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For $k<n$ put $FM_{k\times n}$ for the space of all $k\times n$ full rank matrices with real or complex entries. Note that the permutation group $S_{n}$ has an obvious action on this space (with permuting the columns).

On this space, consider the equivalence relation $A\sim B$ iff they have the same row spaces. Then the Grassmanian $G(k,n)$ is identified with the quotient $FM_{k\times n}/\sim$.

Does every homeomorphism of Grassmanian has a lift to a homeomorphism of $FM_{k\times n}$.

And is it true that this lifted homeomorphism is homotopic or isotopic to a homeomorphism on $FM_{k\times n}$ arising from a permutation of columns?

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    $\begingroup$ For the second question, the answer is no. Consider self-homeomorphism of $G(1,2) \cong S^2$ sending $[z:w]$ to $[\overline z:\overline w]$. This is an orientation-reversing homeomorphism (degree $-1$), and so it does not come from swapping the columns (a complex-analytic homeomorphism, degree $1$). $\endgroup$
    – Tyler Lawson
    Commented Nov 12, 2014 at 3:12
  • $\begingroup$ @TylerLawson Thanks for your answer to the second question. It arise a new question: is there a permutation which is not homotopic to the identity (as a map on grassmanian)? $\endgroup$
    – Ali Taghavi
    Commented Nov 12, 2014 at 18:55

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