Corrections and additions for 2nd Edition of Riemannian Geometry

Transcription

1 Corrections and additions for nd Edition of Riemannian Geometry I d like to thank Victor Alvarez, Igor Belegradek, Gil Cavalcanti, Or Hershkovits, Mayer Amitai Landau, Pablo Lessa, Ciprian Manolescu, Jiayin Pan, Jake Solomon, Fred Wilhelm for finding several mistakes in my book : Replace section 3.4. with the following more general exposition. I have now also added an elementary article about warped products on the webpage. It covers the issues addresses here in an even more elegant fashion. In the rotationally symmetric examples we haven t discussed what happens when '(t) =. In the revolution case, the curve clearly needs to have a vertical tangent in order to look smooth. To be specific, assume that we have dt + ' (t)d, ' :[,b)! [, 1), where '() = and '(t) > for t>. Allother situations can be translated or reflected into this position. More generally we wish to consider metrics on I S n 1 of the type dt + ' (t)ds n 1, where ds n 1 is the canonical metric on S n 1 (1) R n. These are the rotationally symmetric metrics. If we assume that ' () = and ' (t) > for t> we want to check that the metric extends smoothly near t =to give asmoothmetriconr n. The natural coordinate change to make is x = tz where x R n and t> and z S n 1 (1) R n. Thus ds n 1 = nx i=1 dz i, Keeping in mind that the constraint X z i =1 implies the relationship X z i dz i = between the restriction of the differentials to S n 1 (1). The standard metric on R n can now be written as X dx i = X z i dt + tdz i = X z i dt + t dz i +(tdt) z i dz i + z i dz i (tdt) = dt + t ds n 1 when switching to polar coordinates. In the general situation we have to do this calculation in reverse and check that the expression becomes smooth at the origin corresponding to x i =. So 1

2 we have to calculate dt and dz i in terms of x i. In fact we only need to observe that tdt = X x i dx i, dt = 1 X x i dx i t and then from our knowledge of the Euclidean metric that This gives us ds n 1 = P dx i dt t dt + ' (t)ds n 1 = dt + ' (t) = 1 = 1 t ' (t) t ' (t) t 4 Thus we have to ensure that the functions ' (t) 1 t and t P dx i dt t dt + ' (t) X dx i t X x i dx i ' (t) X + dx i t ' (t) t 4 q P are smooth, keeping in mind that t = (xi ) is not differentiable at the origin. Assuming that ' () = is necessary for the first function to be continuous at t =, while we have to additionally assume that ' () = 1 for the second function to be continuous. Next we note that if ' (t) =t + a k t k+1 then the two functions reduce to and 1 t ' (t) t = 1+! a k t k ' (t) t 4 = 1 1! 1 t t + X a k t k 1 = t = a k t k 1 + a k t k +! a k t k 1! a k t k 1

3 Showing that they are smooth. These conditions are all satisfied by the metrics dt +sn k (t)ds n 1, where t [, 1) when k apple and t [, pk ] for k>. It is possible to show that in general the two functions are smooth iff ' (even) () = and '() = 1, but will only need to use the real analytic case we established : extend should be extent 3 9 :(L X ij ) should be (L X ij ) dx i dx j 5 5 : [X, Y ] should be [Y,Z] :(Y,Z) should be (X, Y ) 31 4 : ij,k should be ij,l 33 4 k l : ij should be ij : r Y r X should be r Y r X Z : Z should be V. 4 : Replace the calculation of div (Ric) by div (Ric) = (n 1) div (f I) 4 15 : Replace calculation by = (n 1) df +(n 1) fdiv (I) = (n 1) df dscal (W ) p = D W scal = X (r W R)(E i,e j,e j,e i ) = X r Ej R (E i,w,e j,e i ) X (rei R)(E j,w,e j,e i ) = X r Ej R (E i,w,e j,e i ) = X r Ej R (E j,e i,e i,w) = X r Ej (R (E j,e i,e i,w)) = X r Ej g (Ric (E j ),W) = X g r Ej Ric (W ),E j = div (Ric) (W )(p) : In view of how the exterior derivative is defined in the appendix it is worth mentioning that Theorem 4 can be stated more succinctly as follows: 9 1 : Should be R (X, Y r = d r S (X, Y ). h k (v 1,v,v 3,v 4 ) = h (v 1,v 4 ) k (v,v 3 )+h (v,v 3 ) k (v 1,v 4 ) h (v 1,v 3 ) k (v,v 4 )+h (v,v 4 ) k (v 1,v 3 ) 3

