18
votes
Accepted
What is the cohomological dimension of the commutator subgroup of the pure braid group?
Here is the answer: for $n\geq 2$ we have $\mathrm{cd}([P_n,P_n])=n-2$.
https://arxiv.org/abs/1905.05099
16
votes
Accepted
Cohomology of braid groups with coefficients in the group ring
The braid groups $B_n$ are Bieri-Eckmann duality groups of dimension $n-1$. It follows (either by definition or by a standard result depending on how you set things up) that $H^k(B_n;\mathbb{Z}[B_n])$...
14
votes
Accepted
When do elements in the braid group $B_n$ commute?
Krammer ("The braid group $B_4$ is linear", Invent. Math. 142 (2000), 451–486 (MSN)) constructed
a representation $\rho: B_n \to {\rm GL}_N({\bf Z}[q^{\pm 1}, t^{\pm 1}])$
with $N = {n \choose 2}$, ...
14
votes
Several questions about Gauss's mathematical conception of braids
• Connection with electromagnetism: (see Gauss' linking number revisited for the historical context)
Consider a wire $c$ carrying a current $I$, winding around a closed loop $c'$, as in the ...
14
votes
Accepted
What does $\mathrm{Conf}_n(M)^{h S_n}$ look like?
In general, A nice enough finite dimensional space with a free $\Sigma_i$-action does not admit any homotopy fixed points. This is because a homotopy fixed point provides a splitting of the ...
13
votes
Accepted
Why does the definition of a braided monoidal category not mention the braid equation?
Indeed, the axioms of a braided monoidal category are enough to derive the Yang-Baxter equation. See Braided monoidal categories by Joyal and Street (diagram B7), or 1Lab for a formalised proof.
12
votes
When do elements in the braid group $B_n$ commute?
$\DeclareMathOperator\Aut{Aut}$Here is an easier computation: $B_n$ admits a faithful representation into the automorphism group of a free group of rank $n$, $\rho\colon B_n \to \Aut(F_n)$. I learned ...
12
votes
Accepted
Action of braid groups on regular trees
In the article A group theoretic criterion for property FA, Culler and Vogtmann notice that
for $n \geq 5$, the braid group $B_n$ has property $A \mathbb{R}$,
meaning that every non-trivial (i.e. ...
11
votes
Formality of the 2nd ordered configuration space of a closed Riemann surface
I believe that $F_2 \Sigma_g$ is formal. Here's a sketch of an argument, unfortunately I haven't checked the details. Apply the results of Morgan, "The algebraic topology of smooth algebraic varieties"...
11
votes
Accepted
Homotopy type of the semi-simplicial set of symmetric groups
It is contractible. To see this first observe that it is simply connected. It has 2 arcs $a = (1,2)$ and $b = (2,1)$, however the tirangle $(3,1,2)$ gives us the relation that $ab = b$ and ...
10
votes
Is the configuration space of ordered triples of distinct points in the four-edge banana graph homotopy equivalent to a surface of genus 13?
Since $C_n(G)$ (for any finite $n$ and finite connected $G$ with at least one cycle or vertex of valence at least $3$) and $\Sigma_g$ (for $g\ge 1$) are both $K(\pi,1)$ spaces homeomorphic to CW ...
10
votes
Perfect quotients of braid groups
A useful general strategy to tackle such questions is to use small cancellation theory. For instance, in Small cancellation in acylindrically hyperbolic groups, Michael Hull proved the following ...
9
votes
Accepted
What is known/expected on the co-growth series of the braid group?
I don’t know much about the exponential generating series, so I’ll just address Question 1.
For the braid group on $n=3$ strand, the cogrowth series is $D$-finite by results of Alex Bishop for any ...
8
votes
Accepted
Braid group on 4 strands
This may be resolved by identifying $B_4$ with the mapping class group of the 4-punctured disk (fixing the boundary). I.e. $B_4\cong Mod(D^2-\{p_1,p_2,p_3,p_4\})$. I will consider isotopy classes of ...
8
votes
Accepted
Action of the homotopy braid groups on reduced free groups
The short answer is yes. I will try to explain the intuition behind appearance of the reduced free groups and the link homotopy and why they are related; for the rest, I refer to the great papers of ...
