text:
\section{Introduction}

Lattice simulation is an invaluable tool in our studies of nonperturbative
aspects of quantum field theories.
By the discretization of the spacetime continuum, the original system with
infinitely many degrees of freedom is approximated by one with only a large
but finite number of degrees of freedom, which can be studied numerically.
In practice, the lattice has almost always been chosen as hypercubical since
the original paper by Wilson \cite{Wil}, although some ``uncanonical'' choices
has also been discussed.
Four-dimensional random lattices were discussed by Christ {\it et.~al.}~in
Ref.~\cite{CFL1,CFL2,CFL3}.
The body-centered hypercubical lattices in four dimensions had been first
studied by Celmaster \cite{Cel1,Cel2,Cel3} and subsequently in
\cite{DMP,Neu1,Neu2,Neu3}.
Almost at the same time four dimensional simplicial lattices were studied
in \cite{DM1,DM2,DM3,DM4,DM5}.
(An interesting comparison between the last two uncanonical lattices can be
found in Ref.~\cite{CM}.)

Let's recall the definition of a lattice.
A set of vectors $\{B_i\}$ ($1\le i \le n$) in $m$-dimensional
space ${\bf R}^m$ is {\it linearly independent} if
\begin{equation}
\sum_{i=1}^n x_i B_i = 0 \qquad {\rm iff} \qquad x_i = 0  \quad \forall i.
\end{equation}
With such a set of linearly independent basis vectors $\{B_i\}$ one can
define a lattice as below.

\bigskip

{\sl An $n$-dimensional lattice $\Lambda_n$ is a subset of
$\,{\bf R}^m$ $(n\le m)$ such that
\begin{equation}
\Lambda_n = \{X=\sum_{i=1}^n z_i B_i: z_i\in {\bf Z}\}
\end{equation}}

\noindent The lattice always contains the origin and is symmetric under
reflection $X \to -X$ in ${\bf R}^m$.
Notice that we have not required that $n=m$ as some lattices (like the
triangular lattice in two dimensions) can be more conveniently represented
when embedded in some higher dimensional spaces.
Also note that, under this definition, random lattices are {\it not\/}
lattices.
Still, such a definition encompasses a huge set of lattices, including
extremely skewed ones whose unitcells are long narrow parallelograms or
parallelohedra.
However, such highly asymmetric lattices are rarely used for simulations.
Instead, physicists always use lattices which are ``highly symmetric''
(we will give a mathematically precise meaning to this phrase later).
In particular, there is  a class of lattices called the ``root lattices'',
which contains most of the ``highly symmetric'' lattices, at least in lower
dimensions $n\le4$.
While there is no real justification in neglecting the lattices which do
not belong to this class, root lattices are so well studied that they provide
a natural starting point for our investigation.
As we will see, there are exactly three root lattices in four dimensions,
namely the hypercubical, the body-centered hypercubical and the simplicial
lattices\footnote{This fact has been observed in Ref.~\cite{DM1}.},
{\it i.e.}, exactly the cases which have been studied.

It is important to compare the relative merits of these different lattices
in simulations.
Several papers mentioned above attempted to answer this question by
simulating the same physical system on different lattices and comparing the
results.
These comparisons are very directly relevant to the practical applications
of lattice simulations, but theoretically they are ``unclean'' in the
sense that they are {\it a priori\/} sensitive to the physical problems in
question, the actual algorithms used, and other non-geometrical factors.
This paper attempts to study the same problem in a complementary way.
We will obtain a general prescription on how to construct the Laplacian
operator, the simplest possible differential operator, on any lattice.
While the Laplacian operator is not directly relevant for lattice QCD,
it is used in lattice simulations of strongly coupled scalar $\phi^4$
theories, and it is logical to study the Laplacian case thoroughly before
tackling the more complicated cases of the Dirac and the plaquette
operators.
By studying the discretization errors of the Laplacian operators
mathematically, one can parameterize these errors in ways which depend only
on the geometrical properties of the lattices.
The questions we will be addressing include:

$\bullet$ Are there any lattices on which the discretizations of the
Laplacian operator on the nearest neighbors are free of ${\cal O}(a^2)$
errors?

$\bullet$ What are the merits and demerits of our canonical choice ---
the simple hypercubical lattice?

$\bullet$ What is the role of symmetry?

$\bullet$ In general, how do we eliminate the ${\cal O}(a^2)$ errors?
(Answer: by involving sites which are not nearest neighbors.)
Is it always possible?

$\bullet$ In a given dimension, does the lattice which has the smallest
${\cal O}(a^2)$ errors still work best after the ${\cal O}(a^2)$ errors
are cancelled?

In this paper, we will try to provide answers to all these questions.
The approach is ``clean'' in the sense that all the effects of algorithms
or choices of physics problems are disentangled.
On the other hand, it must be admitted that its relevance to practical
lattice simulations is not immediate.
Roughly speaking, this paper is more about lattice field theory than
about lattice simulations, and the theorems obtained should be regarded as
general ``rules of thumb'' instead of infallible laws in practice.

This paper is organized as follows:
In Section II we present a prescription (one shell discretization)
for approximating the Laplacian on a lattice.
Root lattices are defined in Section III, and the explicit forms of the
discretized Laplacian operators on root lattices are presented in Section IV.
Order $a^2$ errors are studied in Sections V and VI, while the issue of
anisotropy is discussed in Section VII.
In Section VIII a general prescription (two shell discretization) for
eliminating the order $a^2$ errors is presented, and the order $a^4$ errors
in this scheme are estimated in Section IX, followed by the conclusion in
Section X.

\section{Laplacian on a Lattice}

On hypercubical lattices, the discretization of the Laplacian operator is
usually constructed through the trapezoidal rule applied on each lattice
direction.
For example, the second derivative in the $x$ direction of an arbitrary
function $\phi$ at the origin is
\begin{equation}
a^2 \partial_x^2 \phi(0) = \phi(a\hat x) - 2\phi(0) + \phi(-a\hat x)
+ \dots,
\end{equation}
and by summing up over all lattice directions,
\begin{eqnarray}
a^2 \partial^2 \phi(0) &=& \sum_{\hat\mu} \Big(\phi(a\hat\mu) - 2\phi(0)
+ \phi(-a\hat\mu)\Big) + \dots\nonumber\\
&=& \bigg\{ \sum_{\hat\mu} \Big(\exp(a\partial_\mu) -2 + \exp(-a\partial_\mu)
\Big)\bigg\} \phi(0) + \dots,
\label{Z}
\end{eqnarray}
where the $\hat\mu$'s are the unit vectors along the lattice directions.

The construction above depends so heavily on the existence of a set of
mutually orthogonal lattice vectors $\{\hat\mu\}$ that it cannot be naturally
generalized to non-hypercubical lattices.
(Try the two dimensional triangular lattice, for instance.)
To obtain a construction which is equally applicable to hypercubical and
non-hypercubical lattices, we will make use of the coordinate-independent
definition of the Laplacian operator.
Recall that when $\phi$ satisfies the $n$-dimensional Laplace equation
$\partial^2\phi=0$, the value of $\phi$ at any point is the average of its
value on an $(n-1)$-dimensional hypersphere of arbitrary radius centered at
the point in question.
\begin{equation}
\phi(0) = \int_{S^{n-1}} \; d\mu_{n-1} \; \phi(X),
\label{S}
\end{equation}
where $d\mu_{n-1}$ is the unit measure on $S^{n-1}$.
\begin{equation}
\int_{S^{n-1}} \; d\mu_{n-1} \equiv \int_{S^{n-1}} \; d\Omega_{n-1}
{\Gamma(n/2) \over 2\pi^{n/2}} = 1.
\end{equation}
So one can take the Laplacian $\partial^2\phi$ as an indicator of how much
the equality Eq.~(\ref{S}) is violated.
Just from dimensional analysis one knows that
\begin{equation}
\partial^2\phi(0) = N \lim_{r\to0} \bigg( \int_{S^{n-1}} \; d\mu_{n-1} \;
\phi(X) - \phi(0) \bigg)\bigg/r^2,
\label{DefL}
\end{equation}
where $r$ is the radius of the hypersphere $S^{n-1}$, and $N$ is a
proportionality constant one can easily determine to be $2n$.\footnote
{The value of $N$ can be determined by considering the following
electrostatic system satisfying the Poisson equation $\partial^2\phi=\rho$
in $n$ dimensions.
Consider a ball of constant charge density $\rho$ and radius $r$.
One can easily see that the potential difference between the surface and
the center of the ball is $\rho r^2/2n$.
Substitute this back into Eq.~(\ref{DefL}) and $N$ is found to be $2n$.}

On a lattice, however, $\phi$ is only defined on discrete lattice sites,
and one can at best approximate the integral with the sum of $\phi$ over
discrete points.
It is useful to introduce some formalism here and define a shell ${\cal S}_r$
of lattice sites with radius $r>0$ as the set of all sites with $\|X\| = r$.
Obviously these ${\cal S}_r$'s are non-empty only for discrete values of
the $r$'s, and the smallest of these values, $\rho$, is the separation between
each lattice site and its nearest neighbors and is usually called the packing
diameter of the lattice.
We will call ${\cal S}_\rho$ the first shell, and its cardinality, {\it
i.e.}, the number of nearest neighbors, denoted by  $\tau$, is called the
kissing number\footnote
{Also known to chemists as the contact number or the coordination number,
and to historians as the Newton number, after Isaac Newton who studied the
problem in three dimensions.}.
Now we approximate the integral of $\phi$ over $S^{n-1}$ by its average
$\Sigma$ over the first shell
\begin{equation}
\Sigma \phi(0) = {\textstyle{1\over\tau}} \sum_{X_k\in{\cal S}_\rho} \;
\phi(X_k),
\end{equation}
where $\Sigma$ is defined as an operator over $\phi$, which can also be
expressed in terms of the partial derivatives.
\begin{equation}
\Sigma \phi(0) = \bigg\{ {\textstyle{1\over\tau}} \sum_{X_k\in{\cal S}_\rho}
\; \exp(X_k \cdot \partial) \bigg\} \; \phi(0).
\end{equation}

