The Lie algebra of $\Spin(V)$

Recall the linear isomorphism $Q\: \La^\8 V \to \Cl(V, g)$, inverse to $\sg\: a \mapsto c(a)1$. Write

$$b = Q(u \w v) = \half(uv - vu) = uv + g(u, v) 1 \in \Cl^0(V, g).$$

Note in passing that $b^! = \half(vu - uv) = - b$.

Although the algebra $\Cl(V, g)$ is not $\bZ$-graded, it is $\bZ$-filtered: we may write $\Cl^{\leq k}(V,g)$ to denote the vector subspace generated by products of at most $k$ vectors from $V$. With that notation, the subspace $Q(\La^2 V)$ may also be described as the set of all even elements $b \in \Cl^{\leq 2}(V, g)$ with $\tau(b) = 0$.

For $x \in V$, we compute

$$[b, x]= [uv, x] = uvx + uxv - uxv - xuv = 2g(v, x)u - 2g(u, x)v \underline{\mathstrut\in V},$$

so $\ad b\: V\to V$. Also

$$[b, b']= \half [b, u'v' - v'u'] = \half[b, u']v' + \half u'[b, v'] - \half [b, v']u' - \half v'[b, u']$$

so that $[b, b'] \in \Cl^{\leq 2}(V, g)$ with $\tau([b, b']) = 0$. Hence $[b, b']\in Q(\La^2 V)$, and this is a Lie algebra. Next,

$$g(y, [b, x]) = 2g(v, x) g(y, u) - 2g(u, x) g(y, v) = -g([b, y], x),$$

so that $\ad b$ is skewsymmetric: thus $\ad b \in \sol(V)$. By the Jacobi identity,

$$[[b, b'], x] = [b, [b', x]] - [b', [b, x]] \words{for all} x \in V,$$

and so $\ad([b, b']) = [\ad b, \ad b']$. Thus, $\ad\: Q(\La^2 V) \to \sol(V)$ is a Lie algebra homomorphism.

If $\ad b = 0$, so that $[b, x] = 0$ for all $x \in V$, then $b\in Z(\Cl^0(V)) \isom \bC 1$. But $\tau(b) = 0$ then implies $b = 0$, so $\ad$ is injective. Since $\dim\La^2 V = n(n-1)/2 = \dim\sol(V)$, we see that $\ad\: Q(\La^2 V) \to \sol(V)$ is a Lie algebra isomorphism.

There is an important formula for the inverse of $\ad$. For $A \in \sol(V)$, define

$$\dot{\mu}(A) = \frac{1}{4} \sum_{j,k=1}^n g(e_j, Ae_k) e_j e_k = \frac{1}{2} \sum_{j<k} g(e_j, Ae_k) e_j e_k.$$ (5)

Since $\tau(\dot{\mu}(A)) = 0$, we get $\dot{\mu}(A)\in Q(\La^2 V)$. Also
$$\begin{align*}[\dot{\mu}(A), e_r] &= \frac{1}{4}\sum_{j,k} g(e_j, A e_k) \bigl( ... ...} \sum_k g(e_r, Ae_k) e_k = \sum_j g(e_j, Ae_r) e_j \\ &= Ae_r, \end{align*}$$
where we have used the anticommutator notation $\{X,Y\} := XY + YX$. Hence $\ad(\dot{\mu}(A)) = A \in \sol(V)$.

Now consider $u = \exp b := 1 + \sum_{k\geq 1} \frac{1}{k!} b^k \in \Cl^0(V, g)$ for $b \in Q(\La^2 V)$. Then $u^*u = u^!u = \exp(-b)\exp b = 1$ since $b^! = -b$. Also, $u$ is unitary and even, and if $x \in V$ then
$$\begin{align*} uxu^{-1} &= \sum_{k,l\geq 0} \frac{1}{k!l!} b^k x (-b)^l \\ &... ...{r\geq 0} \frac{1}{r!} (\ad b)^r(x) \underline{\mathstrut\in V}, \end{align*}$$
and thus $u = \exp(b)$ lies in $\Spin(V)$. When $b = \dot{\mu}(A)$, we get $\phi(\exp(b)) = \exp(\ad b) = \exp(A)$, and it is known that $\exp\: \sol(V) \to \SO(V)$ is surjective (a property of compact connected matrix groups).

Now $\exp(Q(\La^2 V))$ is a subset of $\Spin(V)$ covering all of $\SO(V)$. If we can show that $-1 = \exp c$ for some $c$, then $- \exp b = (\exp b)(\exp c) = \exp(b + c)$, provided that $c,b$ commute. If $b = \dot{\mu}(A)$, we can express the skewsymmetric matrix $A$ as a direct sum of $2 \x 2$ skewsymmetric blocks in a suitable orthonormal basis:

$$A = \begin{pmatrix} 0 & * \ * & 0 \\ && 0 & * \ && * & 0 ... ... &&&&& 0 & * \ &&&&& * & 0 \\ &&&&&&& \ddots \end{pmatrix} .$$

That is, we can choose the (oriented) orthonormal basis $\{e_1, \dots, e_n\}$ so that

$$b = \half g(e_1, Ae_2) e_1e_2 + \half g(e_3, Ae_4) e_3e_4 +\cdots+ \half g(e_{2r-1}, Ae_{2r}) e_{2r-1}e_{2r}$$

with $r \leq m$. Now this particular $e_1e_2$ commutes with $b$: $(e_1e_2)b = b(e_1e_2)$; take $c := \pi e_1e_2$. Then $\exp c = \exp(\pi e_1e_2) = \cos\pi + \sin\pi e_1e_2 = -1$. We have shown that $\exp\: Q(\La^2 V) \to \Spin(V)$ is surjective.

Note that $t \mapsto \exp(te_1 e_2)$, for $0 \leq t \leq \pi$, is a path in $\Spin(V)$ from $+1$ to $-1$. Since $\pi_1(\SO(V)) \isom \bZ_2$ for $n \geq 3$, the double covering $\Spin(V) \to \SO(V)$ is nontrivial. We get an important consequence.