𝓛-Geodesics and the Variational Formulas

12A.5 Reduced Distance and Reduced Volume defined the $\mathcal L$-length, the reduced distance $\ell$ and the reduced volume $\tilde V$, stated the differential inequalities that $\ell$ satisfies, and showed how they make $\tilde V$ monotone. It took those inequalities on trust. This chapter proves them. The proofs are the content of a single page of Perelman's first preprint, §7.1, which he describes as "natural modifications of those in the classical variational theory of geodesics that can be found in any textbook on Riemannian geometry". That description is accurate and also misleading: every step is classical, but there are many steps, the signs are treacherous, and the time dependence of the metric enters every one of them, producing Ricci terms that must be tracked and, at the end, recognised as Hamilton's Harnack expressions. Morgan and Tian spend twenty pages on the same material. A reader who has done the second variation of arc length once (9A.7 Jacobi Fields and Curvature versus Topology) and the Laplacian comparison for the distance function once (9B.1 Laplacian Comparison) has all the tools; what remains is to see how they adapt.

The chapter follows Perelman's equations in order, (7.1) to (7.15), and gives each its derivation. The one new idea, compared with Riemannian geometry, is the choice of test fields in the second variation: Perelman's fields $Y$ solving $\nabla_XY = -\operatorname{Ric}(Y, \cdot) + \frac{1}{2\tau}Y$, which play the role that the linear Jacobi fields $\frac srE$ play in Euclidean space. With them the second variation of $\mathcal L$ becomes the integral of Hamilton's matrix Harnack expression, and its trace becomes Hamilton's trace Harnack expression $H(X)$. Perelman's quantity $K = \int\tau^{3/2}H(X)\,d\tau$ is then the only unknown, and it cancels from the combinations that matter. 12A.7 Monotonicity of the Reduced Volume uses the result to prove the monotonicity of the reduced volume and the bound $\min\ell \leq \frac n2$ rigorously.

By the end of this chapter you will be able to:

• write the $\mathcal L$-length in the variable $s = \sqrt\tau$, and explain why the $\mathcal L$-geodesic equation is regular at $\tau = 0$;
• derive the first variation formula (7.1) and the $\mathcal L$-geodesic equation (7.2), including the Ricci term that comes from differentiating the metric;
• prove that $\nabla L = 2\sqrt\tau X$ and $\partial_\tau L = 2\sqrt\tau R - \sqrt\tau(R + |X|^2)$;
• derive (7.3)–(7.6), identify $H(X)$ with Hamilton's trace Harnack expression at $t = -\tau$, and express $\partial_\tau\ell$ and $|\nabla\ell|^2$ through $K$;
• derive the second variation formula (7.7) and the $\mathcal L$-Jacobi equation;
• prove the Hessian comparison (7.9) with adapted fields, take its trace to get the Laplacian comparison (7.10), and verify the trace identity that turns the matrix Harnack expression into the trace one;
• combine everything into (7.13)–(7.15), and identify the equality case as a gradient shrinking soliton;
• state Perelman's §7.2 estimates under nonnegative curvature operator, which 12B.3 The Structure of κ-Solutions uses for $\kappa$-solutions.

6.1The setting and the variable $s = \sqrt\tau$

Let $g(t)$ be a Ricci flow on $M$, either with $M$ closed or with each $g(t)$ complete of uniformly bounded curvature, and fix a time $T$. Write $\tau = T - t$ for backward time, so that

A path parameterised by backward time is a curve $\gamma : [\tau_1, \tau_2] \to M$, thought of as the space-time path $\tau \mapsto (\gamma(\tau), T - \tau)$, which moves backward in $t$ as $\tau$ increases. Write $X = \dot\gamma = \frac{d\gamma}{d\tau}$, and measure it with $g(\tau)$. The $\mathcal L$-length is

Morgan and Tian work more generally with generalised Ricci flows, space-times that are locally isomorphic to Ricci flows but may change topology, which is what a flow with surgery is (12B.9 Surgery). Every computation in this chapter is local along a path, so it applies verbatim there; that is why they set it up in that generality.

