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Solved Midterm Exam - Geometry and Topology. | MATHS 441, Exams of Mathematics

Ball State University (BSU)Mathematics

Material Type: Exam; Class: Geometry and Topology.; Subject: MATHEMATICAL SCIENCE; University: Ball State University; Term: Fall 2009;

Typology: Exams

2009/2010

Uploaded on 03/28/2010

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Midterm Exam IV
Part I: TAKE-HOME
MATHS 441
Dr. Fischer
Due: December 11, 2009
1. Let Xbe any space. Recall that two continuous paths αand βin Xare called
equivalent if they both start at the same point, both end at the same point,
and if αcan be continuously deformed into βwithin X, while keeping both
endpoints fixed. Prove the following:
(a) Equivalence of paths is an equivalence relation.
(b) [α]∗[β]=[α·β] is a well-defined operation on equivalence classes which
are represented by paths whose endpoints match appropriately.
(c) ([α]∗[β]) ∗[γ] = [α]∗([β]∗[γ]) for all equivalence classes which are repre-
sented by paths whose endpoints match appropriately.
2. Let Xbe any space and let x0∈X. Denote by π1(X, x0) the set of all equiva-
lence classes [α] of continuous paths α: [0,1] →Xwith α(0) = α(1) = x0.
Prove that π1(X, x0) forms a group under the operation “∗” of Problem 1.
(We call π1(X, x0) the fundamental group of Xbased at x0.)
3. (a) Prove that π1is a topological invariant: homeomorphic spaces have iso-
morphic fundamental groups, provided the basepoints correspond.
(b) Prove that the group π1(R3\K, x0) is a knot invariant for knots Kand
that π1(R3\L, x0) is a link invariant for links L.
4. Suppose the fundamental group G=hgenerators |relations iof a knot com-
plement has been found by the Wirtinger presentation algorithm from class.
(a) Show that any one relation is redundant.
(b) Show that Gis abelian if and only if Gis isomorphic to Z.
5. Each of the two diagrams below features a pair of mountaineering karabiners
with some rope around them. Use the Wirtinger presentation of the link com-
plement of the two karabiners to explain why you cannot slip the rope off the
two karabiners in one of the diagrams, while you can in the other.
[Hints and reminders of some definitions are attached!]
pf3
pf4

Partial preview of the text

Download Solved Midterm Exam - Geometry and Topology. | MATHS 441 and more Exams Mathematics in PDF only on Docsity!

Midterm Exam IV

Part I: TAKE-HOME

MATHS 441 Dr. Fischer

Due: December 11, 2009

  1. Let X be any space. Recall that two continuous paths α and β in X are called equivalent if they both start at the same point, both end at the same point, and if α can be continuously deformed into β within X, while keeping both endpoints fixed. Prove the following:

(a) Equivalence of paths is an equivalence relation. (b) [α] ∗ [β] = [α · β] is a well-defined operation on equivalence classes which are represented by paths whose endpoints match appropriately. (c) ([α] ∗ [β]) ∗ [γ] = [α] ∗ ([β] ∗ [γ]) for all equivalence classes which are repre- sented by paths whose endpoints match appropriately.

  1. Let X be any space and let x 0 ∈ X. Denote by π 1 (X, x 0 ) the set of all equiva- lence classes [α] of continuous paths α : [0, 1] → X with α(0) = α(1) = x 0. Prove that π 1 (X, x 0 ) forms a group under the operation “∗” of Problem 1. (We call π 1 (X, x 0 ) the fundamental group of X based at x 0 .)
  2. (a) Prove that π 1 is a topological invariant: homeomorphic spaces have iso- morphic fundamental groups, provided the basepoints correspond. (b) Prove that the group π 1 (R^3 \ K, x 0 ) is a knot invariant for knots K and that π 1 (R^3 \ L, x 0 ) is a link invariant for links L.
  3. Suppose the fundamental group G = 〈 generators | relations 〉 of a knot com- plement has been found by the Wirtinger presentation algorithm from class.

(a) Show that any one relation is redundant. (b) Show that G is abelian if and only if G is isomorphic to Z.