The corresponding approximation of the Laplacian operator is
\begin{equation}
D^2\phi(0) = 2n \bigg( {\textstyle{1\over\tau}} \sum_{X_k\in{\cal S}_\rho} \;
\phi(X_k)- \phi(0) \bigg)\bigg/\rho^2 = 2n (\Sigma - 1) \phi(0)/\rho^2,
\label{DefD}
\end{equation}
The nonlocal operator $D^2$ defined above, called the {\it one shell
discretization\/} of the Laplacian operator, is the main object of this
study.
How good is $D^2$ as an approximation of $\partial^2$?
Note that
\begin{eqnarray}
\rho^2 D^2\phi(0)/2n &=& {\textstyle{1\over\tau}} \sum_{X_k\in{\cal S}_\rho}
\; \phi(X_k) - \phi(0) = {\textstyle{1\over\tau}} \sum_{X_k\in{\cal S}_\rho}
\; \Big(\phi(X_k) - \phi(0)\Big)\nonumber\\&=&\bigg\{{\textstyle{1\over\tau}}
\sum_{X_k\in{\cal S}_\rho} \; \Big(\exp(X_k \cdot \partial) - 1 \Big) \bigg\}
\; \phi(0).
\end{eqnarray}
So the expression inside the braces is a derivative representation of the
nonlocal operator $\rho^2 D^2/2n$.
By expanding the exponential, which is equivalent to a Taylor expansion in
$\rho$, one can express $\rho^2 D^2/2n$ in terms of powers of the derivative
operator.
Now
\begin{equation}
\exp(X_k \cdot \partial) = 1 + X_k \cdot \partial + \textstyle{1\over2}
(X_k \cdot \partial)^2 + \textstyle{1\over6} (X_k \cdot \partial)^3
+ {\cal O}(\rho^4) .
\end{equation}
The order $\rho^0$ term is just unity, which is cancelled by the ``$-1$''
coming from the $\phi(0)$ term in Eq.~(\ref{DefD}).
The order $\rho^1$ term is proportional to
\begin{equation}
\Big( \sum_{X_k\in{\cal S}_\rho} X_k \Big) \cdot \partial = 0 ,
\end{equation}
due to the reflection symmetry of the lattice.
Similarly, all odd powers of $\rho$ vanish due to reflection symmetry.
On the other hand, the $\rho^2$ term is ${1\over2}\sum_k(X_k \cdot \partial)^2$
which is in general {\it not\/} a Laplacian unless the set of nearest
neighbors ${\cal S}_\rho$ has vanishing quadrapole moment.
So we have the theorem:

\bigskip

{\sl The one shell discretization $D^2$ of the Laplacian operator over the
first shell ${\cal S}_\rho$ is equal to the Laplacian operator $\partial^2$
itself in leading order if and only if the quadrapole moment of
${\cal S}_\rho$ vanishes.}

\bigskip

It must be emphasized that the theorem above applies only for one shell
discretizations as defined in Eq.~(\ref{DefD}).
In general one may give different weights to different lattice sites in
${\cal S}_\rho$ in the sum in Eq.~(\ref{DefD}) or even extend the sum to
sites which are not nearest neighbors, and the theorem does not apply to
such constructions.
However, such alternative discretization schemes contain many free
arbitrary parameters (as in the weights) and they are clearly beyond the
scope of this paper.

\section{Root Lattices}

We have seen that the one shell discretization method provides a good
approximation of the Laplacian operator as long as the nearest shell of
neighbors has vanishing quadrapole moment.
Unfortunately, the author is not aware of any classification of the
lattices satisfying this condition.
However, it is well known that root lattices have been completely
classified, and all of them have vanishing quadrapole moment.
Moreover, both the body-centered hypercubical and the simplicial lattices
fall into this class, and so does the usual hypercubical lattice.
So it is useful to study these root lattices closely.

\bigskip

{\sl A root lattice is one whose basis vectors can all be chosen to be nearest
neighbors and the angles between any two of them are either $\pi/2$ or
$2\pi/3$.
In other words, when the basis vectors are scaled to have unit length, the
inner products of any two of them are either $0$ or $-{1\over2}$.  }

\bigskip

\noindent This definition depends solely on the relative directions of
the nearest neighbors (as opposed to definitions in textbooks on Lie groups,
which may involve next-to-nearest neighbors) is more useful for the study
of one shell discretizations, which also just depend on the first shell.

The classification theorem for root lattices is as follows:

\bigskip

{\sl An $n$-dimensional root lattice is either isometric to one of the
Cartan--Dynkin root lattices $Z_n$, $A_n$, $D_n$ and $E_n$ for some
value of $a$, or is an outer product of several of them with the same $a$.  }

\bigskip

\noindent The Cartan--Dynkin lattices $Z_n$, $A_n$ and $D_n$ are defined
as below for any $n \in {\bf Z}$:
\begin{eqnarray}
Z_n \equiv \{X=(x_1,\dots,x_n)& &\in a{\bf Z}^n \}, \nonumber\\
A_n \equiv \{X=(x_1,\dots,x_n,&x_{n+1})\;&\in a{\bf Z}^{n+1}; \;
\sum_k x_k = 0\},\label{ZAD} \\
D_n \equiv \{X=(x_1,\dots,x_n)& &\in a{\bf Z}^n; \;\sum_k x_k \in
2a{\bf Z}\},
\nonumber
\end{eqnarray}
These representations have the nice features that the coordinates of the
lattice sites are always integer multiples of the lattice spacing $a$, and
each lattice is symmetric under permutations of the coordinates.
It is obvious that $Z_n$ are the square, cubic and hypercubical lattices for
$n=2$, 3 and 4 respectively.
The simplicial lattices $A_n$ are for $n=2$ the hexagonal lattices, for
$n=3$ the face-centered cubic lattices, and for $n=4$ the four dimensional
simplicial lattices studied in Ref.~\cite{DM1,DM2,DM3,DM4,DM5}.
The checkerboard lattices $D_n$ are also square lattices for $n=2$
$(D_2 \simeq Z_2)$, face-centered cubic lattices for $n=3$ $(D_3
\simeq A_3)$ and Celmaster lattices \cite{Cel1,Cel2,Cel3} for $n=4$.
Lastly, the $E_n$ lattices are defined only for $n=6$, 7 or 8.
Since we are mainly interested in lattices in low dimensions, we will
not discuss $E_n$ and refer the interested reader to
Ref.~\cite{Con} instead.

Several questions immediately come to mind.
Firstly, are the $A_n$ and $D_n$ really the same Cartan--Dynkin root
lattices one uses for the representations of Lie groups\footnote{
Readers who are not familiar with the Cartan--Dynkin classification
scheme for Lie groups can consult Slansky \cite{Sla} and Georgi \cite{Geo}.}?
What is $Z_n$ in that language, and where are the $B_n$, $C_n$, and the
exceptional groups?
The first question can be easily answered by evaluating the angles between
the basis vectors of $A_{n-1}$:
\begin{equation}
(x_1, \dots, x_n) = (1,-1,0,\dots,0), (0,1,-1,0,\dots,0),
\dots, (0,\dots,0,1,-1),
\end{equation}
where each vector is rotated by an angle $2\pi/3$ from its predecessor as
described by the $A_n$ Dynkin diagrams.
The $D_n$ lattice has one more basis vector than $A_{n-1}$, namely $(-1,-1,0,
\dots,0)$, which is orthogonal to the first basis vector of $A_{n-1}$ but
makes an angle of $2\pi/3$ with the second, exactly as decreed by the $D_n$
Dynkin diagrams.
Lastly, the $Z_n$ lattice also has one more basis vector than $A_{n-1}$,
namely $(-1,0,\dots,0)$, which makes an angle $3\pi/4$ with the first
basis vector of $A_{n-1}$, is orthogonal to the rest, and is shorter than the
rest of the basis vectors by a factor of $1/\sqrt{2}$, {\it i.e.}, it has
exactly the properties of the ``double bond'' of the Dynkin diagram of $B_n$.
In other words, $Z_n$ is the root lattice of the Lie group $B_n$.

The last exercise illustrates the fact that a given root lattice can be
spanned by different sets of basis vectors (and hence different Lie groups),
and the choice most convenient for the study of Lie groups is not
necessarily the most convenient for our purposes.
As we have seen, the set of nearest neighbors ${\cal S}_\rho$ plays a
special role in our discussion.
As a result, we have chosen to define root lattices in a way which depends
only on the relative directions of the nearest neighbors.
The unique Lie group defined by the nearest neighbors on a given root
lattice is always simply laced (no multiple bonds, all basis vectors have
the same length).
For $A_n$ it is SU$(n+1)$, for $D_n$ it is SO$(2n)$ and for $Z_n$ it is
SU$(2)^n$ (not SO$(2n-1)$).\footnote
{This is why we avoid calling the hypercubical lattice $B_n$ --- to emphasize
that we are studying the root lattice, not the Lie group.}

The remaining Dynkin diagrams ($C_n$, $F_4$ and $G_2$) are all multiply
laced (with multiple bonds, and basis vectors of different lengths).
One can easily check that the $C_n$ and $D_n$ groups have the same root
lattices, but the former is doubly laced while the latter is simply laced.
Similarly, the simply laced counterparts of $G_2$ and $F_4$ are $A_2$ and
$D_4$ respectively\footnote{The authors of Ref.~\cite{Neu1,Neu2,Neu3}
actually referred to the Celmaster lattice $D_4$ as the $F_4$ lattice.}.

So we have clarified the relationship between the Lie groups (which have
little to do with this paper) and the root lattices (our real concern).
The reader should be warned that, for the purposes of this paper, it is
advisable to temporarily forget about the Lie group structures of the
lattices and regard the root lattices as geometrical constructions in their
own right.

