What follows is a very rough transcript taken during the lecture. It has only been edited to ensure the TeX compiles. No effort has been expended as yet to ensure accuracy. There are surely many errors. Please send any comments, suggestions, typos etc. to Paul Bryan
Download transcript PDF\((M^n, g)\) is a complete, non-compact Riemannian manifold with \(\Ric \geq 0\).
If \(u > 0\), \(\Delta u = 0\) on \(B_r(x_0) \subset M\) then \[ (r^2 -\rho(x)^2) \abs{\nabla \ln u(x)} \leq c(n) r \] on \(B_r(x_0)\).
WLOG \(r = 1\). We use \(\Delta u = 0\), \(u > 0\) and let \(v = \ln u\).
\begin{equation} \label{eq:Deltav} \Delta v = \nabla_i \left(\frac{\nabla_i u}{u}\right) = \frac{\Delta u}{u} - \abs{\frac{\nabla u}{u}}^2 = - \abs{\nabla v}^2. \end{equation}Let \(F = \abs{\nabla v}^2\). Then differentiating \eqref{eq:Deltav}, \[ 0 = \nabla_i (\Delta v + F) = \Delta \nabla_i v - R^p_i \nabla_p u + \nabla_i F \] Note at a maximum point \(\abs{\nabla F} = 2\abs{\tfrac{\nabla \varphi}{\varphi}} F\).
Therefore
\begin{align} \label{eq:DeltaF} 0 &= \Delta F - 2 \abs{\nabla^2 v}^2 - 2\operatorname{Ric}(\nabla v, \nabla v) + 2 \nabla v \cdot \nabla F \\ &\leq \Delta F - \frac{2}{n} (\Delta v)^2 + 2 \sqrt{F}\abs{\nabla F} \notag \\ &= \Delta F - \frac{2}{n} F^2 + 2\sqrt{F} \abs{\nabla F} \notag \end{align}Write \(\varphi = 1 - \rho^2\) with \(\rho = d(\cdot, x_0)\). Let \(G = \varphi^2 F\). On \(\partial B_1(x_0)\), \(G = 0\).
At \(x\),
\begin{align*} 0 = \nabla G = \varphi^2 \nabla_i F = 2 \varphi \nabla_i \varphi F \\ = \varphi^2\left(\nabla_i F + \frac{\nabla \varphi}{\varphi^3} G\right) \end{align*}At \(x\), using \eqref{eq:DeltaF}
\begin{align*} 0 &\geq \Delta G = \varphi^2(\Delta F + (2 \frac{\Delta \varphi}{\varphi^2} - 6 \frac{\abs{\nabla \varphi}^2}{\varphi^4})\varphi^2 F) \\ &\geq \varphi^2(\frac{2}{n} F^2 - 2\sqrt{F} 2 \frac{\abs{\nabla \varphi}}{\varphi} F + (2 \frac{\Delta \varphi}{\varphi} - 6 \frac{\abs{\nabla \varphi^2}}{\varphi^2} F)) \\ &= F(\frac{2}{n} G - 4 \sqrt{G} \abs{\varphi} + 2 \varphi\Delta\varphi - 6 \abs{\nabla \varphi^2}) \\ &\geq F(\frac{1}{n} G - C \abs{\nabla \varphi}^2 + 2\varphi\Delta\varphi) \end{align*}using Cauchy-Schwarz in the last line.
From \(\varphi = 1 - \rho^2\), \(\nabla \varphi = -2\rho\abs{\nabla \rho} \Rightarrow \abs{\nabla \varphi} \leq 4\rho^2 \leq 4\) and \(\Delta \varphi = - \Delta \rho^2 \geq -2n\). Thus \[ G \leq C(n). \]
If \(u > 0\), \(\Delta u = 0\) on \(B_r(x_0) \subseteq M\) then \[ \sup_{B_{r_1}(x_0)} u \leq C(n) \inf B_{r_2}(x_0) u \]
WLOG \(r = 1\). By the gradient estimate, \(\abs{\nabla \ln u} \leq C(n)\), \[ \frac{u(y)}{u(x)} \leq e^{C(n)} d(x, y) \leq e^{C(n)}. \]
Assume \((M, g)\) is complete, non-compact with \(\operatorname{Ric} \geq 0\).
If \(v\) is a non-negative, subharmonic function \((\Delta v \geq 0)\) on \(B_r(x_0)\), then \[ v(x_0) \leq c(n) \frac{1}{\abs{B_r(x_0)}} \int_{B_r(x_0)} v. \]
[Proof in the case \(v = u^2\), \(\Delta u = 0\).] \[ \Delta v = 2 \abs{\nabla u}^2 \geq 0 \]
using volume comparison.
