THE FUNDAMENTAL GROUP AND COVERING SPACES


 Marcia Thornton
 2 years ago
 Views:
Transcription
1 THE FUNDAMENTAL GROUP AND COVERING SPACES JESPER M. MØLLER Astrct. These notes, from first course in lgeric topology, introduce the fundmentl group nd the fundmentl groupoid of topologicl spce nd use them to clssify covering spces. Contents 1. Homotopy theory of pths nd loops Chnge of se point nd unsed homotopies 4 2. Covering spces 5 3. The fundmentl group of the circle, spheres, nd lense spces Applictions of π 1 (S 1 ) 9 4. The vn Kmpen theorem Fundmentl groups of knot nd link complements Ctegories Ctegories of right Gsets Trnsitive right ctions The clssifiction theorem Cyley tles, Cyley grphs, nd Cyley complexes Norml covering mps Sections in covering mps Universl covering spces of topologicl groups 27 References Homotopy theory of pths nd loops Definition 1.1. A pth in topologicl spce X from x 0 X to x 1 X is mp u: I X of the unit intervl into X with u(0) = x 0 nd u(1) = x 1. Two pths, u 0 nd u 1, from x 0 to x 1 re pth homotopic, nd we write simply u 0 u 1, if u 0 u 1 rel I, ie if u 0 nd u 1 re re homotopic reltive to the endpoints {0, 1} of the unit intervl I. The constnt pth t x 0 is the pth x 0 (s) = x 0 for ll s I The inverse pth to u is the pth from x 1 to x 0 given y u(s) = u(1 s) If v is pth from v(0) = u(1) then the product pth pth u v given y { u(2s) 0 s 1 (u v)(s) = 2 1 v(2s 1) 2 s 1 where we first run long u with doule speed nd then long v with doule speed is pth from u(0) to v(1). In greter detil, u 0 u 1 if there exists homotopy h: I I X such tht h(s, 0) = u 0 (s), h(s, 1) = u 1 (s) nd h(0, t) = x 0, h(1, t) = x 1 for ll s, t I. All pths in homotopy clss hve the sme strt point nd the sme end point. Note the following rules for products of pths x 0 u u u x 1 Dte: Decemer 10, I would like to thnk Morten Poulsen for supplying the grphics nd Yokun Wu for spotting some errors in n erlier version. The uthor is prtilly supported y the DNRF through the Centre for Symmetry nd Deformtion in Copenhgen. /home/moller/underv/lgtop/notes/chp1/clsscover.tex. 1
2 2 J.M. MØLLER u 1 (s) x 0 x 1 h(s, t) u 0 (s) Figure 1. A pth homototopy etween two pths u u x 0, u u x 1 (u v) w u (v w) u 0 u 1, v 0 v 1 = u 0 v 0 u 1 v 1 These drwings re ment to suggest proofs for the first three sttements: u(s) x 0 u v w x 0 x 1 x 0 x 0 x 0 x 1 x 2 x 3 u(2s) x 1 u(2s) u(1 2s) u v w In the first cse, we first run long u with doule speed nd then stnd still t the end point x 1 for hlf the time. The homotopy consists in slowing down on the pth u nd spending less time just stnding still t x 1. In the second cse, we first run long u ll the wy to x 1 with doule speed nd then ck gin long u lso with doule speed. The homotopy consists in running out long u, stnding still for n incresing length of time (nmely for 1 2 (1 t) s 1 2 (1 + t)), nd running ck long u. In the third cse we first run long u with speed 4, then long v with speed 4, then long w with speed 2, nd we must show tht cn deformed into the cse where we run long u with speed 2, long v with speed 4, long w with speed 4. This cn e chieved y slowing down on u, keeping the sme speed long v, nd speeding up on w. The fourth of the ove rules is proved y this picture, u 1 (2s) v 1 (2s 1) x 0 x 1 x 2 u 0 (2s) v 0 (2s 1) We write π(x)(x 0, x 1 ) for the set of l homotopy clsses of pths in X from x 0 to x 1 nd we write [u] π(x)(x 0, x 1 ) for the homotopy clss contining the pth u. The fourth rule implies tht the product opertion on pths induces product opertion on homotopy clsses of pths (1.2) π(x)(x 0, x 1 ) π(x)(x 1, x 2 ) π(x)(x 0, x 2 ): ([u], [v]) [u] [v] = [u v] nd the other three rules trnslte to similr rules [x 0 ] [u] = [u] = [u] [x 1 ] (neutrl elements) [u] [u] = [x 0 ], [u] [u] = [x 1 ] (inverse elements) ([u] [v]) [w] = [u] ([v] [w]) (ssocitivity)
3 THE FUNDAMENTAL GROUP AND COVERING SPACES 3 for this product opertion. We next look t the functoril properties of this construction. Suppose tht f : X Y is mp of spces. If u is pth in X from x 0 to x 1, then the imge fu is pth in Y from f(x 0 ) to f(x 1 ). Since homotopic pths hve homotopic imges there is n induced mp π(x)(x 0, x 1 ) π(f) π(y )(f(x 0 ), f(x 1 )): [u] [fu] on the set of homotopy clsses of pths. Oserve tht this mp π(f) does not chnge if we chnge f y homotopy reltive to {x 0, x 1 }, π(f) respects the pth product opertion in the sense tht π(f)([u] [v]) = π(f)([u]) π(f)([v]) when u(1) = v(0), π(id X ) = id π(x)(x1,x 2), π(g f) = π(g) π(f) for mps g : Y Z We now summrize our findings. Proposition 1.3. For ny spce X, π(x) is groupoid, nd for ny mp f : X Y etween spces the induced mp π(f): π(x) π(y ) is groupoid homomorphism. In fct, π is functor from the ctegory of topologicl spces to the ctegory of groupoids. Definition 1.4. π(x) is clled the fundmentl groupid of X. The fundmentl group sed t x 0 X is the group π 1 (X, x 0 ) = π(x)(x 0, x 0 ) of homotopy clsses of loops in X sed t x 0. The pth product (1.2) specilizes to product opertion nd to trnsitive free group ctions π 1 (X, x 0 ) π 1 (X, x 0 ) π 1 (X, x 0 ) (1.5) π 1 (X, x 0 ) π(x)(x 0, x 1 ) π(x)(x 0, x 1 ) π(x)(x 0, x 1 ) π 1 (X, x 1 ) so tht π 1 (X, x 0 ) is indeed group nd π(x)(x 0, x 1 ) is n ffine group from the left nd from the right. For fundmentl groups, in prticulr, ny sed mp f : (X, x 0 ) (Y, y 0 ) induces group homomorphism π(f) = f : π 1 (X, x 0 ) π 1 (Y, y 0 ), given y π 1 (f) = f ([u]) = [fu], tht only depends on the sed homotopy clss of the sed mp f. Proposition 1.6. The fundmentl group is functor π 1 : hotop Grp from the homotopy ctegory of sed topologicl spces into the ctegory of groups. This mens tht π 1 (id (X,x0)) = id π1(x,x 0) nd π 1 (g f) = π 1 (g) π 1 (f). It follows immeditely tht if f : X Y is homotopy equivlence of sed spces then the induced mp π 1 (f) = f : π 1 (X, x 0 ) π 1 (Y, y 0 ) is n isomorphism of groups. (See Section 5 for more informtion out ctegories nd functors.) Corollry 1.7. Let X e spce, A suspce, nd i : π 1 (A, 0 ) π 1 (X, 0 ) the group homomorphism induced y the inclusion mp i: A X. (1) If A is retrct of X then i hs left inverse (so it is monomorphism). (2) If A is deformtion retrct of X then i hs n inverse (so it is n isomorphism). Proof. (1) Let r : X A e mp such tht ri = 1 A. Then r i is the identity isomorphism of π 1 (A, 0 ). (2) Let r : X A e mp such tht ri = 1 A nd ir 1 X rel A. Then r i is the identity isomorphism of π 1 (A, 0 ) nd i r is the identity isomorphism of π 1 (X, 0 ) so i nd r re ech others inverses. Corollry 1.8. Let X nd Y e spces. There is n isomorphism (p X ) (p Y ) : π 1 (X Y, x 0 y 0 ) π 1 (X, x 0 ) π 1 (Y, y 0 ) induced y the projection mps p X : X Y X nd p Y : X Y Y. Proof. The loops in X Y hve the form u v where u nd v re loops in X nd Y, respectively (Generl Topology, 2.63). The ove homomorphism hs the form [u v] [u] [v]. The inverse homomorphism is [u] [v] [u v]. Note tht this is welldefined. We cn now compute our first fundmentl group. Exmple 1.9. π 1 (R n, 0) is the trivil group with just one element ecuse R n contins the suspce {0} consisting of one point s deformtion retrct. Any spce tht deformtion retrct onto on of its points hs trivil fundmentl group. Is it true tht ny contrctile spce hs trivil fundmentl group?
