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In a 1984 paper, Squier wrote down an explicit Hermitian form $J$ relative to which the reduced Burau representation is unitary. The form is valued in the ring $\mathbb{Z}[s^{\pm 1}]$, where $s^2 = t$ and $t$ is the standard variable appearing in the Burau representation. In this context, the involution on $\mathbb{Z}[s^{\pm 1}]$ is induced from the map $s \mapsto s^{-1}$. Later it was understood that the reduced Burau representation is also skew-Hermitian with respect to a sort of symmetrized intersection form on the Burau cover of the disk, valued in $\mathbb{Z}[t^{\pm}]$.

I am wondering if there is an extension of either of these forms to the unreduced Burau representation. In light of the fact that the unreduced Burau rep is given by action on a relative homology group, for which there doesn't seem to be a good notion of an intersection pairing, I'm not optimistic that the unreduced Burau rep can be made to be skew-Hermitian.

On the other hand, I don't know if it's possible to define something like Squier's form for unreduced Burau. Squier pulls his form out of thin air, and I haven't come across any discussion in the literature of a topological definition of Squier's form (I've looked in the papers of Gambaudo-Ghys, McMullen, and Venkataramana that study the Burau representation at roots of unity; in each of these the intersection form plays a crucial role, but none seem to explain where Squier's version of the form comes from).

In summary,

Is there a Hermitian form on the domain of the unreduced Burau representation for which Burau is unitary? Can one extend Squier's form in some way?

Is there a topological construction of Squier's intersection form? How did he write his form down?

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  • $\begingroup$ Do you have a reference for the skew-Hermitian property? $\endgroup$ Jul 1, 2019 at 1:43

2 Answers 2

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Regarding your second question.

Squier's form doesn't come out of thin air. It's induced by Poincare duality + universal coefficients in the relevant covering space of a punctured disc.

I suspect Squier suppressed how the form is constructed to make his paper look more mysterious. I don't know why mathematicians do this, but you see this kind of behaviour in many papers.

If you check out my paper where I construct the corresponding intersection form for the Lawrence-Krammer representation, I point this fact out there.

Of course, I could be wrong about Squier's motivations as I have not talked with him. I have heard several mathematicians admit to this kind of behaviour. It can simplify a paper, I suppose. My paper would have been much less complicated if I excluded all the Morse theory I used to compute Poincare duality. But it would also appear to be a black box -- far less enabling.

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Let $P_n$ be the $n$-times punctured disc. Let $\tilde P_n$ be the abelian cover corresponding to the sum of the winding numbers about the punctures. We take a CW-decomposition homotopy-equivalent to $P_n$ having $n$ 1-cells and one $0$-cell. Call the $1$-cells $e_1, e_2, \cdots, e_n$.

Let $\Lambda(t)$ be the Laurent polynomial ring with one variable, over $\Bbb Z$. $\Lambda(t) = \Bbb Z[t^{\pm 1}]$.

Then the chain complex for the homology of the abelian cover has the form

$$\Lambda(t)^n \to \Lambda(t)$$

with the only nontrivial map given by the matrix

$$[t-1, \ t-1, \ \cdots \ t-1]$$

Giving $H_1 \tilde P_n \simeq \Lambda(t)^{n-1}$ with free generators $$E_2 = e_2-e_1, \ E_3 = e_3-e_1, \ \cdots \ E_n = e_n-e_1$$

Let's call the total intersection form on $H_1 \tilde P_n$

$$\langle a, b \rangle = \sum_{i \in \Bbb Z} \mu(t^i a, b) t^i$$

where $\mu(\cdot, \cdot)$ is the regular intersection form on $H_1 \tilde P_n$.

The total intersection form is anti-symmetric sesquilinear

$$\langle ta, b\rangle = t^{-1} \langle a, b \rangle = \langle a, t^{-1}b\rangle$$

$$\overline{ \langle a, b \rangle } = - \langle b, a \rangle$$

With the cell structure I describe above we can compute

$$\langle E_i, E_i \rangle = t-t^{-1}$$ $$\langle E_i, E_j \rangle = t-1 \hskip 1cm i<j$$

There is a standard procedure to turn anti-symmetric sesquilinear forms into symmetric sesquilinear forms -- you multiply by $i$. Define

$$[a,b] = i\langle a, b\rangle$$

then this is sesquilinear, plus

$$[b,a] = i\langle b, a\rangle = -i\overline{\langle a, b\rangle}= \overline{[a,b]}$$

This isn't quite the way Squier expressed the form as this is taking values in $\Lambda(t)$ adjoin $i$. I think this is where his change of basis matrix $P$ and his $t^2=s$ idea kicks in, allowing him to reduce the form to take values in $\Lambda(s)$. But the form above is symmetric sesquilinear. Perhaps that's all you need?

I suspect if you just follow his ideas from here with $P$ and $s^2=t$ you'll get the rest of the way and recover the form exactly.

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  • $\begingroup$ Is it not equally likely that it was suppressed in the editorial process for the sake of space? (This is not a Socratic question: I do not know and am interested in your thoughts.) $\endgroup$
    – mme
    Dec 6, 2018 at 12:36
  • $\begingroup$ Thanks for your answer! Could you explain a little more how one can obtain Squier's form in the way you describe? I'm unclear on two (possibly related) points: how does Poincare duality plus universal coefficients give rise to a symmetric form? And why is it necessary to introduce the square root s? Does this have topological meaning in this context? $\endgroup$ Dec 6, 2018 at 14:24
  • $\begingroup$ @MikeMiller: it's certainly true that could be the rationale. But if such was the case you'd think the author would include one sentence that would give the reader a clue. $\endgroup$ Dec 6, 2018 at 16:13
  • $\begingroup$ @NickSalter: I'll edit my answer to fill in those details. Currently suffering from jetlag. . . Will nap then reply. $\endgroup$ Dec 6, 2018 at 16:45
  • $\begingroup$ @RyanBudney Yeah, fair enough. I agree that is quite frustrating. $\endgroup$
    – mme
    Dec 6, 2018 at 17:03
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After lots of trial and error, it looks like I can give an explicit construction of such a Hermitian form, although it's utterly mysterious to me where such a thing comes from!

Let $e_1, \dots, e_n$ be the standard basis vectors for the domain of the unreduced Burau rep. We seek to define a $\mathbb{Z}[t^{\pm}]$-valued form $(\cdot, \cdot)$ that is symmetric, bilinear, and Hermitian with respect to the involution $t \mapsto t^{-1}$. If you experiment with the conditions on the values $(e_i, e_j)$ imposed by the action of Burau (assuming that it acts unitarily), you can quickly deduce that all values $(e_i, e_i)$ are equal for $1 \le i \le n$, and that all values $(e_i, e_j)$ are equal for $1 \le i < j \le n$.

An additional consideration gives you a relationship between $(e_1,e_1)$ and $(e_1,e_2)$ which turns out to be satisfied for $(e_1,e_1) = -1$ and $(e_1,e_2) = t$. At this point it's just a matter of checking that the form determined by these conditions is indeed preserved by the generators of the unreduced Burau rep. This is an easy computation that reduces to checking the preservation on the relevant $2 \times 2$ block.

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