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Triangular basis changes diagonalize an invertible differential block

Statement

Let $X^{∙}$ be a cochain complex in an additive category and let $d^{n} = (\begin{array}{cc} a & b \\ c & φ \end{array}) : A \oplus U \to B \oplus V$ be an invertible-block decomposition as in An invertible cochain differential block and its candidate reduction: the pivot $φ : U \to V$ is an isomorphism, the neighbouring components are $d^{n - 1} = (\begin{array}{c} p \\ q \end{array})$ and $d^{n + 1} = (\begin{array}{cc} r & s \end{array})$ , and $\overset{ˉ}{d} : = a - b φ^{- 1} c$ is the Schur complement. Put $L = (\begin{array}{cc} 1 & - b φ^{- 1} \\ 0 & 1 \end{array}) : B \oplus V \to B \oplus V, R = (\begin{array}{cc} 1 & 0 \\ - φ^{- 1} c & 1 \end{array}) : A \oplus U \to A \oplus U .$ Then:

$L$ and $R$ are isomorphisms, with inverses $L^{- 1} = (\begin{array}{cc} 1 & b φ^{- 1} \\ 0 & 1 \end{array})$ and $R^{- 1} = (\begin{array}{cc} 1 & 0 \\ φ^{- 1} c & 1 \end{array})$ .
$L d^{n} R = (\begin{array}{cc} \overset{ˉ}{d} & 0 \\ 0 & φ \end{array})$ .
$c p + φ q = 0$ and $r b + s φ = 0$ ; consequently $R^{- 1} (\begin{array}{c} p \\ q \end{array}) = (\begin{array}{c} p \\ 0 \end{array})$ and $(\begin{array}{cc} r & s \end{array}) L^{- 1} = (\begin{array}{cc} r & 0 \end{array})$ .
$\overset{ˉ}{d} p = 0$ and $r \overset{ˉ}{d} = 0$ .
The candidate reduction ${\overset{ˉ}{X}}^{∙}$ of An invertible cochain differential block and its candidate reduction is a cochain complex: all composites of consecutive reduced differentials vanish.

Facts & Assumptions

Given: A cochain complex $X^{∙}$ in an additive category, an integer $n$ , a pivot decomposition $X^{n} = A \oplus U$ , $X^{n + 1} = B \oplus V$ with invertible block $φ : U \to V$ , neighbouring components $p, q, r, s$ , the Schur complement $\overset{ˉ}{d} = a - b φ^{- 1} c$ , and the morphisms $L, R, L^{- 1}, R^{- 1}$ displayed above.

[L1]

The components of $d^{n}, d^{n - 1}, d^{n + 1}$ are the blocks $a, b, c, φ$ and $p, q, r, s$ , the pivot satisfies $φ φ^{- 1} = 1_{V}$ and $φ^{- 1} φ = 1_{U}$ , and composition of morphisms between finite biproducts is matrix multiplication (An invertible cochain differential block and its candidate reduction, Composition of morphisms between finite biproducts is matrix multiplication).

[L2]

$X^{∙}$ is a cochain complex, so $d^{n} d^{n - 1} = 0$ and $d^{n + 1} d^{n} = 0$ (Complexes, homotopies and contractibility in an additive category).

Proof

technique · direct

1.1

Multiplying the two matrices with [L1], the first column of $d^{n} R$ is $d^{n} (1_{A}; - φ^{- 1} c) = (a - b φ^{- 1} c; c - φ φ^{- 1} c) = (\overset{ˉ}{d}; 0)$ and the second is $d^{n} (0; 1_{U}) = (b; φ)$ , so $d^{n} R = (\begin{array}{cc} \overset{ˉ}{d} & b \\ 0 & φ \end{array})$ .

L1algebra

1.2

The displayed inverses work: $L L^{- 1} = (\begin{array}{cc} 1 & - b φ^{- 1} \\ 0 & 1 \end{array}) (\begin{array}{cc} 1 & b φ^{- 1} \\ 0 & 1 \end{array}) = (\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array})$ , and symmetrically $L^{- 1} L = 1$ ; likewise $R R^{- 1} = (\begin{array}{cc} 1 & 0 \\ - φ^{- 1} c & 1 \end{array}) (\begin{array}{cc} 1 & 0 \\ φ^{- 1} c & 1 \end{array}) = (\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array})$ and $R^{- 1} R = 1$ , all uses of $φ φ^{- 1} = 1_{V}$ cancelling the middle terms.

