$\Spinc$ and $\Spin$ groups

Let $v$ be a unit vector, $g(v, v) = 1$. Then $v^2 = 1$ in $\Cl(V, g)$, so $v = v^*$ and $v^*v = vv^* = 1$ in $\bCl(V)$. If $w = \la v \in V^\bC$ with $\vert\la\vert = 1$, then $ww^* = w^*w = 1$ in $\bCl(V)$ also. Now
$$\begin{align*} \braket{wa}{wb} &= \tau(a^*w^*wb) = \tau(a^*b) = \braket{a}{b}, \... ...w} &= \tau(w^*a^*bw) = \tau(ww^*a^*b) = \tau(a^*b) = \braket{a}{b}, \end{align*}$$
so $a \mapsto wa$, $a\mapsto aw$ are unitary operators in $\sL(\bCl(V))$.

$[\![$Exercise: Conversely, if $u \in \bCl(V)$ and $a \mapsto ua$ and $a \mapsto au$ are both unitary, then $u^*u = uu^* = 1$.$]\!]$

If $v, x\in V$, with $g(v, v) = 1$, then

$$-vxv^{-1} = -vxv = (xv - 2g(v,x))v = x - 2g(v,x)v \underline{\mathstrut\in V}.$$

This is a reflection of $x$ in the hyperplane orthogonal to $v$. For $w = \la v$, $\vert\la\vert = 1$ we also get $- wxw^{-1} = - \la \bar\la vxv^{-1} = -vxv^{-1}$, which is the same as above. If $a = w_1\dots w_r$ is a product of unit vectors in $V^\bC$, then

$$\chi(a) x a^{-1} = (-1)^r w_1 \dots w_r x w_r^{-1} \dots w_1^{-1}$$

is a product of $r$ reflections of $x \in V$. If $r = 2k$ is even, and $a = w_1 \dots w_{2k}$, then $axa^{-1} \in V$ after $k$ rotations. Thus $\phi(a) \: x \mapsto axa^{-1}$ lies in $\SO(V) = \SO(V, g)$.

The inverse of $u = w_1 \dots w_{2k}$ is $u^{-1} = u^* = \bar w_{2k} \dots \bar w_1$.

Suppose $u \in \ker\phi$, which means that $uxu^{-1} = x$ for all $x \in V$. Thus $\ker\phi \subset Z(\bCl(V))$ for $n$ even, and $\ker\phi \subset Z(\Cl^0(V))$ for $n$ odd; in both cases, $u$ lies in $\bC 1$. It follows that $\ker\phi \isom \set{\la \in \bC : \vert\la\vert = 1} = \bT = \rU(1)$. Therefore, there is a short exact sequence (SES) of groups:

$$1 \to \bT \to \Spinc(V) \xrightarrow{\phi} \SO(V) \to 1.$$ (3)

If $u = w_1\dots w_{2k} \in \Spinc(V)$ with $w_j = \la_j v_j$ where $\la_j \in \bT$ and $v_j \in V$, then $u^! = w_{2k}\dots w_1$, and $u^!u = \la_1^2 \la_2^2 \dots \la_n^2 \in \bT$. Thus, $u^!u$ is central, so $(u_1u_2)^! u_1u_2 = u_2^!u_1^!u_1u_2 = u_1^!u_1 u_2^!u_2$, so that $u \mapsto u^!u$ is a homomorphism $\nu\: \Spinc(V) \to \bT$, which restricts to $\bT \subset \Spinc(V)$ as $\la \mapsto \la^2$. The combined $(\phi,\nu)\: \Spinc(V) \to \SO(V)\times \bT$ is a homomorphism with kernel $\{\pm 1\}$.

Indeed $\Spin(V)$ is included in (the even part of) the real Clifford algebra $\Cl^0(V, g)$:

$$u^*u = 1, u^!u = 1 \implies u^* = u^! \implies \bar u = u \implies u \in \Cl^0(V, g).$$

The SES () now becomes

$$1 \to \{\pm 1\} \to \Spin(V) \xrightarrow{\phi} \SO(V) \to 1,$$ (4)

so that $\phi$ is a double covering of $\SO(V)$. Furthermore, $\Spinc(V) \isom \Spin(V) \x_{\bZ_2} \bT$.