The variable $s$. Set $s = \sqrt\tau$ and $A(s) = \frac{d\gamma}{ds}$. Then $d\tau = 2s\,ds$ and $X = \frac{A}{2s}$, that is, $A = 2\sqrt\tau X$. Substituting,

so

This is an energy plus a potential, and in this form the theory looks like the classical calculus of variations (6A.11 Calculus of Variations and Gradient Flows): the weight $\sqrt\tau$, which seemed strange, is exactly what makes $\mathcal L$ a standard energy in $s$. On static flat space, where $R = 0$, minimisers are straight lines at constant $s$-speed, $\gamma(s) = p + 2sv$, that is, $\gamma(\tau) = p + 2\sqrt\tau\,v$: in $\tau$ they leave $p$ with infinite speed and slow down, like the fan of curves in 12A.5 Reduced Distance and Reduced Volume.

6.2The first variation and the $\mathcal L$-geodesic equation

Let $\gamma_u(\tau) = \gamma(\tau, u)$ be a smooth family of paths parameterised by backward time, with $\gamma_0 = \gamma$, and let $Y = \partial_u\gamma$ and $X = \partial_\tau\gamma$ on the surface swept out. Because these are coordinate vector fields on that surface, $[X, Y] = 0$, so $\nabla_YX = \nabla_XY$ (the connection being torsion-free; the connection of $g(\tau)$ is used at each $\tau$, and its time dependence does not affect this identity, which holds at each fixed $\tau$).

One new identity. For vector fields $V$, $W$ along $\gamma$, the derivative of their inner product along the path has an extra term, because the metric changes:

The first two terms are the usual Leibniz rule for the Levi-Civita connection at fixed $\tau$; the last is $\frac{\partial g}{\partial\tau}(V, W) = 2\operatorname{Ric}(V, W)$. Every Ricci term in this chapter comes from this line.

Now compute, following Perelman's (7.1). Differentiating under the integral,

By the identity, $\langle\nabla_XY, X\rangle = \frac{d}{d\tau}\langle Y, X\rangle - \langle Y, \nabla_XX\rangle - 2\operatorname{Ric}(Y, X)$. Substituting, and integrating $2\sqrt\tau\frac{d}{d\tau}\langle Y, X\rangle$ by parts (with $\frac{d}{d\tau}2\sqrt\tau = \frac{1}{\sqrt\tau}$):

Here $\operatorname{Ric}(X, \cdot)$ denotes the vector dual to the one-form $\operatorname{Ric}(X, \cdot)$. For variations fixing the endpoints the boundary term vanishes, and $\gamma$ is critical for all of them exactly when the bracket vanishes:

This is the $\mathcal L$-geodesic equation. Compare the Riemannian geodesic equation $\nabla_XX = 0$: the term $\frac12\nabla R$ is a force pushing the path toward larger scalar curvature (the potential $2s^2R$ in the $s$-form, with the sign of a Lagrangian), the term $\frac{1}{2\tau}X$ is the "slowing down" that makes $\sqrt\tau X$ rather than $X$ the natural velocity, and $2\operatorname{Ric}(X, \cdot)$ is the correction from the moving metric.

Regularity at $\tau = 0$. In the variable $s$, with $X = \frac{A}{2s}$ and $\frac{d}{d\tau} = \frac{1}{2s}\frac{d}{ds}$, one finds $\nabla_XX = \frac{1}{4s^2}\nabla_AA - \frac{1}{4s^3}A$, and (7.2) becomes, after multiplying by $4s^2$ (Exercise 6.1),

This ODE has smooth coefficients at $s = 0$. So for every $v \in T_pM$ there is a unique $\mathcal L$-geodesic with $\gamma(0) = p$ and $A(0) = 2v$, that is, $\lim_{\tau\to0}\sqrt\tau X(\tau) = v$ (Morgan–Tian, Lemma 6.8). The $\mathcal L$-exponential map is $\mathcal L\exp_\tau(v) = \gamma_v(\tau)$. On a complete flow of bounded curvature, $\mathcal L$-geodesics extend to the whole time interval (their Lemma 7.3), and minimising curves exist between any two space-time points with $\tau_1 < \tau_2$ (their Lemma 7.2): a minimising sequence has bounded energy $\int|A|^2ds$ in the $s$-variable, hence is uniformly Hölder continuous, a subsequence converges weakly in $H^1$, and $\mathcal L$ is lower semicontinuous, exactly the direct method of 6A.11 Calculus of Variations and Gradient Flows and 4A.7 Weak Convergence and the Direct Method. The minimiser satisfies (7.2) and is smooth.

