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

Question

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

Answer 1

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 $e_i\wedge e_j$ ($i<j$) 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):

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

Answer 2

Following the answer of @user10354138 I implemented a concrete map from SU(4) to SO(6) which I tried to write as symmetrically as possible.