Estimates on the Metric Tensor

David Glickenstein February 14, 2000

1. Introduction. This is a summary of some easy estimates on the metric tensor of a manifold if the sectional curvatures are bounded. It generally follows [1, section 1.8].

2. Definitions. We will estimate derivatives of the metric in geodesic polar coordinates. Suppose we have a Riemannian manifold (M, g). We consider geodesic polar coordinates, and in these coordinates the metric can be written as g = dr^{2} + r^{2}g_{θ^{i}θ^{j}}dθ^{i}dθ^{j}.

We also are going to need to take a geodesic γ(t) on M which is parametrized by arclength, and we will say that P = γ(0), Q = γ(r), and T (t) = γ′(t).

In the sequel, we shall talk about vector fields V along the curve γ. When we write V′(t), it is understood to mean ∇_{T (t)}V (t), where ∇ is the Riemannian connection on M.

We shall also restrict ourselves to r such that γ(r) is in the ball around P where exp_{P} is a diffeomorphism (inside the injectivity radius).

3. Relevant Theorems. We will use Jacobi fields and the Index Inequality. We first should define the Index of a vector field.

Definition 3.1. The index at r of a vector field V (t) along a curve γ is

I_{r}(V, V) = ∫_{0}^{r} [g(V′(t), V′(t)) + g(R(T (t), V (t))T (t), V (t))] dt

where T (t) = \frac{dγ}{dt} is the tangential vector field to γ.

This is, I think, essentially the second variation of arclength.

The important theorem is the following:

Theorem 3.2 (Index Inequality. See, for instance, [2]). Let γ(t) be a geodesic between P = γ(0) and Q = γ(r) such that there are no conjugate points between P and Q, and let J(t) be a Jacobi field along γ such that J(0) = 0 and V (t) be another smooth vector field along γ such that both J and V are orthogonal to T (t) = \frac{dγ}{dt}, V (0) = 0, and J(r) = V (r), then I_{r}(J, J) ≤ I_{r}(V, V).

In order to use the Index Inequality, we will need the following two propositions.

Proposition 3.3 (See [1], 1.46 ). Q = exp_{P} X_{0} is a conjugate point if and only if exp_{P} is singular at X_{0}.

Thus if we stay within the injectivity radius on M, we will not encounter any conjugate points. We will also need to know:

Proposition 3.4. The injectivity radius for a space of constant curvature b^{2} is π/b.

Finally, we will use the fact that Jacobi fields are the infinitesimal variations of geodesics. In geodesic normal coordinates at P, the geodesics through p are given by exp_{p}(tv). Thus if we have a Jacobi field along a geodesic γ such that J(0) = 0, then we see that in our geodesic polar coordinates we have Jacobi fields J(t) = t∥J′(0)∥ \frac{∂}{∂θ} |_{tv} where ∥w∥ for a vector w means \sqrt{g(w, w)}. Thus we see that:

Proposition 3.5. For a Jacobi field J along a geodesic γ(t), t ∈ [0, r] with γ(r) = Q, such that J(0) = 0 and J(r) = \frac{∂}{∂θ}|_{Q}, we have g(J(r), J(r)) = r^{2}g_{θθ}∥J′(0)∥^{2}.

4. Lower Bounds for the Components of the Metric Tensor. We shall prove the following:

Theorem 4.1. If the sectional curvature of M is bounded above by b^{2} then we have the following two estimates, provided r < π/b:

• \frac{∂}{∂r} log \sqrt{g_{θθ}} ≥ b cot(br) − \frac{1}{r}

• g_{θθ}(r, Θ) ≥ \frac{sin br}{br}

where θ = θ^{i} for any i (as defined in the polar representation of the metric).

Now suppose the sectional curvature of M is bounded above by b^{2}, so g(R(J, T)T, J) ≤ b^{2}, or g(R(T, J)T, J) ≥ −b^{2}.

