text:
\begin{document}

\allowdisplaybreaks

\newcommand{\arXivNumber}{1406.4645}

\renewcommand{\PaperNumber}{075}

\FirstPageHeading

\ShortArticleName{An Asymmetric Noncommutative Torus}

\ArticleName{An Asymmetric Noncommutative Torus}

\Author{Ludwik D\c{A}BROWSKI~$^\dag$ and Andrzej SITARZ~$^{\ddag\S}$}

\AuthorNameForHeading{L.~D\c{a}browski and A.~Sitarz}

\Address{$^\dag$~SISSA (Scuola Internazionale Superiore di Studi Avanzati),\\
\hphantom{$^\dag$}~via Bonomea 265, 34136 Trieste, Italy}
\EmailD{\href{mailto:dabrow@sissa.it}{dabrow@sissa.it}}

\Address{$^\ddag$~Institute of Physics, Jagiellonian University,\\
\hphantom{$^\ddag$}~Stanis\l{}awa \L{}ojasiewicza 11, 30-348 Krak\'ow, Poland}
\EmailD{\href{mailto:andrzej.sitarz@uj.edu.pl}{andrzej.sitarz@uj.edu.pl}}

\Address{$^\S$~Institute of Mathematics of the Polish Academy of Sciences,\\
\hphantom{$^\S$}~\'Sniadeckich 8, 00-656 Warszawa,  Poland}

\ArticleDates{Received December 09, 2014, in f\/inal form September 17, 2015; Published online September 26, 2015}

\Abstract{We introduce a family of spectral triples that describe the curved noncommutative two-torus. The relevant family of new Dirac operators is given by rescaling one of two terms in the f\/lat Dirac operator. We compute the dressed scalar curvature and show that the Gauss--Bonnet theorem holds (which is not covered by the general result of Connes and Moscovici).}

\Keywords{noncommutative geometry; Gauss--Bonnet; spectral triple}
\Classification{58B34; 46L87}

\section{Introduction}

In the seminal works \cite{cohcon, CoTr} Connes and Tretkof\/f initiated the investigation of curvature
aspects on the noncommutative two torus ${\mathbb T}_\theta^2$ and have shown the analogue of the Gauss--Bonnet theorem
for the conformally rescaled Dirac operator~$D$ and the related spin Laplacian corresponding to the standard
conformal structure. In~\cite{fatkha} these studies were extended to arbitrary conformal structure. The
scalar curvature itself was def\/ined and computed in~\cite{conmos2} and independently in~\cite{fatkha1}.

The methods used in these papers build on  Connes' pseudodif\/ferential calculus~\cite{CoTr}
and heat kernel short-time asymptotic expansion. The novelty therein is the employment of
{\em twisted} (or {\em non-unimodular}) spectral triples~\cite{cm2}, and as a consequence non-tracial weights and the modular operator. For some related papers see~\cite{bhumar12,fatkha2,fatwon}.

In the present paper we introduce a family of {\em usual} spectral triples in the sense of \cite{cm1},
known also as $K$-cycles~\cite[Def\/inition IV.2.11]{Co94}, that describe a curved~${\mathbb T}_\theta^2$.
The relevant new Dirac operators correspond to the rescaling one of two terms in the f\/lat Dirac operator by a positive element $k$ from the opposite coordinate algebra in the commutant.
This assures that the commutators of~$D$ with
the elements of the algebra are bounded.
Thus, to the best of our knowledge, this provides the f\/irst instance of usual spectral triples which describe non f\/lat (curved) noncommutative geometry.
It should also be stressed that partial rescalings by~$k$ from the algebra itself go outside the realm of spectral triples (even twisted), with geometric meaning (to the best of our knowledge) not yet fully clarif\/ied (in this respect see however, e.g.,~\cite{pw15}).