Note \(\abs{u}\) is subharmonic hence \(\abs{u}(x_0) \leq h(x_0)\) so
\begin{align*} u^2(x_0) &\leq h^2(x_0) \leq C(n) \left(\frac{1}{B_{r/2}(x_0)} \int_{B_{r/2}(x_0)} h\right)^2 \\ &\leq C(n) \frac{1}{B_{r/2} (x_0)} \int_{B_{r/2} (x_0)} h^2 \\ \end{align*}by Hölder's inequality.
Write \[ \int_{B_{r/2}} h^2 = \int_{B_r} (h - \abs{u} + \abs{u})^2 \leq 2 \int_{B_{r/2}} (h - \abs{u})^2 + 2 \int_{B_{r/2}} u^2 \] noting that \(h - \abs{u} = 0\) on the boundary. The using the Poincare inequality, \[ \int_{B_{r/2}} (h - \abs{u})^2 \leq C(n) \int_{B_{r/2}} \abs{\nabla h - \nabla \abs{u}}^2 \leq C \int_{B_{r/2}} \abs{\nabla h}^2 + \abs{\nabla u}^2 \] Since \(h\) is harmonic, it minimises the Dirichlet energy among maps with the same energy and hence \[ \int_{B_{r/2}} \abs{\nabla h}^2 \leq \int_{B_{r/2}} \abs{\nabla{\abs{u}}}^2 \leq \int_{B_{r/2}} \abs{\nabla u}^2 \] where we note \(u\) is smooth hence in \(W^{2,2}\).
Let \(\Phi \in C^{\infty}_c(B_1)\), \(\Phi \equiv 1\) on \(B_{r/2}\), \(\abs{\nabla \Phi} \leq C\).
\begin{align*} \int_{B_1} \Phi \abs{\nabla u}^2 &= - \int_{B_1} \Phi^2 u \Delta u - \int_{B_1} 2 \Phi u \nabla \Phi \cdot \nabla u \\ &= 2\left(\int_{B_1} \Phi^2 \abs{\nabla u}^2\right)^{1/2} \left(\int_{B_1} u^2 \abs{\nabla \Phi}^2\right)^{1/2} \end{align*}Thus \[ \int_{B_{1/2}} \abs{\nabla u}^2 \leq \int_{B_1} \Phi^2 \abs{\nabla u}^2 \leq 4 \int_{B_1} u^2 \abs{\nabla \Phi}^2 \leq C \int_{B_1} u^2 \] giving \[ \int_{B_{r/2}} h^2 \leq C \int_{B_1} u^2 \] Using volume comparison, \[ u^2(x_0) \leq \frac{C}{\abs{B_{r/2} (x_0)}} \int_{B_1} u^2 \leq 2^n C \frac{1}{\abs{B_1}} \int_{B_1} u^2. \]
\[ \operatorname{dim} \mathcal{H}_p (M) \leq C(n) p^{n-1} \] where \[ \mathcal{H}_p(M) = \{u \in C^{\infty}(M) : \Delta u = 0, \abs{u(x)} \leq C(1 + d(x, x_0)^p\} \]
We aim to bound the dimension of any finite dimensional subspace, \(K\) of \(\mathcal{H}_p(M)\) which will give the result. First we have an estimate on how harmonic functions can be "packed" into a ball.
Let \(K\) be any finite dimensional subspace of \(\mathcal{H}_p(M)\) of \(\{u \in C^{\infty}(M) : \Delta u = 0\}\). Let \(\{u_i\}_{i=1}^k\) be any orthonormal basis of \(K\) with respect to \(L^2(B_r(x))\).
Then for any \(0 < \epsilon < 1/2\), \[ \int_{B_{(1-\epsilon)r}(x)} \sum_{i=1}^k u_i^2 \leq C(n) \epsilon^{-(n-1)}. \]
WLOG \(r = 1\).
