1 Normal (geodesic) coordinates

Transcription

1 Riemannian Geometry The Bochner- Weitzenbock formula If we need to verify some tensor identity (or inequality) on Riemannina manifolds, we only need to choose, at every point, a suitable local coordinate, and verify the identity (or inequality) at the given point. In this handout, we will discuss how to make the choice of the local coordinate and prove (or reprove) some useful formulas for the differential operators on the Riemannina manifolds M. 1 Normal (geodesic) coordinates We first prove the following statement about the exitence of the normal coordinates: For each point p M, there exists the normal coordinate (U, x 1,..., x m ) at p, i.e. we have x i (p) = 0, g ij (p) = δ ij and Γ k ij(p) = 0 for 1 i, j, k m. Proof: According the theorem about existence of the solution for (system) ODEs, for every point P M and v T P (M), there exists a unique geodesics C(t, P, v) (or we just write C v (t)) such that C v (0) = P, C v(0) = v. Before we continue, we make the following very important remark: we always have C λv (t) = C v (λt), becuase if we set α(u) = C v (λu), then dα du = λdc v dt, d 2 α du 2 = λ2 d2 C v dt 2. Thus, since C v is a geodesic, α is also a geodesic and stasfies that α(0) = P, α (0) = λv, so by the uniquness, C v (λt) = C λv (t). This proves the remark. The exponential mapping exp P (v): T P (M) M is defined by v C v (1). Then, we can prove that there exists ɛ > 0 such that exp P : B(O P, ɛ) T P (M) M is a diffeomorphism onto its image. Let U = exp P (B(O P, ɛ)). Take an orthonormal basis {e 1,..., e m } for T P (M), this determines an associated coordinate (x 1,..., x m ). We verify that the coordinates (U, x 1,..., x m ) is a normal coordinate for P. First, we claim that x i P = e i. 1

2 In fact, let φ = (x 1,..., x m ), and take any function f around P, we have (only veify for i = 1) f x 1 P = f φ 1 x 1 P = d dt f φ 1 (t, 0,..., 0) t=0 = d dt f C te 1 (1) t=0 = d dt f C e 1 (t) t=0 = e 1 (f). So x i P = e i. This impiles that g ij (p) = δ ij. Next, we show that Γ k ij(p) = 0 for 1 i, j, k m. Let q = C v (1) U and let C i (t) be the coordinate components of C v (t). Write φ(q) = (ξ 1,..., ξ m ). Then, since C v (t) = C tv (1), we have that C i (t) = tξ i. Since C v (t) is geodesic, the geodesic equation implies that Let t 0, we get Γ k ij(c v (t))ξ i ξ j = 0. Γ k ij(c v (0))ξ i ξ j = 0. Because ξ i are arbitray, and Γ k ij = Γ k ji, this implies that Γ k ij(p) = 0. So the statement is proved. 2 Reproof of some formulas for Differential operators In this notes, we always assume that M is a Riemannian manifold with the Levi-Civita connection, and X, Y,... are smooth vecor fields. A basis formula for d. Let (M, g) be a Riemannian manifold with the Levi-Civita connection. Let {e i } be a local frame field on M (i.e. a basis for Γ(U, T M)) and {ω i } be its dual, i.e. ω j (e i ) = δ j i. Then, for every smooth differential r-form θ, dθ = i ω i ei θ. (1) 2

3 Proof: First notice that it is independent of the choice of coordinates. So we choose normal coordinates x i, i.e. we have x i (p) = 0, g ij (p) = δ ij and Γ k ij(p) = 0 for 1 i, j, k m. Let {e i } is a local orthonormal frame field on M with e i (p) = x i p, and let {ω j } be the dual to {e i }. We claim that ( ei ω j )(p) = 0. In fact, since δ jk = ω j (e k ) = (e k, ω j ), we have i.e. ω j (e k ) = ω j ( e k ), hence Thus we get, using Γ j ik (p) = 0, Therefore the claim holds. dδ jk = 0 = ( e k, ω j ) + (e k, ω j ), ei ω j = Γ j ik ωk. k=1 ( ei ω j )(p) = 0. Now, since / x idx j = 0, we have, for θ = fdx i 1 dx ir, ω i ei θ = i i f x i dx i dx i 1 dx ir = dθ. The divergence operator. A (1, 1)-tensor T can be viewed as an endomorphism T : V V, and the trace of T is called the contraction of T, denoted by trace(t ) or C 1 1(T ). Let X Γ(T M). Then X is a smooth (1, 1)-tensor field on M. Take a contraction of X, we get a smooth function on M. It is called the divergenece of X, i.e. div(x) = C 1 1( X). The map div : Γ(T M) C (M) given by X div(x) is called the divergence operator. In temrs of local coordinate (U; x i ), div(x) = 1 m G x ( GX i ), i where G = det(g ij ), g ij = g( / x i, / x j ), X i = dx i (X). 3

