Kusner's manifold exercises (Math 704, Spring 2003) All material is Copyleft 2003 by Rob Kusner* ======================================================================== 1/28 [In class, we gave a brief overview of where this course -- and the topic of differential topology -- is heading. In addition to problems mentioned in class, here are a few from my notes:] Let V be an n-dimensional vector space over R. Compute the dimension of the exterior algebra V. [Hint: if {e1, e2, ... , en} is a basis for V, show that {1, e1, e2, ... , en, e1e2, ... , en-1en, ... , e1e2e3...en} is a basis for the exterior algebra. ------------------------------------------------------------------------ From the above problem, you should have found the dimension of the degree k vectors (or k-vectors, for short) in the exterior algebra of V to be n \choose k = n!/k!(n-k)! Since n \choose k = n \choose n-k, there is an isomorphism between k- and (n-k)-vectors. Here's a way to define such an isomorphism, "Hodge duality": if {e1, e2, ... , en} is a basis for V, then the n-vectors e1e2...en and -e1e2...en are ORIENTATIONS for V; pick an orientation (say, the + one); then any k-vector is paired with its Hodge dual (n-k)-vector according to this rule for basis vectors: *(ei1ei2...eik) := (+/-) e1...ei1^...eik^...en where the v^ means v is omitted, and where the (+/-) is chosen to make *(x)x a nonnegative multiple of the (+) orientation chosen, i.e. *(ei1ei2...eik)ei1ei2...eik = e1e2...en. Show that, on k-vectors, *^2 = (+/-)Id - how does the sign depend on k? In case V = R^2 and k=1, *^2 = -Id, and * is "rotation by 90 degrees". In case V = R^4 and k=2, *^2 = Id; compute the (+/-)1 eigenspaces for *, i.e. the self- and anti-self-dual 2-vectors on R^4. ------------------------------------------------------------------------ If L:V -> V is a linear map, then L induces a linear map on the exterior algebra. In particular, show that the induced map on the 1-dimensional space of n-vectors is multiplication by det(L). ------------------------------------------------------------------------ If V* is the dual space of V, i.e. the space of all linear maps from V to R, then we can define exterior algebra of V* as well. Show that the k-vectors on V* can be regarded as skew-symmetric k-linear maps from V to R (or k-forms, for short). Show that L as above induces a linear map on k-forms - what is this map? ======================================================================== 1/30 [No class: Rob lectures at Rutgers for Jean Taylor's retirement] ======================================================================== 2/3 We have seen how to define Hodge duality for smooth forms: extend it to commute with multiplication by smooth functions. For example, let \theta = (x dy - y dx)/(x^2 + y^2) be a 1-form on R^2 \ {0}; then *dx = -dy and *dy = dx extend to *\theta = (x dx + y dy)/(x^2 + y^2) = dr/r = d(ln r) where r^2 = x^2 + y^2. Compute d\theta and d(*theta). Interpret \theta in polar coordinates. Do these forms remind you of anything from complex analysis?! ----------------------------------------------------------------------- If we write dz = dx +i dy and d\barz = dx - i dy, then they span the complexified (tensor with C) cotangent space of R^2. The complexified tangent space is spanned by their duals \d_z and \dbar_z. Write \d_z and \dbar_z in terms of \d_x and \d_y. Show that the Cauchy-Riemann equations for f = f1 + i f2 can be expressed \dbar_z f = 0. Also, express the exterior derivative operator d in terms of the complex operators \d and \dbar defined by \df = \d_z f dz and \dbar f = \dbar_z d\barz and conversely. ======================================================================= 2/5 Show that the boundary operator \d for a polyhedral (simplicial) complex satisfied \d^2 = 0, at least for a tetrahedron. (This is Rob's KOAN: a boundary has no boundary.... Ooooooooooooommmmmmmmmmmmmmmmmmmm....) ----------------------------------------------------------------------- Verify that exterior derivative defines a co-complex, that is, d^2 = 0. Thus im(d) is contained in ker(d), and the quotient defines a graded vector space, the deRham co-homology. ----------------------------------------------------------------------- For S^1, show that \theta (defined above) satisfies d\theta = 0, and that the class of \theta is a basis for the deRham co-homology in degree 1. (Hint: show that every closed 1-form must be a multiple of \theta, but \theta is not exact.) ----------------------------------------------------------------------- The differential operators grad, curl and div from advanced calculus have a nice interpretation in terms of exterior derivative d and Hodge duality *, as follows. Let I, J and K be the standard constant vector fields on R^3. Then any vectorfield A = aI + bJ + cK on R^3 is determined by smooth functions a, b and c. We may identify A with a 1-form #A = a dx + b dy + c dz (here # or "sharp" is what Marcel Berger calls a "musical isomorphism" - its inverse should be "flat", but I will use # for it also in what comes next, so ## = Id). Considering the spaces of forms on R^3, the operator d, and Hodge *: <---*---> \Omega^0 -d-> \Omega^1 -d-> \Omega^2 -d-> \Omega^3 <---------------*-----------------> please check the following (up to sign): grad f = #df curl A = #(*d(#A)) div A = *(d*(#A)) and in particular, since d^2 = 0, we have the calculus facts curl(grad f) = 0 and div(curl A) = 0. ----------------------------------------------------------------------- Think about how the deRham co-homology of various open subsets of R^3 (such as R^3\{0}) relate to classical gravity and electro/magnetostatics. ======================================================================= 2/11 We observed that