Mirror Symmetry p.188
Eq. 10.109 states that $$H \left\vert \alpha\right> = 0 \Longleftrightarrow Q \left\vert\alpha\right> = \overline{Q} \left\vert\alpha\right> =0. \tag{10.109}$$ I dont see why this is true, should not just $$Q\overline{Q}\left\vert \alpha\right> = - \overline{Q} Q \left\vert \alpha\right>$$ be true?
The book states on top of p. 188 that $\overline{Q}$ is the Hermitian adjoint operator $Q^{\dagger}$. Recall that a supernumber consists of body and soul. If $$0~=~2\langle \alpha|H|\alpha \rangle ~\equiv~ \langle \alpha|Q^{\dagger}Q|\alpha \rangle+\langle \alpha|QQ^{\dagger}|\alpha \rangle~\equiv~\underbrace{||Q|\alpha \rangle||^2}_{\geq 0}+\underbrace{||Q^{\dagger}|\alpha \rangle||^2}_{\geq 0},$$ then we must have both $Q|\alpha \rangle = 0$ and $Q^{\dagger}|\alpha \rangle = 0$.
Suppose that $H = A^\dagger A$ for some matrix $A$ and assume that $H | v \rangle = 0$. Then $$\langle v | H | v \rangle = 0.$$ So the norm of the vector $| v' \rangle \equiv A | v \rangle$ vanishes. But in a Hilbert space$\ldots$ (finish the argument yourself).