4 4 J.M. MØLLER Our tools to compute π 1 in more interesting cses re covering spce theory nd vn Kmpen s theorem Chnge of se point nd unsed homotopies. Wht hppens if we chnge the se point? In cse, the new se point lies in nother pthcomponent of X, there is no reltion t ll etween the fundmentl groups. But if the two se points lie in the sme pthcomponent, the fundmentl groups re isomorphic. Lemm If u is pth from x 0 to x 1 then conjugtion with [u] is group isomorphism. π 1 (X, x 1 ) π 1 (X, x 0 ): [v] [u] [v] [u] Proof. This is immedite from the rules for products of pths nd specil cse of (1.5). We lredy noted tht if two mps re homotopic reltive to the se point then they induce the sme group homomorphism etween the fundmentl groups. We shll now investigte how the fundmentl group ehves with respect to free mps nd free homotopies, ie mps nd homotopies tht do not preserve the se point. Lemm Suppose tht f 0 f 1 : X Y re homotopic mps nd h: X I Y homotopy. For ny point x X, let h(x) π(y )(f 0 (x), f 1 (x)) e the pth homotopy clss of t h(x, t). For ny u π(x)(x 0, x 1 ) there is commuttive digrm f 0 (x 0 ) h(x 0) f 1 (x 0 ) in π(y ). f 0(u) f 1(u) f 0 (x 1 ) f 1 (x 1 ) h(x1) Proof. Let u e ny pth from x 0 to x 1 in X. If we push the left nd upper edge of the homotopy I I Y : (s, t) h(u(s), t) into the lower nd right edge f 1 (u) h(x 0 ) h(x 1 ) f 0 (u) we otin pth homotopy h(x 0 ) f 1 (u) f 0 (u) h(x 1 ). Corollry In the sitution of Lemm 1.12, the digrm commutes. π 1 (Y, f 1 (x 0 )) (f 1) π 1 (X, x 0 ) (f 0) π 1 (Y, f 0 (x 0 )) = [h(x0)] [h(x 0)] Proof. For ny loop u sed t x 0, f 0 (u)h(x 0 ) = h(x 0 )f 1 (u) or f 0 (u) = h(x 0 )f 1 (u)h(x 0 ). Corollry (1) If f : X Y is homotopy equivlence (possily unsed) then the induced homomorphism f : π 1 (X, x 0 ) π 1 (Y, f(x 0 )) is group isomorphism. (2) If f : X Y is nullhomotopic (possily unsed) then f : π 1 (X, x 0 ) π 1 (Y, f(x 0 )) is the trivil homomorphism.
5 THE FUNDAMENTAL GROUP AND COVERING SPACES 5 Proof. (1) Let g e homotopy inverse to f so tht gf 1 X nd fg 1 Y. By Lemm 1.12 there is commuttive digrm π 1 (X, x 0 ) π 1 (Y, f(x 0 )) = f f π 1 (Y, f(x 0 )) g π 1 (X, gf(x 0 )) π 1 (Y, fgf(x 0 )) = π 1 (X, x 0 ) which shows tht g is oth injective nd surjective, ie g is ijective. Then lso f is ijective. (2) If f homotopic to constnt mp c then f followed y n isomorphism equls c which is trivil. Thuse lso f is trivil. We cn now nswer question from Exmple 1.9 nd sy tht ny contrctile spce hs trivil fundmentl group. Definition A spce is simply connected if there is unique pth homotopy clss etween ny two of its points. The spce X is simply connected if π(x)(x 1, x 2 ) = for ll x 1, x 2 X, or, equivlently, X is pth connected nd π 1 (X, x) = t ll points or t one point of X. 2. Covering spces A covering mp over X is mp tht loclly looks like the projection mp X F X for some discrete spce F. Definition 2.1. A covering mp is continuous surjective mp p: Y X with the property tht for ny point x X there is neighorhood U (n evenly covered neighorhood), discrete set F, nd homeomorphism U F p 1 (U) such tht the digrm U F p 1 (U) commutes. pr 1 p p 1 (U) U Some covering spces, ut not ll (7.22), rise from left group ctions. Consider left ction G Y Y of group G on spce Y. Let p G : Y G\Y e the quotient mp of Y onto the orit spce G\Y. The quotient mp p G is open ecuse open susets U Y hve open sturtions GU = g G gu = p 1 G p G(U) (Generl Topology 2.82). The open sets in G\Y correspond ijectively to sturted open sets in Y. We now single out the left ctions G Y Y for which the quotient mp p G : Y G\Y of Y onto its orit spce is covering mp. Definition 2.2. [5, (*) p. 72] A covering spce ction is group ction G Y Y where ny point y Y hs neighorhood U such tht the trnslted neighorhoods gu, g G, re disjoint. (In other words, the ction mp G U GU is homeomorphism.) Exmple 2.3. The ctions Z R R: (n, t) n + t Z/2 S n S n : (±1, x) ±x Z/m S 2n+1 S 2n+1 : (ζ, x) ζx, where ζ C is n mth root of unity, ζ m = 1, {±1, ±i, ±j, ±k} S 3 S 3, quternion multipliction [5, Exmple 1.43], re covering spce ctions nd the orit spces re Z\R = S 1 (the circle), Z/2\S n = RP n (rel projective spce), nd Z/m\S 2n+1 = L 2n+1 (m) (lense spce). The ction Z S 1 S 1 : (n, z) e πi 2n z is not covering spce ction for the orits re dense.
6 6 J.M. MØLLER Exmple 2.4. The mps p n : S 1 S 1, n Z, nd p : R S 1 given y p n (z) = z n, nd p (s) = e 2πs = (cos(2πs), sin(2πs)) re covering mps of the circle with fire p 1 n (1) = Z/nZ nd p 1 (1) = Z. There re mny covering mps of S 1 S 1. The mp S n C 2 \S n = RP n, n 1, is covering mp of rel projective nspce. The mp S 2n+1 C m \S 2n+1 = L 2n+1 (m) is covering mp of the lens spce. M g N g+1 doule covering mp of the unorientle surfce of genus g + 1 with F = Z/2Z. Cn you find covering mp of M g? Cn you find covering mp of R? Theorem 2.5 (Unique HLP for covering mps). [5, 1.30] Let p: Y X e covering mp, B e ny spce, nd h: B I X homotopy into the se spce. If one end of the homotopy lifts to mp B {0} Y then the whole homotopy dmits unique lift B I Y such tht the digrm commutes. e h0 B {0} Y e h B I X Proof. We consider first the cse where B is point. The sttement is then tht in the sitution h {0} y0 Y eu I X u there is unique mp ũ: I Y such tht pũ = u nd ũ(0) = y 0. For uniqueness of lifts from I see Theorem 2.12.(1). We need to prove existence. The Leesgue lemm (Generl Topology, 2.158) pplied to the compct spce I sys tht there is sudivision 0 = t 0 < t 1 < < t n = 1 of I such tht u mps ech of the closed suintervls [t i 1, t i ] into n evenly covered neighorhood in X. Suppose tht we hve lifted u to ũ defined on [0, t i 1 ]. Let U e n evenly covered neighorhood of u(t i 1 ). Suppose tht the lift ũ(t i 1 ) elongs to U {l} for some l F. Continue the given ũ with (p (U {l})) 1 u [t i 1, t i ]. After finitely mny steps we hve the unique lift on I. We now turn to the generl sitution. Uniqueness is cler for we hve just seen tht lifts re uniquely determined on the verticl slices {} I B I for ny point of B. Existence is lso cler except tht continuity of the lift is not cler. We now prove tht the lift is continuous. Let e ny point of X. By compctness, there is neighorhood N of nd sudivision 0 = t 0 < t 1 < < t n = 1 of I such tht h mps ech of the sets N [t i 1, t i ] into n evenly covered neighorhood of X. Suppose tht h(n [0, t 1 ]) is contined in the evenly covered neighorhood U X nd let Ũ p 1 (U) Y e neighorhood such tht p Ũ : Ũ U is homeomorphism nd h 0 (, 0) Ũ. We cn not e sure tht h 0 (N {0}) Ũ; only if N is connected. Replce N y 1 N h 0 (Ũ). Then h 0 (N {0}) Ũ. Then (p Ũ) 1 h N [0, t 1 ] is lift of h N [0, t 1 ] extending h 0. After finitely mny steps we hve lift defined on N I (where N is possily smller thn the N we strted with). Do this for every point of B. These mps must gree on their overlp y uniqueness. So they define lift B I Y. This lift is continuous since it is continuous on ech of the open tues N I. We emphsize the specil cse where B is point. Let y 0 Y e point in Y nd x 0 = p(y 0 ) X its imge in X. Corollry 2.6 (Unique pth lifting). Let x 0 nd x 1 e two points in X nd let y 0 e point in the fire p 1 (x 0 ) Y over x 0. For ny pth u: I X from x 0 to x 1, the exists unique pth ũ: I Y in Y strting t ũ(0) = y 0. Moreover, homotopic pths hve homotopic lifts: If v : I X is pth in X tht is pth homotopic to u then the lifts ũ nd ṽ re lso pth homotopic. Proof. First, in Theorem 2.5, tke B to e point. Next, tke B to e I nd use the HLP to see tht homotopic pths hve homotopic lifts. Corollry 2.7. Let p: Y X e covering mp nd let y 0, y 1, y 2 Y, x 0 = py 0, x 1 = py 1, x 2 = py 2. p p