4 9 16 : Should be R = scal n (n 1) g g + 1 n Ric scal n g g + W :Replace v() = v with v(a) =v : Replace by v and () with v () : Lemma 1 can also be proven by showing that dr apple1 (with equality only on vectors proportional to rr) and then noting `( ) = ˆ b ˆ b ˆ b dt dr ( ) dt d (r ) dt = r(q) r(p) : Instead of the sentence: "A careful...proved" Insert: Note that in coordinates this can be written as g x i,v j = ij x i v j Since dr (v) = ij x i v j we note that this can only be true if rr r = x i, which is our version of the Gauss lemma : The proof of Theorem 15 should be simplified as follows: Proof. We use polar coordinates around p M and the asymptotic behavior of g r and Hessr near p that was just established. We shall also use the fundamental equations g = Hessr Hessr)(X, Y ) Hess r (X, Y ) = R r,@ r,y). that were introduced in chapter. Our assumption that the curvature is constant implies that the second equation becomes Hessr Hess r = kg r. We wish to show that g = dr + g r = dr +sn k (r) ds n 1, Hessr = sn k (r) sn k (r) g r. 4

5 This is almost automatic as the right hand sides both have the same initial conditions as g and Hessr as we approach p and also simultaneously solve the two equations g = Hessr, Hessr Hess r = kg r :...integralcurvesforr... should be...integral curves r : The last = in the display is incorrect. Use the law of cosines to find cos \ (v, w). 149: Exercise 1: Assume that (M,g) has the property that all unit speed geodesics : Exercise 4: Incorrect as stated. Should read: Let r : U! R be a distance function on an open set U (M,g) and r = rr. OnU change the metric g to a new metric ĝ such that g (@ r,v)=ĝ (@ r,v) for all v TU. Show that r is still a distance function with respect to the new metric. 149: Exercise 5: Also incorrectly stated. It is T that vanishes at p, the vector X can be anything : There is a sign mistake in the last line of the displayed inequality: (d(p, (t)) (d(p, ()) +g (rf,p, ()) t + t : The display should read (d(p, ()) +(d( (), (t)) + (d(p, ()) (d( (), (t)) cos \ (rf,p, ()) inj (M) p or inj (M) = 1 (length of shortest closed geodesic) K 187 : Much in Chapter 7 has been completely revised. Look at the link to Bochner-Lichnerowicz-Weitzenböck Formulas. 4: Index mistakes in Lemma 7. The claim is that k 1 =( 1) (k+1)(n k)+1 d n k And the proof is 5

6 If! 1 k (M),! k 1 (M), then ( k 1! 1,! ) = (! 1,d k 1! ) = ( 1) k(n k) (! 1,d k 1! ) ˆ = ( 1) k(n k)! 1 ^ d k 1! M ˆ = ( 1) n k+k(n k) d n 1 (! 1 ^! ) Mˆ ( 1) n k+k(n k) (d n k! 1 ) ^! M ˆ = ( 1) n k+k(n k) (d n k! 1 ) ^! = ( 1) (n k)(k+1)+1 ( d n k! 1,! ). 41 :Moreprecisely:Fix ( ) Iso p Fix ( ) 56 7 : The display should be R (v, w) =lim t! I P t t 6 :replace closed by exact : Change the proof of Theorem 63 to be Proof. First observe that for k> there is nothing to prove, as we know that b 1 =from Myers theorem. Suppose we have chosen a covering M of M with torsion-free Abelian Galois group of deck transformations =h 1,..., b 1 i such that for some x M we have M d (x, i (x)) apple diam (M), d (x, (x)) > diam (M), 6= 1. Then we clearly have that all of the balls B (x), diam(m) are disjoint. Now set I r = { : = l l b1 b 1, l 1,..., l b1 appler}. Note that for B I r we have diam (M) (x), B x, r diam (M)+ diam (M). All of these balls are disjoint and have the same volume, as acts isometrically. We can therefore use the relative volume comparison theorem to conclude that the cardinality of I r is bounded from above by volb x, r diam (M)+ diam(m) v n, k, r diam (M)+ diam(m) apple. volb x, diam(m) v n, k, diam(m) 6

