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As far as I can tell, $\mathbb{CP}^2\mathbin{\#}\mathbb{CP}^2$ is not homeomorphic to $\mathbb{CP}^2\mathbin{\#}\overline{\mathbb{CP}^2}$. This is because $\mathbb{CP}^2\#\mathbb{CP}^2$ has a definite intersection form but the intersection form of $\mathbb{CP}^2\mathbin{\#}\overline{\mathbb{CP}^2}$ is indefinite, and any homeomorphism can only change the signature of the intersection form by $\pm{1}$ according to whether it preserves or changes the orientation.

(Question 1: Is this argument okay?)

Now, on some level I find this completely flummoxing. The manifolds $\mathbb{CP}^2$ and $\overline{\mathbb{CP}^2}$ $\it{are}$ homeomorphic (not as manifolds with orientation however, since $\mathbb{CP}^2$ has no orientation-reversing self-homeomorphisms). So you take two copies of the same manifold and glue them together, but the second time you do the gluing of the second manifold in the mirror (but this image is still homeomorphic to the original!), and you get two different results. Huh?

I think I must be missing something subtle which happens in the connected sum operation.

Question 2: Does anyone know what I'm missing, and have a way to make this discrepancy between the resulting manifolds seem less unintuitive?

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  • $\begingroup$ I don't know if this makes it less unintuitive or not, but: for even simple objects like $S^n$ rather than $\Bbb{CP}^2$, connected sum of two $S^n$'s might not be diffeomorphic to $S^n$. There are things called exotic spheres, which are homeomorphic but not diffeomorphic to $S^n$: such things can be obtained from gluing two $n$-disks along the boundary by a self-diffeomorphism of $S^{n-1}$ (Lee Mosher's analysis below tells you that $S^{n-1}$ must have a lot of different diffeomorphisms upto isotopy for this to happen). The examples are all high dimensional $(n \geq 7)$ though. $\endgroup$ – Balarka Sen Aug 16 '16 at 9:28
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Let's break down the connected sum operation which, given connected manifolds $X,Y$, produces a manifold $X \# Y$.

  1. Choose closed balls $B_X \subset X$, $B_Y \subset Y$ whose boundaries $S_X = \partial B_X$, $S_Y = \partial B_Y$ are smoothly embedded.
  2. Consider the manifold-with-boundary $X - \text{interior}(B_X)$, whose boundary is $S_X$.
  3. Consider also the manifold-with-boundary $Y - \text{interior}(B_Y)$, whose boundary is $S_Y$.
  4. Choose a diffeomorphism $f : S_X \to S_Y$.
  5. Let $X \# Y$ be the quotient of the disjoint union of $X - \text{interior}(B_X)$ and $Y - \text{interior}(B_Y)$, where each $x \in S_X$ is identified to the point $f(x) \in S_Y$.

The choices made in this construction have been highlighted, and there is a definite not-so-subtle issue here: Is the construction well-defined independent of the choices made?

It is possible to prove that the construction is independent of the choices of $B_X$ and $B_Y$. So let's set that aside.

Also, it is possible to prove that the construction is independent of the choice of $f_X$ up to smooth isotopy. This is a good exercise which you should try, in particular to see exactly how the hypothesis of "smooth isotopy" is used, and to get an inkling of why it is reasonable that the resulting connected sums might actually differ when smooth isotopy fails.

In general there is no reason to expect that if $f_X$, $f'_X$ are not smoothly isotopic then $X \# Y$ constructed using $f_X$ is the "same" as $X \#' Y$ constructed using $f'_X$.

In your particular example, since $\overline{\mathbb{CP}^2}$ and $\mathbb{CP}^2$ have opposite orientations, the two gluing maps are not even homotopic, let along smoothly isotopic.

Added comments in the oriented category: In the case where $X,Y$ are both oriented, then there is a natural concept of "oriented connected sum". In this case, each of $X - \text{interior}(B_X)$ and $Y - \text{interior}(B_Y)$ inherit orientations by restriction, and then each of $S_X,S_Y$ inherit natural boundary orientations. One then requires that the gluing map $f : S_X \to S_Y$ be orientation reversing. With this requirement, one obtains a natural orientation on $X \# Y$ whose restrictions to $X - \text{interior}(X)$ and $Y - \text{interior}(Y)$ agree with the orientations restricted from $X$ and $Y$.