the kernel of a 1-form \alpha on R^n can be viewed as a field of (n-1) dimensional planes on R^n. If \alpha is EXACT, that is, if \alpha = df for some function (0-form) f on R^n, these planes will be tangent planes to the level sets of f. Every 1-form on R^1 is exact. (Hint: Fundamental Theorem of Calculus.) In general, there is no such f (even locally). Poincare's Lemma asserts that if \alpha is CLOSED (d\alpha = 0) on R^n, then it is exact. (Hint: FTOC, in the guise of line integrals and Stokes' Theorem.) ----------------------------------------------------------------------- On R^2 there are plenty of non-exact 1-forms \alpha (x dy, for example), but we can always find a pair of nonvanishing functions f and g so that g \alpha = df and thus ker(\alpha) is still tangent to the curves {f = constant}. (In ODE classes, g is called a "multiplying factor".) Consider the 1-form \alpha = x dy - dz on R^3. Sketch the field of planes on R^3 defined by ker(\alpha). Compute d\alpha to conclude there are no level surfaces {f = constant} for which these are tangent planes. Is there ANY family of surfaces to which these planes are tangent? (That is, can you find the analog of a multiplying factor g as above?) ======================================================================= 2/13 [Class abbreviated by candidate visit] We sketched the bilinear pairing between k-forms and smooth k-chains on a manifold M, given by integration. For k=1, this can also be viewed as a function from smooth paths or loops on M to R. The fundamental theorem of calculus implies that the integral of an exact 1-form along a path is the difference between the terminal and initial values of its primitive; in particular, the integral of an exact 1-form around a loop must vanish. Prove the converse: if a 1-form integrates to zero on all loops, it is exact. --------------------------------------------------------------------- On R^2-{0}, relate the integral of the closed 1-form \theta around a loop to the winding number of the loop around 0. --------------------------------------------------------------------- Stokes' theorem for surfaces shows that the integral of a closed 1-form around a loop depends only on the homotopy (or even weaker, the homology) class of the loop. More generally, Stokes' asserts that, under the pairing between k-forms and k-chains, the exterior derviative d is dual to the boundary operator \d. Thus (first show this pairing is non-degenerate) we have the deRham theorem: Integration gives an isomorphism between deRham cohomology of forms and (real, smooth) singular homology on a manifold. (We'll offer a Mayer-Vietoris argument for this as well, later....) [This leads to deRham's theory of integral currents, which generalizes the notion of a Schwartz distribution (cf. deRham, Federer or Morgan).] ======================================================================= 2/20 The last of the 2/11 problems should remind you of the Frobenius integrability theorem. If \alpha1, ... , \alphak are (pointwise) linearly independent 1-forms on R^n (i.e. their k-fold wedge product is non-vanishing), then the intersections of their kernels defines a codimension-k distribution on R^n. The distribution is integrable provided the derivatives of the \alpha's are 2-forms lying in the ideal generated by the \alpha's: for each i = 1, ... , k d\alphai = \alpha1\thetai1 + ... + ak\alphak\thetaik (F) for some 1-forms \thetai1, ... , \thetaik. --------------------------------------------------------------------- To see this condition agrees with the version using vector fields, first show that the exterior derivative of a 1-form \omega applied to a pair of general vector fields X and Y satisfies d\omega(X,Y) = X\omega(Y) - Y\omega(X) - \omega([X,Y]). (Hint: the "correction term" involving Lie bracket [X,Y] is needed to make both sides bilinear with respect to smooth functions.) ---------------------------------------------------------------------- On R^3, the distribution defined by ker(\alpha) is integrable iff \alpha d\alpha = 0. Conversely, a distribution defines an ideal I generated by the linear space of 1-forms I(1) which annihilates the distribution. (If the \alpha's are given in advance, their span - with smooth function coefficients - defines I(1), and I is as above.) ----------------------------------------------------------------------- To check that condition (F) above agrees with vector field version of Frobenius condition - namely, if X and Y are tangent to the distribution, then so must their Lie bracket [X,Y] - we need to know that a 2-form \Omega can be written as \omega \theta if and only if \Omega(X,Y) = 0 for all X, Y in ker(\omega). (The "only if" direction is clear - please check the "if" direction.) ----------------------------------------------------------------------- If we insist upon the Leibniz rule d(\al \om) = d\al \om + (-1)^deg(\al) \al d\om the above formula for the exterior derivative of a 1-form \om applied to a pair of vector fields X and Y, along with the formula for exterior derivative of a function f df(X) = Xf d\om(X,Y) = X\om(Y) - Y\om(X) - \om([X,Y]) has a unique extension to k-forms. For example, consider a 2-form \Om = \al \om (the general 2-form is a sum of these SIMPLE 2-forms, though in R^3 observe that all 2-forms are simple); then we must have d\Om(X,Y,Z) = (d\om \al)(X,Y,Z) - (\om d\al)(X,Y,Z) = d\om(X,Y) \al(Z) - \om(X) d\al(Y,Z) + [other permutations of X, Y and Z] = [a formula which uses only the definition of exterior derivative d on lower degree forms] which suggests what d of a k-form applied to X0, X1, ... , Xk is. ======================================================================= 2/25 Today we gave the briefest hint that there is a more