\section{Discretized Laplacian Operators on Root Lattices}

In this section, we will give explicit formulae for $\Sigma$ and $D^2$ on
root lattices $Z_n$, $A_n$ and $D_n$.
Recall that, since all lattices are symmetrical under reflection with
respect to the origin ($X_m \to -X_m$ for all $m$, $m$ being the
dimensionality of the coordinate space, {\it i.e.}, $n$ for $Z_n$ and $D_N$
and $n+1$ for $A_n$), in the Taylor expansion of $\Sigma$, only terms with
even powers of $\rho$ survive.
Hence
\begin{equation}
\Sigma = {\textstyle{1\over\tau}} \sum_{X_k\in{\cal S}_\rho} \;
\exp(X_k \cdot \partial) \equiv {\cal O}_0 + a^2 {\cal O}_2 + a^4 {\cal O}_4 +
a^6 {\cal O}_6 + \dots .
\end{equation}
The operators ${\cal O}_l$ are $l$-th order derivatives, their forms
constrained by the symmetries of the lattices.
As mentioned before, all three root lattices defined in Eq.~(\ref{ZAD})
are invariant under permutations of the coordinates.
Moreover, $Z_n$ and $D_n$ are also symmetric under reflection of individual
coordinates ($X_m \to -X_m$ for a given $m$).
These symmetry constraints limit the number of possible operators appearing
at each order.

To enumerate the number of possible operators appearing at each order, the
following definitions are useful.
\begin{eqnarray}
{\cal B}_{j_1j_2\dots j_p}^{(m)} &=& \sum_{1\le i_1 \neq
i_2\neq\dots\neq i_p\le m} \partial_{i_1}^{j_1} \partial_{i_2}^{j_2} \dots
\partial_{i_p}^{j_p}, \quad \hbox{all $j$'s even,}\nonumber\\
{\cal A}_{j_1j_2\dots j_p}^{(m)} &=& \sum_{1\le i_1 \neq
i_2\neq\dots\neq i_p\le m} \partial_{i_1}^{j_1} \partial_{i_2}^{j_2} \dots
\partial_{i_p}^{j_p}, \quad \hbox{some $j$'s odd.}
\end{eqnarray}
The definitions of the even operators $\cal B$ and the odd operators
$\cal A$ are actually identical except that $\cal A$ may contain odd powers
of $\partial_i$, which is not invariant under reflection of coordinate $x_i$.
As a result, ${\cal O}_l$ for $Z_n$ and $D_n$ contains only terms like
${\cal B}_{j_1j_2\dots j_p}^{(n)}$ with $\sum_p j_p = l$ but no $\cal A$
terms.
On the other hand, for $A_n$, both ${\cal B}_{j_1j_2\dots j_p}^{(n+1)}$ and
${\cal A}_{j_1j_2\dots j_p}^{(n+1)}$ contribute.

It is easy to see that ${\cal O}_0=1$ for any lattice.
There are two operators which appear at order $a^2$, namely
\begin{eqnarray}
{\cal B}_2^{(m)} =& \sum_{1\le i \le m} &\partial_i^2, \nonumber\\
{\cal A}_{11}^{(m)} =& \sum_{1\le i\neq j \le m} &\partial_i \partial_j .
\end{eqnarray}
As mentioned before, for $Z_n$ and $D_n$, $\cal A$ does not contribute and
${\cal O}_2$ must be proportional to the even operator ${\cal B}_2^{(m)}$,
which is exactly the Laplacian operator in $m$ dimensions, and $m=n$ for
$Z_n$ and $D_n$.
On the other hand, for $A_n$, ${\cal O}_2$ cannot be simply proportional to
${\cal B}_2^{(m)}$, as we want something proportional to the Laplacian
operator in $n$, not $m=n+1$, dimensions.
However, for $A_n$ the odd operator ${\cal A}_{11}^{(m)}$ does contribute, and
as we will see below, exactly in the right amount to give the Laplacian
operator in an $n$ dimensional subspace.

When one goes to order $a^4$, more operators appear.
\begin{eqnarray}
{\cal B}_4^{(m)} =& \sum_{1\le i \le m} &\partial_i^4, \nonumber\\
{\cal B}_{22}^{(m)} =& \sum_{1\le i\neq j \le m} &\partial_i^2 \partial_j^2,
\nonumber\\
{\cal A}_{31}^{(m)} =& \sum_{1\le i\neq j \le m} &\partial_i^3\partial_j,
\nonumber\\
{\cal A}_{211}^{(m)} =& \sum_{1\le i\neq j\neq k \le m} &\partial_i^2
\partial_j \partial_k, \nonumber\\
{\cal A}_{1111}^{(m)} =& \sum_{1\le i\neq j\neq k\neq l \le m} &\partial_i
\partial_j \partial_k \partial_l.
\end{eqnarray}
As before, the odd operators only contribute in the $A_n$ case while the
even ones in general may contribute for all three classes of root lattices.

After discussing the generalities, we will move on to specific cases and
calculate $\Sigma$, and hence $D^2 = 2n(\Sigma-1)/ \rho^2$, for all three
classes of root lattices.
We will also list the kissing numbers $\tau$, packing diameters $\rho$ and
densities of lattice sites $\delta$ of the lattices.
Many of the properties listed below are extracted from the excellent book
by Conway and Sloane \cite{Con}, although the definitions of $\rho$ and
$\delta$ here are different from those in the book.

\subsection{The Hypercubical Lattice $Z_n$}

The hypercubical lattice $Z_n$, defined as
\begin{equation}
Z_n \equiv \{X=(x_1,\dots,x_n) \in a{\bf Z}^n \},
\end{equation}
has kissing number $\tau=2n$, packing diameter $\rho=a$ and density
$\delta = 1/a^n$.
One can easily evaluate $\Sigma$, which takes the form
\begin{equation}
\Sigma = 1 + \textstyle{1\over2!n} a^2 {\cal B}_2^{(n)} +
\textstyle{1\over4!n} a^4 {\cal B}_4^{(n)} + {\cal O}(a^6) .
\end{equation}
One can easy show that, due to the orthogonal nature of $Z_n$,
\begin{equation}
{\cal O}_l = {\textstyle{1\over l!n}}\sum_{1\le i \le n}\partial_i^l .
\end{equation}
Lastly, the one shell discretization $D^2$ is
\begin{equation}
D^2 = {\cal B}_2^{(n)} + {\textstyle{2\over 4!}} a^2 {\cal B}_4^{(n)} + \dots
= \partial^2 + {\textstyle{1\over 12}} a^2 \sum_{1\leq i\leq n} \partial_i^4
+ {\cal O}(a^4),
\end{equation}
and as predicted, $D^2 = \partial^2$ in the leading order.

\subsection{The Checkerboard Lattice $D_n$}

The checkerboard lattice $D_n$, defined as
\begin{equation}
D_n \equiv \{X=(x_1,\dots,x_n) \in a{\bf Z}^n; \sum_k x_k \in 2a{\bf Z}\},
\end{equation}
has kissing number $\tau=2n(n-1)$ and packing diameter $\rho = \sqrt{2}a$.
Only half of the lattice sites of $Z_n$ is occupied in $D_n$, so its
density $\delta = 1/2a^n$ is half of that of $Z_n$.
It is straightforward to calculate $\Sigma$,
\begin{equation}
\Sigma = 1 + \textstyle{2\over 2!n} a^2 {\cal B}_2^{(n)} +
\textstyle{2\over 4!n} a^4 {\cal B}_4^{(n)} +
\textstyle{6\over 4!n(n-1)} a^4 {\cal B}_{22}^{(n)} + {\cal O}(a^6),
\end{equation}
and bearing in mind that now $\rho^2 = 2 a^2$,
\begin{eqnarray}
D^2 &=& {\cal B}_2^{(n)} + {\textstyle{2\over 4!}} a^2 {\cal B}_4^{(n)} +
{\textstyle{6\over 4!(n-1)}} a^2 {\cal B}_{22}^{(n)} + {\cal O}(a^4)\nonumber\\
&=& \partial^2 + {\textstyle{1\over 12}} a^2 \sum_{1\leq i\leq n} \partial_i^4
+ {\textstyle{1\over 4(n-1)}} a^2 \sum_{1\leq i\neq j\leq n} \partial_i^2
\partial_j^2 + {\cal O}(a^4).
\end{eqnarray}
Again, we verify that $D^2 = \partial^2$ in the leading order\footnote{
This discretization is equivalent to that used in Ref.~\cite{Neu1,Neu2,Neu3}}.

The four dimensional checkerboard lattice $D_4$ is of special importance
as it has been extensively studied by Celmaster \cite{Cel1,Cel2,Cel3} and
is usually known as the Celmaster, or body-centered hypercubical lattice.
\begin{equation}
D^*_4 \equiv \{X=(x_1,x_2,x_3,x_4) \in a{\bf Z}^4 \; \cup \;
a({\bf Z}^4 + (\textstyle{1\over2}, {1\over2}, {1\over2}, {1\over2})) \},
\end{equation}
{\it i.e.}, the coordinates are either all integers or all half integers.
The Celmaster lattice $D^*_4$ is actually the dual, or reciprocal lattice,
of $D_4$.
Reciprocal lattices are in general quite different: in three dimensions,
the face-centered cubic and body-centered cubic lattices are reciprocal,
and are not similar at all.
In this special case, however, $D_4$ and $D^*_4$ are actually similar to
each other.
First note that both have 24 sites in the first shell:
\begin{eqnarray}
D_4&:& \quad (\pm1,\pm1,0,0) + \hbox{perm.}; \nonumber\\
D^*_4&:& \quad (\pm1,0,0,0) + \hbox{perm.}, \qquad (\textstyle
{\pm{1\over2},\pm{1\over2},\pm{1\over2},\pm{1\over2}}) + \hbox{perm.};
\label{D1}
\end{eqnarray}
and 24 sites on the second shell
\begin{eqnarray}
D_4&:& \quad (\pm2,0,0,0) + \hbox{perm.}, \qquad (\pm1,\pm1,\pm1,\pm1) +
\hbox{perm.}; \label{D2}\\ D^*_4&:& \quad (\pm1,\pm1,0,0) + \hbox{perm.}\quad.
\nonumber
\end{eqnarray}
It is clear that the first shell of $D_4$ is exactly the second shell
of $D^*_4$ and the second shell of $D_4$ is the first shell of $D^*_4$
scaled up by a factor of 2.
One can actually show that one can start out with $D_4$, perform a global
rotation and overall rescaling by a factor of $1/\sqrt{2}$ and end up with
$D^*_4$.
So $D_4$ and $D^*_4$ are geometrically similar in the sense that similar
triangles are similar --- they can be mapped onto each other through
global rotation and overall scaling.
As we will see below, the first and second shell of $D_4$ being similar
will be important when we try to cancel ${\cal O}(a^2)$ errors by going
beyond one shell discretization.

