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.
One of the most important examples of a Lie group we’ve already seen: the general linear group of a finite dimensional vector space . Of course for the vector space this is the same as — or at least isomorphic to — the group of all invertible real matrices, so that’s a Lie group we can really get our hands on. And if has dimension , then , and thus .
So, how do we know that it’s a Lie group? Well, obviously it’s a group, but what about the topology? The matrix group sits inside the algebra of all matrices, which is an -dimensional vector space. Even better, it’s an open subset, which we can see by considering the (continuous) map . Since is the preimage of — which is an open subset of — is an open subset of .
So we can conclude that is an open submanifold of , which comes equipped with the standard differentiable structure on . Matrix multiplication is clearly smooth, since we can write each component of a product matrix as a (quadratic) polynomial in the entries of and . As for inversion, Cramer’s rule expresses the entries of the inverse matrix as the quotient of a (degree ) polynomial in the entries of and the determinant of . So long as is invertible these are two nonzero smooth functions, and thus their quotient is smooth at .