Fixed point property for the total space of the canonical line bundle over $\mathbb{C}P^{2n}$ 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? 
 A: 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.)