Unlike the classical case the new family of Dirac operators is not conformally related to the standard family of ``f\/lat'' (equivariant) Dirac operators.
Since they are not an overall rescaling of any of the so far known Dirac operators one cannot apply the general argument used by Connes and Moscovici~\cite[Theorem~2.1]{conmos2} to establish the Gauss--Bonnet theorem. Thus it is the f\/irst instance when one needs to ef\/fectively compute it, without certainty that it will be satisf\/ied.

This has been shown {\em perturbatively} up to second order in~\cite{DS}
for small modif\/ications of the f\/lat Dirac operator depending on four elements in the opposite algebra.
In the present paper we perform the {\em exact} computation, which shows that the Gauss--Bonnet theorem indeed extends to a much larger class of spectral triples.

Moreover, we compute the expression for the
dressed
scalar curvature as an element of the opposite algebra of~${\mathbb T}_\theta^2$.
In the classical case it corresponds to the product of the scalar curvature by the coef\/f\/icient of the metric volume form with respect to the Lebesgue volume.
It should be mentioned that such a dressing is actually tacitly present in all the previous computations.

The results obtained here are possible due to a recent neat generalisation by Lesch~\cite{lesch}
of the ``rearrangement lemma'', which is an important technical tool in~\cite{CoTr}.

The paper is organised as follows: in Section~\ref{section2} we brief\/ly present the
asymmetric rescaling of the classical f\/lat 2-torus and its curvature. Next we introduce a quantum (noncommutative) analogue of the associated Dirac operator, and the corresponding new family of spectral triples on the noncommutative torus.
In Section~\ref{section3}, which is the main part of the paper, we compute the value of the respective zeta function of the new Dirac operators at $z=0$ following the heat trace techniques and pseudodif\/ferential calculus elaborated in~\cite{conmos2, CoTr}, and an adaptation of the Lesch rearrangement lemma~\cite{lesch}.
In Section~\ref{section4} we comment on some algebra computations of the (undressed) curvature.

\section{Dirac operator}\label{section2}

Let  ${\mathbb T}^2$ be the classical torus with coordinates $(x, y)\in [0, 2\pi]^2$,
equipped with the metric
\begin{gather}
\label{metric}
dx^2 + k^{-2}(x,y) dy^2,
\end{gather}
where $k$ is a  positive function.
For example such an ``asymmetric torus'' metric with
\begin{gather*}
k^{-1} = c + \cos y
\end{gather*}
comes as the induced metric on the usual realisation of~${\mathbb T}^2$ as an embedded surface in~${\mathbb R}^3$
with the following parametrisation
\begin{gather*}
X	=	(c+\cos  y)  \cos  x, \qquad
Y	=	(c+ \cos y)  \sin  x, \qquad
Z	=	\sin y .
\end{gather*}
\begin{figure}[h!]\centering
\begin{minipage}[b]{50mm}\centering
\includegraphics[width=4cm]{torus}\\
Torus embedded in ${\mathbb R}^3$.
\end{minipage}\qquad
\begin{minipage}[b]{50mm}\centering
\includegraphics[width=4cm]{torus_as} \\
 Asymmetric torus in ${\mathbb R}^3$.
\end{minipage}
\end{figure}

The scalar curvature of the torus with the metric \eqref{metric} reads
\begin{gather}
\label{classcurv}
R = 2k^{-1} \partial_x^2 (k) - 4 k^{-2} (\partial_x (k) )^2 .
\end{gather}

In the commutative case such metric is, of course, conformally equivalent to some f\/lat metric on the
torus even though the explicit formula for the curvature depends on the chosen coordinate system.
However, when passing to the noncommutative torus we are entering a new unexplored land, where
one does not know what is a {\em metric} and what is the exact meaning of {\em conformally equivalent}.
As in the approach of Connes the natural object is the Dirac operator rather than the metric itself.
For this reason we propose a new Dirac operator, which generalises to the noncommutative situation
the classical case of asymmetric torus.

