Saturday, December 22, 2012

Axiomatic definition of the center of mass

This  was prompted by a nice Mathoverflow question.    $\newcommand{\bR}{\mathbb{R}}$  $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\bp}{{\boldsymbol{p}}}$ $\newcommand{\Div}{\mathrm{Div}}$ $\newcommand{\supp}{\mathrm{supp}}$ $\newcommand{\bm}{\boldsymbol{m}}$ $\newcommand{\eC}{\mathscr{C}}$ $\newcommand{\bc}{\boldsymbol{c}}$ $\newcommand{\bq}{{\boldsymbol{q}}}$

We define an effective divisor   on $\bR^N$ to be a   function with finite support $\mu:\bR^N\to\bZ_{\geq 0}$. Its mass, denoted by $\bm(\mu)$, is the  nonnegative integer

$$\bm(\mu)=\sum_{\bp\in\bR^N} \mu(\bp).$$

We denote by $\Div_+(\bR^N)$ the set of effective divisors.  Note that $\Div_+(\bR^N)$ has a natural structure of Abelian semigroup.

For any $\bp\in\bR^N$ we denote by $\delta_\bp$ the Dirac divisor of mass $1$ and supported at  $\bp$.   The  Dirac divisors generate  the  semigroup $\Div_+(\bR^N)$.     We have a natural  topology on  $\Div_+(\bR^N)$ where $\mu_n\to \mu$ if and only if

$$\bm(\mu_n)\to \bm(\mu),\;\; {\rm dist}\,\bigr(\;\supp(\mu_n),\; \supp(\mu)\;\bigr)\to 0,$$

where ${\rm dist}$ denotes the Haudorff distance.

A center of mass   is a map

$$\eC:\Div_+(\bR^N)\to\Div_+(\bR^N)$$

satisfying the following conditions.

1. (Localization) For any divisor $\mu$ the support of $\eC(\mu)$ consists of  a single point $\bc(\mu)$.

2.  (Conservation of mass)

$$\bm(\mu)=\bm\bigl(\;\eC(\mu)\;\bigr),\;\;\forall\mu \in\Div_+(\bR^N),$$

so that

$$\eC(\mu)=\bm(\mu)\delta_{\bc(\mu)},\;\;\forall\mu \in\Div_+(\bR^N).$$

3. (Normalization)

$$\bc(m\delta_\bp)=\bp,\;\;\bc(\delta_\bp+\delta_\bq)=\frac{1}{2}(\bp+\bq),\;\;\forall \bp,\bq\in\bR^N,\;\;m\in\bZ_{>0}.$$

$$\eC(\mu_1+\mu_2)= \eC\bigl(\,\eC(\mu_1)+\eC(\mu_2)\,\bigr),\;\;\forall \mu_1,\mu_2\in \Div_+(\bR^N).$$

For example, the   correspondence

$$\Div_+ \ni \mu\mapsto \eC_0(\mu)=\bm(\mu)\delta_{\bc_0(\mu)}\in\Div_+,\;\;\bc_0(\mu):=\frac{1}{\bm(\mu)}\sum_\bp \mu(\bp)\bp$$

is a center-of-mass  map.  I want to show that this is the only center of mass map.

Proposition  If $\eC:\Div_+(\bR^N)\to \Div_+(\bR^N)$ is a  center-of-mass map, then $\eC=\eC_0$.

Proof.      We carry the proof in several steps.

Step 1 (Rescaling).   We can write the additivity property as

$$\bc(\mu_1+\mu_2) =\bc\bigl(\; \bm(\mu_1)\delta_{\bc(\mu_1)} +\bm(\mu_2)\delta_{\bc(\mu_2)}\;\bigr).$$

In particular, this implies that the rescaling property

$$\bc( k\mu)=\bc(\mu),\;\;\forall\mu \in\Div_+,\;\; k\in\bZ_{>0}. \tag{R}\label{R}$$

This follows by induction $k$. For $k=1$ it is obviously true.  In general

$$\bc( k\mu)=\bc\bigl(\;(k-1)\bm(\mu)\delta_{\bc(\;(k-1)\mu)}+\bm(\mu)\delta_{\bc(\mu)}\;\bigr) =\bc\bigl(\; k\bm(\mu)\delta_{\bc(\mu)}\;\bigr)={\bc(\mu)}$$

