How to use the Killing form to write down a Lie group isomorphism, and what is the induced Lie algebra isomorphism?

Question

This is a follow up question on my previous question here.

One of the answers suggests that I find a map $$SL_2(\mathbb{C}) \to SO_3(\mathbb{C})$$

and then the map induced by this map will be a Lie algebra isomorphism from the Lie algebra of $SL_2(\mathbb C)$ to the Lie algebra of $ O(3,\mathbb C)$.

But there are some things I don't understand. For the sake of this answer let's assume that I successfully determined a group homomorphism $$\varphi: SL_2(\mathbb{C}) \to SO_3(\mathbb{C}).$$

How does it induce a map on the corresponding Lie algebras?

(For this I assume that the Lie algebra of $O(3,\mathbb C)$ and of $SO(3,\mathbb C)$ are equal.)

Where and how do I use the Killing form in all of this?

Answer 1

Any Lie group homomorphism $\varphi: G \to H$ must in particular map the identity $I$ of $G$ to the identity $I$ of $H$, and so taking the tangent map at the identity gives a linear map $T_I \varphi : T_I G \to T_I H$, but we can identify $T_I G \leftrightarrow \mathfrak{g}$ and likewise for $T_I H$. One can show that the tangent map of a Lie group homomorphism is automatically a Lie algebra homomorphism (in fact, one can motivate the definition of Lie algebra this way).

But in this case we can avoid such a construction and simply build all of our objects directly at the Lie algebra level. Constructing the given isomorphism roughly means that we want to show we can think of $\mathfrak{sl}_2 := \mathfrak{sl}_2(\mathbb{C})$ as the Lie algebra that acts on a (complex) $3$-dimensional vector space and preserves a (complex) inner product there.

There is a natural candidate for this vector space: Since $\dim_{\mathbb{C}} \mathfrak{sl}_2 = 3$, $\mathfrak{sl}_2$ is itself such a vector space, and we have a natural Lie algebra action of $\mathfrak{sl}_2$ on itself, namely, just Lie bracket: $$A \cdot B := [A, B].$$ (This is called the adjoint action, and this Lie algebra representation here is the adjoint representation.) In the usual basis $$ H := \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}, \qquad X := \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}, \qquad Y := \begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix} $$ the nontrivial brackets on $\mathfrak{sl}_2$ are those determined by $$ [H, X] = 2 X, \qquad [H, Y] = -2 Y, \qquad [X, Y] = H. $$ So, in the basis $(H, X, Y)$, the action of $H$, that is, the linear map $Z \mapsto \text{ad}(H)(Z) := [H, Z]$, has matrix $$ \begin{pmatrix} 0 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -2 \end{pmatrix}. $$ Computing likewise the $3 \times 3$ matrices for the action of $X$ and $Y$ on $\mathfrak{sl}_2$ defines a map $$ \phi: \mathfrak{sl}_2 \to \mathfrak{gl}(\mathfrak{sl}_2) \cong \mathfrak{gl}_3, \qquad \phi: W \mapsto \text{ad}(W) $$ and it's easy to see that this map is injective. In fact, for any Lie algebra this map is a Lie algebra homomorphism.

So, it remains to show that $\phi$ preserves an inner product on $\mathfrak{sl}_2$. One can do this in two ways: Firstly, one can write down a general bilinear form on $\mathfrak{sl}_2$ (say, for convenience, in the basis dual to $(H, X, Y)$); this depends on $6$ parameters, and we can determine them all (up to an overall multiplicative constant) using the fact that it is invariant under $\phi(\mathfrak{sl}_2)$. Alternatively, as Qiaochu suggested in a previous thread, one can compute the Killing form $K(Z, W) = \text{tr}(\text{ad } Z \circ \text{ad } W)$ (which is, up to an overall constant that doesn't matter for our purposes, $K(Z, W) = \text{tr}(Z W)$) for $\mathfrak{sl}_2$ directly, and this is guaranteed to be invariant under the adjoint action. This shows that $\phi(\mathfrak{sl}_2) \leq \mathfrak{so}_3$, but we saw that $\phi$ is injective and both algebras here have dimension $3$, so the containment is an equality, and hence $\phi$ is a Lie algebra isomorphism.

(In fact, we only need that the Killing form is preserved by the adjoint action, which is automatic, and that it is nondegenerate, which holds because $\mathfrak{sl}_2$ is semisimple; seeing this latter fact is more involved than the other ingredients here, though, and it makes the proof more elementary to compute explicitly.) Since all of the coefficients so far are real, all of our computations formally apply to $\mathfrak{sl}(2, \mathbb{R})$; in this setting the Killing form has signature $(1, 2)$, so this also establishes a Lie algebra isomorphism $\mathfrak{sl}_2 \cong \mathfrak{so}_{1, 2}$.

Note that the realization $\mathfrak{so}_3 = \phi(\mathfrak{sl}_2)$ is not the standard one, i.e., the one comprising the skew-symmetric matrices, which corresponds to the standard inner product $(e^1)^2 + (e^2)^2 + (e^3)^2$. This can be remedied by an appropriate change of basis, but the change-of-matrix basis cannot be real for reasons described in an earlier answer.

