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Introductory Analysis II

by: Dorothea Bode

Introductory Analysis II MA 22400

Marketplace > Purdue University > Mathematics (M) > MA 22400 > Introductory Analysis II
Dorothea Bode
GPA 3.97

Salvador Barone

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Salvador Barone
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This 18 page Class Notes was uploaded by Dorothea Bode on Saturday September 19, 2015. The Class Notes belongs to MA 22400 at Purdue University taught by Salvador Barone in Fall. Since its upload, it has received 45 views. For similar materials see /class/208115/ma-22400-purdue-university in Mathematics (M) at Purdue University.

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Date Created: 09/19/15
Find webpage Grader Peter Weigel of ce hours on the webpage Monday 12 1 pm Tuesday 1 2 pm Also syllabus online There will be a midterm and a nal and there may or may not be a quiz The rest of the grade distribution is online Weekly homework is assigned on Thursday and will be due the next Thursday Parts will be graded and returned but other problems that you want to see the solution you will have to go to of ce hours Of ce hours Monday Tuesday 1 2 PM Book Topology from a Differential Viewpoint Milnor Also another one webpage7 01 August 25th Day 1 The main topic of the course are manifolds In particular we want to study Differential Topology which is the study of sets in R which are invariant under diffeomorphisms A diffeomorphism b R a R is a smooth 0 map with a smooth inverse We can weaken the invariance to locally invariant7 but that is a different question since you lose the information about the global view For example two manifolds might be locally the same but one compact and one unbounded Theorem 011 Implicit Function Theorem FT Let f W a Rk be a 01 map where W C R is open Assume that p E W is a quotregular point that is thfp rh igmfirgtlltrltk 7 727 M WLOG Dfp AlB where A is n gtlt h and B is h gtlt h matrices with detB 71510 Write p ab E R gtlt Rk Rnk and fab c Then under the above assumptions 3a 6 U C R open 3b 6 V C Rk open and Hg U a V g 6 01 with 9a b st E U xylfzy c Note that the set on the right is a level set whose elements belong to U gtlt V Thus the map U 9 z gt gt fzgx c const 239 0 g If my 6 71 U gtlt V C R gtlt Rk then we have x 7 ltgt An application f R a R f0 0 Df0 invertible then f is a local diffeomor phism at zero ie 30 E U C R open st f U a fU is a diffeomorphism Proof of application Let Fzy fy7x Rn a R Then DF00 lllDf0 gt rkDF0 0 is maximal So 00 is a regular point for the map F so we can apply the implicit function theorem to this map F at the point 00 Applying lFT to F we have that 30 E U C R 0 E V C R Hg U a V with 90 0 such that E U xy E U gtlt VlFzy 0 Using the last remark of the theorem Fz is constant on U and hence at p 0 0 we have 0 g and even better is the equality 0 7 1 which implies 0 7Df0 1 ill gt Dg0 Df0 1 is non degenerate Now do the same thing for the function g and then you7re done since inverses are unique We now rephrase lFT7 and we prove this new formulation Suppose f Bk a Bk f0 07 and Df0 has maximum rank Then7 3 local diffeo o U i with U7 V C R and 0 0 such that f o 1a7b b7 where ab 6 R gtlt Rk Rn l U 0 B which is invertible Claim 2 f o 1a7b b on V We can write a7b 15s7y7 since o U i V gt f o 61075 f o 45 1 0 Way Way and note that W MI 96 WHO Proof of rephrasing of IFT De ne sy sfsy Rnk a Rn ll Claim 1 o is a local diffeo Note that D 00 lt0 D Note