$\newcommand{\GL}{\mathrm{GL}}\newcommand{\SO}{\mathrm{SO}}\newcommand{\SU}{\mathrm{SU}}\newcommand{\Spin}{\mathrm{Spin}}\renewcommand{\O}{\mathrm
O}\newcommand{\R}{\mathbb
R}\newcommand\Z{\mathbb Z}$If $\rho\colon \SU\_8\to\GL\_n(\R)$ is a representation, then we can lift $\rho$ to land in the spin group
$\Spin\_n$ in three steps:


* Since $\SU\_8$ is compact, $\GL\_n(\R)$ admits an $\SU\_8$-invariant product, and the image of
$\rho$ is contained in the subgroup of orthogonal matrices, which is isomorphic to $\O\_n$.
* Since $\SU\_8$ is connected, $\rho\colon\SU\_8\to\O\_n$ lands in the connected component of the
identity in $\O\_n$, which is $\SO\_n$.
* Now we want to lift $\rho$ across the double cover $\Spin\_n\to\SO\_n$. Such a lift exists because
$\SU\_8$ is simply connected.


I want to know whether this is always possible for the quotient $\SU\_8/\{\pm 1\}$. The first two steps are
unchanged, but this group is not simply connected, so a priori there could be a representation
$\rho\colon\SU\_8/\{\pm 1\}\to\SO\_n$ that does not lift to $\Spin\_n$. As additional evidence, the obstruction $w\_2(\rho)$ for lifting
from $\SO\_n$ to $\Spin\_n$ lives in $H^2(B(\SU\_8/\{\pm 1\});\Z/2)\cong\Z/2$, so it could be nonzero in
principle. But on all the examples I've calculated, the obstruction vanishes, and $\rho$ lifts to $\Spin\_n$.


This seems like the sort of thing that may have been worked out already, or that would follow from a clever
application of highest weight theory. But the fact that $\SU\_8/\{\pm 1\}$ isn't simply connected, and that I
want to look at real representations, makes the combinatorics more confusing for me. Is an example of a non-spin
representation of this group known, and if not, how would I go about working this out?



Is there a representation of $\mathrm{SU}_8/\{\pm 1\}$ that doesn't lift to a spin group?
$\def\ZZ{\mathbb{Z}}\def\RR{\mathbb{R}}\def\CC\mathbb{C}\def\SU{\text{SU}}\def\SO{\text{SO}}\def\Spin{\text{Spin}}\def\diag{\text{diag}}\def\Id{\text{Id}}$Such a lift is always possible. We will show:


**Theorem** Let $n \equiv 0 \bmod 8$ be a positive integer and let $\rho : \SU\_n/\{ \pm \Id\_n \} \to \SO\_m$ be a (continuous) representation. Then $\rho$ lifts to a map $\SU\_n/\{ \pm \Id\_n \} \to \Spin\_m$.


We have $\pi\_1(\SU\_n/ \{ \pm \Id\_n \}) \cong \pi\_1(\SO\_m) \cong \ZZ/ 2 \ZZ$ (for $m \geq 3$), and such a lift exists if and only if $\rho\_{\ast} : \pi\_1(\SU\_n/ \{ \pm 1 \}) \longrightarrow \pi\_1(\SO\_m)$ is $0$, so that is what we will be trying to show.




---


**Part one: Tori** Our first task will be to write down some useful tori. Let $S$ be the circle group $\{ z \in \CC : |z| = 1 \}$; we will also write $z$ as $e^{i \theta}$.


Map the interval $[0, 2 \pi]$ to $\SU\_n$ by $\theta \mapsto \diag(e^{i\theta/2}, e^{- i \theta/2}, e^{i\theta/2}, e^{- i \theta/2}, \dots, e^{i\theta/2}, e^{- i \theta/2})$. This is a path from $\Id\_n$ to $- \Id\_n$. In the quotient $\SU\_n/\{ \pm \Id\_n \}$, it becomes a closed loop and a group homorphism $\gamma: S \to \SU\_n/\{ \pm \Id\_n \}$. Moreover, $\gamma\_{\ast}: \pi\_1(S) \to \pi\_1(\SU\_n/\{ \pm \Id\_n \})$ is surjective. So it is enough to show that the induced map $\rho\_{\ast} \circ \gamma\_{\ast}$ from $\pi\_1(S)$ to $\pi\_1(\SO\_m)$ is $0$.


Put $Z(\theta) = \left[ \begin{smallmatrix} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \\ \end{smallmatrix} \right]$. If $m$ is even, let $T$ be the maximal torus in $\SO\_m$ of block diagonal matrices of the form $\diag(Z(\theta\_1), Z(\theta\_2), \ldots, Z(\theta\_{m/2}))$; if $m$ is odd, let $T$ be the block diagonal matrices of the form $\diag(Z(\theta\_1), Z(\theta\_2), \ldots, Z(\theta\_{m/2}), 1)$. We write $\delta$ for the inclusion $T \hookrightarrow \SO\_m$.


