《黎曼几何》课程教学资源（英文讲义）Lec08 THE RIEMANNIAN CURVATURE.pdf_大学文库

4 LECTURE 8: THE RIEMANNIAN CURVATURE 2. Decomposition of the Riemann curvature tensor ¶ Some tensor algebra: symmetric tensors. Let V be any vector space. Recall that ∧ 2V ∗ ⊂ ⊗2V ∗ represents the space of anti-symmetric 2-tensors on V , while S 2V ∗ ⊂ ⊗2V ∗ represents the space of symmetric 2-tensors on V . Any 2-tensor T on V can be decomposed uniquely as the summation of a symmetric 2-tensor and an anti-symmetric 2-tensor as T(u, v) = T(u, v) + T(v, u) 2 + T(u, v) − T(v, u) 2 . If dim V = m, then we have dim ∧ 2V ∗ = m(m − 1) 2 , and dim S 2V ∗ = m(m + 1) 2 . Note that by definition, S 2 (∧ 2V ∗ ) contains (0,4)-tensors that are symmetric with respect to (1, 2) ↔ (3, 4) and anti-symmetric with respect to 1 ↔ 2 and 3 ↔ 4, i.e. T(X, Y, Z, W) = −T(Y, X, Z, W) = −T(X, Y, W, Z) = T(Z, W, X, Y ). For example, one can easily check that for any two symmetric (0,2)-tensor S, T ∈ S 2 (V ∗ ), their Kulkarni-Nomizu product S○∧ T ∈ S 2 (∧ 2V ∗ ). The set S 2 (∧ 2V ∗ ) is a vector space of dimension (6) dim S 2 (∧ 2V ∗ ) = m(m − 1)(m2 − m + 2) 8 . Moreover, the space of 4-forms, ∧ 4V ∗ , is a subspace of S 2 (∧ 2V ∗ ) with dimension (7) dim ∧ 4V ∗ = m 4 . Let α, β ∈ ∧2V ∗ be any two linear 2-forms, both viewed as skew-symmetric 2-tensors on V . Define the symmetric product of α and β to be the (0, 4)-tensor (8) (α ⊙ β)(X, Y, Z, W) := α(X, Y )β(Z, W) + α(Z, W)β(X, Y ). In local coordinates one can write (α ⊙ β)ijkl = αijβkl + αklβij . Obviously each α⊙β is in S 2 (∧ 2V ∗ ). It turns out that these (0,4)-tensors generates the whole space S 2 (∧ 2V ∗ ): Lemma 2.1. Any element in S 2 (∧ 2V ∗ ) can be written as a linear combination of elements of the form α ⊙ β. Proof. Let e 1 , · · · , em be a basis of V ∗ , then E 1 = e 1 ∧ e 2 , E2 = e 1 ∧ e 3 , · · · , Em(m−1)/2 = e m−1 ∧ e m is a basis of Λ2V ∗ , and E i⊙E j (i ≤ j) are linearly independent in S 2 (∧ 2V ∗ ) (check). By dimension counting, we see these elements form a basis of S 2 (∧ 2V ∗ ). □

LECTURE 8: THE RIEMANNIAN CURVATURE 5 ¶ Some tensor algebra: Curvature-like tensors. To explore the cyclic symmetry that arise in the first Bianchi identity, we define Definition 2.2. The Bianchi symmetrization of any T ∈ S 2 (∧ 2V ∗ ) is the 4-tensor (9) bT(X, Y, Z, W) = 1 3 (T(X, Y, Z, W) + T(Y, Z, X, W) + T(Z, X, Y, W)). So the first Bianchi identity for Rm now becomes the simple equation b(Rm) = 0. Definition 2.3. If T ∈ S 2 (∧ 2V ∗ ) and b(T) = 0, we call T a curvature-like tensor. The set of all curvature-like tensors is denoted by C . Example. For any S, T ∈ S 2V ∗ , we have b(S○∧ T) = 0 since 3b(S○∧ T)ijkl =SikTjl + SjlTik − SilTjk − SjkTil + SjkTli + SliTjk − SjiTlk − SlkTji + SlkTij + SijTlk − SljTik − SikTlj =0. As a result, S○∧ T is a curvature-like tensor. Note that by definition, curvature-like tensors are exactly those tensors satisfying all the algebraic symmetries that Rm admit, namely (1), (4), (5) and (2). To study the set C of all curvature-like tensors, we first study the image of the Bianchi symmetrization b. Note that for any α, β ∈ Λ 2 (V ∗ ), 3(b(α ⊙ β))ijkl = αijβkl + αklβij + αjkβil + αilβjk + αkiβjl + αjlβki = αijβkl − αikβjl + αilβjk + αjkβil − αjlβik + αklβij = 4! 2!2! 1 4! X π∈S4 (α ⊗ β) π ijkl = (α ∧ β)ijkl. In other words, we have Lemma 2.4. For any α, β ∈ Λ 2 (V ∗ ), b(α ⊙ β) = 1 3 α ∧ β. In view of Lemma 2.1, we conclude that the map b has image Im(b) = ∧ 4V ∗ ⊂ Λ 2 (V ∗ ). In particular, for any T ∈ S 2 (∧ 2V ∗ ) we have bT ∈ S 2 (∧ 2V ∗ ). By definition it is straightforward to check Lemma 2.5. For any T ∈ S 2 (∧ 2V ∗ ), one has b(b(T)) = T. So the Bianchi symmetrization map b, as a linear map b : S 2 (∧ 2V ∗ ) → S 2 (∧ 2V ∗ ), is a projection. It follows from the standard linear algebra that S 2 (∧ 2V ∗ ) = Ker(b) ⊕ Im(b) = C ⊕ ∧4V ∗ . As a consequence, C is a vector space of dimension (10) dim C = m(m − 1)(m2 − m + 2) 8 − m 4 = 1 12 m2 (m2 − 1)

