text:
\begin{document}

\newtheorem{thm}{Theorem}
\newtheorem{prop}{Proposition}
\newtheorem{coroll}{Corollary}
\newtheorem{defn}{Definition}
\newtheorem{lemma}{Lemma}
\title{Infinitesimal aspects of the Laplace operator}
\author{Anders Kock}
\date{}

\maketitle

In the context of synthetic differential
geometry, we study the Laplace ope\-rator an a Riemannian
manifold. The main new aspect is a neighbourhood of the
diagonal, smaller than the  second neighbourhood usually
required as support for second order differential operators.
The new neighbourhood has the property that a function is affine on it
if and only if it is harmonic.

 \section*{Introduction} Recall \cite{FM}, \cite{SDG}, \cite{Levi-Civita},
\cite{DFIC}  that any manifold $M$, when seen in a
model of Synthetic Differential Geometry (SDG), carries a reflexive
symmetric relation $\sim _k$ ($k=0,1,2,...$), where $x\sim _k y$ reads
``$x$ and $y$ are $k$-neighbours''; $x \sim _0 y$ means $x=y$; $x\sim _k y
$ implies $x\sim _{k+1} y$.  Also, $x\sim _k y , y\sim _l z$ implies
$x\sim_{k+l} z$.  The set of $(x,y)\in
M\times M$ with $x\sim _k y$ is denoted $M_{(k)}$, the ``$k$'th
neighbourhood of the diagonal", and for fixed $x$, the set $\{y \in M
\mid y\sim _k x \}$ is denoted ${\cal M}_k (x)$ (``the $k$-monad
around $x$").
In $R^n$, ${\cal M}_k (0)$ is denoted $D_k (n)$.
Its elements $u$ are characterized by the condition that any
homogeneous polynomial of degree $k+1$ vansihes on $u$.
--- In this context, a {\em Riemannian metric} on $M$ can
be given in terms of a map
$$g: M_{(2)} \to R$$
with $g(x,x)=0$, and with $g(x,y)= g(y,x)$ to be thought of as
the ``square-distance between $x$ and $y$", se
\cite{Levi-Civita}, \cite{VF}. (Also $g$ should be
positive-definite, in a certain sense.)

Given a Riemannian metric $g$ on $M$, in this sense, one can
construct the Levi-Civita connection \cite{Levi-Civita},
volume form \cite{VF}, and hence also a notion of {\em
divergence} of a vector field. And to a function $f:M \to R$,
one can construct its {\em gradient vector field}, and hence
one can construct the Laplacian $\Delta$ by $\Delta (f) = div
(grad (f))$. This is what we shall {\em not} do here, rather,
we shall exploit the richness of synthetic language to give a
more economic and more geometric construction of $\Delta$. The
construction is more economic in the sense that the definition
of $\Delta f (x)$ only depends on knowing $f$ on a certain
subset ${\cal M}_{L} (x) \subseteq {\cal M}_{2}(x)$, where ${\cal
M}_{2}(x)$ is what is required to make the usual $div \; grad $
construction work, or for defining the individual terms in the formula
$\Delta f (x) = \sum \partial ^2 f /\partial x_i ^2 (x)$.

The description of ${\cal M}_L  (x) \subseteq M$, or equivalently, the
description of the $L$-neighbour relation $\sim _L$, is
coordinate free, see Definition \ref{crucdef} below, and therefore,
too, is the description of $\Delta f$ and of the notion of harmonic
function.  We get a characterization of harmonic functions, in terms
of an average-value property, which is infinitesimal in character and
does not involve integration, see Theorem \ref{Delta} and Proposition
\ref{seven}.

\medskip

In Section 3 we prove that diffeomorphisms which preserve the
$L$-neigh\-bour relation are precisely the conformal ones.  Section 4
deals with the special case of the complex plane, and Section 5
explains the ``support of the Laplacian'' in systematic algebraic
terms.

\section{Preliminaries}
Although the notions we use are introduced in a coordinate
free way, we have no intention of avoiding use of coordinates
as a tool of proof. This Section contains in fact mainly
certain coordinate calculations, which we believe will be
useful also in other contexts where Riemannian geometry is
treated in the present synthetic manner.

Working in coordinates in $M$ means that we are identifying (an open
subset of) $M$ with (an open subset of) $R^n$; for simplicity, we talk
about thse open subsets as if they were all of $M$ and $R^n$,
respectively; all our considerations are anyway only {\em local}.  The
Riemannian metric $g$ on $M$ then becomes identified with a Riemannian
metric on $R^n$, likewise denoted $g$, and it may be written (for
$x\sim _2 y$) in the form of a matrix product,
$$g(x,y)= (y-x)^T \cdot G(x)\cdot  (y-x),$$
where $x-y\in R^n$ is viewed as a column matrix, and $G(x)$,
for each $x$, is a symmetric positive definite $n\times n$ matrix.

Using coordinates, we may form affine combinations of (the coordinate
sets of) points of $M$, at least for sufficiently nearby points, and
such combinations will in general have only little geometric
significance, since they depend on the choice of the coordinate
system.  However, we have the following useful fact:

\begin{prop} Assume $y_1 \sim _1 x$ and $y_2 \sim _1 x$ (so
$(x+ y_2 - y_1 )
\sim _2 x $).  Then
$$g(x, x+ y_2 - y_1 ) = g(y_1 , y_2 );$$
in particular, for $x=0$,
$$ g(0, y_2 - y_1 ) = g(y_1 , y_2 ).$$
\label{simple}\end{prop}

{\bf Proof.} We may assume $x=0$.  Then $g(0, y_2 - y_1 )$ and $g(y_1
,y_2 )$ are given, respectively, by
$$(y_2 - y_1 )^T \cdot G(0) \cdot (y_2 - y_1 ) = -2 y_1 ^T \cdot G(0) \cdot y_2$$
and
$$(y_2 - y_1 )^T \cdot G(y_1 ) \cdot (y_2 - y_1 ) = -2 y_1 ^T \cdot
G(y_1 ) \cdot y_2 .$$
Now expand $G(y ) $ as $G(0) + H(y )$ where $H$ depends linearily on
$y \sim _1 0$.  The difference between our two expressions is then
$-2 y_1 ^T \cdot H(y_1 ) \cdot y_2$, which depends bilinearily on
$y_1$ and therefore vanishes.

\medskip

We shall see below (Proposition \ref{good}) that the ``coordinatewise'' affine
combination considered in Proposition \ref{simple} {\em does } have an
invariant geometric meaning, provided the coordinate system is {\em
geodesic}:

We say that
the metric $g$ on $R^n$ (or equivalently, the coordinate
system around $x_0 \in M$) is {\em geodesic} at $0\in R^n$ (or
at $x_0 \in M$, respectively), if the first partial derivatives of
$G(x)$, as  functions of $x\in R^n$, vanish at $0$;
equivalently, if $G(x)= G(0)$ for every $x\sim _1 0$. (This is
in turn equivalent to the vanishing at $x_0$ of the Christoffel symbols
of the metric, in the given  coordinate system.) It is
classical that for every point $x_0$, there exists a
coordinate system which is geodesic at $x_0$. If
$G(0) $ is the identity matrix, one talks about a
{\em normal} coordinate system at $x$, and such also exist.
Cf.\ e.g.\ \cite{CCL} for such notions.

Recall from \cite{Levi-Civita} formula (2) that any Riemannian metric $g: M_{(2)}
\to R$ admits a unique symmetric extension $\overline{g} : M_{(3)} \to R$;
in coordinates it is given by
\begin{equation} \overline{g} (x,y) = (y-x)^T \cdot ( G(x) + 1/2
(D_{(y-x)}G)(x))\cdot (y-x).\label{corr}\end{equation}

 Recall also from \cite{Levi-Civita} Theorem 3.6 that for $x\sim _2 z$ in a
 Riemannian manifold, and for $t\in R$, there exists a unique $y_0$
 with $y_0 \sim _2 x$ and $y_0 \sim _2 z$ which is a critical point for the
 function of $y$ given by  \begin{equation}t\overline{g} (x,y) + (1-t)
 \overline{g} (z,y);
\label{crit}\end{equation}

We call this $y_0$ an (intrinsic) {\em affine combination} of $x$ and $z$.  We
write it $tx + (1-t)z$; this raises a compatibility problem in case we
are working in coordinates, since we can then also form the
``algebraic'' affine combination of two coordinate $n$-tuples.
However, in geodesic coordinates at $x$, there is no problem,
according to the following Proposition, which extends Proposition 3.7
in \cite{Levi-Civita}.  Let us consider a coordinate system with $x$
identified with $0$.

\begin{prop} The critical point $y_0$ for the function in
(\ref{crit}) is the algebraic affine combination $tx +
(1-t)z$, if either $x\sim _1 z$, or if the coordinate system is
geodesic at $x$.
\label{alg}\end{prop}

{\bf Proof.} Since $x$ is identified with $0$ in the
coordinate system, the affine combination in question is just
$(1-t)z$. To show that it is a critical value for (\ref{crit})
means that
\begin{equation}t\overline{g}(0,(1-t)z +v ) +(1-t)
\overline{g}(z,(1-t)z +v)\label{hh}\end{equation}
 is independent of $v\sim _1
0$.  We write $g$ in terms of the symmetric matrices $G$, as above.
Let us take a Taylor expansion of the function $G(y)$, writing
$$G(y) = G(0) + H(y),$$
where the entries of the matrix $H(y)$ are of degree $\geq 1$
in $y$; and if the coordinate system is geodesic at $x=0$,
$H(y)$ is even of degree $\geq 2$ in $y$. We then calculate.
We get a ``significant'' part from each of the two terms in (\ref{hh}), and then
some ``error'' terms, each of which will vanish for degree reasons, as
we shall argue.

