# Every continuous map $f:\mathbb{C}P(2) \to \mathbb{C}P(2)$ has a fixed point, without Lefschetz theorem.

I would like to know if there is a nice proof of the fact that every continuous map $$f:\mathbb{C}P(2) \to \mathbb{C}P(2)$$ has a fixed point, without use of the Lefschetz fixed point theorem.

• I don’t know a single example of a proof that a non-contractible space has the fixed point property that doesn’t use Lefschetz... sounds tricky! Oct 27 '20 at 18:21
• I believe this might work: note that having a fixed point is a closed property so if one is arbitrarily close we are done. Any map is arbitrarily close to a smooth map. Every smooth map from a manifold of dimension n or less into a 2n manifold is arbitrarily close to an immersion (I think). Under connect sum regular homotopy classes of immersions of spheres into your manifold form a group. For n this has a bilinear form given by the algebraic intersection. It has a quadratic refinement given by self intersection. This self intersection is calculated by perturbing an arbitrarily small amount. Oct 27 '20 at 18:24
• The algebraic self intersection number can be calculated via the self intersection number, so if the geometric number is 0 so is the algebraic. From this one can deduce that any map without a fixed point must be trivial on $H^2$ because of the cup product on $CP(2)$ and the generator being represented by a sphere. Not sure where to go from there but perhaps since you are essentially collapsing a cell you must have a fixed point in that case. Oct 27 '20 at 18:25

## 2 Answers

Through any two distinct points in $$\mathbb{C}P^n$$ there is a unique (complex) geodesic. Therefore from any fix point free self map $$f$$ , we have a 1 dimensional complex subbundle of the tangent bundle by taking the subspace above a point $$p$$ to be the tangent space of the geodesic from $$p$$ to $$f(p)$$ at $$p$$.

This implies that the total Chern class of $$\mathbb{C}P^n$$ has a linear factor. If $$n=2$$ this implies that the total Chern class $$1+3x+3x^2$$ has two real (integer) roots. However, this is easily checked to be false since the discriminant is negative.

As a sanity check, this should be different if $$n=3$$. In that case the total Chern class is $$1+4x +6x^2 +4x^3$$ and this equals $$(2 x + 1) (2 x^2 + 2 x + 1)$$, as expected.

I imagine if you are better with polynomials than I, you can get this to work for any even $$n$$.

• I like this answer because the original question struck me as a "do I need to use a hammer here?" question, which I think implies "can't I just use scissors?". And Connor's reply is "How about this tank?". Oct 28 '20 at 17:05
• I think it’s neat to point out that this doesn’t even use the Euler characteristic! Oct 28 '20 at 19:07
• @JonathanZsupportsMonicaC The top Chern class of a complex vector bundle coincides with the Euler class. The Euler class of the tangent bundle evaluated on the fundamental class gives you the Euler characteristic. So this means that the coefficient of the highest degree term in my polynomial should be the Euler characteristic. However, none of this is used when calculating the Chern classes of Complex projective space. Oct 28 '20 at 19:28
• @Connor: that first sentence seems key and I have no intuition for it at all so can you clarify what this complex geodesic is and why it's unique? The analogous statement for real geodesics is certainly false, e.g. for $n = 1$ we have $\mathbb{CP}^1 \cong S^2$ and between two antipodal points there are uncountably many real geodesics. Oct 28 '20 at 22:35
• Since all they need is a complex 1d subbundle, i guess you can replace complex geodesic to the complex line joining $p$ to $f(p)$ @qiaochu yuan Oct 28 '20 at 22:56

I don't understand the first paragraph of Connor's answer at all, but assuming it checks out, in general the total Chern class of $$\mathbb{CP}^n$$ is

$$(1 + \alpha)^{n+1} = \sum_{k=0}^n {n+1 \choose k} \alpha^k \in \mathbb{Z}[\alpha]/\alpha^{n+1}$$

(note that the $$k = n+1$$ term vanishes). If the (complex) tangent bundle of $$\mathbb{CP}^n$$ has a (complex) line subbundle then the total Chern class must factor as

$$(1 + c \alpha)(1 + c_1 \alpha + \dots + c_{n-1} \alpha^{n-1})$$

and since we never get a coefficient of $$\alpha^{n+1}$$ or higher, the problem of determining whether this is possible is equivalent to the problem of determining when $$(1 + \alpha)^{n+1} - \alpha^{n+1}$$ has a linear factor of the form $$(1 + c \alpha)$$, as an ordinary polynomial.

It will be easier to reverse the order of the coefficients: this is equivalent to determining when $$\frac{(x + 1)^{n+1} - 1}{x}$$ has a linear factor of the form $$(x + c)$$, where $$c$$ is an integer. This implies

$$(1-c)^{n+1} = 1$$

and if $$n$$ is even this gives $$c = 0$$, but $$x + c = x$$ is not a factor of the above polynomial because its constant coefficient is $$n+1$$ (this coefficient corresponds to the top Chern class and hence to the Euler characteristic $$\chi(\mathbb{CP}^n) = n+1$$ so we really are using that the Euler characteristic doesn't vanish). We conclude:

Claim: $$\mathbb{CP}^{2m}$$ has the fixed point property.

This is usually proven by Lefschetz (as far as I know anyway). If $$n$$ is odd then this gives either $$c = 0$$ or $$c = 2$$ so we don't get a contradiction, and I guess the existence of fixed-point-free maps in this case implies that the tangent bundle has a line subbundle with first Chern class $$2$$. I wonder if anyone knows a more explicit description of it.

• From the identification of $\mathbb{C}^{2n}$ with $\mathbb{H}^n$, one gets a map $\mathbb{C}P^{2n-1}\rightarrow \mathbb{H}P^{n-1}$. It turns out this map is a bundle with fiber $S^2$. The vectors tangent to the $S^2$ fibers give the complex line bundle of $T\mathbb{C}P^{2n-1}$ you mention at the end of your post. Oct 29 '20 at 3:42
• Another description of the same thing: seeing $\mathbb{C}^{2n}$ as $\mathbb{H}^n$ equips it with the antilinear automorphism given by multiplication by $j$ on the left. Applying this to the tautological bundle $\tau$ of $\mathbb{C}P^{2n-1}$ sitting inside the trivial rank $2n$ bundle, we get a subbundle of the orthogonal complement. Now $T\mathbb{C}P^{2n-1}$ is isomorphic to $Hom(\tau, \tau^{\perp})$, so this gives a line subbundle of the tangent. This bundle is isomorphic to $\tau^* \otimes \tau^*$ hence has first Chern number $2$, which is coherent with your computation. Oct 29 '20 at 3:58