Irreducible representations of $\bCl(V)$

We continue to suppose that $n = 2m$ is even.

This is a complex Hilbert space of dimension $2^m$. Choose and fix a unit vector $\Om \in \La^0 W_J$: it is unique up to a factor $\la \in \bT$. For $w \in W_J$ (so that $\bar w \in \overline{W}_J = W_J^\perp$), we write
$$\begin{align*} \eps(w) : z_1 \wyw z_k & \mapsto w \w z_1 \wyw z_k, \\ \iota(\ba... ...1}^k (-1)^{k-1} \bbraket{w}{z_j} z_1 \wyw \widehat{z_j} \wyw z_k. \end{align*}$$
For $v\in V$, write $w = \half(v - iJv) = P_J v \in W_J$ and define

$$c_J(v) := \eps(w) + \iota(\bar w) = \eps(P_Jv) + \iota(P_{-J}v).$$

Then

$$c_J^2(v) := \bbraket{w}{w} 1 = \braket{v}{v}_J 1 = g(v,v) 1,$$

so that $c_J\: V \to \End_\bC(\sF_J V) \equiv \sL(\sF_J V)$. That is to say, $c_J$ is a representation of $\bCl(V)$ on the Hilbert space $\sF_J V$.

Note that we complexify the representation of $\Cl(V, g)$, given by universality. One can check that

$$c_J(w) = \eps(w) \word{if} w \in W_j; \quad c_J(\bar z) = \iota(\bar z) \word{if} \bar z \in \overline{W}_J.$$

From () and the properties of determinants, it is easy to check that the operators $\eps(w)$ and $\iota(\bar w)$ are adjoint to one another, that is, $\eps(w)^\7 = \iota(\bar{w})$ for $w \in W_J$; in particular, $c_J(v)^\7 = c_J(v)$ for $v\in V$. (This is a consequence of our choice of $g$ to have positive definite signature: were we to have taken $g$ to be negative definite, as in done in many books, then the operators $c_J(v)$ would have been skewadjoint.) More generally, we get $c_J(a)^\7 = c_J(a^*)$ for $a \in \bCl(V)$: we say that $c_J$ is a selfadjoint representation of the $*$-algebra $\bCl(V)$ on the Fock space $\sF_J(V)$.

Now, if $T \in \sL(\sF_J(V))$ commutes with $c_J(V^\bC)$, then in particular $\iota(\bar z)T\Om = T\iota(\bar z)\Om = T(0) = 0$ for $\bar z \in \overline{W}_J$. Therefore $T\Om \in \La^0 W_J$, i.e., $T\Om = t\Om$ for some $t \in \bC$. Now

$$T(w_1 \wyw w_k) = T \eps(w_1) \dots \eps(w_k) \Om = \eps(w_1) \dots \eps(w_k) T \Om = t w_1 \wyw w_k$$

for $w_1,\dots,w_k \in W_J$. Thus $T = t 1 \in \sL(\La^\8 W_J)$. By Schur's lemma, the representation $c_J$ is irreducible.

Suppose $K \in \sJ(V,g)$ with $K = hJh^{-1}$ for $h \in \OO(2m)$. Then $hJ = Kh$, $h P_{\pm J} = P_{\pm K} h$, and so $c_K(hv) = (\La^\8 h) c_J(v)$. By universality again, we get $c_K \circ \La^\8 h = \La^\8 h \circ c_J$, so that the irreducible representations $c_K$ and $c_J$ are equivalent.

The Fock space is $\bZ_2$-graded as $\La^\even W_J \oplus \La^\odd W_J$. ¿What operator determines its $\bZ_2$-grading? In fact, this operator is $c_J(\ga)$. To see that, write $\ga = (-1)^m e_1 e_2\dots e_{2m}$, where $e_{2j} = Je_{2j-1}$ for $j = 1,\dots,m$. If $z_1 := P_J e_1 = \half(e_1 - ie_2)$ we get

$$\bar z_1 z_1 - z_1 \bar z_1 = \quarter(e_1 + ie_2)(e_1 - ie_2) - \quarter(e_1 - ie_2)(e_1 + ie_2) = - e_1e_2.$$

With $z_j := P_J e_{2j-1} = \half(e_{2j-1} - ie_{2j})$, this gives

$$\ga = (\bar z_1 z_1 - z_1\bar z_1)\dots (\bar z_m z_m - z_m\bar z_m) \word{in} \bCl^0(V).$$

Now $c_J(\bar z_j z_j - z_j\bar z_j) = \iota(\bar z_j) \eps(z_j) - \eps(z_j) \iota(\bar z_j)$ is the operator

$$z_{k_1} \wyw z_{k_r} \longmapsto \begin{cases} - z_{k_1} \wy... ...\\ + z_{k_1} \wyw z_{k_r} & \text{if } j \notin K. \end{cases}$$

Thus $c_J(\ga)$ acts as $(-1)^k$ on $\La^k W_J$: this is indeed the $\bZ_2$-grading operator.

Finally, the odd case is treated as follows. Let $U := \bR\text{-}\spn\{e_1,\dots,e_{2m}\} \leq V$. Then $\bCl(U) \isom \bCl^0(V)$ via $u \mapsto i u e_{em+1}$, extended to $\bCl(U)$. Now $\sF_J(U)$ is an irreducible $\bCl^0(V)$-module, while $Z(\bCl(V)) = \bC 1 \oplus \bC\ga$. Since $\ga^2 = 1$, we can extend the action of $\bCl^0(V)$ on $\sF_J(U)$ to the full $\bCl(V)$ by setting either $c_J(\ga) := +1$ or $c'_J(\ga) := -1$ on $\sF_J(U)$.

These representations $c_J$, $c'_J$ are inequivalent, since $T(1) = (-1)T$ is not possible unless $T = 0$, using Schur's lemma again. Thus $\bCl(V)$ has two irreducible Fock representations of dimension $2^m$ in the odd case.