We furthermore observe that if J is a Jacobi field, then, by definition, J′′(t) = R(T, J)T, so we get

I_{r}(J, J) = ∫_{0}^{r} [g(J′(t), J′(t)) + g(J′′(t), J(t))] dt = g(J′(r), J(r))  (4-1)

We are going to compare the metric on M with the metric on a space of constant curvature b^{2}. To do this, we need to make a frame {e_{1}(t), ..., e_{n}(t)} on M by taking an orthonormal basis {e_{1}, ..., e_{n}} of T_{γ(0)}M such that e_{1} = γ′(0) and then parallel translating it along γ. Notice that since γ is a geodesic and parametrized by arclength, e_{1}(t) = γ′(t). We also need to take a geodesic γ̃ on another manifold M̃ which is the same length as γ and do the same thing to get a frame {ẽ_{1}(t), ..., ẽ_{n}(t)} along γ̃. We then have the map ∼: c^{i}(t)e_{i}(t) ↦ c^{i}(t)ẽ_{i}(t), or V ↦ Ṽ. Notice the following:

Lemma 4.2. The map ∼: (M, g) → (M̃, g̃) has the following properties for vector fields V and W along γ and Ṽ and W̃ along γ̃:

• g(V, W) = g̃(Ṽ , W̃)

• g(V′, W) = g̃(Ṽ′, W̃)

• g(V′, W′) = g̃(Ṽ′, W̃′)

Proof: Use the frames we have already set up. Then if V = v^{i}e_{i} and W = w^{i}e_{i} where v^{i}, w^{i} are real-valued functions, then

g(V, W) = v^{i}w^{j}g(e_{i}, e_{j}) = ∑ v^{i}w^{i} = v^{i}w^{j}g̃(ẽ_{i}, ẽ_{j}) = g̃(Ṽ , W̃)

(Notice that it is important that we use frames instead of coordinates.) Furthermore, since the frames are parallel, i.e. ∇_{T} e_{i} = 0,, we get V′ = v^{i}′e_{i} + v^{i}∇_{T} e_{i} = v^{i}′e_{i} and Ṽ′ = v^{i}′ẽ_{i} so the last items follow. □

Now,

I(J, J) = ∫_{0}^{r} [g(J′, J′) + g(R(T, J)T, J)] dt
≥ ∫_{0}^{r} [g(J′, J′) − b^{2}g(J, J)] dt
= I^{b}(J̃, J̃)

where I^{b} is the index on a manifold of constant curvature b^{2} and where we have chosen a frame {e_{i}} along γ and a corresponding frame {ẽ_{i}} along some curve γ̃ of the same length in the manifold of constant curvature.

In constant curvature, we have a Jacobi field which vanishes at t = 0 and ends at the vector J̃(r) given by \frac{sin bt}{sin br} J̃(r).

What do we really mean by this? Suppose J(r) = ∑ c^{i}e_{i}(r) with our frame as above. Let our new field be defined by Ṽ (t) = \frac{sin bt}{sin br}c^{i}(r)ẽ_{i}(t). Notice that Ṽ (r) = J̃(r), Ṽ (0) = 0, Ṽ′(t) = b\frac{cos bt}{sin br}c^{i}(r)ẽ_{i}(t) since the frame is parallel, and Ṽ is orthogonal to γ̃ at every point since J(r) is and we have a parallel frame along a geodesic, so ẽ_{1}(t) = T̃(t) = γ̃′(t) for all t. Now compute (letting c_{i} = c_{i}(r)) using the Index Inequality (3.2) on the manifold of constant curvature b^{2} (which we can do since we are inside the injectivity radius by assumption that r ≤ π/b):

I_{r}^{b}(J̃, J̃) ≥ I_{r}^{b} (\frac{sin(bt)}{sin(br)}c^{i}ẽ_{i}(t))
= \frac{1}{sin^{2} br} ∫_{0}^{r} [b^{2} cos^{2}(bt) ∑(c^{i})^{2} − b^{2} sin^{2}(bt) ∑(c^{i})^{2}] dt
= \frac{b}{sin^{2} br} sin(br) cos(br) ∑(c^{i})^{2}
= b cot(br)g(J(r), J(r))

We have just about proven an estimate on the metric tensor. We use 4-1. We now know that

g(J′(r), J(r)) ≥ b cot(br)g(J(r), J(r))  (4-2)