6.3The gradient and the time derivative of $L$

Fix a base point $p$ at $\tau = 0$ and write $L(q, \bar\tau)$ for the infimum of $\mathcal L$ over paths from $(p, 0)$ to $(q, \bar\tau)$. Pretend, as Perelman does, that the minimiser $\gamma$ is unique and depends smoothly on $(q, \bar\tau)$; 12A.7 Monotonicity of the Reduced Volume shows that this is true on an open set of full measure (the analogue of the complement of the cut locus), and that elsewhere the inequalities hold in the barrier sense.

The gradient. Vary the endpoint $q$ with velocity $Y(\bar\tau)$ through minimisers. The integral term of (7.1) vanishes ($\gamma$ is an $\mathcal L$-geodesic) and the boundary term at $\tau = 0$ vanishes ($Y(0) = 0$, and $\sqrt\tau X$ stays bounded), so $\delta_YL = 2\sqrt{\bar\tau}\langle X(\bar\tau), Y(\bar\tau)\rangle$, that is,

This is the analogue of $\nabla d(p, \cdot) = \dot\gamma$ for the Riemannian distance: the gradient of the length function at the endpoint is the velocity of the minimiser, here rescaled.

The time derivative. Along the minimiser, $\frac{d}{d\tau}L(\gamma(\tau), \tau) = \sqrt\tau(R + |X|^2)$ by the fundamental theorem of calculus (an initial segment of a minimiser is a minimiser, as for geodesics). On the other hand, by the chain rule, $\frac{d}{d\tau}L(\gamma(\tau), \tau) = \partial_\tau L + \langle\nabla L, X\rangle$. Hence

(Morgan–Tian, Lemma 6.32.) Both formulas involve the unknown quantity $R + |X|^2$ at the endpoint, and the next step computes it.

6.4Hamilton's trace Harnack expression appears

Differentiate $R + |X|^2$ along an $\mathcal L$-geodesic. The scalar curvature contributes $\frac{d}{d\tau}R(\gamma(\tau), \tau) = R_\tau + \langle\nabla R, X\rangle$. The speed contributes, by the new identity, $\frac{d}{d\tau}|X|^2 = 2\langle\nabla_XX, X\rangle + 2\operatorname{Ric}(X, X)$, and by (7.2), $2\langle\nabla_XX, X\rangle = \langle\nabla R, X\rangle - \frac1\tau|X|^2 - 4\operatorname{Ric}(X, X)$. Adding, and adding and subtracting $\frac1\tau R$,

where

Perelman identifies $H(X)$ as Hamilton's expression for the trace Harnack inequality with $t = -\tau$. Indeed, Hamilton's trace Harnack quantity for a forward Ricci flow is $R_t + \frac Rt + 2\langle\nabla R, V\rangle + 2\operatorname{Ric}(V, V)$ (11B.2 Ancient Solutions and the Harnack Inequality); substituting $t = -\tau$ (so $R_t = -R_\tau$ and $\frac Rt = -\frac R\tau$) and $V = -X$ (the velocity in forward time) gives exactly $H(X)$. On a Ricci flow with nonnegative curvature operator defined for all earlier times, Hamilton's theorem says this expression is nonnegative for every $V$ when $t$ is measured from the initial time; here the base of $\tau$ is the final time, which is why Perelman's §7.2 needs a correction term (below).

