DIFFERENTIAL GEOMETRY. LECTURES 9-10,

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1 DIFFERENTIAL GEOMETRY. LECTURES 9-10, Let us rovide some more details to the definintion of the de Rham differential. Let V, W be two vector bundles and assume we want to define an oerator (1) a : Γ(V) Γ(W). Assume s Γ(V). In order to define a section a(s) of W, it is enough to define a collection of sections a(s) i of W over oen subsets U i of X covering X, so that the restrictions a(s) i Ui,j and a(s) j Ui,j coincide. This is usually what one does in order to construct an oerator (1). Let us construct in this way the de Rham differential d : C (X) Ω 1 (X). In our case V = 1, W = T X and we choose the cover of X by the charts of an atlas. If φ : D U X is a ma and s C (X) the function s U is a function of the standard variables x 1,..., x n in R n and we ut ds = (sφ) i dx i. Let us check that the section of T X over U so obtained coincide at the intersections of the charts. If φ 1 and φ 2 are two charts as above, we will have (2) (3) ω 1 = j ω 2 = j (s φ 1 ) x j, (s φ 2 ), x j two sections of T X over U 1 and U 2 resectively, written in the coordinates of φ 1 and φ 2. Comatibility means that the transition function θ = φ 1 2 φ 1 exressing the deendence of x j on x j, carries ω 1 to ω 2. Since s φ 1 = s φ 2 θ, one has ω 1 = (s φ 2 θ) dx j == (s φ 2 ) x i dx x j j x j j i i x j and this is exactly what is suosed to be More about tensors. Similarly to the above, one can check that the oerator d : Ω k Ω k+1 defined in the revious lecture, makes sense. It is, however, more clear what haens, if one uses a multilicative structure of the de Rham comlex. 1

2 2 Note two algebraic oerations with tensors. Let us agree to use the following notation. Tq (X) or just Tq is the bundle of (, q)-tensors. The sections of this bundle are called the (, q)-tensors; we denote Tq = Γ(Tq ). We have already had secial cases of this notion: T0 1 is just T and T1 0 is Ω Product. Since Tq = T (T X) T q (T X), there is an obvious ma Tq T q T + q+q. This gives a roduct oeration T q T q T + q+q. 2. Contraction. The ma T X T X 1 is defined by airing Tx (X) with T x X. This allows one to define a collection of mas Tq T 1 q 1, for each choice of a vector and a covector argument. Passing to the sections, we have contraction oerations c : T q T 1 q 1. The exterior ower k V can be identified with the antisymmetric art of T k V : the natural rojection T k V k V is slit by the ma (4) x 1... x k σ S k sign(σ)x σ(1)... x σ(k). The natural ma k V l V k+l V is defined so that the rojection from T k V to k V reserves the multilication. Exercise. Check that the ma (4) does not reserve the multilication. Now we can formulate the roerty characterizing the de Rham differential Theorem. There is a unique collection of linear mas d : Ω k Ω k+1 satisfying the following roerties. Its restriction to Ω 0 = C is as described above. It satisfies a (skew) Leibniz formula where ω Ω k. d d = 0. d(ω ω ) = d(ω) ω + ( 1) k ω d(ω ), Proof. Let us rove first of all uniqueness. Assume such collection of oerators exists. Then in local coordinates one has to have (5) d(fdx i1... dx ik ) = df dx i1... dx ik because of Leibniz formula and the condition dd = 0. This roves uniqueness. Let us rove existence. It consists of two arts. First of all, we resent formulas in local coordinates satisfying the required roerties. Then we have to rove that the formulas agree at the intersections of the charts. Fortunately, we do not have to do this since the formulas have to agree because of the uniqueness roerty we have already roven.