If \(y \in B_{1-\epsilon}(x)\) by Gram-Schmidt, choose a rotation, \(\Theta\) of \(\mathbb{R}^k\) such that \[ \Theta \begin{pmatrix} u_1(y) \\ \vdots \\ u_k(y) \end{pmatrix} = \sqrt{\sum_{i=1}^k u_i^2(y)} \begin{pmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix} \]
Let \(u = \sum_i \Theta_i^i u_i\) (harmonic!). The MVI gives \[ u^2(y) \leq C \frac{1}{\abs{B_{1-\rho(y)}}} \int_{B_{1-\rho(y)}} u^2 \leq C \frac{1}{\abs{B_1(x)}} \int_{B_1(x)} u^2 \frac{\abs{B_1(x)}}{\abs{B_{1-\rho(y)}(y)}}. \] Volume comparison doesn't directly apply since we have different centres. But we get \[ u^2(y) \leq C \frac{\abs{B_{1+\rho}(y)}}{\abs{B_{1-\rho}(y)}} \frac{1}{\abs{B_1(x)}} \int_{B_1(x)} u^2 \leq C\left(\frac{1+\rho}{1-\rho}\right)^n \frac{1}{\abs{B_1(x)}} \int_{B_1(x)} u^2. \] Easy estimate: on \(B_{1-\epsilon}\), \(\tfrac{1+\rho}{1-\rho} \leq \tfrac{2}{\epsilon}\) giving \[ \sum_{i=1}^k u_i^2 (y) = u^2(y) \leq \frac{C \epsilon^{-n}}{\abs{B_1(x)}} \] and hence \[ \int_{B_{1-\epsilon}(x)} \sum_{i=1}^k u_i^2 \leq C \epsilon^{-n} \frac{\abs{B_{1-\epsilon}}}{\abs{B_1}}. \]
More work is required to bump up the estimate to \(\epsilon^{-(n-1)}\).
What follows is a very rough transcript taken during the lecture. It has only been edited to ensure the TeX compiles. No effort has been expended as yet to ensure accuracy. There are surely many errors. Please send any comments, suggestions, typos etc. to Paul Bryan
Download transcript PDFWe continue proving the lemma from last week.
\label{upperbound} Let \(K\) be any finite dimensional subspace of \(\mathcal{H}(M)\) of \(\{u \in C^{\infty}(M) : \Delta u = 0\}\). Let \(\{u_i\}_{i=1}^k\) be any orthonormal basis of \(K\) with respect to \(L^2(B_r(x))\).
Then for any \(0 < \epsilon < 1/2\), \[ \int_{B_{(1-\epsilon)r}(x)} \sum_{i=1}^k u_i^2 \leq C(n) \epsilon^{-(n-1)}. \]
Recall
WLOG, \(r = 1\). If \(y \in B_{(1-\epsilon)} (x)\), then \[ \sum_{i=1}^k u_i^2(y) \leq c(n) \left(\frac{1+\rho(y)}{1-\rho(y)}\right)^n \avgint_{B_1} u^2. \] Recall we rotated in \(k\) so \(u^2(y) = \sum_{i=1}^k u_i^2(y)\) and we had the normalisation \(\int_{B_1} u^2 = 1\). Thus \[ \int_{B_{1-\epsilon}} \sum_{i=1}^k u_i^2 \leq \frac{C(n)}{\abs{B_{1-\epsilon}(x)}} \int_{B_{(1-\epsilon)} (x)} \left(\frac{1+\rho(y)}{1-\rho(y)}\right)^n. \]
Now we use the Laplace comparison theorem to bump the estimate up to the desired power \(\epsilon^{-(1-n)}\). We integrate by parts on the annulus,
\begin{align*} \int_{B_{1-\epsilon} \backslash B_{1/2}} (1-\rho)^{-n} &= \int_{B_{1-\epsilon} \backslash B_{1/2}} (1-\rho)^{-n} \abs{\nabla \rho}^2 \\ &= \frac{1}{n-1} \int_{B_{1-\epsilon} \backslash B_{1/2}} \nabla\left((1-\rho)^{-(n-1)} - \epsilon^{-(1-n)}\right) \cdot \nabla \rho \\ &= -\frac{1}{n-1} \int_{\partial B_{1/2}} \left((1-\rho)^{-(n-1)} - \epsilon^{-(1-n)}\right) \\ &\quad - \frac{1}{n-1} \int_{B_{1-\epsilon} \backslash B_{1/2}} \left((1-\rho)^{-(n-1)} - \epsilon^{-(1-n)} \Delta \rho \right) \\ &\leq C\epsilon^{-(n-1)} \frac{n-1}{\rho} \leq C\epsilon^{-(n-1)} \frac{n-1}{2} \end{align*}where we use the Bishop-Gromov again to control the size of the region of integration.
Note that lemma \ref{upperbound} did not use the polynomial growth hypothesis. Now we make use of the hypothesis for the next lemma. The assumption means that such harmonic functions cannot grow too much "all the time" - a sort of converse to the previous lemma.
\label{lowerbound} Let \(K\) be any finite dimensional subspace of \(\mathcal{H}_p(M)\) of \(\{u \in C^{\infty}(M) : \Delta u = 0, \abs{u(y)} \leq C(1 + d(y, x_0))^p\}\).
