanifold without boundary (in the interior of $M$) of dimension $n$. The question is when we can realize a framed submanifold $N$ in $M$ as the zero-set of a smooth map $f : M \to \mathbb R^k$ transverse to $0$, $N = f^{-1}(0)$.

$N$ being the pre-image of a regular value gives a canonical choice of trivialization of the normal bundle of $N$, the framing. Let $\nu N$ be an open normal bundle for $N$. Then the map $f$ on $M \setminus \nu N$ does not have $0$ in its image, so we can normalize it to a map $M \setminus \nu N \to S^{k-1}$.

Up until this point we have an if and only if statement. Moreover, if we take a point $p \in S^{k-1}$ that is a regular value of this map and take its pre-image we get a framed manifold $W$ such that $\partial W = N$.

Observation: a framed manifold $N \subset M$ is the pre-image of a regular value of a function to some $\mathbb R^k$ if and only if there is a framed manifold $W \subset M$ with $\partial W = N \cup N'$ where $N' \subset \partial M$.

Unfortunately, it's possible that $W \cap \partial M \neq \emptyset$. For example, let $M = S^1 \times [0,1]$ and $N = S^1 \times \{1/2\}$, then $f : M \to \mathbb R$ is $f(z,t) = t-1/2$, and either $W = S^1 \times [0,1/2]$ or $W = S^1 \times [1/2,1]$ depending on how you chose $p$ above.

This is something that's pretty inescapable, but I'm not sure if you were trying to avoid that.

To answer your question in the bounty box, no, not every submanifold is the zero-set of a function transverse to $0$. For example, take the real projective plane $\mathbb RP^2$ as $N$, and let $M$ be any sphere $S^m$, we need $m \geq 4$ for $N$ to be a submanifold of $M$. If $\mathbb RP^2$ were the zero-set of a function, it would have an orientable normal bundle. Since the sphere has an orientable normal bundle, that would give us an orientation of the tangent bundle to $\mathbb RP^2$. But $\mathbb RP^2$ has a non-orientable tangent bundle.

This kind of obstruction is fairly general. Basically the only kinds of manifolds $N$ that can be realized would be ones that for some framing of its normal bundle, it (together with the framing) is the boundary of a framed manifold.

  • $\begingroup$ I agree, but this is very general, it's hard to show extendability to a sphere. Is there no hope to have some more tractable condition? For example, a classification of which framed surgeries I'm allowed to do in order not to loose the property of "being the zero set", resp. "being the zero set and equal to $g$ on $\partial M$"? $\endgroup$ May 18, 2014 at 17:03
  • $\begingroup$ It's a delicate condition. The extensibility could be phrased as a sequence of obstructions being zero. It will be some cohomology computations but the coefficients will be in the homotopy groups of $S^{k-1}$. I guess I'm not seeing anything near as clean as the Pontriagin-Thom construction. $\endgroup$ May 18, 2014 at 17:09
  • $\begingroup$ I know the basics of obstruction theory but somehow hoped that it can be avoided here, at least in some range of dimensions.. :( $\endgroup$ May 18, 2014 at 17:10
  • $\begingroup$ (1) "allow you to do this if "$m>2n$" shouldn't be "$m<2n-1$" ($n$ is the codimension)? (2) So, the boundary of your $F^{-1}(1)$ is not exactly $N$ but Goodwillie's $N'$, is that right? (3) Please, could you recommend me some reference for why the last construction is possible in the stable dimension range? I hope I'm not asking too much, thanks. $\endgroup$ May 20, 2014 at 8:31
  • $\begingroup$ Do you think, Ryan, that if $\dim N < (\dim M)/2 \,-1$ and $M$ is orientable, then for any framed submanifold $N$ there exists a framed $W$ s.t. $\partial W$ consists of $N$ and something in $\partial M$? I cannot find any counter-example. $\endgroup$ May 27, 2014 at 11:21

If $N$ is framed and of codimension $n>0$ then a normal vector field gives you a manifold $N'$ isomorphic to $N$ (near $N$, running parallel to $N$). For $N$ to be of the form $f^{-1}(0)$ with $f:M\to \mathbb R^n$ transverse to $0$, it seems to be necessary and sufficient that $N$ is framed cobordant to $\emptyset$ in $M-N'$. If $n$ is less than one half the dimension of $M$ then this condition can be strictly stronger than "$N$ is framed cobordant to $\emptyset$ in $M$".

  • $\begingroup$ If $n$ is no more than $\dim M/2$, then it is equivalent to $N$ framed cobordant to $\emptyset$ in $M$? Could you give me a hint, what to study in order to see these facts? Thank you! $\endgroup$ May 18, 2014 at 17:26
  • $\begingroup$ Sorry. I wrote "more" when I should have written "less". I have corrected it now. A nullcobordism of $N$ in $M\times I$ implies a nullcobordism disjoint from $N'\times I$ if the dimension of the cobordism ($m+1-n$) is less than the codimension of $N'\times I$ ($n$). $\endgroup$ May 18, 2014 at 20:30
  • $\begingroup$ Nice. Please, is there a good reference for the claim that the nullcobordism can avoid $N'$ in that dimension range? Or is it obvious? $\endgroup$ May 19, 2014 at 17:11

Take $D^3\times K$ for some manifold $K$ and attach a 4 handle to the boundary via a map homotopic to the Hopf map $S^3\to S^2\times pt \subset S^2\times K$. You can't extend projection to $D^3\times K\to D^3$ to the handle without new zeroes (this is just a concrete example of Ryan Budney's reference to obstructions, here the obstruction comes from the fact that the Hopf map is not null homotopic.) Note that there is no framed $W$ so $\partial W = 0\times K \cup$ something on the boundary.


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