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It is well known that the even dimensional complex projective pace $\mathbb{C}P^{2n}$ has the fixed point property.

What about the total space of the canonical line bundle over $\mathbb{C}P^{2n}$? Does it have the fixed point property?

Ali Taghavi
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    @PaulSinclair: I don't understand this comment. Let's do that with the tautological bundle over $\Bbb{CP}^1$. Then it should send the zero section, $\sigma_0$, to another section that does not intersect the zero section. But this is impossible, because the tautological bundle has no nonvanishing sections. –  Sep 05 '15 at 21:02
  • @MikeMiller - true. I was too cavalier in my comment. – Paul Sinclair Sep 05 '15 at 21:24

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No. To reduce notation, I'll put $m=2n$. Let $(z_0:z_1:\cdots:z_m)$ be homogenous coordinates on $\mathbb{CP}^m$. Let $q_0$, $q_1$, ..., $q_m$ be distinct nonzero scalars. Let $f: \mathbb{CP}^m \to \mathbb{CP}^m$ be the map $(z_0:\cdots :z_m) \mapsto (q_0 z_0: \cdots: q_m z_m)$. The map $f$ fixes $m+1$ points, let's call them $p_0$, ..., $p_m$.

Let $X$ be the total space of the bundle you are interested in, and $\pi: X \to \mathbb{CP}^m$ the projection map. Let $f^{\ast}$ be the induced map $X \to X$. So $f^{\ast}$ acts on the fiber $\pi^{-1}(p_i)$ by some nonzero scalar $a_i$. (Explicitly, $a_i = \prod_{j \neq i} (q_j/q_i)$, or possibly the reciprocal, depending on your conventions.) Let $\alpha: \mathbb{CP}^m \to \mathbb{C}^{\ast}$ be any continuous map with $\alpha(p_i) = a_i$. (To see that such a map exists, take an arbitrary logarithm $\log a_i$ of each $a_i$, choose a bump function $\phi_i$ around each $p_i$ with $\phi_i(p_i)=1$, and put $\alpha(z) = \exp(\sum (\log a_i) \phi_i(z))$.) Let $\sigma: \mathbb{CP}^m \to X$ be any smooth section which is nonvanishing at the $p_i$.

Let $g: X \to X$ be the map which takes each fiber $\pi^{-1}(z)$ to itself, acting on $\pi^{-1}(z)$ by $v \mapsto \alpha(z)^{-1} v + \sigma(z)$.

Now, I claim that $g \circ f^{\ast}$ has no fixed points. If $g(f^{\ast}(z))=z$, then $\pi(g(f^{\ast}(z)) = \pi(z)$ so $f(\pi(z)) = \pi(z)$ and we deduce that $\pi(z)$ is one of the $p_i$. But $g \circ f^{\ast}$ acts on the fiber $\pi^{-1}(p_i)$ by the translation $v \mapsto v+\sigma(z)$, which has no fixed points.

After writing this solution, I just remembered that a connected non-compact manifold always has a nonvanishing vector field. So a shorter solution is just to flow along that vector field. (There are some details to fill in here.)

David E Speyer
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  • @DavidSpayer Thank you for your very interesting answer. What is a proof of existence of a non vanishing vector field. However we should be carefull about periodic orbits(which are fixed point of the flow). This is a motivation to ask: Is there always a non vanishing gradient vector field on a non compact manifold? A gradient vector field does not have a periodic solution. – Ali Taghavi Sep 10 '15 at 18:09
  • @AliTaghavi Regarding the question about gradient vector fields: I think so, but I don't know for sure. That's a good question. The periodic flows were the "details to fill in" that I referred to. See http://math.stackexchange.com/a/48583/448 for related discussion, but that question doesn't exactly address this set up. – David E Speyer Sep 10 '15 at 18:13
  • PS: To the anonymous editor who wanted $f^{\ast}$ to act by the identity on the fibers $\pi^{-1}(z)$: If $z$ isn't a fixed point, then this doesn't make sense. If $z$ is a fixed point we can arrange that this is true, which is what the trick with $\alpha$ is for, but we need some other action to start with. That action is the natural action of $f^{\ast}$ on the canonical bundle. – David E Speyer Sep 10 '15 at 18:14
  • Assume that $E$ is a bundle over $Y$ and $f:Y \to Y$ is a map which fixes a point $p$. Accordinig the the definition of pull back, is not true that $f^{*}$ restrict to $E_{p}$ as identity map?.(it is the definition of pull back) – Ali Taghavi Sep 10 '15 at 18:14
  • Why $\alpha(z)^{-1}$ is necessary, in your argument? – Ali Taghavi Sep 10 '15 at 18:17
  • (Re first question) Either I'm missing something or you are. I don't think there is any functorial map $f^{\ast} : E \to E$ in that level of generality. Would you write out what you think the definition is? I am using the functorial property of $f^{\ast}$ that comes from the fact that $E$ is the top wedge of the cotangent bundle. – David E Speyer Sep 10 '15 at 18:18
  • Re second question: Because the linear map $v \mapsto av+\sigma$ will have a fixed point at $\sigma/(1-a)$ if $a$ is not $1$. – David E Speyer Sep 10 '15 at 18:19
  • I think I am missing some thing. It was my fault that I was thinking about $f^{}:f^{}E \to E$. So I need to understand what is $f^{*}$ in your main answer? – Ali Taghavi Sep 10 '15 at 18:50
  • The canonical bundle is $\bigwedge^n \Omega^1$. A map $f$ induces a map on $\Omega^1$ by pulling back covectors, and then on $\bigwedge^n \Omega^1$ by functoriality of wedge. I don't want to get to hung up on this, though. All I need is some way to lift $f$ to some $f^{\ast}$, so that the action on $\pi^{-1}(p_i)$ is multiplication by a nonzero scalar. – David E Speyer Sep 10 '15 at 19:15
  • For example, if you were working with the tautological bundle instead of the canonical bundle, I would just use the map $(z_0, \ldots, z_m) \mapsto (q_0 z_0, \ldots, q_m z_m)$ from $\mathbb{C}^{m+1}$ to itself, and then embed $E$ into $\mathbb{CP}^m \times \mathbb{C}^{m+1}$. – David E Speyer Sep 10 '15 at 19:16
  • What is the reason that a non compact manifold admit a non vanishing vector field? then How can we use the flow map to prove non fix point property?(what we should do with closed orbit, and control the period at infinity)? It seems that we should not expect to have a non vanishing gradient field:http://math.stackexchange.com/questions/1431087/does-every-open-manifold-admit-a-function-without-critical-point – Ali Taghavi Sep 11 '15 at 17:01
  • I apologize for my previous comment. Now I realiz that there is a non vanishing gradient vector field. – Ali Taghavi Sep 11 '15 at 20:58