We start with the commutative case of Dirac operator on $L^2({\mathbb T}^2, k^{-1} dx dy)\otimes {\mathbb C}^2$
for the metric~\eqref{metric}
\begin{gather*}
\tilde{D} = -i \sigma^1 \left(\partial_x -\tfrac{1}{2} k^{-1} \partial_x (k)  \right) -  i \sigma^2   k\  \partial_y ,
\end{gather*}
where
\begin{gather*}
\sigma^1= \left(
\begin{matrix}
0      & 1   \\
 1     &  0
\end{matrix}
\right),\qquad
\sigma^2= \left(
\begin{matrix}
0      & -i  \\
 i     &  0
\end{matrix}
\right).
\end{gather*}
Using the multiplication  by $\sqrt{k}$ we obtain a unitarily equivalent self-adjoint Dirac operator
\begin{gather}
\label{dirac}
\tilde{D} =  - i \sigma^1 \partial_x  -  i \sigma^2 \left(  k \partial_y + \tfrac{1}{2}  \partial_y(k)  \right)
\end{gather}
on the dense domain $H^{1,2}(T)$ in the Hilbert space $L^2(T,  dx dy)\otimes {\mathbb C}^2$.

Next we pass to the noncommutative torus ${\mathbb T}_\theta^2$ \cite{Co94}, about which we
collected the needed information in Appendix~\ref{appendixA}.
In particular we shall use therein
the usual trace $\mathfrak{t}$,
the usual self-adjoint derivations $\delta_1$, $\delta_2$,
and the standard real structure
(Tomita--Takesaki conjugation)~$J$ on the coordinate algebra~$A({\mathbb T}_\theta^2)$. Then we introduce the following new family of spectral triples, which is an immediate noncommutative analogue of~\eqref{dirac}.

\begin{dfn}
For $k \in J A({\mathbb T}_\theta^2) J^{-1}$ we set
\begin{gather*}

D_k =   \sigma^1 \delta_1 +   \sigma^2 \left(  k\, \delta_2 + \tfrac{1}{2}  \delta_2(k)  \right)
\end{gather*}
acting on ${\mathcal H}= L^2({\mathbb T}_\theta^2, \mathfrak{t})\otimes {\mathbb C}^2 $,
obtaining a family of spectral triples $({\mathbb T}_\theta^2, {\mathcal H},  D_k)$.
\end{dfn}

The form of the operator $D_k$ resembles the classical Dirac operator  $D$  \eqref{dirac}, however with the partial derivatives replaced by~$i\delta_1$,~$i\delta_2$.
Since we take~$k$ in the algebra
$J A({\mathbb T}_\theta^2) J^{-1}$ commuting with $A({\mathbb T}_\theta^2)$
due to the property~\eqref{eq:Jzero-LRcomm},
$D_k$ has bounded commutators with $a\in A({\mathbb T}_\theta^2)$.
Next the self-adjointness and the compact resolvent property follow from the fact that~$D_k$ is a perturbation of the f\/lat Dirac operator~$D$ (corresponding to $k=1$) by a (self-adjoint) bounded operator. Thus it def\/ines a spectral triple in the usual sense.

Furthermore, $D_k$ is a dif\/ferential operator in the sense of~\cite{CoTr} and we can extract the
associated (dressed) scalar curvature following the methods introduced and applied to
the global conformal rescaling of the noncommutative~\cite{conmos2,CoTr}. The idea
is to look for the functional:
\begin{gather*}
f \mapsto a_2(D_k; f),
\end{gather*}
where $f$ is an element of $J A({\mathbb T}_\theta^2) J^{-1}$ (or $A({\mathbb T}_\theta^2)$) and $a_2(D_k; f)$
is the second heat trace coef\/f\/icient of $\operatorname{Tr} (f e^{- t D_k^2})$. Since the computations
which use the pseudodif\/ferential operators allow us to express the results in terms of $\tau(f \widetilde{R})$
for an element~$\widetilde{R}$ of the algebra~$J A({\mathbb T}_\theta^2) J^{-1}$, one can interpret it as the
{\em dressed scalar curvature}, which in the classical case equals $\widetilde{R} = \sqrt{g} R$.

