目录
Definitions and notations
- \(M\) is a complete Riemannian manifold.
- \(H^d(M):=\{u\in C^{\infty}(M)|\Delta u=0 ~\text{and}~ u(x)=\Omicron (r^d(x)) \,\text{as}\, x \to \infty \}\)
- \(h^d(M)=\dim (H^d(M))\)
Iintroduction
Let us start from Yau's conjecture:
Conjecture (Yau) If \(M\) has non-negative Ricci curvature, then $$h^d(M)<\infty. $$
- Yau’s conjecture was first proved by Colding and Minicozzi in 1996. Indeed, they proved the dimension estimate of the form
where \(C_1\) depends on the Neumann Poincaré inequality and the volume doubling constant. In particular, it gives a sharp order estimate of the form
\[h^d(M)\leq C_1 d^{m-1 } \]for manifolds with nonnegative Ricci curvature.
2. Meanwhile, in 1997, Peter Li gave a much simplified proof of this estimate which holds for a larger class of manifolds. In this paper, Li only required the manifold to satisfy a volume comparison condition \((\mathcal{V}_{\mu})\) (see Definition 28.1) and a mean value inequality \((\mathcal{M})\) (see Definition 28.2).
Note that the volume comparison condition and the volume doubling condition can easily be seen to be equivalent. However, the mean value inequality is weaker than the Neumann Poincaré inequality. Moreover, the author’s argument can be applied to sections of vector bundles.
Finally, it is important to note that Li also proved that finite dimensionality and estimates of \(h^d(M)\) actually hold for an even more general class of manifolds (Theorem 28.7), namely those satisfying a weak mean value inequality \((\mathcal{W}\mathcal{M})\) and with polynomial volume growth. However, in this case, the estimate is not sharp.
Main results
Def 28.1 Condition\((\mathcal{V}_{\mu})\): $$V_x(\rho_2) \leq C_{\mathcal{V}} V_x(\rho_1) \left( \frac{\rho_2}{\rho_1} \right)^\mu.$$
Def 28.2 Condition \((\mathcal{M})\): For any non-negative subharmonic function \(f\) on \(M\), it must satisfy
\[f^2(x) \leq C_{\mathcal{M}} V_x^{-1}(\rho) \int_{B_x(\rho)} f^2(y) \, dy. \]
Remark: Note that if \(M\) has nonnegative Ricci curvature, then it satisfies both conditions \((\mathcal{V}_{\mu})\) and \((\mathcal{M})\).
\[\left\{ \begin{align*}V_p(\rho) &\leq C \rho^\mu\\|u(x)| &\leq C r^d(x) \end{align*} \right. \Rightarrow \operatorname{tr}_{\beta \rho} A_{\rho} \geq k \beta^{-(2d+\mu+\delta)}. \]Lemma 28.3 (Li)
Proof. Denote \(A_{\rho}(u, v) = \int_{B_{\rho}(\rho)} \langle u, v \rangle.\)
Then \(\lambda\) being an eigenvalue of \(A_{\rho}\) w.r.t. \(A_{\beta\rho}\) iff $$A_{\rho}(u, v)=\lambda A_{\beta\rho}(u, v),, \forall v,$$
where \(u\) is the corresponding eigenvector.
Suppose
then one can see
by using variation formula (\(\lambda\) is the eigenvalue of \(A_{\rho}\) w.r.t. \(A_{\beta\rho}\))
\[\lambda=\inf \frac{A_{\rho}(u, v)}{A_{\rho'}(u, v)}. \](one may need the polarization formula.)
Using
\[\lambda_1=\inf \frac{A_{\rho}(u, u)}{A_{\beta\rho}(u, u)}=\inf \frac{A_{\rho}(u, u)}{A_{\beta^2\rho}(u, u)}\frac{A_{\beta^2\rho}(u, u)}{A_{\beta\rho}(u, u)}\geq \lambda_3 \lambda_2^{-1}. \]Theorem 28.4 (Li) Suppose \(M\) satisfies
\[\left\{ \begin{align*} &(\mathcal{V}_{\mu} )\\ &(\mathcal{M}) , \end{align*} \right. \]
and subspace \(S_d(E,M)\subset \Gamma(E)\) satisfies
\[\left\{ \begin{align*} &\Delta|u| \geq 0 \\ &|u(x)| \leq O(r^d(x)) . \end{align*} \right. \]
Then the dimension of \(\mathcal{S}_d(M, E)\) is finite. Moreover, for all \(d \geq 1\), there exists a constant \(C \geq 0\) depending only on \(\mu\) such that
\[\dim \mathcal{S}_d(M, E) \leq n C C_M d^{\mu-1}. \]Proof. Let \(K\subset S_d\) be any finite dimensional subspace with \(\dim(K)=k\), and \(\{u_i\}_{i=1}^k\) is a basis. Consider the function
\[\begin{equation} \sum_{i=1}^k|u_i|^2(x) . \end{equation} \]Note that at a fixed point \(x\in B_p(\rho)\), we can rewrite (1) as
\[\begin{equation} \sum_{i=1}^k|u_i|^2(x) = \sum_i^n|u_i|^2(x) \end{equation} \]after an orthonormal changing of basis.
