Figure 1: Closed surface, surface with boundary, or not a surface?

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Edith Collins
5 years ago
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1 QUESTION 1 (10 marks) Two o th topological spacs shown in Figur 1 ar closd suracs, two ar suracs with boundary, and two ar not suracs. Dtrmin which is which. You ar not rquird to justiy your answr, but, i you do, it might giv you som partial crdit i you mad a mistak with idntiication. (a) Ic-cram surac (b) Sausag link surac (c) Disks with strips in th middl (d) Squars with strips on th dg () Squar with two pairs o dgs glud () Squar with on pair o dgs glud Figur 1: Closd surac, surac with boundary, or not a surac? Solution (a) and () closd suracs; (d) and () suracs with boundary; (b) and (c) not suracs. :) 1

2 QUESTION 2 (15 marks) Idntiy th surac whos plan modl is shown in Figur 2. a a d d b b c c Figur 2: Idntiy this surac Solution First, notic that th surac is non-orintabl (two dgs lablld a ar both clockwis). Thus it s th sphr with k cross-caps. Furthr, w nd to labl vrtics and to mark path-connctd componnts o th boundary. It s don in Figur 3. B A E E a a B d d A 2 D D 1 2 A B C C b b 2 c c B A E E Figur 3: Th surac rom Figur 2 with vrtics lablld by capital lttrs and boundary componnts lablld by numbrs. Thus w hav V = 5, E = 10 (6 lablld and 4 unlablld dgs), F = 1, and h = 2 (th numbr o hols). From th Eulr charactristic, w must gt 2 k h = V E + F 2 k 2 = k = 4 and hnc th surac is th sphr with 4 cross-caps and 2 hols. :) 2

3 QUESTION 3 (15 marks) Prov that th graph shown in Figur 4 cannot b mbddd into th orintabl surac o gnus 3. Hr, thr ar our groups with 3 vrtics in ach group and only vrtics rom dirnt groups ar joind by dgs. Figur 4: Show that this graph cannot b mbddd into th sphr with 3 handls Solution Th graph has 36 dgs, 12 vrtics, and th girth o 4. Th Eulr charactristic o th sphr with 3 handls is 4. I th graph could b bmddd into th surac, thn, by an inquality rom tutorials, w would hav 36 4 (12 ( 4)) 36 32, 4 2 which is als. Thus th graph cannot b mbddd into th sphr with 3 handls. :) 3

4 QUESTION 4 (15 marks) Apply Sirt s algorithm to construct a Sirt surac o th link shown in Figur 5. Sktch th surac you constructd and ind its gnus. Figur 5: Construct a Sirt surac o this link Solution Thr ar two ways to introduc orintation, both with 4 Sirt cycls. Th Eulr charactristic o th Sirt surac is, according to a ormula rom tutorials, is 4 8 = 4. Without th 2 hols, it would hav bn 2 and hnc th gnus is 2. Th orintd link and th corrsponding Sirt surac ar shown in Figur 6. :) Figur 6: Th orintd link and its Sirt surac in Qustion 4. 4

5 QUESTION 5 (15 marks) Vriy that th ollowing rlation holds or th Conway polynomial L (z): ( ) + = z Solution Applying th skin rlation at th lt crossing, w hav = z and = z. Adding ths two quations togthr, w gt + 2 = z Applying th dining skin rlation to th RHS, w gt + 2 = z 2, which is xactly what w wr rquird to show. :) 5

QUESTION 6 (15 marks) For ach o th two substs o R 2 shown in Figur 7 (a cross mad out o two closd intrvals and a rctangl with two hols), dtrmin, with xplanation, i it satisis th ixd point proprty.

6 QUESTION 6 (15 marks) For ach o th two substs o R 2 shown in Figur 7 (a cross mad out o two closd intrvals and a rctangl with two hols), dtrmin, with xplanation, i it satisis th ixd point proprty. Figur 7: Do ths igurs satisy th ixd point proprty? Solution Th cross is a rtract o a squar and hnc satisis th FPP sinc th squar satisis th FPP (Brouwr s Fixd Point Thorm). Th othr igur rtracts onto a circl and hnc dosn t satisy th FPP sinc th circl dosn t. Such a rtraction can b constructd, or xampl, as a projction rom th cntr o th lt hol onto th boundary o th igur. :) 6

7 QUESTION 7 (15 marks) According to Alxandr s Thorm, any link is th closur o som braid. Find two non-isotopic braids that th link shown in Figur 8 is th closur o and xplain why thy ar non-isotopic. Figur 8: Find two braids that this link is th closur o Solution s 1 3 s 2s 1 3 s 1 1 s 1 3 s 1 1 s 1 2 s 3s 1 B 4 and s 1 3 s 2s 1 3 s 1 1 s 1 3 s 1 1 s 1 2 s 3s 1 s 4 B 5. :) 7

Strongly Connected Components

Strongly Connected Components Strongly Connctd Componnts Lt G = (V, E) b a dirctd graph Writ if thr is a path from to in G Writ if and is an quivalnc rlation: implis and implis s quivalnc classs ar calld th strongly connctd componnts