\subsection{The Simplicial Lattice $A_n$}

Lastly, we will discuss the simplicial lattice $A_n$, which is the hardest
to study as it is embedded in an $n+1$-dimensional coordinate space.
\begin{equation}
A_n \equiv \{X=(x_1,\dots,x_n,x_{n+1})\;\in a{\bf Z}^{n+1}; \sum_k x_k = 0\}.
\end{equation}
it has kissing number $\tau=n(n+1)$, packing diameter $\rho = \sqrt{2}a$,
and density\footnote{
The easiest way to show this is by observing that $Z_{n+1}$ has density
$1/a^{n+1}$, and the $A_n$-like sublattices $\{X \in a{\bf Z}^{n+1};
\sum_k x_k = C\}$ where $C$ is a fixed integer, have density $\sqrt{n+1}/a$
in $Z_{n+1}$.
The quotient of the two gives the density of $A_n$.}
$\delta = 1/\sqrt{n+1}a^n$, and
\begin{eqnarray}
\Sigma &=& 1 + \textstyle{2\over 2!(n+1)} a^2 {\cal B}_2^{(n+1)} +
\textstyle{-2\over 2!n(n+1)} a^2 {\cal A}_{11}^{(n+1)} \nonumber\\ & &\quad +
\textstyle{2\over 4!(n+1)} a^4 {\cal B}_4^{(n+1)} +
\textstyle{6\over 4!n(n+1)} a^4 {\cal B}_{22}^{(n+1)} +
\textstyle{-6\over 4!n(n+1)} a^4 {\cal A}_{31}^{(n+1)} + {\cal O}(a^6),
\end{eqnarray}
without any terms proportional to ${\cal A}_{211}$ or ${\cal A}_{1111}$.
One can then read off $D^2$, which takes a rather complicated form.
\begin{eqnarray}
D^2 &=& \textstyle{n\over (n+1)} {\cal B}_2^{(n+1)} -
\textstyle{1\over (n+1)} {\cal A}_{11}^{(n+1)} \nonumber \\& &\quad +
\textstyle{2n\over 4!(n+1)} a^2 {\cal B}_4^{(n+1)} +
\textstyle{6n\over 4!n(n+1)} a^2 {\cal B}_{22}^{(n+1)} -
\textstyle{6n\over 4!n(n+1)} a^2 {\cal A}_{31}^{(n+1)} + {\cal O}(a^4)\\
&=& {\textstyle{n\over (n+1)}} \sum_{1\le i \le n+1} \partial_i^2 -
{\textstyle{1\over (n+1)}} \sum_{1\le i \neq j\le n+1} \partial_i\partial_j
\nonumber\\ & &\quad + {\textstyle{n\over 12(n+1)}} a^2 \sum_{1\le i \le n+1}
\partial_i^4 + {\textstyle{1\over 4(n+1)}} a^2 \sum_{1\le i \neq j\le n+1}
\partial_i^2\partial_j^2 - \textstyle{{1\over 4(n+1)}} a^2
\sum_{1\le i \neq j\le n+1} \partial_i^3\partial_j^1 + {\cal O}(a^4).
\nonumber
\end{eqnarray}
Does the sum of the first two terms give the Laplacian operator in the
$n$-dimensional subspace described by $\sum x_i = 0$?
Note the following relation between the Laplacian operators in $n$ and $n+1$
dimensions.
\begin{equation}
\partial^2\Big|_{{\bf R}^{n+1}} = \partial^2\Big|_{{\bf R}^n} +
\partial_{\bar x}^2,
\end{equation}
where $\bar x=(1,1,\dots,1)/\sqrt{n+1}$ is the unit normal vector of the
subspace $\sum x_i = 0$.
One can easily check that
\begin{equation}
\partial_{\bar x}^2 = \bigg( \sum_{1\le i \le n+1} \partial_i^2
+ \sum_{1\le i \neq j\le n+1} \partial_i\partial_j \bigg) \bigg/(n+1),
\end{equation}
and hence
\begin{equation}
\partial^2\Big|_{{\bf R}^n} = \partial^2\Big|_{{\bf R}^{n+1}} -
\partial_{\bar x}^2 = {\textstyle{n\over (n+1)}} \sum_{1\le i \le n+1}
\partial_i^2 - {\textstyle{1\over (n+1)}} \sum_{1\le i \neq j\le n+1}
\partial_i\partial_j ,
\end{equation}
exactly as advertised.

\section{Discretization Errors}

We have seen that $D^2$ does reproduce $\partial^2$ exactly in leading order
for all three classes of root lattices we have studied.
In this section, we will analyze the order $a^2$ errors of $D^2$, which
we will denote by $E$ and which is related to ${\cal O}_4$ by
\begin{equation}
E = 2n {\cal O}_4 a^4/ \rho^2.
\end{equation}
Its forms for all three classes of root lattices are summarized below.
\begin{eqnarray}
Z_n &: E= {\textstyle{1\over 4!}} a^2 &2 {\cal B}_4^{(n)} , \nonumber\\
D_n &: E= {\textstyle{1\over 4!}} a^2 &\big(2 {\cal B}_4^{(n)}
+ \textstyle{6\over n-1}  {\cal B}_{22}^{(n)} \big) , \\
A_n &: E= {\textstyle{1\over 4!}} a^2 &\big(\textstyle{2n\over n+1}
{\cal B}_4^{(n+1)} + \textstyle{6n\over n(n+1)} {\cal B}_{22}^{(n+1)} -
\textstyle{6n\over n(n+1)} {\cal A}_{31}^{(n+1)} \big) . \nonumber
\end{eqnarray}
At first sight the coefficients may look random and without any discernible
patterns, but as we will see below they are actually not as random as they
look.

All the expressions for $E$ above are quartic polynomials of $\partial_i$.
Naively, one may estimate their size by $p^4$, where $p$ is the typical
momentum scale in question.
However, such an estimate is sloppy, since as an operator $p^4 = \partial^4
= (\partial^2)^2$, {\it i.e.}, the square of the Laplacian operator, which
is a scalar operator in $n$-dimensional Euclidean space, while ${\cal B}_4$,
${\cal B}_{22}$ and ${\cal A}_{31}$ are all non-scalar operators as defined.
Thus motivated, we will define $E_0$, the scalar or isotropic part of $E$, as
\begin{equation}
E_0 = \int d \mu_{n-1}\;E =\;\hbox{(dimensionless numerical factor)}\;a^2
\partial^4,
\end{equation}
where $d\mu_{n-1}$ is again the unit measure on $S^{n-1}$.
In other words, $E_0$ is $E$ integrated over all directions in $n$-dimensional
momentum space.
When one evaluates $E_0$, it turns out that
\begin{equation}
E_0 = {\textstyle{1\over 4!}} \partial^4 a^2\times\cases{6/(n+2), &for $Z_n$;
\cr 12/(n+2), &for $D_n$ or $A_n$, \cr}
\end{equation}
which is surprisingly simple (when contrasted with the expressions before
the integration).
Moreover, for all cases one has
\begin{equation}
E_0 = \rho^2 \partial^4/4(n+2).
\label{E0}
\end{equation}

The above relation is not obvious (at least to the author) from the raw
expressions for $E$, but once motivated it can actually be proved very
easily for all lattices (not necessarily root lattices).
Recall that $\Sigma$ is defined as
\begin{equation}
\Sigma = \bigg\{ {\textstyle{1\over\tau}} \sum_{X_k\in{\cal S}_\rho}
\; \exp(X_k \cdot \partial) \bigg\}.
\end{equation}
For each site $X_k\in{\cal S}_\rho$, one can expand the exponential as
\begin{equation}
\exp(X_k \cdot \partial) = 1 + \textstyle{1\over2!} (X_k \cdot \partial)^2
+ \textstyle{1\over4!} (X_k \cdot \partial)^4 + \;\hbox{odd terms}\;+
{\cal O}(\rho^6),
\end{equation}
where the odd terms will eventually be cancelled and do not concern us here.
The quadratic and quartic terms can be rearranged in the following way.
\begin{eqnarray}
(X_k \cdot \partial)^2 &=& \rho^2 \partial^2 /n +
\Big[(X_k \cdot \partial)^2 -\rho^2 \partial^2 /n\Big], \nonumber\\
(X_k \cdot \partial)^4 &=& 3 \rho^4 \partial^4 /n(n+2) +
\Big[(X_k \cdot \partial)^4 - 3 \rho^4 \partial^4 /n(n+2)\Big].
\label{Sym}
\end{eqnarray}
The integrals
\begin{equation}
\int d\mu_{n-1} \; (X_k \cdot \partial)^2 = \rho^2 \partial^2 /n, \qquad
\int d\mu_{n-1} \; (X_k \cdot \partial)^4 = 3 \rho^2 \partial^4 /n(n+2),
\end{equation}
decree that the terms in square brackets in Eqs.~(\ref{Sym}) vanish under
integration over all directions.
Hence
\begin{equation}
\int d\mu_{n-1} \; \Sigma = {\textstyle{1\over\tau}}
\sum_{X_k\in{\cal S}_\rho} \big(1 + \textstyle{1\over2!n} \rho^2 \partial^2 +
\textstyle{3\over4!n(n+2)} \rho^4 \partial^4 \big) + {\cal O}(\rho^6) ,
\end{equation}
which is a sum over $\tau$ identical expressions and then divided by $\tau$,
and is equal to the expression itself,
\begin{equation}
\int d\mu_{n-1} \; \Sigma = 1 + \textstyle{1\over2!n} \rho^2 \partial^2 +
\textstyle{3\over4!n(n+2)} \rho^4 \partial^4 + {\cal O}(\rho^6).
\end{equation}
Lastly, we have the scalar part of $D^2$.
\begin{equation}
\int d\mu_{n-1} \; D^2 = \int d\mu_{n-1} \; 2n (\Sigma - 1) /\rho^2 =
\partial^2 + \textstyle{6\over4!(n+2)} \rho^2 \partial^4 + {\cal O}(\rho^4),
\end{equation}
and $E_0$ can be read off simply as the $\rho^2$ term.

