Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
1  
@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
show 2 more comments

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.

share|improve this answer
1  
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
2  
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
1  
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
1  
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
show 2 more comments

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) $\chi(X)=1+\dim H_2(X)+1>1$.

share|improve this answer
1  
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
1  
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
6  
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
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.