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This is one of those perhaps rare occasions when someone takes the advice of the FAQ and asks a question to which they already know the answer. This puzzle took me a while, but I found it both simple and satisfying. It's also great because the proof doesn't use anything fancy at all but it's still a very nice little result.

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Could you please clarify the question a bit? Are you looking for a counterexample? Why should $\mathbb{CP}^2$ be a covering space for any other manifold? –  Rasmus Mar 3 '11 at 11:35
I couldn't tell you exactly why $\mathbb{CP}^2$ should cover any other manifolds, other than that it's quite symmetric and also simply-connected and so it's totally believable that it might admit a free action of a finite group. But I'm not looking for a counterexample, because the statement just so happens to be true. –  Aaron Mazel-Gee Mar 3 '11 at 17:16
How do you know the statement just so happens to be true? –  Mariano Suárez-Alvarez Apr 1 '11 at 18:13
What happens with $\mathbb{C} \mathbb{P}^n$ for other $n$? I know the answer for $n = 1$: $\mathbb{C} \mathbb{P}^1$ covers $\mathbb{R} \mathbb{P}^2$... –  Pete L. Clark Apr 1 '11 at 19:26
@Mariano: I was asking because I knew the answer but I thought it was a fun puzzle and wanted to share it. –  Aaron Mazel-Gee Apr 2 '11 at 5:44

2 Answers 2

up vote 12 down vote accepted

Here's another argument that has the disadvantage of being less elementary, but the advantage of working on all $\mathbb{C}P^{2k}$ simultaneously. (This also answers Pete's question in the comments).

We're going to apply the Lefshetz fixed point theorem which states the following: Suppose $f:M\rightarrow M$ with $M$ "nice enough" (certainly, this applies to compact manifolds - I think it applies to all compact CW complexes). Then $f$ induces a (linear) map $f_*:H_*(M)/Torsion\rightarrow H_*(M)/Torsion$. Let $Tr(f)\in\mathbb{Z}$ denote the trace of this map. If $Tr(f)\neq 0$, then $f$ has a fixed point.

Now, we'll show that every diffeomorphism $f:\mathbb{C}P^{2k}\rightarrow \mathbb{C}P^{2k}$ has trace $\neq 0$, so that every diffeomorphism has a fixed point. Believing this for a second, note that every element of $\pi_1(X)$ for a hypothetical space $X$ covered by $\mathbb{C}P^{2k}$ acts by diffeomorphisms, and thus has a fixed point. But it is easy to show that the only element of the deck group which fixes any point must be the identity. It follows that $\pi_1(X)$ is trivial, so $X=\mathbb{C}P^{2k}$.

So, why does every diffeomorphism of $\mathbb{C}P^{2k}$ have a fixed point? Well, every diffeomorphism (or even homotopy equivalence!) must act as multiplication by $\pm 1$ on each of the $2k+1$ $\mathbb{Z}$s in the cohomology ring of $\mathbb{C}P^{2k}$ and the trace of the induced map is the sum of all the $\pm 1$s. But since there is an odd number of $\pm 1$s, they can't sum to 0 (by, say, checking the parity), so by the Lefshetz fixed point theorem, every diffeomorphism (or even homotopy equivalence) must have a fixed point.

What about $\mathbb{C}P^{2k+1}$? Now we must investigate using the ring structure of $\mathbb{C}P^{2k+1}$. Since there is a single multiplicative generator, once we know what happens on $H^2(\mathbb{C}P^{2k+1})$ we know what happens everywhere. It's easy too see that every orientation preserving homotopy equivalence must have a fixed point: if $f$ is orientation preserving, it's the identity on $H^{4k+2}(\mathbb{C}P^{2k+1})$, which implies it must have been the identity on $H^2(\mathbb{C}P^2)$ so it's the identity on all cohomology groups. Thus, the trace of such an $f$ is $2k+1\neq 0$, and so, by the Lefshetz theorem, this map has a fixed point.

As an immediate corollary, if $\mathbb{C}P^{2k+1}$ covers anything, it can only double cover it. For the product of any two nontrivial elements in the deck group must be trivial: any nontrivial map must be orientation reversing and the composition of two orientation reversing maps is orientation preserving, hence has a fixed point, hence is the identity. That is, any two nontrivial elements product to $e$. It's easy to show that this implies the Deck group is $\mathbb{Z}/2\mathbb{Z}$ (or trivial).

