Let $M \subset \mathbf R^n$ be a (smooth) submanifold of dimension $d$. Under which conditions does there exist global equations defining $M$? By global equations I mean : does there exist a smooth function $f: \mathbf R^n \to \mathbf R^{n-d}$, submersive at each point of $M$ and such that $M=f^{-1}(0)$.

Of course there are two necessary conditions: 1) $M$ must be a closed subset of $\mathbf R ^n$. 2) The normal bundle of $M$ in $\mathbf R^n$ must be trivial.

At first, I would have guessed that these conditions are sufficient, but I can't prove it.

I have partial answers, however.

1)The first natural thing to do is to take a tubular neighbourhood $U$ of $M$ in $\mathbf R^n$. The indentification $U \simeq M \times \mathbf R^{n-d}$ allows to define a function $f : U \to \mathbf R^{n-d}$ which has the required properties. But it is not clear to me whether $f$ can be extended to the whole $\mathbf R^n$.

2) There is a way to give an answer if we change a bit the problem: the Pontryagin-Thom construction gives a function $f: \mathbf R^n \to \widehat{\mathbf R^{n-d}} \simeq \mathrm S^{n-d}$ by sending all the points outside a tubular neighbourhood at infinity. This maybe means that this is the good formulation of the problem, but I am still curious about the original one.

3) If $M$ has codimension $1$, then the function $f$ defined on a tubular neighbourhood $U$ of $M$ as in 1) can actually be extended to $\mathbf R ^n$ by a constant function (using the fact that the complement of $M$ has two connected components).

  • $\begingroup$ For 3), how do you know that codimension 1 implies that the complement is not connected? I think this is not trivial. But I am not an expert so I might be wrong.. $\endgroup$ Oct 15, 2011 at 17:07
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    $\begingroup$ @Tommaso: Alexander duality. $\endgroup$
    – Jim Conant
    Oct 15, 2011 at 17:42
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    $\begingroup$ For $S^1$, what about $f(x, y, z) = (x^2 + y^2 - 1, z)$? But is it possible to do a knot? $\endgroup$ Oct 15, 2011 at 19:42
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    $\begingroup$ Martin, knots in $\mathbb R^3$ are all do-able. In my response, the map $W \to S^1$ is the classifying map of $\pi_1 W \to \mathbb Z$, the abelianization map. More generally, you can do any co-dimension two knot in $\mathbb R^n$. $\endgroup$ Oct 15, 2011 at 21:20

4 Answers 4


An example of a compact $16$-dimensional submanifold of $\mathbb{R}^{30}$ with trivial normal bundle that is not defined by such global equations may be found in [Akbulut-King, Submanifolds and homology of nonsingular real algebraic varieties, Am J Math 107 45 (1985), Theorem 5.8].

  • $\begingroup$ While I love the questions I do not fully understand the answer. Theorem 5.8 states that this manifold is not isotopic to a complete intersection in $\mathbb{R}^n$ and I do not know what it means. It would be great if you could add an explanation what it means so people with limited knowledge in topology, like me, could appreciate the result. $\endgroup$ May 11, 2020 at 13:12
  • $\begingroup$ Complete intersection means "transversal intersection of hypersurfaces defined as zero loci of functions", exactly as in the question asked (take the coordinates of $f$ as functions). In the Akbulut-King paper, the functions are assumed to be polynomial (this is why they use the term "complete intersection", which is common in algebraic geometry), but their argument works as well if the functions are smooth ($\mathcal{C}^{\infty}$) as in the question. $\endgroup$ May 11, 2020 at 15:24
  • $\begingroup$ Does the submersion property play a role here, or can you not write it as the zero locus of any $C^\infty$ function $\mathbf R^{30} \to \mathbf R^{14}$? $\endgroup$ May 11, 2020 at 16:58
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    $\begingroup$ @R.vanDobbendeBruyn All compact $\mathcal{C}^{\infty}$ submanifolds $M\subset\mathbb{R}^n$ are the zero locus of a single $\mathcal{C}^{\infty}$ function: write $M$ as the zero locus of finitely many $\mathcal{C}^{\infty}$ functions (idea: extend local equations to global ones using a bump function), and sum their squares. So the submersion property is indeed important. $\endgroup$ May 11, 2020 at 17:05
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    $\begingroup$ Right. Real numbers are weird. $\endgroup$ May 11, 2020 at 17:33

I no longer think there are further obstructions. Here's why.

