Relationship between $\mathfrak{sl}(2,\mathbb{C})$ and $\mathfrak{so}(1,3)$

Question

i'm very new to Lie Theory and in general to differential geometry so this might be a very naive question.

The question is: what is the relationship between $\mathfrak{sl}(2,\mathbb{C})$ and $\mathfrak{so}(1,3)$?

When studying the Lie Algebras of the two spaces $SL(2,\mathbb{C})$, $SO(1,3)$ I arrive to the conclusion that $\mathfrak{so}(1,3)$ is generated (as a real vector space) by the six matrices \begin{equation} A_1=\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{pmatrix} A_2=\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{pmatrix} A_3=\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \end{equation} \begin{equation} B_1=\begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} B_2=\begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} B_3=\begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix} \end{equation} while $\mathfrak{sl}(2,\mathbb{C})$ is generated as a complex vector spaces by the three matrices: \begin{equation} X_1=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} X_2=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} X_3=\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \end{equation}

If I'm not mistaken the Lie Bracket in $\mathfrak{so}(1,3)$ is defined as follows: \begin{equation} [A_i,B_j] = \sum_k\epsilon_{ijk}B_k \hspace{1cm} [A_i,A_j] =\sum_k\epsilon_{ijk}A_k \hspace{1cm}[B_i,B_j]=-\sum_k\epsilon_{ijk}A_k \end{equation} While the Lie Bracket in $\mathfrak{sl}(2,\mathbb{C})$ is: \begin{equation} [X_1,X_2]=2X_2 \hspace{1cm} [X_1,X_3]=-2X_3 \hspace{1cm} [X_2,X_3]=X_1 \end{equation}

Now, firstly I want to ask if what I stated so far is correct. If so, these two spaces are 6-dimensional vector spaces and so they are isomorphic as vector spaces, my question is: are they isomorphic also as Lie Algebras? If so, what is the map that realize such isomorphism? Moreover, is one of the two starting groups ($SO(1,3)$, $SL(2,\mathbb{C})$) simply connected? If that were the case, alongside with the (yet to prove) fact that they share the same Lie Algebra (up to isomorphism) would make one the universal covering group of the other.

Thank you in advance :)

Answer 1

Edit: changed some of the wording around the real form to make it clear what I mean

Here's a nice way to see this isomorphism with no explicit choice of basis and that extends to all similar ones.

Firstly, we note this isomorphism is as real Lie algebras even though you may be more familiar with $\mathfrak{sl}(2,\mathbb{C})$ as a complex Lie algebra. Any complex Lie algebra $\mathfrak{g}$ can be thought of as a real one with complexification $\mathfrak{g}\oplus\mathfrak{g}$.

So on the level of complex Lie algebras we have an isomorphism $\mathfrak{so}(4,\mathbb{C})\cong \mathfrak{sl}(2,\mathbb{C})\oplus\mathfrak{sl}(2,\mathbb{C})$. This can be conceived naturally as follows. Let $V_1,V_2$ be the 2 dimensional representations of the two copies of $\mathfrak{sl}(2,\mathbb{C})$ respectively. Then $\mathfrak{sl}(2,\mathbb{C})\oplus\mathfrak{sl}(2,\mathbb{C})$ acts naturally on $V_1 \otimes V_2$ by $(X_1,X_2)(v\otimes w) = X_1(v)\otimes w +v\otimes X_2(w) $ (this is the natural way to define representations of a direct sum of Lie algebras).

Then we note there is a natural symmetric bilinear form on $V_1 \otimes V_2 \cong \mathbb{C}^4$ given by treating it as a subset of $\bigwedge^2 (V_1 \oplus V_2)$ (i.e. treating the tensor product as if it was a wedge product) and letting the bilinear form be the wedge product: $$(v_1 \otimes v_2, w_1 \otimes w_2) := v_1 \wedge v_2 \wedge w_1 \wedge w_2 \in \bigwedge\nolimits^4(V_1 \oplus V_2) \cong \mathbb{C}.$$

Now we can quickly observe that this bilinear form is invariant for the action of our Lie algebra forcing $\mathfrak{sl}(2,\mathbb{C})\oplus\mathfrak{sl}(2,\mathbb{C})\subset\mathfrak{so}(4,\mathbb{C}) $ but they have the same dimension so they must be equal.

Finally we should discuss real forms. There are several ways real forms of a Lie algebra can manifest, in terms of what happens to their representations. There can be a real structure on the representation, a Hermitian structure (i.e. an inner product) or a quaternionic structure. In each case restricting to the elements that preserve the structure gives a real form of the Lie algebra.

