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The four entries and Koszul signs in a two-term tensor bicomplex

Example

Fix a commutative ring $k$ and unital graded $k$ -algebras $B, A, C$ . Let $F$ be a two-term cochain complex of graded $(B, A)$ -bimodules $F^{0} \overset{u}{\to} F^{1}$ , and let $X$ be a two-term cochain complex of graded $(A, C)$ -bimodules $X^{0} \overset{v}{\to} X^{1}$ . Assume $u$ is right $A$ -linear and $v$ is left $A$ -linear, and both preserve internal degree. The four summands of the signed total tensor complex $T = F \otimes_{A} X$ are

T^{0} = F^{0} \otimes_{A} X^{0}, T^{1} = (F^{1} \otimes_{A} X^{0}) \oplus (F^{0} \otimes_{A} X^{1}), T^{2} = F^{1} \otimes_{A} X^{1},

with $T^{n} = 0$ for $n \notin {0, 1, 2}$ . In the indicated order of the middle summands, its differentials are

d_{T}^{0} = (\begin{array}{c} u \otimes_{A} 1 \\ 1 \otimes_{A} v \end{array}), d_{T}^{1} = (\begin{array}{cc} - (1 \otimes_{A} v) & u \otimes_{A} 1 \end{array}) .

The minus sign occurs on the $F^{1} \otimes_{A} X^{0}$ summand because its first cochain degree is $1$ . The two composites from $F^{0} \otimes_{A} X^{0}$ to $F^{1} \otimes_{A} X^{1}$ cancel.

For a concrete instance, take $A = B = C = Z$ concentrated in internal degree zero, $F = (Z \overset{2}{\to} Z)$ , and $X = (Z \overset{3}{\to} Z)$ . Then

T^{0} = Z, T^{1} = Z^{2}, T^{2} = Z, d_{T}^{0} = (\begin{array}{c} 2 \\ 3 \end{array}), d_{T}^{1} = (\begin{array}{cc} - 3 & 2 \end{array}),

so $d_{T}^{1} d_{T}^{0} = - 3 \cdot 2 + 2 \cdot 3 = - 6 + 6 = 0$ .

Facts & Assumptions

Given: Two-term cochain complexes of graded bimodules and degree-zero bimodule-linear differentials $u$ and $v$ .

[L1]

The total degree is $p + q$ and the signed differential is $d (f \otimes x) = d_{F} (f) \otimes x + (- 1)^{p} f \otimes d_{X} (x)$ for $f \in F^{p}$ (Bounded graded bimodule complexes and signed tensor totalization).

[L2]

The total differential descends to the balanced tensor, preserves internal degree, commutes with outer actions, and tensoring bimodule chain maps gives chain maps (Bimodule tensor totalization respects differentials and homotopies).

Verification

Proof technique: expand the signed total differential on the four summands and specialize the resulting matrices over $Z$ .

1.1L1algebra

Since the only nonzero pairs $(p, q)$ have $p, q \in {0, 1}$ , their total degrees are $0, 1, 1, 2$ , giving exactly the four displayed summands. Formula [L1] sends $f^{0} \otimes x^{0}$ to $(u (f) \otimes x, f \otimes v (x))$ , which is the displayed $d_{T}^{0}$ .

1.2L1L2algebra

On $F^{1} \otimes_{A} X^{0}$ the second-factor term in [L1] has sign $(- 1)^{1} = - 1$ , while on $F^{0} \otimes_{A} X^{1}$ the first-factor term has sign $+ 1$ ; hence $d_{T}^{1}$ is the displayed row. The maps are well-defined on the balanced tensor and preserve the outer $B, C$ -actions and internal grading by [L2].

1.3L1L2algebra

For an elementary tensor $f \otimes x \in F^{0} \otimes_{A} X^{0}$ , the two paths give $- (u (f) \otimes v (x))$ and $u (f) \otimes v (x)$ , respectively, because $u \otimes_{A} 1$ and $1 \otimes_{A} v$ act on separate factors. Therefore $d_{T}^{1} d_{T}^{0} (f \otimes x) = 0$ , and additivity proves $d_{T}^{1} d_{T}^{0} = 0$ on all of $T^{0}$ .

2.1L1algebra∎

In the stated integer example the displayed maps have matrices $(\begin{array}{c} 2 \\ 3 \end{array})$ and $(\begin{array}{cc} - 3 & 2 \end{array})$ , whose product is $- 6 + 6 = 0$ . If $f$ has internal degree $r$ and $x$ has internal degree $s$ , every nonzero matrix entry preserves degree $r + s$ ; the sign is determined only by $p$ . With $F$ concentrated in degree $0$ the surviving tensor differential is $1 \otimes v$ , and with $F$ concentrated in degree $1$ it is $- 1 \otimes v$ , as [L1] prescribes.

Depends on

Used by

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Dependency tree · two levels

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Sources

Khovanov and Seidel, Quivers, Floer Cohomology, and Braid Group Actions, §2c and Proposition 2.4 (standard reference, not scraped)