7 THE FUNDAMENTAL GROUP AND COVERING SPACES 7 (1) By recording end points of lifts we otin mps p 1 (x 1 ) π(x)(x 1, x 2 ) p 1 (x 2 ), p 1 (x 0 ) π 1 (X, x 0 ) p 1 (x 0 ) given y y [u] = ũ y (1) where ũ y is the lift of u strting t y. Multipliction y pth u from x 1 to x 2 slides the fire over x 1 ijectively into the fire over x 2. (2) The covering mp p: X Y induces injective mps π(y )(y 1, y 2 ) p π(x)(x 1, x 2 ), π 1 (Y, y 0 ) p π 1 (X, x 0 ) The suset p π(y )(y 1, y 2 ) π(x)(x 1, x 2 ) consists of ll pths from x 1 to x 2 tht lift to pths from y 1 to y 2. The sugroup p π 1 (Y, y 0 ) π 1 (X, x 0 ) consists of ll loops t x 0 tht lloft to loops t y 0. Definition 2.8. The monodromy functor of the covering mp p: X Y is functor F (p): π(x) Set of the fundmentl groupoid of the se spce into the ctegory Set of sets. This functor tkes point in x X to the fire F (p)(x) = p 1 (x) over tht point nd it tkes pth homotopy clss u π(x)(x 0, x 1 ) to F (p)(x 0 ) = p 1 (x 0 ) p 1 (x 1 ) = F (p)(x 1 ): y y u. (The nottion here is such tht F (p)(uv) = F (p)(v) F (p)(u) for pths u π(x)(x 0, x 1 ), v π(x)(x 1, x 2 ).) In prticulr, the fire F (p)(x) = p 1 (x) over ny point x X is right π 1 (X, x)set. Corollry 2.9 (The fundmentl groupoid of covering spce). The fundmentl groupoid of Y, π(y ) = π(x) F (p) is the Grothendieck construction of the fier functor (2.8). In other words, the mp π(p): π(y )(y 0, y 1 ) π(x)(x 0, x 1 ) is injective nd the imge is the set of pth homotopy clsses from x 0 to x 1 tht tke y 0 to y 1. In prticulr, the homomorphism p : π 1 (Y, y 0 ) π 1 (X, x 0 ) is injective nd its imge is the set of loops t x 0 tht lift to loops t y 0. Proof. We consider the functor F (p) s tking vlues in discrete ctegories. The ojects of π(x) F (p) re pirs (x, y) where x X nd y F (p)(x) Y. A morphism (x 1, y 1 ) (x 2, y 2 ) is pir (u, v) where u is morphism in π(x) from x 1 to x 2 nd v is morphism in F (p)(x 2 ) from F (p)(u)(x 1 ) = x 1 u to y 2. As F (p)(x 2 ) hve no morphisms ut identities, the set of morphisms (x 1, y 1 ) (x 2, y 2 ) is the set of u π(x)(x 1, x 2 ) such tht y 1 u = y 2. This is precisely π(y )(y 1, y 2 ). Definition For spce X, let π 0 (X) e the set of pth components of X. Lemm Let p: X Y e covering mp. (1) Suppose tht X is pth connected. The inclusion p 1 (x 0 ) Y induces ijection p 1 (x 0 )/π 1 (X, x 0 ) π 0 (Y ). In prticulr, Y is pth connected π 1 (X, x 0 ) cts trnsitively on the fire p 1 (x 0 ) (2) Suppose tht X nd Y re pth connected. The mps π 1 (Y, y 1 )\π(x)(x 1, x 2 ) p 1 (x 2 ) π 1 (Y, y 0 )\π 1 (X, x 0 ) p 1 (x 0 ) π 1 (Y, y 1 )u y 1 u [pu y ] y π 1 (Y, y 0 )u y 0 u [pu y ] y re ijections. Here, u y is ny pth in Y from y 1 or y 0 to y. In prticulr, π 1 (X, x 0 ): π 1 (Y, y 0 ) = p 1 (x 0 ). Proof. The mp p 1 (x 0 ) π 0 (Y ), induced y the inclusion of the fire into the totl spce, is onto ecuse X is pth connected so tht ny point in the totl spce is connected y pth to point in the fire. Two points in the fire re in the sme pth component of Y if nd only if re in the sme π 1 (X, x 0 )orit. If Y is pth connected, then π 1 (X, x 0 ) cts trnsitively on the fire p 1 (x 0 ) with isotropy sugroup π 1 (Y, y 0 ) t y 0.
8 8 J.M. MØLLER Theorem 2.12 (Lifting Theorem). Let p: Y X e covering mp nd f : B X mp into the se spce. Choose se points such tht f( 0 ) = x 0 = p(y 0 ) nd consider the lifting prolem (B, 0 ) f f (Y, y 0 ) p (X, x 0 ) (1) If B is connected, then there exists t most one lift f : (B, 0 ) (Y, y 0 ) of f over p. (2) If B is pth connected nd loclly pth connected then There is mp f : (B, 0 ) (Y, y 0 ) such tht f = p f f π 1 (B, 0 ) p π 1 (Y, y 0 ) Proof. (1) Suppose tht f 1 nd f 2 re lifts of the sme mp f : B X. We clim tht the sets { B f 1 () = f 2 ()} nd { B f 1 () f 2 ()} re open. Let e ny point of B where the two lifts gree. Let U X e n evenly neighorhood of f(). Choose Ũ p 1 (U) = U F so tht the restriction of p to Ũ is homoemorphism nd f 1 () = f 2 () elongs to Ũ. f 1 1 Then f 1 nd f 1 2 gree on the neighorhood (Ũ) f 2 (Ũ) of. Let e ny point of B where the two lifts do not gree. Let U X e n evenly neighorhood of f(). Choose disjoint open sets Ũ1, Ũ2 p 1 (U) = U F so tht the restrictions of p to Ũ1 nd Ũ2 re homoemorphisms nd f 1 () elongs to Ũ1 nd f 2 () to Ũ2. Then f 1 nd f 2 do not gree on the neighorhood f 1 f 1 2 (Ũ2) of. 1 (Ũ1) (2) It is cler tht if the lift exists, then the condition is stisfied. Conversely, suppose tht the condition holds. For ny point in B, define lift f y f() = y 0 [fu ] where u is ny pth from 0 to. (Here we use tht B is pth connected.) If v is ny other pth from 0 to then y 0 [fu ] = y 0 [fv ] ecuse y 0 [fu fv ] = y 0 s the loop [fu fv ] π 1 (Y, y 0 ) fixes the point y 0 y Lemm We need to see tht f is continuous. Note tht ny point B hs pth connected neighorhood tht is mpped into n evenly covered neighorhood of f() in X. It is evident wht f does on this neighorhood of. A mp f : B S 1 C {0} into the circle hs n nth root if nd only if the induced homomorphism f : π 1 (B) Z is divisile y n. 3. The fundmentl group of the circle, spheres, nd lense spces For ech n Z, let ω n e the loop ω n (s) = (cos(2πns), sin(2πns), s I, on the circle. Theorem 3.1. The mp Φ: Z π 1 (S 1, 1): n [ω n ] is group isomorphism. Proof. Let p: R S 1 e the covering mp p(t) = (cos(2πt), sin(2πt)), t R. Rememer tht the totl spce R is simply connected s we sw in Exmple 1.9. The fire over 1 is p 1 (1) = Z. Let u n (t) = nt e the ovious pth from 0 to n Z. By Lemm 2.11 the mp Z π 1 (S 1, 1): n [pu n ] = [ω n ] is ijective. We need to verify tht Φ is group homomorphism. Let m nd n e integers. Then u m (m + u n ) is pth from 0 to m + n so it cn e used insted of u m+n when computing Φ(m + n). We find tht Φ(m + n) = [p(u m (m + u n ))] = [p(u m ) p(m + u n )] = [p(u m )][p(m + u n )] = [p(u m )][p(u n )] = Φ(m)Φ(n) ecuse p(m + u n ) = pu n s p hs period 1. Theorem 3.2. The nsphere S n is simply connected when n > 1.
