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It is fairly easy to see that the set of non invertible matrices is path connected. Are they simply connected? If not what is their fundamental group? What are their homotopy and homology groups. I'm looking for the answer to any of these questions. Any examples for particular (non 0 or 1) dimensions are welcome as well, as well as for real or complex coefficients. (Sorry about the phrasing, it is currently 3:30 am and I will edit the question in the morning)

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    $\begingroup$ Matrices over what field or ring? And of what dimension? $\endgroup$ May 2, 2019 at 7:37
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    $\begingroup$ With real or complex coefficients, and for any dimension you can show. $\endgroup$
    – user565844
    May 2, 2019 at 7:40
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    $\begingroup$ I'm mostly curious to know if they're simply connected, as my intuition is that the set looks like a bunch of lower dimensional vector spaces glued together at the origin $\endgroup$
    – user565844
    May 2, 2019 at 7:42

1 Answer 1

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The space of non-invertible matrices (with real or complex coefficients) is contractible. There is an explicit homotopy between the identity map and the constant map equal to the zero matrix, simply given by: $$H(A,t) = tA.$$ In particular it is simply connected, and all its higher homotopy and homology groups vanish.

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    $\begingroup$ I didn't realize the space was star convex, I guess posting questions at 3am isn't the best idea. $\endgroup$
    – user565844
    May 2, 2019 at 7:45
  • $\begingroup$ @Liquid No big deal. When you learn all these sophisticated tools, the first impulse becomes to try using them, even though the answer is sometimes much simpler... $\endgroup$ May 2, 2019 at 7:46
  • $\begingroup$ You are meaning the $0$ matrix and not the identity if I follow you well? $\endgroup$ May 2, 2019 at 7:47
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    $\begingroup$ @mathcounterexamples.net I mean that the identity $H(A,1) = A$ is homotopic to the constant map equal to the zero matrix $H(A,0) = 0$. $\endgroup$ May 2, 2019 at 7:48
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    $\begingroup$ It appears that we can conclude that either mathcounterexamples.net is not familiar with homotopy and contractibility, or your assertion is false. $\endgroup$ May 2, 2019 at 16:04

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