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Identify $ \mathbb{R}^4$ with the space of $2×2$ matrices $M(2×2,\mathbb{R})$.

The set $M$ of matrices with determinant $3$ is a smooth manifold of dimension $3$.

Prove that the tangent space to M at I ( identity matrix) may be identified with the set of matrices with zero trace .

How can i show this , i think i need to show every matrices with zero trace can be seen as tangent vector but how ? In addition i can't see necessity of matrices with zero trace why only matrices with zero trace.

Hints will be better to start for me .

bytrz
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  • There is an error in your question. The manifold or rather Lie group should be $SL_n(\mathbb{R})$ and its lie algebra $sl_n(\mathbb{R})$ is the set of matrices with trace $0$ – Sandeep Thilakan Apr 05 '14 at 14:10
  • @SandeepThilakan i can't understand your comment, which part of my question is wrong could you explain it more ? – bytrz Apr 05 '14 at 14:30

1 Answers1

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What Sandeep is saying is that the statement, "The tangent space to M is the trace 0 matrices" is not correct. Rather than M, you need to look at a subspace of M, the set of 2 x 2 matrices with determinant 1. If you do that, parameterize a curve, and see what you can say about the determinant of a tangent vector to that curve.

csappenf
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  • i see so sorry it is my big mistake i am editing the question now thanks for your warning – bytrz Apr 05 '14 at 14:51