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Suppose that $M$ is a smooth manifold and $g_0, g_1$ are Riemann metrics on $M$. $\newcommand{\eH}{\mathscr{H}}$ Denote by $\eH_{g_i}$, $i=0,1$ and the sheaf of $g_i$-harmonic functions. More precisely for any open set $U\subset M$

$$\eH_{g_i}(U)=\big\{\; f\in C^\infty(U):\;\;\Delta_{g_i} u=0\;\big\}, $$

where $\Delta_{g_i}$ denotes the scalar Laplacian of the metric $g_i$.

Long time ago I proved the following result.

Suppose that $\eH_{g_0}(U)=\eH_{g_1}(U)$, for any open set $U\subset M$.

  • If $\dim M\geq 3$, then there exists $c\in (0,\infty)$ such that $g_1=c g_0$.

    If $\dim M\geq 3$, then there exists $c\in (0,\infty)$ such that $g_1=c g_0$.

     
  • If $\dim M=2$, then there exists a smooth function $f: M\to (0,\infty)$ such that $g_1=fg_0$, i.e., the metrics $g_0$ and $g_1$ live in the same conformal class.

    If $\dim M=2$, then there exists a smooth function $f: M\to (0,\infty)$ such that $g_1=fg_0$, i.e., the metrics $g_0$ and $g_1$ live in the same conformal class.

The strong unique continuation property of harmonic functions shows that this statement is really a statement about the stalks of the sheaves $\eH_{g_i}$. Note that these are sheaves of vector spaces, not rings. In dimension $\geq 3$ these sheaves determine the metric up to a multiplicative positive constant.

Suppose that $M$ is a smooth manifold and $g_0, g_1$ are Riemann metrics on $M$. $\newcommand{\eH}{\mathscr{H}}$ Denote by $\eH_{g_i}$, $i=0,1$ and the sheaf of $g_i$-harmonic functions. More precisely for any open set $U\subset M$

$$\eH_{g_i}(U)=\big\{\; f\in C^\infty(U):\;\;\Delta_{g_i} u=0\;\big\}, $$

where $\Delta_{g_i}$ denotes the scalar Laplacian of the metric $g_i$.

Long time ago I proved the following result.

Suppose that $\eH_{g_0}(U)=\eH_{g_1}(U)$, for any open set $U\subset M$.

  • If $\dim M\geq 3$, then there exists $c\in (0,\infty)$ such that $g_1=c g_0$.
     
  • If $\dim M=2$, then there exists a smooth function $f: M\to (0,\infty)$ such that $g_1=fg_0$, i.e., the metrics $g_0$ and $g_1$ live in the same conformal class.

The strong unique continuation property of harmonic functions shows that this statement is really a statement about the stalks of the sheaves $\eH_{g_i}$. Note that these are sheaves of vector spaces, not rings. In dimension $\geq 3$ these sheaves determine the metric up to a multiplicative positive constant.

Suppose that $M$ is a smooth manifold and $g_0, g_1$ are Riemann metrics on $M$. $\newcommand{\eH}{\mathscr{H}}$ Denote by $\eH_{g_i}$, $i=0,1$ and the sheaf of $g_i$-harmonic functions. More precisely for any open set $U\subset M$

$$\eH_{g_i}(U)=\big\{\; f\in C^\infty(U):\;\;\Delta_{g_i} u=0\;\big\}, $$

where $\Delta_{g_i}$ denotes the scalar Laplacian of the metric $g_i$.

Long time ago I proved the following result.

Suppose that $\eH_{g_0}(U)=\eH_{g_1}(U)$, for any open set $U\subset M$.

  • If $\dim M\geq 3$, then there exists $c\in (0,\infty)$ such that $g_1=c g_0$.

  • If $\dim M=2$, then there exists a smooth function $f: M\to (0,\infty)$ such that $g_1=fg_0$, i.e., the metrics $g_0$ and $g_1$ live in the same conformal class.

The strong unique continuation property of harmonic functions shows that this statement is really a statement about the stalks of the sheaves $\eH_{g_i}$. Note that these are sheaves of vector spaces, not rings. In dimension $\geq 3$ these sheaves determine the metric up to a multiplicative positive constant.

