Atiyah, Bott, and Shapiro paper on Clifford Modules around page 10 shows two facts.

## 1 - There is a lift $U(n) \to Spin^c(2n)$ from $U(n) \to SO(2n)\times U(1)$. Also an embedding (injective group homomorphism) $ U(n) \subset Spin^c(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.

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 $$ \tilde{l}(T) = \prod_{j=1}^n \left( \cos (t_j/2) + \sin (t_j/2) e_{2j-1}e_{2j} \right) \times \exp( i \sum t_j /2).$$

## 2 - There is a lift $SU(n) \to Spin(2n)$ from $SU(n) \to SO(2n)$. Also an embedding (injective group homomorphism) $ SU(n) \subset Spin(2n) $:

Another way to say is this valid fact: "Does the homomorphism $SU(n) \to SO(2n)$ lift to $SU(n) \to Spin(2n)$?"

We can 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 $Spin(2n)$.

## My questions

The above we had shown $ U(n) \subset Spin^c(2n) = \frac{Spin(2n) \times U(1)}{\mathbf{Z}/2}$. However, when $n=2k+1$, the $Spin^c(2n)$ has a ${\mathbf{Z}/4}$ center. So the $Spin(2n)$ and $U(1)$ can share a common normal subgroup $\mathbf{Z}/4$, more than just a $\mathbf{Z}/2$. I want to prove or disprove the following fact

When $n=2k+1$, is there any valid group homomorphism $$U(n) \to \frac{Spin(2n) \times U(1)}{\mathbf{Z}/4}:=\frac{Spin(4k+2) \times U(1)}{\mathbf{Z}/4} \tag{1}$$ that is also the embedding $ U(n) \subset \frac{Spin(2n) \times U(1)}{\mathbf{Z}/4}?$ Namely, $$ U(2k+1)=\frac{SU(2k+1) \times U(1)}{\mathbf{Z}/(2k+1)} \subset \frac{Spin(4k+2) \times U(1)}{\mathbf{Z}/4}? \tag{2}$$

p.s. If this relation does not hold for general $n=2k+1$, it will be great to know whether certain $n=3,5,7,\dots$, my relations eq.(1) and eq.(2) still hold.

8more comments