The noncommutative case is, in fact,  much richer, since the chiral heat trace functional
$\operatorname{Tr} (f\gamma e^{- t D_k^2})$ does not vanish identically. Using similar arguments
as above we can introduce the {\em dressed chiral scalar curvature}, $\widetilde{R_\gamma}$ as
an element of  $J A({\mathbb T}_\theta^2) J^{-1}$ such that the heat trace coef\/f\/icient~$a_2(D_k, f \gamma)$
is  expressed as~$\tau(f \widetilde{R}_\gamma)$.

\section{The curvature}\label{section3}
The square of $D$ reads
\begin{gather*}
(D_k)^2 = (\delta_1)^2 + k^2 (\delta_2)^2
           + \left( \tfrac{3}{2} k \delta_2(k) + \tfrac{1}{2} \delta_2(k) k + \sigma^1\sigma^2 \delta_1(k) \right) \delta_2 \\
\hphantom{(D_k)^2 =}{}
           + \tfrac{1}{4}  (\delta_2(k))^2 + \tfrac{1}{2} \sigma^1\sigma^2 \delta_{12}(k) + \tfrac{1}{2} k \delta_{22}(k),
\end{gather*}
and its symbol is
\begin{gather*}
\sigma((D_k)^2) = a_0 + a_1 +a_2,
\end{gather*}
where
\begin{gather*}
a_0 =  \xi_1^2 + k^2 \xi_2^2,  \qquad
a_1 = \left( \tfrac{3}{2} k \delta_2(k) + \tfrac{1}{2} \delta_2(k) k + \sigma^1\sigma^2 \delta_1(k) \right) \xi_2, \\
a_2 =  \tfrac{1}{4}  (\delta_2(k))^2 + \tfrac{1}{2} \sigma^1\sigma^2 \delta_{12}(k) + \tfrac{1}{2} k \delta_{22}(k).
\end{gather*}
As was demonstrated f\/irst in \cite{CoTr} the value $\zeta(0)$ at the origin of the zeta function
of the operator $(D_k)^2$ is given by
\begin{gather*}
\zeta(0)= - \int \mathfrak{t} (b_2(\xi))  d \xi,
\end{gather*}
where $b_2(\xi)$ is a symbol of order $-4$ of the pseudodif\/ferential operator $((D_k)^2+1)^{-1}$.
Here we assume that $(D_k)$ has no zero eigenvalues, which is satisf\/ied for generic $k$.
Otherwise, in the case that the Dirac operator has $0$ as an eigenvalue,
a constant term equal to $\dim (\operatorname{ker} (D_k))$ is present.
The term on the right hand side is interpreted as the quantum analogue of the integrated scalar curvature dressed by the square root of the determinant of the metric tensor.

The symbol $b_2(\xi)$ can be computed by pseudodif\/ferential calculus of symbols from the symbol
$a_2(\xi) + a_1(\xi) + a_0(\xi)$ of
$(D_k)^2$ as follows
\begin{gather}
b_0 =  (a_2+1)^{-1},\qquad
b_1 = -(b_0 a_1 b_0  + \partial_1(b_0) \delta_1(a_2) b_0 +
\partial_2(b_0)\delta_2(a_2) b_0), \nonumber\\
b_2=  -(b_0 a_0 b_0 + b_1 a_1 b_0 + \partial_1(b_0) \delta_1(a_1) b_0 +
\partial_2(b_0)\delta_2(a_1)b_0+ \partial_1(b_1)\delta_1(a_2)b_0
\nonumber \\
\hphantom{b_2=}{}  + \partial_2(b_1) \delta_2(a_2)b_0 + \tfrac{1}{2}\partial_{11}(b_0)\delta_1^2(a_2)b_0
  + \tfrac{1}{2}\partial_{22}(b_0)\delta_2^2(a_2)b_0  + \partial_{12}(b_0)\delta_{12}(a_2)b_0).
\label{b2}
\end{gather}

