Solutions to Exercises in Notes 1., multiplying byg kl, and using [ ]

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1 Fo any vecto fields X, Y and Z that Solutions to Execises in Notes. 2g ( X Y, Z) = X (g (Y, Z)) + Y (g (X, Z)) Z (g (X, Y )) () + g ([X, Y ], Z) g ([X, Z], Y ) g ([Y, Z], X). Applying this to X =, Y =, Z =, multiplying byg kl, and using [ ], x i x j x l x i x = 0, we obtain j equation (). Solution to Execise. The deivation of this fomula illustates a nice tick in computing evolutions of vaious geometic quantities such as the connection and the cuvatues. We compute at an abitaily chosen point p in nomal coodinates centeed at p so that Γ k ij (p) = 0. Note that g x i jk (p) = 0. In such coodinates, k a j j q i i (p) = have s Γk ij = ( 2 gkl x i s g jl + x k a j j q x j i i (p) fo any (, q)-tenso a. Thus, at p we s g il ) x l s g ij and s Γk ij = 2 gkl ( i v jl + j v il l v ij ) (2) follows since i v jl (p) = v x i jl (p). Finally we note that since both sides of (2) ae the components of tensos, equation (2) in fact holds as a tenso equation, that is, it is tue fo any coodinate system, not just nomal coodinates. Hee used the hint that the diffeence of two connections is a tenso and hence the vaiation of a family of connections is a tenso. Let α be a covaiant -tenso and X be a vecto field. Lie deivative of α with espect to X is denoted by L X α. Fomula: (L X α) (Y,..., Y ) = ( X α) (Y,..., Y ) (3) + α (Y,..., Y i, Yi X, Y i+,..., Y ). i= Solution to Execise 2. () Use X g = 0. (2) Apply (3) with Y =, Y x i 2 =. To obtain x j equation (4), let X i = i f. Note that i j f = 2 f ( ), x i x and 2 f (X, Y ) = X (Y f) ( j X Y ) f. Solution to Execise 3. () ultiplying (6) by g im, i.e., tacing, we obtain 0 = g im ( i R jklm + j R kilm + k R ijlm ) = g im i R jklm j R kl + k R jl. Tacing again, i.e., multiplying by g kl, we obtain g im i R jm j R + g kl k R jl = 0, which yields (7). (2) Taking the divegence of R jk = n Rg jk yields 2 kr = g ij i R jk = g ij i ( ) n Rg jk = n kr

2 and the esult follows fom n 2. Solution to Execise 4. These ae special cases of (8). Acting on functions, the Laplacian is ( ) f = div f = tace g ( 2 f) = g ij i j f = g ij 2 f x i x f j Γk ij. x k One can show that f = g i,j= whee g det (g ij ). Solution to Execise 5. () This follows fom whee we used the Ricci identities. (2) We compute ( g g ij f ), (4) x i x j i f = j i j f = i j j f R jijk k f, f 2 = i i j f 2 = 2 i ( i j f j f) = 2 i j f j f j f and obtain pat (2) fom pat (). (3) If Rc 0 and f = 0, then pat (2) implies f 2 = 2 2 f Rc ( f, f) 2 f 2. Hence, if also f = 0, then 2 f = 0 and Rc ( f, f) = 0. Divegence theoem: Poof. Define the (n )-fom α by div (X) dµ = α ι X (dµ). Using d 2 = 0 and L X (dµ) = (d ι X + ι X d) (dµ), we compute X, ν dσ. (5) dα = d ι X (dµ) = (d ι X + ι X d) (dµ) = L X (dµ) = div(x)dµ, whee to obtain the last equality, we may compute in an othonomal fame e,..., e n : L X (dµ) (e,..., e n ) = Now Stokes s theoem implies that div(x)dµ = dα = dµ (e,..., ei X,..., e n ) i= = div(x)dµ (e,..., e n ). α = 2 ι X (dµ) = X, ν dσ.

