Demonstrate $\mathrm{Spin}(3, 3) = \mathrm{SL}(4, R)$ It seems that there is an accidental isomorphism
$$
\mathrm{Spin}(3, 3) = \mathrm{SL}(4, R).
$$
Some facts I am aware is that

*

*$\mathrm{SL}(4, R)$ has the Lie algebra of 15 generators.


*the homotopy group of $\pi_1(\mathrm{SL}(4, R))=\pi_1(\mathrm{SO}(4))=\mathbf{Z}/2$.
How to show or convince the accidental isomorphism
$
\mathrm{Spin}(3, 3) = \mathrm{SL}(4, R)?$
 A: $SL_4(\mathbb{R})$ has a defining $4$-dimensional representation on $V = \mathbb{R}^4$. Its exterior square is a $6$-dimensional representation $\wedge^2(V) = \wedge^2(\mathbb{R}^4)$, and the action of $SL_4(\mathbb{R})$ on this representation preserves the wedge product $\wedge^2(V) \otimes \wedge^2(V) \to \wedge^4(V) \cong \mathbb{R}$ (the last isomorphism is an isomorphism of $SL_4(\mathbb{R})$-representations), which is a nondegenerate symmetric bilinear form in this degree.
Now we need to compute the signature of the wedge product. If $e_1, \dots e_4$ is the standard basis of $\mathbb{R}^4$ then $e_i \wedge e_j, i < j$ is a basis of $\wedge^2(\mathbb{R}^4)$. The wedge product acting on this basis has the property that $(e_i \wedge e_j) \wedge (e_k \wedge e_{\ell}) = 0$ unless $i, j, k, \ell$ is a permutation of $1, 2, 3, 4$, in which case it is the sign of this permutation times the volume form $e_1 \wedge e_2 \wedge e_3 \wedge e_4$. It follows that as a symmetric bilinear form the wedge product is the direct sum of $3$ copies of the "hyperbolic" form, which has signature $(1, 1)$, so altogether it has signature $(3, 3)$. This gives a map
$$SL_4(\mathbb{R}) \to SO(3, 3).$$
The action of $SL_4(\mathbb{R})$ is faithful so the induced map on Lie algbras is injective, and since the two have the same dimension ($15$) the induced map on Lie algebras is an isomorphism, so this map is a covering map. We have $\pi_1(SO(3, 3)) \cong \mathbb{Z}_2 \times \mathbb{Z}_2$, and in fact the block-diagonal inclusion of the connected component of the maximal compact $SO(3) \times SO(3)$ is an isomorphism on $\pi_1$. Now it suffices to show that the image of $\pi_1(SL_4(\mathbb{R}))$ in $\pi_1(SO(3, 3))$ is the diagonal copy of $\mathbb{Z}_2$, and it will then follow that the map above lifts to $\text{Spin}(3, 3)$. The inclusion of $SO(4)$ is an isomorphism on $\pi_1$, so in turn it suffices to show that the induced map
$$SO(4) \to SO(3) \times SO(3)$$
exhibits $\pi_1(SO(4)) \cong \mathbb{Z}_2$ as the diagonal copy of $\mathbb{Z}_2$ in $\pi_1(SO(3) \times SO(3))$. I can't actually think of a really slick way to do this off the top of my head but it ought to be doable by a direct if unenlightening computation, or by a consideration of the other two double covers of $SO(3) \times SO(3)$, which are $SU(2) \times SO(3)$ and $SO(3) \times SU(2)$. Maybe someone else has a clever idea here.