  1. Each of the two diagrams below features a pair of mountaineering karabiners with some rope around them. Use the Wirtinger presentation of the link com- plement of the two karabiners to explain why you cannot slip the rope off the two karabiners in one of the diagrams, while you can in the other.

[Hints and reminders of some definitions are attached!]

Definitions: Here is a more mathematical definition of equivalence of paths: two continuous functions α : [0, 1] → X and β : [0, 1] → X are called equivalent, denoted α ∼ β, if there is a continuous function H : [0, 1] × [0, 1] → X, defining a path Ht(s) = H(s, t) at each time t, such that (i) H(s, 0) = α(s) for all s ∈ [0, 1]; (ii) H(s, 1) = β(s) for all s ∈ [0, 1]; (iii) H(0, t) = α(0) = β(0) at all times t; (iv) H(1, t) = α(1) = β(1) at all times t. Concatenation of paths is defined as follows: if the path α : [0, 1] → X ends where the path β : [0, 1] → X begins, i.e., if α(1) = β(0), we may define the concatenation α · β of α and β to be the path which first runs through α and then through β (each twice as fast):

(α · β)(s) =

  

α(2s) for 0 6 s 6 1 / 2

β(2s − 1) for 1/ 2 6 s 6 1

Hints:

  1. (a) In case you have already sold your MATHS 215 book, here are the three things that need to be checked: (i) α ∼ α; (ii) if α ∼ β, then β ∼ α; (iii) if α ∼ β and β ∼ γ, then α ∼ γ. For (i), you can take H(s, t) = α(s). For (ii), start with H(s, t) as in the definition of α ∼ β and define F (s, t) = H(s, 1 −t). Then check that if you replace F (s, t) with H(s, t) in the definition of equivalence, this shows that β ∼ α. For (iii), suppose H(s, t) deforms α into β and F (s, t) deforms β into γ. Define G(s, t) = H(s, 2 t) for 0 6 t 6 1 /2 and G(s, t) = F (s, 2 t − 1) for 1 / 2 6 t 6 1. Show that G(s, t) deforms α into γ. The attached Figure 1 depicts a schematic of the definition of G(s, t). Shown is the domain [0, 1] × [0, 1] of the functions H, F , and G. Explain why the pieces fit together at the line t = 1/2. Then show that G(s, t) satisfies all characteristics for a deformation of α into γ. (b) You have to show that the answer of the operation [α] ∗ [β] does not depend on the choice of representative for the equivalence classes [α] and [β]. That is, you need to show that if [α] = [γ] and [β] = [δ], then [α · β] = [γ · δ]. So, from α ∼ γ and β ∼ δ, it should follow that α · β ∼ γ · δ. Say H(s, t) deforms α into γ and F (s, t) deforms β into δ. Define G(s, t) = H(2s, t) for 0 6 s 6 1 /2 and G(s, t) = F (2s − 1 , t) for 1/ 2 6 s 6 1. Show that G(s, t) deforms α · β into γ · δ. See attached Figure 2. Again, do not neglect to explain why the pieces fit together at the line s = 1/2. (c) You need to find a deformation H(s, t) to show that (α · β) · γ ∼ α · (β · γ). Proceed as suggested in Figure 3. The domain of your formula should have three parts, one for each region of the figure. Start by finding an equation for each of the two slanted lines through the domain square, solved for s as a function of t. Then draw some random horizontal line at a fixed height t and define H(s, t) in three intervals for s, which depend on t, so that H(s, t) runs through α, β and γ at the appropriate speed. (Answers are displayed in Figure 3. Show your work!) Explain why the pieces fit together on the overlap and check all required properties of your deformation H(s, t).
  2. For a point x ∈ X, let ex : [0, 1] → X denote the constant path that is always equal to x, i.e., ex(s) = x for all s ∈ [0, 1]. Prove that [eα(0)] ∗ [α] = [α] and [α]∗[eα(1)] = [α] for every continuous path α : [0, 1] → X, whatsoever. Likewise, let α denote the reverse of the path α, that is, α(s) = α(1 − s). Prove that