Now if we take J(t) to be a Jacobi field such that J(r) = \frac{∂}{∂θ}|_{Q} for θ = θ^{i} for some i, then g(J(r), J(r)) = r^{2}g_{θθ}∥J′(0)∥^{2} by Proposition 3.5, so we can compute:

b cot(br)g(J(r), J(r)) ≤ g(J′(r), J(r))
b cot(br) ≤ \frac{g(J′(r), J(r))}{g(J(r), J(r))}
= \frac{\frac{∂}{∂r}g(J(r), J(r))}{2g(J(r), J(r))}

= \frac{\frac{∂}{∂r} (r^{2}g_{θθ}∥J′(0)∥^{2})}{2r^{2}g_{θθ}∥J′(0)∥^{2}}
= \frac{1}{r} + \frac{∂}{∂r} log \sqrt{g_{θθ}}

And thus we have our first estimate on the metric tensor in polar coordinates:

\frac{∂}{∂r} log \sqrt{g_{θθ}} ≥ b cot(br) − \frac{1}{r}

Notice that

b cot(br) − \frac{1}{r} = \frac{d}{dr} log \frac{sin(br)}{r} = \frac{d}{dr} (log \frac{sin(br)}{br} + log b)

so if we integrate from 0 to r_{0} we get

log \sqrt{g_{θθ}(r_{0}, Θ)} ≥ log \frac{sin br_{0}}{br_{0}}

since at 0, the metric is Euclidean, so g_{θθ}(0, Θ) = 1 (since in ℝ^{2} we have g(\frac{∂}{∂θ}, \frac{∂}{∂θ}) = r^{2}) and lim_{r→0} \frac{sin br}{br} = 1. Thus we have

g_{θθ}(r, Θ) ≥ \frac{sin br}{br}

for any θ, a coordinate on S^{n−1}.

5. Lower Bounds on Determinant of Metric. We shall now use what we proved above to prove:

Theorem 5.1. If the sectional curvature of M is bounded above by b^{2} then we have the following two estimates, provided r < π/b:

• \frac{∂}{∂r} log \sqrt{|g(r, θ)|} ≥ (n − 1)b cot(br) − \frac{n−1}{r}

• \sqrt{|g(r, θ)|} ≥ (\frac{sin br}{br})^{n−1}

where |g| = det g_{θ^{i}θ^{j}}.

We first take a bunch of Jacobi Fields J_{1}, ..., J_{n−1} such that together with T at r they form a basis for T_{Q}M. We now let |g| = det g_{θ^{i}θ^{j}} and suppose that in coordinates we have J_{i}(t) = c_{i}^{k} \frac{∂}{∂x^{k}} and compute:

\frac{∂}{∂r} log \sqrt{|g|} = \frac{1}{2} \frac{∂}{∂r}tr log |g|
= \frac{1}{2}tr (\frac{∂}{∂r}g_{θ^{i}θ^{j}}g^{θ^{j}θ^{k}})
= \frac{1}{2} \frac{∂}{∂r} \frac{g(J_{i}(r), J_{j}(r))}{r^{2}}g^{θ^{i}θ^{j}}
= g(J′_{i}(r), J_{j}(r))g^{θ^{i}θ^{j}} − \frac{n − 1}{r}
= ∑_{i} g(J′_{i}(r), J_{i}(r)) − \frac{n − 1}{r}

We get the last equality seen above by letting J′_{i}(r) = c_{i}^{k}J_{k}(r) (since the J′_{k}(r) form an orthonormal basis) and compute:

g(J′_{i}(r), J_{j}(r))g^{θ^{i}θ^{j}} = c_{i}^{k}r^{2}g_{θ^{k}θ^{j}}g^{θ^{i}θ^{j}}
= r^{2}c_{i}^{k}δ_{k}^{i}
= ∑_{i} r^{2}c_{i}^{i}
= ∑_{i} g(J′_{i}(r), J_{i}(r))