In fact $\mathbb{C}P^{2k+1}$ does double cover something (though, to my knowledge, it doesn't have a more common name, except in the case of $\mathbb{C}P^1 = S^2$ double covering $\mathbb{R}P^2$). In homogeneous coordinates, the involution maps $[z_0:z_1:...:z_{2k+1}:z_{2k+2}]$ to $[-z_1:z_0:...:-z_{2k+2}:z_{2k+1}]$. This involution acts freely, and the quotient of $\mathbb{C}P^{2k+1}$ by the involution is a space which $\mathbb{C}P^{2k+1}$ double covers.

I do not know if $\mathbb{C}P^{2k+1}$ covers anything else.

Incidentally, just to preempt a bit, the space $\mathbb{H}P^{n}$ doesn't cover anything unless $n=1$. The proof is much more complicated in general (though the case where $n$ is even follows precisely as it did in the $\mathbb{C}P^{2k}$ case).

In general, one needs to compute Pontrjagin classes and note that they are preserved by diffeomorphisms.

We have $p_1(\mathbb{H}P^n) = 2(n-1)x$ where $x$ is a particular choice of generator for $H^4(\mathbb{H}P^n)$. Since any diffeomorphism must preserve $p_1$, it follows that so long as $n\neq 1$, we must have $x\rightarrow x$ on $H^4$. Then, the Lefshetz theorem once again guarantees a fixed point.

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This is a really cool application of Lefschetz! As for the question of whether $\mathbb{C}P^{2k+1}$ double-covers anything else, is this just a question of whether there are any other, non-conjugate automorphisms in $Aut(\mathbb{C}P^{2k+1}$? –  Aaron Mazel-Gee Apr 3 '11 at 16:53
Well, not just automorphisms, but involutions with no fixed points, but yes. –  Jason DeVito Apr 3 '11 at 20:14
Slightly relatedly, I just learned today that the action of $\mathbb{Z}/2$ on $\mathbb{C}P^2$ given by $[z_0:z_1:z_2]\mapsto [\overline{z_0}: \overline{z_1} : \overline{z_2} ]$ has quotient homeomorphic to $S^4$. Of course this isn't exactly what we're talking about since the action isn't free, but I think it's pretty cool nonetheless. This was proved by Arnold. The professor who mentioned this didn't know offhand whether this was known to be diffeomorphic to $S^4$, though... –  Aaron Mazel-Gee Apr 6 '11 at 2:28
That's pretty interesting! Since it's not a free action, the quotient doesn't inherit a smooth structure from $\mathbb{C} P^2$, so I'm not sure what it means for this $S^4$ to be diffeomorphic to the standard one. –  Jason DeVito Apr 6 '11 at 3:11
Well, the quotient $S^4$ will certainly be an orbifold and hence smooth almost everywhere. Said another way, there is an open dense set of points in $\mathbb{C} P^2$ so that the action preserves these points and is free on them. The quotient will be an open dense subset of the $S^4$. I think the set of "bad" points is diffeomorphic to $\mathbb{R}P^2$ in both $\mathbb{C}P^2$ and the quotient. –  Jason DeVito Apr 6 '11 at 4:21

Euler characteristic is multiplicative, so (since $\chi(P^2)=3$ is a prime number) if $P^2\to X$ is a cover, $\chi(X)=1$ and $\pi_1(X)=\mathbb Z/3\mathbb Z$ (in particular, X is orientable). But in this case $H_1(X)$ is torsion, so (using Poincare duality) $\chi(X)=1+\dim H_2(X)+1>1$.

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I like this argument, but it may be worth pointing out that your'e also implicitly using the fact that $\pi_1(X) = \mathbb{Z}/3\mathbb{Z}$ implies $X$ is orientable. –  Jason DeVito Apr 1 '11 at 19:24
this is a nice answer. If I am understand it properly, to apply PD we need to know that $X$ is orientable. For this, you could e.g. observe that a nonorientable manifold has a connected degree 2 "orientation cover" and thus an index 2 subgroup of its fundamental group, which obviously $X$ does not. –  Pete L. Clark Apr 1 '11 at 19:25
@Jason, @Pete Oh, you're right, of course. –  Grigory M Apr 1 '11 at 19:28
Or do it $\mathrm{mod }2$, no? –  Mariano Suárez-Alvarez Apr 1 '11 at 19:34
:) This is the proof I had in mind. I like that it draws on a bunch of pieces of basic algebraic topology. I myself used the fact that you need orientability for Poincare duality, I like Mariano's idea a lot too. –  Aaron Mazel-Gee Apr 2 '11 at 7:06

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