9 THE FUNDAMENTAL GROUP AND COVERING SPACES 9 Proof. Let N e the North nd S the South Pole (or ny other two distinct points on S n ). The prolem is tht there re pths in S n tht visit every point of S n. But, in fct, ny loop sed t N is homotopic to loop tht voids S (Prolem nd Solution). This mens tht π 1 (S n {S}, N) π 1 (S n, N) is surjective. The result follows s S n {S} is homeomorphic to the simply connected spce R n. Corollry 3.3. The fundmentl group of rel projective nspce RP n is π 1 (RP n ) = C 2 for n > 1. The fundmentl group of the lense spce L 2n+1 (m) is π 1 (L 2n+1 (m)) = C m for n > 0. Proof. We proceed s in Theorem 3.1. Consider the cse of the the covering mp p: S 2n+1 L 2n+1 (m) over the lense spce L 2n+1 (m). Let N = (1, 0,..., 0) S 2n+1 C n+1. The cyclic group C m = ζ of mth roots of unity is generted y ζ = e 2πi/m. The mp ζ j ζ j N, j Z, is ijection C m p 1 pn etween the set C m nd the fire over pn. As S 2n+1 is simply connected there is ijection Φ: p 1 pn = C m π 1 (L 2n+1 (m), pn): ζ j [pω j ] where ω j is the pth in S 2n+1 from N to ζ j N given y ω j (s) = (e 2πisj/m, 0,..., 0). Since ω i+j ω i (ζ i ω j ), it follows just s in Theorem 3.1 tht Φ is group homomorphism. For the projective spces, use the pths ω j (s) = (cos(2πjs), sin(2πjs), 0,..., 0) from N to ( 1) j N, to see tht Φ: p 1 pn = C 2 π 1 (RP n, pn): (±1) j [pω j ] is ijection Applictions of π 1 (S 1 ). Here re some stndrd pplictions of Theorem 3.1. Corollry 3.5. The nth power homomorphism p n : (S 1, 1) (S 1, 1): z z n induces the nth power homomorphism π 1 (S 1, 1) π 1 (S 1, 1): [ω] [ω] n. Proof. (p n ) Φ(1) = (p n ) [ω 1 ] = [p n ω 1 ] = [ω n 1 ] = [ω n ] = Φ(n) = Φ(1) n. Theorem 3.6 (Brouwer s fixed point theorem). (1) The circle S 1 is not retrct of the disc D 2. (2) Any mp selfmp of the disc D 2 hs fixed point. Proof. (1) Let i: S 1 D 2 e the inclusion mp. The induced mp i : Z = π 1 (S 1 ) π 1 (D 2 ) = 0 is not injective so S 1 cn not e retrct y 1.7. (2) With the help of fixedpoint free self mp of D 2 one cn construct retrction of D 2 onto S 1. But they don t exist. Theorem 3.7 (The fundmentl theorem of lger). Let p(z) = z n + n 1 z n z + 0 e normed complex polynomil of degree n. If n > 0, then p hs root. Proof. Any normed polynomil p(z) = z n + n 1 z n z + 0 is nonzero when z is lrge: When z > 1 + n , then p(z) 0 ecuse n 1 z n n 1 z n < n 1 z n z n 1 = ( n ) z n 1 < z n Therefore ny normed polynomil p(z) defines mp S 1 (R) C {0} where S 1 (R) is the circle of rdius R nd R > 1 + n In fct, ll the normed polynomils p t (z) = z n + t( n 1 z n z + 0 ), t I, tke S 1 (R) into C {0} so tht we hve homotopy S 1 (R) I C {0}: (z, t) z n + t( n 1 z n z + 0 ) etween p 1 (z) = p(z) S 1 (R) nd p 0 (z) = z n. If p(z) hs no roots t ll, the mp p S 1 (R) fctors through the complex plne C nd is therefore nullhomotopic (s C is contrctile) nd so is the homotopic mp S 1 (R) C {0}: z z n nd the composite mp S 1 z Rz S 1 (R) z zn C {0} z z/ z S 1 But this is simply the mp S 1 S 1 : z z n which we know induces multipliction y n (3.5). However, nullhomotopic mp induces multipliction y 0 (1.14). So n = 0. A mp f : S 1 S 1 is odd if f( x) = f(x) for ll x S 1. Any rottion (or reflection) of the circle is odd (ecuse it is liner).
10 10 J.M. MØLLER Lemm 3.8. Let f : S 1 S 1 e n odd mp. Compose f with rottion R so tht Rf(1) = 1. The induced mp (Rf) : π 1 (S 1, 1) π 1 (S 1, 1) is multipliction y n odd integer. In prticulr, f is not nullhomotopic. Proof. We must compute (Rf) [ω 1 ]. The HLP gives lift 0 {0} R eω I ω 1 S 1 S 1 nd we hve (Rf) [ω 1 ] = [p ω]. When 0 s 1/2, ω 1 (s+1/2) = ω 1 (s) nd lso Rfω 1 (s+1/2) = Rfω 1 (s) s Rf is odd. The lift, ω of Rfω 1, then stisfies the eqution Rf ω(s + 1/2) = ω(s) + q/2 for some odd integer q. By continuity nd connectedness of the intervl [0, 1/2], q does not depend on s. Now ω(1) = ω(1/2) + q/2 = ω(0) + q/2 + q/2 = q nd therefore (Rf) [ω 1 ] = [p ω] = [ω q ] = [ω 1 ] q. We conclude tht (Rf) is multipliction y the odd integer q. Since nullhomopotic mp induces the trivil group homomorphism (1.14), f is not nullhomotopic. Theorem 3.9 (Borsuk Ulm theorem for n = 2). Let f : S 2 R 2 e ny continuous mp. Then there exists point x S 2 such tht f(x) = f( x). Proof. Suppose tht f : S 2 R 2 is mp such tht f(x) f( x) for ll x S 2. The composite mp f(x) f( x) S 1 x incl f(x) f( x) 2 S is odd so it is not nullhomotopic. But the first mp S 1 S 2 is nullhomotopic ecuse it fctors through the contrctile spce D+ 2 = {(x 1, x 2, x 3 ) S 2 x 3 0}. This is contrdiction. This implies tht ther re no injective mps of S 2 R 2 ; in prticulr S 2 does not emed in R 2. Proposition 3.10 (Borsuk Ulm theorem for n = 1). Let f : S 1 R e ny continuous mp. Then there exists point x S 1 such tht f(x) = f( x). Proof. Look t the mp g(x) = f(x) f( x). If g is identiclly 0, f(x) = f( x) for ll x S 1. Otherwise, g is n odd function, g( x) = g(x), nd g hs oth positive nd negtive vlues. By connectedness, g must ssume the vlue 0 t some point. This implies tht there re no injective mps S 1 R; in prticulr S 1 does not emed in R. 4. The vn Kmpen theorem Let G j, j J, e set of groups indexed y the set J. The coproduct (or free product) of these groups is group i J G j with group homomorphisms ϕ j : G j j J G j such tht (4.1) Hom( j J S 1 G j, H) = j J Hom(G j, H): ϕ (ϕ ϕ j ) j J is ijection for ny group H. The group j J G j contins ech group G j s sugroup nd these sugroups do not commute with ech other. If the groups hve presenttions G j = L j R j then j J L j R j = j J L j j J R j s this group hs the universl property. See [9, 6.2] for the construction of the free product. The chrcteristic property (4.1) pplied to H = j J G j shows tht there is group homomorphism Gj G j from the free product to the direct product whose restriction to ech G j is the inclusion into the product. Exmple 4.2. [9, Exmple II III p 171] Z/2 Z/2 = Z Z/2 nd Z/2 Z/3 = PSL(2, Z). We cn prove the first ssertion: Z/2 Z/2 =, 2, 2 =,, c 2, 2, c = =, 2, cc, c =, 2, c, c = c 1 ut the second one is more difficult.