Let $f : \mathbb R^n \to S^k$ be smooth with $p \in S^k$ a regular value, and $M=f^{-1}(p)$. You get this map from your conditions (1) and (2) + the Pontriagin construction.


$$ g : \mathbb R^n \times \mathbb R \to \mathbb R^{k+1}$$

given by $g(v,t) = e^tf(v)$ is smooth, has $p$ also as a regular value, and $M=g^{-1}(p)$.

I'm using the convention that $S^k$ is the unit sphere in $\mathbb R^{k+1}$.

If you're not happy including $M$ into $\mathbb R^{n+1}$ like above, then the original argument is all I have. That is, $M$ is the pre-image of the regular value of a smooth function $f : \mathbb R^n \to \mathbb R^k$ if and only if your conditions (1), (2) and further that the complement $W$ of an open tubular neighbourhood of $M$ in $\mathbb R^n$ has this property: unit normal spheres (from the normal bundle of $M$ in $\mathbb R^n$) are retracts of $W$. I suspect this is a non-trivial restriction to your original question but I haven't found a concrete example.

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    $\begingroup$ In your Hopf link example, can't we make a different choice of trivialization of the normal bundles of the two circles so that the map to $S^1$ extends over all of $T^2 \times I$ minus a point? i.e. make the linking numbers of both trivializations either $+1$ or $-1$, the two slopes which are fixed by the meridian $\leftrightarrow$ longitude involution. $\endgroup$ Oct 15, 2011 at 21:49
  • $\begingroup$ Yes, you're right. And I think all links in $\mathbb R^3$ have such maps $W \to S^1$ -- you can think of this map as basically being the sum of the linking numbers for all the components of the link. So if this constraint $W \to S^{n-m-1}$ is non-trivial it will be for a more complicated example. $\endgroup$ Oct 15, 2011 at 22:07

I have two comments which don't fit into the comments field so I'm posting this as an answer.

  1. As Ryan Budney mentioned above the problem becomes trivial if we compose the original embedding $M^d\to \mathbb R^n$ with the canonical inclusion $\mathbb R^n\times\{0\}\subset \mathbb R^{n+1}$. That means in particular that in the stable range (when $n>2d+1$) there are no obstructions because in that range any two embeddings of $M\to \mathbb R^n$ are ambiently isotopic so if the problem is solvable for one then also for the other. Since $n>2d+1$ we can always isotope our original embedding into $\mathbb R^{n-1}\times \{0\}\subset \mathbb R^n$ and the claim follows.

  2. When $k=n-d-1$ is odd then $S^k$ is an Eilenberg-Maclane space over $\mathbb Q$ which means that rationally the same exact argument that worked for $k=1$ also works here. To elaborate further, given a trivialization of the normal bundle we have a tubular neighborhood $U\cong D^{k+1}\times M$ and we have an obvious map $f: U\to D^{k+1}$ given by the projection onto the second factor. On the boundary of $U$ the map takes values in $S^k$ and we want to extend it to a map from $W=\mathbb R^n\backslash U$ to $S^k$. Since $S^k$ is rationally equivalent to $K(\mathbb Q,k)$, rationally the homotopy type of $f|_{\partial U}$ is determined by $f^*([S^k]) \in H^*(\partial U, \mathbb Q)$ and the question becomes whether or not this class in the image of $i^*:H^k(W,\mathbb Q)\to H^k(\partial W\cong M\times S^k,\mathbb Q)$. By Alexander duality $H_k(W,\mathbb Q)$ is isomorphic to $H^d(M,\mathbb Q)\cong \mathbb Q)$ with the generator given by $i_*([S^k])$ where $S^k$ is the normal $S^k$ in $\partial W\cong S^k\times M$. Let $\alpha\in H^k(W,\mathbb Q)$ be the dual generator to $i_*([S^k])$. Now, a priori, the original trivialization may have been wrong so that $i^*(\alpha)$ is not equal to $f^*([S^k])$ if $H^k(M)\ne 0$. However, since the evaluation map $SO(k+1) \to S^k$ is a rational isomorphism on $H^k$ we can modify the original trivialization by an appropriate map $M\to SO(k+1)$ which does make $i^*(\alpha)=f^*([S^k])$ meaning that the map extends. What this means is that whatever (if any) obstructions are present in this case they are all torsion.

When $k=1$ then $S^1$ is already a $K(\mathbb Z,1)$ space and the above works on the nose without tensoring with $\mathbb Q$ as Ryan mentioned in a comment above.

