# A question on fixed point theory

I asked this question in MSE, but I did not received any answer, so I repeat it here:

https://math.stackexchange.com/questions/858238/a-question-on-fixed-point-property

Assume that $0<k<n-1$, Note that $\mathbb{C}P^{k}$ can be considered as a closed subset of $\mathbb{C}P^{n}$, in a natural way. We collapse $\mathbb{C}P^{k}$ to a point. The resulting space is denoted by $\mathbb{C}P^{n}/\mathbb{C}P^{k}$

My fixed point question:

Does $\mathbb{C}P^{n}/\mathbb{C}P^{k}$ satisfies fixed point property?(At least when $n$ is even)

This question is motivated by:

https://math.stackexchange.com/questions/845057/show-mathbbcp2-cp1-is-not-a-retract-of-mathbbcp4-cp1#comment1754879_845057

• An slight variation of the usual cohomological argument should give an easy positive answer when $n\gg k$ and $k$ is odd or $n$ is even. For instance, for $k=1$ and $n\geq 6$ every endomorphism of the cohomology ring of $\mathbb{C}P^n/\mathbb{C}P^1$ extends to the cohomology ring of $\mathbb{C}P^n$, and so the fixed point property follows by an easy computation. – Eric Wofsey Jul 14 '14 at 7:37
• @EricWofsey: could you elaborate on your answer? – Michael Jul 18 '14 at 0:32
• Consider, for instance, $\mathbb{C}P^6/\mathbb{C}P^1$. Its cohomology is the subring of $\mathbb{Z}[x]/x^7$ generated by $s=x^2$ and $t=x^3$. We have a relation $s^3=t^2$, and this forces any endomorphism of the ring to be of the form $s\mapsto d^2s$, $t\mapsto d^3 t$ for some $d\in\mathbb{Z}$. Since $1+d^2+d^3+\dots+d^6$ can never be zero, every map has a fixed point by the Lefschetz fixed point theorem. – Eric Wofsey Jul 18 '14 at 4:08
• More generally, $n$ needs to be large enough compared to $k$ to ensure the existence of an integer $d$ as in the argument above (I haven't checked carefully, but I think $n\geq (k+1)(2k+1)$ should suffice). The condition that either $k$ is odd or $n$ is even is just so that it is impossible for $1+d^{k+1}+\dots+d^n$ to be zero. – Eric Wofsey Jul 18 '14 at 4:09

Here is a partial affirmative answer using mod 2 Steenrod operations; the simplest case of this (for $n$ and $k$ even) is just a correction of the slightly incorrect answer originally posted by Włodzimierz Holsztyński. The result is that if $k+1$ and $n+1$ are both odd multiples of $2^d$ for some integer $d\geq 0$, then $\mathbb{C}P^n/\mathbb{C}P^k$ has the fixed point property. In particular, for $d=0$ we get the fixed point property whenever $n$ and $k$ are both even. All cohomology in this answer will have coefficients in $\mathbb{F}_2$.

Let's start by describing the action of the Steenrod squares on the cohomology $H^*(\mathbb{C}P^n)=\mathbb{F}_2[x]/(x^{n+1})$. The following formulas are easy to prove by induction using the Cartan formula (induct on $d$ and for fixed $d$ induct on $m$):

$$Sq^{2^{d+1}}\left(x^{2^dm}\right)=x^{2^d(m+1)} \text{ if m is odd}$$ $$Sq^{2^{d+1}}\left(x^{2^dm}\right)=0 \text{ if m is even}$$

From these, we deduce the following for all $0\leq \ell<2^d$: $$Sq^{2^{d+1}}\left(x^{2^dm+\ell}\right)=x^{2^d(m+1)+\ell} \text{ if m is odd}$$ $$Sq^{2^{d+1}}\left(x^{2^dm+\ell}\right)=0 \text{ if m is even}$$

The quotient map $\mathbb{C}P^n\to\mathbb{C}P^n/\mathbb{C}P^k$ identifies $H^*(\mathbb{C}P^n/\mathbb{C}P^k)$ with the subring of $H^*(\mathbb{C}P^n)$ which has as a basis $\{1,x^{k+1},x^{k+2},\dots, x^n\}$, and so the same relations hold in $H^*(\mathbb{C}P^n/\mathbb{C}P^k)$.

Suppose now that $f:\mathbb{C}P^n/\mathbb{C}P^k\to\mathbb{C}P^n/\mathbb{C}P^k$ is any map. For $k<i\leq n$, let $a_i\in \mathbb{F}_2$ be such that $f^*(x^i)=a_ix^i$. By the Lefschetz fixed point theorem, $f$ must have a fixed point if $1+\sum_{k+1}^n a_i\neq 0$ (the $1$ coming from $H^0$), or equivalently if $\sum a_i=0$.

Since $f^*$ must commute with Steenrod operations, we must have $a_{2^dm+\ell}=a_{2^d(m+1)+\ell}$ for $m$ odd and $0\leq \ell<2^d$, as long as $k<2^dm+\ell<2^d(m+1)+\ell\leq n$. Together, these relations imply that if $m$ is odd and $k<2^dm<2^d(m+2)-1\leq n$, then all the $a_i$ for $2^dm\leq i \leq 2^d(m+2)-1$ are equal to each other (everything below $2^d(m+1)$ can be related to $2^d(m+1)$ using the smaller Steenrod squares, and everything above $2^d(m+1)$ can be related to something below it using $Sq^{2^{d+1}}$). That is, the $a_i$ are constant in blocks of length $2^{d+1}$ starting from an odd multiple of $2^d$.

