# The map from SU(4) to SO(6)

We know there is an isomorphism from $$A_3$$ to $$D_3$$, then what is the map from SU(4) to SO(6)?

• $A_3$ and $D_3$ refer to Dynkin diagrams.
– Ted
Commented Nov 5, 2018 at 5:08
• According to one of the very helpful answers here, the Dynkin diagram identity between $A_3$ and $D_3$ on the level of compact real Lie groups rather corresponds to an isomorphism up to isogeny $SU(4) \simeq Spin(6)$. Commented Nov 5, 2018 at 5:26
• Yes, $A_3$ and $D_3$ refer to Dynkin diagrams, what is the relation between SU(4) and SO(6). If SI(4)$\cong$ Spin(6), what is the map? Commented Nov 5, 2018 at 9:15
• Oh, gotcha. I thought you were referring to finite groups for some reason. Commented Nov 5, 2018 at 23:14
• Related: math.stackexchange.com/q/193546/11127 and links therein. Commented Apr 25 at 23:06

The standard representation of $$SU(4)$$ on $$\mathbb{C}^4$$ induces a representation on the second exterior power $$\bigwedge^2\mathbb{C}^4$$ (note it is six-dimensional over $$\mathbb{C}$$). This representation splits, in exactly the same way that $$\bigwedge^2\mathbb{R}^4$$ splits as $$\Lambda^{\pm}\mathbb{R}^4$$ for $$SO(4)$$, into irreducibles $$\Lambda^{\pm}\mathbb{C}^4$$. Now we define $$SU(4)\to SO(6)$$ by the representation $$\Lambda^+\mathbb{C}^4$$ (which is a six-dimension real vector space so it makes sense to say $$SO(6)$$). We could have used $$\Lambda^-$$ instead, but let's leave it at that. An explicit computation shows that it has kernel $$\pm I$$.

More details

For $$g\in SU(4)$$, and $$u\wedge v\in\bigwedge^2\mathbb{C}^4$$, we have

$$g(u\wedge v)=(gu)\wedge(gv)\tag{1}$$

For $$v\in\bigwedge^2\mathbb{C}^4$$, define $$\ast v\in\mathbb{C}^4$$ by

$$u\wedge\ast v=\langle u,v\rangle e_1\wedge e_2\wedge e_3\wedge e_4$$

where $$\langle\cdot,\cdot\rangle$$ is the Hermitian inner product induced by standard Hermitian inner product on $$\mathbb{C}^4$$, with $$e_i\wedge e_j$$ ($$i) an orthonormal basis. Then $$\ast$$ is a $$\mathbb{C}$$-antilinear involution, so we have the $$\pm 1$$ eigenspace of $$\ast$$ as real subrepresentation. Explicitly, we have an orthonormal basis

$$b_{12,34},\bar{b}_{12,34},b_{13,42},\bar{b}_{13,42},b_{14,23},\bar{b}_{14,23}$$

of $$\Lambda^+\mathbb{C}^4$$, where

$$b_{ij,kl}=\frac1{\sqrt{2}}[e_i\wedge e_j+e_k\wedge e_l], \bar{b}_{ij,kl}=\frac{\sqrt{-1}}{\sqrt{2}}[e_i\wedge e_j-e_k\wedge e_l].$$

Clearly $$SU(4)$$ preserves $$\langle\cdot,\cdot\rangle$$ and so maps into $$O(6)$$, and as $$SU(4)$$ is connected the image lies in $$SO(6)$$. You can use equation $$(1)$$ to write out explicitly $$SU(4)\to SO(6)$$ if you want, but it really doesn't do anything more than what is said above.

By complexifying $$\Lambda^+\mathbb{C}^4$$ we can recover the $$SU(4)$$ representation $$\bigwedge^2\mathbb{C}^4$$. Hence the kernel of this $$SU(4)\to SO(6)$$ is precisely the kernel of the representation $$\bigwedge^2\mathbb{C}^4$$, i.e., $$\pm I$$ (since it fixes $$e_1\wedge e_2$$ and $$e_1\wedge e_3$$, it must fix the subspace $$\langle e_4\rangle$$ and hence every 1-dimensional subspace, so is a multiple of identity, and now look at $$e_1\wedge e_2$$ to get only $$\pm I$$). To check this is indeed a covering, we only need to check dimensions (since $$SO(6)$$ is connected):

$$\dim SU(4)=4^2-1=15=\binom{6}{2}=\dim SO(6).$$

• Can you explain the cover map detaily? Commented Nov 5, 2018 at 11:35
• Hopefully all is clear now. Commented Nov 5, 2018 at 23:10
• Yes, thank you very much for your answer. Commented Nov 6, 2018 at 2:20