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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?

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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

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  • $\begingroup$ thank you but this is not my question. My question was the following: $\endgroup$
    – Ali Taghavi
    Commented Oct 31, 2013 at 19:06
  • $\begingroup$ My question was the following:Let E be a complement to the canonical k plane bundle is it true to say that E has no k dimensional sub bundle $\endgroup$
    – Ali Taghavi
    Commented Oct 31, 2013 at 19:07
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    $\begingroup$ Dear Ali Taghavi, You just need to understand that by the canonical (also called universal) bundles Bob Stong meant both the k-plane bundle and its complementary bundle. Best regards, Július Korbaš $\endgroup$
    – Julius Korbas
    Commented Oct 31, 2013 at 22:06
  • $\begingroup$ Thank you very much for your answer and your reference to Stong paper. $\endgroup$
    – Ali Taghavi
    Commented Nov 2, 2013 at 7:31
  • $\begingroup$ Dear Julius Korbas, in the Stong paper, is there a restriction on k and n, when he state his result oabout G(k,n)? If there is no restriction, it seems that some thing is missing. Because a similar argument as above proof of fixed point property of even dimensional projective space, can be repeated to give a proof for fixed point property for odd dimensional projective space. But it is well known that CP^n has no FPP when n is odd $\endgroup$
    – Ali Taghavi
    Commented Nov 4, 2013 at 10:04

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