I am encountering two definitions of the special orthogonal lie algebra, and I would like to know if they are equivalent, and if there are advantages to working with one over the other.
If we begin with an $n$-dimensional vector space $V$ over a field $k$ and a chosen basis, we can define a bilinear form on $V$ by a matrix $S\in M_n(k)$, ie, let $\langle v,w\rangle=v^tSw$ for all $v,w\in V$. Now $g\in GL_n(k)$ preserves the form ($\langle g(v),g(w)\rangle=\langle v,w\rangle$) if and only if $g^tSg=S$, so all such $g$ form a linear algebraic group $G$. The tangent space at the identity of $G$ will be contained in that of $GL_n(k)$, so $T_eG\subset T_eGL_n(k)=M_n(k)$, and in fact, $T_eG=\{B\in M_n(k)\mid B^tS+SB=0\}$. $T_eG$ becomes a lie algebra, $Lie(G)$, if we define the bracket to be the commutator of two matrices.
Now, if $S=I_n$, it follows that $G=O_n(k)$ is the orthogonal group of matrices satisfying $g^tg=I_n$, and $Lie(G)=\mathfrak{so}_n$ is the lie algebra of antisymmetric matrices.
In Humphrey's Introduction to Lie Algebras and Representation Theory, he defines $\mathfrak{so}_n$ to be all matrices $B$ satisyfing $B^tS+SB=0$, where
$$
S=\begin{pmatrix}
1&0&0\\
0&O&I_l\\
0&I_l&O
\end{pmatrix}
\hspace{.5in}\text{or}\hspace{.5in}
S=\begin{pmatrix}
O&I_l\\
I_l&O
\end{pmatrix}
$$
depending on the parity of $n$. The matrices obtained in this way are not antisymmetric, nor is the group $G$ preserving the form defined by $S$ the orthogonal group $O_n(k)$.
Are the two groups obtained from considering different $S$ isomorphic? Are the two lie algebras isomorphic? If so, why would one prefer one form to the other?
Best Answer
As long as $S$ is symmetric, the group of linear maps preserving the inner product induced by $S$ will always be isomorphic to $O(n)$ (and so in particular will always have the same Lie algebra). This is because given any inner product you can find an orthornormal basis and with respect to this basis $S$ is just the identity matrix.
The reason I'm familiar with for choosing $S$ to be one of the above matrices is that then the root space decomposition of the Lie algebra is a lot easier. For example, when choosing a Cartan subalgebra of a matrix Lie algebra, it is nice to be able to choose these to consist of only diagonal matrices. This doesn't work for the usual definition of $so(n)$ but does if you choose $S$ appropriately.