Integrating (7.3). Multiply by $\tau^{3/2}$: $\frac{d}{d\tau}\big(\tau^{3/2}(R + |X|^2)\big) = \frac32\sqrt\tau(R + |X|^2) + \tau^{3/2}\frac{d}{d\tau}(R + |X|^2) = \frac12\sqrt\tau(R + |X|^2) - \tau^{3/2}H(X)$. Integrate from $0$ to $\bar\tau$; the boundary term at $0$ vanishes because $\tau^{3/2}|X|^2 = \sqrt\tau|\sqrt\tau X|^2 \to 0$. With

the result is

Substituting into the formulas for $\partial_{\bar\tau}L$ and $|\nabla L|^2$:

In terms of the reduced distance $\ell = \frac{L}{2\sqrt{\bar\tau}}$ these become (Exercise 6.3)

On flat space $H \equiv 0$, $K = 0$, and these are $\partial_\tau\ell = -\frac{\ell}{\tau}$ and $|\nabla\ell|^2 = \frac\ell\tau$, which $\ell = \frac{|x|^2}{4\tau}$ satisfies. These are first-order identities: they involve $K$, which depends on the whole geodesic, not just its endpoint. Second-order information comes next.

6.5The second variation and $\mathcal L$-Jacobi fields

Take a variation with $Y(0) = 0$ and differentiate (7.1) once more. Perelman's (7.7) is the result after the same manipulations as in Riemannian geometry: commute $\nabla_Y$ and $\nabla_X$ using the curvature tensor, move derivatives of $Y$ along the curve using the new identity (which now produces derivatives of $\operatorname{Ric}$, since the metric's derivative is $2\operatorname{Ric}$ and that changes along the curve), and discard terms proportional to the geodesic equation. The two ingredients are:

• $\frac{d}{d\tau}\langle\nabla_YY, X\rangle = \langle\nabla_X\nabla_YY, X\rangle + \langle\nabla_YY, \nabla_XX\rangle + 2Y\cdot\operatorname{Ric}(Y, X) - X\cdot\operatorname{Ric}(Y, Y)$, the identity for the derivative of $\langle\nabla_YY, X\rangle$ along the curve, in which the last two terms come from the $\tau$-derivative of the Christoffel symbols: $\partial_\tau\Gamma^k_{ij} = g^{kl}(\nabla_i\operatorname{Ric}_{jl} + \nabla_j\operatorname{Ric}_{il} - \nabla_l\operatorname{Ric}_{ij})$ (11A.2 How Curvature Evolves; Morgan–Tian, Claim 6.34, do the computation in coordinates);
• the commutation $\nabla_Y\nabla_XY = \nabla_X\nabla_YY + R(Y, X)Y$, with the curvature convention of Perelman's text.

The outcome, for any variation with $Y(0) = 0$, is

Compare the second variation of Riemannian energy, $\int\big(|\nabla_XY|^2 - \langle R(Y, X)X, Y\rangle\big)$ (9A.7 Jacobi Fields and Curvature versus Topology): the curvature term and the $|\nabla_XY|^2$ term are the same up to factors; the Hessian of $R$ is the second variation of the potential; and the two derivatives of $\operatorname{Ric}$ are new, the price of a moving metric.

$\mathcal L$-Jacobi fields. The variation field of a family of $\mathcal L$-geodesics satisfies the linearisation of (7.2), the $\mathcal L$-Jacobi equation (Morgan–Tian, Lemma 6.10):

As for Riemannian Jacobi fields: the second variation of a minimising $\mathcal L$-geodesic is a nonnegative quadratic form on fields vanishing at both ends, it vanishes exactly on $\mathcal L$-Jacobi fields (their Proposition 6.13), and the differential of $\mathcal L\exp_{\bar\tau}$ at $v$ sends $w$ to the value at $\bar\tau$ of the $\mathcal L$-Jacobi field with $Y(0) = 0$ and $\lim\sqrt\tau\nabla_XY = w$ (their Lemma 6.19). So $\mathcal L\exp_{\bar\tau}$ is singular at $v$ exactly when there is a nonzero $\mathcal L$-Jacobi field vanishing at both ends: the $\mathcal L$-conjugate points. An $\mathcal L$-Jacobi field along a minimiser also computes the Hessian of $L$ exactly (their Lemma 6.40):

