LECTURE 14: NORMAL COORDINATES 1. the Normal

LECTURE 14: NORMAL COORDINATES

1. The Normal Coordinates Now we see for any p ∈ M, there exists a neighborhood U ⊂ M of p and a neighborhood V ⊂ TpM of 0 so that the exponential map

expp : V → U −1 is a diffeomorphism. But TpM, and thus V , is Euclidian, so the triple {expp , U, V } form a local chart of M near p. Usually one takes such U’s to be geodesic balls (and thus V to be Euclidean balls). We will fix an orthonormal basis {ei} of TpM, which gives us linear coordinates for V . We denote the corresponding coordinate functions on U by {xi}. Definition 1.1. The local chart {U; x1, ··· , xm} described above is called normal coordinate system at p. This is a particularly nice coordinate system. For example, the parametric i equation for the geodesic γ(t) = expp(tXp), where Xp = x ∂i, is given by γ : x(t) = (tx1, tx2, ··· , txm), Moreover, we have Lemma 1.2. Let {U; x1, ··· , xm} be a normal coordinate system at p. Then

(1) For all 1 ≤ i, j ≤ m, gij(p) = δij. k (2) For all 1 ≤ i, j, k ≤ m, Γij(p) = 0. (3) For all 1 ≤ i, j, k ≤ m, ∂kgij(p) = 0.

Proof. (1) This is obvious, since ∂i|p = d(expp)0ei = ei. (2) Put the geodesic parametric equation x(t) = (tx1, tx2, ··· , txm). into the geodesic equation, we get for 1 ≤ k ≤ m, k i j k i j k 0 =x ¨ (t) +x ˙ (t)x ˙ (t)Γij(γ(t)) = x x Γij(γ(t)). Letting t = 0, we conclude that for any 1 ≤ k ≤ m, i j k x x Γij(p) = 0 for all xi, xj. It follows that for any fixed k, all eigenvalues of the symmetric matrix k k (Γij)1≤i,j≤m are 0. So Γij(p) = 0 for all i, j, k. (3) This is a consequence of (2) and the metric compatibility.  1 2 LECTURE 14: NORMAL COORDINATES

From (1) and (3) above, we see that the Taylor’s expansion of gij(x) at p using normal coordinates start with the constant term is δij, and has no linear term. What is the next term?

Theorem 1.3. With respect to normal coordinates near p, the functions gij’s admit the following Taylor expansion at x = 0, 1 (1) g (x) = δ + R (p)xkxl + O(|x3|). ij ij 3 iklj Proof. Let (U; x1, ··· , xm) be a normal coordinate system near p. Fix xi’s and let i γ be the geodesic starting at p in the direction Xp = x ∂i, γ(t) = (tx1, ··· , txm), 0 ≤ t ≤ 1. For each 1 ≤ i ≤ m, consider a geodesic variation

1 i m fi(t, s) = (tx , ··· , t(x + s), ··· , tx ).

Its variation field Vi = t∂i is thus a Jacobi field along γ, which satisfies

Vi(0) = 0, ∇γ˙ (0)Vi = ∂i. In view of the first equation and the Jacobi field equation

∇γ˙ ∇γ˙ V + R(γ, ˙ V )γ ˙ = 0 we also have (2) ∇γ˙ (0)Vi := ∇γ˙ (0)∇γ˙ Vi = 0. Moreover, if we take the (k − 2)th covariant derivative of the Jacobi field equation, we get k−2   (k) X k − 2 (k−2−l) (l) ∇ V + (∇ R)(γ, ˙ ∇ V )γ ˙ = 0, γ˙ i l γ˙ γ˙ i l=0 where we used the facts

∇W (R(X,Y )Z) = (∇W R)(X,Y )Z +R(∇W X,Y )Z +R(X, ∇W Y )Z +R(X,Y )∇W Z and ∇γ˙ γ˙ = 0. Taking k = 3, we get

(3) ∇γ˙ Vi + (∇γ˙ R)(γ, ˙ Vi)γ ˙ + R(γ, ˙ ∇γ˙ Vi)γ ˙ = 0.

Evaluating at t = 0, and using Vi(0) = 0, we get

(3) ∇γ˙ (0)Vi = −R(Xp, ∂i)Xp = R(∂i,Xp)Xp. So if we let 2 1 m h(t) = t gij(tx , ··· , tx ) = hVi(t),Vj(t)i, LECTURE 14: NORMAL COORDINATES 3 then

h(0) = hVi(0),Vj(0)i = 0, 0 h (0) = h∇γ˙ (0)Vi,Vj(0)i + hVi, ∇γ˙ (0)Vj(0)i = 0, 00 (2) (2) h (0) = h∇γ˙ (0)Vi,Vj(0)i + 2h∇γ˙ (0)Vi, ∇γ˙ (0)Vji + hVi, ∇γ˙ (0)Vj(0)i = 2δij, 3   X 3 (3−l) (l) h000(0) = h∇ V , ∇ V i = 0, l γ˙ (0) i γ˙ (0) j l=0 4   X 4 (4−l) (l) h0000(0) = h∇ V , ∇ V i = 8R(∂ ,X ,X , ∂ ). l γ˙ (0) i γ˙ (0) j i p p j l=0 As a consequence, we get 1 g (tx1, ··· , txm) = hV (t),V (t)i ij t2 i j 1  8  = δ t2 + R(∂ ,X ,X , ∂ )t4 + O(t5) t2 ij 4! i p p j 1 = δ + R(∂ ,X ,X , ∂ )t2 + O(t3) ij 3 i p p j 1 = δ + R(∂ , ∂ , ∂ , ∂ )(txk)(txl) + O(t3). ij 3 i k l j This proves the theorem.  (k) Remark. One can continue to calculate ∇γ˙ (0)Vi’s and get a full expansion of gij in normal coordinates. For example, the next two terms are 1 1 2 R xkxlxr + ( R + R mR )xkxlxrxs. 6 iklj;r 20 iklj;rs 45 kil rjsm Taking derivative of (1), we get 1 1 ∂ g = R xl + R xk + O(|x|2). r ij 3 irlj 3 ikrj Taking derivative again and evaluate at p, we get 1 1 ∂ ∂ g (0) = R (p) + R (p). s r ij 3 irsj 3 isrj As a consequence, we get Corollary 1.4. With respect to normal coordinates, one has 1 R (p) = (∂ ∂ g + ∂ ∂ g − ∂ ∂ g − ∂ ∂ g )(0). ijkl 2 i l jk j k il i k jl j l ik Proof. The right hand side equals 1 (R + R + R + R − R − R − R − R )(p), 6 jlik jilk ikjl ijkl jkil jikl iljk ijlk 4 LECTURE 14: NORMAL COORDINATES