Then for any \(x \in M\), \(\epsilon \in (0, 1/2]\), \(r_0 > 0\), \(\delta > 0\), there exists \(r > r_0\) such that if \(\{u_i\}_{i=1}^k\) is an orthonormal basis of \(K\) with respect to \(L^2(B_r(x))\), we have \[ \sum_{i=1}^k \int_{B_{(1-\epsilon)r}(x)} \abs{u_i}^2 \geq k(1-\epsilon)^{2p + n + \delta} = (1-\epsilon)^{2p+n+\delta} \int_{B_r(x)} \sum_{i=1}^k u_i^2. \]
Fix \(x \in M\), \(\epsilon \in (0, 1/2]\), \(r_0 > 0\), \(\delta > 0\). The \(\delta\) is here just to give a little extra room to obtain a strict inequality.
Define \(r_{\alpha} = r_0(1-\epsilon)^{-\alpha}\) for \(\alpha \in \NN\).
To obtain a contradiction, suppose the claimed inequality does not hold for \(r = r_{\alpha}\). Let \(G_{\alpha}\) be the inner-product on \(K\) coming from \(L^2(B_{r_{\alpha}}(x)\). If \(\{u_i\}_{i=1}^k\) is an orthonormal basis for \(K\) with respect to \(L^2(B_{r_{\alpha}}(x)\) so \((G_{\alpha})_{ij} = \delta_{ij}\). Then \[ (G_{\alpha-1})_{ij} = \int_{B_{(1-\epsilon)r_{\alpha}(x)}} u_i u_j \] hence \[ \Tr (G_{\alpha}^{-1} \circ G_{\alpha-1}) = \int_{B_{(1-\epsilon)r_{\alpha}}} \sum_{i=1}^k \abs{u_i}^2 < (1-\epsilon)^{2p+n+\delta} \] by the previous lemma, \ref{upperbound}. By the arithmetic-geometric inequality, \[ \det (G_{\alpha}^{-1} \circ G_{\alpha-1}) \leq (\tfrac{1}{k} \Tr (G_{\alpha}^{-1} \circ G_{\alpha-1}))^k < (1-\epsilon)^{(2\rho+n+\delta)k}. \] Thus
\begin{equation} \label{det} \det (G_{\alpha}^{-1} \circ G_0) = \det (G_{\alpha}^{-1} \circ G_{\alpha-1}) \det (G_{\alpha-1}^{-1} \circ G_{\alpha-2}) \det (G_{1}^{-1} \circ G_{0}) < (1-\epsilon)^{(2\rho + n+\delta)k\alpha}. \end{equation}Fix an o/n basis for \(G_0\), \(\{u_i\}_{i=1}^k\) so \(\int_{B_{r_0}(x)} u_i u_j = \delta_{ij}\). Since \(u_i \in K \subseteq \mathcal{H}_p\), \[ \abs{u_i(y)} \leq C (1 + d(x,y))^p. \]
Again using the arithmetic-geometric inequality and Bishop-Gromov,
\begin{align*} k \det (G_0^{-1} \circ G_{\alpha})^{1/k} &\leq \Tr (G_0^{-1} \circ G_{\alpha}) = \sum_{i=1}^k \int_{B_{r_0} (x)} \abs{u_i}^2 \\ &\leq C r_{\alpha}^{2p} k \abs{B_{r_{\alpha}}} \\ &\leq C k\abs{B_{r_0}} (1-\epsilon)^{-n\alpha} r_0^{2p}(1-\epsilon)^{1-2p} \end{align*}But by \eqref{det} the left hand side satisfies \[ k \det (G_0^{-1} \circ G_{\alpha})^{1/k} > k(1-\epsilon)^{-(2p+n+\delta)\alpha} \] giving \[ k(1-\epsilon)^{-(2p+n+\delta)\alpha} < C k\abs{B_{r_0}} (1-\epsilon)^{-n\alpha} r_0^{2p}(1-\epsilon)^{1-2p}. \] Simplifying gives for any \(\alpha\), \[ (1-\epsilon)^{-\alpha\delta} \leq C \] with \(C\) independent of \(\alpha\). Sending \(\alpha \to \infty\) gives a contradiction.
We see in the last inequality where the \(\delta > 0\) is required and that the proof does not work with \(\delta = 0\).
