Let $M$ be a connected smooth manifold. Let $S$ be a connected embedded submanifold of positive dimension and co-dimension, which is also a closed subset of $M$.

Is it true that the quotient space $M /S$ (i.e the space obtained by identifying all $S$ to a single point) is never a manifold?

Here is an attempt to show this:

If we denote the equivalence relation by $E \subseteq M \times M$, then it is proven in Bourbaki (see here), that the quotient $M /S$ inherits from $M$ a smooths structure if and only if:

  • $E$ is a closed submanifold of $M \times M$, and

  • The first projection $\pi_1: E \to M$ is a submersion

In our case $E=\{(x,x)|x \notin S \}\cup S \times S=\{(x,x)|x \in M \}\cup S \times S$

Now assume $E$ is a submanifold of $M \times M$.

Then, since $(S\times S)^c$ is open in $M \times M$, $E \cap (S\times S)^c=\{(x,x)|x \in S^c \}$ is open in $E$.

Since $S$ is closed in $M$, $S^c$ is open in $M$, so $\dim E =\dim \big( E \cap (S\times S)^c \big)=\dim \big( \operatorname{diag}(S^c) \big)=\dim(S^c)=\dim M:=n$

If I am not mistaken, then for every point $s \in S$, $\dim_{(s,s)}E=(n-s)+2s > n$ which is a contradiction.

The part of $(n-s)$ comes from paths spanning the complementary directions to $T_sS$ in the diagonal, i.e $\alpha(t)=(\beta(t),\beta(t))$ starting from $(s,s)$, such that $\dot \beta(0)=v \in T_sM\setminus T_sS$. The part of $2s$ comes from the freedom of movement inside the manifold $S \times S$.

Is this true? Is there a shorter (direct) argument? (perhaps one which does not use Bourbaki's result)

  • $\begingroup$ I have added the assumption $M$ is connected. Otherwise one could take $M=M_1 \cup M_2$ to be a disjoint union of two connected components, and choose $S=M_2$. Then $M/S = M_1 \cup \{ \cdot \}$ is a manifold. Another possible way to eliminate these kind of trivial counter-exmaples, is by requiring the dimension of $M/S$ to be constant, and that of $M$ to be positive. $\endgroup$ – Asaf Shachar May 13 '16 at 17:31
  • $\begingroup$ Great question, I had a lot of fun with it. $\endgroup$ – user98602 May 13 '16 at 18:29
  • $\begingroup$ Doesn't taking $M = \Bbb R P^2$ and $S = \textrm{(some projective line)}$ make $M / S$ a $2$-sphere? $\endgroup$ – Travis Willse May 13 '16 at 18:30
  • $\begingroup$ Reading @MikeMiller's quite good answer, I see that this is basically a real version of his first example. $\endgroup$ – Travis Willse May 13 '16 at 18:33

I haven't carefully read the proof in your post; I am more interested in whether this subset is a topological manifold, rather than a smooth one such that the quotient map is a submersion, which as you say is not true, and straightforward to see; the differential must necessarily kill the tangent space of $S$, but the quotient "manifold" is still of the same dimension as $M$, because $S$ has positive codimension.

And yes, this is indeed possible. Take $\Bbb{CP}^1 \subset \Bbb{CP}^2$. Collapsing this to a point gives you a copy of $S^4$. (To see this straightforwardly, note that $\Bbb{CP}^2$ is obtained by gluing a $D^4$ to $S^2$ by the Hopf map on the boundary; collapsing the $S^2$ gives us a space obtained by collapsing the boundary of $D^4$ to a point: $S^4$.)

Here is how I came up with the example, and the general principle at play here. The tubular neighborhood theorem says there is a neighborhood of a submanifold $S$ diffeomorphic to a vector bundle (the normal bundle) over $S$. So it suffices to answer it for this case. Now if $E$ is a bundle, and $S$ the zero section, $E/S$ is contractible, so has trivial homology. This implies that $H_{*-1}(E/S - [S]) \cong H_*(E/S,E/S-[S])$ by the relative long exact sequence. The first group is isomorphic to $H_*(E-S)$ (they're homeomorphic!), which deformation retracts onto the unit sphere bundle of the vector bundle. Now we can conclude: If $E/S$ was a manifold, then the local homology at $[S]$ $H_*(E/S,E/S-[S])$ would have the same homology of the appropriate-dimensional sphere. This is false for many, many vector bundles (for instance, if you collapse a circle in a manifold, the result is never a manifold) but not always false; the way the above example came to mind is that the circle bundle over $S^2$ with Euler class 1 is $S^3$, and hence of course does have the same homology as $S^3$...