The two significant terms are the two terms in

$$t (((1-t)z +v)^T \cdot G(0)\cdot  ((1-t)z +v)) +
(1-t)( (-tz +v)^T \cdot G(0) \cdot (-tz+v) ).$$

Expanding out by bilinearity and symmetry, the terms involving
$v$ linearly cancel each other; and the terms involving $v$ quadratically
vanish because $v\sim _1 0$.  So the significant terms, jointly, do
not depend on $v\sim _1 0$.

The ``error'' terms are of two kinds: partly, arising from the
replacement of $G(z)$ by $G(0)$; here, $H(z)$ enters; and partly there are
correction terms when passing from $g$ to $\overline{g}$ defined on
pairs of third order neighbours.  The error term of the first kind is
a multiple of
$$(-tz+v)^T \cdot H(z)\cdot  (-tz+v);$$
we expand this out by bilinearity, and use that $H(z)$ is
of degree $\geq 1$, and $v\sim _1 0$.  We get four terms each
of  which vanish for
degree reasons if {\em either} $z\sim _1 0$ {\em or} if $H(z)$ is of degree
$\geq 2$.

Finally, the correction  terms for upgrading $g$ to
$\overline{g}$ don't occur if $z\sim _1 0$, since then $g$ is
only applied to pairs of second order neighbours. Thus the
assertion of the Proposition about the case $z\sim _1 x$ is
already proved. In general, the upgrading involves first
partial derivatives of $G$, (see (\ref{corr})), so in the
case the coordinate system is geodesic at $0$, no correction term is
needed for $\overline{g}(0,(1-t)z)$, but only for $\overline{g}(z,
(1-t)z+v)$.  Using the formula (\ref{corr}), we see that the correction needed
is a certain multiple of
$$(-tz + v)^T \cdot ( D_{-tz+v}G)(z) \cdot (-tz+v),$$
hence a linear combination of terms
$$z\cdot D_z G(z)\cdot z,\; v\cdot D_z G(z)\cdot z,\; z \cdot D_v G(z) \cdot z,$$
and something that contains $v$ in a bilinear way.  All these terms
vanish for degree reasons: for, since $H$ vanishes in the first
neighbourhood of $0$, $D_v G(z)$ is of degree $\geq 1$ in $z$, and
$D_z G(z)$ is even of degree $\geq 2$ in $z$.

\medskip

Essentially the same degree counting as in  this
proof gives the following result:
\begin{lemma}Let $y\sim _1 x$ and $z\sim
_2 x$; then using a geodesic coordinate system at $x=0$, the
quantity
$\overline{g}(y,z)$ may be calculated as $(z-y)^T \cdot G(0)
\cdot (z-y)$.
\label{yz}\end{lemma}

\medskip

Given a Riemannian manifold. If $x\sim _2 z$, the {\em mirror
image} $z'$ of $z$ in $x$ is by definition the affine
combination $2x - y$, i.e.\ the $y$ which is critical value for
$2\overline{g}(x,y) - \overline{g}(z,y)$,
\cite{Levi-Civita} Theorem 3.6. Also, the parallelogram
formation $\lambda$ is descibed in \cite{Levi-Civita}.
Finally, if $t$ is a tangent vector $D\to M$, its geodesic
prolongation $\overline{t}:D_2 \to M$ is determined by the validity, for
all $d_1 , d_2 \in D$ of $$\overline{t}(d_1 + d_2 ) = \lambda (t(0), t(d_1 ),
t(d_2
)).$$
(Recall that $D\subseteq R$ are the elements of square zero, $D_2$ the
elements of cube zero.) Now Proposition \ref{alg} has the following Corollary:

\begin{prop}
Let $x\sim _2 z$; then the mirror image $z'$ of $y$ w.r.to $x$
may be calculated as follows: take  a geodesic
coordinate system at $x$ with $x=0$. Then $z' = -z$.

Let $y\sim _1 x$, $z\sim _1 x$. Then $\lambda (x,y,z)$ may be
calculated as follows: take  a geodesic
coordinate system at $x$ with $x=0$. Then $\lambda (x,y,z) =
y+z$.

Let $t$ be a tangent vector $D\to M$ at $x\in M$. Then the
geodesic prolongation $\overline{t} :D_2 \to M$ of $t$ may be
calculated as follows: take  a geodesic
coordinate system at $x$ with $x=0$. Let $u$ be the unique
vector in $R^n$ so that $t(d)= d\cdot u$ for all $d\in D$.  Then for
$\delta \in D_2$, $\overline{t}(\delta) = \delta \cdot u$
\label{good}\end{prop}

(The vector $u \in R^n$ appearing in the last clause is usually called
the {\em principal part} of $t$, relative to the coordinate system.)

If $t$ and $s$ are tangent vectors at the same point $x$ of a
Riemannian manifold $M,g$, we define their inner product
 $<t,s>$
by the validity, for all $d_1 d_2 \in D$, of \begin{equation}d_1 d_2 \;<t,s> =
-\frac{1}{2} g(t(d_1 ), s(d_2 )).
\label{inner}\end{equation}
In this way, the tangent
vector space $T_x M$ is made into an inner product space, (and this is
the contact point with the classical formulation of Riemannian metric).

If $u$ and $v$ are the principal parts of tangent vectors $t$ and $s$
at $x\in M$, in some coordinate system at $x=0$ (not necessarily geodesic),
one has \newline $<t,s> = u^T \cdot G(0) \cdot v$; this follows easily from
Proposition \ref{simple}.

Combining Lemma \ref{yz} and Proposition \ref{good}, one gets

\begin{lemma} Let $t$ be a tangent vector.  Then for $d\in D$, $\delta
\in D_2$, we have $$\overline{g} (t(d), \overline{t}(\delta ))
=(\delta ^2 -
2d\delta )\cdot <t,t>.$$
\label{geo1}\end{lemma}

We are going to define the {\em orthogonal projection} of $z$
($z\sim _2 x$) onto a proper tangent $t$ at $x$.  We first define the {\em
scalar component} of $z$ along $t$; this is unique number $\alpha (z,
t)$ so that
\begin{equation}
d\cdot \alpha (z,t) = \frac{1}{2} \frac{(g(x,z)-
\overline{g}(t(d),z))}{<t,t>}\label{sca}\end{equation} for all $d\in
D$.  Note that if $z =x$, $\alpha (z,t)=0$, and from this follows that
for any $z\sim _2 x$, $\alpha (z,t) \sim _2 0$, in other words $\alpha
(z,t)\in D_2$.  From Lemma \ref{geo1}, applied twice (once with
$d=0$, once with a general $d\in D$), it is immediate to deduce that if
$z$ is of the form $\overline{t}(\delta )$ for a $\delta \in D_2$,
then $\alpha (z, t) = \delta $.

We define the orthogonal projection $proj_{t} (z)$ by $$proj_{t}
(z)=\overline{t} (\alpha (z,t)).$$ Note that it is a second-order
neighbour of $x$.  It follows  from the above that if $z$
is of the form $\overline{t}(\delta )$, then $proj _t (z)=z$.

 \section{Laplacian neighbours}

Here is the crucial definition:

\begin{defn}Let $z\sim _2 x$. We say that $z$ is a {\em  Laplacian
neighbour}  of $x$ (written $z\sim _L x$) if for
every proper tangent $t$ at $x$, we have \begin{equation}g(x,z)= n
\cdot g(x, proj_t (z)),\label{cruceqn}\end{equation} where $n$ is the
dimension of the manifold.
\label{crucdef}\end{defn}

Maybe one of the names ``isotropic, harmonic, or conformal, neighbour"
would be more appropriate.

\medskip

Clearly $z\sim _1 x$ implies $z\sim _L x$; for if $z$ is a
first-order neighbour of $x$, then so is its
orthogonal projection, and hence both the $g$-quantities to be
compared in (\ref{cruceqn}) are zero.  If the dimension $n$ is 1, $\sim _L$
is the same as $\sim _2$ but in general, the set ${\cal M}_{L} (x)$ of
$L$-neighbours of $x $ is much smaller than the set ${\cal M}_2 (x)$ of
second-order neighbours; in fact, the ring of functions on ${\cal M}_{L} (x)$
is a finite dimensional vector space which is just one dimension
bigger than the ring of functions on ${\cal M}_{1} (x)$, as we shall see in
the proof of Proposition \ref{uni-c} below.

  We conjecture that the relation $\sim _L$ is symmetric, but we haven't
  been able to do the necessary calculations, except in the
  case of $R^n$, where the symmetry is easy to prove, using
  Proposition \ref{five} below.

Note the following curious phenomenon in dimension $n\geq 2$:
 if $z\sim _L x$, then $z$ does not connect to $x$ by
{\em any} geodesic $D_2\to M$ (given by a proper tangent vector $t$),
except perhaps in the trivial case when  $g(x,z)=0$.  In
other words, the $L$-neighbours of $x$ are genuinely {\em isotropic}, in the
sense that they are in {\em no} preferred direction $t$ (hence the
alternative name ``isotropic neighbour'' suggested).  Nevertheless, there
are sufficiently many $L$-neighbours of $x$ to define the Laplacian
differential operator $\Delta$, see Theorem \ref{Delta} below.