From now or all maps are smooth C 0 unless otherwise stated De nition 1 M C Rnk is a embedded7 smooth7 sub manifold of dimension n if Vs 6 M7 3U C Rnk open7 s 6 U7 3f U a Rk such that 1 fs 0 2 Dfs has max rank 3 M U f 10 De nition 2 M C Rnk is an n dim7l mfd if Vs E M73U C Rn i s E U39 C R open7 0697 Hg laMCRnMst 1 90 0 2 g Q a U M is a homeomorphism 3 Dg0 has full rank 71 De nition 3 M C Rnk is an n dim7l mfd if Vs 6 M7 3U C Rn i s E U and 3V C Rn i 0 E V and Ho U 3 V diffeo st 1 s 0 2 U M V Rn gtlt 3 o diffeo Remark 012 1 M C Rnk is mfd and 7 Rntk Q Rnk diffeo implies 7M is a mfd 2 in De nition 27 the map 9 is called a local parametrization of M 9 1 is a local coordinate or they7 are local coordinates Examples 1 R C R is an n dim7l manifold 2 L C Rnk af ne subspace L 110 V7 V a subspace is a mfd7 dim L dim V 3 M C Rk N C R1 mfds7 then M gtlt N C R11 is a mfd and dimM gtlt N dim M dim N 4 2 x7y72 E Rslzz y2 22 1 is a 2 dim7l manifold Proof Using Defn 1 Use the map fyz x2 y2 22 7 1 R3 7 R1 which gives the sphere as a level set Just need to check that Dfxyz has maximal rank whenever Lyn E 82 Note Dfxyz 2x72y722 has rank 1 if and only if Dfxyz 31 0 So7 were done since 0 52 Using Defn 2 Use the map 9 Q 7 52 where Q D2 E Rzlzz y2 lt 1 Now gixy Ly7 i 17 2 7 yz Then Dgxy has maximal rk U 02 August 27th Day2 There were three de nitions of manifold given last time A subset M C Rnk is an n dim7l mfd Note the notation M indicates that M is n dim l only if 1 locally M f 10f Rnk 7 Rk and rkDf k for all z E M ltgt 2 locally M gR g R 7 Rn i and rkDgp n for all p E R ltgt 3 locally M R gtlt 0 Rwrk 7 Rnk is a diffeomorphism My question in ii below we will use a map 9 D2 7 R3 as in de nition 2 above For example 52 x7y72 E Ralle y2 22 17 then i Using defn 1 De ne fyz 2 y2 22 7 17 then rkDfp 31 1 if and only ifp 0 So f R3 7 R implies that 52 is a 2 dim7l mfd ii Using defn 2 De ne g D2 7 R3 where D2 my 6 R2le y2 lt 17 and gy Ly7 17 2 7y2 Then 9D2 C 2 and rkDgxy 2 for all my 6 D2 Now with some more local parameters neg square root this will prove that 52 is a 2 dim7l mfd iii Using defn 3 De ne x7y72 Ly7 17 2 7 y z a7b70 Then Ma7 b7 0 ib 1abc ab7 07m Note that o R3 7 R3 is only lo cally de ned7 but it is a local diffeomorphism Also7 Ma7 b7 0 a7 b7 7W E 82 That is7 locally 7R2 gtlt 0 SZ iv U C Rk open set Then U is a k dirn7l rnfd7 where for example 9 id in de nition 2 Also7 if V C M is open and M rnfd7 then V is a n dirn7l rnfd V The space of all invertible linear transformations GLnR A 6 RM R det A 31 0 is a n2 dirn7l manifold Vi The special linear group SLnR A E anl detA 1 is a rnfd proof re quired lndeed7 consider the determinant rnap det RM 7 R Then D detA RM 7 R is a linear map In a special case7 where A I7 we claim Ddet1 B trB this also needs a proof Thus7 at A I the special linear group SLnR is a rnfd In general D detAB detA trA 1B since D detA B lt0 detAA 1A tB d El wetm detI tA 1B det A trA 1B Thus7 if detA 31 0 we have rkD detA 1 implies SLnR is a rnfd Also7 dirn SLnR n2 71 D Proof of Claim about Ddet being trace Assume B is diagonal7 B bL7 7on Then D det1 B 17 th detItB 17 th H11tbi 21 bl trB Now were done since diagonal matrices are dense in SLn7 R using the Euclidean norrn D Next topic Hyperboloids A Hyperboloid H0 x 07 7x 6 RWle 7 a 7 7 xi c f 1c for the fw map f R 7 R Note we have for f that Df satis es Df 2x0772x1 772zn has rank equal to 1 unless z 0 But 0 E H07 thus 1 H0 is a rnfd for all c 31 0 2 H00 might or might not be a rnfd turns out that it is not Pictures Here Tangent Space