We can now use 4-2 and see that

\frac{∂}{∂r} log \sqrt{|g|} ≥ (n − 1)b cot(br) − \frac{n − 1}{r}

Integrating this from 0 to r_{0} we get

∫_{0}^{r_{0}} [(n − 1)b cot(br) − \frac{n − 1}{r}] dr = log (\frac{sin br}{br})^{n−1}|_{0}^{r_{0}}

= log (\frac{sin br_{0}}{br_{0}} )^{n−1}

because lim_{r→0} \frac{sin br}{br} = 1. Now, when r = 0, g is the Euclidean metric, where g(\frac{∂}{∂θ}, \frac{∂}{∂θ}) = r^{2}, so ∥g(0, Θ)∥ = det I_{n−1} = 1 where I_{n−1} is the (n − 1) × (n − 1) identity matrix. Thus we finally get:

log \sqrt{|g(r_{0}, Θ)|} ≥ log (\frac{sin br_{0}}{br_{0}} )^{n−1}
\sqrt{|g(r_{0}, Θ)|} ≥ (\frac{sin br_{0}}{br_{0}} )^{n−1}

6. Upper Bounds on Components of the Metric Tensor. We shall prove the following:

Theorem 6.1. If the sectional curvature of M is bounded below by −a^{2} then we have the following two estimates:

• \frac{∂}{∂r} log \sqrt{g_{θθ}} ≤ a coth(ar) − \frac{1}{r}

• g_{θθ}(r, Θ) ≤ \frac{sinh ar}{ar}

We begin as above. We know that for a Jacobi field J along a geodesic we have g(R(J, T)T, J) ≥ −a^{2}, so g(R(T, J)T, J) ≤ a^{2}.

Let us define a vector field V (t) along the geodesic γ. We first consider a Jacobi field J along γ such that J(0) = 0, g(J′(0), T (0)) = 0, and J(r) = \frac{∂}{∂θ}|_{Q}. We fix a frame \{e_{i}(t)\} along γ as above, so J(t) = c^{i}(t)e_{i}(t). We can now take as our vector field V (t) = \frac{sinh at}{sinh ar}c^{i}(r)e_{i}(t).

Notice that Ṽ would be a Jacobi field in a manifold M̃ of constant curvature −a^{2}. We also see immediately that V (0) = 0 and V (r) = J(r).

The Index Inequality (3.2) gives us

I_{r}(J, J) ≤ I_{r}(V, V)
= ∫_{0}^{r} [g(V′, V′) + g(R(T, V)T, V)] dt
≤ ∫_{0}^{r} [g(V′(t), V′(t)) + a^{2}g(V, V)] dt
= I_{r}^{a}(Ṽ , Ṽ)

where I_{r}^{a} is the index on a manifold of constant curvature −a^{2}. Note we can use the Index Inequality since we are within the injectivity radius and hence there are no conjugate points.

Now, recall 4-1, which handles the left side. We also want to compute the right side (again letting c^{i} = c^{i}(r)):

I_{r}^{a}(Ṽ , Ṽ) = ∫_{0}^{r} [g(V′(t), V′(t)) + a^{2}g(V, V)] dt
= \frac{1}{sinh^{2} ar} ∫_{0}^{r} [a^{2} cosh^{2}(at) ∑(c^{i})^{2} + a^{2} sinh^{2}(at) ∑(c^{i})^{2}] dt
= \frac{a}{sinh^{2} ar} ∫_{0}^{r} [a cosh^{2} at + a sinh^{2} at] dt ∑(c^{i})^{2}
= \frac{a}{sinh^{2} ar}(cosh ar)(sinh ar)g(J(r), J(r))
= a coth(ar)g(J(r), J(r))

Now we can follow the same calculation as above to get the following two estimates:

\frac{∂}{∂r} log \sqrt{g_{θθ}} ≤ a coth(ar) − \frac{1}{r}

g_{θθ}(r, Θ) ≤ \frac{sinh ar}{ar}

References

[1] Thierry Aubin. Some Nonlinear Problems in Riemannian Geometry. Springer-Verlag, Berlin, Germany, 1998.

[2] Manifred Perdigao do Carmo. Riemannian Geometry. Birkhauser, Boston, MA, 1992.