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.

Ref: https://www.maths.ed.ac.uk/~v1ranick/papers/abs.pdf

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    $\begingroup$ What do you mean by "and the lift"? $\endgroup$ Aug 20, 2021 at 19:34
  • $\begingroup$ Also, what is $e_{2j-1}e_{2j}$? $\endgroup$ Aug 20, 2021 at 19:40
  • $\begingroup$ thanks, see ABS paper and also math.stackexchange.com/a/3318221/955245 $\endgroup$ Aug 20, 2021 at 21:03
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    $\begingroup$ Not really. You already have the diagonal and the vertical maps. If you want the triangle to commute, your only option is to take the horizontal map to be their composite. $\endgroup$ Aug 20, 2021 at 21:34
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    $\begingroup$ Sorry, could you please formulate a self-contained question? Embedding cannot be true or false, it is not a statement, it is a map. The combination of words "is there a valid group homomorphism (...) also the embedding (...)" does not parse, neither grammatically nor mathematically. You already have a homomorphism and you want to know whether this homomorphism is an embedding? Or you want to find any embedding? $\endgroup$ Aug 21, 2021 at 5:49

1 Answer 1


Let $\omega = e_1e_2\dots e_{2n-1}e_{2n}$.

For $n > 1$, the center of $Spin(2n)$ is $Z(Spin(2n)) = \{\pm 1, \pm\omega\}$. Note that $\omega^2 = (-1)^n$, so

$$Z(Spin(2n)) = \begin{cases} \langle -1, \omega\rangle & n\ \text{is even}\\ \langle\omega\rangle & n\ \text{is odd} \end{cases} \cong \begin{cases} \mathbb{Z}/2\oplus\mathbb{Z}/2 & n\ \text{is even}\\ \mathbb{Z}/4 & n\ \text{is odd.}\end{cases}$$

We also have the central subgroup $\langle i\rangle < U(1)$ which is isomorphic to $\mathbb{Z}/4$, so we can form the quotient of $Spin(2n)\times U(1)$ by the central subgroup $\langle(\omega, i)\rangle\cong\mathbb{Z}/4$. Denote the quotient by $G$.

As $(\omega, i)^2 = (-1, -1)$, there is a natural map $\varphi : Spin^c(2n) \to G$ which has kernel $\langle[(\omega, i)]\rangle$, so the composite map $\varphi\circ\tilde{l} : U(n) \to G$ has kernel $\ker(\varphi\circ\tilde{l}) = \tilde{l}^{-1}(\langle[(\omega, i)]\rangle)$. Since $\tilde{l}$ is an embedding, $\varphi\circ\tilde{l}$ is injective if and only if $[(\omega, i)] \not\in \tilde{l}(U(n))$.

Suppose $\tilde{l}(T) = [(\omega, i)]$. Note that the coefficient of $\omega$ in $\prod_{j=1}^n \left( \cos (t_j/2) + \sin (t_j/2) e_{2j-1}e_{2j} \right)$ is $\prod_{j=1}^n\sin(t_j/2)$ which is $\pm 1$ if and only if $\sin(t_j/2) = \pm 1$ (and hence $\cos(t_j/2) = 0$) for every $j$. It follows that $e^{it_j} = -1$ for all $j$ so $T = -I$. As $\tilde{l}(-I) = [(\omega, i^n)] = [(\omega, (-1)^ki)]$, we see that $\varphi\circ\tilde{l}$ is injective if and only if $k$ is odd.

Example: When $k = 0$, the group $G$ is the quotient of $U(1)\times U(1)$ by $\langle(i, i)\rangle \cong \mathbb{Z}/4$, and the map $U(1) \to G$ is given by $e^{i\theta} \mapsto [(e^{i\theta/2}, e^{i\theta/2})]$ which is not injective because $-1 \mapsto [(i, i)] = [(1, 1)]$.

We also could have taken the quotient of $Spin(2n)\times U(1)$ by the central subgroup $\langle(\omega, -i)\rangle \cong \mathbb{Z}/4$. Arguing as above, the induced map on $U(n)$ is injective if and only if $k$ is odd, as before. Note that the quotients are isomorphic as the map $Spin(2n)\times U(1) \to Spin(2n)\times U(1)$, $(g, z) \mapsto (g, z^{-1})$ descends to an isomorphism.

I don't know if there is an embedding $U(n) \to G$ for $k$ even, but if there is, the diagram

\begin{array}{ccc} & & Spin^c(2n)\\ & \nearrow & \downarrow\\ U(n) & \longrightarrow & G \end{array}

doesn't commute. For $k = 0$, we have $G \cong U(1)\times U(1)$, so there is an embedding $U(1) \to G$.

  • $\begingroup$ May I confirm your notation, is this ⟨(𝜔,𝑖)⟩≅ℤ/4? and the ⟨[(𝜔,𝑖)]⟩ is the kernel of the map 𝜑:𝑆𝑝𝑖𝑛𝑐(2𝑛)→𝐺 ? Are these notations ⟨(𝜔,𝑖)⟩ and ⟨[(𝜔,𝑖)]⟩ standard? $\endgroup$ Aug 25, 2021 at 16:17
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    $\begingroup$ Yes, $(\omega, i) \in Spin(2n)\times U(1)$, and $\langle(\omega, i)\rangle$, the subgroup generated by $(\omega, i)$, is isomorphic to $\mathbb{Z}/4$. The image of $(\omega, i)$ under the map $Spin(2n)\times U(1) \to Spin^c(2n)$ is denoted $[(\omega, i)]$. The kernel of the map $\varphi : Spin^c(2n) \to G$ is $\langle[(\omega, i)]\rangle$, which is isomorphic to $\mathbb{Z}/2$. I'd say this notation is pretty standard. $\endgroup$ Aug 25, 2021 at 16:27
  • $\begingroup$ Thanks! I think we have: $$𝜑∘𝑙̃ :𝑈(𝑛)→𝐺$$ $$𝜑:𝑆𝑝𝑖𝑛𝑐(2𝑛)→𝐺$$ $$𝑙̃ :𝑈(𝑛)→𝑆𝑝𝑖𝑛𝑐(2𝑛)$$ to begin with. Could you then clarify this sentence: "Since 𝑙̃ is an embedding (based on the ABS paper), 𝜑∘𝑙̃ is injective if and only if [(𝜔,𝑖)]∉𝑙̃ (𝑈(𝑛))." ? Thanks! $\endgroup$ Aug 25, 2021 at 16:35
  • $\begingroup$ Why not the other way around? "if and only if [(𝜔,𝑖)] $\in$ 𝑙̃ (𝑈(𝑛)) " ? $\endgroup$ Aug 25, 2021 at 16:35
  • $\begingroup$ If $[(\omega, i)] = \tilde{l}(A)$, then $(\varphi\circ\tilde{l})(A) = \varphi(\tilde{l}(A)) = \varphi([(\omega, i)]) = [(1, 1)]$ which is the identity of $G$, so $A \in \ker(\varphi\circ\tilde{l})$, so $\varphi\circ\tilde{l}$ is not injective. $\endgroup$ Aug 25, 2021 at 16:53

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