Embedd SU(n) into an enlarged twisted Spin(2n) in terms of Lie groups precisely

It is standard practice to show $$ SU(n) \subset SO(2n). $$

However, some post Is $SU(n) \subset \text{Spin}(2n)$? suggests that $\DeclareMathOperator\Spin{Spin}$ $$ SU(n) \subset \Spin(2n) (?). $$

I am very puzzled, because $1 \to \mathbb{Z}/2 \to \Spin(2n)\to SO(2n) \to 1$, so $SO(2n) \not \subset \Spin(2n) $. The $ SO(2n) $ is only a quotient group not a normal subgroup.

• Is this answer $SU(n) \subset \Spin(2n)$ true of false, Is $SU(n) \subset \text{Spin}(2n)$?? I thought $SU(n) \not\subset \Spin(2n)$

• However, is it possible to show that $$ SU(n) \times U(1) \subset \frac{\Spin(2n) \times \Spin(2)}{\mathbb{Z}/2}= \frac{\Spin(2n) \times U(1)}{\mathbb{Z}/2}? $$

$$ SU(n) \times SU(2) \subset \frac{\Spin(2n) \times \Spin(3)}{\mathbb{Z}/2}=\frac{\Spin(2n) \times SU(2)}{\mathbb{Z}/2}? $$ precisely?

Solution 1:

Since $SU(n)$ is simply connected, any map $SU(n)\rightarrow SO(k)$ lifts to a map $SU(n)\rightarrow Spin(k)$. Further, there is a unique lift mapping the identity in $SU(n)$ to the identity in $Spin(k)$.

Apply this to the inclusion map $i:SU(n)\rightarrow SO(2n)$. Note that $i$ is injective, so any lift must also be injective, so there are always copies of $SU(n)$ in $Spin(2n)$