The simple result Eq.~(\ref{E0}) has important implications.
Firstly, it gives us the following {\it impossibility theorem}:

\bigskip

{\sl One shell discretization of the Laplacian operator always suffers
from order $a^2$ errors, regardless of the choice of the lattice.}

\bigskip

\noindent One might have hope that, by using a lattice with a high degree of
symmetry, all ${\cal O}(a^2)$ terms may vanish identically.
We now see that is an impossible dream.
There is always a scalar part which is not prohibited by any symmetry.
The only way to eliminate order $a^2$ errors is by involving lattice sites
which are not nearest neighbors, as we will discuss later in this paper.

Another corollary of Eq.~(\ref{E0}) is that the scalar discretization error
$E_0$ depends only on a single geometrical parameter --- the packing diameter
$\rho$.
(It also depends on the spacetime dimensionality $n$, but that is usually
determined by the physics problem in question.)
One can then ask, in a given spacetime dimension, which lattice will
minimize $E_0$?
It will be the subject of the next section.

\section{Error Densities of One Shell Discretizations}

As defined, the scalar discretization error $E_0$ is independent of how
we describe the lattice on the coordinate space.
This is certainly a pleasing feature to physicists, who know that all
physical quantities should not depend on the choice of coordinates.
On the other hand, $E_0$ is proportional to $\rho^2$, which is not surprising;
the smaller $\rho^2$, the more lattice sites we are using to describe a
given physical volume, and hence the smaller is the error.
In other words, the error grows with $1/\delta$, the volume of the unitcell
of the lattice.
To determine which class ($Z_n$, $D_n$ or $A_n$) is optimal for our purposes,
one defines the {\it error density\/} $\epsilon$ as the ratio of the $E_0$
to (a certain power of) the unitcell volume.
\begin{equation}
E_0 = {1\over 4(n+2)} \, \epsilon \, \delta^{-2/n} \, \partial^4.
\end{equation}
The $-2/n$ power has been chosen such that $\epsilon$ is $\rho$ independent;
recall that $E_0$ and $\delta$ scale like $\rho^2$ and $\rho^{-n}$
respectively.
Hence $\epsilon$ thus defined is a sheer parameter (not an operator) which
just depends on which lattice which are using.
For root lattices, it carries the value
\begin{equation}
\epsilon = \cases{1,& for $Z_n$; \cr 2^{1-2/n},& for
$D_n$; \cr 2(n+1)^{-1/n},& for $A_n$. \cr}
\end{equation}
Obviously one can check that the error density of $Z_n$ is always smaller than
that of $D_n$ and $A_n$.\footnote{
Moreover, one also has $\epsilon_{Z_2} = \epsilon_{D_2}$ and $\epsilon_{A_3}
= \epsilon_{D_3}$, {\it i.e.}, the error density of a lattice is independent
of how it is named.  }
This is certainly not a surprising result, as we have shown in the previous
section that $E_0$ is proportional just to $\rho^2$, and for our root
lattices with the same $\rho$, $Z_n$ is the one with the smallest density.

We have seen that, {\it among root lattices}, $Z_n$ has the smallest error
density.
We can actually prove the much stronger result that $Z_n$ has the
smallest error density among all lattices whose ${\cal S}_\rho$ have
vanishing quadrapole moment, {\it i.e.}, all the lattices with
$D^2 = \partial^2$ in leading order.
The proof runs in two steps.
First, observe that among all lattices whose basis vectors can be chosen
to be nearest neighbors, $Z_n$ has the lowest density.
(Among all parallelohedra with all sides of unit length, the one with the
largest volume is a cube.)
So $Z_n$ has the lowest error density among them.
What about lattices whose basis vectors {\it cannot\/} all be chosen as
nearest neighbors?
Note that, for such a lattice $\Lambda$, there exists a proper sublattice
$\bar \Lambda \subset \Lambda$ which is generated by the nearest neighbors
of $\Lambda$.
The lattices $\Lambda$ and $\bar \Lambda$ have the same $\rho$ and hence
the same $E_0$, yet being a subset, the density of $\bar\Lambda$ is smaller
than that of $\Lambda$.
As a result, the error density of $\Lambda$ is larger than that of $\bar
\Lambda$, which in turn by the previous argument is larger than that of
$Z_n$.
So we have the important result:

\bigskip

{\sl Under one shell discretization, the hypercubical lattices have the
smallest error densities among all possible choices of lattices in the
same number of spacetime dimensions.}

\bigskip

Most lattice simulations are performed on hypercubical lattices, but usually
the reasons are historical (following Wilson's example in Ref.~\cite{Wil}) or
algorithmic ($D^2$ can be easily written down as in Eq.~(\ref{Z}) by using
the trapezoidal rule).
Here we provide another rationale: the hypercubical lattices have the lowest
error density.

\section{Anisotropy}

While $E_0$ measures the isotropic part of the order $a^2$ errors, $E-E_0$
measures the anisotropy of $D^2$.
The operator $E-E_0$ is a quartic polynomial in $\partial_i$, which in
general can have all $\ell$-th multipole moment up to $\ell=4$.
However, the first and third (dipole and octupole) multipole moments are
identically zero  as demanded by reflection symmetry, the second (quadrapole)
moment is explicitly required to vanish, and the zeroth (monopole) moment
$E_0$ has been subtracted out.
So $E-E_0$ is purely of the fourth (hecadexapole) moment.

It is straightforward to verify that $E-E_0$ is non-vanishing for all root
lattices with one exception: $D_4$, the checkerboard lattice in four
dimensions.
\begin{equation}
E= {\textstyle{1\over 4!}} 2 a^2 \big({\cal B}_4 + {\cal B}_{22} \big)
= {\textstyle{2\over 4!}} a^2 \partial^4 = E_0 \qquad \hbox{for $D_4$.}
\end{equation}
That $D_4$ is more isotropic than $Z_4$ has been noted by Celmaster in
Ref.~\cite{Cel1}, but here the result is stronger.

\bigskip

{\sl The checkerboard lattice in four dimensions $D_4$ is exactly isotropic
at order $a^2$.
It is the only unexceptional root lattice with this property.}

\bigskip

\noindent One can understand the isotropy of the $D_4$ lattice by studying
its geometry.
The $D_4$ lattice has an accidental threefold discrete symmetry (which is
also a symmetry of its Dynkin diagram) which mixes ${\cal B}_4$ and
${\cal B}_{22}$.
The only combination which is invariant under this threefold symmetry is
$\partial^4$.
In other words, rotational symmetry is protected by this accidental
threefold discrete symmetry at order $a^2$.

Since we are mainly interested in lattices in low dimensions ($n\leq 4$),
we have not studied the anisotropy of $E_{6,7,8}$, although the
calculation should be straightforward (though tedious --- the packing
number $\tau$ is 240 for $E_8$).
We also have not studied the issue of anisotropy for non-root lattices.
However, since we have seen that the isotropy of $D_4$ is related to an
accidental discrete symmetry, a feature not apparent in these other cases,
the author would be very surprised if other lattices share this property
of isotropy.

\section{Beyond One Shell Discretization}

We have seen that order $a^2$ errors are unavoidable in one shell
discretization.
The order $a^2$ errors can be substantial on coarse lattices; for example,
as a rule of thumb the order $a^2$ error on a hypercubical lattice is around
10\% if $a$ is chosen to be around a third of the size of the physical
system in question \cite{GPL}.
In order to reduce the error, one may try to use finer lattices.
It is, however, a very costly procedure as the cost of lattice simulations
grows at least as fast as the number of lattice sites, which is proportional
to $a^{-n}$.
(And that is before taking critical slowing down into account.)
For $n=4$, to reduce the error to 1\% would increase the cost a hundredfold.
Another way of seeing the same result is by observing that the error density
$\epsilon$, which is $a$ independent, is the proportionality constant
between $E_0$ and $\delta^{-2/n}$.
To reduce $E_0$ by a factor of 10, $\delta$ must be decreased by a factor
of $10^{n/2}$, {\it i.e.}, a hundredfold in four dimensions.

Clearly it is not practical to control the $a^2$ errors just by using finer
and finer lattices.
Instead, one can completely eliminate the $a^2$ error by involving sites
not in ${\cal S}_\rho$, {\it i.e.}, by going beyond one shell discretization.
As an example, we will review how this is done on hypercubical lattices.

Recall that, for hypercubical lattices, one shell discretization on
${\cal S}_a$ gives
\begin{equation}
D^2[{\cal S}_a] = {\cal B}_2^{(n)} + {\textstyle{2\over 4!}} a^2
{\cal B}_4^{(n)} + {\cal O}(a^4)  = \partial^2 + {\textstyle{1\over 12}} a^2
\sum_{1\leq i\leq n} \partial_i^4 + {\cal O}(a^4),
\end{equation}
The argument in the square brackets is to remind us of which shell we are
using.
One can also discretize on the following set of $2n$ sites:
\begin{equation}
2{\cal S}_a = \{ (\pm 2a,0,\dots,0) + \; \hbox{perm.} \} .
\end{equation}
Notice that $2{\cal S}_a$ is just ${\cal S}_a$ scaled up by a factor of 2.
As a result,
\begin{equation}
D^2[2{\cal S}_a] = \partial^2 + {\textstyle{1\over 12}} (2a)^2
\sum_{1\leq i\leq n} \partial_i^4 + {\cal O}(a^4).
\end{equation}
Consequently, the combination ${4\over 3} D^2[{\cal S}_a] - {1\over 3}
D^2[2{\cal S}_a]$ is free of order $a^2$ errors.