11 THE FUNDAMENTAL GROUP AND COVERING SPACES 11 Suppose tht the spce X = j J X j is the union of open nd pth connected suspces X j nd tht x 0 is point in j J X j. The inclusion of the suspce X j into X induces group homomorphism ι j : π 1 (X j, x 0 ) π 1 (X, x 0 ). The coproduct j J π 1(X j, x 0 ) is group equipped with group homomorphisms ϕ j : π 1 (X j, x 0 ) j J π 1(X j, x 0 ). Let Φ: j J π 1 (X j, x 0 ) π 1 ( j J X j, x 0 ) = π 1 (X, x 0 ) e the group homomorphism determined y Φ ϕ j = ι j. Is Φ surjective? In generl, no. The circle, for instnec, is the union of two contrctile open suspces, so Φ is not onto in tht cse. But, if ny loop in X is homotopic to product of loops in one of the suspces X j, then Φ is surjective. Is Φ injective? It will, in generl, not e injective, ecuse the individul groups π 1 (X i ) in the free product do not intersect ut the suspces do intersect. Any loop in X tht is loop in X i X j will in the free product count s loop oth in π 1 (X i ) nd in π 1 (X j ). We lwys hve commuttive digrms of the form π 1 (X i, x 0 ) ι ij ι i π 1 (X i X j, x 0 ) π 1 (X, x 0 ) ι ji ι j π 1 (X j, x 0 ) where ι ij re inclusion mps. This mens tht Φ(ι ij g) = Φ(ι ji g) for ny g π 1 (X i X j, x 0 ) so tht (4.3) i, j J g π 1 (X i X j ): ι ij (g)ι ji (g) 1 ker Φ Let N j J π 1(X j, x 0 ) e the smllest norml sugroup contining ll the elements of (4.3). The kernel of Φ must contin N ut, of course, the kernel could e igger. The surprising fct is tht often it isn t. Theorem 4.4 (Vn Kmpen s theorem). Suppose tht X = j J X j is the union of open nd pth connected suspces X j nd tht x 0 is point in j J X j. (1) If the intersection of ny two of the open suspces is pth connected then Φ is surjective. (2) If the intersection of ny three of the open suspces is pth connected then the kernel of Φ is N. Corollry 4.5. If the intersection of ny three of the open suspces is pth connected then Φ determines n isomorphism Φ: j J π 1 (X j, x 0 )/N = π 1 (X, x 0 ) Proof of Theorem 4.4. (1) We need to show tht ny loop u π 1 (X) in X is product u 1 u m of loops u i π 1 (X ji ) in one of the suspces. Let u: I X e loop in X. Thnks to the Leesgue lemm (Generl Topology, 2.158) we cn find sudivision 0 = t 0 < t 1 < t m = 1 of the unit intervl so tht u i = u [t i 1, t i ] is pth in (sy) X i. As u(t i ) X i X i+1, nd lso the se point x 0 X i X i+1, nd X i X i+1 is pth connected, there is pth g i in X i X i+1 from the sepoint x 0 to u(t i 1 ). The sitution looks like this: X 1 X 2 u [0, t 1 ] g 1 u [t 1, t 2 ] g 2 X 3 u [t 2, 1] Now u u [0, t 1 ] u [t 1, t 2 ] u [t m 1, 1] (u [0, t 1 ] g 1 ) (g 1 u [t 1, t 2 ] g 2 ) (g m u [t m 1, 1]) is product of loops where ech fctor is inside one of the suspces.
12 12 J.M. MØLLER (2) Let N π 1 (X i ) e the smllest norml sugroup contining ll the elements (4.3). Let u i π 1 (X ji ). For simplicity, let s cll X ji for X i. Consider the product u 1 }{{} u 2 }{{} π 1(X 1) π 1(X 2) u }{{} m j J π 1 (X j ) π 1(X m) nd suppose tht Φ(u 1 u m ) is the unit element of π 1 (X). We wnt to show tht u 1 u m lies in the norml sugroup N or tht u 1 u m is the identity in the quotient group π 1 (X j )/N. Since u 1 u m is homotopic to the constnt loop in X there is homotopy I I X = X j from the loop u 1 u m in X to the constnt loop. Divide the unit squre I I into smller rectngles such tht ech rectngle is mpped into one of the suspces X j. We my ssume tht the sudivision of I {0} is further sudivision of the sudivision t i/m coming from the product u 1 u m. It could e tht one new vertex is (or more new vertices re) inserted etween (i 1)/m nd i/m. X k X l X i Connect the imge of the new vertex with pth g inside X i X k X l to the se point. Now u i is homotopic in X i to the product (u i [(i 1)/m, ] g) (g u i [, i/m]) of two loops in X i. This mens tht we my s well ssume tht no new sudivision points hve een introduced t the ottom line I {0}. Now pertur slightly the smll rectngles, ut not the ones in the ottom nd top row, so tht lso the corner of ech rectngle lies in t most three rectngles. The lower left corner my look like this: X 5 X 6 u 15 u 16 X 1 u 12 X 2 u 1 The loop u 1 in X 1 is homotopic to the product of pths u 15 u 16 u 12 y homotopy s in the proof of Connect the imge of the point to the se point y pth g 156 inside X 1 X 5 X 6 nd connect the imge of the point to the se point y pth g 126 inside X 1 X 2 X 6. Then u 1 is homotopic in X 1 to the product of loops (u 15 g 156 ) (g 156 u 16 g 126 ) (g 126 u 12 ) in X 1. The first of these loops is loop in X 1 X 5, the second is loop in X 1 X 6, nd the third is loop in X 1 X 2. In π 1 (X j ) nd modulo the norml sugroup N we hve tht u }{{} 1 u 2 = u }{{} 15 g 156 g } {{ } 156u 16g 126 g } {{ } 126u 12 u } {{ } 2 = u }{{} 15 g 156 g } {{ } 156u 16g 126 g } {{ } 126u 12 u 2 } {{ } X 1 X 2 X 1 X 1 X 1 X 2 X 5 X 6 X 2 After finitely mny steps we conclude tht modulo N the product u 1 u m equls product of constnt loops, the identity element. Corollry 4.6. Let X j e set of pth connected spces. Then X j ) j J π 1 (X j ) = π 1 ( j J provided tht ech se point x j X j is the deformtion retrct of n open neighorhood U j X j. Proof. Vn Kmpen s theorem does not pply directly to the suspces X j of X j ecuse they re not open. Insted, let X j = X j i J U i. The suspces X j re open nd pth connected nd the intersection of t lest two of them is the contrctile spce i J U i. Moreover, X j is deformtion retrct of X j. For instnce, punctured compct surfces hve free fundmentl groups.
EQUATIONS OF LINES AND PLANES
EQUATIONS OF LINES AND PLANES MATH 195, SECTION 59 (VIPUL NAIK) Corresponding mteril in the ook: Section 12.5. Wht students should definitely get: Prmetric eqution of line given in pointdirection nd twopoint
More informationLINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES
LINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES DAVID WEBB CONTENTS Liner trnsformtions 2 The representing mtrix of liner trnsformtion 3 3 An ppliction: reflections in the plne 6 4 The lgebr of
More informationReasoning to Solve Equations and Inequalities
Lesson4 Resoning to Solve Equtions nd Inequlities In erlier work in this unit, you modeled situtions with severl vriles nd equtions. For exmple, suppose you were given usiness plns for concert showing
More informationPolynomial Functions. Polynomial functions in one variable can be written in expanded form as ( )
Polynomil Functions Polynomil functions in one vrible cn be written in expnded form s n n 1 n 2 2 f x = x + x + x + + x + x+ n n 1 n 2 2 1 0 Exmples of polynomils in expnded form re nd 3 8 7 4 = 5 4 +
More informationGeneralized Inverses: How to Invert a NonInvertible Matrix
Generlized Inverses: How to Invert NonInvertible Mtrix S. Swyer September 7, 2006 rev August 6, 2008. Introduction nd Definition. Let A be generl m n mtrix. Then nturl question is when we cn solve Ax
More informationand thus, they are similar. If k = 3 then the Jordan form of both matrices is
Homework ssignment 11 Section 7. pp. 24925 Exercise 1. Let N 1 nd N 2 be nilpotent mtrices over the field F. Prove tht N 1 nd N 2 re similr if nd only if they hve the sme miniml polynomil. Solution: If
More informationExample 27.1 Draw a Venn diagram to show the relationship between counting numbers, whole numbers, integers, and rational numbers.
2 Rtionl Numbers Integers such s 5 were importnt when solving the eqution x+5 = 0. In similr wy, frctions re importnt for solving equtions like 2x = 1. Wht bout equtions like 2x + 1 = 0? Equtions of this
More informationMath 314, Homework Assignment 1. 1. Prove that two nonvertical lines are perpendicular if and only if the product of their slopes is 1.
Mth 4, Homework Assignment. Prove tht two nonverticl lines re perpendiculr if nd only if the product of their slopes is. Proof. Let l nd l e nonverticl lines in R of slopes m nd m, respectively. Suppose
More informationA.7.1 Trigonometric interpretation of dot product... 324. A.7.2 Geometric interpretation of dot product... 324
A P P E N D I X A Vectors CONTENTS A.1 Scling vector................................................ 321 A.2 Unit or Direction vectors...................................... 321 A.3 Vector ddition.................................................
More informationFactoring Trinomials of the Form. x 2 b x c. Example 1 Factoring Trinomials. The product of 4 and 2 is 8. The sum of 3 and 2 is 5.
Section P.6 Fctoring Trinomils 6 P.6 Fctoring Trinomils Wht you should lern: Fctor trinomils of the form 2 c Fctor trinomils of the form 2 c Fctor trinomils y grouping Fctor perfect squre trinomils Select
More informationExample A rectangular box without lid is to be made from a square cardboard of sides 18 cm by cutting equal squares from each corner and then folding
1 Exmple A rectngulr box without lid is to be mde from squre crdbord of sides 18 cm by cutting equl squres from ech corner nd then folding up the sides. 1 Exmple A rectngulr box without lid is to be mde
More informationExponentiation: Theorems, Proofs, Problems Pre/Calculus 11, Veritas Prep.
