We want to show that for a given $n\in\mathbb{N}$, the set of all the $\infty\times n$ matrices $Mat_{\infty,n}(\mathbb{C})$ of max rank $n$ form a contractible space.

This set is obtained by taking the limit on $k \geq n$ in $Mat_{k,n}(\mathbb{C})$ with max rank $n$, being all $Mat_{j,n}(\mathbb{C}) \subset Mat_{j+1,n}(\mathbb{C})$ by adding a new raw of zeros. Our space is endowed with the Zarisky topology, given by induction on the Zarisky topology of the $Mat_{k,n}(\mathbb{C})$'s .

I don't think this is a CW-complex so I don't know which strategies to use. If we remove the $\infty$ the result is no more true, really similar to $S^n$ and $S^{\infty}$.


  • $\begingroup$ How is the Zariski topology defined? $\endgroup$
    – san
    Aug 11, 2018 at 17:48
  • 1
    $\begingroup$ I'm not sure why you say the result isn't true if you change $\infty$ to some finite $k$. It wouldn't be true for the Euclidean topology but the Zariski topology is very different... $\endgroup$ Aug 11, 2018 at 21:00
  • 1
    $\begingroup$ In particular, my answer at math.stackexchange.com/a/2294516/86856 actually shows that $Mat_{k,n}(\mathbb{C})$ is contractible in the Zariski topology for any $k\geq n$. I haven't worked out the details yet but I think it should be easy to extend the argument to the infinite case. $\endgroup$ Aug 11, 2018 at 21:07
  • 2
    $\begingroup$ Are you sure you want the Zariski topology? Pretty much no one ever does homotopy theory using the Zariski topology. $\endgroup$ Aug 12, 2018 at 1:05
  • 1
    $\begingroup$ In particular, if you actually seriously want to understand vector bundles or something like that, you would not be using the Zariski topology. The question you have written has nothing to do with the geometry of vector bundles and instead is an exercise in pathological pointset topology. $\endgroup$ Aug 12, 2018 at 1:08


You must log in to answer this question.