6.6The Hessian comparison with adapted fields

To bound $\operatorname{Hess}L$ from above, compare $L$ with the length of a family of curves that are not $\mathcal L$-geodesics but are explicit, exactly as one bounds $\operatorname{Hess}d$ by using variations through linear fields. Let $\gamma$ be the minimiser to $(q, \bar\tau)$, and $Y(\bar\tau)$ a unit vector at $q$. Any variation of $\gamma$ through curves from $p$ with variation field $Y$ has length at least $L$ at its endpoints, with equality at $u = 0$; so $\operatorname{Hess}L(Y, Y) \leq \delta^2_Y\mathcal L - 2\sqrt{\bar\tau}\langle\nabla_YY, X\rangle$, the subtracted term being the correction for the endpoint moving along a curve rather than a geodesic of $g(\bar\tau)$ (it cancels the boundary term of (7.7)). So

for every field $Y$ along $\gamma$ with $Y(0) = 0$ and the given value at $\bar\tau$. Perelman now chooses $Y$ to solve

These are adapted fields (Morgan–Tian's term). By the new identity, $\frac{d}{d\tau}|Y|^2 = 2\langle\nabla_XY, Y\rangle + 2\operatorname{Ric}(Y, Y) = \frac1\tau|Y|^2$, so

and in particular $Y(0) = 0$. In flat space they are $Y(\tau) = \sqrt{\tau/\bar\tau}\,E$ with $E$ parallel: the variation fields of the fan of $\mathcal L$-geodesics $2\sqrt\tau v$, which are linear in $s = \sqrt\tau$. The choice is designed so that, in flat space, the comparison is an equality.

Substitute (7.8). The term $2|\nabla_XY|^2$ becomes $2|\operatorname{Ric}(Y, \cdot)|^2 - \frac2\tau\operatorname{Ric}(Y, Y) + \frac{1}{2\tau^2}|Y|^2$, and with $|Y|^2 = \frac\tau{\bar\tau}$ (normalising $|Y(\bar\tau)| = 1$), $\int_0^{\bar\tau}\sqrt\tau\frac{1}{2\tau^2}\frac{\tau}{\bar\tau}d\tau = \frac{1}{\sqrt{\bar\tau}}$. The remaining terms are organised by the derivative of $\operatorname{Ric}(Y, Y)$ along the curve,

which lets one replace $\nabla_X\operatorname{Ric}(Y, Y)$ by a total derivative and lower-order terms. Integrating the total derivative $\frac{d}{d\tau}(2\sqrt\tau\operatorname{Ric}(Y, Y))$ gives the boundary term $2\sqrt{\bar\tau}\operatorname{Ric}(Y(\bar\tau), Y(\bar\tau))$ (Morgan–Tian carry this out as their equation (6.13)). The result is Perelman's (7.9):

where

is Hamilton's expression for the matrix Harnack inequality, at $t = -\tau$. It is quadratic in $Y$ (despite the notation, it depends on $X$ and $Y$). Equality in (7.9) holds exactly when the adapted field $Y$ is also an $\mathcal L$-Jacobi field (Morgan–Tian, Proposition 6.37).

Compare once more with Riemannian geometry. For the distance function $r$ from a point in a manifold with $\operatorname{Ric} \geq 0$, the Hessian comparison along a unit-speed geodesic with the linear fields $\frac srE$ gives $\operatorname{Hess}r(E, E) \leq \frac1r - \frac1{r^2}\int_0^rs^2\operatorname{Rm}(E, \dot\gamma, \dot\gamma, E)\,ds$, and tracing gives $\Delta r \leq \frac{n - 1}{r}$ (9B.1 Laplacian Comparison). Here the role of the linear fields is taken by the adapted fields, the role of the sectional curvature integral by the integral of the matrix Harnack expression, and the curvature hypothesis is replaced by nothing: the Harnack expression is simply carried along.