4 Solution: Choose normal coordinates x i at p M and write Then, at point p, X U = X i x i. 1 m G x ( GX i ) = i where on the other hand, at point p, div(x) = tr{y Y X} = So the formula holds. X i x i, x i Xi = X i x i. The gradient of f Let f C (M), define a tangent vector field grad(f) on M, by g(grad(f), X) = df(x) = X(f), for every smooth tangent vector field X. The tangent vector field grad(f) is called the gradient of f. In local coordinate (U; x i ), ( m ) ij f grad(f) = x i x, j j=1 g where G = det(g ij ), g ij = g( / x i, / x j ), (g ij ) = (g ij ) 1. Beltrami-Laplace operator Let f C (M), define f = div(grad(f)). It is called the Laplace operator. In local coordinate (U; x i ), f = 1 m Gg ij f, G x i x j j=1 where G = det(g ij ), g ij = g( / x i, / x j ), (g ij ) = (g ij ) 1. Interior product For any vector field X, ι(x) sends r-form to r 1 defined by, for every r-form ω and vector fields Y 1,..., Y r 1, (ι(x)ω)(y 1,..., Y r 1 ) = ω(x, Y 1,..., Y r 1 ). 4

5 Let η be its volume form of M. For every smooth tangent vector field X, d(ι(x)η) = div(x)η, where ι(x) is the interior product. Proof: By definition, η = Gdx 1 dx m, we claim ι(x)η = ω where ω = ( 1) i+1 GX i dx 1 d ˆx i dx m. We now prove it. Indep. of the choice of coordinates. Choose normal coordinates x i. Then, at the point p, η = dx 1 dx m and X = mj=1 X j e j, hence ι(e j )(ω 1 ω m ) = ( 1) j+1 ω 1 ˆω j ω m. This proves the claim. The rest of proof follows easily. Divergence theorem: Let (M, g) be a compact oriented Riemannian manifold, then, for every smooth tangent vector field X, where η is the volume form. (divx)η = 0, The expression of the co-differential operator δ: Let {e i } be a local frame field on M compatible with the orientation of M. Let g ij = g(e i, e j ), and (g ij ) = (g ij ) 1. Then the codifferential operator δ can be written as δα = g ij ( ei α), i,j=1 for every α Λ r (M). If {e i } is orthonormal, then we can write δ = ι(e j ) ej, (2) j=1 5

6 where i(x) is the interior product operator, i.e. for every α Λ r (M), and for every tangent vector fields X 1,..., X r 1. δα(x 1,..., X r 1 ) = ι(e j )( ej α)(x 1,..., X r 1 ) j=1 = ( ej α)(e j, X 1,..., X r 1 ). j=1 Proof: For p M, choose the normal coordinate (U, x 1,..., x m ) at p. Let {e i } is a local orthonormal frame field on M with e i (p) = x i p, and let {ω j } be the dual to {e i }. Then To prove ( ei ω j )(p) = 0. δ(α) = ι(e j )( ej α). (3) j=1 We need only to verify it at each point p M. Since the operator is linear, without loss of generality, we assume that Hence α = fω 1 ω r. ej α = ej (f)ω 1 ω r + f ej (ω 1 ω r ) = e j (f)ω 1 ω r + f ej (ω 1 ω r ). Using ( ei ω j )(p) = 0, (only) at the point p, we have Hence, at the point p, we have Because ej α = e j (f)ω 1 ω r. ι(e j )( ej α) = e j (f)(ι(e j )(ω 1 ω r )). ι(e j )(ω 1 ω r ) = ( 1) j+1 ω 1 ˆω j ω r, 6