For x E M C R k and local parameter 9 Q 7 M C Rnk with 90 x then Dg0 R 7 R is a linear map and has rank n 4 Defn2 TEM Dg0R is the tangent space of M at x Rule TmM C Rnk is a linear subspace If M is locally fquotl07 and rkDf k for all z E M Defn1 TEM ker Dfz7 and recall f Rnk a Rk Then TEM C Rnk is a linear subspace and dim TEM dirn M Finally7 if M R gtlt 0 locally7 and o Rwrk i Rnk with 0 Defn3 TmM D x 1R It will be an exercise to check that these de nitions agree A geornetric defn of TEM De ne 731M 7 785 a M C Rnkl y0 Proposition 021 TmM quoty0 y E PmM TPmM where quoty D y01 Proof Fix a parametrization g Q a M C Rn i with 90 x Then de ne U 997 and note that TPmM TPmU Note that g induces a bijection 7309 1 no a l gt g o o and T7309 a TPmU 570 H D90 470 Finally7 note that Dg0 o0 g o o390 To nish7 we obviously have T7309 R Furtherrnore7 TmM DAWN Dg0T7309 TCDEU TCDEM U Corollary 022 Let locally M f 10 for f Rwrk a Rk and rkDfx k Then TmM ker Dfz Proof We show equality of the sets TmM TCDmM by showing both inclusion relationships Q For 1 6 TEM we have to show that Dfzo 0 To that end 37 7575 a M with 70 r and quoty0 o This implies fo y 768 a Rk is the constant zero function 30 f 000 0 and also f O 00 Dfv0 MW Df96 Q We have dirnTmM dirn M dirnker Dfz 5 Conclusion TmM is independent of choice of f7 Q Let L C R be an af ne subspace7 L 110 V for a vector subspace V Then TPL V for all p E L Q Consider SZ x2 y2 22 71 0 Then Dfxyz 2x72y722 R3 a R and v f a Dfxyz b 2az 2by 202 0 So we have TpS2 ker Dfp 7yzla by 02 0 where p a7b70 Note that this set TpS2 is R pi Q SLnR detA 17 then TISLnR ker Ddet0 kertr A E anltrA 07 so the tangent space of SLn7 R at the identity matrix I are the trace free matrices 03 September 1st Day 3 We7ve de ned n dim7l manifold in Rk and also the tangent space TEM Locally M U f 10 means that we can View TmM ker Dfz Locally if 99 M V then TmM has the form Dg0 Finally7 locally U M Rn gtlt V Via the map Ms 0 then nMwwmexw Also7 another possible description is TmM WOW t 7878 n MMO 96 where 70 lt y t Now7 x U C Rk an open set Using this latest de nition7 then U is a k dim7l mfd and TmU Rk For the local parameter de nition 9 U a U with g x then we have that TmU Dg0 Rk Rk For the other de nitions take the identity map as well7 but these two are the most natural approaches in this case Recall for a pair of smooth maps f U a V and g V a W we have 1 gofVaWissmoothand D9f96 0 Df96 D9 O f96 2 For i U gt U an inclusion we have Diz o ii 3 For a linear rnap p R a Rk then Dltp i o Corollary 031 IfU C Rk andV C R are open sets thenfor a di eomorphism f U a V we have that Df Rk a bbrn is a linear isomorphism In particular n k Proof use the chain rule D Lemma 1 Let M be a mfd andg 91 a M and h 92 a M be two local parameters with 90 z h0 Then h 1 o g is a local di eomorphism Proof Note that h92 and 991 are open in M Therefore7 3V C h92 991 which is open in M Now h 1 o g g 1V a hV Now7 we need to extend h l V a h 1V to a smooth rnap de ned on some open set in Rk To that end7 choose a di eomorphisrn o W i W such that W H M Rn gtlt Let 7T1 be the natural projection rnap V 7T1 R gtlt Rk a R Now de ne 6 W a h 1V by 6p h 1 0 fl 0 7T1 0 451 Note that under this rule we have for p E V that 6p h 1p So7 6 is smooth and 6lV h l so this means that h 1 o g 6 o g is a smooth map E De nition 4 lff M a N is a map between manifolds7 then say that f is smooth at z E M if f o g Q a N is smooth at 0 when 9 is any local parametrization Lemma 2 The de nition above is independent of the choice ofg Proof