Now we may prove the main theorem by playing the two lemmas off against each other.
\[ \operatorname{dim} \mathcal{H}_p (M) \leq C(n) p^{n-1} \] where \[ \mathcal{H}_p(M) = \{u \in C^{\infty}(M) : \Delta u = 0, \abs{u(x)} \leq C(1 + d(x, x_0)^p\} \]
Let \(K \subseteq \mathcal{H}_p\) be any finite dimensional subspace, \(\epsilon \in (0, 1/2]\), \(r_0 = 1\) and any \(\delta > 0\) as in Lemma \ref{lowerbound} such that \[ \int_{B_{r(1-\epsilon)}} \sum_{i=1}^k \abs{u_i}^2 \geq k(1-\epsilon)^{2p + n + \delta}. \] On the other hand by Lemma \ref{upperbound} \[ \int_{B_{r(1-\epsilon)}} \sum_{i=1}^k \abs{u_i}^2 \leq C(n) \epsilon^{-(n-1)}. \] Combining these gives, \[ k \leq C(n) \epsilon^{-(n-1)}(1-\epsilon)^{-(2p+n+\delta)}. \] Now we choose \(\epsilon = \tfrac{1}{2p}\) giving, \[ k \leq C(n) 2^{n-1}p^{n-1}(1-\tfrac{1}{2p})^{-2p} s^{n\epsilon\delta} \leq \tilde{C}(n) p^{n-1} \] where we bounded \((1-\tfrac{1}{2p})^{-2p} \leq e\).
Colding-Minicozzi also consider the situation \(M^n \subset \RR^N\) a minimal submanifold with Euclidean volume growth: \(V(B_r\cap M) \leq C_0 r^n\). Then the dimensions of the space of harmonic functions of polynomial growth of order \(p\) is bounded by \(C(n, C_0)p^{n-1}\).
The idea is to use the Michael-Simon inequality of harmonic functions for minimal submanifolds to show that if \(\Delta u = 0\), we have the mean value inequality \[ u(x) \leq \frac{C}{r^n} \int_{B_r \cap M} u^2 \]
By the monotonicity formula, \[ \frac{d}{dr} \frac{\abs{B_r \cap M}}{r^n} \geq 0. \] Combined with Euclidean volume growth, we get \[ \abs{B_r \cap M} \sim r^n. \]
Then the mean value inequality is now a real mean value inequality and \[ \abs{\frac{B_R \cap M}{B_r \cap M}} \leq C\left(\frac{R}{r}\right)^n \] and the proof of the main theorem goes through as before.
In the situation of the remark, \(M\) is contained in a subspace of \(\RR^N\) of dimension depending only on \(n, C_0\).
Since \(M\) is minimal, the coordinate functions are harmonic and of linear growth rate. Thus each \(x^i \in \mathcal{H}_1(M)\). Then apply dimension bound.
What follows is a very rough transcript taken during the lecture. It has only been edited to ensure the TeX compiles. No effort has been expended as yet to ensure accuracy. There are surely many errors. Please send any comments, suggestions, typos etc. to Paul Bryan
Download transcript PDFSo \[ \int_0^{\infty} e^{-ts} h(x, s) ds = \frac{1}{t} u(x, -t). \] Bernstein gives extension of \(t\) into right half complex plane and heat equation estimates give Laplace transform is invertible, \[ h(x, t) = \frac{1}{2\pi i} \int_{-1-i\infty}^{-1+it} e^{st} \frac{u(x, -s)}{s} ds. \]
Recall that \(\Delta f^x(t) + \partial_t f^x(t) = 0\) in \(\RR^n \times [0, \infty)\). Testing against a smooth cut-off function \(\phi \in C^{\infty}_c(\RR^n \to \RR)\),
\begin{align*} \partial_t \int_{\RR^n} f^x(t) \phi(x) dx &= - \int_{\RR^n} \phi(x) \Delta f^x dx \\ &= -\int_{\RR^n} \Delta \phi (x) f^x (t) dx \\ &= -\int_{\RR^n} \int_0^{\infty} s e^{-ts} \phi(x) d\nu(x, s). \end{align*}Thus \[ \int_{\RR^n} \int_0^{\infty} se^{-ts} \phi(x) d\nu(x, s) = \int_{\RR^n}\int_0^{\infty} e^{-ts} \Delta \phi(x) d\nu(x, s) \] which is a Laplace transform again: \[ \int_0^{\infty} e^{-ts} \int_{\RR^n} (\Delta\phi(x) - s\phi(x)) dx d\nu(x, s) = 0 \] for all \(t > 0\) and \(\phi \in C^{\infty}_c(\RR^n \to \RR)\).