Since to obtain the curvature (or~$\zeta(0)$) we need to integrate
with respect to~$\xi_1$,~$\xi_2$, we notice that all the terms which contain odd powers of these variables shall vanish. Therefore, we can neglect them and keep for computations only the relevant parts with even powers. They read then as follows
\begin{gather*}
b_2^{\rm even} = A + B + C ,
\end{gather*}
where
\begin{gather*}
A=
 - 2 k b_0^2 \delta_1(k) k b_0 \delta_1(k) b_0 \xi_2^4
+ 4 k b_0^2 \delta_1(k) k b_0^2 \delta_1(k) b_0 \xi_1^2 \xi_2^4
- 2 k b_0^2 \delta_1(k) b_0 \delta_1(k) k b_0 \xi_2^4 \\
\hphantom{A=}{} + 4 k b_0^2 \delta_1(k) b_0^2 \delta_1(k) k b_0 \xi_1^2 \xi_2^4
+ 8 k b_0^3 \delta_1(k) k b_0 \delta_1(k) b_0 \xi_1^2 \xi_2^4
+ 8 k b_0^3 \delta_1(k) b_0 \delta_1(k) k b_0 \xi_1^2 \xi_2^4 \\
\hphantom{A=}{} - b_0 \delta_1(k) b_0 \delta_1(k) b_0 \xi_2^2
+ 2 b_0^2 \delta_1(k) \delta_1(k) b_0 \xi_2^2
- 2 b_0^2 \delta_1(k) k b_0 \delta_1(k) k b_0 \xi_2^4 \\
\hphantom{A=}{} + 4 b_0^2 \delta_1(k) k b_0^2 \delta_1(k) k b_0 \xi_1^2 \xi_2^4
- 2 b_0^2 \delta_1(k) k^2 b_0 \delta_1(k) b_0 \xi_2^4
+ 4 b_0^2 \delta_1(k) k^2 b_0^2 \delta_1(k) b_0 \xi_1^2 \xi_2^4 \\
\hphantom{A=}{} - 8 b_0^3 \delta_1(k) \delta_1(k) b_0 \xi_1^2 \xi_2^2
+ 8 b_0^3 \delta_1(k) k b_0 \delta_1(k) k b_0 \xi_1^2 \xi_2^4
+ 8 b_0^3 \delta_1(k) k^2 b_0 \delta_1(k) b_0 \xi_1^2 \xi_2^4,
\\
B=
\tfrac{15}{4} k b_0 \delta_2(k) k b_0 \delta_2(k) b_0 \xi_2^2
- 3 k b_0 \delta_2(k) k^2 b_0^2 \delta_2(k) k b_0 \xi_2^4
- 3 k b_0 \delta_2(k) k^3 b_0^2 \delta_2(k) b_0 \xi_2^4 \\
\hphantom{B=}{}+ \tfrac{9}{4} k b_0 \delta_2(k) b_0 \delta_2(k) k b_0 \xi_2^2
+ 6 k^2 b_0^2 \delta_2(k) \delta_2(k) b_0 \xi_2^2
- 8 k^2 b_0^2 \delta_2(k) k b_0 \delta_2(k) k b_0 \xi_2^4 \\
\hphantom{B=}{}- 10 k^2 b_0^2 \delta_2(k) k^2 b_0 \delta_2(k) b_0 \xi_2^4
+ 4 k^2 b_0^2 \delta_2(k) k^3 b_0^2 \delta_2(k) k b_0 \xi_2^6
+ 4 k^2 b_0^2 \delta_2(k) k^4 b_0^2 \delta_2(k) b_0 \xi_2^6 \\
\hphantom{B=}{}- 12 k^3 b_0^2 \delta_2(k) k b_0 \delta_2(k) b_0 \xi_2^4
+ 4 k^3 b_0^2 \delta_2(k) k^2 b_0^2 \delta_2(k) k b_0 \xi_2^6
+ 4 k^3 b_0^2 \delta_2(k) k^3 b_0^2 \delta_2(k) b_0 \xi_2^6 \\
\hphantom{B=}{}- 10 k^3 b_0^2 \delta_2(k) b_0 \delta_2(k) k b_0 \xi_2^4
- 8 k^4 b_0^3 \delta_2(k) \delta_2(k) b_0 \xi_2^4
+ 8 k^4 b_0^3 \delta_2(k) k b_0 \delta_2(k) k b_0 \xi_2^6 \\
\hphantom{B=}{}+ 8 k^4 b_0^3 \delta_2(k) k^2 b_0 \delta_2(k) b_0 \xi_2^6
+ 8 k^5 b_0^3 \delta_2(k) k b_0 \delta_2(k) b_0 \xi_2^6
+ 8 k^5 b_0^3 \delta_2(k) b_0 \delta_2(k) k b_0 \xi_2^6 \\
\hphantom{B=}{}- \tfrac{1}{4} b_0 \delta_2(k) \delta_2(k) b_0
+ \tfrac{3}{4} b_0 \delta_2(k) k b_0 \delta_2(k) k b_0 \xi_2^2
+ \tfrac{5}{4} b_0 \delta_2(k) k^2 b_0 \delta_2(k) b_0 \xi_2^2 \\
\hphantom{B=}{}- b_0 \delta_2(k) k^3 b_0^2 \delta_2(k) k b_0 \xi_2^4
- b_0 \delta_2(k) k^4 b_0^2 \delta_2(k) b_0 \xi_2^4,
\end{gather*}
and
\begin{gather*}
C =  k b_0^2 \delta_{11}(k) b_0 \xi_2^2
- 4 k b_0^3 \delta_{11}(k) b_0 \xi_1^2 \xi_2^2
  + b_0^2 \delta_{11}(k) k b_0 \xi_2^2
- 4 b_0^3 \delta_{11}(k) k b_0 \xi_1^2 \xi_2^2- \tfrac{1}{2} k b_0 \delta_{22}(k) b_0 \\
\hphantom{C=}{}
+ 2 k^2 b_0^2 \delta_{22}(k) k b_0 \xi_2^2
+ 4 k^3 b_0^2 \delta_{22}(k) b_0 \xi_2^2
 - 4 k^4 b_0^3 \delta_{22}(k) k b_0 \xi_2^4
- 4 k^5 b_0^3 \delta_{22}(k) b_0 \xi_2^4.
\end{gather*}