Answer 2

I'll try to go all the way in this answer, so in the end you get an actual map $\def\C{\Bbb C}\def\sl{\mathfrak{sl}}\sl(2, \C) \to \mathfrak{o}(3,\C)$ that is an isomorphism of complex Lie algebras.

The first step is to get an injective Lie algebra morphism $\def\gl{\mathfrak{gl}}\sl(2,\C)\to\gl(3,\C)$. This is done as in the answer by Travis by identifying $\sl(2,\C)\cong\C^3$ though the choice of a basis in $\sl(2,\C)$, for which I will take the matrices $H,X,Y$ of the answer by Travis to this question. Next the Lie algebra morphism arises as the derivative at the identity of the conjugation action of $SL(2,\Bbb C)$ on its own tangent space $\sl(2,\C)$ at the identity. But since this (adjoint) action is what defines the Lie algebra structure on $\mathfrak{sl}(2,\C)$ in the first place, it is clear without computation that this morphism is $\def\ad{\operatorname{ad}}\ad:\sl(2,\C)\to\gl(\sl(2,\C))$ defined by the Lie bracket $$ \ad : Z\in\sl(2,\C)\mapsto(T\in\sl(2,\C)\mapsto[Z,T]\in\sl(2,\C)) $$ With the given identification, we get concretely the matrices $$ \ad H=\begin{pmatrix}0&0&0\\0&2&0\\0&0&-2\end{pmatrix} ,\qquad \ad X=\begin{pmatrix}0&0&1\\-2&0&0\\0&0&0\end{pmatrix} ,\qquad \ad Y=\begin{pmatrix}0&-1&0\\0&0&0\\2&0&0\end{pmatrix} .$$ These matrices being obtained as commutators automatically have trace$~0$ (they are in $\sl(\sl(2,\C))$, still with the same identification $\sl(2,\C)\cong\C^3$). They are not skew symmetric though. Still the adjoint action is bound to preserve a non-degenerate symmetric bilinear form, namely the Killing form. The point is that the chosen identification does not make the Killing form correspond to the standard bilinear form on$~\C^3$.

So the next step is to compute the Killing form, and to adapt our chosen basis to one that is "orthonormal" for the Killing form. The definition of the Killing form interprets $\gl(\sl(2,\C))$ not as a Lie algebra, but as an associative algebra, in other words instead of computing commutators we allow ourselves to compute products of our $3\times3$ matrices. The Killing form is then defined by $$\langle Z,T\rangle_\mathrm K=\operatorname{trace}((\ad Z)\circ(\ad T))$$ Computing the matrix of mutual bilinear form values on the basis $H,X,Y$ gives $$ \begin{pmatrix}8&0&0\\0&0&4\\0&4&0\end{pmatrix}, $$ which shows that while $H$ is orthogonal to $X$ and $Y$, the latter two are orthogonal to themselves but not to the other one. Before looking for an orthonormal basis it is convenient to scale down the bilinear form (which still gives a non-degenerate form invariant under the adjoint action) by a factor$~8$, so define $\langle Z,T\rangle=\frac18\langle Z,T\rangle_\mathrm K=\frac12\operatorname{trace}(ZT)$ (where in the latter expression we treat $\sl(2,\Bbb C)$ itself as an associative algebra; the equality can be checked by explicit computation on the basis $H,X,Y$). Now $H$ has become normalised, but to get an orthonormal basis of the span of $X,Y$ one can take the vectors $\def\ii{\mathbf i}X+Y$ and $\ii X-\ii Y$.

So finally to get our morphism so that its image lies in the set of skew symmetric matrices, we should apply base change to the basis of$~\Bbb\C^3$ corresponding to the basis $H,X+Y, \ii X-\ii Y$. So the change of basis matrix is $$ P=\begin{pmatrix}1&0&0\\0&1&\ii\\0&1&-\ii\end{pmatrix} \qquad\text{with}\qquad P^{-1}=\frac12\begin{pmatrix}2&0&0\\0&1&1\\0&-\ii&\ii\end{pmatrix}, $$ and we shall map $Z$ not to the matrix $\ad Z$ mentioned before, but to $P^{-1}(\ad Z)P$. Now $$ H\mapsto\begin{pmatrix}0&0&0\\0&0&2\ii\\0&-2\ii&0\end{pmatrix} ,\qquad X\mapsto\begin{pmatrix}0&1&-\ii\\-1&0&0\\\ii&0&0\end{pmatrix} ,\qquad Y\mapsto\begin{pmatrix}0&-1&-\ii\\1&0&0\\\ii&0&0\end{pmatrix} .$$ These matrices are indeed skew symmetric, and you can check that they satisfy the required commutation relations (I did not do so for all cases.)

Note that the basis $H,X+Y, \ii X-\ii Y$ of $\sl(2,\C)$ is (up to transposition) that of the Pauli matrices, and that the correspondence above becomes a bit nicer if instead of the elements $H,X,Y$ on the left one also takes $H,X+Y, \ii X-\ii Y$.