Define the signed measure \(\eta_{\phi}\) on \([0, \infty)\) by \[ d\eta_{\phi} = \int_{\RR^n} (\Delta \phi(x) - s \phi(x)) d\nu(x, s). \] We've shown, \[ \operatorname{LT}(\eta_{\phi}) = 0 \Rightarrow \eta_{\phi} = 0. \] That is, \[ 0 = \eta_{\phi}([0, t]) = \int_{\RR^n} \left[\Delta \phi(x) \underbrace{\int_0^t d\nu(x, s)}_{h(x, t)} - \phi(x) \int_0^t sd\nu(x, s) \right]dx. \]
Side note: \[ h(x, u) = \int_0^u d\nu(x, s) \] and so \(h = dx/d\nu(x, s)\) and we can integrate \[ \int_0^t s d\nu(x, s) \] by parts. Thus
\begin{align*} \int_{\RR^n} \Delta \phi(x) \int_0^t d\nu(x, s) &= \int_{\RR^n}\phi(x) \int_0^t s d\nu(x, s) \\ &= \int_{\RR^n} \phi(x) \left(th(x, t) - \int_0^t h(x, s)ds\right)dx \end{align*}Because this is true for all \(\phi \in C^{\infty}_c(\RR^n \to \RR)\), \(h\) satisfies the integro-differential equation, \[ \Delta h(x, t) = t h(x, t) - \int_0^t h(x, s) ds \] hence \(h\) is smooth as the solution of \[ \left(\Delta - t\right) h(x, t) = - \int_0^t h(x, s) ds. \]
Fix \(x\) and consider the measure \(\nu_x\): \[ \nu_x([a, b]) = h(x, b) - h(x, a). \] Aim is to turn the equation for \(h\) into estimates for \(\Delta \nu\).
We have (writing \(I = [a, b]\)),
\begin{align*} \Delta \nu_x (I) &= \Delta h(x, b) - \Delta h(x, a) \\ &= b h(x, b) - \int_0^b h(x, s) ds \\ &\quad - ah(x, a) + \int_0^a h(x, s) ds \\ &= bh(x, b) - ah(x, a) - \int_a^b h(x, s) ds \\ &= b(\underbrace{h(x, b) - h(x, a)}_{\nu_x(I)}) - \int_a^b \underbrace{h(x, s) - h(x, a)}_{\geq 0} ds \end{align*}using side calculation: \((b-a) h(x, a) = \int_a^b h(x, a) ds\) and recalling \(h\) is non-decreasing.
Thus, \[ \Delta \nu_x (I) \leq b \nu_x (I) \] or equivalently, \[ \frac{\Delta \nu_x(I)}{\nu_x(I)} \leq b. \]
Similarly, from below,
\begin{align*} \Delta \nu_x (I) &= b h(x, b) - ah(x, a) - \int_a^b h(x, s) ds \\ &= a(\underbrace{h(x, b) - h(x, a))}_{\nu_x(I)} - \int_a^b \underbrace{h(x, s) - h(x, b)}_{\leq 0} ds \\ &\geq a \nu_x(I). \end{align*}Thus \[ \frac{\Delta \nu_x(I)}{\nu_x(I)} \geq a. \]
Let \[ c(x) = \frac{\Delta \nu_x (I)}{\nu_x (I)} \] satisfies, \[ \abs{c(x)} \leq b \]
The Harnack inequality for \(L = \Delta - C\) with \(I = [0, b]\) gives, \[ \nu_x(I) \leq c(\abs{x}, b) \nu_0(I). \] That is \(\nu_x\) is absolutely continuous with respect to \(\nu_0\).
Radon-Nikodym then gives, \[ \frac{d\nu_x}{d\nu_0}(s) \leq c(\abs{x}, b), \quad s \leq b. \]
From the integral estimate we obtained, \[ \int_0^{\infty} e^{-ts} \int_{\RR^n} (\Delta \phi(x) - s\phi(x)) d\nu_x (s) = 0. \] Now we are able to make a change of variables: \[ 0 = \int_0^{\infty} e^{-ts} \left(\int_{\RR^n} (\Delta \phi(x) - s\phi(x)) \frac{d\nu_x}{d\nu_0} (s) dx\right) d\nu_0(s) \] This is the Laplace transform of the signed measure \(\int_{\RR^n} (\Delta \phi(x) - s\phi(x)) \frac{d\nu_x}{d\nu_0} (s) dx\) so for \(\nu_0\)-a.e. \(s\) we have, \[ \int_{\RR^n} (\Delta \phi(x) - s\phi(x)) \frac{d\nu_x}{d\nu_0} (s) dx = 0 \] for every \(\phi \in C^{\infty}_c(\RR^n \to \RR)\). So, for \(\nu_0\)-a.e. \(s\), \[ (\Delta - s) \frac{d\nu_x}{d\nu_0} = 0 \] and \(\frac{d\nu_x}{d\nu_0}\) is smooth.
See also Kaprelovic `67 and Koranyi `79.
From \[ (\Delta - s) \frac{d\nu_x}{d\nu_0} = 0 \] we have the representation formula, \[ \frac{d\nu_x}{d\nu_0} (s) = \int_{\mathbb{S}^{n-1}} e^{x\cdot \xi\sqrt{s}} d\mu_s(\xi) \] for \(\mu_s\) a Borel measure on \(\mathbb{S}^{n-1}\).