Similarly for the chiral part
$\widetilde{R}_\gamma$ of the curvature, we have
\begin{gather*}
b_{\gamma 2}^{\rm even} =
A_{\gamma } + B_{\gamma } + C_{\gamma },
\end{gather*}
where
\begin{gather*}
A_{\gamma } =
- 2 k^2 b_0^2 \delta_1(k) k b_0 \delta_2(k) b_0 i \xi_2^4
- 2 k^2 b_0^2 \delta_1(k) b_0 \delta_2(k) k b_0 i \xi_2^4 + \tfrac{5}{2} b_0 \delta_1(k) k  b_0 \delta_2(k) b_0 i \xi_2^2\\
\hphantom{A_{\gamma } = }{}
- 2 b_0 \delta_1(k) k^2 b_0^2 \delta_2(k) k b_0 i \xi_2^4
 - 2 b_0 \delta_1(k) k^3 b_0^2 \delta_2(k) b_0 i \xi_2^4
+ \tfrac{3}{2} b_0 \delta_1(k) b_0 \delta_2(k) k b_0 i \xi_2^2,
\\
 B_{\gamma } =
   \tfrac{3}{2} k b_0 \delta_2(k) b_0 \delta_1(k) b_0 i \xi_2^2
- 2 k^2 b_0^2 \delta_2(k) k b_0 \delta_1(k) b_0 i \xi_2^4 \\
\hphantom{B_{\gamma } = }{}
 - 2 k^3 b_0^2 \delta_2(k) b_0 \delta_1(k) b_0 i \xi_2^4
+ \tfrac{1}{2} b_0 \delta_2(k) k b_0 \delta_1(k) b_0 i \xi_2^2,
\end{gather*}
and
\begin{gather*}
C_{\gamma } =
2 k^2 b_0^2 \delta_{12}(k) b_0 i \xi_2^2
- \tfrac{1}{2} b_0 \delta_{12}(k) b_0 i.
\end{gather*}

