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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 $S^{\infty}$.

Thanks!

Maffred
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  • How is the Zariski topology defined? – san Aug 11 '18 at 17:48
  • @san I restated the whole text! Thanks! – Maffred Aug 11 '18 at 18:09
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    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... – Eric Wofsey Aug 11 '18 at 21:00
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    In particular, my answer at https://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. – Eric Wofsey Aug 11 '18 at 21:07
  • Can you be precise about exactly what your space $Mat_{\infty,n}(\mathbb{C})$ is, though? When you say $\infty\times n$ matrices, do you mean that all but finitely many entries in the matrices must be $0$? And you are saying that a subset $U\subseteq Mat_{\infty,n}(\mathbb{C})$ is open iff $U\cap Mat_{k,n}(\mathbb{C})$ is open in the Zariski topology on $Mat_{k,n}(\mathbb{C})$ for all finite $k$? – Eric Wofsey Aug 11 '18 at 21:50
  • @EricWofsey We are trying to construct the universal principle (equivariant) bundle for $GL_n(\mathbb C)$. This is the case iff $E \to B$ is a principal equivariant bundle for $GL_n$ and $E$ is contractible. Since $Mat_{k,n}^{max} \to Gr(n,k)$ is a principle $GL_n$-equivariant bundle (Gr is the Grassmaniann, everything with the Zarisky topology). I don't think $Mat_{k,n}^{max}$ is contractible or the exercise would be useless. Now take the limit of the inclusions $Mat_{k,n}^{max} \to Mat_{k+1,n}^{max}$ and $Gr(n,k) \to Gr(n,k+1)$ and their bundle maps – Maffred Aug 12 '18 at 01:01
  • @EricWofsey , then $Mat_{\infty,n}^{max} \to Gr(n, \infty)$ is universal principle equivariant bundle cause $Mat_{\infty,n}^{max}$ is contractible. – Maffred Aug 12 '18 at 01:01
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    Are you sure you want the Zariski topology? Pretty much no one ever does homotopy theory using the Zariski topology. – Eric Wofsey Aug 12 '18 at 01:05
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    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. – Eric Wofsey Aug 12 '18 at 01:08
  • @EricWofsey I would say that these matrixes have finitely many not-0 entries, and all of them have max rank. I don't know how to describe open sets. I guess there is Zarisky topology cause $Gr(n,k)$ needs it. – Maffred Aug 12 '18 at 01:09
  • @EricWofsey here next to the definition of $Mat_{k,n}^{max}$ it is explicitally stated "with Zarisky topology". I don't know if we mean that $Mat_{\infty,n}^{max}$ is contractible with the Euclidean topology, if it makes sense. – Maffred Aug 12 '18 at 01:17

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