6.7The trace: the Laplacian comparison

Choose an orthonormal basis $\{Y_\alpha(\bar\tau)\}$ at $q$ and extend each by (7.8). Since (7.8) is linear and $\frac{d}{d\tau}\langle Y_\alpha, Y_\beta\rangle = \frac1\tau\langle Y_\alpha, Y_\beta\rangle$, the fields $\sqrt{\bar\tau/\tau}\,Y_\alpha(\tau)$ are orthonormal at every $\tau$. Summing (7.9) over $\alpha$:

The trace identity says that for an orthonormal basis $\{W_\alpha\}$,

the matrix Harnack expression traces to the trace Harnack expression (Exercise 6.4; Morgan–Tian, Claim 6.44). It uses three facts: the contracted second Bianchi identity $\sum_\alpha(\nabla_{W_\alpha}\operatorname{Ric})(W_\alpha, X) = \frac12\langle\nabla R, X\rangle$; the evolution $R_\tau = -\Delta R - 2|\operatorname{Ric}|^2$ of the scalar curvature in backward time (11A.2 How Curvature Evolves); and $R_\tau = \partial_\tau(g^{ij}R_{ij}) = -2|\operatorname{Ric}|^2 + \sum_\alpha\operatorname{Ric}_\tau(W_\alpha, W_\alpha)$, so that $\sum_\alpha\operatorname{Ric}_\tau(W_\alpha, W_\alpha) = -\Delta R$. Since $|Y_\alpha|^2 = \frac\tau{\bar\tau}$, the sum is $\frac{\tau}{\bar\tau}H(X)$, and the integral is $\frac1{\bar\tau}\int_0^{\bar\tau}\tau^{3/2}H(X)\,d\tau = \frac{K}{\bar\tau}$. So

This is the space-time Laplacian comparison, the analogue of $\Delta r^2 \leq 2n$ in a space of nonnegative Ricci curvature. In terms of $\ell$, $\Delta\ell = \frac{\Delta L}{2\sqrt{\bar\tau}} \leq -R + \frac{n}{2\bar\tau} - \frac{K}{2\bar\tau^{3/2}}$, which is the third inequality quoted in 12A.5 Reduced Distance and Reduced Volume.

6.8Combining: the inequalities (7.13)–(7.15)

Now the unknown $K$ can be eliminated. With the two identities for $\partial_\tau\ell$ and $|\nabla\ell|^2$ and the inequality for $\Delta\ell$:

The $K$ terms cancel: $\frac{K}{2} + \frac K2 - K = 0$. Similarly

And for $\bar L = 2\sqrt\tau L = 4\tau\ell$, from (7.5) and (7.10) (12A.5 Reduced Distance and Reduced Volume, exercise on the minimum of $\ell$),

In Morgan and Tian's formulation (their Corollary 6.51), if one sets $\delta = \frac{n}{2\tau} - R - \frac{K}{2\tau^{3/2}} - \Delta\ell \geq 0$, the "defect" in the Laplacian comparison, then (7.13) and (7.14) are the equalities $\partial_\tau\ell - \Delta\ell + |\nabla\ell|^2 - R + \frac{n}{2\tau} = \delta$ and $2\Delta\ell - |\nabla\ell|^2 + R + \frac{\ell - n}{\tau} = -2\delta$. All the inequality in the theory comes from the single Hessian comparison.

The equality case. Equality in (7.13) at a point forces equality in (7.10), hence in (7.9) for every unit $Y$, hence every adapted field is an $\mathcal L$-Jacobi field, and then (Morgan–Tian, Proposition 6.43)

the gradient shrinking soliton equation with potential $\ell$. So if equality holds everywhere, the flow is a gradient shrinking soliton, and the Gaussian soliton on $\mathbb{R}^n$ with $\ell = \frac{|x|^2}{4\tau}$ is the model.

