How to understand $\mathrm{Tr}(-1)^{F}=\mathrm{ker}\, Q-\mathrm{ker}\, Q^{\dagger}$? The formula, $$\mathrm{Tr}(-1)^{F}=\mathrm{ker}\, Q-\mathrm{ker}\, Q^{\dagger}\tag{2.4},$$ appears in Luis Alvarez-Gaume's 'Supersymmetry and the Atiyah-Singer Index Theorem'. If my understand is correct, we can restrict on zero-energy states, which is finite dimensional. In this case $Q$ can be represented as a matrix (actually it is a zero matrix) and $\mathrm{rank}(Q)=\mathrm{rank}(Q^{\dagger})$, which implies $\mathrm{ker}\, Q-\mathrm{ker}\, Q^{\dagger}=0$. Where did I go wrong?
 A: *

*Be aware that Ref. 1 confusingly denotes the operators $$Q:{\cal H}_b\to {\cal H}_f, \qquad Q^{\dagger}:{\cal H}_f\to {\cal H}_b,$$ and their trivial extensions to ${\cal H}={\cal H}_b\oplus {\cal H}_f$ with the same notation. Ref. 2 is better at keeping the distinction. In this answer, we will not use the extensions.


*Let ${\cal H}_0={\cal H}_{0,b}\oplus {\cal H}_{0,f}$ denote the restriction of the Hilbert space ${\cal H}$ to zero-energy, which we split into bosonic and fermionic states. Then the restrictions $Q| :{\cal H}_{0,b}\to {\cal H}_{0,f}$ and $Q^{\dagger}|:{\cal H}_{0,f}\to {\cal H}_{0,b}$ are indeed just zero-maps, finite dimensional, and
$${\rm rank}(Q|)~=~0~=~{\rm rank}(Q^{\dagger}|). $$
However the kernels can be different because of the rectangular form of the matrices, so the index
$$\begin{align} &\dim {\rm ker}(Q)-\dim{\rm ker}(Q^{\dagger})\cr
~=~&\dim{\cal H}_{0,b} - \dim{\cal H}_{0,f}\cr 
~=~&{\rm tr}(-1)^F \end{align}\tag{2.4} $$
does not have to vanish.
References:

*

*L. Alvarez-Gaume, Supersymmetry and the Atiyah-Singer index theorem, Commun.Math.Phys. 90 (1983) 161.


*E. Witten, Constraints on supersymmetry breaking, Nucl. Phys B202 (1982) 253.