\subsection{The classical limit}
At this point we can check the classical (commutative) value of our expressions for $\theta =0$.
The symbol $b_2$ becomes
\begin{gather*}
b_2(\xi) =     48 b_0^5 k^6 \delta_2(k)^2 \xi_2^6 + 48 b_0^5 k^2 \delta_1(k)^2 \xi_1^2 \xi_2^4 - 8 b_0^4 k^5 \delta_{22}(k) \xi_2^4
           - 56 b_0^4 k^4 \delta_2(k)^2 \xi_2^4 \\
\hphantom{b_2(\xi) = }{}
           - 8 b_0^4 k^2 \delta_1(k)^2 \xi_2^4 - 8 b_0^4 k \delta_{11}(k) \xi_1^2 \xi_2^2
           - 8 b_0^4 \delta_1(k)^2 \xi_1^2 \xi_2^2 + 6 b_0^3 k^3 \delta_{22}(k) \xi_2^2 \\
 \hphantom{b_2(\xi) = }{}
           + 14  b_0^3 k^2 \delta_2(k)^2 \xi_2^2
 + 2 b_0^3 k \delta_{11}(k) \xi_2^2 + b_0^3 \delta_1(k)^2 \xi_2^2 - \tfrac{1}{2} b_0^2 k \delta_{22}(k) - \tfrac{1}{4} b_0^2 \delta_2(k)^2,
\end{gather*}
which after integration gives
\begin{gather*}
\int b_2 d\xi_1 d\xi_2
 = - \frac{\pi}{3} \frac{(\delta_1(k))^2}{k^3} + \frac{\pi}{6}  \frac{\delta_{11}(k)}{k^2}.
\end{gather*}
Taking into account that we compute the Gilkey--Seeley--deWitt coef\/f\/icients of the
heat kernel asymptotic for
the square of the Dirac operator and not the Laplace operator itself,
and assuming as above that $D_k$ has no zero eigenvalues, we have
\begin{gather*}
\zeta(0) =  \frac{1}{48 \pi} \int \sqrt{g} R.
\end{gather*}

The assumption that $0$ is not in the spectrum of $D_k$, is satisf\/ied for generic $k$.
Otherwise, in the case that~$D_k$  has $0$ as an eigenvalue, there is
a constant term contribution equal to~$\dim (\operatorname{ker} D_k)$.

Moreover, since
$\mathfrak{t} =  \frac{1}{4\pi^2} \int dx dy$ for $\theta = 0$, taking into account
the appropriate rescaling of the volume form and putting it all together we obtain
\begin{gather*}
\sqrt{g} R = 48 \pi \frac{1}{4\pi^2}
\left( - \frac{\pi}{3} \frac{(\partial_1(k))^2}{k^3} + \frac{\pi}{6}  \frac{\partial_{11}(k)}{k^2} \right)
= \left( 2 k^{-2} \partial_{11}(k) - 4 k^{-3} (\partial_1(k))^2 \right),
\end{gather*}
which agrees with the classical formula~(\ref{classcurv}).
Similarly, the symbol $ b_{\gamma 2}$ becomes
\begin{gather*}
b_{\gamma 2}(\xi) =
       - 12 b_0^4 k^3 \delta_1(k) \delta_2(k) i \xi_2^4 + 2 b_0^3 k^2 \delta_{12}(k) i \xi_2^2 + 6 b_0^3 k
      \delta_1(k) \delta_2(k) i \xi_2^2 - \tfrac{1}{2} b_0^2 \delta_{12}(k) i,
\end{gather*}
which after integration yields
\begin{gather*}
\int d\xi_1 d\xi_2   b_{\gamma 2} = 0
\end{gather*}
and thus
\begin{gather*}  \sqrt{g} R_\gamma = 0.
\end{gather*}