8
votes
Integral homology of braid groups as a ring
Using the Hopf fibration you can show that $\Omega^2S^2\simeq\mathbb{Z}\times\Omega^2S^3$. Also $\pi_1(\Omega^2S^3)=\pi_3(S^3)=\mathbb{Z}$, so the Universal Coefficient Theorem gives $[\Omega^2S^3,S^...
8
votes
Accepted
Loop manipulation subgroup of the braid group
Your group $H_n$ is (I believe) called the wicket group by Brendle and Hatcher in their paper Configuration spaces of rings and wickets. They provide a presentation in Proposition 3.6. They also ...
7
votes
Does the shortest path between two braids pass through string links?
I do not have the reputation to comment, but I'll delete this after you've read it.
There is an idea, mostly from virtual knot theory, called parity (I recommend this reference: arxiv:1211.0403, and ...
7
votes
Accepted
Epimorphisms from the genus $2$ surface braid group to finite groups
I have found a finite image of order $3^9$ in which $\sigma$ has nontrivial image, by using the $\mathtt{pQuotient}$ function which, for a given prime $p$ and class $n$, computes the largest quotient ...
7
votes
When do elements in the braid group $B_n$ commute?
One approach to this question is to try to give a description of the centralizer. This can be done using the Nielsen-Thurston classification of elements in mapping class groups.
To describe this in ...
7
votes
Accepted
Representation stability for systems of braid group representations
A good way to handle systems of braid group representations is to consider the category of functors $\mathcal{C}\to R\textrm{-Mod}$, where $\mathcal{C}$ is a category with the braid groups as ...
7
votes
Accepted
Dehornoy's proof that the application of two elementary embeddings is an elementary embedding
As Monroe Eskew points out in comments: $j_1(j_2|_{V_\gamma})$ really means $j_1$ applied to $j_2|_{V_\gamma}$ (not the pointwise image of $j_2|_{V_\gamma}$ under $j_1$, as you seem to be reading it). ...
6
votes
Is the *unreduced* Burau representation unitary?
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 ...
6
votes
Epimorphisms from the genus $2$ surface braid group to finite groups
Consider $C_2 \times C_2$ minus the diagonal, this is a double cover of your space. Given a symplectic form on $H_1( C_2 \times C_2 - \Delta, \mathbb Z/p)$, we can form an associated extension $1 \to ...
6
votes
How to braid a ribbon knot
If I understand your question correctly, the answer is in the paper
Lee Rudolph, Braided surfaces and Seifert ribbons for closed braids, Comment. Math. Helvetici 58 (1983), 1–37.
In fact, in ...
6
votes
When do two positive braids represent the same link?
This was proven in the Paper "Fibered Transverse Knots and the Bennequin Bound" by John B. Etnyre, Jeremy Van Horn-Morris (https://arxiv.org/abs/0803.0758):
Corollary 1.8. Any two positive braids ...
6
votes
Given a word $w$ in the braid group $B_n$, representing a pure braid, find the image of $w$ in the abelianization of $P_n$
There is a simple geometric way to think about the fact that $P_n^{\text{ab}} \cong \mathbb{Z}^{\binom{n}{2}}$: Draw the braid with strands going from left to right across the page. Pick any two of ...
6
votes
Relations between relations in the positive braid monoid
I found a published reference! This is the main result of:
Fukushi, Takeo, On a braid monoid analogue of a theorem of Tits., SUT J. Math. 47, No. 1, 45-53 (2011). ZBL1235.20036.
I'll keep my write up ...
6
votes
What does $\mathrm{Conf}_n(M)^{h S_n}$ look like?
[UPDATE: Connor Malin's answer is definitely better than mine, but I will leave mine here in case the approach turns out to be useful for some other purpose.]
Put $X=\operatorname{Conf}_n(M)$ and $Y=\...
5
votes
fundamental group of configuration spaces of ordered points on open Riemann surfaces
I can at least answer your second question. I'll be a bit brief, but let me know if you need more details. Let $M$ be an oriented manifold and $M^{(r)}$ the configuration space as in your question. ...
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