This completes the proof because, \[ u(x, -t) = \int_0^{\infty} e^{-st} d\nu_x(s) \] implies \[ u(x, t) = \int_0^{\infty} \int_{\mathbb{S}^{n-1}} e^{st + x \cdot \xi \sqrt{s}} d\mu_s(\xi) d\nu_0(s). \] \qed
Replace \(\RR^n\) with a Riemannian manifold \((M^n, g)\).
For \((M, g)\) complete with \(\Ric \geq 0\) Li-Yau still applies to give \[ u(x, t) = \int_0^{\infty} e^{ts} h(s, x) dx \] where \(h\) solves \[ (\Delta - s) h(s, x) = 0. \]
Classifying solutions of this equation is the hard part.
Key concept: Martin boundary. (Martin '41, Murata ('86, '93, '02, '07)
The general theory gives, \[ h(x) = \int_{\Sigma_0} P_s(x, w) d\mu(w) \] for \(\mu\) a unique, non-negative Borel measure on \(\Sigma\) with \(\mu(\Sigma - \Sigma_0) = 0\).
Here, \[ \Sigma_0 = \{[w] : P(\cdot, w) \text{ is minimal}\} \] where the term minimal means that if \(0 \leq v(x) \leq P(x, w)\) then \(v(x) = P(x, w)\). Also,
\[ P_s(x, y) = \begin{cases} \frac{\Gamma_s(x, y)}{\Gamma_s(x_0, y)}, & y \ne x_0 \\ 0, & y = x_0, x\ne y \\ 1, & x = y = x_0. \end{cases} \]
\(\Gamma_s\) is the minimal Green's function for \(\Delta - s\).
[Lin-Zhang]
Let \((M, g)\) be a complete Riemannian manifold with \(\Ric \geq 0\). Suppose \(u\) is a non-negative, ancient solution of \(\partial_t u = \Delta_g u\). Then
\[ u(x, t) = \int_0^{\infty} \int_{\Sigma_s} e^{ts} P_s(x, w) d\mu(w, s) d\rho(s). \]
The condition \(\Ric\geq 0\) is not strictly necessary. It may be replaced by something like volume doubling and Poincare inequality.
What follows is a very rough transcript taken during the lecture. It has only been edited to ensure the TeX compiles. No effort has been expended as yet to ensure accuracy. There are surely many errors. Please send any comments, suggestions, typos etc. to Paul Bryan
Download transcript PDFA steady gradient soliton is a metric \(g\) such that \[ \Ric = \frac{1}{2}\Lie_V g = \nabla^2 f. \] where \(V = \grad f\).
Given such a metric, let \[ g(t) = \varphi_t^{\ast} g \] where \(\varphi_t\) is the flow of \(-\grad f\) and \(t \in (-\infty, \infty)\). Then \[ \partial_t g = \varphi_t^{\ast} \Lie_{-\grad f} g = -2\varphi_t^{\ast} \Ric(g) = -2 \Ric(g(t)) \] so \(g\) is an eternal RF (Ricci Flow).
Hamilton's Cigar Solition \(\Sigma^2\): \[ g = dr^2 + \varphi^2(r) d\theta^2 \] with \(\theta \in [0, 2\pi)\) and where \(\lim_{r\to\infty}\varphi(r) < \infty\).
\[ (\Sigma, p_i) \to \RR \times \mathbb{S}^1 \] \(\Sigma\) collapsed.
Bryan solition: \(n \geq 3\). \[ g = dr^2 + \varphi^2(r) g_{\mathbb{S}^{n-1}} \] with \[ \text{warping asymptotics: } \varphi(r) \simeq r^{1/2} \] \[ \text{Scalar curvature asymptotics: } R \simeq r^{-1} \]
\[ (\Sigma, R(p_i) g, p_i) \to \RR \times \mathbb{S}^{n-1} \] non-collapsed.
Known 3d SGS (non-flat): \(\RR \times \operatorname{Cigar}\) (gradients) and Bryan solition \(n=3\).
Hamilton's conjecture: there exists a 3d SGS that is a flying wing.
A flying wing is a SGS \((M, g)\) with \(\Rm \geq 0\) such that the blow-down is \(C_{\infty} (g) = \operatorname{Cone}([0, \theta])\), \(\theta \in (0, \pi)\).