L1algebra

1.3

The composite $d^{n} d^{n - 1}$ is the block matrix $(\begin{array}{c} a p + b q \\ c p + φ q \end{array})$ , so $d^{n} d^{n - 1} = 0$ gives $a p + b q = 0$ and $c p + φ q = 0$ ; applying $φ^{- 1}$ on the left to the second equation gives $q = - φ^{- 1} c p$ .

L1L2algebra

1.4

The composite $d^{n + 1} d^{n}$ is the block matrix $(\begin{array}{cc} r a + s c & r b + s φ \end{array})$ , so $d^{n + 1} d^{n} = 0$ gives $r a + s c = 0$ and $r b + s φ = 0$ ; multiplying the second equation on the right by $φ^{- 1}$ gives $r b φ^{- 1} = - s$ .

L1L2algebra

2.1

Applying $L$ to that result, the first column is $L (\overset{ˉ}{d}; 0) = (\overset{ˉ}{d} - b φ^{- 1} 0; 0) = (\overset{ˉ}{d}; 0)$ and the second is $L (b; φ) = (b - b φ^{- 1} φ; φ) = (0; φ)$ , hence $L d^{n} R = (\begin{array}{cc} \overset{ˉ}{d} & 0 \\ 0 & φ \end{array})$ .

step 1.1L1algebra

2.2

The first component of $R^{- 1} (\begin{array}{c} p \\ q \end{array})$ is $1_{A} p + 0 \cdot q = p$ and the second is $φ^{- 1} c p + 1_{U} q = φ^{- 1} c p + q = 0$ by step 1.3, so $R^{- 1} (\begin{array}{c} p \\ q \end{array}) = (\begin{array}{c} p \\ 0 \end{array})$ .

step 1.3L1algebra

2.3

Similarly $(\begin{array}{cc} r & s \end{array}) L^{- 1} = (\begin{array}{cc} r 1_{B} + s \cdot 0 & r b φ^{- 1} + s 1_{V} \end{array}) = (\begin{array}{cc} r & r b φ^{- 1} + s \end{array}) = (\begin{array}{cc} r & 0 \end{array})$ by step 1.4.

step 1.4L1algebra

2.4

$\overset{ˉ}{d} p = (a - b φ^{- 1} c) p = a p - b φ^{- 1} c p = a p + b q = 0$ , substituting $q = - φ^{- 1} c p$ from step 1.3 and then $a p + b q = 0$ .

step 1.3algebra

2.5

$r \overset{ˉ}{d} = r (a - b φ^{- 1} c) = r a - r b φ^{- 1} c = r a + s c = 0$ , substituting $r b φ^{- 1} = - s$ from step 1.4 and then $r a + s c = 0$ .

step 1.4algebra

3.1

Every composite of consecutive differentials of ${\overset{ˉ}{X}}^{∙}$ vanishes. In the two modified degrees these are ${\overset{ˉ}{d}}^{n} d^{n - 1} = \overset{ˉ}{d} p = 0$ by step 2.4 and ${\overset{ˉ}{d}}^{n + 1} {\overset{ˉ}{d}}^{n} = r \overset{ˉ}{d} = 0$ by step 2.5. In the remaining degrees the reduction either keeps the arrows of $X$ or replaces $d^{n - 1}$ by its $A$ -component $p$ and $d^{n + 1}$ by its $B$ -component $r$ : thus $p d^{n - 2} = {pr}_{A} d^{n - 1} d^{n - 2} = 0$ because $d^{n - 1} d^{n - 2} = 0$ , and $d^{n + 2} r = d^{n + 2} d^{n + 1} ı_{B} = 0$ because $d^{n + 2} d^{n + 1} = 0$ , where ${pr}_{A}$ and $ı_{B}$ are the biproduct projection and injection recording the components $p$ and $r$ . All other composites are composites of consecutive differentials of $X$ , hence vanish.

L1L2step 2.4step 2.5algebra

4.1

Step 1.2 proves clause 1; steps 1.1 and 2.1 prove clause 2; steps 1.3, 1.4, 2.2 and 2.3 prove clause 3; steps 2.4 and 2.5 prove clause 4; and step 3.1 proves clause 5. In particular the candidate reduction of the block decomposition is a genuine cochain complex with the neighbouring arrows $p$ and $r$ . ∎

step 1.2step 1.1step 2.1step 1.3step 1.4step 2.2step 2.3step 2.4step 2.5step 3.1

Depends on

Used by

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Sources

Dror Bar-Natan, Fast Khovanov Homology Computations, section 4 Lemma 4.2 and section 5, printed p. 5 (standard reference, not scraped)
David Clark, Scott Morrison and Kevin Walker, Fixing the Functoriality of Khovanov Homology, Appendix A.1, printed pp. 1562-1563 (standard reference, not scraped)