Step 2. (Equidistribution)   For any $n>0$ and any collinear points $\bp_1,\dotsc,\bp_n$ such that

$$|\bp_1-\bp_2|=\cdots=|\bp_{n-1}-\bp_n|$$

we have

$$\eC\Bigl(\sum_{k=1}^n\delta_{\bp_k}\;\Bigr)=\eC_0\Bigl(\sum_{k=1}^n\delta_{\bp_k}\;\Bigr) \tag{E}\label{E}.$$

Equivalently, this means that

$$\bc\Bigl(\sum_{k=1}^n\delta_{\bp_k}\;\Bigr)=\bc_0\bigl(\sum_{k=1}^n\delta_{\bp_k}\;\bigr)={\frac{1}{n}(\bp_1+\cdots+\bp_n)}.$$

We  will prove  (\ref{E}) arguing by induction on $n$. For $n=1,2$ this follows from the normalization property. Assume that (\ref{E}) is true for any $n< m$. We want to prove it is true for $n=m$.

We distinguish two cases.

(a)  $m$ is even, $m= 2m_0$. We set

$$\mu_1=\sum_{j=1}^{m_0} \delta_{\bp_j},\;\;\mu_2=\sum_{j=m_0+1}^{2m_0}\delta_{\bp_j}.$$

Then

$$\bc(\mu_1+\mu_2)= \bc\bigl(\; m_0\delta_{\bc(\mu1)}+m_0\delta{\bc(\mu_2)}\;\bigr) =\bc\bigl( \delta_{\bc(\mu_1)}+\delta_{\bc(\mu_2)}\;\bigr). \tag{1}\label{2}$$

By induction

$$\bc(\mu_1)=\bc_0(\mu_1),\;\;\bc(\mu_2)=\bc_0(\mu_2).$$

The normalization  condition now implies that

$$\bc\bigl( \delta_{\bc(\mu_1)}+\delta_{\bc(\mu_2)}\;\bigr)=\bc_0\bigl( \delta_{\bc_0(\mu_1)}+\delta_{\bc_0(\mu_2)}\;\bigr).$$

Now run the  arguments in (\ref{2}) in reverse, with $\bc$ replaced by $\bc_0$.

(b) $m$ is odd, $m=2m_0+1$.  Define

$$\mu_1=\delta_{\bp_{m_0+1}},\;\;\mu_2'=\sum_{j<m_0+1}\delta_{\bp_j},\;\;\mu_2''=\sum_{j>m_0+1}\delta_{\bp_j},\;\;\mu_2=\mu_2'+\mu_2''.$$

(Observe that $\bp_{m_0+1}$ is the mid-point in the string of equidistant collinear points $\bp_1,\dotsc,\bp_{2m_0+1}$. ) We have

$$\eC(\mu_2'+\mu_2'')=\eC\bigl( \; \eC(\mu_2')+\eC(\mu_2'')\;\bigr).$$

By induction

$$\eC(\mu_2')+\eC(\mu_2'')= \eC_0(\mu_2')+\eC_0(\mu_2'') =m_0\delta_{\bc_0(\mu_2')}+m_0\delta_{\bc_0(\mu_2'')}=m_0\bigl(\;\delta_{\bc_0(\mu_2')}+\delta_{\bc_0(\mu_2'')}\;\bigr).$$

Observing that

$$\frac{1}{2}\bigl(\bc_0(\mu_2')+\bc_0(\mu_2'')\;\bigr)=\bp_{m_0+1}$$

we deduce

$$\eC(\mu_2)= \eC(\mu_2'+\mu_2'')=m_0\eC\bigl( \delta_{\bc_0(\mu_2')}+\delta_{\bc_0(\mu_2'')}\;\bigr)=m_0\eC_0\bigl( \delta_{\bc_0(\mu_2')}+\delta_{\bc_0(\mu_2'')}\;\bigr)=2m_0\delta_{\bp_{m_0+1}}=2m_0\mu_1.$$

Finally  we deduce

$$\eC(\mu)=\eC\bigl(\;\eC(\mu_1)+\eC(\mu_2)\;\bigr)=\eC\bigl(\;\eC(\mu_1)+2m_0\eC(\mu_1)\;\bigr)= (2m_0+1)\delta_{\bp_{m_0+1}}=\eC_0(\mu).$$