Suppose g h Q a M are both local parameterizations at x Then h lg is smooth by the previous Lemma Now7 fogis smooth at0 ltgt fohfogog 10h0 is smooth at 0 W smooth U De nition 5 For a smooth map f M a N we have Df TmM a TEN is de ned by Df o Df o g0 6 Rk where an element of TmM Dg0R is i Dg0 Proof that this map ecoists Recall that TEM TPmM and TyN TCDyN7 and for some i 6 TEM there exists 7 7875 a M such that 70 z and quoty0 1 Claim Df i f og390 To see this then v we Dyltogtzrltogt Dg0 thus Dfo Df o g0o0 Df o g o o 0 7 Df ov0 f ov390 DClaim Now7 obviously Df o E TfwN and this element is f o y390 with fory is 8 a N Now f M a N smooth means Df TmM a TfmN is a linear map D It is easy to check that 1 fMaN7 andgNaPsmoothmapsthengofMaPis smooth and My 0 Mac Dgfz 0 mm TmM TgltfltzgtgtP 2 id M a M the identity map satis es Did TmM a TmM is the identity map Proposition 032 For a di eomorphism f M a N we have that Df Tm a TfwN is a liriear isomorphism Proof Trivial since f o f 1 idN and by the chain rule we have Dff 1y o Df 1y ldTyN D We can then conclude that dimM dimN whenever there exists a di eomorphism between them This should possibly strike you as signi cant since we have not7 until now7 really justi ed that a manifold has a well de ned dimension For example7 in the local chart case7 given a local chart 9 from 91 C R why can there not be another local chart h out of 92 C Rn57 lt7s because h 1 o g is a local di eomorphism Proposition 033 Let f M a N be a smooth map ff satis es Df TmM a TfmN is a liriear isomorphism theri f is a local di eomorphism Proof Homework De nition 6 For a smooth map f M a N7 we call f an immersion if Df TmM a TfmN is an injection7 and we call f a submersion if Df is a surjection We call f an embedding if f is an immersion and a homeomorphism onto its image Remark 034 1 f immersion gt dim M S dim N f submersion gt dim M 2 dim N 2 say f is an immersionsubmersion at x if Df is injectivesurjective 3 f embedding gt f is an injective immersion WARNING an injective immersion is7 in general7 NOT an embedding It might fail that f llZMf is continuous Proposition 035 ff M a Rk is an injectioe immersion which is a homeomorphism onto it s image then fM Q Rk is a manifold Proof Let g Q a M U be a local parametrization Then 9 is a homeomorphism onto its image7 and Dg0 has maximal rank Then we have the map f M a Rk and f o e Q a fM is a local parametrization at Now Dg0 has maximal rank means that Dfog 0 Dfg0 ng0 has maximal rank7 so fM satis es the condition of manifold D Lemma 3 A smooth map F RN a Rq and M Q RN a manifold then f FlM M a Rq is smooth and Df DFxlTwM Also Df i DF i for all i E TmM Proof HW Example 1 Let us again consider Sl Recall that p 81 a R2 de ned by my gt gt Then Dltpy T151 a R2 satis es T011351 uw E Rzlua ob 0 R397ba Now de ne May may and note that Dam and Dltp Do lszi T181 a R2 and furthermore Dsoltabgt jib D ltabgt jib 7 1 b 7b T 0 a a i 7b ab 7 a2 Now Dltp has rank zero if and only if saws 04 September 3 Day 4 We begin with some remarks 0 01 is not a manifold Soon we will see that it is a manifold with boundary o If g Q a M U and rank Dg0 n is maximal with Q Q R then 9 is a homeomorphism onto its image If f M a R and Dfs Df o g0 and recall that f diff7ble implies that f cont Now M C Rn f then M inherits the topology from Rn m A subset V C M is open iff V M UU C R open Now f o 9 continuous implies f f o g o 9 1 is continuous If g is an injective immersion then f o g sd thus f is diff7ble But f 9 1 is NOT continuous Ok7 now we begin with the