Exponentition: Theorems, Proofs, Problems Pre/Clculus, Verits Prep. Our Exponentition Theorems Theorem A: n+m = n m Theorem B: ( n ) m = nm Theorem C: (b) n = n b n ( ) n n Theorem D: = b b n Theorem E:
More informationMODULE 3. 0, y = 0 for all y
Topics: Inner products MOULE 3 The inner product of two vectors: The inner product of two vectors x, y V, denoted by x, y is (in generl) complex vlued function which hs the following four properties: i)
More informationHomework 3 Solutions
CS 341: Foundtions of Computer Science II Prof. Mrvin Nkym Homework 3 Solutions 1. Give NFAs with the specified numer of sttes recognizing ech of the following lnguges. In ll cses, the lphet is Σ = {,1}.
More information4.11 Inner Product Spaces
314 CHAPTER 4 Vector Spces 9. A mtrix of the form 0 0 b c 0 d 0 0 e 0 f g 0 h 0 cnnot be invertible. 10. A mtrix of the form bc d e f ghi such tht e bd = 0 cnnot be invertible. 4.11 Inner Product Spces
More information4: RIEMANN SUMS, RIEMANN INTEGRALS, FUNDAMENTAL THEOREM OF CALCULUS
4: RIEMA SUMS, RIEMA ITEGRALS, FUDAMETAL THEOREM OF CALCULUS STEVE HEILMA Contents 1. Review 1 2. Riemnn Sums 2 3. Riemnn Integrl 3 4. Fundmentl Theorem of Clculus 7 5. Appendix: ottion 10 1. Review Theorem
More informationSquare Roots Teacher Notes
Henri Picciotto Squre Roots Techer Notes This unit is intended to help students develop n understnding of squre roots from visul / geometric point of view, nd lso to develop their numer sense round this
More informationSection 54 Trigonometric Functions
5 Trigonometric Functions Section 5 Trigonometric Functions Definition of the Trigonometric Functions Clcultor Evlution of Trigonometric Functions Definition of the Trigonometric Functions Alternte Form
More information. At first sight a! b seems an unwieldy formula but use of the following mnemonic will possibly help. a 1 a 2 a 3 a 1 a 2
7 CHAPTER THREE. Cross Product Given two vectors = (,, nd = (,, in R, the cross product of nd written! is defined to e: " = (!,!,! Note! clled cross is VECTOR (unlike which is sclr. Exmple (,, " (4,5,6
More informationMATH 150 HOMEWORK 4 SOLUTIONS
MATH 150 HOMEWORK 4 SOLUTIONS Section 1.8 Show tht the product of two of the numbers 65 1000 8 2001 + 3 177, 79 1212 9 2399 + 2 2001, nd 24 4493 5 8192 + 7 1777 is nonnegtive. Is your proof constructive
More informationSection 74 Translation of Axes
62 7 ADDITIONAL TOPICS IN ANALYTIC GEOMETRY Section 74 Trnsltion of Aes Trnsltion of Aes Stndrd Equtions of Trnslted Conics Grphing Equtions of the Form A 2 C 2 D E F 0 Finding Equtions of Conics In the
More informationP.3 Polynomials and Factoring. P.3 an 1. Polynomial STUDY TIP. Example 1 Writing Polynomials in Standard Form. What you should learn
33337_0P03.qp 2/27/06 24 9:3 AM Chpter P Pge 24 Prerequisites P.3 Polynomils nd Fctoring Wht you should lern Polynomils An lgeric epression is collection of vriles nd rel numers. The most common type of
More informationPentominoes. Pentominoes. Bruce Baguley Cascade Math Systems, LLC. The pentominoes are a simplelooking set of objects through which some powerful
Pentominoes Bruce Bguley Cscde Mth Systems, LLC Astrct. Pentominoes nd their reltives the polyominoes, polycues, nd polyhypercues will e used to explore nd pply vrious importnt mthemticl concepts. In this
More informationAppendix D: Completing the Square and the Quadratic Formula. In Appendix A, two special cases of expanding brackets were considered:
Appendi D: Completing the Squre nd the Qudrtic Formul Fctoring qudrtic epressions such s: + 6 + 8 ws one of the topics introduced in Appendi C. Fctoring qudrtic epressions is useful skill tht cn help you
More informationFactoring Polynomials
Fctoring Polynomils Some definitions (not necessrily ll for secondry school mthemtics): A polynomil is the sum of one or more terms, in which ech term consists of product of constnt nd one or more vribles
More informationAssuming all values are initially zero, what are the values of A and B after executing this Verilog code inside an always block? C=1; A <= C; B = C;
B26 Appendix B The Bsics of Logic Design Check Yourself ALU n [Arthritic Logic Unit or (rre) Arithmetic Logic Unit] A rndomnumer genertor supplied s stndrd with ll computer systems Stn KellyBootle,
More information6.2 Volumes of Revolution: The Disk Method
mth ppliction: volumes of revolution, prt ii Volumes of Revolution: The Disk Method One of the simplest pplictions of integrtion (Theorem ) nd the ccumultion process is to determine soclled volumes of
More information5.2. LINE INTEGRALS 265. Let us quickly review the kind of integrals we have studied so far before we introduce a new one.
5.2. LINE INTEGRALS 265 5.2 Line Integrls 5.2.1 Introduction Let us quickly review the kind of integrls we hve studied so fr before we introduce new one. 1. Definite integrl. Given continuous relvlued
More informationPROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY
MAT 0630 INTERNET RESOURCES, REVIEW OF CONCEPTS AND COMMON MISTAKES PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY Contents 1. ACT Compss Prctice Tests 1 2. Common Mistkes 2 3. Distributive
More informationVectors 2. 1. Recap of vectors
Vectors 2. Recp of vectors Vectors re directed line segments  they cn be represented in component form or by direction nd mgnitude. We cn use trigonometry nd Pythgors theorem to switch between the forms
More informationUnit 6: Exponents and Radicals
Eponents nd Rdicls : The Rel Numer Sstem Unit : Eponents nd Rdicls Pure Mth 0 Notes Nturl Numers (N):  counting numers. {,,,,, } Whole Numers (W):  counting numers with 0. {0,,,,,, } Integers (I): 
More informationRegular Sets and Expressions
Regulr Sets nd Expressions Finite utomt re importnt in science, mthemtics, nd engineering. Engineers like them ecuse they re super models for circuits (And, since the dvent of VLSI systems sometimes finite
More informationLecture 15  Curve Fitting Techniques
Lecture 15  Curve Fitting Techniques Topics curve fitting motivtion liner regression Curve fitting  motivtion For root finding, we used given function to identify where it crossed zero where does fx
More information9.3. The Scalar Product. Introduction. Prerequisites. Learning Outcomes
The Sclr Product 9.3 Introduction There re two kinds of multipliction involving vectors. The first is known s the sclr product or dot product. This is soclled becuse when the sclr product of two vectors
More informationSection A4 Rational Expressions: Basic Operations
A Appendi A A BASIC ALGEBRA REVIEW 7. Construction. A rectngulr opentopped bo is to be constructed out of 9 by 6inch sheets of thin crdbord by cutting inch squres out of ech corner nd bending the
More informationLecture 5. Inner Product
Lecture 5 Inner Product Let us strt with the following problem. Given point P R nd line L R, how cn we find the point on the line closest to P? Answer: Drw line segment from P meeting the line in right
More information1.2 The Integers and Rational Numbers
.2. THE INTEGERS AND RATIONAL NUMBERS.2 The Integers n Rtionl Numers The elements of the set of integers: consist of three types of numers: Z {..., 5, 4, 3, 2,, 0,, 2, 3, 4, 5,...} I. The (positive) nturl
More informationMatrix Algebra CHAPTER 1 PREAMBLE 1.1 MATRIX ALGEBRA
CHAPTER 1 Mtrix Algebr PREAMBLE Tody, the importnce of mtrix lgebr is of utmost importnce in the field of physics nd engineering in more thn one wy, wheres before 1925, the mtrices were rrely used by the
More informationUse Geometry Expressions to create a more complex locus of points. Find evidence for equivalence using Geometry Expressions.
Lerning Objectives Loci nd Conics Lesson 3: The Ellipse Level: Preclculus Time required: 120 minutes In this lesson, students will generlize their knowledge of the circle to the ellipse. The prmetric nd
More informationPHY 140A: Solid State Physics. Solution to Homework #2
PHY 140A: Solid Stte Physics Solution to Homework # TA: Xun Ji 1 October 14, 006 1 Emil: jixun@physics.ucl.edu Problem #1 Prove tht the reciprocl lttice for the reciprocl lttice is the originl lttice.
More informationCOMBINATORIAL HOPF ALGEBRAS FROM PROS
COMBINATORIAL HOPF ALGEBRAS FROM PROS JEANPAUL BULTEL AND SAMUELE GIRAUDO Astrct. We introduce generl construction tht tkes s input soclled stiff PRO nd tht outputs Hopf lger. Stiff PROs re prticulr
More informationFormal Languages and Automata Exam
Forml Lnguges nd Automt Exm Fculty of Computers & Informtion Deprtment: Computer Science Grde: Third Course code: CSC 34 Totl Mrk: 8 Dte: 23//2 Time: 3 hours Answer the following questions: ) Consider
More informationAREA OF A SURFACE OF REVOLUTION
AREA OF A SURFACE OF REVOLUTION h cut r πr h A surfce of revolution is formed when curve is rotted bout line. Such surfce is the lterl boundr of solid of revolution of the tpe discussed in Sections 7.