general theory of vector-valued differential forms, or more precisely E-valued k-forms, i.e. smooth sections of \wedge^kT^*M \tensor E, where E is a vector bundle over M. In case E has a global framing (trivialization) by "constant" sections, such a k-form is simply a list of rank(E) ordinary k-forms on M, and exterior derivative is done term by term. Generally, there will be no such framing, and we need to introduce the notion of a _connection_ on E to say how sections of E vary as the basepoint in M varies; then d^2 is no longer necessarily zero, but in fact measures the curvature of this connection (more on this later). Excercise: if rank(E)=1, we still must have d^2=0. ----------------------------------------------------------------------- Determine all smooth maps F:R -> R such that F*(dx) = dx. Describe all smooth maps F:R^2 -> R^2 such that F*(dx dy) = dx dy. Suppose F: R^2 -> R^2 is defined by F(x,y) = (x cosy, x siny). Compute the pullbacks F*(dx), F*(dy) and F*(dx dy). Suppose F: R^3 -> R^3: (x,y,z) -> (x cosy sinz, x siny sinz, x cosz). Compute the pullback F*(dx dy dz). Let F: R^n -> R^m be a smooth map. Verify that pullback commutes with exterior derivatives and products, that is F*\alpha F*\omega = F*(\alpha \omega) d(F*\alpha) = F*(d\alpha) for any k-form \alpha and p-form \omega on R^m. ----------------------------------------------------------------------- Identify R^2 with C via x1 + i y1 = z1, and similarly identify R^4 with C^2. Let \omega = dx1 dy1 + dx2 dy2 be a symplectic form on R^4. Express \omega in terms of dz's and dz_bar's. Which rotations of R^4 preserve \omega, that is, which A in SO(4) satisfy A*(\omega) = \omega? Think of the first coordinate subspace C in C^2 as defining an embedding F: R2 -> R^4, with F*(\omega) = dx1 dy1. Discuss how (AF)*(\omega) = c dx1 dy1 depends on the rotation A. For instance, if A is unitary, c = 1; for what A will c = 0 (that is, which A rotate C = R^2 into a Lagrangian plane in C^2 = R^4)? More generally, describe the Lagrangian n-planes in C^n. (Hint: it is a homogeneous space for U(n) - what is the isotropy subgroup of a particular Lagrangian n-plane, such as R^n?) ----------------------------------------------------------------------- Consider the cotangent bundle T*M with symplectic form \om = d\la, where \la = "-y dx" is the Liouville form on T*M. Determine when, in terms of the 1-form \al on M, the submanifold L = graph(\al) is of T*M Lagrangian, i.e. \om(X,Y)=0 for any X,Y tangent to L. ======================================================================= 2/27 We saw that M^n is orientable iff the R bundle \wedge^nT^*M (or, equivalently \wedge^nTM) has a global (nonvanishing) section (i.e. a volume form) iff this bundle is trivial. Please fill in the details of the partition-of-1 argument needed to go between a collection of charts (with all coordinate changes having positive Jacobian determinant) and this nonvanishing n-form definitions. Check that a 2-sided hypersurface N in an orientable manifold M is itself orientable (recall that 2-sided means the normal bundle to N is a trivial R bundle). Upto diffeormorphism, there are two real line bundles over S^1: the trivial bundle S^1 x R, and the M\"obius strip. How many R-bundles are there over S^2? How many over S^1 x S^1? Investigate the R^2 bundles over S^2 (see Hirsch's _Differential Topology_ for how to classify R^k-bundles over S^n). ----------------------------------------------------------------------- One upshot of the above is this simple example: any orientable surface M is a symplectic manifold, with \om = area form. Any curve in M is Lagrangian. Victor asked: Are there even dimensional orientable manifolds which aren't symplectic? We saw that S^4 was a counterexample (modulo some deRham theory we havn't done in detail - please try to argue directly that any closed 2-form \om on S^4 is exact, and then show that \om^2 must integrate to zero over S^4, so cannot be nondegenerate). Victor also asked: Are there essentially different symplectic structures on the same smooth manifold? There are several ways to make sense of "essentially different". For instance, one could look at the deRham class of the the symplectic form: it is closed, but must it be exact? Or one could ask whether the collection of Lagrangian submanifolds is essentially different.... The simplest example which exhibits both these possibilities is the (co)tangent bundle to a closed surface, such as T*S^2. This carries the exact symplectic form \om = d\la, where \la = "-y dx" is the Liouville form; it also carries the symplectic form \om' = (area form on S^2) + (area form on R^2, where R^2 is a model (co)tangent plane). Check that the latter defines a symplectic form, but it is not exact (hint: integrate \om' over S^2). Furthermore, note that \om has lots of Lagrangian 2-spheres, given by graphs of df in T*S^2 where f is any function on S^2 (see the final 2/25 problem); however \om' has no Lagrangian 2-spheres (but lots of Lagrangian tori). ----------------------------------------------------------------------- The space Lag(R^2n) of Lagrangian n-planes in R^2n is a submanifold of the Grassmannian Gr_n(R^2n). Compare these, starting by computing their dimensions (for n=1 they coincide, but not for n>1). ======================================================================= 3/4 Here's another way to think about orientatability and orientations: Every manifold M has 2-sheeted covering space M~ which is orientable (in the sense that it has an atlas of consistent charts). [Hint: for each chart on M, there is a pair of charts on M~, related by a reflection map.] Assume M is connected. M is orientable iff M~ is disconnected. (In fact, M~ will have exactly two components, each