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A nontrivial five-strand braid lies in the Burau kernel

Statement

Assume AC, inherited from the topological definition of the Burau representations and from the reduced/unreduced same-kernel transfer. There exists a nontrivial element of $B_{5}$ that acts trivially in the unreduced Burau representation $ρ_{5}^{m a t}$ of The unreduced Burau matrices. Explicitly, let $α, β$ be the two embedded arcs on the five-punctured disk displayed in Figure 3 of the source, using the standard Artin labels fixed by its straightening words on p. 403, with $α$ joining the marked points $q_{2}$ and $q_{4}$ and $β$ joining the boundary basepoint $p_{0}$ to the marked point $q_{3}$ , and put $ψ : = [T_{α}, T_{β}] = T_{α}^{- 1} T_{β}^{- 1} T_{α} T_{β},$ the commutator of the clockwise half Dehn twist $T_{α}$ about the boundary of a regular neighbourhood of $α$ (whose induced permutation exchanges $q_{2}$ and $q_{4}$ ) with the full boundary-arc twist $T_{β}$ about the boundary of a regular neighbourhood of $β \cup \partial D$ . Then $ψ \neq 1$ in $B_{5}$ and $ρ_{5}^{m a t} (ψ) = I_{5}$ . This is the explicit kernel element used by the counterexample on the companion page; it is not an instance of $Δ^{2 k}$ . The boundary-arc twist uses the boundary-relative convention of Bigelow, Section 2; changing its representative by a central boundary full twist leaves this commutator unchanged.

Facts & Assumptions

Given: AC; the five-punctured disk $(D, Δ_{5})$ with boundary basepoint $p_{0}$ ; the oriented embedded arcs $α, β$ of Bigelow's Figure 3; the half twist $T_{α}$ and the full twist $T_{β}$ specified in the Statement; the commutator $ψ = T_{α}^{- 1} T_{β}^{- 1} T_{α} T_{β}$ ; and $Λ_{1} = Z [t^{\pm 1}]$ .

[L1]

Write $s_{i} = σ_{i}^{- 1}$ for clockwise generators, since the positive geometric half twists of The elementary geometric half twist, its support disc, and its opposite are anticlockwise. Bigelow's printed p. 403 gives the words $P = s_{3}^{- 1} s_{2} s_{1}^{2} s_{2} s_{4}^{3} s_{3} s_{2}, Q = s_{4}^{- 1} s_{3} s_{2} s_{1}^{- 2} s_{2} s_{1}^{2} s_{2}^{2} s_{1} s_{4}^{5}, R = s_{4} s_{3} s_{2} s_{1}^{2} s_{2} s_{3} s_{4} .$ In the Figure 3 coordinates, $P$ straightens $α$ to the arc between $q_{4}, q_{5}$ and $Q$ straightens $β$ to the boundary arc ending at $q_{5}$ ; hence its kernel witness is $[P^{- 1} s_{4} P, Q^{- 1} R Q]$ . We use these exact words to check the twist argument below, rather than assuming that the source's abbreviated digon check proves nontriviality.

[L2]

Put $S_{i} = ρ_{5}^{m a t} (s_{i}) = B_{i}^{- 1}$ . Its nonidentity block is $(\begin{array}{cc} 0 & 1 \\ t^{- 1} & 1 - t^{- 1} \end{array})$ . For a word $V$ in the $s_{i}^{\pm 1}$ , let $M (V)$ be the ordered product of these blocks or their inverses. This is its unreduced Burau action (The unreduced Burau matrices, The unreduced Burau matrices satisfy the Artin relations).

[L3]

The Artin representation $η : B_{5} \to Aut (F_{5})$ is a homomorphism. In the clockwise convention its substitutions are $η (s_{i}) (x_{i}) = x_{i + 1}, η (s_{i}) (x_{i + 1}) = x_{i + 1}^{- 1} x_{i} x_{i + 1};$ its inverse sends $x_{i}$ to $x_{i} x_{i + 1} x_{i}^{- 1}$ and $x_{i + 1}$ to $x_{i}$ , fixing the other basis letters. Products compose with the rightmost letter first. Free reduction has unique normal forms (Artin automorphisms of the free group, The Artin representation on a free group, Reduced words form the free group on an alphabet). To prove that a braid is nontrivial it suffices that its image is nontrivial; no faithfulness theorem is needed.