7 we have ι(e j )( ej α) = ( 1) j+1 e j (f)ω 1 ˆω j ω r. This tells us, at the point p, that ι(e j )( ej α) = j=1 j ( 1) j+1 e j (f)ω 1 ˆω j ω r. (4) We now calculate the left-hand side. By defintion, δ = ( 1) n(r+1)+1 d. We have, δ(α) = ( 1) n(r+1)+1 d (α) = ( 1) n(r+1)+1 d (fω 1 ω r ) = ( 1) n(r+1)+1 d(fω r+1 ω m ) = ( 1) n(r+1)+1 ( j e j (f)ω j ω r+1 ω m ) Note that (ω j ω r+1 ω m ) = ( 1) (r 1)(n r 1)+(r j) ω 1 ˆω j ω r, hence δ(α) = j = j ( 1) n(r+1)+1 ( 1) (r 1)(n r 1)+(r j) e j (f)ω 1 ˆω j ω r ( 1) j e j (f)ω 1 ˆω j ω r. Comparing the above identity with (4), we conclude that (2) holds at every point p. Hence the theroem holds. The operator δ can be viewed as a generalization of the divergenece: In fact, let X be a vector field, then δ(α X ) = div(x), where α X is the 1-form defined by α X (Y ) = g(x, Y ) for every smooth tangent vector field Y. Proof: Again, they are indep. of the choice of coordinates. Choose normal coordinates x i. Then, at point p. Let {e i } is a local orthonormal frame field on M with e i (p) = x i p, and let {ω j } be the dual to {e i }. 7

8 Let X = X i e j, then α X = X j ω j, div(x) = e i (X j ). On the other hand, δ(α X ) = ei (α X )(e i ). At point p, ei (α X ) = e i (X j )ω j + X j ei ω i = e i (X j )ω j. Hence ei (α X )(e i ) = e i (X i ). This proves the theorem. 3 Bochner-Weitzenbock Formulas The Bochner-Weitzenbock Formulas, sometimes refered to as the Bochner technique, is one of the most important technique in the theory of geometric analysis. We want to express the Hodge-Laplace operator in terms of the Levi- Civita connection. We first consider the function case. For functions, i.e. 0-form f, by defintion, g(grad(f), Y ) = Y (f) and df(y ) = Y (f), so α grad(f) = df. Hence, from above, δ(df) = div(grad(f)). Therefore, f = δd(f) = div(grad(f)) = tr 2 f, where, tr 2 := m ei ei for an orthonormal basis {e i }. In general, let {e i } be a local frame for a Riemannian manifold (M, g), define tr 2 : Λ r (M) Λ r (M) as for every α Λ r (M). For 1-form α, we have tr 2 (α) = g ij ( ei ej ei e j )α, α(x) = tr 2 α(x) + r(α #, X), 8

9 where r is the Ricci tensor of (M, g), and α # is the vector field defined by g(α #, Y ) = α(y ). Proof. The right hand side is independent of the choice of our orthonormal frame field. Therefore, we only need to verify it at every point p M. To do so, we choose normal coordinates centered at p and put at p, e i = x i. Let ω j be its dual frame. Then, always at p, ei e j = 0. This also gives, at p, ei ω j = 0. Using (1) and (2), we then have at p, (δdα)(x) = (δ(ω j ej α))(x) = ( ei (ω j ej α))(e i, X) = (ω j ei ej α)(e i, X) = ei ei α(x) + X j ei ej α(e i ). (dδα)(x) = (ω j ej (δα))(x) = X j ej (δα) = X j ej ei α(e i ). Hence, α(x) = tr 2 α(x)+x j ei ej α(e i ) X j ej ei α)(e i ) = X j R(e i, e j )α(e i ), where where X = X k e k and R(e i, e j )α = ( ei ej ej ei )α. We now claim that X j R(e i, e j )α(e i ) = r(α #, X). 9

10 In fact, write α = α k ω k, then α # = α k e k, and This proves the statement. X j R(e i, e j )α(e i ) = X j α k R(e i, e j )ω k (e i ) = X j α k R(e j, e i )ω k (e i ) = X j α k R kmji ω m (e i ) = X j α k R kiji = r(α #, X). We now derive the formula for a general r-form. For the purpose, we define the second covariant derivative as 2 XY = X Y X Y. Weitzenbock s formula. Let e 1,..., e m be a local orthonormal frame field, with the dual frame field ω 1,..., ω m. Then the Hodge-Laplace operator acting on r-differential forms is given by = 2 e i e i i,j ω i ι(e j )R(e i, e j ). Proof. The right hand side is independent of the choice of our orthonormal frame field. Therefore, we only need to verify it at every point p M. To do so, we choose normal coordinates centered at p and put at p, e i = x i. Then, always at p, Hence, and also ei e j = 0. 2 e i e i = ei ei, [ x i, x j ] = 0. 10

