If we pick a basis of , then we have a matrix for the bilinear form
and one for the endomorphism
So the condition in terms of matrices in comes down to
or, more abstractly, .
So do these form a subalgebra of ? Linearity is easy; we must check that this condition is closed under the bracket. That is, if and both satisfy this condition, what about their commutator ?
So this condition will always give us a linear Lie algebra.
We have three different families of these algebras. First, we consider the case where is odd, and we let be the symmetric, nondegenerate bilinear form with matrix
where is the identity matrix. If we write the matrix of our endomorphism in a similar form
our matrix conditions turn into
From here it’s straightforward to count out basis elements that satisfy the conditions on the first row and column, that satisfy the antisymmetry for , another that satisfy the antisymmetry for , and that satisfy the condition between and , for a total of basis elements. We call this Lie algebra the orthogonal algebra of , and write or . Sometimes we refer to the isomorphism class of this algebra as .
Next up, in the case where is even we let the matrix of look like
A similar approach to that above gives a basis with elements. We also call this the orthogonal algebra of , and write or . Sometimes we refer to the isomorphism class of this algebra as .
Finally, we again take an even-dimensional , but this time we use the skew-symmetric form
This time we get a basis with elements. We call this the symplectic algebra of , and write or . Sometimes we refer to the isomorphism class of this algebra as .
Along with the special linear Lie algebras, these form the “classical” Lie algebras. It’s a tedious but straightforward exercise to check that for any classical Lie algebra , each basis element of can be written as a bracket of two other elements of . That is, we have . Since for some , and since we know that , this establishes that for all classical .