  Let us assume the manifold
in question has dimension $n$.  Then we have

\begin{prop}In any geodesic normal coordinate system at $x=0$, $z=(z_1 ,
\ldots ,z_n )$ is $\sim _L 0$ if and only if
 $$z_i ^2 = z_j ^2 \mbox{ for all $i,j$, and } z_i z_j =0 \mbox{ for }i\neq j$$
 (and $z_i ^3 =0$ for all $i$; this latter condition follows from the
 other two if $n\geq 2$).
 \label{five}\end{prop}
{\bf Proof.} First, if $t$ and $s$ are tangent vectors at $x=0$ with principal
parts $u$ and $v$, respectively (meaning $t(d) = du,
s(d)=dv$), then
$<t,s>= u\bullet v$, where $\bullet$ denotes the usual dot product of
vectors in $R^n$.  Also, if $t$ is a tangent at $x=0$ with principal
part $u$, then $\alpha (z,t ) = (z\bullet u)/(u\bullet u)$; for,
calculating the enumerator in (\ref{sca}) gives (using Proposition
\ref{simple}) $$z\bullet z - \overline{g}(du,z)= z\bullet z -
(z-du)\bullet (z-du) = 2d\; z\bullet u.$$ From the third clause in
Proposition \ref{good}, we then get the familiar looking
\begin{equation}proj _t (z) = \frac{z\bullet u}{u\bullet u} u.
\label{familiar}\end{equation}

In particular, if $t$ is the (proper) tangent vector with principal part $e_i
\in R^n$ ($= (0, \ldots, 1, \ldots 0)$ (with 1 in the $i$'th position,
$0$'s elsewhere), then $proj _t (z_1 , \ldots ,z_n ) = z_i e_i$.  In
particular $g(0, proj _t (z)) = z_i ^2$.  If, on the other hand, $t$
is the tangent vector with principal part $e_{i,j}$ (the vector
with $1$'s in the $i$'th and in the $j$'th position, $i\neq j$, $0$'s
elsewhere), then $proj _t (z)$ has $(z_i +z_j)/2$ in the $i$'th and in
the $j$'th position, and $0$'s elsewhere.  In particular, $$g(0, proj
_t (z))=\frac{1}{2}(z_i ^2 + z_j ^2) + z_i z_j .$$ If $z$ therefore is
an $L$-neighbour of $0$, we conclude that $z_i ^2 = z_j ^2$ for all
$i,j$, and that $z_i z_j =0$ if $i\neq j$.

Conversely, assume that in some geodesic normal coordinate system
at $x=0$, the coordinates $(z_1 , \ldots ,z_n )$ satisfy the equations
$z_i ^2 = z_j ^2$, $z_i z_j =0$ for $i\neq j$, and let $t$ be a proper
tangent vector at $x$ with principal part $u =(u_1 , \ldots ,u_n )$.
Then
 $$proj _t (z) = \frac{z\bullet u}{u\bullet u} u,$$
 and therefore
 $$g(x,proj _t (z)) = (\frac{z\bullet u}{u\bullet u} u)\bullet
( \frac{z\bullet u}{u\bullet u} u),$$
which we calculate by arithmetic to be
$$\frac{(\sum _i u_i z_i )(\sum _j u_j z_j )}{u \bullet u}= \frac{\sum
_{ij } u_i u_j z_i z_j}{u\bullet u},$$
but since $z_i z_j =0$ for $i\neq j$, only the ``diagonal'' terms
survive, and we are left with
$$\frac{\sum _i u_i u_i z_i z_i}{u\bullet u}.$$
But $z_i z_i = z_1 z_1$ for all $i$, so this factor can go outside the
sum sign in the enumerator, and we
get $z_1 ^2 (\sum_i u_i u_i)/u\bullet u = z_1 ^2$, which is $1/n$
times $\sum z_i ^2$ since all the $z_i ^2$ are equal.  This proves the
Proposition.

\medskip

From  Propositions \ref{good} and \ref{five}, one
immediately deduces that if $z\sim _L x$, then also $z'\sim _L x$ for
any affine combination $z' = t x + (1-t )z$ ($t \in R$).

\begin{prop}If two functions $f_1$ and $f_2$: ${\cal M}_L (x)\to R$ agree on
${\cal M}_1 (x)$ there is a unique number $c\in R$ such that for all $z\sim
_L x$
$$f_1 (z)-f_2 (z) =c\cdot g(x,z).$$
\label{uni-c}\end{prop}

{\bf Proof.}
 Using a geodesic normal
coordinate system at $x=0$, it is a matter of analyzing the ring of functions
${\cal M}_L (0) \to R$ for the case where $M= R^n$ with standard
inner-product metric.  The Proposition gives that ${\cal M}_L (0)$ may be
described as $D_L (n) \subseteq R^n$, defined by
\begin{equation}D_L (n):= \{ (d_1 , \ldots , d_n )\in R^n \mid d_i ^2 = d_j ^2
\mbox{, and } d_i d_j =0 \mbox{ for } i\neq j \}
,\label{DLN}\end{equation} (for $n\geq 2$; for $n=1$, $D_L (1) = D_2 =
\{\delta \in R \mid \delta ^3 = 0\}$).  This is (for $n\geq 2$) the
object represented by the Weil algebra ${\cal O}(D_L (n)):=k[Z_1 ,
\ldots , Z_n ]/I$, where $I$ is the ideal generated by the $Z_i ^2 -
Z_j ^2$, and by $Z_i Z_j$ for $i\neq j$.  It is immediate to calculate
that, as a vector space, this ring is $(n+2)$-dimensional, with
linear generators $$1, Z_1 , \ldots ,Z_n , Z_1 ^2 + \ldots + Z_n ^2 .$$
By the general (Kock-Lawvere) axiom scheme for SDG \cite{SDG}, \cite{lav},
this means that any function $f:D_L (n) \to R$ is of the form
$$f(z_1 , \ldots , z_n ) = a + \sum _i b_i z_i + c (\sum _i z_i ^2 ),$$
for unique $a, b_1 , \ldots ,b_n , c \in R$, or equivalently
$$f(z_1 , \ldots , z_n ) = a + \sum _i b_i z_i + c g(0,z).$$
Since the restriction of $f$ to $D(n)$ is given by the data $a, b_1 ,
\ldots ,b_n$, the unique existence of $c$ follows.

\medskip

The following Theorem deals with an arbitrary Riemannian manifold $M,
g$ of dimension $n$, and gives a coordinate free characterization of the
Laplacian operator $\Delta$.

\begin{thm}For any $f:{\cal M}_L (x)\to R$, there is a unique number $L$
with the property that for any $z\sim _L x$
$$f(z) + f(z') -2f(x) = L\cdot g(x,z),$$
where $z'$ denotes the mirror image of $z$ in $x$. We write
$\Delta f (x) := n L$.

\label{Delta}\end{thm}

Put differently,
$$f(z) + f(z') -2f(x) = \frac{\Delta f (x)}{n} g(x,z).$$
If the function $f$ is {\em harmonic} at $x$, meaning that $\Delta f
(x) =0$, it follows that it has a strong {\em average value property}:
the value at $x$ equals the average value of $f$ over any pair of
points $z$ and $z'$ ($L$-neighbours of $x$) which are symmetrically
located around $x$.

\medskip

{\bf Proof.} Again, we pick a normal geodesic coordinate system at
$x=0$, so identify ${\cal M}_L (x)$ with $D_L (n)$; then $z'$ gets identified
with $-z$, by Proposition \ref{good}.  The left hand side of the
expression in the Theorem then has restriction 0 to $D(n)$, being
(with notation as above) $(a + \sum b_i z_i )+(a + \sum b_i (-z_i )) -
2a.$ Hence the unique existence of $L$ follows from
Proposition \ref{uni-c}.

\medskip

The following Proposition serves to as partial justification of the
 use of the name ``Laplacian'' for the $\Delta $ considered in the
Theorem. We consider the standard Riemannian metric on $R^n$,
$g(x,z)= ||z-x||^2$, for $z\sim _2 x$.

\begin{prop}Let $f:R^n \to R$ and let $x\in R^n$. Then
$$\Delta f (x) = \sum _i \frac{\partial ^2 f}{\partial x_i ^2} (x).$$
\label{six}\end{prop}

{\bf Proof.}  For simplicity, let $x=0$, so that $z' = -z$.
We Taylor expand $f(z)$ and $f(-z)$
from $0$, and consider $f(z) + f(-z) -2f(0)$; then terms of degree
$\leq 1$ cancel, and we get
$$\sum _{ij}\frac{\partial ^2 f}{\partial x_i \partial x_j}z_i z_j +
\mbox{ higher terms};$$
 now the calculation proceeds much like the
one in the proof of Proposition \ref{five} above: if $z\sim _L 0$,
only the diagonal terms in the sum survive, all the $z_i ^2$ are equal
to $z_1 ^2$, which we move outside the parenthesis, and get $z_1 ^2 $
times the classical Laplacian $\sum \partial ^2 f/ \partial x_i ^2
(0)$.  But $z_1 ^2 = 1/n \cdot g(0,z)$.

\medskip

Similarly, one proves by Taylor expansion

\begin{prop} If $z\sim _L x$ in $R^n$, then for any $f:R^n \to
R$,
$$f(z)= f(x) + df_x (z-x) + \frac{1}{2n} \Delta f (x)
||z-x||^2 .$$
\label{Taylor} \end{prop}

Recall that any function $M\to R$ looks affine on any 1-monad ${\cal
M}_1 (x)$; functions that look affine on the larger $L$-monads ${\cal
M}_L (x)$ are precisely the harmonic ones:

\begin{prop}Assume $f:M\to R$ is harmonic at $x$.  Then for an $z\sim
_L x$, $f$ preserves affine combinations of $x$ and $z$.
Conversely, if for given $x$, $f$ preserves affine
combinations of $x$ and $z$ for every $z\sim _L x$, then $f$
is harmonic at $x$.  (``Harmonic" at $x$ here in the sense: $\Delta f (x)=0$.)
\label{seven}\end{prop}
(Recall that the affine combination $tx + (1-t)z$ is defined as the
critical point $y_0$ in (\ref{crit}).)