3 To veify the last equality, we used (ι X (dµ)) (e 2,..., e n ) = n (dµ) (X, e 2,..., e n ) = n X, ν (dµ) (e, e 2,..., e n ) = X, ν. (n )! Solution to Execise 6. () This follows fom the divegence theoem and u = div( u). (2) This follows fom the divegence theoem applied to X = u v v u which has div(x) = u v v u. (3) This follows fom the divegence theoem applied to the vecto field fx. Solution to Execise 7. Easy calculation. Solution to Execise 8. This follows fom Execise 5(2) since integation by pats yields f fdµ = ( f)2 dµ. Solution to Execise 9. If u + λu = 0, then 2 u 2 = u 2 + u, u + Rc ( u, u) n ( u)2 λ u 2 + (n ) K u 2 = n λ2 u 2 + ((n ) K λ) u 2. Integating this, we have 0 n λ2 u 2 dµ + ((n ) K λ) u 2 dµ. Dividing by u2 dµ and using the fact that λ = u 2 dµ / u2 dµ implies Solution to Execise 0. Fom 0 n λ2 + ((n ) K λ) λ. R l ijk = i Γ l jk j Γ l ik + Γ p jk Γl ip Γ p ik Γl jp we compute ( ) ( ) s Rl ijk = i s Γl jk j s Γl ik = 2 ( ) i j vk l + k vj l l v jk 2 ( ) j i vk l + k vi l l v ik, which is the fist fomula. The second fomula follows fom this and the commutato fomula i j vk l j i vk l = R q ijk vl q + Rijqv l q k. 3

4 Solution to Execise. () We have s R ij = p ( ) ( ) s Γp ij i s Γp pj. This follows fom computing at the cente in nomal coodinates. Substituting the fomula fo s Γk ij into this, we obtain s R ij = 2 l ( i v jl + j v il l v ij ) 2 i j V. (6) Finally, commute deivatives. (2) Taking the tace, we obtain s R = gij ( ) s R ij s g ij R ij (7) = l i v il V v ij R ij. Solution to Execise 2. () and (2) follow immediately fom the pevious execise. (3) We shall show the equivalent fomula that unde the Ricci flow, the (3, )-Riemann cuvatue tenso evolves by t Rl ijk = Rijk l + g ( ) pq RijpR qk l 2RpikR jq l + 2RpiR l jqk R p i Rl pjk R p j Rl ipk R p k Rl ijp + RpR l p ijk. By applying the second Bianchi identity and commuting covaiant deivatives, we compute ( ) Rijk l = g pq p q Rijk l = g pq p i Rjqk l j Rqik l i p R = g pq jqk l + R pijrqk l + R piqrjk l + R pik Rl jq RpiR l jqk + j p Rqik l R pjirqk l R pjqrik l R pjk Rl iq + RpjR l iqk and then apply the second Bianchi identity again to obtain g pq p R l jqk = g pq g lm ( k R jqmp m R jqpk ) = k R l j l R jk. This lets us ewite the fomula fo R l ijk given above as Rijk l = i k Rj l + i l R jk + j k Ri l j l R ik RpijR + g pq qk l + R piqrjk l + R pik Rl jq RpiR l jqk RpjiR qk l R pjqrik l R pjk Rl iq + RpjR l iqk Now the fist Bianchi identity shows that which implies that R l ijk R pijr l qk R pjir l qk = R ijpr l qk, may be witten as. R l ijk = i k R l j + i l R jk + j k R l i j l R ik Ri Rjk l + RjR ik l RijpR + g pq qk l + R pik Rl jq + RpjR l iqk RpiR l jqk R pjk Rl iq. (8a) (8b) (8c) 4

5 On the othe hand, ewiting the commutato tem g lp ( j i i j ) R kp yields t Rl ijk = i k Rj l + i l R jk + j k Ri l j l R ik (9a) + g ( lp R q ijk R qp + R q ijp R kq). (9b) The eaction-diffusion equation fo the Riemann cuvatue tenso now follows fom compaing tems in fomulas (8) and (9). 5