[L4]

The topological reduced action is the action on $H_{1} (\tilde{X}; Z)$ for the infinite cyclic cover of The Burau infinite cyclic cover. It agrees with the invariant reduced submodule of the matrix action, and the reduced and unreduced integral representations have the same kernel (The topological and matrix Burau representations agree, The reduced and unreduced Burau representations have the same kernel). A boundary full twist $Δ^{2}$ acts on this homology by $t^{5}$ (Bigelow, printed p. 400).

[L5]

Twists are boundary-fixed braid mapping classes, and twists supported on disjoint regular neighbourhoods commute (Boundary-fixed mapping class group of a punctured disk, The braid group by Artin presentation, The elementary geometric half twist, its support disc, and its opposite). Homotopic simple proper arcs are isotopic relative to their endpoints (Homotopic simple proper arcs in the punctured disk are isotopic relative to their endpoints); their geometric intersection number has the meaning in Curves and geometric intersection numbers on the marked disk. The half twist $T_{α}$ exchanges $q_{2}, q_{4}$ , so its induced permutation is an involution. The braid itself has infinite order: on the subgroup preserving these two punctures, forgetting the other strands sends its powers to powers of a generator of $B_{2} ≅ Z$ .

[L6]

For lifts of the oriented arcs to the cyclic cover, the lifted intersection polynomial is $J (α, β) = \sum_{k} (t^{k} \tilde{α}, \tilde{β}) t^{k}$ , where the parentheses denote algebraic intersection. Changing lifts multiplies $J$ by a power of $t$ . Bigelow, Definition 1.3 and Section 3, records the crossing signs and all fifty terms; the exponents are determined by the total winding about punctures between crossings.

Proof

technique · direct

1.1L1L5

Fix the Figure 3 witness and the conventions. Set $a = P^{- 1} s_{4} P$ and $b = Q^{- 1} R Q$ . The puncture permutation of $P$ sends $(1, 2, 3, 4, 5)$ to $(1, 5, 2, 4, 3)$ , and $Q$ sends it to $(2, 1, 5, 4, 3)$ . Thus $P^{- 1} s_{4} P$ exchanges $q_{2}, q_{4}$ and $Q^{- 1} R Q$ uses the boundary arc ending at $q_{3}$ . This fixes the Figure 3 labeling; The source's general arc criterion labels an arbitrary test arc's endpoints $q_{1}, q_{2}$ ; applying that criterion to these words requires a relabeling. By [L1] these are the source's half twist and boundary-arc twist; in particular $ψ = [a, b]$ is the displayed geometric commutator. An ambiguity by a boundary full twist in $b$ has no effect on $[a, b]$ , since a boundary twist is supported in a collar and each boundary-fixed mapping class has a representative that is the identity on a smaller collar, making the two supports disjoint. Every subsequent calculation uses $s_{i} = σ_{i}^{- 1}$ , the source's clockwise convention, and ordinary left actions.