Notice that $2{\cal S}_a$ is {\it not\/} the second shell of $Z_n$.
(In fact it is just part of the fourth shell.)
Since the errors are in general proportional to some high powers of the
radii of the shells involved, can we do better by using just the first and
second shell, ${\cal S}_a$ and ${\cal S}_{\sqrt{2}a}$?
Recall that ${\cal S}_{\sqrt{2}a}$ is the first shell of $D_n$, and $D^2$
has been calculated to be
\begin{equation}
D^2[{\cal S}_{\sqrt{2}a}] = {\cal B}_2^{(n)} + {\textstyle{2\over 4!}} a^2
{\cal B}_4^{(n)} + {\textstyle{6\over 4!(n-1)}} a^2 {\cal B}_{22}^{(n)}
+ {\cal O}(a^4) ,
\end{equation}
which contains both ${\cal B}_4^{(n)}$ and ${\cal B}_{22}^{(n)}$.
In contrast, $D^2[{\cal S}_a]$ involves just the former but not the latter.
Thus one cannot obtain an expression free of both ${\cal B}_4^{(n)}$ and
${\cal B}_{22}^{(n)}$ errors by taking a linear combination of
$D^2[{\cal S}_a]$ and $D^2[{\cal S}_{\sqrt{2}a}]$.
In other words, one cannot eliminate the order $a^2$ order by taking a linear
combination of $D^2$ of any two shells.

Bearing the above examples on the hypercubical lattices in mind, we will
now move on to discuss the framework of elimination of the order $a^2$ errors
on lattices in general.
Consider the one shell discretization on two shells of radii $r_{1,2}$
with $r_1<r_2$.
Then we want to see if the expression
\begin{equation}
D^2[{\cal S}_{r_1},{\cal S}_{r_2}] = c_1 D^2[{\cal S}_{r_1}] - c_2
D^2[{\cal S}_{r_2}],
\end{equation}
is equal to the Laplacian operator without any $a^2$ errors with some choices
of coefficients $c_{1,2}$.
Obviously in order to get the coefficient of the Laplacian operator to be
unity we need $c_1 - c_2 = 1$.
We also want the scalar errors $E_0$ of the two shells, which are
proportional to $r^2$, to cancel, by demanding $c_1 r_1^2 - c_2 r_2^2 =0$,
or $c_1/c_2 = (r_1/r_2)^{-2}$.
Putting these together we have
\begin{equation}
c_1 = r_2^2/(r_2^2 - r_1^2), \qquad c_2 = r_1^2/(r_2^2 - r_1^2).
\end{equation}

So now we can define the {\it two shell discretization\/} of the
Laplacian operator over shells ${\cal S}_{r_1}$ and ${\cal S}_{r_2}$ as
\begin{eqnarray}
D^2[{\cal S}_{r_1},{\cal S}_{r_2}] &=& \bigg(r_2^2 D^2[{\cal S}_{r_1}]
- r_1^2 D^2[{\cal S}_{r_2}] \bigg)\bigg/(r_2^2 - r_1^2) \nonumber\\
&\equiv& \partial^2 + a^2 {\cal H}_2 + a^4 {\cal H}_4 + {\cal O}(a^6).
\end{eqnarray}
Thus defined, the scalar discretization error of order $a^2$ is guaranteed
to vanish, {\it i.e.}, $\int d\mu_{n-1} \; {\cal H}_2 = 0$.
However, ${\cal H}_2$ itself may be non-zero, as shown by the example with
the first and second shell in $Z_n$.

When we defined one shell discretization early in this paper, we had to
explicitly demand that the shell in question have zero quadrapole moment.
Here the situation is similar.
For two shell discretization we have to demand that the two shell
configuration in question, after weighting by $c_{1,2}$, has vanishing
hexadecapole moment.
In other words, the hexadecapole moments of the two shells must cancel
each other.

In principle, one can translate the condition of the cancellation of
hexadecapole moments into mathematical constraints and look for their most
general solutions --- but in practice it is almost impossible.
(We have not even done that for the quadrapole case!)
For example, for $Z_n$ or $D_n$, the shells described by $(y_1, \dots y_n)$
and $(z_1, \dots z_n)$ (plus permutation) will have their fourth multipole
moment cancelled if and only if
\begin{equation}
\sum_i y_i^4 = K \sum_i z_i^4, \quad \hbox{and} \quad
\sum_{i,j} y_i^2 y_j^2 = K \sum_{i,j} z_i^2 z_j^2 \qquad \hbox{for some $K$.}
\end{equation}
These are coupled diophantine-like constraints, which in general cannot be
solved.
The corresponding constraints for $A_n$ are even more hideous.

However, there are special cases which we know the hexadecapole moment
are guaranteed to cancel, namely when the two shells are identical up to
scaling (but no rotation).
To be exact, let's define, for any natural number $p>1$,
\begin{equation}
p{\cal S}_\rho = \{pX: X\in {\cal S}_\rho\}.
\end{equation}
Then the two shell discretization $D^2[{\cal S}_\rho,p{\cal S}_\rho]$ has
vanishing hexadecapole moment and hence is free of order $a^2$ errors.
The proof is immediate.
For $D^2[{\cal S}_\rho] = \partial^2 + \rho^2 {\cal H}_2 + {\cal O}(a^4)$,
one has
\begin{eqnarray}
D^2[{\cal S}_\rho,p{\cal S}_\rho] &=& \bigg((p\rho)^2 D^2[{\cal S}_\rho]
- \rho^2 D^2[p{\cal S}_\rho] \bigg)\bigg/ ((p\rho)^2 - \rho^2)\nonumber\\
&=& \bigg(p^2\rho^2 (\partial^2 + \rho^2 {\cal H}_2) - \rho^2 (\partial^2
+ p^2 \rho^2 {\cal H}_2) \bigg)\bigg/ (p^2-1)\rho^2 + {\cal O}(\rho^4)
\nonumber\\&=& \partial^2 + {\cal O}(\rho^4).
\end{eqnarray}

In practice, the two shells should be chosen as small as possible to
minimize the error.
This suggests setting $p=2$, and
\begin{equation}
D^2[{\cal S}_\rho,2{\cal S}_\rho] = \textstyle{4\over3} D^2[{\cal S}_\rho]
- \textstyle{1\over3} D^2[2{\cal S}_\rho].
\end{equation}

On the other hand, there is a unique case where the hexadecapole moments
do cancel for two shells which are not related simply by scaling.
As the reader may have guessed, this is the case for the first and second
shells of the four dimensional checkerboard lattices, where the hexadecapole
moment of each shell vanishes identically as dictated by the threefold
discrete symmetry.
Recall that both the first shell ${\cal S}_{\sqrt{2}a}$ and the second shell
${\cal S}_{2a}$ contain 24 sites as described in Eqs.~(\ref{D1},\ref{D2}), and
the two shell discretization over these two shells is
\begin{equation}
D^2[{\cal S}_{\sqrt{2}a},{\cal S}_{2a}] = 2 D^2[{\cal S}_{\sqrt{2}a}] -
D^2[{\cal S}_{2a}] \qquad \hbox{for $D_4$},
\end{equation}
which can be checked to be free of order $a^2$ errors.

\section{Error Densities of Two Shell Discretizations}

Consider a two shell discretization with $a^2$ errors completely cancelled.
\begin{equation}
D^2[{\cal S}_{r_1},{\cal S}_{r_2}] = \partial^2 + a^4 {\cal H}_4 + \dots
\end{equation}
To study the order $a^4$ errors, one needs to know ${\cal H}_4$, which in
general is a sextic polynomial of $\partial_i$.
To calculate ${\cal H}_4$ would be a painful exercise.
However, it is easy to evaluate $E'_0$, the scalar part of ${\cal H}_4$,
\begin{equation}
E'_0 = \int d\mu_{n-1} \; a^4 {\cal H}_4 .
\end{equation}
Again we will start with the Taylor expansion of $\exp(X_k \cdot \partial)$.
\begin{equation}
\exp(X_k \cdot \partial) = 1 + \textstyle{1\over2!} (X_k \cdot \partial)^2
+ \textstyle{1\over4!} (X_k \cdot \partial)^4 + \textstyle{1\over6!}
(X_k \cdot \partial)^6 + \hbox{odd terms} + {\cal O}(a^8).
\end{equation}
With $X_k \in {\cal S}_r$ (not necessarily the first shell), it can be
rewritten as
\begin{eqnarray}
\exp(X_k \cdot \partial) &=& 1 + \textstyle{1\over2!} I_2 r^2 \partial^2
+ \textstyle{1\over4!} I_4 r^4 \partial^4 + \textstyle{1\over6!} I_6 r^6
\partial^6 \nonumber\\ & &\quad + \; [\dots] + \hbox{odd terms} + {\cal O}(r^8),
\label{E2}
\end{eqnarray}
where $I_l$ is defined by
\begin{equation}
\int d\mu_{n-1} (X_k \cdot \partial)^l = I_l r^l \partial^l,
\end{equation}
with $I_2 = 1/n$ and $I_4 = 3/n(n+2)$.
We have not calculated $I_6$ which fortunately is not needed for our
purposes.
Lastly, the $[\dots]$ in Eq.~(\ref{E2}), like the terms in square brackets in
Eq.~(\ref{Sym}), vanishes under integration over all directions.

