9
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Eric Wofsey Jul 14 '14 at 7:37
  • $\begingroup$ @EricWofsey: could you elaborate on your answer? $\endgroup$ – Michael Jul 18 '14 at 0:32
  • 2
    $\begingroup$ 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. $\endgroup$ – Eric Wofsey Jul 18 '14 at 4:08
  • 1
    $\begingroup$ 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. $\endgroup$ – Eric Wofsey Jul 18 '14 at 4:09
4
$\begingroup$

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).

$\endgroup$
  • 1
    $\begingroup$ 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 $2$. $\endgroup$ – Eric Wofsey Dec 16 '14 at 4:02
  • $\begingroup$ 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. $\endgroup$ – Włodzimierz Holsztyński Dec 16 '14 at 7:25
  • 1
    $\begingroup$ @EricWofsey thank you so much for your very interesting question. $\endgroup$ – Ali Taghavi Dec 16 '14 at 15:00
  • 1
    $\begingroup$ @WłodzimierzHolsztyński thank you very much for your communication in my question. $\endgroup$ – Ali Taghavi Dec 16 '14 at 15:01
  • $\begingroup$ 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. $\endgroup$ – Eric Wofsey Mar 11 '15 at 23:04
0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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 $k>0$. $\endgroup$ – Eric Wofsey Jul 18 '14 at 4:19
  • $\begingroup$ @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 $\endgroup$ – Ali Taghavi Jul 18 '14 at 13:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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