7 This shows that b 1 apple I 1 v n, k, diam (M)+ diam(m) apple, v n, k, diam(m) which gives us a general bound for b 1. To get a more refined bound we have to use I r for larger r. If r is an integer, then I r =(r + 1) b1. The upper bound for I r can be reduced to v n, k, b 1 r diam (M)+ diam(m) apple v n, k, b 1 r + 1 D v n, k, diam(m) v n, k, D (b 1 r + 1 )D sinh( p n 1 kt) p k dt = 1 D sinh( p n 1 kt) p k dt where in the last step we assume that D p taking logarithms this is equivalent to or (b 1 r + 1 )D p k sinh n 1 (t) dt = 1 Dp k sinh n 1 (t) dt = n b 1 r + 1 n + apple(b 1 5 r) n, b 1 log (r + 1) apple n log (b 1 5 r), or b 1 n apple log (b 1 5 r) log (r + 1) apple log (5b 1) + log r log + log r = log(5b 1) log r +1 log log r +1 apple log (5b 1) log r k is very small relative to r. By If b 1 n +1, this is not possible when r =(6b 1 ) n. Thus select r =(6b 1 ) n and the assume D p k is small enough that (b 1 r + 1 )D p k sinh n 1 (t) dt 1 apple (b 1 5 r) n Dp k sinh n 1 (t) dt 7 +1

8 in order to force b 1 apple n : In the definition of pointed G-H convergence it is necessary to have more flexibility on the radius. Thus we need to have a sequence R i! R such that: B (x i,r i ),x i,d i! B (x, R),x,d 99 8 : The proposition needs an added assumption and should be stated as: For a class C (M,d G H ) of metric spaces all of whose diameters are bounded by D<1, the following statements are equivalent: :Shouldbe Cap X (") Cap Y (" +4 ) : In the definition of pointed C m, convergence we also need to have that F i (p) =p i. 38: In the definition of C m, norms on Riemannian manifolds condition n4 should be deleted as it complicates several formulas and is not strictly necessary. As stated, it also invalidates Proposition 45. It could be replaced by a different type of estimate. However, n and n3 actually imply that ' 1 s ' t C m+1, apple C (n, r, Q) The only place where n4 becomes important is in the proof of Theorem 7. There it is used to conclude that the limit space has smooth transition functions and thus is a smooth manifold (bottom of page 313 and top of 314.) That fact is in fact automatic as a classic Theorem of Calabi and Hartman shows that an isometry between Hölder continuous Riemannian metrics is C 1. The transition functions are as it stands bi-lipschitz with respect to the Euclidean metrics, but also Riemannian isometries if we use the metric coefficients in the respective coordinates. These exist using the arguments in the second paragraph on page : In the definition of of convergence of functions and tensors on a fixed manifold, it is best to assume that the coordinates used are bi-lipschitz :...anysegmentfromx 1 to x... should be...any segment from ' (x 1 ) to ' (x ) :Shouldbe Aslongas"< :Shouldbe Ric appler i Replace with:...to see this consider B (x, R/1) B (p, R/5) B (x.r/) B (p, R) Since B (p, R) is assumed to be incompressible it follows that B (x, R/) does not deformation retract onto B (x, R/1). Otherwise, rank (H (B (x, R/1)! B (x.r/))) = rank (H (B (p, R/5)! B (x.r/))) rank (H (B (p, R/5)! B (p.r))) 8

9 This would imply that contb (p, R) apple contb (x, R/) and thus contradict incompressibility of B (p, R). We can now : Change X n to X p on right hand side. The appendix on de Rham cohomology has been reworked. You can find a corrected version by going to my essay on manifold theory. The only mistakes where some confusion on the degrees of cohomology groups when proving Poincaré duality. Nevertheless I felt that many things should be reworked and hopefully clarified. 9