Now back to the original examples. Each of $\mathbb{CP}^n$ and $\overline{\mathbb{CP}^n}$ are oriented manifolds, and the notations $\mathbb{CP}^n \# \mathbb{CP}^n$ and $\mathbb{CP}^n \# \overline{\mathbb{CP}^n}$ jointly enforce the intention that this connected sum is, indeed, an oriented connected sum (if that were not the intention, why the "overline" symbol on top of one of the $\mathbb{CP}^n$'s?).

I myself like to think of $\mathbb{CP}^n$ and $\overline{\mathbb{CP}^n}$ as being as separate from each other as I can stretch my imagination to think, because that reduces my internal mental confusion (which needs plenty of reduction). Perhaps you will decide to think of them as the "same manifold", but if so then you are obliged to remember at all times that they have opposite orientations. I will compromise: I will think of certain manifolds as "copies" of other manifolds.

So, let's think of $X$ as one copy of $\mathbb{CP}^n$. You might decide to think of $Y$ as another copy of $\mathbb{CP}^n$ but you have to keep track of orientation: either $Y$ has the same orientation, hence $Y$ is a copy of $\mathbb{CP}^n$ itself; or $Y$ has the opposite orientation, hence $Y$ is a copy of $\overline{\mathbb{CP}^n}$.

In the case that $Y$ is a copy of $\mathbb{CP}^n$, even if you decided to think of the removed balls $B_X,B_Y$ as copies of "the same ball", and to think of the spheres $S_X,S_Y$ as copies of "the same sphere", the map $f : S_X \to S_Y$ must reverse the orientation of this sphere. Thus $f$ cannot be chosen to be the identity map.

In the other case that $Y$ is a copy of $\overline{\mathbb{CP}^n}$, you might also decide to think of the removed balls $B_X,B_Y$ as copies of "the same ball", however $B_Y$ has opposite orientation from $B_X$. And then you would probably also decide to think of $S_X,S_Y$ as copies of "the same sphere", however $S_Y$ has opposite orientation from $S_X$. Now when it comes to choosing the homeomorphism $f : S_X \to S_Y$, you are indeed perfectly free to choose the identity map, because that map reverses orientation from $S_X$ to $S_Y$.

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  • $\begingroup$ Do we have to assume something about orientations on $S_X$ and $S_Y$ and how $f$ relates them in the points 1-5? Otherwise, I don't see how the orientation issue is a factor in your last point: we can pick the same ball in both $\mathbb{CP}^2$ and $\overline{\mathbb{CP}^2}$ and identify the boundaries by the identity map. Since the maps are literally equal, they are isotopic. Or am I missing something? $\endgroup$ – Brian Klatt Aug 15 '16 at 22:52
  • $\begingroup$ @BrianKlatt $\Bbb{CP}^2 \# \overline{\Bbb{CP}^2}$ is obtained by gluing the boundary spheres of two removed balls by the identity map relative to the orientation of $\Bbb{CP}^2$ and $\overline{\Bbb{CP}}^2$, yes. But if you identify $\overline{\Bbb{CP}}^2$ with another copy of $\Bbb{CP}^2$ by a orientation reversing diffeomorphism, that identity map becomes the orientation-reversing self diffeomorphism of the sphere. Whereas $\Bbb{CP}^2 \# \Bbb{CP}^2$ is just obtained by gluing the boundary of the removed balls by the identity map. $\endgroup$ – Balarka Sen Aug 16 '16 at 9:16
  • $\begingroup$ I extended my answer to address orientation questions, essentially expanding on the comment of @BalarkaSen. $\endgroup$ – Lee Mosher Aug 16 '16 at 14:05
  • $\begingroup$ In your last paragraph, did you intend to say $\overline{\mathbb{CP}^2}$ rather than $\mathbb{CP}^2$? $\endgroup$ – Brian Klatt Aug 16 '16 at 17:13
  • $\begingroup$ +1 for "...my internal mental confusion (which needs plenty of reduction)." Acutely felt, in my case. $\endgroup$ – Brian Klatt Aug 16 '16 at 17:16

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