\medskip

{\bf Proof.} Assume $z\sim _L x$, and pick a geodesic normal
coordinate system at $x=0$. Without loss of generality, we may
assume $f(0)=0$. Then to say that $f$ preserves
affine combinations of $x$ and $z$ is to say that for all
$s\in R$, $f(sz)=sf(z)$.
For $z \sim _L 0$, we have by Proposition \ref{Taylor} that $f(z) = \sum a_i z_i
+ c\sum z_i ^2$ for unique $a_i$ and $c$ ($c= \Delta f (0) /2n$); $f(sz)$ and $sf(z)$
have the same terms of first order in $z$; their second order terms
are respectively $c s^2 \sum z_i ^2$ and $s c \sum z_i ^2$, and if
these two expressions are to be equal for all $s$ and all $z$, we must
have $c=0$.  This means that $f$ is harmonic at $x$.  Conversely, if
$f$ preserves affine combinations of the kind mentioned, it preserves
the affine combination $2x -z$, or equivalently, the left hand side of
the expression in Theorem \ref{Delta} is 0, hence it follows that $L$
and hence $\Delta f (x) $ is 0.

\medskip

{\bf Remark.} With some hesitation, I propose to call a map $f:M \to N$ between
Riemannian manifolds {\em harmonic} if it preserves affine
combinations of $L$-neighbours in $M$ and if it preserves the property
of being $L$-neighbours.  I have not been able to compare the proposed
definition, with a certain classical concept of harmonic map between
Riemannian manifolds.  But at least: When the codomain is $R$, the
definition is the basic classical one of harmonic function, by
Proposition \ref{seven}.  For, preservation of the $\sim _L$ relation
is automatic when the codomain is $R$, since in $R$, $\sim _L$ is the same
as $\sim _2$.

\section{Conformal maps}

We consider a diffeomorphism $f:M\to N$ between Riemannian
manifolds $(M, g), (N, h)$.
To say that $f$ is an {\em isometry} at $x\in M$ is to say that for all $z\sim
_2 x$,  $g(x,z)=h(f(x),f(z))$.  To say that $f$ is {\em conformal} at
$x\in M$ with constant $k = k(x) > 0$ is to say  that for all
$z\sim _2 x$, $h(f(x),f(z)) = k(x) g(x,z)$ (so if $k(x)=1$, $f$ is an
isometry at $x$).  The terminology agrees with classical usage, as we
shall see below.  We first prove

\begin{prop} Assume $f:M \to N$ is conformal at $x\in M$ with
\newline $h(f(x),f(z)) = k g(x,z)$.  Then for all $y_1 \sim _1 x$, $y_2 \sim
_1 x$,
$$h(f(y_1 ), f(y_2 )) = k g(y_1 ,y_2 ),$$
and conversely.
\label{conf}\end{prop}

{\bf Proof.} We choose coordinates, and assume $x=0$ and $f(x) =0$; the
metrics in $M$ and $N$ are then given by functions $g$ and $h$,
respectively, and they are in turn given by symmetric matrices $G(y)$,
and $H(z)$ for all $y\in M$ and $z\in N$.  We now calculate $k g(y_1
,y_2 )$.  We have, by Proposition \ref{simple} that $$k g(y_1 , y_2 )=
k g (0, y_2 - y_1 ) =
h(0, f(y_2 -y_1 )).$$
Now there is a bilinear $B(-,-)$ such that for all pairs of
$1$-neighbours $y_1 , y_2$ of $0$, we have $f(y_2 - y_1) = f(y_2 ) -
f(y_1 ) + B(y_1 ,y_2 )$.  So the calculation continues $$= h(0 , f(y
_1 ) - f( y_2 ) + B(y_1 ,y_2 ))$$ $$ = (f(y _1 ) - f( y_2 ) + B(y_1 ,y_2
))^T \cdot H(0) \cdot (f(y _1 ) - f( y_2 ) + B(y_1 ,y_2 )).$$ Since
$f$ depends in a linear way of $y_1 \sim _1 0$ and $y_2 \sim _1 0$, this whole
expression multiplies out by linearity, and for degree reasons all terms
involving $B$, as well as some others, vanish, and we are left with
$-2 f(y_1 )^T \cdot H(0) \cdot f(y_2 )$.  On the other hand, $h(f(y _1
), h(f(y_2 )) = h(0, f(y_2 ) - f(y _1 ))$, by Proposition
\ref{simple}, and writing this in terms of $H(0)$ gives the same
expression.

The converse is proved in the same way for $z\sim _2 0$ of the form
$y_2 - y_1 $ with $y_1 \sim _1 0$ and $y_2 \sim _ 1 0$, but this
suffices to get the result for all $z\sim _2 0$, by general principles
of SDG (``$ R$, and hence any manifold, perceives the addition map
$D(n)\times D(n) \to D_2 (n)$ to be epic''.)

\medskip

Call a linear map $F:U\to V$ between inner product vector spaces {\em
conformal} with constant $k > 0$ if for all $u_1 ,u_2 \in U$
$$<F(u _1 ),F(u_2 )> = k <u_1 , u_2 >.$$
It follows immediately from Proposition \ref{conf}, and from the
construction of inner product in the vector space of tangents at $x$,
and at $f(x)$, that if $f$ is conformal at $x$ with constant $k$, then
$ df_x : T_x M \to T_{f(x)} N$ is a conformal linear map with the
same constant $k$.  The converse also holds; for if $df _x$ is
conformal with constant $k$, we deduce that for all pairs of tangents
$t$ and $s$ at $x$
$$g(f(t(d_1 )), f(s(d_2 ))) = k \cdot g(t(d_1 ), s(d_2 )),$$
and hence
\begin{equation}g(f(y_1 )),f(y_2 )) = k \cdot g(y_1 , y_2
)\label{conv}\end{equation}
 for all $y_i$'s of the form $t(d)$ for a
tangent vector $t$ and a $d\in D$.  Again by general principles, any
manifold $N$ ``perceives all 1-neighbours of $x$ to be of this form''.
From Proposition \ref{conf} we therefore deduce that $f$ is conformal
at $x$ with constant $k$.

\begin{thm}A diffeomorphism $f$ is conformal at $x\in M$ if and only if
$f$ maps ${\cal M}_L (x)$ into ${\cal M}_L (f(x))$.
\label{three}\end{thm}

{\bf Proof.} Assume $f$ maps ${\cal M}_L (x) $ into ${\cal M}_L (f(x))$. We may
pick normal coordinates at $x$ as well as at $f(x)$.  The
neighbourhoods ${\cal M}_L (x)$ and ${\cal M}_L (f(x))$ then both get
identified with $D_L (n)$, and $x=0$, $f(x)=0$.  The restriction of
$f$ to $D_2 (n)$, $f:D_2 (n) \to R^n$, takes $0$ to $0$ and is therefore
of the form $f(y) = A\cdot y + B(y)$, where $A$ is an $n\times n$
matrix, and $B(y)$ is a map $R^n \to R^n$ which is homogeneous of
degree 2 in $y\in R^n$, i.e.\ an $n$-tuple of quadratic forms $B_i$.

Assume now that $f$ maps $D_L (n)$ into itself.  For $z\in D_L (n)$,
the $i$'th coordinate of $f(z)$ is
$$f_i (z) = \sum _k a_{ik} z_k + B_i (z).$$
Squaring this, only the terms in $( \sum _k a_{ik} z_k)( \sum _l
a_{il} z_l)$ survive for degree reasons (using that $z\in D_2 (n)$).
But using further that $z_k z_l =0$ for $k\neq l$, only the ``diagonal''
terms survive, and we get
\begin{equation}f_i (z ) ^2 =\sum _k a_{ik} ^2 z_k ^2 = z_1 ^2 \sum _k
a_{ik} ^2 .\label{sq}\end{equation} Similarly for $i\neq j$
 \begin{equation}f_i (z) f_j (z)=z_1 ^2 ( \sum _k a_{ik}
 a_{jk}).\label{ij}\end{equation}
If now $f(z) \in D_L (n)$ for all $z\in D_L (n)$, we get that the
expression in (\ref{sq}) is independent of $i$, and from the
uniqueness assertion in Proposition \ref{uni-c} we therefore conclude
$$ \sum _k a_{ik}^2 = \sum _k a_{jk}^2 \mbox{ for all }i,j ;$$
and similarly we conclude from (\ref{ij}) that
$$\sum _k a_{ik} a_{jk} =0 \mbox{ for } i\neq j.$$
These two equations express that all the rows of the matrix $A$ have the
same square norm $k$, and that they are mutually orthogonal.  This
implies that the linear map $df_x$ represented by the matrix is
conformal, and hence $f$ is conformal at $x$.

The proof that conformality of $f$ at $x$ implies that $f$ maps ${\cal M}_L
(x)$ into ${\cal M}_L (f(x))$ goes essentially through the same calculation,
and is omitted.

 \section{A famous pseudogroup in dimension 2}

 The content of the present section is partly classical, namely the
 equivalence of the various ways of describing the notion of
 holomorphic map from (a region in) the complex plane $C=R^2$ to
 itself.  Synthetic concepts enter essentially in two of the conditions
 in the Theorem below, namely 1) and 7).

 \medskip

An {\em almost complex} structure on a general manifold $M$
consists in giving, for each $x\in M$, a map $I_x :{\cal M}_1 (x)\to
{\cal M}_1 (x)$ with $I_x (x)= x$ and $I_x (I_x (z)) = z'$ for any
$z\sim _1 x$; Here, $z'$ denotes the mirror image of $z$ in
$x$, i.e.\ the affine combination $2x -z$; recall \cite{DFIC}
that affine combinations of 1-neighbours make ``absolutely'' sense,
i.e.\ do not depend on, say, a Riemannian structure.  It is clear what
it means for a map $f$ to {\em preserve} such structure at the point
$x$: $f(I_x (z)) = I_{f(x)} (f(z))$.

