Let $f$ be a continuous function on complex Grassmannian $G(k, 2n+1)$. Is it true to say that there is a $k$-plane $Y$ such that $Y$ has nontrivial intersection with $f(Y)$?


A motivation for this question is the following alternative proof for fixed point
property of $CP^{2n}$:


Assume that $f$ is a map on $CP^{2n}$ without fixed point. Let $l$ be the canonical line bundle on $CP^{2n}$. By $f^\*(l)$ we mean the pull back bundle. Then $l$ has trivial intersection with $f^\*(l)$, since $f$ has no fixed point. This implies that a complement of $l$, in the $2n+1$ trivial bundle, has a sub-bundle $f^\*(l)$. This is a contradiction because the Chern class of each complement of the canonical line bundle is $1-x+x^2-.....+x^{2n}$, which does not have a rational root$.


So our main question has affirmative answer if the answer to the following question 
is affirmative:



> 
> Is it true to say a that a complement of canonical $k$ plane bundle on $G(K, 2N+1)$ in the trivial $2n+1$ bundle does not have a $k$-dimensional sub-bundle?
> 
> 
> 



A weak fixed point property for Grassmannian
As proved by Bob Stong [Robert E. Stong, Splitting the universal bundles over Grassmannians, Algebraic and Differential Topology - Global Differential Geometry, Occas. 90th Anniv. M. Morse’s Birth, Teubner-Texte Math. 70, 275-287 (1984)], over the complex Grassmann manifolds, the canonical bundles never contain proper real or complex subbundles. 


Best regards,
 Július Korbaš