The previous examples are not flying wings: \(\RR \times \operatorname{Cigar}\) has \(\theta = \pi\) while the Bryant solition has \(\theta = 0\). Flying wings also do not fit into the classification results above.
| MCF | RF | |
| \(n=2\) | Grim Reaper | Cigar Soliton |
|
\(n=3\) |
Collapsed: Flying Wings New examples (BLT). Noncollapsed: Bowl soliton |
Collapsed: Flying wing (Thm 1, \(n=3\) and Thm 2) Noncollaped: Bryant soliton |
|
\(n \geq 4\) |
Collapsed: Flying wings. non-collapsed: |
Collapsed: Flying wings? Non colapsed Thm 1 \(n \geq 4\) |
[Thm 1] \(\forall \alpha \in (0, 1)\) \(\exists\) a \(\ZZ_2 \times O(n-1)\)-symmetric SGS \((M, g, f, p)\) with \(\Rm > 0\) such that \[ \lambda_1 = \alpha\lambda_2 = \cdots = \alpha \lambda_n \] where \(\lambda_i\) are eigenvalues of \(\Ric\) at \(p\).
\(n=3\)
Let \(\{X_{iu}, u \in [0, 1]\}_{i=1}^{\infty}\) be a sequence of smooth families of metrics on \(\mathbb{S}^2\) such that
Recall Dernelle's result:
There exists a unique EGS (Expanding Gradient Soliton) \((M_{in}, g_{in}, p_{in})\) with \(\Rm > 0\) such that \(C_{\infty}(g_{in}) = C(X_{in})\) and
A sequence of EGS with \(\ZZ_2 \times O(n-1)\)-symmetry and \(\Rm > 0\) with \(R(p_i) = 1\) and \(AVR(g_i) \to 0\) as \(i \to \infty\) then \[ (M_i, g_i, p_i) \overset{Cheeger-Gromov}{\to} (M, g, p) \quad \text{ SGS \(R(p) = 1\)}. \]
Using Dernelle's result and the lemma \[ (M_{i0}, g_{i0}, p_{i0}) \to \text{Bryant soliton (by rotational symmetry)}, \quad \frac{\lambda_1}{\lambda_3} = 1. \] and \[ (M_{i1}, g_{i1}, p_{i1}) \to \RR \times \operatorname{Cigar}, \quad \frac{\lambda_1}{\lambda_3} = 0. \]
For any \(\alpha_0 \in (0, 1)\) there exists \(u_i \in (0, 1)\) such that \(\tfrac{\lambda_1}{\lambda_3} (g_{iu_i}) = 0\). Applying the lemma \[ (M_{in_i}, g_{in_i}, p_{in_i}) \to (M, g, p) \text{ a SGS} \quad \frac{\lambda_1}{\lambda_3} (g) = \alpha_0 \text{ at } p \]
The next theorem excludes the blow down is a ray so the construction of Thm 1 gives flying wings.
[Thm 2] Let \((M, g, p)\) be a \(\ZZ_2 \times O(2)\)-symmetric 3d SGS, and \(C_{\infty}(g) = \text{a ray} = \operatorname{Cone} (\{0\})\). Then it is a Bryant soliton.
In order to obtain a contradiction, suppose it is not a Bryant soliton.
The profile curve \(\Gamma\) is fixed by the \(O(2)\) action and the section \(\Sigma\) is fixed by the \(\ZZ_2\) action. Take a geodesic in \(\Sigma\) with \(\gamma(0) = p\). Away from the edges \[ g = g_0 + \varphi^2 d\theta, \theta \in [0, \pi) \] with \(g_0\) totally geodesic.
Define
Computing gives \[ \frac{h_2'(2)}{h_2(s)} \leq C R(\gamma(s)) \]
\[ R(\gamma(s)) \leq C h_1^{-2} (s) \]
\[ \frac{h_2(s)}{h_1(s)} \to 0, s \to \infty \]
\[ h_1(s) h_2(s) \geq C s, \quad s \to \infty \]
Using the lemmas, \[ h_2(s) \leq C s^{\epsilon} \forall \epsilon < 1/2 \] and \[ h_2(s) \leq C \to \lim_{s\to\infty} R(\Gamma(s)) > 0 \]
Let \((M, g)\) be a \(\ZZ_2 \times O(2)\)-symmetric SGS. If \[ C_{\infty}(g) = \operatorname{Cone}([0, \alpha]), \alpha \in [0,\pi] \] Then \[ \lim_{s\to\infty} R(\Gamma(s)) = R(p) \sin \tfrac{\alpha}{2} \]
We have
\begin{align*} 0 \simeq \ip{\grad f}{\sigma'(\gamma)}|_{-D}^D &= \int_{-D}^D \partial_r \ip{\grad f}{\sigma'(\gamma)} dr \\ &= \int_{-D}^D \Ric(\sigma'(\gamma), \sigma'(\gamma)) dr \\ &= 2R^{1/2}(\Gamma(s)) \end{align*}Thus \[ \lim_{s\to\infty} R(\Gamma(s)) = 0 \] contradicting \(\Rm > 0\) for the case \(\alpha = 0\).