6.9When the curvature operator is nonnegative

Perelman's §7.2 adds one more estimate, used for $\kappa$-solutions (12B.3 The Structure of κ-Solutions). Suppose the backward flow, defined for $\tau \in [0, \tau_0]$, has nonnegative curvature operator. Then Hamilton's Harnack inequality holds, with time measured from the initial time of the forward flow, $t$-time $T - \tau_0$; translated into $\tau$, it gives

The correction $\frac{1}{\tau_0 - \tau}$ is the price of measuring from the final rather than the initial time; it is harmless as long as $\tau \leq (1 - c)\tau_0$. With these lower bounds on $H$, the identity for $|\nabla\ell|^2$ and (7.11) give, for $\tau \leq (1 - c)\tau_0$,

for $\mathcal L$-Jacobi fields $Y$, with $C$ depending on $c$ (statement; the derivation uses $R \leq R + |X|^2$ and (7.4) to bound $\int\tau^{3/2}R$ in terms of $L$). For an ancient solution, $\tau_0 = \infty$ and the correction disappears entirely. These bounds say that on a $\kappa$-solution, $\ell$ controls both its own gradient and the curvature, at the scale $\sqrt\tau$, which is what makes the rescaled reduced distances compact as $\tau \to \infty$ and produces the asymptotic soliton (12B.1 κ-Solutions, 12B.3 The Structure of κ-Solutions).

6.10History

Perelman's §7.1 appeared in November 2002. He named Li and Yau's 1986 paper as the closest reference: Li and Yau integrated their gradient estimate along paths using a "length" associated with a linear parabolic equation, the same structure with a potential. Hamilton's trace and matrix Harnack inequalities for the Ricci flow date from 1993, and Bennett Chow and Sun-Chin Chu's geometric interpretation of them on space-time from 1995. The first complete written accounts of §7.1 are in Kleiner and Lott's notes (posted from 2003, published 2008), Morgan and Tian's book (2007) and Cao and Zhu's article (2006); Rugang Ye's notes on the reduced distance and reduced volume (2004) also treat it.

6.11Exercises

Solution

(a) $\nabla_XX = \frac{1}{2s}\nabla_A\big(\frac{A}{2s}\big) = \frac{1}{2s}\big(\frac{1}{2s}\nabla_AA - \frac{1}{2s^2}A\big)$. (b) Substituting into (7.2): $\frac{1}{4s^2}\nabla_AA - \frac{1}{4s^3}A - \frac12\nabla R + \frac{1}{2s^2}\cdot\frac{A}{2s} + 2\operatorname{Ric}\big(\frac{A}{2s}, \cdot\big) = 0$. The terms $\mp\frac{1}{4s^3}A$ cancel, and multiplying by $4s^2$ gives the stated equation. (c) With $R = 0$ and $\operatorname{Ric} = 0$, $\frac{d^2\gamma}{ds^2} = 0$, so $\gamma = p + sA_0$, and $A_0 = 2v$ with $v = \lim\sqrt\tau X$.

Solution

$X = v\tau^{-1/2}$, and the integrand of (7.1) is $\sqrt\tau\langle Y, -2\nabla_XX - \frac1\tau X\rangle$ with $\nabla_XX = \dot X = -\frac12v\tau^{-3/2}$, so $-2\dot X - \frac1\tau X = v\tau^{-3/2} - v\tau^{-3/2} = 0$. Only the boundary term remains: $2\sqrt{\bar\tau}\langle X(\bar\tau), Y(\bar\tau)\rangle$, since at $0$, $2\sqrt\tau\langle X, Y\rangle = 2\langle v, Y(0)\rangle = 0$. So $\nabla L = 2\sqrt{\bar\tau}X(\bar\tau) = 2v$, and with $x = 2\sqrt{\bar\tau}v$, $\nabla L = \frac{x}{\sqrt{\bar\tau}}$. Directly, $\nabla\frac{|x|^2}{2\sqrt{\bar\tau}} = \frac{x}{\sqrt{\bar\tau}}$.

