Finding the Tangent Space of SL(n,real) with A(0) being the Identity Matrix

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SUMMARY

The discussion focuses on finding the tangent space of SL(n, real) with A(0) as the identity matrix. For n=2, the tangent space is represented by the matrix A = \(\begin{pmatrix} a & b \\ c & -a \end{pmatrix}\), where a, b, and c are real numbers. The user conjectured that the tangent space for general n consists of traceless matrices but faced challenges in proving the converse. Ultimately, the user resolved the issue independently, indicating a successful conclusion to their inquiry.

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  • Understanding of Lie groups, specifically SL(n, real)
  • Familiarity with matrix calculus and determinants
  • Knowledge of traceless matrices and their properties
  • Experience with differential geometry concepts
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  • Study the properties of traceless matrices in the context of Lie algebras
  • Explore the relationship between determinants and traces in matrix theory
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Diophantus
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Hi,

I am trying to find the tangent space of SL(n,real) where A(0) is defined to be the identity matrix.

First of all I worked on the case when n=2 and found that the tangent space was

[tex]A = \left( \begin{array}{ccc}<br /> a & b \\<br /> c & -a<br /> \end{array} \right)[/tex]

where a,b,c belong to the reals,

so I made the conjecture that for n in general, the tangent space would be the space of traceless matrices.

I attempted to prove this by showing that the tangent space and the space of traceless matrices were subsets of each other. Whilst I could show that an arbitary element of the tangent space is traceless, I could not show the converse.

Do I just need to try harder or is my conjecture just plain wrong?

PS. I used the standard result: d/dt (detA(0)) = tr(dA(0)/dt)

I have reason to believe that det(exp(A)) = exp(tr(A)) may also be important but have not found a way of using this yet.
 
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Done it

Not to worry, I have solved it by myself. It must have just been too hard for you Americans.:smile:
 

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