mh3600 problem set 12

7/21/2019 Mh3600 Problem Set 12

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Problem Set 12

MH3600: Knots and surfaces

November 17, 2015

Easy exercises

Exercise 1.

If braids are multiplied from top to bottom and π : Bn → Sn is the homomorphism that assignspermutation to a braid, then how do we multiply permutations, from left to right or from right to left?

Exercise 2.

Draw the braid α = s3s1s−32 s1s−1

3 s22 ∈ B4 and compute its permutation.

Solution 2.

The braid is shown in Figure 1. The permutation is

σ =

1 2 3 44 2 3 1

Exercise 3.

For Windows users: download SeifertView. Notice that braids there are multiplied from left to right anthat generators are denoted a, b, c, . . . , and A, B, C, . . . . Figure out how to translate it into our notatioi.e., whether A is s1, s−1

1 , sn−1, or s−1n−1. Then plot the Seifert surface of the braid s3s1s−3

2 s1s−13 s2

2.

Exercise 4.

Prove thats3s2s3s2s1s−1

3 s2s−21 = s2

3s1s2s−11

Solution 4.

We have,

s3 (s2s3s2) s1s−13 s2s−21 = s3 (s3s2s3) s1s−13 s2s−21 =s3s3s2 (s3s1) s−1

3 s2s−21 = s3s3s2 (s1s3) s−1

3 s2s−21 =

s23 (s2s1s2) s−2

1 = s23 (s1s2s1) s−2

1

= s23s1s2s−1

1

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Figure 1: Answer to Exercise 2.

Exercise 5.

Let exp : Bn → Z be given by

exp

sk 1i1

· · · sk pi p

= k 1 + · · · + k p.

(a) Prove that exp is a well-defined homomorphism.

(b) Prove that Bn is infinite for n ≥ 2.

Exercise 6.

(a) Prove that s1s2 = s2s1 in B3.

(b) Prove that (s1s2)3 = 1 in B3.

Exercise 7.

(a) How many components does the closure of the braid s25s1s−1

3 s−22 s4 have?

(b) Clearly, the braids α1 = s25s1s−1

3 s−22 s4 ∈ B6 and α2 = s−1

3 s−22 s4s2

5s1 ∈ B6 have the same closurExplain how α2 can be obtained from α1 by Markov moves.

(c) Clearly, the braids α2 = s−13 s−2

2 s4s25s1 ∈ B6 and α3 = s−1

2 s−21 s3s2

4 ∈ B5 have the same closure. Explahow α3 can be obtained from α2 by Markov moves.

Solution 7.

The braid s25s1s−1

3 s−22 s4 is shown in Figure 2.

(a) In the cyclic notation, the permutation π (α1) = (1, 2)(3,5,4)(6) and hence the link α1 has 3 compnents.

(b) α2 =

s25s1

−1α1

s2

5s1

— Markov move I.

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Figure 2: Braid in Exercise 7.

Figure 3: Find a braid that this knot is the closure of.

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(c) Sorry, everyone, this question is much harder than I thought. Anyway, here is how we can do Let f : Bn → Bn be the homomorphism given as follows:

f (s1) = sn−1, f (s2) = sn−2, · · · , f (sn−1) = s1,

inversing the order of generators. Such a homomorphism is, actually, a conjugation with the braiconstructed as follows: imagine that we glue all strands to a strip of paper and then twist this str

by 180◦. This braid is called Garside element and can be expressed in terms of Artin generators by

∆ = s1s2 . . . sn−1 · s1s2 · · · sn−2 · · · s1s2s1

and f (α) = ∆α∆−1. Thus f is a Markov move I.

Now, applying f : B6 → B6, we get

f (α2) = s−13 s−2

4 s2s21s5

Applying Markov move II, we gets−1

3 s−24 s2s2

1 ∈ B5

Applying f : B5 → B5, we gets−1

2 s−21 s3s2

4 = α3

Exercise 8.

Find two non-isotopic braids that the knot shown in Figure 3 is the closure of.