That is because if we denote \(K_x \subset K\) be a subspace such that \(u(x)=0, \forall u\in K_x\), then \(K_x\) has at most codimension \(n\). Otherwise, let \(u_1,\cdots,u_{n+1}\) such that \(u_i(x)\neq0, i=1,...,n+1\), since the rank of the bundle is \(n\), there is a linear combination of \(u_i(x)\) which is zero, a contradition.
\[f(x) \leq C_{\mathcal{WM}} V_x^{-1}(\rho) \int_{B_x(b\rho)} f(y) \, dy. \]Def 28.6 Condition(\(\mathcal{WM}\)) :
A complete manifold \(M\) is said to satisfy a weak mean value inequality \((\mathcal{WM})\) if there exist constants \(C_{\mathcal{WM}} > 0\) and \(b \geq 1\) such that for any nonnegative subharmonic function \(f\) defined on \(M\) it must satisfy
for all \(x \in M\) and \(\rho > 0\).
Theorem 28.7
Let \(M\) be a complete manifold satisfying the weak mean value property \((\mathcal{WM})\). Suppose that the volume growth of \(M\) satisfies \(V_p(\rho) = O(\rho^\mu)\) as \(\rho \to \infty\) for some point \(p \in M\). Then \(H^d(M)\) is finite dimensional for all \(d \geq 0\) and \(\dim H^d(M) \leq C_{WM} (2b + 1)^{2d+\mu}\).
\[\left( \int_{B_p(\rho)} |f|^{2\mu/(\mu-2)} \right)^{(\mu-2)/\mu} \leq C_1 V_p(\rho)^{-2/\mu} \rho^2 \times \int_{B_p(\rho)} |\nabla f|^2 + C_2 V_p(\rho)^{-2/\mu} \int_{B_p(\rho)} f^2. \]Corollary 28.8
Let \(M\) be a complete manifold. Suppose there exist constants \(C_1, C_2 \geq 0\) and \(\mu > 2\) such that for all \(p \in M\), \(\rho > 0\), and for all \(f \in H^1_{c}(M)\) we have
Proof. Verify that the conditions in Theorem 28.7 are satidfied. (The above Sobolev inequality implies both volume growth and weak mean value property.)
\[h^d(M) \leq h^d(\mathbb{R}^m) = \binom{m + d - 1}{d} + \binom{m + d - 2}{d - 1}? \]Question
If \(M\) is a complete manifold with nonnegative sectional curvature, then is it true that
This conjecture seems much harder than the above results, since, in above they only give certain asymptotic estimates. But the conjecture says \(h^d(M)\) has a very precise control.
Moreover
In Li and Wang, they propose the following twe conjectures.
\[V_p(r) = O(r^k) \]Conjecture
Let \(M\) be a complete manifold with non-negative Ricci curvature. Suppose \(M\) has volume growth satisfying
for some \(1 \leq k \leq n\). Then
\[h_d(M) \leq h_d(\mathbb{R}^k) \]for all \(d \geq 0\).
\[h_d(M) \leq h_d(\mathbb{R}^n) \]Conjecture
Let \(M\) be a complete manifold with non-negative Ricci curvature. Then
for all \(d \geq 0\). Moreover, equality holds for some \(d \geq 1\) if and only if \(M = \mathbb{R}^n\).
Also see Colding and Minicozzi.
By the way, they said
Recall that two metric spaces are said to be quasi-isometric if they are bilipschitz.
This gives a quite clear definition of ''quasi-isometric''. I once check it online, but it seems much more complicated, see, for example, wiki.
标签：functions,mu,leq,beta,growth,rho,Polynomial,mathcal,lambda From： https://www.cnblogs.com/crossLH/p/18187964