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Liviu Nicolaescu
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Suppose that $M$ is a smooth manifold and $g_0, g_1$ are Riemann metrics on $M$. $\newcommand{\eH}{\mathscr{H}}$ Denote by $\eH_{g_i}$, $i=0,1$ and the sheaf of $g_i$-harmonic functions. More precisely for any open set $U\subset M$

$$\eH_{g_i}(U)=\big\{\; f\in C^\infty(U):\;\;\Delta_{g_i} u=0\;\big\}, $$

where $\Delta_{g_i}$ denotes the scalar Laplacian of the metric $g_i$.

Long time ago I proved the following result.

Suppose that $\eH_{g_0}(U)=\eH_{g_1}(U)$, for any open set $U\subset M$.

  • If $\dim M\geq 3$, then there exists $c\in (0,\infty)$ such that $g_1=c g_0$.
  • If $\dim M=2$, then there exists a smooth function $f: M\to (0,\infty)$ such that $g_1=fg_0$, i.e., the metrics $g_0$ and $g_1$ live in the same conformal class.

The unique strong strong unique continuation property of harmonic functions shows that this statement is really a statement about the stalks of the sheaves $\eH_{g_i}$. Note that these are sheaves of vector spaces, not rings. In dimension $\geq 3$ these sheaves determine the metric up to a multiplicative positive constant.

Suppose that $M$ is a smooth manifold and $g_0, g_1$ are Riemann metrics on $M$. $\newcommand{\eH}{\mathscr{H}}$ Denote by $\eH_{g_i}$, $i=0,1$ and the sheaf of $g_i$-harmonic functions. More precisely for any open set $U\subset M$

$$\eH_{g_i}(U)=\big\{\; f\in C^\infty(U):\;\;\Delta_{g_i} u=0\;\big\}, $$

where $\Delta_{g_i}$ denotes the scalar Laplacian of the metric $g_i$.

Long time ago I proved the following result.

Suppose that $\eH_{g_0}(U)=\eH_{g_1}(U)$, for any open set $U\subset M$.

  • If $\dim M\geq 3$, then there exists $c\in (0,\infty)$ such that $g_1=c g_0$.
  • If $\dim M=2$, then there exists a smooth function $f: M\to (0,\infty)$ such that $g_1=fg_0$, i.e., the metrics $g_0$ and $g_1$ live in the same conformal class.

The unique strong unique continuation of harmonic functions shows that this statement is really a statement about the stalks of the sheaves $\eH_{g_i}$. Note that these are sheaves of vector spaces, not rings. In dimension $\geq 3$ these sheaves determine the metric up to a multiplicative positive constant.

Suppose that $M$ is a smooth manifold and $g_0, g_1$ are Riemann metrics on $M$. $\newcommand{\eH}{\mathscr{H}}$ Denote by $\eH_{g_i}$, $i=0,1$ and the sheaf of $g_i$-harmonic functions. More precisely for any open set $U\subset M$

$$\eH_{g_i}(U)=\big\{\; f\in C^\infty(U):\;\;\Delta_{g_i} u=0\;\big\}, $$

where $\Delta_{g_i}$ denotes the scalar Laplacian of the metric $g_i$.

Long time ago I proved the following result.

Suppose that $\eH_{g_0}(U)=\eH_{g_1}(U)$, for any open set $U\subset M$.

  • If $\dim M\geq 3$, then there exists $c\in (0,\infty)$ such that $g_1=c g_0$.
  • If $\dim M=2$, then there exists a smooth function $f: M\to (0,\infty)$ such that $g_1=fg_0$, i.e., the metrics $g_0$ and $g_1$ live in the same conformal class.