Now suppose that $k+1$ and $n+1$ are both odd multiples of $2^d$. The numbers from $k+1$ to $n$ can be broken into blocks of length $2^{d+1}$, each starting with an odd multiple of $2^d$. All of the $a_i$ in each block are equal to each other, and hence their sum is zero since there are an even number of them. Thus the sum of all of the $a_i$ is zero, and so $f$ must have a fixed point.

Let me conclude with a couple remarks on this result. First, as Włodzimierz observed, this argument works equally well for projective spaces over $\mathbb{R}$ or $\mathbb{H}$ (for $\mathbb{R}$, replace $Sq^{2^{d+1}}$ with $Sq^{2^d}$ and for $\mathbb{H}$ replace it with $Sq^{2^{d+2}}$). Second, the condition obtained here is sufficient but not necessary for $\mathbb{C}P^n/\mathbb{C}P^k$ to have the fixed point property. Indeed, in the comments I sketched an argument using cup products and integer coefficients rather than Steenrod squares and mod 2 coefficients which shows that the fixed point property holds when $n\gg k$ as long as either $n$ is even or $k$ is odd (note that in fact, using only mod 2 coefficients there is no hope of proving the fixed point property in cases where $n$ and $k$ have different parity).

• Actually, the hypothesis that $k+1$ and $n+1$ are divisible by $2$ the same number of times is far stronger than what is needed for this argument to work, though the actual necessary and sufficient condition seems a lot more complicated to state. For instance, if I'm not mistaken, if $k+1=2^d$ for some $d>0$ then $n+1$ can be any even number greater than $2^{d+1}$ that is not a power of $2$. – Eric Wofsey Dec 16 '14 at 4:02
• Eric, you're much too kind to me. My answer (which I have just removed) was not slightly but grossly incorrect. I am very glad for your meaningful post. – Włodzimierz Holsztyński Dec 16 '14 at 7:25
• @EricWofsey thank you so much for your very interesting question. – Ali Taghavi Dec 16 '14 at 15:00
• @WłodzimierzHolsztyński thank you very much for your communication in my question. – Ali Taghavi Dec 16 '14 at 15:01
• In case anyone cares, I believe I've worked out that the necessary and sufficient conditions for this argument to work when $k$ is odd are: (1) $n$ is odd, (2) $n>k+2^d$ if $2^d$ is the least power of $2$ dividing $k+1$, and (3) $n$ is not the least integer greater than $k$ which is $1$ less than a multiple of $2^d$ for any $d$. The proof is a messy but straightforward induction on $n$ with $k$ fixed. – Eric Wofsey Mar 11 '15 at 23:04

I don't think $\mathbb{C}P^n/\mathbb{C}P^k$ (or $\mathbb{R}P^n/\mathbb{R}P^k$) ever has the fixed point property in the range you describe. I haven't thought about this very long so could be wrong, but on first look I think you can construct an endomorphism with no fixed points in the following way:

Consider $\mathbb{C}P^n$ as the space of 1-dimensional complex subspaces of $\mathbb{C}^{n+1}$, and $\mathbb{C}P^k$ as 1-dim subspaces in some $k+1$ dimensional subspace. Now every 1-dim subspace in $\mathbb{C}^{n+1}$ has $n$ naturally associated other subspaces, namely those orthogonal to it. So to construct a continuous endomorphism, we could try to find a continuous choice of orthogonal subspace, and in order for it to pass to an endomorphism of the quotient, we just need to make sure that all the subspaces contained in $\mathbb{C}P^{k+1}$ are sent to a common subspace. Since $k<n$, there is at least one choice of 1-dim subspace which is orthogonal to all of $\mathbb{C}^{k+1}$, so pick one and call it $V$. Now given any 1-dim subspace $U$ which is not in $\mathbb{C}^{k+1}$, $U$ has an orthogonal projection $\bar{U}$ onto $\mathbb{C}^{k+1}$, and so we can apply the unique unitary operator which fixes the orthogonal complement $W$ of $U+\bar{U}$ ($W$ is codimension 2) and sends $\bar{U}$ to $U$. Under this map, $V$ is sent to a subspace orthogonal to $U$. So we have assigned to every 1-dim subspace of $\mathbb{C}^{n+1}$ an orthogonal one, in a manner which is clearly continuous on passage to $\mathbb{C}P^n$, and so gives a map which by construction passes to the quotient $\mathbb{C}P^n/\mathbb{C}P^k$ and has no fixed points.

• I don't entirely follow how you're defining your map: what does the map send $U$ to? What if $U$ is orthogonal to $\mathbb{C}^{k+1}$? How is the unitary you're specifying unique? In any case, it doesn't appear that your argument ever uses that $k>0$. – Eric Wofsey Jul 18 '14 at 4:19
• @user56137 in line 7 of your answer you wrote "continuous choice". When $n$ is even, such choice is impossible. the reason is identical to the argument in the following: sciencedirect.com/science/article/pii/S0723086914000036 – Ali Taghavi Jul 18 '14 at 13:42