1.2L2algebra

The two block calculations. Let $u_{0} = - e_{4} + t^{- 1} e_{5}$ and $v_{0} = e_{4}^{*} - e_{5}^{*}$ , so $S_{4} = I + u_{0} v_{0}$ . Multiplying the eight blocks of $R$ gives $M (R) = (\begin{array}{ccccc} t^{- 1} & 0 & 0 & 0 & 1 - t^{- 1} \\ 0 & t^{- 1} & 0 & 0 & 1 - t^{- 1} \\ 0 & 0 & t^{- 1} & 0 & 1 - t^{- 1} \\ 0 & 0 & 0 & t^{- 1} & 1 - t^{- 1} \\ t^{- 4} - t^{- 5} & t^{- 3} - t^{- 4} & t^{- 2} - t^{- 3} & t^{- 1} - t^{- 2} & 1 - t^{- 1} + t^{- 5} \end{array}) .$ Put $u = M (P)^{- 1} u_{0}$ and $v = v_{0} M (P)$ . The ten blocks of $P$ give $u = (t^{3} - t^{2}, t - 2 t^{2} + 2 t^{3} - t^{4} + t^{5}, t^{3} - t^{2}, - t^{2}, t^{- 1} - 1 + t - t^{2})^{T},$ $v = (t^{- 2} - t^{- 3}, - t^{- 5} + t^{- 4} - t^{- 3}, t^{- 5} - 2 t^{- 4} + 2 t^{- 3} - 2 t^{- 2} + t^{- 1}, t^{- 4} - t^{- 3} + 2 t^{- 2} - 2 t^{- 1} + 1, t^{- 3} - t^{- 2} + t^{- 1} - 1) .$ Thus $M (a) = I + u v$ . These computations use only the displayed two-by-two blocks; in particular they take place over the integral Laurent ring.

1.3L6algebra

The source's lifted intersection calculation. Normalize lifts so the first crossing along $β$ contributes $+ t^{0}$ . Upward crossings are positive and downward crossings negative. When successive crossing subarcs bound a disk containing $k$ punctures, the exponent changes by $k$ with the sign of the orientation around that disk. In the fifty-term calculation on Bigelow's printed p. 403, the positive terms at exponents $- 3, - 2, - 1, 0, 1, 2, 3, 4, 5$ have respective multiplicities $(1, 2, 3, 4, 5, 4, 3, 2, 1)$ , and the negative terms have exactly the same multiplicities. Therefore every coefficient cancels and $J (α, β) = 0$ , independently of the choice of lifts. The explicit matrix calculation below verifies the resulting commuting twist action in the frozen convention.

2.1L2step 1.2algebra

The intersection cancellation in matrix coordinates. Put $y = M (Q) u$ and $z = v M (Q)^{- 1}$ . Multiplication by the sixteen blocks of $Q$ gives the four scalar identities $y_{5} = 0$ , $y_{1} + t y_{2} + t^{2} y_{3} + t^{3} y_{4} = 0$ , $z_{5} = 0$ , and $z_{1} + z_{2} + z_{3} + z_{4} = 0$ . These can be checked without forming any full matrix: a positive letter $s_{i}$ changes a column pair $(c_{i}, c_{i + 1})$ to $(c_{i + 1}, t^{- 1} c_{i} + (1 - t^{- 1}) c_{i + 1})$ and a row pair $(d_{i}, d_{i + 1})$ to $(t^{- 1} d_{i + 1}, d_{i} + (1 - t^{- 1}) d_{i + 1})$ ; a negative letter uses the inverse pair operations. For columns apply the word from right to left, and for rows from left to right. With those four identities, the displayed matrix $M (R)$ gives $M (R) y = t^{- 1} y$ and $z M (R) = t^{- 1} z$ : its fifth column pairs to zero with $z$ , and its fifth row pairs to zero with $y$ . Consequently $M (b) u = t^{- 1} u$ and $v M (b) = t^{- 1} v$ . Hence $M (b) (I + u v) = M (b) + t^{- 1} u v = (I + u v) M (b)$ , proving $ρ_{5}^{m a t} (ψ) = I_{5}$ . This explicitly checks the boundary-arc case of the source's intersection/twist argument.

