Defining an isomorphism that respects the Lie bracket: is my work correct?

Question

I previously determined that if $\mathfrak{sl}$ denotes the Lie algebra of $SL_2(\mathbb C)$ and $\mathfrak o$ denotes the Lie algebra of $O(3,\mathbb C)$ then a basis for $\mathfrak{sl}$ is given by

$$ b_1 = \begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix} \text{ and } b_2 = \begin{pmatrix}0 & 0 \\ 1 & 0 \end{pmatrix}, b_3 = \begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix} , ib_1, ib_2, ib_3$$

and a basis for $\mathfrak o$ is given by

$$ B_1 = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}, B_2 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{pmatrix}, B_3 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}, iB_1, iB_2, iB_3$$

It is my goal to prove that $\mathfrak{sl}$ and $\mathfrak o$ are isomorphic.

My idea is to define what values a map $\varphi: \mathfrak{sl}\to \mathfrak o$ takes on the basis elements. By trial and error I found that the following assignment

$$ b_i \mapsto B_i, ib_i \mapsto iB_i$$

satisfies $[\varphi(b_1),\varphi(b_2)] = B_3 = \varphi([b_1,b_2])$ and $[\varphi(ib_1),\varphi(ib_2)] = iB_3\varphi([ib_1, ib_2])$.

My main question is: is there a way to prove that is satisfies the bracket relation for the other basis elements without doing all the other computations?

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My other question is: is my work correct?

Answer 1

(1) Presumably you are looking for a isomorphism $\mathfrak{sl}(2, \mathbb{C}) \stackrel{\cong}{\to} \mathfrak{o}(3, \mathbb{C})$ as complex Lie algebras, which means in particular that we think of the vector spaces underlying the two Lie algebras as complex. (It's conventional to write the latter Lie algebra here as $\mathfrak{so}(3, \mathbb{C})$) As such, the two given lists of vectors are not bases, as, e.g., $b_1$ and $ib_1$ are scalar multiples of one another, and $\{b_1, b_2, b_3\}$ is a basis for $\mathfrak{sl}(2, \mathbb{C})$.

(2) The matrix Lie algebra spanned by the matrices $B_1$, $B_2$, and $B_3$ is not a copy of $\mathfrak{o}(3, \mathbb{C})$---one way to see this is to note that the Lie algebra they span is nilpotent, whereas $\mathfrak{o}(3, \mathbb{C})$ is semisimple. The usual representation of $\mathfrak{o}(3, \mathbb{C})$ is as the matrix Lie algebra of skew-symmetric $3 \times 3$ matrices.

(3) There are certainly invariant/noncomputational ways to approach the problem, and Qiaochu's answer describes one efficient method. But note that with only three basis elements, one need only verify that a candidate isomorphism $\varphi$ respect just three brackets, e.g., $[b_2, b_3]$, $[b_3, b_1]$, $[b_1, b_2]$.

Answer 2

Yes, there is definitely a way to do this without explicit computations. The point is that this map of Lie algebras in fact comes from a map of Lie groups $SL_2(\mathbb{C}) \to SO_3(\mathbb{C})$, and you should be trying to construct this map.

Here's one way to do it. Every Lie group $G$ has a distinguished representation called the adjoint representation $\mathfrak{g}$ on its own Lie algebra given by differentiating the conjugation action of $G$ on itself at the identity. For $SL_2(\mathbb{C})$ the adjoint action on $\mathfrak{sl}_2(\mathbb{C})$ is just by conjugation of matrices. Now, $\dim \mathfrak{sl}_2(\mathbb{C}) = 3$, so this gives a natural $3$-dimensional representation of $SL_2(\mathbb{C})$, and hence a map

$$SL_2(\mathbb{C}) \to GL_3(\mathbb{C}).$$

Our goal will be to show that this map in fact lands in $SO_3(\mathbb{C})$; equivalently, we are going to find an invariant nondegenerate symmetric bilinear form on the adjoint representation of $SL_2(\mathbb{C})$. This form will be, up to some scalar, the Killing form: a short description of it in this case is

$$\mathfrak{sl}_2(\mathbb{C}) \times \mathfrak{sl}_2(\mathbb{C}) \ni (X, Y) \mapsto \langle X, Y \rangle = \text{tr}(XY).$$

This is clearly a symmetric bilinear form; invariance comes from the conjugacy invariance of the trace, while nondegeneracy is a nice exercise. It follows that we have a map

$$SL_2(\mathbb{C}) \to O_3(\mathbb{C})$$

of (complex) Lie groups, but since $SL_2(\mathbb{C})$ is connected and $SO_3(\mathbb{C})$ is the connected component of the identity of $O_3(\mathbb{C})$ this map naturally lands in $SO_3(\mathbb{C})$. From here there are a few things you could do to show that this map induces an isomorphism on Lie algebras; it suffices to show that it induces an injection on Lie algebras, for example, since the two Lie algebras have the same dimension.

You can tell this story without any mention of Lie groups by looking at the adjoint representation of $\mathfrak{sl}_2(\mathbb{C})$ on itself directly, which is given by

$$\text{ad}_X(Y) = [X, Y].$$

In general, the kernel of the adjoint representation of a Lie algebra $\mathfrak{g}$ is its center, and so to show that the adjoint representation is faithful it suffices to show that $\mathfrak{sl}_2(\mathbb{C})$ has trivial center. This is also a nice exercise.