6 LECTURE 8: THE RIEMANNIAN CURVATURE ¶ Some tensor algebra: metric contractions. Now suppose the vector space V is endowed with an inner product g(·, ·) = ⟨·, ·⟩, so that one can identify V ∗ with V using the musical isomorphisms ♭ and ♯. In particular, for any (0,4)-tensor T and for any vectors X, Y, Z, the linear map T(Z, X, ·, Y ) : V → R, is in V ∗ and thus be identified with a vector ♯T(Z, X, ·, Y ) ∈ V . Definition 2.6. For any (0,4)-tensor T on an inner product space (V,⟨·, ·⟩), the Ricci contraction c(T) of T is the following (0,2)-tensor:1 (11) c(T)(X, Y ) := Tr(Z 7→ ♯T(Z, X, ·, Y )). For a Riemannian manifold, we call Rc := c(Rm) its Ricci curvature tensor. It turns out that c(T) is symmetric if T is curvature-like: Lemma 2.7. If T ∈ C , then c(T)(X, Y ) = c(T)(Y, X). Proof. Fix X and Y . Let K : V → V and Ke : V → V be the maps K(Z) = ♯T(Z, X, ·, Y ) and Ke(Z) = ♯T(·, X, Z, Y ). Then for any Z, W ∈ V , ⟨K(Z), W⟩=⟨♯T(Z, X, ·, Y ), W⟩=T(Z, X, W, Y )=⟨Z, ♯T(·X, W, Y )⟩= ⟨Z, Ke(W)⟩. So Ke is the transpose of K, and thus they have the same trace. But by definition Tr(K) = c(T)(X, Y ), while Tr(Ke) = c(T)(Y, X) (since T ∈ C ⊂ S 2 (∧ 2V ∗ )). □ Similarly one can define the trace of a (0,2)-tensor T using the metric: the map T(X, ·) : V → R is an element in V ∗ , and thus can be identified with a vector ♯T(X, ·) in V . Definition 2.8. The trace of a (0,2)-tensor T on (V,⟨·, ·⟩) is (12) Tr(T) := trace(X 7→ ♯T(X, ·)). For a Riemannian manifold (M, g), we call S(g) = Tr(Rc) the scalar curvature. Example. For the metric tensor g, by definition ♯g(X, ·) = X and thus Tr(g) = m. Locally if v 1 , · · · , vn be a basis of V ∗ , and if we denote g pq = g ∗ (v p , vq ) (where g ∗ is the dual metric on V ∗ ), then one can check [exercise] c(T)ij = g pqTipjq and Tr(T) = g ijTij . Note that ⟨T, g⟩ = Tijgklg ikg jl = Tijδ i l g jl = Tijg ij , one has (13) Tr(T) = ⟨T, g⟩. 1One can define metric contractions between other pairs of indices. However, for curvature-like tensors, what one can get are either Ricci contraction, or zero