2.2L3step 1.1algebra

The geometric twists do not commute. Conjugate by $P$ , so the two twists become $s_{4}$ and $w = P Q^{- 1} R Q P^{- 1}$ . Compute their actions on $x_{1}$ using [L3]. Successively applying $P^{- 1}, Q, R, Q^{- 1}, P$ and then $s_{4}$ gives freely reduced lengths $13, 83, 185, 1993, 14095, 19199$ . The reduced word $η (w) (x_{1})$ begins $x_{5}^{- 1} x_{3}^{- 1} x_{5}^{- 1}$ , while $η (s_{4} w) (x_{1})$ begins $x_{5}^{- 1} x_{4}^{- 1} x_{5}$ . The full substitutions and reductions, including the two different prefixes, are given by the finite certificate below. Since $s_{4}$ fixes $x_{1}$ , $η (w s_{4}) (x_{1}) = η (w) (x_{1})$ ; the two different reduced prefixes show $η (s_{4} w) \neq η (w s_{4})$ . Thus $s_{4} w \neq w s_{4}$ , and after conjugating back, $a b \neq b a$ and $ψ \neq 1$ .

3.1L5step 2.2

Why Figure 3 has essential intersection. If $α$ could be homotoped off $β$ relative to endpoints, the proper-arc homotopy/isotopy identification in [L5] would allow disjoint representatives. Their regular neighbourhoods, including the boundary collar in the boundary-arc construction, could then be chosen disjoint, so their supported twists would commute. This contradicts step 2.2. Thus the Figure 3 arcs cannot be homotoped apart. This gives a local proof of the geometric conclusion, including the boundary-arc case, through their explicit actions rather than the source's abbreviated digon argument. The induced permutation of $T_{α}$ is the involution in [L5], and the half twist itself is not an order-two braid.

4.1L4step 2.1step 2.2algebra∎

Conclusion and exclusion of boundary full twists. Steps 2.1 and 2.2 prove the claimed nontrivial kernel element. By [L4] its reduced action is also the identity. If $ψ = Δ^{2 k}$ , that reduced action would be $t^{5 k} I_{4}$ , so $t^{5 k} = 1$ in $Z [t^{\pm 1}]$ and $k = 0$ ; this contradicts $ψ \neq 1$ . The statement retains AC through its topological suppliers; the explicit matrix and reduced-word computations are finite and require no choice.

Remarks

Here is the complete finite free-word certificate for step 2.2. A signed integer $j$ denotes $x_{j}$ and $- j$ denotes $x_{j}^{- 1}$ ; signed braid integers refer to $s_{j}$ . The stack cancels adjacent inverse letters, and therefore returns the unique free-group reduced word. No truncation of an intermediate word occurs. The displayed assertions follow by these explicit substitutions.

P = [-3, 2, 1, 1, 2, 4, 4, 4, 3, 2]
Q = [-4, 3, 2, -1, -1, 2, 1, 1, 2, 2, 1, 4, 4, 4, 4, 4]
R = [4, 3, 2, 1, 1, 2, 3, 4]

def inverse(word):
    return [-j for j in reversed(word)]

def reduce(word):
    stack = []
    for j in word:
        if stack and stack[-1] == -j:
            stack.pop()
        else:
            stack.append(j)
    return stack

def substitute(word, braid_letter):
    i = abs(braid_letter)
    if braid_letter > 0:
        images = {i: [i+1], i+1: [-i-1, i, i+1]}
    else:
        images = {i: [i, i+1, -i], i+1: [i]}
    expanded = []
    for j in word:
        image = images.get(abs(j), [abs(j)])
        expanded.extend(image if j > 0 else inverse(image))
    return reduce(expanded)

def act(braid_word, word):
    for j in reversed(braid_word):
        word = substitute(word, j)
    return word

word = [1]
for factor, length in zip(
    [inverse(P), Q, R, inverse(Q), P], [13, 83, 185, 1993, 14095]
):
    word = act(factor, word)
    assert len(word) == length
assert word[:3] == [-5, -3, -5]
other = act([4], word)
assert len(other) == 19199 and other[:3] == [-5, -4, 5]
assert act([4], [1]) == [1]

Depends on

Used by

Equal actions on K₀ do not imply isomorphic derived autoequivalences Counterexample

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83 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Stephen J. Bigelow, The Burau representation is not faithful for n=5, Geometry & Topology 3 (1999) 397-404 (standard reference, not scraped)
Joan S. Birman and Tara E. Brendle, Braids: A Survey, Sections 4.2 and 4.4 (standard reference, not scraped)