lecture for today Recall that f M a N then we say that o f immersion ltgt Vs Dfs is injective gt m S n o f submersion ltgt Vs Dfs is surjective gt m 2 n o f is an embedding ltgt Vs Dfs is injective and f is a homeomorphism onto its image For an example7 consider p 81 a R2 de ned by ltpsy Then p My where sy s7sy is the map from R2 to itself Note that Dltps D slTw51 has rank 1 for all s E 81 Thus7 p is an immersion Lastly7 note that 07i1 00 and also that ltpi10 i10 In a more general case7 5 rl so7 7snlrk E Rn ll 1 Then p Snfk a Rk de ned by 10s07 7snlrk s07 7sk the natural projection onto the rst k1 coor dinates7 ands S a Snk the natural inclusion Then p is a submersion at 00 7071 N the north pole and s39 is an immersion Next7 we de ne a submanifold De nition 7 A subset E C M with M C Rq a manifold is called a submanifold of dimension d7 if Vs E 2 3s 6 U C Rq and there exists a local parametrization of M open 99aM U 0gt gts such that gQ Rdgtlt0 E U Note that glmquwD is a local parametrization of E and is a map from 9 Qq gtlt 0 to 2 U Thus7 E is a manifold Lemma 4 We have Ed C M C Rq and E a submfd of M if and only if Hob R a Rq di eomorphism such that M U Rn gtlt V and also o U Rq gtlt V PTOOf lt77 g billmnxmp v Use the construction to turn g into 15 ie7 use the appropriate alternate de nition D Remark 041 o M i N diffeomorphism HE C M is asubmanifold of M7 then 2 C N is a submanifold of N Question If Ed M 7 are some manifolds then 1 ifdgtncanitbethatECM7 ANS consider the inclusion map i E gt M7 then i idl implies Dix DidxlTw ilexg Also7 iplTw Dix TmE a TmM and the source has dimension d and the target has dimension n7 and the map is injective7 so d S u So the answer is NO 2 ifdnthen can 2 C M What about 2 g M ANS Yes7 any open subset of a manifold is a submanifold of the same dimension Something is being brushed under the rug here in particular all of this is easy because all of our manifolds are living inside of R space and hence they are all getting induced metrics from the R topology In general it is harder when you have to relate a lot of di erent manifolds where the concept of open is more general Now 2 M mfds7 and E C M C Rq ls E automatically a submanifold of M Proposition 042 Yes Proof 2 C M implies that Vp E X T12 C TEM Choose b Kg 3 Kg such that M U Rn gtlt V7 then we have 2 gtE U C Rn gtlt V Claim The projection map ME E C R is a manifold Proof of Claim g Q a 2 U local parametrization of 2 De ne g 7r 0 t o g which is a map from Q to f Then g is a homeomorphism onto its image7 and Dg0 has maximal rank This implies that g is a local parametrization of f and so i is a manifold This nishes the Claim D Claim proued Continuing with the proof of the proposition We have that i C R is a d dim7l mfd7 which implies directly that 31 R i R with ib Rd gtlt Now extend 7 to KI R gtlt Rq a Rq Then KI is a diffeomorphism and KT 0 b is the map which shows that E is a submanifold Recall f M a Rk an immersion and homeomorphism onto its image7 then fM C Rk is a manifold Corollary 043 IfN C Rq then for an embedding f M a N we have that fM C N is a submanifold Additionally ifE is a submanifold of M then fE C N is a submanifold Proof Regard f M a Rq and then fM C Rq is a submanifold by the proposition Also7 fM C N since f was a map with target N So fM is a submanifold D Proposition 044 For a submersion f M a N and for every p E N we have 2 f 1p is a submanifold The dimension ofE is dimE dimM 7 dim N Proof Let x0 6 E C M C Rk De ne K ker Df C TmOM7 and note that dimK m 7 n Choose a linear map