corresponding to a choice of orientation on M.) Consequently, any simply connected manifold is orientable. Suppose now that M is the M\"obius band. What is the orientation double covering M~? Deduce from this that the M\"obius band is not orientable. ----------------------------------------------------------------------- The associated orientation line bundle and sections: The orientation double cover M~ can be regarded as a principal Z/2-bundle over M; in fact, a typical fiber is simply the unit vectors {+1,-1} in the associated line bundle, which is given by the top exterior power of T*M. As we have seen, this line bundle is trivial - which is another way of saying M admits a non-vanishing n-form - iff M is orientable. If we take the tensor product of the line bundle with this Z/2 bundle (Z/2 acting on R by {+1,-1}), the resulting "twisted" line bundle is always trivial. In this way, a Z/2-orientation can always be represented by a nonvanising n-form, viewed as section of this twisted line bundle. ======================================================================= 3/6 Show that the 2-sheeted covering spaces of M are in bijective correspondence with the homomorphisms from the fundamental group of M to Z/2 (this is also H^1(M, Z/2), the first cohomology of M with Z/2 coefficients). Show the R-bundles over M are also in bijective correspondence. In particular, there are exactly 2^n different R-bundles over the n-torus S^1 \times ... \times S^1. ----------------------------------------------------------------------- Verify that the Mayer-Vietoris sequence of cocomplexes 0 -> \Om^*(M) -> \Om^*(U) \+ \Om^*(V) -> \Om^(U \cap V) -> 0 is exact, where the last non-trivial arrow is the difference between the restrictions to U and V. ======================================================================= 3/11 We computed the deRham cohomology H*(S^1) using the Mayer-Vietoris exact sequence for the cover of S^1 by two (interval) charts. Use this as the beginning of an induction to compute H*(S^n), where S^n is again covered by N- and S-polar charts. [Hint: argue - informally, for now - that the intersection of these charts has the same deRham cohomology as S^{n-1}] Can you compute H*(S^1 \times ... \times S^1) this way? ----------------------------------------------------------------------- We showed that the "connecting homomorphism" H^k(C) -d*-> H^k+1(A) in the long exact cohomology sequence (associated to a short exact sequence 0 -> A -f*-> B -g*-> C -> 0 of co-complexes) is well-defined, as are H^k(A) -f*-> H^k(B) and H^k(B) -g*-> H^k(C), and we checked exact exactness at H^k(B). Also check exactness at H^k(A) and H^k(C). Explicitly describe the above homomorphisms in the case of the Mayer-Vietoris exact sequence. ======================================================================= 3/13 [Class may not meet today, and no problem session tomorrow] ======================================================================= Spring Break Project A few weeks back we discussed the interior product (or contraction) of a vector field X with a k-form \omega: \i_X\omega is the (k-1)-form satisfying \i_X\omega(X-1,...,X_k-1) = \omega(X,X_1,...,X_k-1) for any vector fields X_1,...,X_k-1. Explore the properties of interior product (for instance, \i_X is a derivation with square 0, just like d). Furthermore, show that \i_Xd\omega + d\i_X\omega = d/dt|_t=0(\Phi_t*\omega) =: L_X\omega the Lie derivative of \omega with respect to the vector field X (here \Phi_t is the flow with d/dt|_t=0(\Phi_t) = X). (This is an example of a "cochain homotopy".) ---------------------------------------------------------------------------- Let \omega be a volume form (i.e. non-vanishing n-form) and let X be a vector field on M. The divergence of X is the function div X defined by L_X \omega = div X \omega. Check that this agrees with the usual definition of divergence on R^n. For general M, show that the flow \Phi_t of X is volume preserving (\Phi_t*\omega = \omega) iff div X = 0. ----------------------------------------------------------------------- Suppose M also has boundary \delM. We know how to induce an orientation on \delM using an outward pointing vector field along \delM; but in order to get a volume (n-1)-form \alpha on \delM from the volume n-form \omega we need to choose a particular vector field Y along \delM and form the interior product \alpha = \i_Y\omega|_\delM. If M has a Riemannian metric <.,.>, then a canonical choice of Y is the outward unit normal vector field along \delM. Use Stokes' Theorem to derive the Divergence Theorem: \int_M div X \omega = \int_\delM \alpha ----------------------------------------------------------------------- Formulate and prove a version of Stokes' Theorem for a manifold "with corners", whose charts correspond to open subsets of a polyhedron in R^n. (Note that a manifold with boundary is a special case, where the polyhedron is the upper-half-space H^n in R^n.) ======================================================================= 3/25 Using the Mayer-Vietoris connecting homomorphism, we found that a deRham representative of H^1(S^1) is a 1-form supported in a small neighborhood of a point (more precisely, a function multiple of "d\theta", where the function is non-negative and supported in such a neighborhood). Use M-V inductively to show a representative for H^n(S^n) for any n is like the one for n = 1. ----------------------------------------------------------------------- What are the homotopy classes of maps from the zero sphere S^0 = {-1,+1} to a manifold M which send +1 to a basepoint q on M? In particular, how is this set [(S^0,+1), (M,q)] related to the zeroth deRham cohomology group H^0(M)? ----------------------------------------------------------------------- Let 