I just realized that algori's answer below can not be correct. His(hers?) idea was that if we write $M$ as $(f_1,\ldots, f_{k})=0$ then after a small perturbation we can assume that $0$ is still a regular value $(f_1,\ldots, f_{k-1})=0$ and hence $M$ frame bounds in $N=(f_1,\ldots, f_{k-1})=0$ because it clearly separates $N$. However, this argument does not work because the level set $(f_1,\ldots, f_{k-1})=0$ might not be compact so the fact that $M$ separates it does not imply that it's cobordant to zero. This is not a fake issue since otherwise by Ryan's observation above it would imply that every framed cobordance class is stably trivial which is known not to be the case.

Also, as far as I can tell the effect of the change of the trivialization by a map $M\to SO(n-d)$ changes the framed cobordance class by something in the image of the $J$-homomorphism $J\colon \pi_d(SO(n-d))\to \pi_n(S^{n-d})\cong\Omega^{fr}_d(\mathbb R^n)$.

More explicitly it seems to me that it works as follows. Given any $\alpha\in \pi_d(SO(n-d))$ and $M^d$ as above, take a degree one map $f:M\to S^d$. Then twisting the trivialization by $\alpha\circ f:M\to SO(n-d)$ should give a new framed cobordism class which is different from the original one by $J(\alpha)$. This should be very well-known I'm sure so could somebody in the know please comment on this?

So it would seem that the group $\pi_n(S^{n-d})/\mathrm{Im }(J)$ is relevant here but I'm having trouble phrasing our obstruction problem in cobordism terms. In particular, is it clear that if we have two frame cobordant manifolds in $\mathbb R^n$ and one can be given by a single equation then so is the other?

  • $\begingroup$ Dear Vitaly, I think the compactness issue can be dealt with. I'll amend my answer to address that. However, I don't think that what I say implies that every framed cobordism class is stably trivial. Rather, that every framed cobordism class is killed by the first suspension after a change of framing. $\endgroup$
    – algori
    Oct 20, 2011 at 10:16
  • $\begingroup$ .. e.g. the unknotted circle in the 3-space bounds a disk, but the same circle equipped with the "Hopf" framing is non-trivial, even stably: it generates $\pi_{k+1}(S^k)$. $\endgroup$
    – algori
    Oct 20, 2011 at 11:14
  • $\begingroup$ I believe the compactness issue is essential even after gauging by trivialization. It it wasn't then the group $\pi_n(S^{n-d})/\mathrm{Im }(J)$ would be stably trivial which it is not. $\endgroup$ Oct 20, 2011 at 14:24
  • $\begingroup$ Dear Vitali -- re framing changes acting as the image of the J-homomorphism: could you perhaps elaborate a bit on this? The image of the J-homomorphism in $\pi_n(S^{n−d})\cong\Omega^{fr}_d(\mathbb{R}^n)$ is formed by the standard sphere $S^d\subset S^n$ with an arbitrary framing, so the image of the J-homomorphism acts transitively on the framings of $S^d$ and hence it acts on the framings of an arbitrary $M$. However it is not clear (to me) why for arbitrary $M$ the action is transitive. $\endgroup$
    – algori
    Oct 20, 2011 at 15:18
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    $\begingroup$ Dear Vitali -- $Im(J)$ acts on $\Omega^{fr}_d(\mathbb{R}^n)$ by twisting a given framing by a framing of the standard $d$-sphere. So the image of $Im(J)$ in $\pi_0(Map(M,SO(n-d))$ can be identified with the image of the set of the maps that are identity outside some fixed ball $U$ around some $x\in M$. (In your construction, take the degree 1 map $M\to S^d$ to be the map that collapses the complement of $U$ to a point.) This is indeed a subgroup of $\pi_0(Map(M,SO(n-d))$, but I do not see any reason it should be the whole of $\pi_0(Map(M,SO(n-d))$. $\endgroup$
    – algori
    Oct 20, 2011 at 22:02

It seems to me now that there may be obstructions after all. Here is a potential source of counter-examples.

Let $M$ be a $d$-dimensional complete intersection in $\mathbb{R}^n$, i.e. $M$ is given as the zero locus of $f_1,\ldots, f_k,k=n-d$ defined globally on $\mathbb{R}^n$ and the differentials of $f_i$'s are linearly independent at each point of $M$.