11 Therefore R(e i, e j ) = ei ej ej ei. Using (1) and (2), we then have at p, δd = ι(e j ) ej (ω i ei ) = ι(e j )(ω i ej ei ) = ej ej + ω i ι(e j ) ej ei. To calcluale dδ, we note that, since ei ω j = 0, ι(e j ) ei = ei ι(e j ). Hence, dδ = ω i ei (ι(e j ) ej ) = ω i ι(e j ) ei ej. So = dδ + δd = 2 e i e i i,j This proves the statement. ω i ι(e j )R(e i, e j ). Remak: On functions, i.e. 0-form f, we have R(e i, e j )f = fr(e i, e j )1 = 0. Hence, we have, for a local orthonormal frame field, f = 2 e i e i f. Theorem. For any smooth differential form η (r-forms), η 2 = 2 η < i 2 e i e i η, η >, where η 2 = i ei η 2. 11

12 Remark: Consider the Eucliean case R 2 and η = f. Then f 2 = 2f x f x + 2f y f y + 2f xx f + 2f yy f = 2 f < f xx + f yy, f >. The theorem is motivated by it, and can be derived by direct computation. We only verify it for 1 form η. We only need to verify it at every point p M. Let {e 1,..., e m } be a local orthonormal frame fields and {ω 1,..., ω m } be the cofarme fields. Write η = i η i ω i, then, at point p, using the formula of for smooth functions, 1 2 η 2 = i < ei ei η, η > + i ei η 2. This finishes the proof. Combining the both, we get Bochner s formula. Let e 1,..., e m be a local orthonormal frame field, with the dual frame field ω 1,, ω m. Let η be a r-form on M. Then 1 2 η 2 =< η, η > η 2 + < ω i ι(e j )R(e i, e j )η, η >. Proof: Choose normal coordinate. From above, 1 2 η 2 = η 2 i And from the Weitzenbock s formula, < 2 e i e i η, η >. Hcece, η = 2 e i e i η ω i ι(e j )R(e i, e j )(η). i,j 2 e i e i η = (η) + ω i ι(e j )R(e i, e j )(η). i,j 12

13 Sbustituting it into above, we get 1 2 η 2 =< η, η > η 2 + < ω i ι(e j )R(e i, e j )η, η >. For any smooth one-form, we get Corollary Let e 1,..., e m be a local orthonormal frame field, with the dual frame field ω 1,, ω m. Let η be a smooth one-form, then 1 2 η 2 =< η, η > η 2 r(η, η), where η 2 := i < ei η, ei η >, and writing η = f i ω i, r(η, η) := i,j r(f i e i, f j e j ) = i,j f i f j r(e i, e j ). Proof: We only need to compute, for 1-form η, < η, ω i ι(e j )R(e i, e j )η > = < f l ω l, ω i ι(e j )R(e i, e j )f k ω k > = f l f k < ω l, ω i ι(e j )R(e i, e j )ω k > = f l f k < ω l, ω i ι(e j )R kmij ω m > = f l f k < ω l, R kjij ω i > = f l f k R kjlj = f l f k R kl = r(η, η). Theorem(Bochner) Let M be a compact Riemannian manifold. If M has positive Ricci curvautre, then M has no nontrivial harmonic 1-form, thus, H 1 dr(m, R) = {0}. Proof. We integrate the formula above, and using the divergence theorem, 0 = ω η = 2 ( ω 2 + r(ω, ω))η. M M By the assuption, the integrand on the right hand side is pointwise nonnegative. It therefore has to vanish identically. Hence r(ω, ω) = 0, which implies that ω = 0 since the Ricci curvautre on M is positive. This proves the statement. 13

14 4 Proof of Garding s inequality Theorem Garding s inequality: There exist constant c 1, c 2 > 0, such that for every f (M), we have ( f, f) c 1 f 2 1 c 2 f 2 0. Proof. For every f (M), from Bochner s formula, < f, f > = 1 2 f 2 + f 2 < ω i ι(e j )R(e i, e j )f, f > 1 2 f 2 + f 2 a 1 f 2, where a 1 is a constant independent of f. Note that the last inequality holds because < ω i ι(e j )R(e i, e j )f, f > does not depend on the derivative(differential) of f (Note: although R(e i, e j ) depends on the derivative(differential), but since R(e i, e j )(αf) = αr(e i, e j )f for every α C inf (M), so < ω i ι(e j )R(e i, e j )f, f > is a quadratic form on r (M), its corfficients only depend on M, and since M is compact, such constant a 1 exists). Taking the integration on M and by definition, we get But by Stokes theorem, ( f, f) c 1 f 2 1 c 2 f This proves the Garding s inequality. M f 2 η = 0. M f 2 η. 14