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The Krammer fraction-field generator matrices for B three

Example

For $n = 3$ Krammer's fraction-field model of the The Lawrence-Krammer-Bigelow representation is a free module with basis $x_{12}, x_{13}, x_{23}$ over $Λ = Z [q^{\pm 1}, t^{\pm 1}]$ , on which the generators act by the seven-case formula $σ_{k} x_{k, k + 1} = t q^{2} x_{k, k + 1}; σ_{k} x_{i k} = (1 - q) x_{i k} + q x_{i, k + 1} (i < k);$ $σ_{k} x_{i, k + 1} = x_{i k} + t q^{k - i + 1} (q - 1) x_{k, k + 1} (i < k);$ $σ_{k} x_{k j} = t q (q - 1) x_{k, k + 1} + q x_{k + 1, j} (k + 1 < j);$ $σ_{k} x_{k + 1, j} = x_{k j} + (1 - q) x_{k + 1, j} (k + 1 < j);$ $σ_{k} x_{i j} = x_{i j} (i < j < k or k + 1 < i < j); σ_{k} x_{i j} = x_{i j} + t q^{k - i} (q - 1)^{2} x_{k, k + 1} (i < k < k + 1 < j) .$

This example records the two resulting $3 \times 3$ matrices over $Λ$ , checks the braid relation by direct multiplication, and checks the full-twist value $q^{2 n} t^{2} = q^{6} t^{2}$ at $n = 3$ . The model is a fraction-field model: by The integral LKB module is free of rank n choose two it is not integrally identified with the closed-surface basis ${v_{i, j}}$ when $n \geq 3$ , and the parameter translation between Krammer's and Bigelow's conventions is $t_{K r a m m e r} = - t_{B i g e l o w}$ .

Verification

Given: the seven-case formula displayed above with $n = 3$ (so $k \in {1, 2}$ ), the basis ordered as $(x_{12}, x_{13}, x_{23})$ , and matrices acting on column vectors, the columns being the images of the basis vectors.

1.1givenalgebra

The columns for $σ_{1}$ . Every case with $i < k$ is vacuous for $k = 1$ . The case $σ_{1} x_{1, 2} = t q^{2} x_{1, 2}$ gives the first column $(t q^{2}, 0, 0)$ ; the case $k + 1 < j$ applied to $(k, j) = (1, 3)$ gives $σ_{1} x_{1, 3} = t q (q - 1) x_{1, 2} + q x_{2, 3}$ ; and the case for $x_{k + 1, j}$ with $(k + 1, j) = (2, 3)$ gives $σ_{1} x_{2, 3} = x_{1, 3} + (1 - q) x_{2, 3}$ . Hence $M_{1} = (\begin{array}{ccc} t q^{2} & t q (q - 1) & 0 \\ 0 & 0 & 1 \\ 0 & q & 1 - q \end{array}) .$

1.2givenalgebra

The columns for $σ_{2}$ . For $k = 2$ the case $i < k$ with $i = 1$ gives $σ_{2} x_{1, 2} = (1 - q) x_{1, 2} + q x_{1, 3}$ ; the case $x_{i, k + 1}$ with $i = 1$ gives $σ_{2} x_{1, 3} = x_{1, 2} + t q^{2} (q - 1) x_{2, 3}$ ; and the case $x_{k, k + 1}$ gives $σ_{2} x_{2, 3} = t q^{2} x_{2, 3}$ . Hence $M_{2} = (\begin{array}{ccc} 1 - q & 1 & 0 \\ q & 0 & 0 \\ 0 & t q^{2} (q - 1) & t q^{2} \end{array}) .$

2.1step 1.1step 1.2algebra

The braid relation by direct multiplication. Multiplying the two matrices gives $M_{1} M_{2} = (\begin{array}{ccc} 0 & t q^{2} & 0 \\ 0 & t q^{2} (q - 1) & t q^{2} \\ q^{2} & - q^{2} t (q - 1)^{2} & - q^{2} t (q - 1) \end{array}) .$ Multiplying this product on the right by $M_{1}$ and on the left by $M_{2}$ gives, by expansion of the nine entries of each of the two products, $M_{1} M_{2} M_{1} = M_{2} M_{1} M_{2} = (\begin{array}{ccc} 0 & 0 & t q^{2} \\ 0 & t q^{3} & 0 \\ t q^{4} & 0 & 0 \end{array}) = : Δ .$ Each entry of the two triple products is a sum of at most three Laurent monomials; collecting the terms in each of the nine positions gives the displayed common value, so the braid relation $ρ (σ_{1}) ρ (σ_{2}) ρ (σ_{1}) = ρ (σ_{2}) ρ (σ_{1}) ρ (σ_{2})$ holds.

3.1step 2.1algebra

The full twist. The matrix $Δ$ has a single nonzero entry in each row and column, so squaring it multiplies the diagonal entries $t q^{2} \cdot t q^{4} = t^{2} q^{6}$ , $(t q^{3})^{2} = t^{2} q^{6}$ , $t q^{4} \cdot t q^{2} = t^{2} q^{6}$ and kills all off-diagonal entries: $(ρ (σ_{1}) ρ (σ_{2}) ρ (σ_{1}))^{2} = Δ^{2} = q^{6} t^{2} I_{3},$ which is the value $q^{2 n} t^{2}$ of the full-twist scalar at $n = 3$ . The columns of $Δ$ are exactly the values $Δ x_{2, 3} = t q^{2} x_{1, 2}$ , $Δ x_{1, 3} = t q^{3} x_{1, 3}$ , $Δ x_{1, 2} = t q^{4} x_{2, 3}$ predicted by the half-twist identity of the source Section 3, $Δ x_{n + 1 - j, n + 1 - i} = t q^{i + j - 1} x_{i j}$ .

4.1step 2.1step 3.1algebra∎

Invertibility and conventions. Since $\det M_{1} = \det M_{2} = - q^{3} t$ is a unit of $Λ$ , both matrices lie in ${G L}_{3} (Λ)$ . The matrices above are those of Krammer's fraction-field model with basis $x_{12}, x_{13}, x_{23}$ , not matrices in Bigelow's integral closed-surface basis; the two models are isomorphic as $B_{n}$ -representations only after fraction-field extension for $n \geq 3$ , and the translation between the parameters is $t_{K r a m m e r} = - t_{B i g e l o w}$ , so the displayed formulas record Krammer's normalization and not Bigelow's.

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Sources

Krammer, Braid groups are linear, Ann. of Math. 155 (2002) 131-156 (standard reference, not scraped)
Bigelow, The Lawrence-Krammer representation, arXiv:math/0204057v1 (standard reference, not scraped)