The strong unique continuation property of harmonic functions shows that this statement is really a statement about the stalks of the sheaves $\eH_{g_i}$. Note that these are sheaves of vector spaces, not rings. In dimension $\geq 3$ these sheaves determine the metric up to a multiplicative positive constant.

added 28 characters in body
Source Link
Liviu Nicolaescu
  • 34.7k
  • 2
  • 91
  • 165

Suppose that $M$ is a smooth manifold and $g_0, g_1$ are Riemann metrics on $M$. $\newcommand{\eH}{\mathscr{H}}$ Denote by $\eH_{g_i}$, $i=0,1$ and the presheafsheaf of $g_i$-harmonic functions. More precisely for any open set $U\subset M$

$$\eH_{g_i}(U)=\big\{\; f\in C^\infty(U):\;\;\Delta_{g_i} u=0\;\big\}, $$

where $\Delta_{g_i}$ denotes the scalar Laplacian pfof the metric $g_i$.

Long time ago I proved the following result.

Suppose that $\eH_{g_0}(U)=\eH_{g_1}(U)$, for any open set $U\subset M$.

  • If $\dim M\geq 3$, then there exists $c\in (0,\infty)$ such that $g_1=c g_0$.
  • If $\dim M=2$, then there exists a smooth function $f: M\to (0,\infty)$ such that $g_1=fg_0$, i.e., the metrics $g_0$ and $g_1$ live in the same conformal class.

An inspectionThe unique strong unique continuation of the proofharmonic functions shows that this statement is really a statement about the stalks of the sheaves associated to $\eH_{g_i}$. Note that these are sheaves of vector spaces, not rings. In dimension $\geq 3$ these sheaves determine the metric up to a multiplicative positive constant.

Suppose that $M$ is a smooth manifold and $g_0, g_1$ are Riemann metrics on $M$. $\newcommand{\eH}{\mathscr{H}}$ Denote by $\eH_{g_i}$, $i=0,1$ and the presheaf of $g_i$-harmonic functions. More precisely for any open set $U\subset M$

$$\eH_{g_i}(U)=\big\{\; f\in C^\infty(U):\;\;\Delta_{g_i} u=0\;\big\}, $$

where $\Delta_{g_i}$ denotes the scalar Laplacian pf the metric $g_i$.

Long time ago I proved the following result.

Suppose that $\eH_{g_0}(U)=\eH_{g_1}(U)$, for any open set $U\subset M$.

  • If $\dim M\geq 3$, then there exists $c\in (0,\infty)$ such that $g_1=c g_0$.
  • If $\dim M=2$, then there exists a smooth function $f: M\to (0,\infty)$ such that $g_1=fg_0$, i.e., the metrics $g_0$ and $g_1$ live in the same conformal class.

An inspection of the proof shows that this statement is really a statement about the sheaves associated to $\eH_{g_i}$. Note that these are sheaves of vector spaces, not rings. In dimension $\geq 3$ these sheaves determine the metric up to a multiplicative positive constant.

Suppose that $M$ is a smooth manifold and $g_0, g_1$ are Riemann metrics on $M$. $\newcommand{\eH}{\mathscr{H}}$ Denote by $\eH_{g_i}$, $i=0,1$ and the sheaf of $g_i$-harmonic functions. More precisely for any open set $U\subset M$

$$\eH_{g_i}(U)=\big\{\; f\in C^\infty(U):\;\;\Delta_{g_i} u=0\;\big\}, $$

where $\Delta_{g_i}$ denotes the scalar Laplacian of the metric $g_i$.

Long time ago I proved the following result.

Suppose that $\eH_{g_0}(U)=\eH_{g_1}(U)$, for any open set $U\subset M$.

  • If $\dim M\geq 3$, then there exists $c\in (0,\infty)$ such that $g_1=c g_0$.
  • If $\dim M=2$, then there exists a smooth function $f: M\to (0,\infty)$ such that $g_1=fg_0$, i.e., the metrics $g_0$ and $g_1$ live in the same conformal class.

The unique strong unique continuation of harmonic functions shows that this statement is really a statement about the stalks of the sheaves $\eH_{g_i}$. Note that these are sheaves of vector spaces, not rings. In dimension $\geq 3$ these sheaves determine the metric up to a multiplicative positive constant.

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Liviu Nicolaescu
  • 34.7k
  • 2
  • 91
  • 165
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