3 Let us now define the differential in the local coordinates by the formula (5) and let us check it satisfies the required roerties. First of all, Leibniz formula. One has fdx i1... dx ik gdx j1... dx jl = fgdx i1... dx ik dx j1... dx jl so that it is enough to check the Leibniz formula for k = l = 0, that is to check d(fg) = df g + fdg. This is a standard calculus claim. Exercise. Find out why did we need the mysterious sign ( 1)k to aear. Now, let us check dd = 0. Recall that df = f dx i so (6) (7) dd(fdx i1... dx ik ) = d( f dx i dx i1... dx ik ) = 2 f dx j dx i dx i1... dx ik = 0, x j since dx i dx i = 0 and i,j 2 f x j = 2 f x j Morhisms of vector bundles, continued. This is a second attemt to start telling this toic. Let F : V W be a morhism of vector bundles over X. Recall that this means that F is a smooth ma commuting with the rojection to X and such that for each x X the ma of fibers V x W x is linear (recall that the fibers are vector saces). The ma F defines by comosition the ma of the corresonding sections which we will denote Γ(F ) : Γ(V) Γ(W). Note that one can multily sections by smooth functions, and that Γ(F ) obviously reserves this structure (8) Γ(F )(fs) = fγ(f )s. As we already know, there are interesting mas Γ(V) Γ(W) which do not satisfy this roerty, and, therefore, they do not come from morhisms of vector bundles. Such is, for instance, the de Rham differential. Today we will see more examles of this sort. Still, interesting examles of oerators satisfy the following roerty which we call locality Definition. An oerator f : Γ(V) Γ(W) is local if for each s Γ(V) and for each oen subset U X the equality s U = 0 imlies f(s) U = 0. 3

4 4 De Rham differential is obviously local since zero function has zero derivatives. In what follows we need the following basic fact which we leave without roof Lemma. Let U, U be two oen subsets of X such that Ū U. Then there exists a smooth function α on X having suort in U such that α U = 1. Recall that suort of a function f is defined as the closure of the set {x f(x) 0} Corollary. Let U U be oen subsets of X, so that Ū U and let V be a vector bundle on X. For any section s Γ(V U ) there exists a section t Γ(V) such that s U = t U. Proof. Choose a function α as in Lemma and define t(y) = α(y)s(y) for y U and t(y) = 0 for y U Lemma. Assume that an oerator f : Γ(V) Γ(W) is C -linear. Then f is local. Proof. Let s U = 0. We want to check that f(s) U = 0. It is sufficient to rove that for each x U there exists a neighborhood U x such that f(s) U = 0. We can assume Ū U. Then there exists a smooth function α on X such that α U = 1 and the suort of α is in U. Under these conditions αs = 0 therefore f(αs) = αf(s) = 0. Thus, f(s) U = 0 as required. Local oerators f : Γ(V) Γ(W) satisfy a very secial roerty: one can uniquely extend them to sections over oen subsets, see the following theorem Theorem. Let f : Γ(V) Γ(W) be a local oerator. There is a unique collection of oerators comatible with the restrictions. f U : Γ(V U ) Γ(W U ) Of course, this is recisely the roerty which makes local oerators so imortant. Proof. Let U, U be two oen subsets of X such that Ū U. We define an oerator f U,U : Γ(V U ) Γ(W U ) as follows. Let s Γ(V U ). By there exists a section t Γ(V) such that t U = s U. We define f U,U (s) := f(t) U. The definition makes sense since, if t U = t U, then by locality of f one has f(t ) U = f(t) U. By the construction, f U,U is the only oerator comatible

5 5 with f, that is making the diagram below Γ(V) f Γ(W) (9) fu,u Γ(V U ) Γ(W U ) commutative. We can now resent U as a union of subsets U as above. The mas f U,U 1 and f U,U 2 coincide on the intersection U 1 U 2 because of uniqueness roerty. Gluing the sections f U,U (s) for various U we get a section over U which we denote f U (s). The uniqueness of the construction is obvious. We will now rove the following Theorem. An oerator f : Γ(V) Γ(W) comes from a morhism of vector bundles if and only if it is C -linear, that is it satisfies (8). Proof. We will first of all rove the claim in case V is a trivial bundle, and then will use a trivialization of V. In the second art we will use Theorem Assume V is trivial, that is isomorhic to X R k for some k. Let s i, i = 1,..., k, be constant sections of V corresonding to a fixed basis of R k. Any section of V can be uniquely resented as a linear comonation s = a i s i where a i C (X). Thus, a C -linear ma f : Γ(V) Γ(W) is uniquely defined by k sections t i = f(s i ) Γ(W). These sections uniquely define a morhism X R k W by the formula F (x, c i s i (x)) = (x, c i t i (x)). Choose a trivialization X = U i of V. On each one of U i we have by Theorem an oerator f Ui : V Ui W Ui. By the first art of the roof this gives rise to a unique ma F i : V Ui W Ui for each i. The mas F i are comatible on the intersections by uniqueness. That gives the result Integral curves. Vector fields are just sections of the tangent bundle. However, they have a secial meaning (since the tangent bundle is a very secial bundle). Let s be a vector field on a manifold X. One can look for a curve γ : (a, b) X satisfying the equation (10) γ (t) = s(γ(t)). The theory of ODE claims the following.