1 be the basepoint on the circle S^1. The homotopy classes [(S^1,1), (M,q)] define a group \pi_1(M,q) -- the fundamental group of M (at q) -- under the operation of concatenation. Integration of a closed 1-form around a loop based at q in M defines a homomorphism h from \pi_1(M,q) to the (dual space of) the first deRham group H^1(M). What is the image under h of any commutator of elements in \pi_1(M,q)? ----------------------------------------------------------------------- Show that x -> x/|x| defines a deformation retraction from R^n+1\{0} to S^n. Is there a deformation retraction from B^n+1 to S^n? ======================================================================= 3/27 Verify that the existence of a (co)chain homotopy K between two (co)chain maps f and g (i.e. f - g = (+/-) Kd + (+/-)dK) means that f* = g* on (co)homology. Is the converse true? ----------------------------------------------------------------------- We saw that interior product with a vector field X on M provides an "infinitesimal" chain homotopy from 1=identity^* on H^*(M) to the actiom \Phi_t^* on H^*(M) of the stages \Phi_t^* of the flow \Phi: M x R -> M. In particular, \Phi_t* = 1 on H^*(M). Give examples of diffeomorphisms of M which do NOT induce 1 on H^*(M). ----------------------------------------------------------------------- Recall the trivial line bundle \pi: M x R -> M and its zero section \sigma: M -> M x R. Use the fact that \pi* and \sigma* are inverses on cohomology to deduce the Poincare' Lemma. [Take M = R^n-1 and use induction on n from the known case n=1.] ======================================================================= 4/1 Check the signs in the chain homotopy K (integration along the fibre R) which we constructed between \pi*\after\s_0* and the identity on \Omega*(M x R). ----------------------------------------------------------------------- The map F: M x R -> N provides a homotopy between f = F\after\s_0 and g = F\after\s_1, using the sections \s_0 and \s_1: M -> M x R. Since both \s_0* and \s_1* invert \pi* on H^*, deduce g* = f* on H^*. (Can you express g* - f* on the level of forms, using F* and dK - Kd?) ----------------------------------------------------------------------- Compare the construction of the chain homotopies for the flow of a vector field with the one above. ----------------------------------------------------------------------- Try to understand Bott-Tu's argument for the compactly supported case. Note that the chain homotopy K is a little bit different here. ----------------------------------------------------------------------- Poincare's Lemma implies H_c^n(R^n) = R. Later we'll see (by Poincare' duality) that the same is true for any connected n-manifold N in place of R^n. The isomorphism is by integration of a compactly supported n-form over N. Exercise: what if N is not connected? ======================================================================= 4/3 [Reprise of Thurston Geometrization and the Poincare' Conjecture: Any compact 3-manifold decomposes into a connected sum of primes P; P futher decomposes along incompressible tori to give components Q which are Siefert-fibered or contain no further incompressible tori; each Q carries one of the 8 homogeneous geometries in dimension 3 (S^3, R^3, H^3, S^2 x R, H^2 x R, SL(2,R), Nil [Heisenberg] and Solv [dilations and translations of R^2]. In case the fundamental group G of Q is finite, this gives a spherical space-form S^3/G, with G represented in SO(4); in particular, for G trivial, the Poincare' Conjecture!] ----------------------------------------------------------------------- [Next week at MIT (4/7-9-11 at 4:30 in 35-225) Grisha Perelman reports on his approach to proving Geometrization by Ricci flow.] ======================================================================= 4/8 [So Okada presented a recap of Perelman's first talk. The notion of Seifert fibred space came up....] S^3 is fibred by S^1's via the Hopf fibration: a neighborhood of any S^1 fiber F_0 is foliated by S^1's each of which has a 1-to-1 projection back to F_0. In a Seifert M^3, there are a finite number of exceptional fibers F_1, ... , F_n whose complement in M is an S^1 bundle over a (punctured) surface. The fibers near any F_j project to F_j as a k_j-sheeted covering S^1 -> S^1, and the base surface completes as an orbifold, with cone points replacing the punctures. S^1 acts on S^3 in various ways. Given relatively prime (p,q) we get an action by (e^it)*(z,w) = (e^ipt z, e^iqt w). This gives a Seifert fibration of S^3 with two exceptional fibers F_z = {w=0} and F_w = {z=0}. The lens space L(p,q) is a quotient of S^3 by the cyclic group Z/p acting as above by restriction of S^1 to the p-roots of 1. Describe the Seifert fibration of L(p,q) and its base orbifold. ======================================================================= 4/10 We defined the (mod 2 and the integer) degree for a proper smooth map f: M -> N between orientable (or orientable mod 2 - this is an empty condition, since any manifold is mod 2 orientable) manifolds of the same dimension as a (unsigned and signed, resp.) sum over the preimage of any regular value p in N. Check that for connected N, these are well defined (i.e. independent of p). ----------------------------------------------------------------------- For manifolds of different dimensions, there are generalizations of the degree. For instance, the Hopf invariant H(f) is an integer defined for any map f: S^3 -> S^2, as follows: take a pair of regular values p and q and consider their preimages in S^3, which are oriented links P = f^-1(p) and Q = f^-1(q); then H(f) = the linking number of P and Q. Show that H(the Hopf map = ((z,w) -> z/w)) = 1. [See