Now here's a full proof that the quotient is a manifold if and only if the sphere bundle is a sphere. (It's quite clean, cleaner than the above, but I thought I would leave it in to show my thought process.) The quotient $E/S$ is homeomorphic to the cone on the sphere bundle. The above argument shows that whenever the sphere bundle has the wrong homology, the quotient is not a manifold. When it has the right homology, but is not simply connected, follow the argument here. Now any closed simply connected manifold with the same homology as a sphere is a sphere, by the Poincare conjecture.

OK, when is a sphere bundle a sphere? Suppose $S^n$ fits into a fibration with fiber $S^k, k>0$, and base $M$. Then $M$ is necessarily simply connected, and a straightforward spectral sequence argument shows that necessarily either the cohomology of the base is $\Bbb Z[x]/(x^n)$ where $|x|$ is even. This is only possible if $|x|$ is 2 or 4 (or $|x|=8$ and $n \leq 3$); see proposition 4L.10 of Hatcher. So necessarily the base has the same cohomology ring as $\Bbb{CP}^n$ or $\Bbb{HP}^n$ and is simply connected (or the base is $S^8 = \Bbb{OP}^1$ or has the cohomology ring of $\Bbb{OP}^2$) In the case $k=0$ the same holds with $\Bbb{RP}^n$ and $\Bbb Z/2$-coefficients. Unfortunately, this is as far as we go: it is entirely possible that this is not the obvious fibration. For instance, there are fake projective spaces, both real and complex, all of which (by the argument given on the latter page) support sphere bundles with total space $S^n$. I'm sure there's also a classification of fake quaternionic projective spaces. This is probably a complete classification of sphere bundles with appropriate total space: all are $S^0$, $S^1$, or $S^3$ bundles over a fake projective space (or again the case of $\Bbb{OP}^2$). I am not going to carry out the details that this is the case. (EDIT: Yes, this is the classification; simply connected manifolds with this cohomology ring are homotopy equivalent to the appropriate projective space, because they have a CW decomposition with a cell in each degree $kn$, $k=2,4,8$; then these are homotopy equivalent to the projective spaces by an inductive argument.)

  • $\begingroup$ (+1) Blowing up points (or rather, blowing down after blowing up) furnishes more examples. :) $\endgroup$ – Andrew D. Hwang May 13 '16 at 18:11
  • $\begingroup$ @AndrewD.Hwang And all of the examples "look like this", by the argument I've just edited in. $\endgroup$ – user98602 May 13 '16 at 18:27

This is not true as stated trivially if we let $S=M$, since $M/M$ is a single point and a zero dimensional manifold.

But for a non-trivial example, consider the circle, $M=S^1$, and let $S$ be any connected proper submanifold, a.k.a. an arc of the circle. Then $M/S$ is still going to be a circle.

Edit below

Additional example with $M$ having boundary and $S$ is a closed submanifold: Let $M$ be the closed ball of dimension $n$, $B^n$. Then let the boundary be $S=S^{n-1}$. Then $M/S=S^n$.

I would not be surprised if this is true for $M$ and $S$ closed as Mike suggests though.

  • $\begingroup$ They presumably want closed manifolds. $\endgroup$ – user98602 May 13 '16 at 17:37
  • $\begingroup$ An arc is not closed (if it is then it is a submanifold with boundary). Also, let's assume $S$ has a positive co-dimension, so the question won't become trivial. $\endgroup$ – Asaf Shachar May 13 '16 at 17:40
  • $\begingroup$ +1 Your first example violates the positive codimension requirement, but I note that it was not added until after your post, so you cannot be blamed for lacking prescience. The final example is excellent. $\endgroup$ – Paul Sinclair May 14 '16 at 3:22
  • $\begingroup$ @PaulSinclair thank you! $\endgroup$ – N. Owad May 15 '16 at 0:15

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