9LECTURE8:THERIEMANNIANCURVATUREDecomposition of a curvature-like tensor via the metric: Step 2.We may continue to decompose any R e s2v* into orthogonal ones via the mapTr : s2* → IR. Again we need the dual of Tr, which, in view of (13), is simplyTr*:R→S?V*,t→tg.By repeating the same arguments, again we get an orthogonal decompositionS2V* = ker(Tr) @ Im(Tr*).So there exist E e ker(Tr) and t e R so that R can be decomposed orthogonally toR=E+tg,Applying the map Tr to both sides we get Tr(R)=tTr(g)=mt, and thusTr(R)mNote that if R = Rc is the Rici curvature tensor of (M, g), then t = S/m.Definition 2.15. For a Riemannian manifold (M,g), we callSE := Rc-mgthe traceless Ricci tensor of (M,g)Wepoint outthatthetwo decompositions arecompatible,inthe sensethatIm(亚) =亚(S2V*) = (ker(Tr))④ (Im(Tr*))is an orthogonal decomposition of Im(), since for t e R and any E eker(Tr),((tg), 亚(E) = (c((tg)), E) = (2m - 2)(tg), E) = 2(m - 2)tTr(E) = 0.Thus we end up with an orthogonal decomposition6 = ker(c) (ker(Tr)) ④(Im(Tr*)),so that any T e can be decomposed into three curvature-like tensors which areorthogonal to each other:the"Weyl part"W that lies in ker(c),Egfor atrace-lesssymmetric 2-tensor E, and a multiple of ggRemark.A more algebraic way to understand the two decompositions: Consider thenatural action of O(m) on V that preserves (-.-).which induces natural actions ofO(m)on 2v*and on that preserve the induced inner products.: For the case of s2*, since g (and thus the 1-dimensional space Rg) is invari-ant under the O(m)-action, one may decompose $2v* = Rg (IRg). Thisdecomposition can be explained via the O(m)-invariant map Tr as above,and thus (Rg)-=ker(Tr)consists ofall traceless symmetric (o,2)-tensors.. Similarly since c is O(m)-equivariant, it induce a decomposition of intoker(c)@(ker(c))+, and we have seen (ker(c))+= (S2V*) since c is surjective.Can we decompose further? The answer is no, since one can prove the O(m)-actionon ker(Tr) and on ker(c) are transitive(i.e. they are irreducible representations of O(m)

LECTURE 8: THE RIEMANNIAN CURVATURE 9 ¶ Decomposition of a curvature-like tensor via the metric: Step 2. We may continue to decompose any R ∈ S 2V ∗ into orthogonal ones via the map Tr : S 2V ∗ → R. Again we need the dual of Tr, which, in view of (13), is simply Tr∗ : R → S 2V ∗ , t 7→ tg. By repeating the same arguments, again we get an orthogonal decomposition S 2V ∗ = ker(Tr) ⊕ Im(Tr∗ ). So there exist E ∈ ker(Tr) and t ∈ R so that R can be decomposed orthogonally to R = E + tg, Applying the map Tr to both sides we get Tr(R) = tTr(g) = mt, and thus t = Tr(R) m . Note that if R = Rc is the Ricci curvature tensor of (M, g), then t = S/m. Definition 2.15. For a Riemannian manifold (M, g), we call E := Rc − S m g the traceless Ricci tensor of (M, g). We point out that the two decompositions are compatible, in the sense that Im(Ψ) = Ψ(S 2V ∗ ) = Ψ(ker(Tr)) ⊕ Ψ(Im(Tr∗ )) is an orthogonal decomposition of Im(Ψ), since for t ∈ R and any E ∈ ker(Tr), ⟨Ψ(tg), Ψ(E)⟩ = ⟨c(Ψ(tg)), E⟩ = ⟨(2m − 2)(tg), E⟩ = 2(m − 2)tTr(E) = 0. Thus we end up with an orthogonal decomposition C = ker(c) ⊕ Ψ(ker(Tr)) ⊕ Ψ(Im(Tr∗ )), so that any T ∈ C can be decomposed into three curvature-like tensors which are orthogonal to each other: the “Weyl part” W that lies in ker(c), E○∧ g for a trace-less symmetric 2-tensor E, and a multiple of g○∧ g. Remark. A more algebraic way to understand the two decompositions: Consider the natural action of O(m) on V that preserves ⟨·, ·⟩, which induces natural actions of O(m) on S 2V ∗ and on C that preserve the induced inner products. • For the case of S 2V ∗ , since g (and thus the 1-dimensional space Rg) is invariant under the O(m)-action, one may decompose S 2V ∗ = Rg ⊕ (Rg) ⊥. This decomposition can be explained via the O(m)-invariant map Tr as above, and thus (Rg) ⊥ = ker(Tr) consists of all traceless symmetric (0,2)-tensors. • Similarly since c is O(m)-equivariant, it induce a decomposition of C into ker(c)⊕(ker(c))⊥, and we have seen (ker(c))⊥ = Ψ(S 2V ∗ ) since c is surjective. Can we decompose further? The answer is no, since one can prove the O(m)-action on ker(Tr) and on ker(c) are transitive(i.e. they are irreducible representations of O(m))