The Lie Algebra of a General Linear Group
Since is an open submanifold of , the tangent space of at any matrix is the same as the tangent space to at . And since is (isomorphic to) a Euclidean space, we can identify with using the canonical isomorphism . In particular, we can identify it with the tangent space at the identity matrix , and thus with the Lie algebra of :
But this only covers the vector space structures. Since is an associative algebra it automatically has a bracket: the commutator. Is this the same as the bracket on under this vector space isomorphism? Indeed it is.
To see this, let be a matrix in and assign . This specifies the value of the vector field at the identity in . We extend this to a left-invariant vector field by setting
where we subtly slip from left-translation by within to left-translation within the larger manifold . We do the same thing to go from another matrix to another left-invariant vector field .
Now that we have our hands on two left-invariant vector fields and coming from two matrices and . We will calculate the Lie bracket — we know that it must be left-invariant — and verify that its value at indeed corresponds to the commutator .
Let be the function sending an matrix to its entry. We hit it with one of our vector fields:
That is, , where is right-translation by . To apply the vector to this function, we must take its derivative at in the direction of . If we consider the curve through defined by we find that
Similarly, we find that . And thus
Of course, for any we have the decomposition
Therefore, since we’ve calculated we know these two vectors have all the same components, and thus are the same vector. And so we conclude that the Lie bracket on agrees with the commutator on , and thus that these two are isomorphic as Lie algebras.