Question about the expression of Witten Index I am studying supersymmetry by myself. I do not understand the expression of Witten index, which is ${\rm Tr}(-1)^{F}$. What does it mean by writing $-1$ to the power of an operator $F$? Is this expression related with the parity in $\Bbb{Z}_{2}$-graded algebra?
 A: *

*The idea behind the notation is that the operator $F$ is supposed to count the number of fermions in an expression, i.e. $$[F,A_n]= n A_n$$ if the operator $A_n$ contains $n$ fermions, what that means. 

*Then $$[f(F),A_n]= f(n) A_n$$ for a sufficiently well-behaved function $f:\mathbb{C}\to \mathbb{C}$. 

*In particular, for $f(x)=(-1)^x $, one has
$$[(-1)^F,A_n]= (-1)^n A_n.$$ 

*The operator $(-1)^F$ has eigenvalue $+1$ ($-1$) for Grassmann-even (Grassmann-odd) operators, respectively. 

*The notation $(-1)^F$ is used even if the operator $F$ itself is not well-defined. 
A: This is probably motivated by differential geometry. In this paper, Witten and Bar-Natan denoted the operator $a\rightarrow dx^{i}\wedge a$ on differential forms as a fermion $\psi^{i}$. The opposite contraction is denoted as the canonical conjugate spinor $\chi_{j}$. They satisfy the canonical relations
$$\left\{\psi^{i},\psi^{j}\right\}=\left\{\chi_{i},\chi_{j}\right\}=0,\quad\left\{\psi_{i},\chi_{j}\right\}=g_{ij}.$$
Let $\ast$ be the Hodge star operator, then one finds
$$\ast\psi_{i}\ast=(-1)^{F}\chi^{i},\quad\ast\chi_{j}\ast=\psi_{j}(-1)^{F},$$
where $(-1)^{F}a\equiv(-1)^{q}a$, for $\forall a\in\Omega^{q}$. This $(-1)^{F}$ operator on differential forms is exact the Witten index.