[upd 2: Then the normal bundle of $M$ is framed. Let us show that, unless $\dim M=0$, $M$ is framed cobordant to 0 bounds a submanifold of $\mathbb{R}^n$. Notice that when $\dim M=0$, this is not necessarily so: take e.g. $M$ to be one point in $\mathbb{R}$.

For each fixed $d>0$ we proceed by induction on $n\geq d+1$. The case $n=d+1$ (i.e., $M$ is a hypersurface) is clear. Suppose $n>d+1$ and embed $\mathbb{R}^n$ in $S^n$ as the complement of a point, $\infty$. Take a function $\phi:\mathbb{R}^n\to\mathbb{R}$ that is equal 1 in some ball containing $M$ and decreases sufficiently fast at the infinity, so that we can extend the functions $g_i=\phi f_i$ to $S^n$ by setting $g_i(\infty)=0$. Let $\bar g_i,i=1,\ldots,k$ be a slight perturbation of $g_i$ such that

  1. both $\bar g_1=\cdots=\bar g_{k}=0$ and $\bar g_1=\cdots=\bar g_{k-1}=0$ define smooth complete intersections; denote these as $M'$ and $N$ respectively;

  2. the components of $M'$ are $M$ and possibly some submanifolds of a small open $n$-ball $U$ such that $\infty\in U$ and $M\cap \bar U=\varnothing$;

  3. $N$ intersects $S=\partial \bar U$ transversally.

Then $M$ is framed cobordant to $N\cap S$. A cobordism can be obtained by taking $N_+=\{x\in N\mid \bar g_k(x)\geq 0\}$ and intersecting with the exterior of $U$. Now, $N\cap S$ is a smooth complete intersection of dimension $d$ in $\mathbb{R}^{n-1}$, and hence is framed cobordant to 0 bounds a submanifold of $S$ by the induction hypothesis.]

So if we find a manifold $M\subset \mathbb{R}^n$ with trivial normal bundle but such that no framing of this bundle makes $M$ framed cobordant to 0 that does not bound any manifold in $\mathbb{R}^n$, we have a counter-example. The Pontrjagin-Thom construction gives an isomorpfhism $$\Omega^{fr}_{d}(\mathbb{R}^{n})\cong \pi_{n}(S^{n-d}).$$

Each choice of a framing $f$ of the normal bundle of $M$ gives some element of $\pi_{n}(S^{n-d})$. What is not clear (to me) is how this element changes when $M$ is fixed and $f$ varies under the action of the gauge group $Map(M,O(n-d))$.

Remark: from Ryan's answer it follows that if there are counterexamples, then the corresponding elements of $\pi_{n}(S^{n-d})$ are killed by the suspension map $\pi_{n}(S^{n-d})\to \pi_{n+1}(S^{n+1-d})$ after a change of framing. From the tables of the homotopy groups there are plenty of elements killed by suspension maps, but I don't know how to describe them explicitly.

On the positive side: if one takes any framing of the normal bundle and doubles $M$ using one of the sections, them the resulting "two copies" of $M$ can be given by global equations.

  • $\begingroup$ I just realized that this can not be correct. The problem is that the level set $(f_1,\ldots, f_{k-1})=0$ might not be compact so the fact that $M$ separates it does not imply that it's cobordant to zero. This problem is not fixable since otherwise by Ryan's observation above we'd have that any framed cobordance class is stably trivial which is definitely not true. ok, this is getting too long for the comment field so I'm going to add the rest to my answer above. $\endgroup$ Oct 19, 2011 at 13:06
  • $\begingroup$ Sorry but this is still wrong. You can not extend $f_1,\ldots, f_{k-1}$ from $\mathbb R^n$ to $S^n$ continuously. $\endgroup$ Oct 21, 2011 at 1:41
  • $\begingroup$ Maybe not, but one can restrict to a tubular neighborhood of $M$ and then extend. $\endgroup$
    – algori
    Oct 21, 2011 at 2:10
  • $\begingroup$ That completely breaks the argument about $M$ separating $(f_1,\ldots,f_{k-1})=0$. I don't believe there is an easy way to fix this. $\endgroup$ Oct 21, 2011 at 2:32
  • $\begingroup$ Yes, you are right, extending from a tubular neighborhood was not a very good idea. However, I still think that $M$ in framed null-cobordant. A proof of that has been added. $\endgroup$
    – algori
    Oct 24, 2011 at 18:58

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