Completed Hall wall crossing for finite acyclic quivers

Fix distinct phases ${\varphi }_1>\cdots >{\varphi }_r$. Iterating the defining formula $(f\star g)(M)={\sum }_{U\subseteq M}f(U)g(M\mathrm{/}U)$ from [F1], with left-associated products, shows that the coefficient of ${\delta }_M$ in $S_Z({\varphi }_1)\star \cdots \star S_Z({\varphi }_r)$ is the number of weak flags $0=M_0\subseteq M_1\subseteq \cdots \subseteq M_r=M$ for which every nonzero factor $M_j\mathrm{/}{M}_{j-1}$ is semistable of phase ${\varphi }_j$. The induction step chooses $M_{r-1}\subseteq M$ in the last binary product and then uses the previous flag inside $M_{r-1}$; zero quotients exactly represent skipped phases.

For a fixed dimension vector $\mathbf{d}$, any contributing nonzero factor has dimension vector $\mathbf{e}$ with $0<\mathbf{e}\le \mathbf{d}$. There are finitely many such vectors and therefore finitely many possible phases. A contributing flag has at most $\mathrm{\mid }\mathbf{d}\mathrm{\mid }$ nonzero factors. Thus the ordered product over all phases is a finite computation in each degree and defines an element of ${\widehat{H}}_Q$.

By [F2], for every $M$ exactly one flag counted by the preceding flag-count argument has strictly decreasing semistable factors: its unique HN filtration. Therefore the coefficient of each ${\delta }_M$ in the ordered product is $1$, which is the coefficient of ${\delta }_M$ in $A_{\mathrm{a}\mathrm{l}\mathrm{l}}$. The zero object contributes the common degree-zero unit. This proves the Hall identity.

Apply $I$ and $J$ from [F3] to the degreewise truncations of the identity in step 2.1. Their continuity shows that the limits preserve the identity, giving both equalities of completed Euler and skew ordered products for $Z_{-}$ and $Z_{+}$.

Write each integrated phase factor as $1+g_{\varphi }$, where every term of $g_{\varphi }$ has positive total dimension. Then $(1+g_{\varphi }{)}^{-1}=1-g_{\varphi }+g_{\varphi }^2-\cdots$, since for any fixed degree only finitely many powers contribute. By step 1.2 only finitely many phases affect a fixed degree, so the ordered product $P_Z$ and the reverse-ordered product of these inverses both converge degreewise. Their finite truncations are mutual inverses, and continuity of multiplication gives $P_Z{P}_Z^{-1}=P_Z^{-1}{P}_Z=1$. Thus $a\mapsto P_{Z}a{P}_Z^{-1}$ is a continuous automorphism with inverse $a\mapsto P_Z^{-1}a{P}_Z$. Step 3.1 gives the same $P_Z$ for $Z_{-}$ and $Z_{+}$, so these conjugations agree. All limits are formal in the degree completion.