This may seem like an easy question, but my differential geometry is a little rusty. I'm trying to find the tangent space to the Lie group [itex] U(n) [/itex]; that is, for an arbitrary [itex] X \in U(n) [/itex] I'm trying to find an expression for [itex] T_X U(n) [/itex].(adsbygoogle = window.adsbygoogle || []).push({});

I can't quite remember how to do this. I've been playing around with the idea that if we define a function

[tex] F(X) = X^\dagger X - I [/tex]

where [itex] \dagger [/itex] is the conjugate transpose, then [itex] U(n) = F^{-1}(0) [/itex]. I think [itex] F: M_n(\mathbb C) \to Sp(n) [/itex] where the domain is nxn matrices and the codomain is the symplectic group, though I don't think this is too important.

So if [itex] U(n) = F^{-1}(0) [/itex], can I compute the tangent space as

[tex] T_X U(n) = DF^{-1}(0)[X] = \left\{ Z \in M_n(\mathbb C) : X^\dagger Z + Z^\dagger X = 0 \right\} [/tex]

This feels like it would be a suitable candidate, since if X is the identity matrix, this reduces to

[tex] T_I U(n) = \left\{ Z \in M_n(\mathbb C) : Z + Z^\dagger = 0 \right\} = \fraktur{u}(n) [/itex]

the corresponding Lie Algebra of traceless skew-Hermitian matrices.

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# Tangent Space to Unitary Group

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