Solution 8.

s22s−1

3 s−12 s3s−1

2 s1s−12 s−1

3 ∈ B4 and s22s−1

3 s−12 s3s−1

2 s1s−12 s−1

3 s4 ∈ B5.

Exercise 9.

One of the links shown in Figure 4 is a circular link. Determine which one and find a braid that it is thclosure of.

Solution 9.

It’s link (c). One of possible words is s−31 s2s−2

1 s2.

Exercise 10.

Prove that the Burau representation

ϕ : Bn → Mn×n(Z[t, t−1])

defined in the lecture as

ϕ(si) =

I n−1

1 − t t

1 0I n−i−1

is indeed, a homomorphism from the braid group to the group of nonsingular matrices with entries the ring of Laurent polynomials. In other words,

(a) find ϕ(s−1i ),

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(a) (b)

(c) (d)

Figure 4: Which of these links is circular?

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(b) prove that ϕ(si)ϕ(s j) = ϕ(s j)ϕ(si) for |i − j| ≥ 2,

(c) prove that ϕ(si)ϕ(si+1)ϕ(si) = ϕ(si+1)ϕ(si)ϕ(si+1).

Exercise 11.

For a surface S (with or without boundary), what is B1(S)?

Exercise 12.

If an element of the group π 1

F(D2, n)

can be visualized as a braid sandwiched between two discs, hocan we visualize a braid over the 2-sphere, i.e., an element of the group π 1

F(S2, n)

= Bn(S2)? Expla

why a braid over the 2-sphere can be always expressed in terms of Artin generators, just like a braid ovethe disk.

Exercise 13.

Explain why the relation s1s2 · · · sn−1sn−1 · · · s2s1 = 1 holds in Bn(S2).

Remark in fact, Bn(S2) is the quotient-group of Bn by the minimal normal subgroup that contains th braid

R = s1s2 · · · sn−1sn−1 · · · s2s1.

In other words, Bn(S2) can be presented with the same generators and same relations as Bn but with onextra relation that tells us that R = 1.

Exercise 14.

Let A be an annulus (cylinder, sphere with 2 holes). The braid group Bn( A) can be presented witArtin generators s1, . . . , sn−1 and with n new generators, denoted a1, a2, . . . , an, where each a i is a loop F( A, n) that goes as follows: the i th point goes once around the hole and the others don’t move.

Of course, Artin’s relation for s1, . . . , sn−1 will still hold but there will be new relations that invoai and a mixture of ai and s j. What are the new relations? You are not required to prove that there

nothing else.

Exercise 15.

A Brunnian braid is a braid that becomes trivial if any of its strands is removed.

(a) Show that the braid

s1s−12

3∈ B3 is Brunnian.

(b) Show that all Brunnian braids are pure if n ≥ 3.

(c) Show that Brunnian braids form a normal subgroup in Bn.

(d) Prove that

s1s−123

= 1 in B3. Hence the group of Brunnian braids is, indeed, non-trivial (in fac

it is isomorphic to the free group with infinitely many generators).

Hard exercises

Remark: most of these exercises are not conceptually hard, but they either require either a significacomputational effort or good spatial imagination.

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Figure 5: Find a braid that this link is the closure of.

Exercise 16.

This is a continuation of Exercise 15 on Brunnian braids. Just like a Brunnian braid, a Brunnian link islink that becomes trivial if any of its components is removed. Obviously, the closure of a Brunnian braiis a Brunnian link. Construct a Brunnian link that is not the closure of a Brunnian braid.

Exercise 17.

Find a braid that the link shown in Figure 5 is the closure of. It’s possible to do it in 3 steps of Vogelalgorithm presented in the lecture.

Exercise 18.

Explain why (s1s2 · · · sn−1)n commutes with any element of the group Bn. Your explanation doesn’t havto be a rigorous proof, but it needs to be convincing.

Remark In fact, the centre of the braid group Bn is the cyclic group generated by (s1s2 · · · sn−1)n.

Exercise 19.Explain why (s1s2 · · · sn−1)2n = 1 in Bn(S2).

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mh3600 problem set 12

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