L Rk a R such that LlK K a R is a linear isomorphism Next FMmgtRmingtltNn 96 H L957f96 then F is a local diffeomorphism Now DFx0 LDfx0 TEOM a R gtlt TyON and TEOM KEB Kc notation K0 is the linear complement This implies that Dfzo K0 a TyON is an iso Now use L is linear iso between K and R and also that Dfzo is a linear iso between K0 and TyO N7 this gives that DFx0 is a linear iso and implies order F is a local diffeomorphism at do By de nition 3x0 6 U C M7 and Lzoyo E V C Rm gtlt N open open such that F U i V diffeomorphism By construction FE U C R gtlt yo so 2 is a submanifold and T102 ker Dfzo D 05 September 8 Day 5 Recall Defn We will use the following de nitions most for a sub manifold We say that Ed C M C Rq is a submanifold if any of the following equivalent statements hold 0 3 local parametrization g Q a M U of M such that 99 Rd gtlt 2 U o Ho Rq a Rq diffeomorphism such that M U Rn gtlt V AND gtE U Rd gtlt 0 m V o E is a d dim7l manifold and also is a subset of M Note recall that f M a N is an embedding if f is an immersion7 Df has rank in for all z 6 M7 and also f is a homeomorphism onto its image In this case fM C N is a submanifold and fE C N is a submanifold for all submanifolds E C M Remark 051 if f M a N is a submersion7 rkDf 71 Va 6 M7 then f 1p is a submanifold of M for all p E N To prove this last statement which we did in the last lecture7 we only needed that rkDf n for all z E f 1p We use this notion to generalize the concept to a larger class of sets than submersions ln particular7 we need to de ne critical points De nition 8 Let f M a N be a smooth map between manifolds 1 z E M is a critical poirit of f is rkDf lt ii Let C x E Ml rkDf lt n be the set of critical points 2 y E fC C N is called a critical value 3 y E N7fc is called a regular value Remark If m lt ii7 then C M and fC fM imf In general we will have N7fC will be non empty7 but this is a theorem and requires proof The important thing to note right now is that a regular value might very well be an element of N which is not even in the image of f so7 not a value in the typical sense Proposition 052 For a smooth map f M a N ify E N is a Tegulma value then f 1y C M is a submariifold Proof Since y is a regular value we have y E N7fC If y fM7 then f 1y Q is a submanifold If y E fM then use the Remark 051 B Note the properties of f 1p depends a lot on p When f is not a submersion even ifp is perturbed very little7 we might get very different bers f 1p depending on what p is Question How about if f is a submersion Does f 1p depend on p ANS No7 unless everything is compact Let f M a N be a smooth map between manifolds with dimM dim N Let y be a regular value of f7 then f 1y is a 0 dim7l manifold7 ie7 its a discrete set That is Valwg E f 1y we can nd U17U2 open separation of the points 1723 U1 U2 0 1 E 17 2 6 U2 To see this7 if z E f 1y then rkDf n dimM dim N as Df TmM a TEN is a linear isomorphism So f is a local diffeomorphism near x So Ba 6 U C M neighborhood of z and also 3y 6 V C N such that f U 3 V is a diffeomorphism So f 1y is discrete Now7 assume that M is compact Then the number of elements in f 1y is a well de ned value for any regular value y Consider V N7fO a N which counts this number 13 Vy f 1y Then V is locally constant That is Vy E N7fC there is a neighborhood y E V C N7fC open such that Vy Vy for all y E V To see this let 1 zn f 1y then f is a local diffeomorphism near each xi So 3U C M V C N such that f 3 Ui W is a diffeomorphism Set V V1 o o Vn fM U1 U U U We could work with this set