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 $n\ge 2$.

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

1 Answer 1


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.

  • $\begingroup$ may you know this math.stackexchange.com/q/3607058/141334 ? $\endgroup$ Apr 2, 2020 at 21:14
  • $\begingroup$ vote up nice +1 $\endgroup$ Apr 2, 2020 at 21:14
  • $\begingroup$ 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 $Spin(2n)$? $\endgroup$ Aug 30, 2021 at 16:03

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