below for more on the linking number.] ----------------------------------------------------------------------- If f: M -> N is a proper smooth map between orientable manifolds of the same dimension, if N is connected, and if \alpha is a compactly supported n-form on N with integral 1, then deg(f) = \int_M f*\alpha. In other words, deg(f) is the induced homomorphism on cohomology H_c^n(N) = R -f*-> H_c^n(M) -> R composed with "summation" over the components of M. It follows that degree is a homotopy invariant: if f \~ g then deg(f) = deg(g). It also follows that deg(f g) = deg(f) deg(g). Try also to see these directly via the approach of (4/11). ----------------------------------------------------------------------- For each integer k there exists a map S^n -> S^n of degree k. (Hint: start with n = 1 and use cylidrical polar coordinates to "suspend".) ----------------------------------------------------------------------- For any M^n, there is a map M^n -> S^n of degree 1, and hence of arbitrary integer degree. When is there a such a map from M -> N? ----------------------------------------------------------------------- When is it true that deg(f) = deg(g) implies f \~ g? ======================================================================= 4/15 The linking number lk(L,M) of two (disjoint, simple, closed, oriented) curves L, M in R^3 is the degree of the the secant map L x M -> S^2. Show that this is also the number of crossings of L over M, counted with an appropriate sign, when L and M are projected to a general 2-plane in R^3. (Hint: the unit normal q to this 2-plane is a regular value of the secant map - use the orientations in the degree to work out a sign convention for the crossings picture.) Finally, argue that lk(L,M) is the intersection number of L with any surface Q with boundary M (use the fact that if Q and Q' bound M, their difference is the boundary of a 3-chain, hence intersection number of L with Q-Q' is zero, and thus lk(L,M) is independent of the choice of Q). ----------------------------------------------------------------------- Suppose P is a closed, oriented n-manifold. We defined the geometric (mod 2 and Z) intersection number for two oriented submanifolds M, N of P with complementary dimensions k, l with k+l=n. Similarly, the intersection number of two deRham classes [\al] and [\om] of complementary degree is defined by integrating the form \al \wedge \om over P. These two ideas are related via Poincare Duality!!!!!!!!!!!!!!!!!!!! What does this mean? As we saw today on a torus P = S^1 x S^1 = R^2/Z^2, the longitude L = S^1 x 1 and meridian M = 1 x S^1 intersect once. Each can be represented in H^1(P) by the forms \th_1 = "dt_1" and \th_2 = "dt_2" which integrate to 1 along L and M, respectively; and the integral of their wedge product over P is also 1. The form \th_2 is a Poincare dual to L, often denoted \eta_L in general, that is \eta_L is a closed form on P such that for any closed form \om on L \int_L \om = \int_P \om \wedge \eta_L. Show that \eta_M = - \th_1 in this torus case. What about the general case of a k-manifold L in an n-manifold P? The same formula applies, except now \eta_L is an (l=n-k)-form on P. The basic lemma is that the Poincare dual of a (transverse) intersection is the wedge of their duals: \eta_{L\capM} = \eta_L \wedge \eta_M. ----------------------------------------------------------------------- Note that intersection (or equivalently \wedge) gives rise to a multiplicative structure on (co)homology - work this out for the (deRham co)homology of the n-torus H*(S^1 x ... x S^1) - it's a (skew) polynomial ring on n generators (the 1-forms\th_i, i=1,...,n). ======================================================================= 4/17 [Recapitulation - we discussed some of this earlier in the term] The singular k-chains on a manifold M comprise the vector space C_k(M) of all (finite) linear combinations of smooth maps \sigma: \Delta_k -> M, where \Delta_k is the standard k-simplex. There is a natural boundary operator \d which is defined on each simplex and extended linearly. The image and kernel of \del, denoted B_k(M) and Z_k(M), are the boundaries and cycles. Check that \d\d = 0, and hence we get a complex C_*(M). Its homology, H_*(M) = Z_*(M)/B_*(M) is known as the singular homology of M. Show that H_0(M) = R^t, where t is the number of (path) compoments of M. Compare with the deRham cohomology H^n(M) - what's a nice basis for this? We can also define singular cochains C^k(M) by taking the dual space to the C_k(M). We get a (dual co)complex, with coboundary operator \delta, cocyles Z^k(M) and coboundaries B^k(M), whose (co)homology is singular cohmology H^*(M) = Z^*(M)/B^*(M). Verify that the singular cohomology groups of R^n are the same as the deRham cohomology groups of R^n. ----------------------------------------------------------------------- The singular (co)homology groups of a manifold also satisfy the Mayer-Vietoris theorem: given a covering of M by two open sets U and V, there is a short exact sequences of (co) chain complexes, and therefore along exact (co)homology sequence which computes H*(M) in terms of H*(U), H*(V) and H*(U\intV). It follows - from the above exercise, and the fact (see Lemma below) that an n-manifold can be covered by charts which are diffeomorphic to R^n, and all of whose intersections are also diffeomorphic to R^n (or emepty) - that the deRham cohomology and singular cohomology of a manifold agree - this is the deRham Isomorphism Theorem. ----------------------------------------------------------------------- Lemma. A compact manifold M can be covered by a finite number of charts which are convex with respect to a Riemannian metric g. The idea is to show there is a