V or use the following simplifying assumption If V V1 V2 V then V V7fM7U1 o u U so that alternate nish Set V V1 Vn Then f 1V C U1 U U U and on each U f 1V we have that f is bijective since it is a local diffeomorphism on each Ui But then if y E V then 3le E U with y since each U is diffeomorphic to V D V So f 1y 2 n and it is exactly 71 if there is no z E M 7 U1 U NU with f y But such an x can not exist since f 1V C U1 U U U A proof of the fundamental theorem of algebra Let p C a C be a non constant poly nomial want to show that p has a zero We can extendp to a smooth map f 52 a 52 using the identi cation CUoo 82 and poo 00 This is standard one point compacti cation of the topological space C picture here homework assigned Homework The map f 52 a 52 satis es the following rkDfxyz 31 2 if and only if p hNyz 0 lfp is non constant then the critical points correspond to zeros of p which implies that the number of critical points is nite So fC C 2 is nite and here7s the punchline hence we have SLfC is connected We conclude that the counting function V SzifO a N is constant Say V E no Then there is certainly a non critical point z E Szif 1fC which makes rkDf 2 which implies f is a regular value so no Vfz 2 1 This implies that f is a surjection which then implies that p is a surjection so 1040 31 0 D How many critical points are there Let f M a N be a smooth map between manifolds and C x E Ml rkDf lt Assume m 2 n so that its not triVial that every point is critical Theorem 053 Sard Let f U a R be a smooth function U C R open subset Recall C x E Ul thfx lt Then fC C R has Lebesgue measure zero Corollary 054 The set of regular values Rni C is dense Remark 055 Rni C is of the 2nd Baire Category it is contained in a countable intersection of open and dense sets ii Rni C has full measure iii this is also true in oo dimensional case 3A1A2 dense such that A1 A2 0 but if B1 B2 are of 2nd Baire category then B1 B2 also is of 2nd Baire category iv 3A C 01 of 2nd Biare category with Lebesgue measure zero V m lt n implies C U Then Sard7s Theorem gives that fC has measure zero so the image of f has measure zero For example how about Peano curves 0 01 a 012 continuous and surjective Corollary 056 If f M a N is a smooth map between manifolds then the set of non critical values is dense in N Proof Use Sard in local parametrization Homework 06 September 10 Day 6 Last time we introduced notions of critical points critical values and regular values If f M a N is a smooth map then x E M is a critical point ltgt rkDf lt 71 dim N Set 0 the set of critical points Then y E fC is a critical value and y E N7fC is a regular value Note that some regular values7 are not values at all That is y E N7fM is a regular value that is not in the image of f Proposition 061 For any regular value p f 1p C M is a submanifold Remark 062 If p is a critical value then f 1p might or might not be a manifold we might just get lucky Theorem 063 Sard Let f U C R a R be a smooth map of manifolds then open fC C R has measure zero Corollary 064 The following hold 1 Ifm lt n then fC imf C R has measure zero 2 Rni C is dense 3 ff M a N then the set of regular values is dense in N 061 Manifolds with boundary Note that 01 is not a manifold In general the half plane Hm x1xm 6 RM zm 2 0 is not a manifold Denote 6H its boundary 6H Rm l gtlt Consider D7quot z 6 RM lt 1 it is a fact that there does not exist a homeomorphism g Hm D a D7quot which maps 90 0 De nition 9 The map 9 H a Rk is smooth at z 6 6H if exists neighborhood U C R and exists 