The manifold $R^2$ carries a canonical almost-complex
structure, given by
$$I_{(x_1 ,x_2 )} (z_1 , z_2 ) = (x_1 - (z_2 - x_2 ), x_2 +
(z_1 - x_1 )).$$
Identifying $R^2$ with the complex plane $C$, this is just
$$I_x (z) = x + i(z-x).$$
Utilizing the multiplication of the complex plane $C$, we may
consider the set $D_C$ of elements of square zero in $C$ (recalling
the fundamental role which the set $D $ of elements of square
zero in $R$ plays in SDG). We have, by trivial calculation,

\begin{prop} Under the identification of $C$ with $R^2$,
$$D_C = D_L (2).$$
\label{eleven}\end{prop}
Having $D_C$, we may mimick the basics of SDG and declare a
function $f: C\to C$ to be {\em complex differentiable} at
$x\in C$ if there is a number $f'(x)\in C$ so that
$$f(z) = f(x) + f'(x)\cdot (z-x) \mbox{ for all }z \mbox{ with }
z-x \in D_C .$$
(The uniqueness of such $f'(x)$, justifying the notation,
follows from the general axiom scheme of SDG, applied to
$D_1 (2)$, the 1-jet classifier in $R^2$.  Note $D_1 (2) \subseteq
D_L (2)$.)  --- The notion of course makes sense for functions $f$ which are
just defined {\em locally} around $x$.

\begin{thm}Let $f:R^2 \to R^2$ be a (local) orientation
preserving diffeomorphism.  Let $x\in R^2 =C$.  Then the following conditions
are equivalent:

1) $f$ maps ${\cal M}_L (x)$ into ${\cal M}_L (f(x))$

2) $f$ is conformal at $x$

3) $f$ satisfies Cauchy-Riemann equations at $x$

4) $f$ preserves almost complex structure at $x$.

\noindent Also the following conditions are equivalent, and they imply 1)-4):

5) $f$ is complex-differentiable at $x$

6) $f$ maps ${\cal M}_L (x)$ into ${\cal M}_L (f(x))$, and $f$ preserves affine
combinations of $x$ and $z$ for any $z\sim _L x$.

\noindent Finally, if 1)-4) hold for all $x$, 5) and 6) hold for all $x$.
\label{famous}\end{thm}
(Note that 6) says that $f$ is {\em harmonic}
at $x$, in the sense of Remark at the end of Section 2.)

\medskip

{\bf Proof.} The equivalence of 1) and 2) is already in
Theorem \ref{three}, and this in turn is, as we have seen, equivalent to
conformality of the linear $df _x$.  But conformal orientation
preserving $2\times 2$ matrices are of the form
\begin{equation}\left[ \begin{array}{rr}a& -b\\b & a \end{array}
\right] .\label{CR}\end{equation}
 Since the entries of the matrix for $df _x$ are
$\partial f_i / \partial x_j$, this form (\ref{CR}) of the matrix
therefore expresses that the Cauchy-Riemann equations hold at
$x$, i.e. is equivalent to 3). On the other hand, a simple
calculation with $2\times 2$ matrices give that a matrix
commutes with the matrix $I = \left[ \begin{array}{rr}
0&-1\\1&0 \end{array} \right]$  for the almost complex
structure iff it has the above ``Cauchy-Riemann" form (\ref{CR}).

Now assume 5). If $f$ is complex differentiable at $x$, we prove that condition 1)
holds at $x$ as follows.  Let $z\sim _L x$.  Then $(z-x)^2 =0$
Proposition \ref{eleven}, and by complex differentiability
\begin{equation}f(z)-f(x)= f'(x)(z-x),\label{co}\end{equation}
 so since the right hand side has
square zero, then so does the left hand side, but again by Proposition
\ref{eleven}, this means that $f(z)\sim _L f(x)$,  proving 1),
and hence also the first part of 6).  But also,  if $f$ is
complex-differentiable at $x$, $f$ preserves affine
combinations of the form $tx + (1-t)z$ for $z\in {\cal M}_L (x)$; this
follows from (\ref{co}), since $z-x \in D_C (x)$ by Proposition
\ref{eleven}, so also the second part of 6) is proved.  Conversely, if
6) holds, $f$ is conformal at $x$ by Theorem \ref{three}, so $df_x$ is
if the form (\ref{CR}).  Then $f'(x) = a+ib$ will serve as the complex
derivative; for, since $f$ preserves affine combinations of $x$ and
$z$ for $z\sim _L x$, we have the first equality sign in
$$f(z)= f(x)+ df_x (z-x) = f(x) + f'(x)\cdot (z-x)$$
for such $z$, i.e.  for $z-x \in D_L (2)= D_C $.

Finally, assume 1)-4) hold for all $x$.  Then we may differentiate the Cauchy-Riemann
equations for $f = (f_1 , f_2 )$   by
$\partial /\partial x_1$ and $\partial / \partial x_2$
and compare, arriving in the standard way to $\Delta f_1 \equiv 0$ and
$\Delta f_2 \equiv 0$.  From the ``Taylor expansion'' in Proposition
\ref{Taylor}, applied to $f_1$, we conclude that $f_1 (z) = f_1 (x) +
(df_1 )_x (z-x)$ for $z\sim _L x$, and similarly for $f_2$, so $f(z)=
f(x)+ df_x (z-x)$ for such $z$, i.e.\ for $z-x \in D_C$.  Since $df_x$
is given by a conformal matrix (\ref{CR}), by 2), this proves that
$a+ib$ will serve as the complex derivative of $f$ at $x$.

\section{Support of the Laplacian}
We arrived at $D_L (n)$ from the geometric side, namely as the
$\sim _L$-neighbours of $0$ in the Riemannian manifold
$M=R^n$; the differential operator $\Delta$ was then seen to provide
the top term in the Taylor expansion of functions defined on
$D_L (n)$.

Here, we briefly indicate how to arrive at $D_L (n)$ from the
algebraic side, starting with $\Delta = \sum \partial ^2  /\partial
x_i ^2$. More precisely, we consider $\Delta$ as a {\em
distribution} at $0\in R^n$. So $\Delta$ is the linear map
$$k[X_1 , ... ,X_n ] \to R$$
given by
\begin{equation}f\mapsto \sum _i \frac{ \partial ^2 f} {\partial
x_i ^2}(0).
\label{star}\end{equation}
The algebraic concept that will give $D_L (n)$ out of this
data is the notion of {\em coalgebra}, and {\em subcoalgebra}, as in
\cite{swe}.  If we let $A$ denote the algebra $k[X_1 , ...  ,X_n ]$,
then the distribution $\Delta$ of (\ref{star}) factors
$$A\to B \to k,$$
where $A\to B$ is an {\em algebra} map, and $B$ is finite
dimensional (take e.g.\ $B=A/J$ where $J$ is the ideal
generated by monomials of degree $\geq 3$). The set $A^{o}$ of
linear maps $A\to k$ having such a factorization property
constitute a coalgebra, \cite{swe} Proposition 6.0.2. Every
element in a coalgebra generates a finite dimensional
subcoalgebra, by \cite{swe} Theorem 2.2.1. In particular,
$\Delta \in A^{o}$ generates a finite dimensional coalgebra
$[\Delta ]$ of $A^{o}$, and this coalgebra {\em ``is"} $D_L
(n)$. More specifically, the dual algebra of $[\Delta ]$ is
the coordinate ring ${\cal O} (D_L (n))$ of $D_L (n)$, i.e.\ the Weil
algebra ${\cal O} (D_L (n)):= k[Z_1 , ...  ,Z_n ]/I$ considered in the
proof of Proposition \ref{uni-c}, as we shall now argue.

The following ``Leibniz rule" for $\Delta$ is well known,
$$\Delta (f\cdot g) = \Delta f \cdot g + 2\sum _i \frac{\partial
f}{\partial x_i} \cdot \frac{\partial g}{\partial x_i} +
f\cdot \Delta g .$$
This means that in the coalgebra $A^{o}$, we have the
following formula for $\psi (\Delta )$ ($\psi =$ the
comultiplication of the coalgebra; $\delta $ the Dirac distribution
``evaluate at 0''): \begin{equation}\psi (\Delta ) = \Delta \otimes
\delta  +2\sum _i \frac{\partial }{\partial x_i} \otimes \frac{\partial
}{\partial x_i} + \delta \otimes \Delta ,
\label{twostar}\end{equation}
where now $\Delta$, $\partial d /\partial x_i$, $\delta$ are viewed as
distributions at $0$, like in (\ref{star}), meaning that one evaluates
in $0$ after application, $$f\mapsto (\Delta f)(0)\mbox{, }f\mapsto
\frac{\partial
f}{\partial x_i}(0)\mbox{, } f\mapsto f(0).$$
From (\ref{twostar}), (and from $\psi (\partial /\partial x_i
)=\partial /\partial x_i \otimes \delta + \delta \otimes
\partial /\partial x_i$, which expresses the Leibniz rule for
$\partial /\partial x_i$)
we see that the subcoalgebra $[\Delta ]$
generated by $\Delta$ is generated as a vector space by the
elements
$$\delta \mbox{, }\frac{\partial}{\partial x_1} , ...
,\frac{\partial}{\partial x_n}\mbox{, } \Delta ,$$
and since these are clearly linearly independent, we see that $[\Delta ]
\subseteq A^{o}$ is $(n+2)$-dimensional.  The dual algebra of $[\Delta
]$ is a quotient algebra $A/I$ of $A$, where $I$ is the ideal of those
$f\in A$ which are annihilated by the elements of $[\Delta ]$.  This
ideal $I$ contains $x_i ^2 - x_j ^2$, and $x_i x_j$ for $i\neq j$.
Since the quotient of $A$ by the ideal generated by $x_i ^2 - x_j ^2$,
and $x_i x_j$ for $i\neq j$ is already $(n+2)$-dimensional, as
calculated in the proof of Proposition \ref{uni-c}, it follows that
the quotient algebra there is actually the dual of $[\Delta ]$.