More informationSPECIAL PRODUCTS AND FACTORIZATION
MODULE  Specil Products nd Fctoriztion 4 SPECIAL PRODUCTS AND FACTORIZATION In n erlier lesson you hve lernt multipliction of lgebric epressions, prticulrly polynomils. In the study of lgebr, we come
More informationIntegration. 148 Chapter 7 Integration
48 Chpter 7 Integrtion 7 Integrtion t ech, by supposing tht during ech tenth of second the object is going t constnt speed Since the object initilly hs speed, we gin suppose it mintins this speed, but
More informationCS99S Laboratory 2 Preparation Copyright W. J. Dally 2001 October 1, 2001
CS99S Lortory 2 Preprtion Copyright W. J. Dlly 2 Octoer, 2 Ojectives:. Understnd the principle of sttic CMOS gte circuits 2. Build simple logic gtes from MOS trnsistors 3. Evlute these gtes to oserve logic
More informationFUNCTIONS AND EQUATIONS. xεs. The simplest way to represent a set is by listing its members. We use the notation
FUNCTIONS AND EQUATIONS. SETS AND SUBSETS.. Definition of set. A set is ny collection of objects which re clled its elements. If x is n element of the set S, we sy tht x belongs to S nd write If y does
More information1 Numerical Solution to Quadratic Equations
cs42: introduction to numericl nlysis 09/4/0 Lecture 2: Introduction Prt II nd Solving Equtions Instructor: Professor Amos Ron Scribes: Yunpeng Li, Mrk Cowlishw Numericl Solution to Qudrtic Equtions Recll
More informationWarmup for Differential Calculus
Summer Assignment Wrmup for Differentil Clculus Who should complete this pcket? Students who hve completed Functions or Honors Functions nd will be tking Differentil Clculus in the fll of 015. Due Dte:
More informationOperations with Polynomials
38 Chpter P Prerequisites P.4 Opertions with Polynomils Wht you should lern: Write polynomils in stndrd form nd identify the leding coefficients nd degrees of polynomils Add nd subtrct polynomils Multiply
More informationVolumes of solids of revolution
Volumes of solids of revolution We sometimes need to clculte the volume of solid which cn be obtined by rotting curve bout the xxis. There is strightforwrd technique which enbles this to be done, using
More informationGraphs on Logarithmic and Semilogarithmic Paper
0CH_PHClter_TMSETE_ 3//00 :3 PM Pge Grphs on Logrithmic nd Semilogrithmic Pper OBJECTIVES When ou hve completed this chpter, ou should be ble to: Mke grphs on logrithmic nd semilogrithmic pper. Grph empiricl
More informationGeometry and Measure. 12am 1am 2am 3am 4am 5am 6am 7am 8am 9am 10am 11am 12pm
Reding Scles There re two things to do when reding scle. 1. Mke sure you know wht ech division on the scle represents. 2. Mke sure you red in the right direction. Mesure Length metres (m), kilometres (km),
More informationA Note on Complement of Trapezoidal Fuzzy Numbers Using the αcut Method
Interntionl Journl of Applictions of Fuzzy Sets nd Artificil Intelligence ISSN  Vol.  A Note on Complement of Trpezoidl Fuzzy Numers Using the αcut Method D. Stephen Dingr K. Jivgn PG nd Reserch Deprtment
More informationLecture 2: Matrix Algebra. General
Lecture 2: Mtrix Algebr Generl Definitions Algebric Opertions Vector Spces, Liner Independence nd Rnk of Mtrix Inverse Mtrix Liner Eqution Systems, the Inverse Mtrix nd Crmer s Rule Chrcteristic Roots
More informationEcon 4721 Money and Banking Problem Set 2 Answer Key
Econ 472 Money nd Bnking Problem Set 2 Answer Key Problem (35 points) Consider n overlpping genertions model in which consumers live for two periods. The number of people born in ech genertion grows in
More information200506 Second Term MAT2060B 1. Supplementary Notes 3 Interchange of Differentiation and Integration
Source: http://www.mth.cuhk.edu.hk/~mt26/mt26b/notes/notes3.pdf 256 Second Term MAT26B 1 Supplementry Notes 3 Interchnge of Differentition nd Integrtion The theme of this course is bout vrious limiting
More informationBayesian Updating with Continuous Priors Class 13, 18.05, Spring 2014 Jeremy Orloff and Jonathan Bloom
Byesin Updting with Continuous Priors Clss 3, 8.05, Spring 04 Jeremy Orloff nd Jonthn Bloom Lerning Gols. Understnd prmeterized fmily of distriutions s representing continuous rnge of hypotheses for the
More informationMath 135 Circles and Completing the Square Examples
Mth 135 Circles nd Completing the Squre Exmples A perfect squre is number such tht = b 2 for some rel number b. Some exmples of perfect squres re 4 = 2 2, 16 = 4 2, 169 = 13 2. We wish to hve method for
More informationMathematics. Vectors. hsn.uk.net. Higher. Contents. Vectors 128 HSN23100
hsn.uk.net Higher Mthemtics UNIT 3 OUTCOME 1 Vectors Contents Vectors 18 1 Vectors nd Sclrs 18 Components 18 3 Mgnitude 130 4 Equl Vectors 131 5 Addition nd Subtrction of Vectors 13 6 Multipliction by
More information0.1 Basic Set Theory and Interval Notation
0.1 Bsic Set Theory nd Intervl Nottion 3 0.1 Bsic Set Theory nd Intervl Nottion 0.1.1 Some Bsic Set Theory Notions Like ll good Mth ooks, we egin with definition. Definition 0.1. A set is welldefined
More informationApplications to Physics and Engineering
Section 7.5 Applictions to Physics nd Engineering Applictions to Physics nd Engineering Work The term work is used in everydy lnguge to men the totl mount of effort required to perform tsk. In physics
More informationCurve Sketching. 96 Chapter 5 Curve Sketching
96 Chpter 5 Curve Sketching 5 Curve Sketching A B A B A Figure 51 Some locl mximum points (A) nd minimum points (B) If (x, f(x)) is point where f(x) reches locl mximum or minimum, nd if the derivtive of
More informationExperiment 6: Friction
Experiment 6: Friction In previous lbs we studied Newton s lws in n idel setting, tht is, one where friction nd ir resistnce were ignored. However, from our everydy experience with motion, we know tht
More information2 DIODE CLIPPING and CLAMPING CIRCUITS
2 DIODE CLIPPING nd CLAMPING CIRCUITS 2.1 Ojectives Understnding the operting principle of diode clipping circuit Understnding the operting principle of clmping circuit Understnding the wveform chnge of
More informationModule Summary Sheets. C3, Methods for Advanced Mathematics (Version B reference to new book) Topic 2: Natural Logarithms and Exponentials
MEI Mthemtics in Ection nd Instry Topic : Proof MEI Structured Mthemtics Mole Summry Sheets C, Methods for Anced Mthemtics (Version B reference to new book) Topic : Nturl Logrithms nd Eponentils Topic
More informationINTERCHANGING TWO LIMITS. Zoran Kadelburg and Milosav M. Marjanović
THE TEACHING OF MATHEMATICS 2005, Vol. VIII, 1, pp. 15 29 INTERCHANGING TWO LIMITS Zorn Kdelburg nd Milosv M. Mrjnović This pper is dedicted to the memory of our illustrious professor of nlysis Slobodn
More informationExponents base exponent power exponentiation
Exonents We hve seen counting s reeted successors ddition s reeted counting multiliction s reeted ddition so it is nturl to sk wht we would get by reeting multiliction. For exmle, suose we reetedly multily
More informationNull Similar Curves with Variable Transformations in Minkowski 3space
Null Similr Curves with Vrile Trnsformtions in Minkowski spce Mehmet Önder Cell Byr University, Fculty of Science nd Arts, Deprtment of Mthemtics, Murdiye Cmpus, 45047 Murdiye, Mnis, Turkey. mil: mehmet.onder@yr.edu.tr
More informationBabylonian Method of Computing the Square Root: Justifications Based on Fuzzy Techniques and on Computational Complexity
Bbylonin Method of Computing the Squre Root: Justifictions Bsed on Fuzzy Techniques nd on Computtionl Complexity Olg Koshelev Deprtment of Mthemtics Eduction University of Texs t El Pso 500 W. University
More informationPure C4. Revision Notes
Pure C4 Revision Notes Mrch 0 Contents Core 4 Alger Prtil frctions Coordinte Geometry 5 Prmetric equtions 5 Conversion from prmetric to Crtesin form 6 Are under curve given prmetriclly 7 Sequences nd
More informationwww.mathsbox.org.uk e.g. f(x) = x domain x 0 (cannot find the square root of negative values)
www.mthsbo.org.uk CORE SUMMARY NOTES Functions A function is rule which genertes ectl ONE OUTPUT for EVERY INPUT. To be defined full the function hs RULE tells ou how to clculte the output from the input
More informationMechanics Cycle 1 Chapter 5. Chapter 5
Chpter 5 Contct orces: ree Body Digrms nd Idel Ropes Pushes nd Pulls in 1D, nd Newton s Second Lw Neglecting riction ree Body Digrms Tension Along Idel Ropes (i.e., Mssless Ropes) Newton s Third Lw Bodies
More informationOr more simply put, when adding or subtracting quantities, their uncertainties add.