positive number i(M,g), the injectivity radius of the metric g, such that geodesic balls of radius less than half i(M,g) are convex. The injectivity radius is largest number r so that geodesics emanating from any point on M are minimizing out to distance r (this can also be described in terms of the exponential map). ----------------------------------------------------------------------- Integration gives a pairing between smooth k-forms and singular k-chains on M. Stokes Theorem extends linearly to this situation, and shows that the pairing descends to (co)homology. Specifically, we have = <\del c, \al> by Stokes, so = 0 = . Thus = for any cycle a in Z_k and any closed form \om in Z^k. The deRham Isomorphism Theorem asserts this pairing is nondegenerate. ----------------------------------------------------------------------- To prove deRham's theorem use the above to define a homomorphism from deRham cohomology to singular cohmomology. By Mayer-Vietoris set up, it suffices to check this homomorphism is an isomorphism on convex charts - that is, on \R^n (please check this!) - and then apply the 5-lemma to get the isomorphism on the union. ======================================================================= 4/22 Carry out the details of our unknotting lemma: If K in \R^3 is an embedded circle with a height function having exactly 2 critical points (max and min) then K bounds an embedded disk (and is thus unknotted). ----------------------------------------------------------------------- Let f: M -> R be smooth and suppose p is a critical point of f. Verify that Hess(f) is a well-defined symmetric bilinear form on the tangent space T_p M. By linear algebra (check this) we can thus define the rank, nullity, signature, index and co-index of Hess(f) at p. If the nullity is 0, we say p is nondegenerate. Prove the Morse Lemma: around a nondegenerate critical point p of index k we have f(x) = f(p) - x_1^2 - ... - x_k^2 + x_k+1^2 + ... + x_n^2 with respect to some coordinate system (centered at p). ----------------------------------------------------------------------- We say f is a Morse function if all its critical points are nondegenerate. In particular, a Morse function has isolated critical points, finite in number if M is compact. A good supply of Morse functions on M is obtained by embedding M in R^N, and taking (almost) any height function. [Hint: use Sard's theorem to check this!] ----------------------------------------------------------------------- Explore the geometry of the sublevel sets of a Morse function f near a critical point p. Without loss, we may assume the critical values of f are all distinct. The quadratic form of f near p shows that passing an index k critcal point p corresponds to attaching a k-handle: (B^{n-k} x D^k) is pasted along its boundary (B^{n-k} x S^{k-1}) onto the boundary of the sublevel set just below f(p) in order to get the sublevel set just above. Try to see this explicitly for small n and k (less than 3, say) by looking at various quadratic cones and nearby hyperboloids. ======================================================================= 4/24 The gradient vector field grad f defines a flow on M whose fixed points are the critical points of f. The Hopf-Morse Index formula relates the number of fixed points of a flow to the Euler number of M: \Sum (-1)^k = Euler(M), where k is the Morse index, and the sum is over all critical points. This can be seen from the handle attaching picture (previous and next problems). Bott extended the definition - now called Morse-Bott - to include a function M -f-> R which is critical along a collection of embedded submanifolds M_1, ... , M_t with Hess(f) nondegenerate on the normal bundles to these M_j's. (The M_j's can have differing dimensions; when all are 0-dimensional, we recover the usual definition of Morse, since the normal bundle to {p} is simply T_p M.) For such an f, there is a Hopf-Morse index formula for the Euler number of M, but now the coefficients include the Euler numbers of the M_j's as well. Try to figure out this formula by considering examples of functions invariant under a Lie group action (for example, consider a surface of revolution and a height function along the axis). ----------------------------------------------------------------------- Describe handle decompositions of S^1, S^1 x S^1, a surface of genus g, a sphere S^n and a general product of spheres with respect to a Morse function on each (you might wish to pick an embedding of these manifolds into R^N and take a generic height function). ----------------------------------------------------------------------- Consider the function |z_0|^2 + 2|z_1|^2 + ... + 2^n|z_n|^2 on C^n+1. It restricts to an S^1-invariant function on S^2n-1, and thus a function CP^n -f-> R. Show that f is Morse with critical points of indices 0, 2, ... 2n. Describe the resulting handle decomposition. Try the analogous problem for RP^n, first finding a Morse function invariant under the antipodal map of S^n. ----------------------------------------------------------------------- Let M a compact n-manifold which carries a Morse function with exactly two critical points. Prove that M is homeomorphic to S^n. (It need not be diffeomorphic once n > 6....) ======================================================================= 4/29 We defined the Lefschetz number of a map M -f-> M and related it to the number of fixed points counted algebraically with a local degree of f. We also saw several corollaries, including the Hopf-Morse index theorems. Prove the Lefschetz formula. [Hint: consider the (transverse) intersection of graph(f) with the diagonal in M x M; see Guillemin-Pollack for details; in case this intersection is transverse, all the