# Why these two complex representations of $Cl(V)$ are not isomorphic?

Let $(V,g)$ be a complex vector space of dimension $2n+1$ with symmetric bilinear form $g$. There exists a basis $a_1$, $\dots$, $a_n$, $b_1$, $\dots$, $b_n$, $c$ of $V$ such that $g(a_i,a_j) = g(b_i,b_j)=g(a_i,c)=g(b_i,c)=0$, $g(a_i,b_j)= \tfrac 1 2 \delta_{ij}$, $g(c,c)=1$ (just note that the corresponding matrix of $g$ is non-degenerate and symmetric and all such matrices are congruent to each other over $\mathbb C$).

Let $S = \wedge (a_1,\dots,a_n)$. We can define the action of $Cl(V)$ on $S$ by: $$a_i(\omega) = a_i \wedge \omega, \\ b_i(a_i) = 1, \; b_i(a_j) = 0 \; (i \neq j), \; b_i(a_s \wedge \omega) = b_i(a_s) \wedge \omega - a_s \wedge b_i(\omega), \\ c(1) = 1, \; c(a_i) = -a_i, \; c(a_i\wedge\cdots \wedge a_k)=c(a_i)\wedge \cdots \wedge c(a_k).$$ This defines a representation of $Cl(V)$ on $S$ which we call $S_+$. It is irreducible since for any basis elements $a_i \wedge \cdots \wedge a_k$ and $a_s \wedge \cdots \wedge a_l$ there is an element of $Cl(V)$ which sends the first one into the second one.

We can also define an irreducible representation $S_-$ of $Cl(V)$ on $S$ by defining the action of $a_i$ and $b_i$ as above and by putting $$c(1)=-1, \; c(a_i)=a_i, \; c(a_i \wedge a_j) = -a_i \wedge a_j, \; \dots.$$ It is asked in Problem 2.25 at p. 31 of these notes by P. Etingof et al. to show that $S_+$ and $S_-$ are not isomorphic. But if we compute the traces of operators $a_i$, $b_i$, $c$ in $S_+$ and in $S_-$ it turns out that they are the same so that the representations must be isomorphic. Indeed, $a_i$ and $b_i$ act in the same way in $S_+$ and $S_-$ and the trace of $c$ in $S_+$ is $$1 - {n \choose 1}+ {n \choose 2} - \cdots = (1-1)^n = 0,$$ whereas in $S_-$ it is $$-1 + {n \choose 1} - {n \choose 2} + \cdots = -(1-1)^n = 0.$$ Thus, the characters of $S_+$ and $S_-$ are the same and they must be isomorphic. Could you please point out where am I mistaken?

It is not enough to show that the characters of the $a_i$'s, the $b_i$'s and $c$ agree. The $a_i$'s, $b_i$'s and $c$ do not span $Cl(V)$ as a complex vector space.
Consider $b_n\dots b_1a_1 \dots a_n\in Cl(V)$. This is a projection onto the $1$ vector. $c$ acts on $1$ by multiplication by $+1$ if we're in $S_+$, or by multiplication by $-1$ if we're in $S_-$. So the character of $cb_n\dots b_1a_1 \dots a_n$ is $+1$ or $-1$ depending on whether we're in $S_+$, or $-1$ or $S_-$.
• Thank you, yes! Now I see that it is not enough to show this for characters of $a_i$, $b_i$, $c$. The trace of $c b_1 \cdots b_n$ is different in $S_+$ and $S_-$ – Appliqué Mar 4 '17 at 14:02
• I think $cb_1 \dots b_n$ has character zero. But $cb_1 \dots b_n a_n \dots a_1$ should work. (see my edit of the answer) – Kenny Wong Mar 4 '17 at 14:03