Step 3. (Replacement) We will show that for any distinct points $\bq_1,\bq_2$ and any positive integers $m_1,m_2$ we  can find  $(m_1+m_2)$ equidistant points  $\bp_1,\dotsc,\bp_{m_1+m_2}$ on the line determined by $\bq_1$ and $\bq_2$  such that

$$m_1\delta_{\bq_1}=\eC_0\Bigl(\sum_{j=1}^{m_1} \delta_{\bp_j}\Bigr)=\eC\Bigl(\sum_{j=1}^{m_1} \delta_{\bp_j}\Bigr),\;\;\;m_2\delta_{\bq_2}=\eC_0\Bigl(\sum_{j=m_1+1}^{m_1+m_2} \delta_{\bp_j}\Bigr)=\eC\Bigl(\sum_{j=m_1+1}^{m_1+m_2} \delta_{\bp_j}\Bigr).$$

This is elementary. Without  restricting the generality we can assume  that $\bq_1$ and $\bq_2$ lie on an axis (or geodesic) $\bR$ of $\bR^N$, $\bq_0=0$ and $\bq_2=q>0$.       Clearly we can find  real numbers $x_0, r$, $r>0$, such that
$$\frac{1}{m_1}\sum_{j=1}^{m_1}(x_0+j r)=0,\;\;\frac{1}{m_2}\sum_{j=m_1+1}^{m_1+m_2}(x_0+jr)=q.$$

Indeed, the above two equalities can be rewritten as

$$x_0+\frac{m_1+1}{2} r=0,$$

$$q=x_0 +m_1 r+\frac{m_2+1}{2}=x_0+\frac{m_1+1}{2} r+\frac{m_1+m_2}{2} r.$$

Now place the points  $\bp_j$ at the locations $x_0+jr$.

Step 4. (Conclusion)   We argue on  by induction on mass that

$$\eC(\mu)=\eC_0(\mu),\;\;\forall \mu\in \Div_+\tag{2}\label{3}$$

Clearly, the normalization condition  shows that (\ref{3})  is true if $\supp\mu$ consists of a single point, or if  $\bm(\mu)\leq 2$.

In general if $\bm(\mu)>2$ we write $\mu=\mu_1+\mu_2$ where $m_1=\bm(\mu_1),m_2=\bm(\mu_2)<\bm(\mu)$.

By induction we have

$$\eC(\mu)= \eC\bigl( \eC(\mu_1)+\eC(\mu_2)\bigr)=\eC(\;\eC_0(\mu_1)+\eC_0(\mu_2)\;\bigr).$$

If $\bc_0(\mu_1)=\bc_0(\mu_2)$  the divisors $\eC_0(\mu_1)$ $\eC_0(\mu_2)$ are supported at the same point and we are done. Suppose that  $\bq_1=\bc_0(\mu_1)\neq\bc_0(\mu_2)=\bq_2$. By Step 3, we can  find     equidistant points $\bp_1,\dotsc,\bp_{m_1+m_2}$ such that

$$m_1\delta_{\bq_1}=\eC\Bigl(\sum_{j=1}^{m_1} \delta_{\bp_j}\Bigr)= \eC_0\Bigl(\sum_{j=1}^{m_1} \delta_{\bp_j}\Bigr)$$
$$m_2\delta_{\bq_2}=\eC\Bigl(\sum_{j=m_1+1}^{m_1+m_2} \delta_{\bp_j}\Bigr)=\eC_0\Bigl(\sum_{j=m_1+1}^{m_1+m_2} \delta_{\bp_j}\Bigr).$$

We deduce that

$$\eC(\mu)=\eC\Bigl(\sum_{k=1}^{m+1+m_2}\delta_{\bp_k}\Bigr),\;\; \eC_0(\mu)=\eC_0\Bigl(\sum_{k=1}^{m+1+m_2}\delta_{\bp_k}\Bigr).$$

The conclusion now follows from  (\ref{E}).  q.e.d

Remark.    The above proof    does not really use the  linear structure. If we uses only the fact that any two points in $\bR^N$ determine a unique geodesic.  The  Normalization condition can be replaced by the equivalent one

$$\bc(\delta_\bp+\delta_\bq)= \mbox{the midpoint of the geodesic segment [\bp,\bq]}.$$

If we replace $\bR^N$ with a hyperbolic space the same arguments show that  there exists at most  one center of mass map.