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Decategorification is the unreduced Burau action

Statement

Each $X \in C_{m}$ has a class $[X] \in G (A_{m})$ (The graded Grothendieck group of A_m), and the class $[R_{σ} \otimes_{A_{m}} -]$ depends only on $σ \in B_{m + 1}$ by The Khovanov-Seidel complexes give a weak derived braid action, so the braid action induces a representation of $B_{m + 1}$ by $Z [q, q^{- 1}]$ -linear maps on $G (A_{m})$ . On the basis $[P_{0}], \dots, [P_{m}]$ of The graded Grothendieck group is free on the vertex-projective classes the generators act by $[R_{i}] [P_{i}] = - q [P_{i}], [R_{i}] [P_{i + 1}] = [P_{i + 1}] - [P_{i}], [R_{i}] [P_{i - 1}] = [P_{i - 1}] - q [P_{i}], [R_{i}] [P_{j}] = [P_{j}] (∣ i - j ∣ > 1),$ with the terms $[P_{i + 1}]$ , respectively $[P_{i - 1}]$ , omitted when the index leaves ${0, \dots, m}$ . Moreover, let $C$ be the $(m + 1) \times (m + 1)$ matrix over $Z [q, q^{- 1}]$ with $C_{r, r} = C_{r, r + 1} = (- q)^{m - r} (0 \leq r \leq m - 1), C_{m, m} = 1,$ all other entries zero. Then $C$ is invertible and $C [R_{i}] C^{- 1} = B_{i} ∣_{t = q} (1 \leq i \leq m),$ where $B_{1}, \dots, B_{m}$ are the unreduced Burau matrices of The unreduced Burau matrices in their column-vector convention with parameter $t$ . Consequently, after the explicit change of basis $C$ and the parameter identification $q = t$ , the decategorified action is exactly the unreduced Burau representation of $B_{m + 1}$ ; no identification is left implicit, and the comparison uses the same generator indexing and word order as the Burau matrices.

Facts & Assumptions

Given: An integer $m \geq 1$ , the free $Z [q, q^{- 1}]$ -module $G (A_{m})$ with basis $[P_{0}], \dots, [P_{m}]$ , the twist complexes $R_{i}$ , the functors $U_{i}$ , and the unreduced Burau matrices $B_{i}$ with parameter $t$ .

[L1]

$[R_{i}] = [A_{m}] - [U_{i}]$ : the twist complex is the cone of $β_{i} : U_{i} \to A_{m}$ between complexes concentrated in degree $0$ , its class in $K_{0}$ is $[A_{m}] - [U_{i}]$ by the triangle relation $[A_{m}] = [U_{i}] + [R_{i}]$ , and $[A_{m}] = [P_{0}] + \dots + [P_{m}]$ (The twist complexes R_i and R_i^{-1}, The triangulated K_0 of the Khovanov–Seidel projective category, Homological and internal shifts on K_0(C_m)).

[L2]

$U_{i} \otimes_{A_{m}} P_{j} ≅ P_{i} \otimes_{Z} e_{i} A_{m} e_{j}$ with the corner bases $e_{i} A_{m} e_{i} = Z e_{i} \oplus Z (i ∣ i - 1 ∣ i)$ in degrees $0, 1$ , $e_{i} A_{m} e_{i + 1} = Z (i ∣ i + 1)$ in degree $0$ , $e_{i} A_{m} e_{i - 1} = Z (i ∣ i - 1)$ in degree $1$ , and $e_{i} A_{m} e_{j} = 0$ for $∣ i - j ∣ > 1$ (Corner computations: the U_i satisfy the Temperley-Lieb relations, The 4m+1 path basis).

[L3]

$G (A_{m})$ is free over $Z [q, q^{- 1}]$ with basis $[P_{0}], \dots, [P_{m}]$ , and the class map is additive over direct sums with $[X {r}] = q^{r} [X]$ (The graded Grothendieck group is free on the vertex-projective classes, Homological and internal shifts on K_0(C_m)).

[L4]

The unreduced Burau matrices act on column vectors by the identity except for the block $B_{i} = (\begin{array}{cc} 1 - t & t \\ 1 & 0 \end{array})$ at rows and columns $i - 1, i$ , and satisfy the Artin relations (The unreduced Burau matrices).

Proof

technique · direct

1.1L1L2L3

The class of the twist. By [L1] the operator $[R_{i}]$ on $G (A_{m})$ is $Id - [U_{i}]$ , where $[U_{i}]$ is induced by the exact functor $U_{i}$ , and the class of $[U_{i}]$ is computed on the basis by [L2]: $[U_{i}] [P_{i}] = [P_{i}] + q [P_{i}]$ , $[U_{i}] [P_{i + 1}] = [P_{i}]$ , $[U_{i}] [P_{i - 1}] = q [P_{i}]$ and $[U_{i}] [P_{j}] = 0$ for $∣ i - j ∣ > 1$ , where the degrees of the corner generators turn the tensor shifts into the powers of $q$ by [L3].