Choose coordinates on $\SO\_m$ so that the homomorphism $\rho \circ \gamma : S \to \SO\_m$ factors as $\delta \circ \beta$.


$$\begin{matrix}
S & \overset{\beta}{\longrightarrow} & T \\
\gamma \downarrow && \delta \downarrow \\
\SU\_n/{\pm \Id\_n} & \overset{\rho}{\longrightarrow} & \SO\_m.
\end{matrix}$$
It is well known that $\delta\_{\ast} : \pi\_1(T) \to \pi\_1(\SO\_m)$ is the map $\ZZ^{\lfloor m/2 \rfloor} \to \ZZ/2 \ZZ$ given by $(b\_1, b\_2, \ldots, b\_{\lfloor m/2 \rfloor}) \mapsto \left( \sum b\_i \bmod 2 \right)$. The map $\beta$, being a group homomorphism, must be $\theta \mapsto \diag(Z(b\_1 \theta), Z(b\_2 \theta), \ldots, Z(b\_{\lfloor m/2 \rfloor} \theta))$ (or the same thing with $1$ tagged on the end) for some integers $b\_1$, $b\_2$, ..., $b\_{\lfloor m/2 \rfloor}$. So our goal is to show that $\sum b\_j$ is even.




---


**Part two: Combinatorics** Consider the character of the group homomorphism $\rho \circ \gamma = \delta \circ \beta : S \to SO\_m$; it is a Laurent polynomial in the coordinate $z$, and we will write it $f(z)$.
Explicitly, we have
$$f(z) = \sum\_{j=1}^{\lfloor m/2} \left( z^{b\_j} + z^{-b\_j} \right) + \begin{cases} 0 & m \equiv 0 \bmod 2 \\ 1 & m \equiv 1 \bmod 2 \end{cases}.$$


**Lemma** We have $f(1) - f(-1) \equiv 4 \sum b\_j \bmod 8$.


**Proof** If $b\_j$ is even, then $1^{b\_j} + 1^{-b\_j} = (-1)^{b\_j} + (-1)^{-b\_j}$. If $b\_j$ is odd, then $1^{b\_j} + 1^{-b\_j} = (-1)^{b\_j} + (-1)^{-b\_j}+4$ $\square$


Therefore, our goal is to show that $f(1) \equiv f(-1) \bmod 8$.


Now, recall the representation $\rho : \SU\_n/\{ \pm 1 \} \to \SO\_m$ and restrict it to the torus $D := \{ \diag(z\_1, z\_2, \ldots, z\_n) : z\_j \in S,\ \prod z\_j = 1 \}$ inside $\SU\_n$. The character of $\rho$ is a symmetric polynomial in the $z\_j$ with integer coefficients, call it $g(z\_1, z\_2, \ldots, z\_n)$.


The point $-1$ in $S$ is mapped to $\diag(i, -i, i, -i, \ldots, i, -i)$ in $\SU\_n/(\pm \Id\_n)$. So $f(-1) = g(i, -i, i, -i, \ldots, i, -i)$. Of course, $1 \in S$ maps to $\diag(1,1,\ldots,1)$, so $f(1) = g(1,1,\ldots, 1)$. Thus, we are done if we prove the following lemma:


**Lemma** Let $n \equiv 0 \bmod 8$ and let $g(z\_1, z\_2, \ldots, z\_n)$ be a
symmetric polynomial with integer coefficients. Then
$$g(1,1,\ldots,1) \equiv g(i, -i, i, -i, \ldots, i, -i) \bmod 8.$$


**Proof** By the fundamental theorem of symmetric polynomials, $g$ is a polynomial in the elementary symmetric polynomials $e\_1, e\_2, \dotsc, e\_n$ with $\ZZ$-coefficients. So it is enough to show that
$$e\_k(1,1,\dotsc,1) \equiv e\_k(i, -i, i, -i, \dotsc, i, -i) \bmod 8$$
for $1 \leq k \leq n$. Organizing this into a single generating function, we want to show the polynomial congruence
$$(1+t^2)^{n/2} \equiv (1+t)^n \bmod{8 \ZZ[t]}.$$
To this end, it is enough to show that $(1+t^2)^{4} \equiv (1+t)^8 \bmod{8 \ZZ[t]}$ and then raise both sides to the $n/8$ power. This last condition can be checked by hand; we do indeed have
$$\begin{multline\*}
t^8 + 8 t^7 + 28 t^6 + 56 t^5 + 70 t^4 + 56 t^3 + 28 t^2 + 8 t + 1 \\ \equiv
t^8 + 4 t^6 + 6 t^4 + 4 t^1 + 1 \bmod{8 \ZZ[t]}. \qquad \square\end{multline\*}$$




---


**Remark:** The fact that $\rho$ is a representation of $\SU\_n/(\pm \Id\_n)$ means that $g$ has even degree. The condition that $\rho$ is a real (rather than complex) representation of $\SU\_n$ also imposes nontrivial conditions on $g$. But the lemma is right for all symmetric polynomials.


**Remark:** For the reader who doesn't like computing binomial coefficients, the formula $(1+t)^{2^k} \equiv (1+t^2)^{2^{k-1}} \bmod{2^k \ZZ[t]}$ can also be proved by induction on $k$.