6 Theorem. For any x X there exists an interval (a, b) containing zero and a unique curve γ : (a, b) X satisfying the equation (10) and γ(0) = x. The size of the segment (a, b) may deend on x. However, for each x there exists a neighborhood U x and a common segment (a, b) such that the the differental equation (10) has solution in (a, b) for any initial condition γ(0) = y, y U. Fix such U and (a, b). Uniqueness of the solution of the differential equation allows one to define a smooth ma Θ : (a, b) U X. We will write Θ t (x) instead of Θ(t, x) Theorem. Assume t, s, t + s belong to (a, b). Then the mas Θ t+s and Θ t Θ s coincide at the common doman of definition. In articular, Θ t is a local diffeomorhism for small t. Proof. Θ s (x) is defined as γ(s) where γ satisfes the differential equation 10 with the initial condition γ(0) = x. Therefore Θ t Θ s (x) is δ(t) where δ satisfies the same equation with initial condition δ(0) = γ(s). By the uniqueness of the solution of ODE the functions δ(t) and γ(t + s) coincide. In articular, for small t one has Θ t Θ t = Θ 1 = id Lie derivative. In this subsection we define, for each vector field s T, a local oerator L s : Tq Tq. This oerator is called Lie derivative The case = q = 0 As we know, any vector field s T defines a derivation on the sace of smooth functions. This is L s in the case = q = The case = 1, q = 0 We have T 1 0 = T and the oerator L s is defined by the formula L s (t) = [s, t] Theorem. There exists a unique collection of oerators L s : Tq Tq satisfying the following conditions L s is as defined above in the cases = q = 0 (on functions) and = 1, q = 0 (on vector fields). L s satisfies Leibniz rule: L s (u v) = L s (u) v + u L s (v). L s commutes with (all) contractions: if c : Tq one has L s (c(u)) = c(l s (u)). T 1 q 1 is a contraction,

7 Proof. As usual, we will first rove uniqueness, and then take care of existence. Let us first of all show that the contraction axiom allows one to determine the action of L s on 1-forms. Let ω T 0 1 = Ω 1 and let t T. One has to have L s (t ω) = [s, t] ω + t L s (ω), so by contraction condition s( t, ω ) = [s, t], ω + t, L s (ω) We claim that L s (ω) is uniquely defined by this formula. In fact, L s (ω) is a section of T X = Hom(T X, 1) that is (by Exercise 2 of Lecture 7) a ma T X 1 that is a C -linear ma from T to C. The formula recisely defines this ma. Now, having defined L s on C, T and Ω 1, we have it (locally) for all tensors. This roves L s (if exists) is defined uniquely. The best way to rove existence of L s is to resent another definition and then check that it satisfies the required roerties. We will sketch this definition without entering into details. The vector field s T defines a collection of diffeomorhisms Θ t of X for small t. Any diffeomorhism Θ : X Y defines a ma Θ : Tq(Y ) Tq(X). In articular, for u Tq the assignment t Θ t (u) defines a ath in Tq. The tangent vector at t = 0 is an element of Tq which is denoted L s (u). Let us calculate L s (f) where f is a smooth function. By definitnion, Θ t (f) is a function assigning to x X the value f(θ(t, x)). By definition this coincides with f, s (x). In this way one can check that L s satisfies all the requirements (we will not do this). Alternatively, one can follow the way we used in dealing with de Rham differential. This means that one has to make exlicit calculations in order to check that in the local coordinates the oerators L s defined as in the first art of the roof, satisfy all requirements. Then uniqueness would imly that the formulas coincide on the intersection of the charts. We leave this to an interested reader. We will just indicate that if s = s i s j then L s ( ) = j x j and L s (dx i ) = s i j x j dx j. This allows one (after very lengthy calculations) to rove the existence locally, and, because of the uniqueness, also globally. 7