# Is $SU(n) \subset \text{Spin}(2n)$?

I believe this should be true for the following reason:

The injection $$SU(n) \hookrightarrow SO(2n)$$ induces a map of Lie algebras $$\mathfrak{su}(n) \hookrightarrow \mathfrak{so}(2n) \cong \mathfrak{spin}(2n)$$. This should allow us to exponentiate $$\mathfrak{su}(n)$$ within $$Spin(2n)$$ to obtain $$SU(n)$$ as a subgroup, since $$SU(n)$$ is simply connected (let's say $$n\ge 2$$).

Is that solid reasoning or have I missed something? I am aware that one can do better in low dimensions (such as $$SU(2) \cong Spin(3)$$ ), but I am after the statement for general $$n\ge 2$$.

• See here. Aug 9, 2019 at 10:08
• Danke! I'll turn that into an answer Aug 9, 2019 at 11:51
• I am puzzled, see here math.stackexchange.com/q/3605319/141334 Apr 1, 2020 at 18:26
• 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. Apr 1, 2020 at 18:26

The answer is Yes, $$SU(n) \subset Spin(2n)$$. This is addressed in greater generality by Atiyah, Bott, and Shapiro in the paper Clifford Modules on page 10. I'll reproduce their answer here:

My question can be rephrased as, "Does the homomorphism $$SU(n) \to SO(2n)$$ lift to $$Spin(2n)$$?" ABS show that a homomorphism $$U(n) \to SO(2n)\times U(1)$$ lifts to $$Spin^c(2n)$$ and give an explicit description of the lifting in terms of matrices. As a corollary, the answer to my question is yes.

Here is the homomorphism they wish to lift:

$$l: U(n) \to SO(2n)\times U(1)$$ given by $$T \mapsto j(T) \times \det(T)$$. (Here $$j: U(n) \to SO(2n)$$).

Here is their lift $$\tilde{l}: U(n) \to Spin^c(2n)$$ :

Let $$T \in U(n)$$ be expressed relative to an orthonormal basis $$f_1, \ldots, f_n$$ of $$\mathbb{C}^n$$ by a diagonal matrix with diagonal entries $$e^{it_1}, e^{it_2} , \ldots e^{it_n}$$. Let $$e_1,\ldots,e_{2n}$$ be the corresponding basis of $$\mathbb{R}^{2n}$$, so that $$e_{2j-1} = f_j$$ and $$e_{2j} = i f_j$$. Then the corresponding element of $$Spin^c(2n)$$ is $$\tilde{l}(T) = \prod_{j=1}^n \left( \cos (t_j/2) + \sin (t_j/2) e_{2j-1}e_{2j} \right) \times \exp( i \sum t_j /2).$$

(Let me repeat: this is all taken directly from the above-referenced paper)

To answer my original question, take $$T$$ to be in $$SU(n)$$, i.e. take $$\prod e^{it_j} =1$$. Then $$\exp( i \sum t_j /2) = \pm 1$$, so $$\tilde l (T)$$ is actually in $$Spin(2n)$$.

I would still be grateful if anyone could comment on whether my original reasoning for this fact is valid.

• may you know this math.stackexchange.com/q/3607058/141334 ? Apr 2, 2020 at 21:14
• vote up nice +1 Apr 2, 2020 at 21:14
• If you restrict $\prod e^{it_j} =1$ to $\exp( i \sum t_j /2) = \pm 1$, could it be the restriction from $𝑆𝑝𝑖𝑛^𝑐(2𝑛)$ to $SO(2n) \times \mathbf{Z}_2$ not $Spin(2n)$? Aug 30, 2021 at 16:03