Is $\operatorname{Spin}(n-1)$ a subgroup of ${\rm Spin}(n)$?

Question

Just to double check that this statement, which looks true, is indeed true:

Is $\operatorname{Spin}(n-1)$ always a subgroup of $\operatorname{Spin}(n)$?

Any reference or quick argument would be appreciated.

Thanks!

Answer 1

This is elementary covering space business. Consider the composite $\mathrm{Spin}(n) \to \mathrm{SO}(n) \hookrightarrow \mathrm{SO}(n+1)$ where the first map is the universal double cover. Now lift this up to $\mathrm{Spin}(n+1)$ and note that the resulting lift $\mathrm{Spin}(n) \to \mathrm{Spin}(n+1)$ must be injective (because of how you realize $\mathrm{SO}(n)$ inside $\mathrm{SO}(n+1)$).

This argument applies in the case $n \geq 3$, since we need $\mathrm{Spin}(n)$ to be simply connected to get the lift. Since it is common to take $\mathrm{Spin}(2) = S^1$ and $\mathrm{Spin}(3)=S^3$, it is still true at $n=2$ as we can realize $S^1$ as the equatorial subgroup of $S^3$.