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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)?$

PrincessEev
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wonderich
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1 Answers1

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$SL_4(\mathbb{R})$ 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.

Qiaochu Yuan
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  • Is https://math.stackexchange.com/questions/459675/the-quaternions-and-so4 related? (I feel like this representation of SO(4) by a pair of quaternions is pretty well-known under various other guises...) – Steven Stadnicki Jan 09 '21 at 04:32
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    @Steven: well, everything's related when it comes to these sorts of things. One way to say it is that you can identify $M_4(\mathbb{R}) \cong \mathbb{H} \otimes \mathbb{H}^{op}$ in the sense that $\mathbb{H}$ acts on itself by left and right multiplication and this induces a map $\mathbb{H} \otimes \mathbb{H}^{op} \to \text{End}(\mathbb{H}) \cong M_4(\mathbb{R})$ which is an isomorphism. So you can identify $SL_4(\mathbb{R})$ as a certain subgroup of the unit group of $\mathbb{H} \otimes \mathbb{H}^{op}$, and this induces an identification of maximal compact subgroups also... – Qiaochu Yuan Jan 09 '21 at 04:57
  • ...which should end up exhibiting an isomorphism between $SO(4)$ and the quotient of $Sp(1) \times Sp(1)$ by $(-1, -1)$. I don't know off the top of my head how this relates to the wedge product on $\wedge^2(\mathbb{R}^4)$ but I wouldn't be surprised if there was some relationship. You can also do all this exceptional isomorphism business using Clifford algebras and spin representations but I'm less familiar with that. – Qiaochu Yuan Jan 09 '21 at 04:58
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    Maybe it's worth pointing out that just the identity of root systems $D_3=A_3$ immediately gives, on the Lie algebra level, $\mathfrak{so}6(\mathbb C) \simeq \mathfrak{sl}_4(\mathbb C)$ and $\mathfrak{so}{3,3}(\mathbb R) \simeq \mathfrak{sl}_4(\mathbb R)$ (split forms) as well as $\mathfrak{so}_6(\mathbb R) \simeq \mathfrak{su}_4$ (compact forms); and likewise, $D_2=A_1 \times A_1 =B_1 \times B_1$ gives ... – Torsten Schoeneberg Jan 09 '21 at 05:57
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    ... $\mathfrak{so}4(\mathbb C) \simeq \mathfrak{sl}_2(\mathbb C) \times \mathfrak{sl}_2(\mathbb C) \simeq \mathfrak{so}_3(\mathbb C) \times \mathfrak{so}_3(\mathbb C)$, $\mathfrak{so}{2,2}(\mathbb R) \simeq \mathfrak{sl}2(\mathbb R) \times \mathfrak{sl}_2(\mathbb R) \simeq \mathfrak{so}{2,1}(\mathbb R) \times \mathfrak{so}_{2,1}(\mathbb R)$, and $\mathfrak{so}_4(\mathbb R) \simeq \mathfrak{su}_2 \times \mathfrak{su}_2 \simeq \mathfrak{so}_3(\mathbb R) \times \mathfrak{so}_3(\mathbb R)$. On the group level, one has to fine-tune these things, but it always gives something to look for. – Torsten Schoeneberg Jan 09 '21 at 05:58