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 $n\ge 2$.  
 A: 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.
