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?


My other question is: is my work correct?

  • $\begingroup$ Your map isn't a Lie algebra homomorphism: $\phi([b_3,b_1])=2\phi b_1=2B_1$ but $[\phi b_3,\phi b_1]=[B_3,B_1]=0$. $\endgroup$ – Matthew Towers Feb 2 '15 at 8:09

(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., $[b_2, b_3]$, $[b_3, b_1]$, $[b_1, b_2]$.

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    $\begingroup$ Thank you for your help, I made a mistake. What I really meant was $$ B_1 = \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}, B_2 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0\end{pmatrix}, B_3 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0\end{pmatrix}, iB_1, iB_2, iB_3$$ I did intend for it to be a basis over $\mathbb R$. Do you agree with my corrected matrices $B_i$? $\endgroup$ – learner Feb 2 '15 at 9:15
  • $\begingroup$ Or is there any reason why I have to consider these two algebras complex Lie algebras and considering them over $\mathbb R$ is wrong? $\endgroup$ – learner Feb 2 '15 at 9:18
  • $\begingroup$ I can't seem to make it work. I got $$ [B_1, B_3]=-B_2$$ but because $$ [b_1,b_3]=-2b_2$$ I need $$ [B_1, B_3]=- 2B_2$$ if my map $b_i \mapsto B_i$ is supposed to work. What am I doing wrong? $\endgroup$ – learner Feb 2 '15 at 9:27
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    $\begingroup$ Your new matrices $B_i$ area perfectly good basis (over $\mathbb{C}$) of $\mathfrak{o}$. It's not "wrong" to consider the Lie algebras as real algebras rather than as complex ones, it's simply a different statement. Note that if you have an isomorphism $\varphi$ of complex Lie algebras, it is also an isomorphism of the underlying real Lie algebras. Also, if the dimension of a complex Lie algebra over $\mathbb{C}$ is $n$, then the dimension of its underlying Lie algebra over $\mathbb{R}$ is $2n$, so when checking whether a map is an isomorphism using a basis, the former is less work. $\endgroup$ – Travis Feb 2 '15 at 19:23
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    $\begingroup$ Alternatively, we can fix any bases $(b_a)$ and $(B_a)$ that we want, but then a general vector space isomorphism $\varphi$ maps $b_a \mapsto c_{1a} B_1 + c_{2a} B_2 + c_{3a} B_3$, and in general only for certain choices of the coefficients $c_{da}$ will $\varphi$ respect the bracket, i.e., will make $\varphi$ a Lie algebra isomorphism. $\endgroup$ – Travis Feb 3 '15 at 1:20

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.

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    $\begingroup$ One of the many reasons to prefer this approach is that it tells you a general fact about, say, semisimple Lie algebras: they all have adjoint representations and Killing forms, so they all admit natural homomorphisms into orthogonal Lie algebras, although the Killing form is indefinite in general, and this map won't be an isomorphism in general for dimension reasons. But the low dimensions are nice. We get in this way natural homomorphisms $\mathfrak{su}(2) \to \mathfrak{so}(3)$ and $\mathfrak{sl}_2(\mathbb{R}) \to \mathfrak{so}(2, 1)$ which are also isomorphisms, for example. $\endgroup$ – Qiaochu Yuan Feb 2 '15 at 7:21
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    $\begingroup$ By "Killing forms" in your comment, I believe you mean to specify "nondegenerate Killing forms". $\endgroup$ – Travis Feb 2 '15 at 7:25
  • $\begingroup$ Right, sorry about that. I can't edit the comment anymore. $\endgroup$ – Qiaochu Yuan Feb 2 '15 at 7:46
  • $\begingroup$ Why is it equivalent for the image to land in $SO(3(\mathbb C)$ to finding an invariant nondegenerate symmetric bilinear form on the adjoint representation? $\endgroup$ – learner Feb 3 '15 at 8:36
  • $\begingroup$ @learner: $O_3(\mathbb{C})$ is, by definition, the Lie group of automorphisms of $\mathbb{C}^3$ preserving a nondegenerate symmetric bilinear form on $\mathbb{C}^3$. (If you learned a more concrete definition, it's a good exercise to prove it's equivalent to this one.) $SO_3(\mathbb{C})$ is its connected component containing the the identity. $\endgroup$ – Qiaochu Yuan Feb 3 '15 at 8:50

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