\medskip

The idea that a coalgebra like $[\Delta ]$ {\em is} itself an
infinitesimal geometric object goes back to Gavin Wraith in
the early seventies, \cite{wraith}.  The specific way of generating Weil
algebras from differential operators was considered by Emsalem
\cite{Emsalem} (without coalgebras).


TITLE: The conductor and minimal moduli of abelian extensions
QUESTION [6 upvotes]: Assume that $L/K$ is a finite abelian extension of global fields and $S$ the set of primes of $K$ ramifying in $L$. Then the conductor $\mathfrak{f}(L/K)$ is the smallest modulus s.t. the Artin map
$$\psi_{L/K}:I_K^S \to \operatorname{Gal}(L/K)$$
factors through the ray class group $C_{\mathfrak{f}(L/K)}$. I'm looking at Milne's notes on CFT and he seems to sort of skip this whole existence and uniqueness thing to a remark without proving anything. My question is that given two different moduli $\mathfrak{m}$ and $\mathfrak{n}$, both containing the ramifying primes of $K$, then to prove the uniqueness of the conductor, we just need to show that the Artin map factors through the ray class group of the moduli
$$\prod_{\mathfrak{p}} \mathfrak{p}^{\min\{\mathfrak{n}(\mathfrak{p}),\mathfrak{m}(\mathfrak{p})\}}.$$
Is there a slick way of proving this? I just seem to get into a mess of multiplicative congruences while trying to figure out what's in the kernel of $\psi_{L/K}$. Maybe I'm just missing the completely obvious.

REPLY [3 votes]: What are you giving yourself?  If you don't assume that Artin reciprocity law, then you don't know that there is any modulus at all.  If you assume Artin reciprocity, then a standard (but very nice, and very important) argument (explained well in Tate's article in Cassels and Frolich) allows one to pass from the ideal-theoretic to the adelic point of view, and from the latter point of view, the existence of a minimal modulus is clear.
With regard to getting "into a mess of multiplicative congruences", this is
precisely what the ideles are good at dealing with, so I think that the idelic approach provides at least one "slick way of proving this".


TITLE: How to integrate this integral using Green's theorem?
QUESTION [0 upvotes]: $\int_{L} (x+2xy)dy + y^{2}dx$ in the circle $L=\{(x,y)|x^2+y^2=y\}$. I already know that $P_y'=2y$ and that $Q_x'=1+2y$ where $P$ is the function multiplied by $dx$ and $Q$ is the function multiplied by $dy$. So I figured that since the given circle has $x$ in $[-0.5,0.5]$ and $y$ is in $[0,1]$ then I should integrate twice $\int_{0}^{1}(\int_{-0.5}^{0.5}1+4ydx)dy$. Is this correct?

REPLY [1 votes]: It'll look cluttered in a comment, so I'll try to address the main points here.
You're correct in deducing that $L$ traces out a circle with radius $1/2$ and center $(0,1/2)$. Green's theorem can be used for your integral. There is a sign error in your expression for the curl of your vector field $\mathbf{F}(x,y) = (y^2, x+2xy)$. It should be
$$ \text{curl}(\mathbf{F}) = \left( \dfrac{\partial Q}{\partial x} \ {\color{red}-} \ \dfrac{\partial P}{\partial y} \right) = 1+2y-2y = 1 $$
Let $R$ denote the region enclosed by $L$. By Green's theorem, we can express our integral as
$$ \oint_{L} \mathbf{F} \cdot \text{d}s = \iint_{R} \text{curl}(\mathbf{F}) \text{ d}A $$
For the purpose of this question, it should be easy to see that $ \iint_{R} \text{d}A = \text{area}(R) $.

 If you wanted to actually compute such an integral, the bounds you've selected, namely $x \in [-1/2,1/2]$ and $y \in [0,1]$, actually trace out a square in $\mathbb{R}^{2}$. You can try parameterising a circular region with center $(0,0)$ and radius $1$ using Cartesian coordinates. The you can just shift and scale this region to $R$.)


TITLE: Showing that a proposition is a substitute of the Euclidean fifth postulate.
QUESTION [0 upvotes]: I'm needing some guidance in order to prove that the following proposition:
Any point in the interior of an (not $\pi$) angle $\angle AÔB$ is in a segment whose extremes are in the rays $\vec{OA}$ and $\vec{OB}$.
is a substitute to Euclid's fifth postulate, which I can use the Playfair substitute, since that equivalence I already proved:
Given a line $r$ and a point $P \notin r$, there is exactly one line $s$ such that $P \in s$ and $r \cap s = \phi $.
Here is my attempt:
Suppose that there is a point $P$ in the interior of $ \angle AÔB$. Since $P$ is not in the support line of $OA$, there is exactly one line through $P$ not intersecting sup. line $OA$. consider the perpendicular line to said line. My aim is to show that this perpendicular will intersect the ray $OB$, but I'm having trouble doing this.
Can someone say if this is a good approach and how to fix it? I would also accept different solutions to this problem.
Thanks.

REPLY [0 votes]: Let's recall first of all that an immediate consequence of Euclid's fifth postulate is the following: if two lines are intersected by a common transversal and form with it non supplementary consecutive interior angles, then these lines meet on that side of the transversal where the sum of consecutive interior angles is less than $\pi$ (as a matter of fact, this is exactly how the postulate was presented by Euclid himself).
But even without Euclid's fifth postulate we can state a weaker result:
LEMMA - If two lines are intersected by a common transversal and form with it non supplementary consecutive interior angles, then these lines cannot meet on that side of the transversal where the sum of consecutive interior angles is greater than $\pi$.
This can be proven by the exterior angle theorem: if $\alpha+\beta>\pi$ (see picture below) then $\gamma=\pi-\beta<\alpha$; if $AA'$ and $BB'$ would meet at $P$ on the right, then triangle $PAB$ would have an external angle $\gamma$ greater then an internal angle, which is impossible.

Let's now prove that Euclid's fifth postulate implies your proposition. Consider a convex angle $\angle AOB\ne\pi$ and a point $P$ in its interior. Draw from $P$ the parallel to $OB$, which meets ray $OA$ at $A$ (see picture below), and take a point $Q$ on $OA$ such that $A$ is between $O$ and $Q$. We have $\alpha>\phi$ and $\alpha=\pi-\theta$, whence $\phi+\theta<\pi$: it follows that line $PQ$ meets ray $OB$.

Let's now show the converse, in its contrapositive form: supposing that uniqueness of a parallel line is not guaranteed, I'll prove that one can construct an angle, and a point inside it, for which your proposition does not hold.
Let $a$ be a line in the plane and $P$ a point not belonging to $a$ and $O$ a point on $a$. Through $P$ we can construct a line $a'$ forming equal alternating interior angles with $OP$, so we know that $a'\parallel a$. Suppose now that uniqueness of parallel line is violated, so that there exists another line $b'$ through $P$ parallel to $a$. We can construct a line $b$ through $O$ forming equal alternating interior angles with $OP$ and thus parallel to $b'$. Of the four convex angles formed by $a$ and $b$ choose the one containing $P$: I'll call this $\angle ab$ and from now on I will use $a$ and $b$ to denote those rays of the lines which are the sides of $\angle ab$. I want to prove that there exists no segment with endpoints on $a$, $b$ and containing $P$.
To this end, take any point $A$ on $a$, and let $\angle AOP=\alpha$ (see picture below). Denote then by $\beta$ the angle formed by $b$ with $OP$ and notice that $\alpha+\beta<\pi$ (because the angle is convex). Moreover, we have $\angle APO<\beta$, because if $\angle APO=\beta$ then $A$ would belong to $b'$ by construction (impossible, as $a$ and $b'$ are parallel) and if $\angle APO>\beta$ then $b'$ would intersect $AO$ (impossible again).
It follows that line $AP$ and ray $b$ form with $OP$ consecutive interior angles greater than $\pi$: by the lemma stated above, that proves the assertion.


TITLE: Stuck on this proof that $ord(f) = ord(g)$
QUESTION [0 upvotes]: Let $f, g: \mathbb R \to \mathbb R$ be smooth maps such that $f(a) = g(a') = 0$ and let $\tau, \sigma : \mathbb R \to \mathbb R$ be diffeomorphisms such that
$$ \tau \circ f = g \circ \sigma$$
Define $ord(f) = \min \{ n \in \mathbb N_{>0} \mid f^{(n)} (a) \neq 0\}$.
I am trying to show that $ord(f) = ord(g)$. I wanted to do induction but the induction step is broken since $ \tau \circ f = g \circ \sigma$ does not imply that there are diffeomorphisms $\phi, \pi$ such that
$ \phi \circ f' = g' \circ \pi$
Next I thought I could derive a closed form for the $n$-th derivative of $ \tau \circ f = g \circ \sigma$ but I can't see a pattern. The sums get rather long.
Now I am starting to wonder whether the thing I'm trying to show is even true. If it is I think the proof should be short and simple very unlike what I've tried so far.

Does anyone have any ideas on whether this is true and if it is on how
  to prove it?