which equals Rijkl(p) by using symmetries of the Riemann curvature tensor.  Remark. This is Riemann’s original definition of the curvature tensor.

2. Geometric Meaning of Curvatures Now we are ready to give geometric interpretations of sectional, Ricci and scalar curvatures.

0 Theorem 2.1. Let Πp ⊂ TpM be a 2-dimensional plane. Denote by Cr the circle of 0 radius r in Πp centered at p, and Cr = expp(Cr ). Let Lr be the length of Cr. Then

2πr − Lr π lim = K(Πp). r→0 r3 3

Proof. Take an orthonormal basis {e1, ··· , em} of TpM so that Πp is spanned by e1, e2, and consider the normal coordinate system with respect to {ei}. Then for r small, the cricel Cr has equation 1 2 k Cr : x (t) = r cos t, x (t) = r sin t, x (t) = 0 (k ≥ 3). It follows that ˙ 2 i j |Cr(t)| = gij(Cr(t))x ˙ (t)x ˙ (t)  1   1  1 = 1 + R x2x2 x˙ 1x˙ 1 + 1 + R x1x1 x˙ 2x˙ 2 + 2 R x1x2x˙ 1x˙ 2 + O(r5) 3 1221 3 2112 3 1212 r4 = r2 − K(Π ) + O(r5). 3 p So r Z 2π Z 2π r2 ˙ 3 Lr = Length(Cr) = |Cr|dt = r 1 − K(Πp) + O(r )dt 0 0 3 π = 2πr − K(Π )r3 + O(r4). 3 p This implies the theorem.  So the sectional curvature measures the deviation of the geodesic circle to the standard circle in Euclidean space. To give a geometric interpretation of the Ricci curvature, we first prove Lemma 2.2. In a normal coordinate system near p, we have 1 det(g ) = 1 − Ric (p)xkxl + O(|x|3). ij 3 kl 1 k l 3
Proof. Let A = ln(gij). Since (gij) = I + 3 Riklj(p)x x + O(|x |) , and B2 Bk ln(I + B) = B + + ··· + + ··· , 2 k LECTURE 14: NORMAL COORDINATES 5 we see 1  A = R (p)xkxl + O(|x3|) . 3 iklj In particular, 1 1 tr(A) = R (p)xkxl + O(|x|3) = − Ric (p)xkxl + O(|x|3), 3 ikli 3 kl where in the last step we used the fact δij = gij. As a consequence, 1 det(g ) = det(exp(A)) = exp(tr(A)) = 1 − R (p)xkxl + O(|x|3). ij 3 kl  As an immediate consequence, we get p Corollary 2.3. The volume element det(gij) in normal coordinates has the ex- pansion q 1 det(g ) = 1 − Ric (p)xkxl + O(|x|3). ij 6 kl In particular, we get

Corollary 2.4. Let up ∈ SpM be a unit tangent vector at p, and let γ(t) be the geodesic starting at p with γ˙ (0) = up. Then q Ric(u ) det(g (γ(t))) = 1 − p t2 + O(t3). ij 6

Proof. Take an orthonormal basis of TpM so that up = e1. Then with respect to the corresponding normal coordinates, the geodesic γ(t) = expp(tup) can be written as γ : x1(t) = t, x2 = ··· = xm = 0. It follows q 1 1 det(g (γ(t))) = 1 − Ric (p)t2 + O(t3) = 1 − Ric(u )t2 + O(t3). ij 6 11 3 p  So the Ricci curvature measures the change of the volume element in the given direction. Finally let’s study the scalar curvature S. We have Theorem 2.5. For r small enough,  S(p)  Vol(B (p)) = ω rm 1 − r2 + O(|r3|) , r m 6(m + 2) m where ωm is the volume of the unit ball in R . 6 LECTURE 14: NORMAL COORDINATES

Proof. By definition Z q 1 m Vol(Br(p)) = det(gij)dx ··· dx Br(0) Z   1 k l 1 m 3 = 1 − Rickl(p)x x dx ··· dx + O(r ) Br(0) 6 Z m Rickl(p) k l 1 m 3 = ωmr − x x dx ··· dx + O(r ). 6 Br(0) An elementary computation shows Z ω xkxldx1 ··· dxm = m rm+2δkl Br(0) m + 2 and the theorem follows.  So the scalar curvature measures the deviation of the volume of a geodesic ball to the standard ball.

Corollary 2.6. The surface area of geodesic sphere Sr(p) is S(p) Area(S (p)) = mω rm−1 − ω rm+1 + O(rm+2). r m 6 m