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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?

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2 Answers

up vote 4 down vote accepted

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.

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Thanks Eric. A quick question about your answer. Are you assuming something about the field? I'm having a difficult time showing I can find an orthonormal basis with respect to a symmetric form over an arbitrary field. What are some other conditions I may need on the form to show this fact, if any? –  Jared Mar 7 '13 at 18:41
    
I'm used to just $\mathbb R$ and $\mathbb C$ but I think everything I said holds for arbitrary fields. Maybe for characteristic 2 one could get a nice root space decomposition taking $S = Id$ since then all diagonal matrices are skew-symmetric. –  Eric O. Korman Mar 7 '13 at 18:57
    
I now see that all symmetric nondegenerate bilinear forms over algebraically closed fields of characteristic not 2 are equivalent. As you say, this is because we can always find a basis such that the form gives the identity. Also, the groups preserving these forms are all isomorphic via conjugation by a change of basis matrix, so that as you said, their lie algebras are also isomorphic. I still have some thinking to do about the root space decomposition, but your responses have been very helpful. Thank you! –  Jared Mar 7 '13 at 23:26
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Wikipedia says that over reals, the Lie groups are different. They are infact determined by the signature of S. See the wiki article http://en.wikipedia.org/wiki/Generalized_orthogonal_group

As the Lie groups are different over reals, the corresponding Lie algebras should also be different.

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