REPLY [0 votes]: First, you do not define $ord(g)$; I shall presume it to be like $ord(f)$
but with $a$ replaced by $a'$.
Second, your statement is true if and only if $\sigma(a) = a'$ (read the proof for $org(g) = 1$ to understand why).
Third, we shall make essential use of the following formula which generalizes Leibniz's formula:
$$(f_1 f_2 \dots f_n) ^{(p)} = \sum \limits _{i_1 + i_2 + \dots + i_n = p} \frac {p!} {i_1 ! i_2 ! \dots i_n !} f_1 ^{(i_1)} f_2 ^{(i_2)} \dots f_n ^{(i_n)}$$
(if you want, it is easy to prove it by induction).
Note that since $\tau$ is a diffeomorphism, we may invert it, getting $f = \tau ^{-1} \circ g \circ \sigma$.
Let us see what happens when $ord(g) = 1$ (i.e. $g'(a') \ne 0$). We have
$$f'(a) = (\tau ^{-1})' \circ g \circ \sigma (a) \cdot g' \circ \sigma (a) \cdot \sigma ' (a) = (\tau ^{-1})' (g(a')) \cdot g' (a') \cdot \sigma ' (a) = \\ (\tau ^{-1})' (0) \cdot g' (a') \cdot \sigma ' (a) .$$
Note that since $\tau ^{-1}$ and $\sigma$ are diffeomorphisms, their derivatives cannot cancel anywhere; since $g'(a') \ne 0$ by assumption, this gives $f'(a) \ne 0$, i.e. $ord(f) = 1$.
Now let us see the induction step: assume $ord(g) = n \ge 2$. We shall prove that $f^{(k)} (a) = 0 \; \forall k<n$ and $f^{(n)} (a) \ne 0$.
Note that
$$f^{(k)} (a) = (f')^{(k-1)} (a) = ((\tau ^{-1})' \circ g \circ \sigma \cdot g' \circ \sigma \cdot \sigma ') ^{(k-1)} (a) = \sum \limits _{i_1 + i_2 + i_3 = k-1} \frac {(k-1)!} {i_1 ! i_2 ! i_3 !} \big( (\tau ^{-1})' \circ g \circ \sigma \big) ^{(i_1)} \cdot \big( g' \circ \sigma \big) ^{(i_2)} \cdot \big( \sigma ' \big) ^{(i_3)} .$$
Now stop calculating and think: in the above sum, the only derivatives of $g$ showing up will be of order at most $k$, because $i_1$ and $i_2$ can go up to at most $k-1$. Since $ord(g) = n > k$, all these derivatives are $0$, from which follows $f^{(k)} (a) = 0 \; \forall k<n$.
In order to see that $f^{(n)} (a)\ne 0$ we shall use the same formula above, but with $k$ replaced by $n$:
$$f^{(n)} (a) = (f')^{(n-1)} (a) = ((\tau ^{-1})' \circ g \circ \sigma \cdot g' \circ \sigma \cdot \sigma ') ^{(n-1)} (a) = \sum \limits _{i_1 + i_2 + i_3 = n-1} \frac {(n-1)!} {i_1 ! i_2 ! i_3 !} \big( (\tau ^{-1})' \circ g \circ \sigma \big) ^{(i_1)} \cdot \big( g' \circ \sigma \big) ^{(i_2)} \cdot \big( \sigma ' \big) ^{(i_3)} .$$
The only non-zero term will be the one with $i_2 = n-1$ (which implies $i_1 = i_3 = 0$), which is $(g' \circ \sigma) ^{(n-1)} (a)$. Now, you either understand at a glance what happens here, or you can use induction once more to show that, in fact, this is equal to $( g^{(n)} \circ \sigma ) (a) = g^{(n)} (a') \ne 0 $ by assumption. Therefore, $f^{(n)} (a) = g^{(n)} (a') \ne 0$, so $ord(f) = n$.


TITLE: Unimodal discrete distribution
QUESTION [0 upvotes]: How do I prove that the following discrete distribution is unimodal (in the sense that it has only one local maxima)?
\begin{align}
  & P(X=k)=\left. \frac{1}{k!}\frac{{{d}^{(k)}}}{dz}\prod\limits_{i=1}^{n}{(p{_i}+q{_i}\cdot z+r{_i}\cdot {{z}^{2}})\ } \right|\ z=0 \
 & for\quad k:0,1,...,2n \
\end{align}
for any $n\ge 1$, where for each $i ,\quad 0<p{_i},q{_i},r{_i}<1,\quad p{_i}+q{_i}+r{_i}=1,\quad$
and it is never the case that $p{_i}>q{_i}<r{_i}$.

REPLY [1 votes]: Work in progress Uncompleted proof of a special case of the desired result.
Let us call a sequence $(a_0, \ldots, a_n)$ as unimodal with mode
$M$ if it enjoys the property that
$$a_0 < a_1 < \cdots < a_{M-1} < a_M > a_{M+1} > \cdots > a_n$$
where $1 \leq M < n$.
Let $\{Y_i\}$, denote a sequence of  be n i.i.d. discrete random variables
taking on values $0,1,2$ with probabilities $p$, $q$, and $r=1−p−q$
respectively where we assume that $p < q > r$ so that $(p,q,r)$ is
a unimodal sequence with mode $1$.  Define $X_n = \sum_{i=1}^n Y_i$.
Then, $X_n$ is a discrete random variable taking on value
$i$ with probability $p_n(i)$, $0 \leq i \leq 2n$.
Claim: The sequence
$(p_n(0), p_n(1), \ldots, p_n(2n))$ is unimodal.
We prove the claim by induction.
Basis:  The result holds for $n = 1$ since we have assumed that
$(p_1(0), p_1(1), p_1(2)) = (p,q,r)$ is unimodal.
Induction hypothesis: $(p_n(0), p_n(1), \ldots, p_n(2n))$ satisfies
$$p_n(k) > p_n(k-1) ~\text{for}~ 1 \leq k \leq M_n, ~~
p_n(k-1) > p_n(k)~\text{for}~  M_n < k < 2n.$$
Now consider the sequence $(p_{n+1}(0), p_{n+1}(1), \ldots, p_{n+1}(2{n+2}))$.
Since
$$p_{n+1}(k) = p_n(k)\cdot p + p_n(k-1)\cdot q + p_n(k-2)\cdot r$$
(where we use $p_n(-2) = p_n(-1) = p_n(2n+1) = p_n(2n+2) = 0$), we have
that for $1 \leq k \leq M_n$,
$p_{n+1}(k) -p_{n+1}(k-1)$ is the sum of three positive terms and therefore
$(p_{n+1}(0), p_{n+1}(1), \ldots, p_{n+1}(M_n))$ is an increasing
sequence.  Similarly, for $M_n +2 \leq k \leq 2n+2$,
$p_{n+1}(k) -p_{n+1}(k-1)$ is the sum of three negative terms and therefore
$(p_{n+1}(M_n+2), p_{n+1}(M_n+3), \ldots, p_{n+1}(2n+2))$ is a decreasing
sequence.

If $p_{n+1}(M_n) < p_{n+1}(M_n+1) < p_{n+1}(M_n+2)$, then
$(p_{n+1}(0), p_{n+1}(1), \ldots, p_{n+1}(2{n+2}))$ is unimodal with
mode $M_n+2$.
If $p_{n+1}(M_n) < p_{n+1}(M_n+1) > p_{n+1}(M_n+2)$, then
$(p_{n+1}(0), p_{n+1}(1), \ldots, p_{n+1}(2{n+2}))$ is unimodal with
mode $M_n+1$.
If $p_{n+1}(M_n) > p_{n+1}(M_n+1) > p_{n+1}(M_n+2)$, then
$(p_{n+1}(0), p_{n+1}(1), \ldots, p_{n+1}(2{n+2}))$ is unimodal with
mode $M_n$.

All that remains to prove unimodality of $(p_{n+1}(0), p_{n+1}(1), \ldots, p_{n+1}(2{n+2}))$ is to show that the case
$$p_{n+1}(M_n) < p_{n+1}(M_n+1) > p_{n+1}(M_n+2)$$
cannot occur.  Compare to the OP's $p > q < r$ is not allowed.


TITLE: How impulse momentum theorem is put to use?
QUESTION [2 upvotes]: Guys the below link is a question already asked in this website and I have referred the answers given by the users and some of them were good ,but I have few doubts and I wanted to ask it in a comment but my reputation wasn't 50,so i'm posting it as a question.
Sorry for the duplication!
the link is this:
Why isn't momentum conserved in this pulley problem?.
My doubt is this:
In the solution give in the users question the impulse imparted to the partice should be mV-mv right?
What I mean is that it should be final momentum -initial momentum right?
And,impulse imparted for block should be -mV right? i.e it should negative because the block moves up.
Please forgive my bad presentation of the problem.

REPLY [0 votes]: The outlined solution has up positive, but $v$ and $V$ are implicitly positive in a downward direction. With up, $v$, and $V$ positive, the ball's initial momentum is $-mv$ and it's final momentum is $-mV$. The change in momentum is thus $(-mV) - (-mv)$ or $mv-mV$.
Note that with up positive, it would perhaps have been better to write equation (ii) as
$$\int(T-N)\,dt = -mV$$
Negating both sides results in equation (ii) as written in the outlined solution.


text:
\DOC REAL_POLY_POW_CONV

\TYPE {REAL_POLY_POW_CONV : term -> thm}

\SYNOPSIS
Raise real polynomial to numeral power while retaining canonical form.

\DESCRIBE
For many purposes it is useful to retain polynomials in a canonical form. For
more information on the usual normal form in HOL Light, see the function
{REAL_POLY_CONV}, which converts a polynomial to normal form while proving the
equivalence of the original and normalized forms. The function
{REAL_POLY_POW_CONV} is a more delicate conversion that, given a term
{p1 pow n} where {p} is a real polynomial in normal form and {n} a numeral,
returns a theorem {|- p pow n = p'} where {p'} is in normal form.

\FAILURE
Fails if applied to a term that is not a real term raised to a numeral power.
If the subterm is not a polynomial in normal form, the overall normalization is
not guaranteed.

\EXAMPLE
{
  # REAL_POLY_POW_CONV `(x + &1) pow 4`;;
  val it : thm =
    |- (x + &1) pow 4 = x pow 4 + &4 * x pow 3 + &6 * x pow 2 + &4 * x + &1
}

\USES
More delicate polynomial operations that simply the direct normalization with
{REAL_POLY_CONV}.

\SEEALSO
REAL_ARITH, REAL_POLY_ADD_CONV, REAL_POLY_CONV, REAL_POLY_MUL_CONV,
REAL_POLY_NEG_CONV, REAL_POLY_SUB_CONV, REAL_RING.