Solution

$\partial_\tau\ell = \frac{\partial_\tau L}{2\sqrt\tau} - \frac{L}{4\tau^{3/2}} = R - \frac{L}{4\tau^{3/2}} + \frac{K}{2\tau^{3/2}} - \frac{L}{4\tau^{3/2}} = R - \frac{L}{2\tau^{3/2}} + \frac{K}{2\tau^{3/2}}$, and $\frac{L}{2\tau^{3/2}} = \frac\ell\tau$. Also $|\nabla\ell|^2 = \frac{|\nabla L|^2}{4\tau} = -R + \frac{L}{2\tau^{3/2}} - \frac{K}{\tau^{3/2}} = -R + \frac\ell\tau - \frac{K}{\tau^{3/2}}$.

Solution

Sum the terms of $H(X, W_\alpha)$: $-\sum\nabla_{W_\alpha}\nabla_{W_\alpha}R = -\Delta R$; $-2\sum\langle R(W_\alpha, X)W_\alpha, X\rangle = 2\operatorname{Ric}(X, X)$; $-4\sum\nabla_X\operatorname{Ric}(W_\alpha, W_\alpha) = -4\langle\nabla R, X\rangle$ (the trace of $\nabla_X\operatorname{Ric}$ is $\nabla_XR$); $+4\sum\nabla_{W_\alpha}\operatorname{Ric}(W_\alpha, X) = 2\langle\nabla R, X\rangle$ (Bianchi); $-2\sum\operatorname{Ric}_\tau(W_\alpha, W_\alpha) = 2\Delta R$; $+2\sum|\operatorname{Ric}(W_\alpha, \cdot)|^2 = 2|\operatorname{Ric}|^2$; $-\frac1\tau\sum\operatorname{Ric}(W_\alpha, W_\alpha) = -\frac R\tau$. Total: $\Delta R + 2|\operatorname{Ric}|^2 + 2\operatorname{Ric}(X, X) - 2\langle\nabla R, X\rangle - \frac R\tau = -R_\tau - \frac R\tau - 2\langle\nabla R, X\rangle + 2\operatorname{Ric}(X, X) = H(X)$.

Solution

(7.8) with $\operatorname{Ric} = 0$ is $\dot Y = \frac{1}{2\tau}Y$, so $Y = c\sqrt\tau E$, and $Y(\bar\tau) = E$ gives $c = \bar\tau^{-1/2}$. The variation $\gamma_u(\tau) = 2\sqrt\tau(v + uE/(2\sqrt{\bar\tau}))$ is a family of $\mathcal L$-geodesics with variation field $\sqrt{\tau/\bar\tau}E$, so $Y$ is $\mathcal L$-Jacobi. All curvature terms vanish and $H = 0$, so (7.9) reads $\operatorname{Hess}L \leq \frac{1}{\sqrt{\bar\tau}}$, with equality since $Y$ is $\mathcal L$-Jacobi. Directly, $\operatorname{Hess}\frac{|x|^2}{2\sqrt{\bar\tau}} = \frac{1}{\sqrt{\bar\tau}}I$.

Solution

$\partial_\tau g = 2\operatorname{Ric} = 2(n - 1)g_{S^n}$, so the factor is $1 + 2(n - 1)\tau$ and $R = \frac{n(n - 1)}{1 + 2(n - 1)\tau}$, $R_\tau = -\frac{2n(n - 1)^2}{(1 + 2(n - 1)\tau)^2}$. With $X = 0$, $H(0) = -R_\tau - \frac R\tau = \frac{n(n - 1)}{1 + 2(n - 1)\tau}\Big(\frac{2(n - 1)}{1 + 2(n - 1)\tau} - \frac1\tau\Big) = -\frac{n(n - 1)}{\tau(1 + 2(n - 1)\tau)^2} < 0$. But $H(0) + \frac R\tau = -R_\tau > 0$. For an ancient solution with nonnegative curvature operator, Hamilton's trace Harnack inequality (with the initial time sent to $-\infty$) gives $R_t \geq 0$, that is $-R_\tau \geq 0$, which is exactly $H(0) \geq -\frac R\tau$. The negative term $-\frac R\tau$ in $H$ is there because $H$ measures time from the base point $\tau = 0$, not from the beginning of the flow.