Lie group embeddings $SO(10) \supset SU(3) \times SU(2) \times U(1)?$

Question

Let $G$ be $SO(10)$ or $Spin(10)$.

Does either of them $G=SO(10)$ or $G=Spin(10)$ contain $SU(3) \times SU(2) \times U(1)$ as a subgroup?

Can one show which of the following embeddings are possible rigorously: $$G\supset SU(3) \times SU(2) \times U(1)?$$ $$G \supset \frac{SU(3) \times SU(2) \times U(1)}{\mathbb{Z}_2}?$$ $$G \supset \frac{SU(3) \times SU(2) \times U(1)}{\mathbb{Z}_3}?$$ $$G \supset \frac{SU(3) \times SU(2) \times U(1)}{\mathbb{Z}_3 \times \mathbb{Z}_2}?$$

Here $\mathbb{Z}_n$ means the cyclic group of order $n$.

it looks to me that the first one is impossible, can we prove it is a no go? for $G=SO(10)$ or $G=Spin(10)$?

Answer 1

I think there is an embedding of $SU(3)\times SU(2)\times U(1)$ into $SO(10)$, but I am not sure whether this is what you want. You simply view $\mathbb C^5$ as $\mathbb R^{10}$ and recall that the standard inner product on $\mathbb R^{10}$ is the real part of the standard Hermitian inner product on $\mathbb C^5$. Splitting $\mathbb C^5=\mathbb C^3\times\mathbb C^2$ you can than define a representation of $SU(3)\times SU(2)\times U(1)$ by $(A,B,\lambda)\cdot (z,w):=(\lambda Az,\lambda Bw)$. Each of theses maps is visibly unitary and thus orthogonal so you obtain a homomorphims to $O(10)$, which has values in $SO(10)$ since $SU(3)\times SU(2)\times U(1)$ is connected. Thus one only has to check injectivity. But if you assume that $(\lambda Az,\lambda Bw)=(z,w)$ for all $z$ and $w$, then looking at the second component, you must have $(B.\lambda)=(\mathbb I,1)$ or $(-\mathbb I,-1)$. But since minus the identity is not contained in $SU(3)$, you conclude that only the identity element of $SU(3)\times SU(2)\times U(1)$ acts as the identity on $\mathbb C^5$.

Edit (addressing your quesiont about SU(5)): As stated above, the homomorphism has values $U(5)\subset SO(10)$. The determinant of the action of $(A,B,\lambda)$ by construction is $\lambda^5\det(A)\det(B)=\lambda^5$. Guided by this, there is an obvious modification which does have values in $SU(5)$, namely $(A,B,\lambda)\cdot (z,w):=(\bar\lambda^2Az,\lambda^3Bz)$, or in matrix language, sending $(A,B,\lambda)$ to the block matrix $\begin{pmatrix} \bar\lambda^2A & 0 \\ 0 & \lambda^3B\end{pmatrix}$. Now this has values in $SU(5)$, but it is not injective any more. I haven't checked very carefully, but I think the kernel should be $\mathbb Z_3\times\mathbb Z_2$ with generators $(\zeta^2 \mathbb I,\mathbb I,\zeta)$ for $\zeta$ a third root of unity and $(\mathbb I,-\mathbb I,-1)$. I don't know whether one can do better.