Lefschetz degrees at the fixed points are +/-1, depending on the sign of det(df-I).] ----------------------------------------------------------------------- Give examples of vector fields on R^2 (and thus, R^n) with zeros of arbitrary Hopf index. [Hint: R^2 = C; see Milnor, Topology from the Differentiable Viewpoint for some pictures of examples.] ----------------------------------------------------------------------- We also defined the stable and unstable manifolds S_p and U_p associated to each critical point p of a Morse function f on a Riemannian manifold M. These depend on the metric on M, but there is chain complex (the Morse-Smale-Floer-Witten-... Complex) whose homology is independent of the metric (and of f): let C_k(M,f) = \R[{p \in M | index(f,p) = k}] (k = 0, 1, ... , n) be the \R vector space with basis the critical points of Morse index k. (The field \R can be replaced with \Z or \Z/2 as well....) ======================================================================= 5/1 These give a chain complex with boundary operators \d_k: C_k -> C_k-1 defined as follows (assume M is oriented): 1) Pick orientations on S_p and U_p so that the tangent spaces T_pS_p+T_pU_p are oriented as M is. (If M is not orientable, forget signs and just consider Z/2 coefficients.) 2) For q \in C_k, let \d q = \Sum p where p \in C_k-1 and counts the number of gradient flow lines from p to q with a sign, depending on whether the orientation of U_q \int S_p agrees with the direction of flow. ----------------------------------------------------------------------- Complete the details of the "toothy" S^2 example with 6 critical points, and work out the example of torus S^1 x S^1 with 4 critical points (here you will need to wiggle the metric to get a generic gradient flow). ----------------------------------------------------------------------- Now try the general case (you may want to forget signs, and do just a Z/2 version for a start): Show that \d_k-1 \d_k = 0, and thus there is a well defined homology; show that this homology is independent of the metric and the choice of f (in fact, it computes the singular homology of M, by the Mayer-Vietoris principle). ======================================================================= 5/6 The Morse inequalities are a consequence of the fact that the Morse-... complex computes the singular homology of M: i) c_k >= b_k for k = 0, 1, ..., n ii) c_k - c_k-1 + ... +/- c_0 >= b_k - b_k-1 + ... +/- b_0 where c_k = dim C_k are the Morse numbers, and b_k = dim H_k = dim Z_k - dim B_k are the Betti numbers. Both i) and ii) are purely algebraic facts about a complex C_* and its homology H_*; in fact 1) is rather obvious since Z_k is a subspace of C_k. It's nice to think of these in terms of attaching handles to make M: i) there are at least as many k-handles as b_k ii) 1) the number of 1-handles is at least as many as b_1 plus the number of 1-handles needed to attach the 0-handles: c_1 >= b_1 + [c_0 - b_0] ii) 2) the number of 2-handles is at least as many as b_2 plus the number of 2-handles needed to attach the 1-handles: c_2 >= b_2 + [(c_1 - b_1) - (c_0 - b_0)] and so forth, noting that the preceding inequality implies [...] >= 0. ----------------------------------------------------------------------- The Morse inequalities are used both ways: to bound the topology of M, and prove the existence of critical points. For instance, a Morse function on RP^n must have n+1 critical points. (We noted in an earlier problem that a quadratic form on R^n+1 defines a function on RP^n, and it is Morse when it has distinct eigenvalues - the critical points correspond to the eigenlines.) ----------------------------------------------------------------------- How many critical points must a Morse function have on a surface of genus g? On the n-torus S^1 x ... S^1? On CP^n? ======================================================================= 5/8 Fill in the details of the Gauss-Bonnet-Kroenecker-Lipschitz-Killing- Allendoerfer-Weil-Chern-Lashof theorem for an embedded hypersurface e:M^n -> R^n+1: deg(\nu:M -> S^n) = Euler(M)/2. You need to see that the Gauss (or rather, the Lipschitz-Killing) curvature det(d\nu) at p has sign which agrees with (-1)^k where k is the index of the critical point p for the height function f_v(p) = v.e(p) in the direction v, and use the change of variables formula to express deg(\nu) as (half) the average over v in S^n of the (signed) number of critical points for f_v (because, the critical points of f_v are the same as those for f_-v = -f_v, this double counts). ----------------------------------------------------------------------- Note that the integral of |det(d\nu)|, the absolute total curvature, is bounded below by the (no longer alternating) sum of the Betti numbers. ======================================================================= Reading/Exam Period Short Course on Riemannian Geometry (extrinsic and intrinsic): Covariant derivatives, connections and parallel transport** Second fundamental form and extrinsic curvature Intrinsic curvature, holonomy and the Gauss/Coddazzi equation Geodesics and curvature Harmonic forms and curvature Curvature (Ricci, etc.) flows to canonical metrics (**not to be confused with alternative transport ;) ======================================================================= Have a great summer, year, and career! *"Copyleft" means that you have permission to use this material for non-commercial purposes as long as you acknowledge its source. Any use, in whole or in part, must include the "Copyleft 2003 by Rob Kusner" message or its equivalent. Commercial use without prior approval of the copyleft holder is strictly forbidden, and is punishable by the holder in any manner he sees fit!