For example, the real forms of $\mathfrak{so}(4,\mathbb{C}) $ are $\mathfrak{so}(4), \mathfrak{so}(3,1), \mathfrak{so}(2,2)$ all coming from real structures with different signatures together with $\mathfrak{so}^*(4)$ coming from a quaternionic structure. On the other side, $\mathfrak{sl}(2,\mathbb{C})$ has only two real forms $\mathfrak{sl}(2,\mathbb{R})$ and $\mathfrak{su}(2)$ from a real and Hermitian structure respectively. Mixing and matching these two gives the following isomorphisms:

$$ \mathfrak{su}(2)\oplus \mathfrak{su}(2) \cong \mathfrak{so}(4)$$ $$ \mathfrak{sl}(2,\mathbb{R})\oplus \mathfrak{sl}(2,\mathbb{R}) \cong \mathfrak{so}(2,2)$$ $$ \mathfrak{su}(2)\oplus \mathfrak{sl}(2,\mathbb{R}) \cong \mathfrak{so}^*(4)$$

The final real form is the one we are looking for and that comes from a 'real structure' that swaps $V_1$ and $V_2$. A concrete way to implement this is: choose a real structure $\sigma$ (an anti-linear involution) on $V_1 \oplus V_2$ such that $\sigma(V_1) = V_2$. Then the corresponding real form of $\mathfrak{sl}(2,\mathbb{C})\oplus\mathfrak{sl}(2,\mathbb{C})$ is: $$\{(X,\sigma X \sigma)| X\in\mathfrak{sl}(2,\mathbb{C})\} \cong \mathfrak{sl}(2,\mathbb{C})$$

Then $\sigma(v\otimes w):= \sigma(v)\otimes \sigma(w)$. Note they are the wrong way round now but we already identified this with the wedge product so we can do that again here to swap them round so $\sigma(v\otimes w) = -\sigma(w) \otimes \sigma(v)$. This defines a real structure which you can check fixes a space of signature $(3,1)$ (easiest way to check that is that the only null subspaces are lines).

Obviously, we can choose our bases so that $v_2 = \sigma(v_1)$ and $w_2 = \sigma(w_1)$ so for example we get $\sigma(v_1 \otimes v_2) = -v_1 \otimes v_2$ and $\sigma(v_1 \otimes w_2) = -w_1 \otimes v_2$. Then our real subspace in this basis is: $$ \langle iv_1 \otimes v_2, iw_1 \otimes w_2, v_1\otimes w_2 - w_1 \otimes v_2, i(v_1\otimes w_2 + w_1 \otimes v_2)\rangle.$$

Answer 2

Callum's answer is the right way to look at this. Yet I want to point out an alternative interpretation of what is going on:

Consider $X :=$ the space of Hermitian $2\times 2$-matrices. This is an $\mathbb R$-vector space of dimension $4$. A convenient basis of it is made up of:

$$\pmatrix{1&0\\0&1}, \pmatrix{1&0\\0&-1}, \pmatrix{0&1\\1&0}, \pmatrix{0&i\\-i&0}$$

There is a (surprising) quadratic form on this vector space given by $q(x) = \det(x)$ (as for a $2\times 2$-matrix, $\det(ax) = a^2 \det(x)$ for scalars $a$). This quadratic form (or more precisely, its corresponding symmetric bilinear form $b(\cdot, \cdot)$) has signature $(1,3)$; in fact, the basis vectors above are mutually orthogonal w.r.t. this form, and it's plain the determinant of the first is $1$, while all the others have determinant $-1$.

Now the joke (due to Weyl, according to Wikipedia) is that the group $SL_2(\mathbb C)$ acts on $X$, via $g.x := gxg^\dagger$, and this action (almost trivially) leaves that bilinear form invariant. This means we get a nontrivial homomorphism of real Lie groups

$$SL_2(\mathbb C) \rightarrow SO(1,3)$$

and the rest is exercises (namely, to show its image is ... rather big, and its kernel is $\{\pm Id \}$), anyway this is close enough to produce an isomorphism on the level of Lie algebras.

Also, that one can be seen directly by "differentiating" the above action: Unless I'm mistaken, in fact $\mathfrak{sl}_2(\mathbb C)$ acts on the same space $X$ via the (unusual)

$$\rho(g)(x) := gx + xg^\dagger$$

I found it rather straightforward to show that this does define a Lie algebra representation, only the fact that this is again "invariant" for that bilinear form (now in the sense of Lie algebra representations, i.e. $b(\rho(g)x, y) = -b(x, \rho(g)(y))$) seems to only come from cumbersome computations. But that taken for granted, this gives a non-trivial Lie algebra homomorphism

$$\mathfrak{sl}_2(\mathbb C) \rightarrow \mathfrak{so}(1,3)$$

and by simplicity of both sides, it must be an iso.

It might be fun to see how this construction is a concrete manifestation of things in Callum's answer.