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?


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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    $\begingroup$ 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? $\endgroup$ – Jared Mar 7 '13 at 18:41
  • $\begingroup$ 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. $\endgroup$ – Eric O. Korman Mar 7 '13 at 18:57
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    $\begingroup$ 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! $\endgroup$ – Jared Mar 7 '13 at 23:26
  • $\begingroup$ This is true over an algebraically closed field, but, as @Siddhant points out, it's not true over $\mathbb{R}$. Specifically, the identity matrix for $S$ gives a quadratic form of signature $(n,0)$ which gives the compact real form of $\mathfrak{so}_n$; whereas an $S$ like Humphreys' corresponds to a quadratic form of signature $(l,l)$ or $(l,l+1)$ which gives the split real form. (So $S :=$the matrix with $l$ 1's and $l$ (resp. $l+1$) -1's on the diagonal would give the same form as Humphreys, but $S=Id$ would not. It is still true that Humphreys' $S$ makes it easier to write down a CSA.) $\endgroup$ – Torsten Schoeneberg May 10 '18 at 20:12

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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    $\begingroup$ No, the Lie algebras are isomorphic. $\endgroup$ – Dietrich Burde Sep 12 '16 at 13:03
  • $\begingroup$ @DietrichBurde: I don't think that for different signatures $(p,q)$, the real Lie algebras $\mathfrak{so(p,q)}$ are isomorphic in general. E.g. $\mathfrak{so}(2n,0)$ is the compact real form, but $\mathfrak{so}(n,n)$ is the split real form of the complex Lie algebra $\mathfrak{so}_{2n}(\mathbb{C})$. $\endgroup$ – Torsten Schoeneberg May 10 '18 at 19:54
  • $\begingroup$ The point is rather that the isomorphism class of the Lie algebra depends only on the signature of the quadratic form (actually, on less than that), but not on which matrix represents the quadratic form. Now over $\mathbb{C}$, there is basically only one quadratic form anyway, which can be represented by both matrices $S$ in the OP. Over the reals on the other hand, as Siddhant points out, there are non-isomorphic Lie algebras for quadratic forms with different signatures. Although each of them might still be represented by various choices of $S$. $\endgroup$ – Torsten Schoeneberg May 10 '18 at 19:59
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    $\begingroup$ Sorry, I was thinking of $O_n(\Bbb{R})$ and $SO_n(\Bbb{R})$, which are different as Lie groups, but have isomorphic Lie algebras. So the claim "As the Lie groups are different ... Lie algebras should be different" is not true in general. $\endgroup$ – Dietrich Burde May 11 '18 at 8:13
  • $\begingroup$ That's a correct general observation of course, but I think the OP asks if, and this answer correctly points out that, over the field $\Bbb R$, the choices a) $S=I_n$ vs. b) $\text{(depending on parity,) } S= \begin{pmatrix} 1&0&0\\ 0&O&I_l\\ 0&I_l&O \end{pmatrix} \text{ resp. } S=\begin{pmatrix} O&I_l\\ I_l&O \end{pmatrix}, $ give non-isomorphic Lie algebras. $\endgroup$ – Torsten Schoeneberg May 11 '18 at 19:41

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