Confusion over 2 different basis for $\mathfrak{s}\mathfrak{o}(3,\mathbb{R})$.

Question

I am confused about two different basis I have found for the orthogonal lie algebra $\mathfrak{s}\mathfrak{o}(3,\mathbb{R})$. In Humphrey's "Introduction to Lie Algebras and Representation Theory", it defines the orthogonal lie algebra to be the collection of matrices $X\in \mathfrak{g}\mathfrak{l}(3,\mathbb{R})$ satisfying $\beta(X(v), w) = -\beta(v, X(w))$ for $\beta$ a non-degenerate symmetric bilinear form and $v,w\in \mathbb{R}$. Computing from the definition, this can be written more explicitly as matrices with the following form $M = \begin{bmatrix} 0 & b & c\\ -c & d & 0\\ -b & 0 & -d \end{bmatrix}$ for any $b,c,d\in \mathbb{R}$. Hence, a basis would look like $$ \begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 0\\ -1 & 0 & 0 \end{bmatrix}, \quad \begin{bmatrix} 0 & 0 & 1\\ -1 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix},\quad \begin{bmatrix} 0 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & -1 \end{bmatrix}.$$ However, I have seen elsewhere (in particular the wikipedia page for the orthogonal Lie group which discuss the orthogonal Lie algebra) $\mathfrak{s}\mathfrak{o}(3,\mathbb{R})$ is skew-symmetric matrices, which would have the following basis: $$ \begin{bmatrix} 0 & 1 & 0\\ -1 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}, \quad \begin{bmatrix} 0 & 0 & 1\\ 0 & 0 & 0\\ -1 & 0 & 0 \end{bmatrix},\quad \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 1\\ 0 & -1 & 0 \end{bmatrix}.$$ To me, it does not seem as though these Lie algebras with these bases would be isomorphic, since the first basis has a matrix that acts as a scalar, while the second does not. Is there a way to reconcile these two definitions of the orthogonal Lie algebra (i.e. show they are the same Lie algebra), or am I missing something entirely?

Answer 1

Assume a finite dimensional nondegenerate symmetric bilinear form $\beta(x,y) = x^T A y; A^T = A, x,y\in \mathbb{R}^3.$ Let $X$ satisfy your condition $$\beta(Xx,y) = -\beta(x,Xy),$$ or more concretely $$(Xx)^TAy = -x^TAXy.$$ Manipulating the left side we get $(Xx)^TAy = x^TX^TA y.$ Therefore, we will satisfy the condition if $X^TA = -AX.$

However, $A$ is often taken to be $I$, which means that $$X^T = -X,$$ producing the standard skew-symmetric matrices. Ultimately, Lie algebras have several representations, and need not be married to a specific representation.

Answer 2

Thought it might be worthwhile to turn my comment into a full answer as this is an area I have seen people confused about before.

If we don't worry about matrices for a moment we can see more fundamentally that for a non-degenerate symmetric bilinear form $\beta$ on $V$, $\mathfrak{so}(\beta,V)$ is the set of endomorphisms which are skew for this form i.e.

$$ \beta(Xv,w) + \beta(v,Xw) = 0$$

In effect, this is the derivative of the fact that $\operatorname{SO}(\beta,V)$ preserves $\beta$.

Over the complex numbers all non-degenerate symmetric bilinear forms are equivalent so it doesn't matter which we choose, we get the same Lie algebra which we denote $\mathfrak{so}(n,\mathbb{C})$. Thus we are free to make a choice of representing $\beta$ by: $$ \begin{pmatrix}1&0&0\\0&0&I\\0&I&0\end{pmatrix} \text{or} \begin{pmatrix}0&I\\I&0\end{pmatrix}$$

These don't result in skew-symmetric matrices (for that we take $\beta = I$) but gives an isomorphic Lie algebra. The advantage of this choice is that the diagonal matrices form a Cartan subalgebra so it is a little easier to see the root decomposition (or at least it lines up neatly with the ideas for $\mathfrak{sl}(n,\mathbb{C})$) and so on which is why Humphreys takes this approach.

If we are looking at real Lie algebras this gets a bit more complicated however. Over the reals there are several inequivalent classes of non-degenerate symmetric bilinear forms. Indeed by Sylvester's law of inertia there are $n+1$ where $n$ is the dimension of the vector space. They can be distinguished by their signature $(p,q)$ which denotes that in an orthogonal basis $\{x_i\}$ we have $(x_i,x_i)$ positive $p$ times and negative $q$ times ($0\leq p,q \leq n$, $p+q=n$). For equivalent bilinear forms we have isomorphic Lie algebras so we denote can denote them by $\mathfrak{so}(p,q)$. These are called the indefinite orthogonal Lie algebras.

Additionally, even though the bilinear forms are inequivalent we have an isomorphism $\mathfrak{so}(p,q) \cong \mathfrak{so}(q,p)$. Thus we have $\left\lfloor \frac{n}{2} \right\rfloor + 1$ different Lie algebras here up to isomorphism. In fact, these are almost all of the real forms of $\mathfrak{so}(n,\mathbb{C})$ (there is an extra one for $n$ even).

Humphreys' construction taken over the reals always produces the split real form which is $\mathfrak{so}(p,p)$ or $\mathfrak{so}(p,p+1)$ depending on the parity of of $n$.