\ENDDOC

subset_name: curated


TITLE: the probability for exactly two consecutive successs
QUESTION [0 upvotes]: Calculate the probability of getting exactly 50 heads and 50 tails after flipping a fair coin 100 times.
for this question we can easily apply the the binomial distribution formula, as $100 \choose 50$  $\frac{1}{2}^{100}$
What if we are asked
Calculate the probability of getting exactly consecutive 10 heads or 10 tails after flipping a fair coin 100 times?  would binomial distribution still work. thank you

REPLY [0 votes]: The probability of getting exactly consecutive 10 heads and 10 tails is $\frac{1}{2}^{20}$.
Then you have to consider that it can happen from zero to five times.
So you have to use the Bernoulli’s Formula to calculate the total probability:
$$\sum_{k=0}^{5} \binom{5}{k}  \left( \frac{1}{2}^{20} \right)^{5}  \left(1 - \frac{1}{2}^{20} \right)^{5-k}$$


TITLE: What determines the probability of a pair of photons interacting, and producing a positron and an electron?
QUESTION [4 upvotes]: The second answer to this question describes how this process might occur, and I'm curious for more details about it:

What is the probability distribution of the interaction producing electron-positron pairs (and what's the general process for calculating it)?
Is it possible to produce beams of positrons and/or electrons, through this process?

I would love some references that describe this kind of interaction in depth.

REPLY [1 votes]: What is the probability distribution of the interaction producing electron-positron pairs (and what's the general process for calculating it)?

The probability can be calculated using the Feynman diagrams

In this case the wiggly lines are the photons and the solid the electron positron ones.

Is it possible to produce beams of positrons and/or electrons, through this process?

It might be possible but not very efficient for electrons, there are much more efficient ways. Positron beams are also generated with general scatterings and isolating the positrons into beam lines. The crossections are much higher. Also nuclear beta+ sources are more efficient than two photon production.
Gamma gamma colliders have been proposed for studying the higgs, as the crossection rises with energy.


TITLE: Points $P_i$ on an ellipse such that angle $P_iOP_{i+1}=\frac{\pi}{n}$
QUESTION [1 upvotes]: Consider an ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ with $O$ as the origin. $n$ points denoted as $P_1,P_2,\cdots$ are taken on the ellipse such that angle $P_iOP_{i+1}=\frac{\pi}{n}$ where $i\in(1,n-1)$. Find the value of: $$\sum_{i=1}^n\frac{1}{OP_i^2}$$
I took $P_1$ as $(3,0)$. Let other points be $P_i(3\cos\theta_i,2\sin\theta_i)$.
$$\tan\frac{\pi}{n}=\frac{2}{3}\tan\theta_2$$
$$\tan\frac{2\pi}{n}=\frac{2}{3}\tan\theta_3$$
and so on.
$$OP_1^2=a^2$$
$$OP_2^2=9\cos^2\theta_2+4\sin^2\theta_2=9\cos^2\theta_2\left(1+\tan^2\frac{\pi}{n}\right)$$
$$OP_3^2=9\cos^2\theta_3+4\sin^2\theta_3=9\cos^2\theta_3\left(1+\tan^2\frac{2\pi}{n}\right)$$
and so on.
But I am not able to compute the required sum.

REPLY [2 votes]: If we set $P_i=(x_i,y_i)=\left(3\cos\theta_i,2\sin\theta_i\right)$ we have
$$\vartheta_0+\frac{\pi(i-1)}{n}=\arctan\frac{y_1}{x_i}=\arctan\left(\frac{2}{3}\tan\theta_i\right)$$
from which:
$$ \tan\theta_i = \frac{3}{2}\tan\left(\vartheta_0+\frac{\pi(i-1)}{n}\right) $$
and
$$ OP_i^2 = 9\cos^2\theta_i+4\sin^2\theta_i = 4+5\cos^2\theta_i=\frac{9+4 \tan^2\theta_i}{1+\tan^2\theta_i}$$
gives that the wanted sum is:
$$ \sum_{i=1}^{n}\frac{1+\tan^2\theta_i}{9+4\tan^2\theta_i}=\sum_{i=1}^{n}\frac{1+\frac{9}{4}\tan^2\left(\vartheta_0+\frac{\pi(i-1)}{n}\right) }{9+9\tan^2\left(\vartheta_0+\frac{\pi(i-1)}{n}\right) } $$
or:
$$ \frac{1}{36}\sum_{i=1}^{n}\left[4\cos^2\left(\vartheta_0+\frac{\pi(i-1)}{n}\right) +9\sin^2\left(\vartheta_0+\frac{\pi(i-1)}{n}\right) \right]$$
that can be computed by exploiting the duplication formulas for $\sin$/$\cos$, De Moivre's formula and geometric series, or simply by noticing that it does not really depend on $\vartheta_0$ by differentiation with respect to $\vartheta_0$:
$$\forall\, n,\vartheta_0,\qquad \sum_{i=1}^{n}\sin^2\left(\vartheta_0+\frac{\pi(i-1)}{n}\right)=
\sum_{i=1}^{n}\cos^2\left(\vartheta_0+\frac{\pi(i-1)}{n}\right)=\color{red}{\frac{n}{2}}\tag{1}$$
gives:

$$ \sum_{i=1}^{n}\frac{1}{OP_i^2} = \color{red}{\frac{13n}{72}}.\tag{2}$$


TITLE: Independent Spaces
QUESTION [1 upvotes]: What does it mean for spaces to be independent?

REPLY [1 votes]: Given a vector space $V$, we say that a collection of subspaces $\{W_1, \dots, W_k\}$ is independent if for every $1 \le i \le k$,
$$W_i \cap \sum_{j \ne i} W_j = \{0\}$$
In other words, no nonzero vector in $W_i$ can be a linear combination of vectors from the other subspaces.


TITLE: Unbiased Estimate of Variance
QUESTION [0 upvotes]: Consider a simple linear regression model for $n$ observations where
$$Y_i = \beta_1 X_i + \epsilon_i$$
where $\epsilon_i \sim N(0,\sigma^2).$ I want to show that
$$\hat{\sigma}^2 = \frac{1}{n-2} \sum_{i=1}^n \hat{\epsilon}_i^2,$$
is an unbiased estimate of $\sigma^2$,
where $\hat{\epsilon}_i$ is an unbiased estimate of $\epsilon_i.$
The $n-2$ confused me enough, but when it came to calculating
$E[\hat{\sigma}^2]$, I found that
$$E[\hat{\sigma}^2] = E\left[ \frac{1}{n-2} \sum_{i=1}^n \hat{\epsilon}_i^2
\right]$$
$$=\frac{1}{n-2} \sum_{i=1}^n E[\hat{\epsilon}_i^2]
= \frac{1}{n-2} \sum_{i=1}^n Var[\hat{\epsilon}_i] + (E[\hat{\epsilon}_i])^2,$$
which uses the definition of variance. However, I am having some difficulty
figuring out $Var[\hat{\epsilon}_i]$, although I am rather sure that
$E[\hat{\epsilon}_i] = \epsilon_i$ given unbiasedness. Note that
we have defined $\hat{\epsilon}_i = (Y_i-\hat{Y}_i)^2,$ where
$\hat{Y}_i = \hat{\beta}X_i$ $ \forall i \in [n].$ Any recommendations on this?

REPLY [0 votes]: The standard assumptions on the noise/error term are $\epsilon \sim \mathcal{N} (0, \sigma^2)$. Hence, as was mentioned by @Ian in the comment,
$$
E\hat{\epsilon}_i = E(y_i - \hat{y}_i) = E(y_i) - E(\hat{\beta} x_i)=\beta x_i - \beta x_i=0.
$$


TITLE: Find distance from point to line using trigonometry
QUESTION [1 upvotes]: I have a scene with center $C$ and outer corners A and B. I have a camera with a fixed focal length that I want to place at a fixed height, at a distance $d$ such that the entire scene is in view, while the camera looks at center point $C$. The situation is described in the picture above. Black labels are fixed, blue labels are variables.
I am trying to use trigonometric relations to find the formula for $d$, but I am failing miserably. I realize that if I find $\theta$ I can easily find $d$, but I'm not sure how to get $\theta$.
Could someone point me in the right direction?

REPLY [0 votes]: This is not as complete an answer as I'd like, but it's too much to put in comments.
For any given value of $\alpha$ we can construct two circles, one with arc angle $2\alpha$ from $A$ to $C$, and another with arc angle $2\alpha$ from $B$ to $C$. Then $X$ would be at the second intersection of those circles
(not $C$) if there is one and if it is not on the $2\alpha$ arc of either circle.
Consider the locus of $X$ for $\alpha = 0$ up to $\alpha$ equal to half the field of view of your camera. (If $\alpha$ is larger, you won't see $A$ or $B$ in the picture frame.) This is some curve in the plane. If it intersects the given horizontal line at least once, each intersection is a possible solution. Otherwise there is no solution.
We could solve this numerically by taking the distance of $X$ above the desired line as a function $f(\alpha)$, and solve for $f(\alpha) = 0$.
But maybe there's a way to do this directly that I haven't seen yet.
If the best we can do is a numeric solution, I think it's simpler to find
the subtended angle of $A$ and $C$ (call this $\alpha_1$)
and the subtended angle of $B$ and $C$ (call this $\alpha_2$) both
as functions of the $x$-coordinate of $X$ on the given line.
Then solve for $x$ such that $\alpha_1(x) - \alpha_2(x) = 0$,
if such an $x$ exists.
But if you have a particular fixed value of $\alpha$ in mind, there will
be only two places (if any) where you can put $X$ in the plane.
Unless one of those points happens to be exactly on the line you want,
you have no solution. In other words, for a fixed $\alpha$ I think your
problem is overconstrained.