1.2L3

The matrix $C$ is invertible. $C$ is upper triangular, with diagonal entries $(- q)^{m}, (- q)^{m - 1}, \dots, (- q), 1$ , all units of $Z [q, q^{- 1}]$ ; hence $\det C = (- q)^{m + (m - 1) + \dots + 1}$ is a unit and $C \in {G L}_{m + 1} (Z [q, q^{- 1}])$ .

2.1step 1.1L3

The displayed action. Subtracting the four formulas of step 1.1 from $[P_{j}]$ gives the four displayed rules: $[R_{i}] [P_{i}] = [P_{i}] - [P_{i}] - q [P_{i}] = - q [P_{i}]$ , $[R_{i}] [P_{i + 1}] = [P_{i + 1}] - [P_{i}]$ , $[R_{i}] [P_{i - 1}] = [P_{i - 1}] - q [P_{i}]$ , and $[R_{i}] [P_{j}] = [P_{j}]$ for $∣ i - j ∣ > 1$ ; indices outside ${0, \dots, m}$ do not occur. Hence the representation is well defined on the whole basis.

2.2step 1.2L2L4

The intertwining identity. Write $A_{i}$ for the matrix of $[R_{i}]$ and $U$ for the matrix of $[U_{i}]$ in the basis $[P_{0}], \dots, [P_{m}]$ , so that $A_{i} = I - U$ by step 1.1; the columns of $U$ are $U e_{i} = (1 + q) e_{i}$ , $U e_{i + 1} = e_{i}$ , $U e_{i - 1} = q e_{i}$ and $U e_{j} = 0$ otherwise. Decompose $C = D (I + N)$ , where $D$ is the diagonal matrix with entries $d_{r} = (- q)^{m - r}$ for $r < m$ and $d_{m} = 1$ , and $N$ is the matrix with ones on the superdiagonal and zeros elsewhere, so that $N e_{j} = e_{j - 1}$ ; then $(I + N)^{- 1} = I - N + N^{2} - \dots + (- 1)^{m} N^{m}$ and the $j$ -th column of $(I + N)^{- 1}$ is $\sum_{k \geq 0} (- 1)^{k} e_{j - k}$ . Compute $(I + N) U (I + N)^{- 1}$ column by column. In column $i$ the alternating sum of the $U$ -images is $(1 + q) e_{i} - q e_{i} = e_{i}$ , and applying $(I + N)$ gives $e_{i} + e_{i - 1}$ . In column $i - 1$ only $U e_{i - 1} = q e_{i}$ survives, and applying $(I + N)$ gives $q (e_{i} + e_{i - 1})$ . In every other column the finitely many nonzero contributions $U e_{i + 1}, U e_{i}, U e_{i - 1}$ occur with consecutive alternating signs and cancel to $0$ . Hence $(I + N) U (I + N)^{- 1}$ has columns $e_{i} + e_{i - 1}$ at $i$ , $q (e_{i} + e_{i - 1})$ at $i - 1$ and $0$ elsewhere. Conjugating by the diagonal $D$ multiplies each entry $(r, j)$ by $d_{r} d_{j}^{- 1}$ , so $C U C^{- 1}$ has entries $1$ at $(i, i)$ , $- q$ at $(i - 1, i)$ , $- 1$ at $(i, i - 1)$ and $q$ at $(i - 1, i - 1)$ , and $0$ elsewhere; that is, $C U C^{- 1} = I - B_{i}$ with $B_{i}$ the unreduced Burau matrix at $t = q$ , whose block at rows and columns $i - 1, i$ is $(\begin{array}{cc} 1 - q & q \\ 1 & 0 \end{array})$ [L4]. Hence $C A_{i} C^{- 1} = I - C U C^{- 1} = B_{i}$ .

3.1step 2.1step 2.2∎

Conclusion. The decategorified action of the generators is the displayed four-case action, and the single invertible matrix $C$ , independent of $i$ , conjugates every generator to the unreduced Burau matrix at $t = q$ ; since both sides are representations of the presented group $B_{m + 1}$ [L4], the change of basis $C$ and the parameter identification $q = t$ identify the whole representation with the unreduced Burau representation, with the same indexing and word order. The action on the reduced quotient is not claimed here. No choice principle is used beyond the inputs already recorded.

Depends on

Used by

Equal actions on K₀ do not imply isomorphic derived autoequivalences Counterexample
Decategorifying a generator on the vertex-projective basis Example

Dependency tree · two levels

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Sources

Mikhail Khovanov and Paul Seidel, Quivers, Floer Cohomology, and Braid Group Actions, J. Amer. Math. Soc. 15 (2002) 203-271, Section 2e.1 (standard reference, not scraped)
Joan S. Birman and Tara E. Brendle, Braids: A Survey, Section 4.2 (standard reference, not scraped)