The scalar part of the one shell discretization on ${\cal S}_r$ is
\begin{equation}
\int d\mu_{n-1} \; D^2[{\cal S}_r] = \partial^2 + \textstyle{2!\over4!}
(I_4/I_2) r^2 \partial^4 + \textstyle{2!\over6!} (I_6/I_2) r^4 \partial^6
+ {\cal O}(r^6).
\end{equation}
and the scalar part of the two shell discretization on ${\cal S}_{r_1}$
and ${\cal S}_{r_2}$ is
\begin{equation}
D^2[{\cal S}_{r_1},{\cal S}_{r_2}] = \partial^2 - \textstyle{2!\over6!}
(I_6/I_2) r_1^2 r_2^2 \partial^6 + {\cal O}(r^6),
\end{equation}
and hence
\begin{equation}
E'_0 = - \textstyle{2!\over6!} \, (I_6/I_2) \, r_1^2 r_2^2 \, \partial^6 .
\end{equation}

Now one can define the error density $\epsilon'$ of a two shell
discretization as
\begin{equation}
E'_0 = - \textstyle{2!\over6!} \, (I_6/I_2) \, \epsilon' \, \delta^{-4/n}
\, \partial^6 .
\end{equation}
As expected, using larger shells will increase the error density.
For root lattices, and using shells ${\cal S}_\rho$ and $2{\cal S}_\rho$,
the error densities are
\begin{equation}
\epsilon' = \cases{4,& for $Z_n$; \cr 2^{4-4/n},& for
$D_n$; \cr 16(n+1)^{-2/n},& for $A_n$. \cr}
\end{equation}
Again, the error densities are the smallest for hypercubical lattices.
The reason is the same as the one shell case: for the same $\rho$, $Z_n$
has the lowest density.
And the result can be as easily generalized to non-root lattices as before.

On the other hand, for $D_4$ we can simply use the first and second shells,
and the error density is
\begin{equation}
\epsilon = 4, \quad \hbox{for} \quad D^2[{\cal S}_{\sqrt{2}a},{\cal S}_{2a}],
\end{equation}
just as low as that of $Z_n$.
Hence we come to the following {\it conjecture}:

\bigskip

{\sl Under two shell discretization, the hypercubical lattices have the
smallest error densities among all possible choices of lattices in the
same number of spacetime dimensions.
The only other case with an error density as low is two shell discretization
on the first and second shells of the four dimensional checkerboard lattice.}

\bigskip

\noindent The statement above is just a conjecture, not a theorem, as there
may be two shells configurations other than ${\cal S}_\rho$ and
$2{\cal S}_\rho$ which have even lower error densities, just as the $D_4$
case we have seen.
However, exact cancellations of the hexadecapole moments of two shells
are very unlikely to happen without a symmetry rationale (like the threefold
discrete symmetry of $D_4$), and since we do not see such extra symmetries
in other lattices, the author concludes that it is very unlikely that
counterexamples to the above conjecture will be found.

\section{Conclusion}

Let's recapitulate what we have learned through this study:

$\bullet$ One shell discretization faithfully reproduces the Laplacian
operator (with order $a^2$ errors) if and only if the shell has vanishing
quadrapole moment.

$\bullet$ The order $a^2$ error for one shell discretization is non-zero
for any choice of lattice.
The hypercubical lattices have the smallest error densities in any spacetime
dimensions.

$\bullet$ Due to its threefold discrete symmetry, the four dimensional
checkerboard lattice is isotropic up to order $a^2$.
It is the only root lattice, and also probably the only lattice (root or
non-root), with this property.

$\bullet$ Two shell discretization reproduces the Laplacian operator without
order $a^2$ errors if and only if the hexadecapole moments of the two shells
cancel each other.

$\bullet$ The order $a^2$ error for one shell discretization is non-zero
for any choice of lattice.
It is likely (though not yet proved) that the hypercubical lattices have
the smallest error densities in any spacetime dimensions.
In four dimensions, the error density of the checkerboard lattice is just
as small as the hypercubical.

One can see a clear pattern recurring throughout this discussion.
The scalar (isotropic) properties of discretizations can be very cleanly
studied by Taylor expanding $\exp(X_k \cdot \partial)$ and then integrating
over all directions.
By involving more shells, one can eliminate the errors order by order.
On the other hand, the anisotropic properties are in general very difficult
to study.
The isotropy of $D_4$ is more like an accident (at least in our framework)
--- and it is straightforward to show that that the symmetry does not prevent
higher multipoles appearing at order $a^4$.
As a result, the most tractable way of eliminating anisotropic errors is
by cancellation between shells related just by a simple scaling.

We have discussed the root lattices in detail and calculated their $\Sigma$
explicitly.
However, most of our results are equally valid for root and non-root lattices.
However, root lattices are so tractable in theory, and useful in practice,
that it makes a very natural starting point.
The only important non-root lattice in low dimensions is $D_3^*$, the three
dimensional body-centered cubic lattice, which is the dual of $D_3$.
\begin{equation}
D^*_3 \equiv \{X=(x_1,x_2,x_3) \in a{\bf Z}^3 \; \cup \;
a({\bf Z}^3 + (\textstyle{1\over2}, {1\over2}, {1\over2})) \},
\end{equation}
Its coordinates are either all integers or all half integers.
Its nearest neighbors are $(\pm{1\over2}, \pm{1\over2}, \pm{1\over2})$,
and angles between them are not rational fractions of $\pi$.
One can study $D_3^*$ just as we study the root lattices, but we
do not find any properties of interest.

What are the implications of this study for real lattice simulations?
In no way does it challenge the canonical choice --- the hypercubical lattice,
which always has the smallest error density.
The only case when one may consider otherwise is when isotropy is
important; in such scenarios the four dimensional checkerboard lattice may
be a tempting alternative.
Still, as disclaimed in the introduction, the geometry of the lattice is
only one of many factors which affect the efficiency of lattice simulations.
Algorithmic considerations may overwhelm geometrical ones; remember the two
shell discretization of $D_4$ involves 48 neighbors, opposed to 16 for $Z_4$.
Still, the author expects this study is useful in providing a general
framework in which different lattices may be discussed on equal footing,
and the results may serve as ``rules of thumb'' for lattice simulations.

In this paper, we have studied how the geometries of non-hypercubical
lattices affect the discretization of the Laplacian operator.
Possible generalization to the case of the Dirac operator is possible but
not straightforward, due to the problem of parity doubling.
Without doing a careful analysis, it is difficult to foresee whether the
results we have obtained in this paper can be extended to the Dirac case,
However, the author is willing to make the following {\it conjecture}:
the order $a^2$ discretization error of the Dirac operator does not vanish
for any choice of lattice.
Just as the discretization of the Laplacian operator is always plagued with
errors like $a^2 \partial^4$ which have the same tranformation properties
as the Laplacian $\partial^2$, it is probable that the discretization of
the Dirac operator $\gamma \cdot \partial$ will suffer from errors like
$a^2 (\gamma \cdot \partial) \partial^2$.
Hopefully the author can extend the present framework to the Dirac case in
the near future and determine the correctness of this conjecture.

\acknowledgements
I would like to thank Silas Beane, Tom Cohen, Jim Griffin and Steve Wallace
for discussions.
This work is supported by the U.S.~Department of Energy grant
DE-FG02-93ER-40762.

redpajama_set_name: RedPajamaArXiv


Here you will find listings of where you can buy Dr Optimiser's music. Click on the album covers below to listen to the albums on SoundCloud. Some of the earlier tracks are free to download on SoundCloud. Music videos can be viewed on Dr Optimiser's YouTube Channel.


text:
Joe Dingwall (born 2 October 1988) is a former professional footballer who played for Scottish Football League club East Stirlingshire, as a left back. Joe retired from football to pursue an acting career,  where notable appearances include Burniston and Trainspotting 2.

Career
Born in Edinburgh, Dingwall began his youth career with Scottish Premier League club Hibernian, before signing with Raith Rovers in August 2006. Originally signed for the youth team Dingwall exceeded expectations and was promoted to the first team, where he was often asked to form a defensive partnership with Marvin Andrews. It was rumoured at the time that Dingwall was brought into the squad after proving his aerial prowess, during a training session where he out-headered the much more physically dominant Andrews. Ironically Andrews had bet Dingwall his luxury car that he would not win one header in the training session, Dingwall is still waiting for the car.

After making 17 league appearances for Raith Rovers in a two-year spell, he was released and signed for Arbroath. Dingwall struggled to find opportunities during his six-month spell at the club, managing only one league cup appearance.  Looking for more playing time, Dingwall signed for Berwick Rangers in January 2009, where unfortunately was unable to make an appearance being released in June 2009.  After a spell away from the football league, Dingwall signed for East Stirlingshire in July 2011.

In his short time at East Stirlingshire, Dingwall has become a cult favourite of the shire fans due to his tendency to sport an outrageous moustache during games. He only strengthened his cult status after his 25-yard cross found the back of the net via the post against Berwick Rangers on 25 February, with East Stirlingshire running out 2–0 victors. The goal was Dingwall's first-ever in the Football League and the grand stature of such an event prompted the defender to share his memory with fellow footballer and Scotland international David Goodwillie, with whom he made his début with for Raith back in 2007.

