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In Loring Tu's Introduction to Manifolds, the author states:

"The tangent space of $S_n$ [the space of symmetric $n\times n$ matrices] at any point is canonically isomorphic to $S_n$ itself, because $S_n$ is a vector space."

There are several questions on Math stackexchange, but in all of those that I've seen, the answers are either incomplete, leaving aside details that shouldn't - specially for someone who is learning this by themselves - or they are too terse and confusing. Also, they usually assume that we can differentiate the determinant. When Tu writes the sentence above, he still hasn't talked about what's the differential of a determinant.

So, I'm looking for a complete proof, with an effort from the author to be as clear as possible, and without using the differential of the determinant. This last condition is not binding.



marked as duplicate by José Carlos Santos, Hans Engler, Yanior Weg, Leucippus, Lord Shark the Unknown May 3 at 4:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ What does determinant have to do with anything in your question? $\endgroup$ – Ted Shifrin Apr 30 at 18:45
  • $\begingroup$ @TedShifrin Check this question and some of the links therein : math.stackexchange.com/questions/2231893/… $\endgroup$ – An old man in the sea. Apr 30 at 21:41
  • $\begingroup$ Well, $SL(n)$ is a very different question. Unlike the space of symmetric spaces, it is not a vector space unto itself, and it is defined by an equation $\det(A)=1$. So you are mixing apples and oranges. $\endgroup$ – Ted Shifrin Apr 30 at 21:54
  • $\begingroup$ @TedShifrin Ups... you're right... =D $\endgroup$ – An old man in the sea. Apr 30 at 22:02
  • $\begingroup$ It the tangent space to a vector space were not isomorphic to the vector space, that could only mean that we'd gotten the definition of :"Tangent space" wrong. The tangent space represents all the different directions (and speeds) you can leave the point. In a vector space, the vectors of that space are exactly how you can leave points in it. $\endgroup$ – Paul Sinclair May 1 at 2:12