Gts common line versus independent pair framing

An independent auxiliary source has its own scalar gauge action and an independent arrow to the old pair vertex. Its arrow to the old object need not equal the composite through that vertex. In contrast, a tuple of $m\ge 1$ vectors with one common source line has one shared source gauge. In the explicitly completed associative algebra below, conjugating the auxiliary exponential gives the nested-commutator expansion with coefficients $1\mathrm{/}n!$. This formal identity supplies no geometric uncapped seed or wall-crossing theorem.

Bojko, Wall-crossing for Calabi–Yau fourfolds: framework, tools, and applications, Assumptions 5.1 and 5.10; Definition 5.3; §6.2; §6.3; Assumption 6.9; Remark 6.10; Example 6.11; Theorem 6.12; Lemma 6.13; Equation (6.14); §7.2.

Bojko, Kuhn, Liu, and Thimm, Wall-crossing for equivariant DT4 invariants, §2.2.8, §3.2, Assumption 4.2.4, and Theorem 1.2.1; used only under their stated hypotheses.

Take an old pair source $p$ with arrow $s:p\to v$ and a new auxiliary source $q$ with arrows $u:q\to p$ and $w:q\to v$:

For one-dimensional spaces at all vertices, $w$ and the composite $s\circ u$ have the same gauge weight but are distinct coordinates. A tuple of $m$ vectors sharing one common line instead has a single source gauge scalar for all $m$ components.

For the formal calculation fix $d\ge 1$ and old charges in ${\mathbb{N}}^d$. Start with the free unital associative $\mathbb{C}$-algebra on symbols $z_{\beta }$, one for each nonzero $\beta \in {\mathbb{N}}^d$, and $\delta$. Give $z_{\beta }$ bidegree $(\beta ,0)$ and $\delta$ bidegree $(0,1)$, the second component counting auxiliary symbols. Quotient by the two-sided ideal spanned by all words containing at least two $\delta$'s, and complete in old charge: the resulting algebra ${\widehat{U}}_{\le 1}$ is the product of its homogeneous word spaces of bidegrees $(\alpha ,0)$ and $(\alpha ,1)$. Multiplication is concatenation followed by the quotient, and $[x,y]=xy-yx$. For each fixed $\alpha$ there are only finitely many decompositions into nonzero charges, and at most $\mathrm{\mid }\alpha \mathrm{\mid }={\sum }_i{\alpha }_i$ old letters; hence all coefficients below are finite sums. Put $A={\sum }_{\beta \mathrm{\ne }0}{z}_{\beta }$, and let $\tau$ be a central formal scalar, independent of the arrow $s$. Work in ${\widehat{U}}_{\le 1}[\tau ]\mathrm{/}({\tau }^2)$; $e^{\pm A}={\sum }_{n\ge 0}(\pm A{)}^n\mathrm{/}n!$ are defined charge by charge.

lem-gts-original-k-independent-pair-master-quadratic-model and the separate actual uncapped-seed interface name the held construction interfaces discussed by this diagnostic. Their conclusions are not assumed or used to establish the calculations below; this example does not prove those interfaces.