9 U a Rk smooth such that ngm U g Hm U De nition 10 A subset M C Rk is a manifold with boundary of dimension in if Vd E M there either exists a local parametrization in the old sense or Hg D H a M U smooth where 0 E D C R is open U C Rk is open and 9 satis es 0 90 96 ii rkDg0 m and iii 9 is a homeomorphism onto its image In the latter case we call x a boundary point Moreover we de ne 3M z E Ml z is a boundary point De nition 11 In both cases manifold and manifold with boundary we have TmM Dg0R DQ0R In particular dim TmM dimM even at z 6 3M Lemma 5 Let M be a manifold with boundary Then 3M is a manifold with dim 3M dim M 7 1 Remark 065 The following hold 1 3M has no boundary That is 66M 0 This is related to ddw 0 for a differential form 2 3M is NOT a submanifold of M 3 i 3M gt M the inclusion map is smooth which implies that Tw6M C TmM for all z 6 3M Proof of lemma Pick a local boundary parametrization g D Hm a M U where M U C Rq De ne D C R 1 as D D H Rm l gtlt Then g gl D a 3M U and in fact g is a local interior parametrization of 3M 1 D 0 D90lRm1x0 Rm l a Rquot has maximal rank in 71 2 g is a homeomorphism onto its image Remark 066 If t Rk a Rk is a smooth map and M C Rk is a manifold with boundary then M is a manifold with boundary 6M 6 M In fact we summarily extend all previous results up to this point to the case where our manifold is a manifold with boundary This should be obvious how to do this Lemma 6 Let f M a R be a smooth function and 3M Q IfO E R is a regular value off then 2 x E Ml fx 2 0 E f 2 0 is a manifold with boundary 32 f 0 Note dimE dimM since f gt 0 is an open subset of M Proof of lemma We set 52 f 10 f 0 and denote 2 2752 f gt 0 the interior of 2 Since 2quot C M is open every point in 2quot has a local parametrization For x E 32 in the proof of f 1reg pt submfd77 we constructed a map 3 Lf M a R 1 gtlt R such that 3 U a V is a diffeo with z E U and 00 6 V Also 35E U Rl quotl gtlt 0 V We set 9 1 3lUQEU E a Hm V g Hm V a U E is a local boundary parametrization D Remark 067 For f M a R we have that o f 0 is a regular level set 0 f S 0 is a regular sub level set 0 f 2 0 is a regular super level set All of the above are manifolds the rst without boundary and the second and third with boundary Example 2 Rm x1xm E le S 1 is a manifold with boundary Rm f S 1 where fx1zm Then Df 2x12m and for z E 3W f 1 we get Lemma 7 Let X be a manifold with boundary and N a manifold without boundary dim X gt dim N Then f M a N smooth p a regular value off AND ALSO of flax 6X a N then 2 f 1p C X is a manifold with boundary and 32 2 3X Proof Note that we can already assume 7 that is p f3X implies 32 2 3X 0 well nish the proof next time An immediate application is Brower7s Fixed Point Theorem If g Rm a Rm is a continuous map then 9 has a xed point ie 3x 6 Rm such that 9a x Lemma 8 Let X be a compact manifold with boundary then there CCElStS no smooth map f X a 3X which res the boundary flax idaX Proof Assume f Xl a 3Xl exists Pick a regular value y 6 3X of f Note that y is also a regular value of flax z d since the identity map has only regular values So f 1y is a smooth7 1 dimensional manifold with boundary Also7 6f 1y f 1y 3X7 and this set is just since f is identity on the boundary Now f 1y is closed7 thus compact since f was continuous and X was compact Now f 1y E 1 Sl U01 U01 is a fact7 but then the number of boundary points is even7 but this number is one since the set of boundary points is This completes the proof


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