Propgtion of Uncertint through Mthemticl Opertions Since the untit of interest in n eperiment is rrel otined mesuring tht untit directl, we must understnd how error propgtes when mthemticl opertions re
More informationIntegration by Substitution
Integrtion by Substitution Dr. Philippe B. Lvl Kennesw Stte University August, 8 Abstrct This hndout contins mteril on very importnt integrtion method clled integrtion by substitution. Substitution is
More informationCalculus of variations with fractional derivatives and fractional integrals
Anis do CNMAC v.2 ISSN 1984820X Clculus of vritions with frctionl derivtives nd frctionl integrls Ricrdo Almeid, Delfim F. M. Torres Deprtment of Mthemtics, University of Aveiro 3810193 Aveiro, Portugl
More informationDouble Integrals over General Regions
Double Integrls over Generl egions. Let be the region in the plne bounded b the lines, x, nd x. Evlute the double integrl x dx d. Solution. We cn either slice the region verticll or horizontll. ( x x Slicing
More information1. Find the zeros Find roots. Set function = 0, factor or use quadratic equation if quadratic, graph to find zeros on calculator
AP Clculus Finl Review Sheet When you see the words. This is wht you think of doing. Find the zeros Find roots. Set function =, fctor or use qudrtic eqution if qudrtic, grph to find zeros on clcultor.
More informationDETERMINANTS. ] of order n, we can associate a number (real or complex) called determinant of the matrix A, written as det A, where a ij. = ad bc.
Chpter 4 DETERMINANTS 4 Overview To every squre mtrix A = [ ij ] of order n, we cn ssocite number (rel or complex) clled determinnt of the mtrix A, written s det A, where ij is the (i, j)th element of
More informationThe Velocity Factor of an Insulated TwoWire Transmission Line
The Velocity Fctor of n Insulted TwoWire Trnsmission Line Problem Kirk T. McDonld Joseph Henry Lbortories, Princeton University, Princeton, NJ 08544 Mrch 7, 008 Estimte the velocity fctor F = v/c nd the
More informationPROBLEMS 13  APPLICATIONS OF DERIVATIVES Page 1
PROBLEMS  APPLICATIONS OF DERIVATIVES Pge ( ) Wter seeps out of conicl filter t the constnt rte of 5 cc / sec. When the height of wter level in the cone is 5 cm, find the rte t which the height decreses.
More informationThe remaining two sides of the right triangle are called the legs of the right triangle.
10 MODULE 6. RADICAL EXPRESSIONS 6 Pythgoren Theorem The Pythgoren Theorem An ngle tht mesures 90 degrees is lled right ngle. If one of the ngles of tringle is right ngle, then the tringle is lled right
More informationThe Chain Rule. rf dx. t t lim " (x) dt " (0) dx. df dt = df. dt dt. f (r) = rf v (1) df dx
The Chin Rule The Chin Rule In this section, we generlize the chin rule to functions of more thn one vrible. In prticulr, we will show tht the product in the singlevrible chin rule extends to n inner
More informationGeometry 71 Geometric Mean and the Pythagorean Theorem
Geometry 71 Geometric Men nd the Pythgoren Theorem. Geometric Men 1. Def: The geometric men etween two positive numers nd is the positive numer x where: = x. x Ex 1: Find the geometric men etween the
More informationPoint Groups and Space Groups in Geometric Algebra
Point Groups nd Spce Groups in Geometric Alger Dvid Hestenes Deprtment of Physics nd Astronomy Arizon Stte University, Tempe, Arizon, USA Astrct. Geometric lger provides the essentil foundtion for new
More informationLecture 3 Gaussian Probability Distribution
Lecture 3 Gussin Probbility Distribution Introduction l Gussin probbility distribution is perhps the most used distribution in ll of science. u lso clled bell shped curve or norml distribution l Unlike
More informationMultiplication and Division  Left to Right. Addition and Subtraction  Left to Right.
Order of Opertions r of Opertions Alger P lese Prenthesis  Do ll grouped opertions first. E cuse Eponents  Second M D er Multipliction nd Division  Left to Right. A unt S hniqu Addition nd Sutrction
More informationSolving BAMO Problems
Solving BAMO Problems Tom Dvis tomrdvis@erthlink.net http://www.geometer.org/mthcircles Februry 20, 2000 Abstrct Strtegies for solving problems in the BAMO contest (the By Are Mthemticl Olympid). Only
More information4 Geometry: Shapes. 4.1 Circumference and area of a circle. FM Functional Maths AU (AO2) Assessing Understanding PS (AO3) Problem Solving HOMEWORK 4A
Geometry: Shpes. Circumference nd re of circle HOMEWORK D C 3 5 6 7 8 9 0 3 U Find the circumference of ech of the following circles, round off your nswers to dp. Dimeter 3 cm Rdius c Rdius 8 m d Dimeter
More informationexcenters and excircles
21 onurrene IIi 2 lesson 21 exenters nd exirles In the first lesson on onurrene, we sw tht the isetors of the interior ngles of tringle onur t the inenter. If you did the exerise in the lst lesson deling
More informationDAGmaps: Space Filling Visualization of Directed Acyclic Graphs
Journl of Grph Algorithms nd Applictions http://jg.info/ vol. 13, no. 3, pp. 319 347 (2009) DAGmps: Spce Filling Visuliztion of Directed Acyclic Grphs Vssilis Tsirs 1,2 Sofi Trintfilou 1,2 Ionnis G. Tollis
More informationAll pay auctions with certain and uncertain prizes a comment
CENTER FOR RESEARC IN ECONOMICS AND MANAGEMENT CREAM Publiction No. 12015 All py uctions with certin nd uncertin prizes comment Christin Riis All py uctions with certin nd uncertin prizes comment Christin
More informationModule 2. Analysis of Statically Indeterminate Structures by the Matrix Force Method. Version 2 CE IIT, Kharagpur
Module Anlysis of Stticlly Indeterminte Structures by the Mtrix Force Method Version CE IIT, Khrgpur esson 9 The Force Method of Anlysis: Bems (Continued) Version CE IIT, Khrgpur Instructionl Objectives
More informationBasic Research in Computer Science BRICS RS0213 Brodal et al.: Solving the String Statistics Problem in Time O(n log n)
BRICS Bsic Reserch in Computer Science BRICS RS0213 Brodl et l.: Solving the String Sttistics Prolem in Time O(n log n) Solving the String Sttistics Prolem in Time O(n log n) Gerth Stølting Brodl Rune
More information9 CONTINUOUS DISTRIBUTIONS
9 CONTINUOUS DISTIBUTIONS A rndom vrible whose vlue my fll nywhere in rnge of vlues is continuous rndom vrible nd will be ssocited with some continuous distribution. Continuous distributions re to discrete
More informationFAULT TREES AND RELIABILITY BLOCK DIAGRAMS. Harry G. Kwatny. Department of Mechanical Engineering & Mechanics Drexel University
SYSTEM FAULT AND Hrry G. Kwtny Deprtment of Mechnicl Engineering & Mechnics Drexel University OUTLINE SYSTEM RBD Definition RBDs nd Fult Trees System Structure Structure Functions Pths nd Cutsets Reliility
More informationAlgebra Review. How well do you remember your algebra?
Algebr Review How well do you remember your lgebr? 1 The Order of Opertions Wht do we men when we write + 4? If we multiply we get 6 nd dding 4 gives 10. But, if we dd + 4 = 7 first, then multiply by then
More informationAnswer, Key Homework 10 David McIntyre 1
Answer, Key Homework 10 Dvid McIntyre 1 This printout should hve 22 questions, check tht it is complete. Multiplechoice questions my continue on the next column or pge: find ll choices efore mking your
More informationWeek 11  Inductance
Week  Inductnce November 6, 202 Exercise.: Discussion Questions ) A trnsformer consists bsiclly of two coils in close proximity but not in electricl contct. A current in one coil mgneticlly induces n
More informationOstrowski Type Inequalities and Applications in Numerical Integration. Edited By: Sever S. Dragomir. and. Themistocles M. Rassias
Ostrowski Type Inequlities nd Applictions in Numericl Integrtion Edited By: Sever S Drgomir nd Themistocles M Rssis SS Drgomir) School nd Communictions nd Informtics, Victori University of Technology,
More information