Toric checks decompose into joint syndrome eigenspaces

A product of retained checks with identity binary support must have zero X support and zero Z support separately. By [F3], this forces the selected star subset and plaquette subset both to be empty, so the retained checks are independent. By [F1] and [F2] they are commuting Hermitian involutions. For a retained check $g_a$ and sign $s_a$, put $p_{a,s_a}=(I+(-1{)}^{s_a}{g}_a)\mathrm{/}2$. Direct expansion using $g_a^{\ast }=g_a$ and $g_a^2=I$ gives $p_{a,s_a}^{\ast }=p_{a,s_a}$ and $p_{a,s_a}^2=p_{a,s_a}$. The checks commute, so their factors commute; hence $P_s$ is a self-adjoint idempotent, thus an orthogonal projection.

If $s\mathrm{\ne }t$, choose an index $a$ with $s_a\mathrm{\ne }{t}_a$. Then $p_{a,s_a}{p}_{a,t_a}=(I-g_a^2)\mathrm{/}4=0$, so $P_s{P}_t=0$. Summing over all sign strings and distributing the finite product gives ${\sum }_s{P}_s={\prod }_a(p_{a,0}+p_{a,1})=I$. Thus the ranges are pairwise orthogonal and their direct sum is the whole Hilbert space; zero projections are allowed.

By [F2], the omitted star is the product of all retained stars, so on $\mathrm{ran}{P}_s$ its eigenvalue is the product of their signs; the product of the eigenvalues over all stars is therefore $+1$. A finite list of signs in $\{+1,-1\}$ has product $+1$ exactly when it contains an even number of minus signs. The same argument for the omitted plaquette gives even parity for the negative plaquette checks. By [F4], these are the e and m defect parities. The projectors therefore decompose the register into the stated full-check syndrome sectors.