References

1988 births
Living people
Scottish footballers
East Stirlingshire F.C. players
Scottish Football League players
Hibernian F.C. players
Raith Rovers F.C. players
Arbroath F.C. players
Berwick Rangers F.C. players
Footballers from Edinburgh
Association football fullbacks

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Status of Parents in 2019
Oct 9, 2019 Oct 11, 2019
By Stephanie Reid
In today's legal climate, parental rights cases are increasingly evolving across a spectrum of niche areas, including paternity, parental fitness, third-party visitation and cases involving children in foster care.
A proposed Constitutional amendment on the horizon will formally be conferring the highest honor upon the oldest familial tie. Moreover, the definition of "parent" continues to evolve and change, while its common-law status as an unwavering fundamental liberty interest remains intact.
Parent Defined
The status of "parent" is a legal term of art that carries many duties and obligations. Also known as "parentage," a person with this status is responsible for any of the following:
Care, custody, and control of the child
Legal, medical and educational decision-making
Deciding who may visit or have contact with the child
Feeding, clothing and bathing the child
Providing safe shelter and protecting the child from harm
There are several ways this legal status is created. The first is through biological conception and childbirth. To determine paternity, there are generally three methods:
(1) the mother and one holding himself out to be the father is married at the time of birth, which creates a rebuttable presumption that the husband and father are one and the same
(2) for unmarried parents, the father may sign a "voluntary acknowledgment of paternity" (note: this is different than signing the birth certificate)
(3) DNA testing
Another way parentage can be established is through adoption, which can also be accomplished in several ways.
One or both natural parents may voluntarily relinquish their parental rights in favor of a chosen adoptive family.
Parental rights may be involuntarily stripped through a court process that requires a high burden of proof of parental unfitness (i.e., adoption through foster care).
Some children are unfortunately abandoned – or a father is unknown – thereby freeing the child for adoptive permanency
For same-sex couples, adoption by one or both partners is recognized and legal in all 50 states following a 2017 decision challenging an Arkansas regulation refusing to acknowledge same-sex partners on a birth certificate. However, private adoption and foster care agencies continue to discriminate against same-sex couples citing competing religious liberty interests. These cases are being met with a legal challenge, and a full resolution on this issue has not reached nationwide consensus.
Parental Rights as a Fundamental Liberty Interest
Regardless of the pathway to achieving parenthood status, the law is clear: the right to parent one's child is a fundamental right akin to free speech, religion and all other First Amendment guarantees. This means that governmental interference in the right to parent is only acceptable in cases in which the state has a compelling interest to intervene such as child safety.
This fundamental liberty interest cannot be interrupted by court orders demanding third-party visitation or third-party contact over a fit parent's objection. If a parent is fit and does not wish a certain person to have access to the child (including a grandparent or extended family member), a court cannot overrule this objection and override the parent's wishes.
Emphasis on Family Preservation
In 2018, Congress placed a great deal of emphasis on preserving the family structure, particularly given the unrelenting effects of the opioid crisis and ongoing nationwide mental health issues. Fiscal analysis gleaned that it costs states less money to invest in prevention than it does to try and reunify families after children are traumatically placed in foster care. For parents struggling with addiction, mental health issues, homelessness or disabilities, the Family First and Prevention Act infuses the states with additional funds and resources to keep parents and children together and avoid family separation if at all possible.
Constitutional Implications
Given the national spotlight on fundamental parental rights and keeping families together, lawmakers have proposed a full Constitutional amendment to ensure there are no further doubts as to the importance of the parent-child bond. However, a formal federal amendment permanently solidifies this notion in the annals of Constitutional law.
If adopted as proposed, the amendment would specifically guarantee parents the right to direct the upbringing of their children. Supporters of the initiative assert that this measure is an essential step toward ensuring the strength of fundamental parental rights cannot be undone or reduced by state law. Opponents, however, worry that parents could be cloaked with too much power and that controversial or dangerous parenting decisions would have to be honored despite the potential for harm to the child.
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Women's History Month: What Laws Protect Women In The Workplace?
Resolve to protect your loved ones by writing your will
Changing lives: Adopting a child from foster care
About Stephanie Reid
Stephanie Reid obtained her J.D. from Regent University School of Law and her Bachelor of Arts degree from Florida State University. After two years in private practice, Stephanie has opened her own law firm, Stephanie Reid Law. Her practice offers innovative web-based legal services for estate planning, family law and business clients. Stephanie also volunteers her time serving children engaged in family court proceedings, including those involved in the foster care system. She is admitted to practice law in Delaware and Maryland, and lives in Southern Delaware with her husband and two young sons.
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Home Relationships Family/Kids Status of Parents in 2019
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Q: Select TOP Elements in a Group with EF I have following entities, basically User and Message have M2M relationship via MsgGroup. Also first message in a thread (or group) has ThreadId=MessageId, where others share the same ThreadId
public class User
{
    public int UserId { get; set; }
    [System.ComponentModel.DataAnnotations.Schema.InverseProperty("Received")]
    public virtual System.Collections.Generic.ICollection<MsgGroup> ReceivedGroups { get; set; }
}

public class Message
{
    public int MessageId { get; set; }
    public int ThreadId { get; set; }
    public virtual System.Collections.Generic.ICollection<MsgGroup> MsgGroups { get; set; }
}

public class MsgGroup
{
    public int MsgGroupId { get; set; }

    public int UserId { get; set; }
    public virtual User Received { get; set; }

    public int MessageId { get; set; }
    public virtual Message Message { get; set; }
}

I want to get lates messages in groups for a given user. Particularly the following SQL query:
select * from messages where MessageId IN (
    SELECT MaxId FROM (
        select ThreadId , MAX(MessageId) as MaxId from messages where ThreadId in (
            select distinct MessageId from msggroups where UserId = 1
        )
        GROUP BY ThreadId
    ) AS t1
)

I have tried:
var query = from grp in db.MsgGroups.Where(g => g.UserId == userId)
select new
{
    f = (from msg in db.Messages where (msg.ThreadId == grp.MessageId) select msg).OrderByDescending(m => m.Date).Take(1)
};

however it creates a very complex query where there are subqueries for each field of message.
Is there a solution? Also can I ask the same question in method based format and query expression format (since I can't figure out in query expression format at all)

A: I think you may be looking to do a join, maybe something like this, this will get you the most recent messages for a particular user, ordered by thread ID, you can replace 100 in the Take section with how many messages you want:
var query = (from grp in db.MsgGroups
    .Join(
        db.Messages,
        g => g.MessageId,
        m => m.ThreadId,
        (g, m) => new { grp = g, mgs = m })
        .Where(x => x.grp.UserId == userId)
    select new
    {
        Message = grp.mgs
    })
    .Take(100)
    .OrderByDescending(t => t.Message.ThreadId);

This will give you a collection of anonymous objects, each one having a single property called Message, which is the Message object you want.

A: Use the navigation properties and let EF take care of the SQL joins:
var query = from rg in db.MsgGroups
            where rg.UserId == userId
            group rg by rg.ThreadId into thread
            select new {
                         Thread = thread.Key,
                         LastMessage = thread.Select(t => t.Message)
                                             .OrderByDescending(m => m.Date)
                                             .FirstOrDefault()
                       };

This gives you the last message for each thread the user has messages in.

A: For anyone interested below query exactly did what I requested, only difference is it used JOINs instead of INs in my question:
var query =
from mm in
(
    from grp in db.MsgGroups.Where(g => g.UserId == userId)
    from msg in db.Messages
    where msg.ThreadId == grp.MessageId
    group msg by msg.ThreadId into thr
    select new { t = thr.Key, m = thr.Max(t => t.MessageId) }
)
join omsg in db.Messages
on mm.m equals omsg.MessageId
select new { t = mm.t, m = omsg.Text } ;

redpajama_set_name: RedPajamaStackExchange


text:
The Much Beloved Kamuning Bakery Will Rise Again
Nana Ozaeta
Breads and pastries at the counter
Last February 6, we woke up to the alarming news that a treasured 79-year-old institution, Kamuning Bakery, was on fire. While the fire was quickly put out and no one was harmed, the two-story building sustained major damage, with all of the baking equipment destroyed in the fire. The two original pugon or wood-fired ovens were water damaged but can still be rehabilitated. This landmark bakery has been baking its famous pan de suelo and other traditional Filipino breads since 1939 from its original site at the corner of Judge Jimenez Street and K-1st Street in Kamuning, Quezon City.
Baking pan de suelo in the pugon oven
Just baked pan de Espana or Spanish bread
Amy Uy, author of the award-winning book Panaderia, related the story of Kamuning Bakery in FOOD Magazine's February 2015 issue. She writes, "The breads here are still handmade using time-honored techniques, without preservatives, and baked in the trusty pugon. They still bear old names like pan de España (Spanish bread) and bicho-bicho (doughnut twist) and there are still the otap, lubid-lubid, ugoy-ugoy, faborita, pan de coco, pan de Rizal, pinagong, and pan de limon that have delighted generations." The photos featured here show the bakery in its former glory, with pan de suelo baking in its wood-fired oven, its age-old equipment still in use, and its bakers shaping the dough all by hand.
Freshly baked pandesal
Kamuning Bakery's present owner, columnist-entrepreneur Wilson Lee Flores relates his dismay at the destruction, "We're trying to find ways to recover from the loss." What Flores didn't expect was people's emotional reaction to the tragic news. He shares, "Firemen told me during the fire, there were customers crying beside them." On TV, he saw scriptwriter-director Bibeth Orteza Siguion-Reyna saying that she and her two brothers cried upon hearing about the fire. The painter Aba Lluch Dalena also cried while passing by the fire site, according to another TV news report. The granddaughter of rock icon Pepe Smith, a Kamuning suki or regular, sent him a touching message. Third-generation customer, 79-year-old Joaquin "Joaqui" Roces, son of Chino Roces and grandson of the bakery's first suki Don Alejandro Roces, visited the bakery as well.
Cinnamon rolls ready to be baked
Despite the heavy loss, Flores is nonetheless thankful for all the support he has received since the news of the fire hit. He says, "We thank all the unprecedented outpouring of support from many people from all walks of life, from families, to writers, artists, politicians, activists, and even many other bakery owners here and overseas." He also made sure to thank the Quezon City firemen and Filipino-Chinese volunteer fire brigades from other parts of Metro Manila and even as far as Bulacan, as well as QC Administrator Aldrin Cuña who personally helped save paintings and memorabilia from the fire.
While the bakery needs extensive renovation, Flores is committed to rebuilding the site, restoring its old façade, and mindful of the history of the place. Thankfully, the bakery still sells its breads (except the loaf bread), cakes, pies, and cookies at its branches in Scout Delgado corner Tomas Morato Avenue, and at SM North Edsa (2/F East Bridge). Flores hopes to open several more branches as a way to provide work to the bakers and staff while the bakery is being renovated. While we wait for the old Kamuning Bakery to rise from its ashes, we're happy to continue enjoying its old-style breads, cakes, and pastries as our small yet meaningful way of honoring this much beloved institution.
Follow Kamuning Bakery Café on Facebook
Photos by Justin de Jesus
Kamuning
Kamuning Bakery
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