Before we can proceed with the noncommutative computation let us recall the general
framework of computations as shown recently by Lesch~\cite{lesch}.

\subsection{Rearrangement lemma}

In \cite{lesch} Lesch  proved the following formula
(with a slightly modif\/ied notation)
\begin{gather*}
\int_0^\infty  f_0\big(uk^2\big)   a_1
f_1\big(uk^2\big)  a_2 \cdots a_p   f_p\big(uk^2\big) du \nonumber \\
\qquad {} = k^{-2} m\, F\big(\md{1},\md{1}\md{2},\ldots, \md{1} \cdots \md{p}\big)(a_1 \otimes a_2 \otimes \cdots\otimes a_p),
\end{gather*}
where $m$ is the $p$-fold multiplication, the function $F$ of $p$ variables reads
\begin{gather*}
F(s_1,\ldots,s_p) = \int_0^\infty f_0(u)f_1(us_1)\cdots f_p(u s_p) du
\end{gather*}
and $\md{j}$ signif\/ies the square of the modular operator~$\Delta$, $\Delta (a) = k^{-1} a k$,
acting on the $j$-th factor $a_j$ in the $p$-fold
tensor product product $a_1 \otimes a_2 \otimes \cdots\otimes a_p$.

In our case we need to adapt the formula to a slightly dif\/ferent setting, when we integrate over two
variables~$\xi_1$ and~$\xi_2$. All the integrals we have are of the form
\begin{gather*}
{\mathcal J} = \int_{- \infty}^\infty d\xi_1 \int_{- \infty}^\infty d\xi_2
k^{n_1} b_0^{m_1}(\xi_1, \xi_2)   X   k^{n_2} b_0^{m_2}(\xi_1, \xi_2)   Y   k^{n_3} b_0^{m_3}(\xi_1, \xi_2) \xi_1^{2 k_1} \xi_2^{2 k_2}, \end{gather*}
where
$X$, $Y$ are some derivatives of $k$ and
\begin{gather*}
b_0(\xi_1, \xi_2) = \frac{1}{1 + \xi_1^2 + k^2 \xi^2}.
\end{gather*}

Extending the result of Lesch we see that
\begin{gather*}
{\mathcal J} = F\big(\mo{1},\mo{1} \mo{2}\big)(X \cdot Y),
\end{gather*}
where after change of variables we obtain
\begin{gather*}
F(s,t) =  2 \int_{0}^\infty \!d v \int_{0}^\infty\! d u     k^{n_1+ n_2+ n3 -1 -2k_2}  \frac{u^{k_2 - \frac{1}{2}} v^{2 k_1} }{(1\!+\! v^2\! +\! u)^{m_1}}
 \frac{s^{n_2}}{(1\!+\! v^2\! + \!u s^2 )^{m_2}}  \frac{t^{n_3}}{(1\!+\! v^2\! + \!u t^2 )^{m_3}} .
 \end{gather*}

In case $Y=1$ the resulting function depends only on $s$.
\subsection{The curvature and its trace}

In order to compute explicitly the expressions for the curvature we shall use the following lemma.
\begin{lem}
Under the trace an entire function $F$ of two variables satisfies
\begin{gather*}
\mathfrak{t} \big(  F\big(\mo{1},\mo{1} \mo{2}\big)(X \cdot Y) \big) =
\mathfrak{t} \big(  F\big(\mo{1}, {\rm id} \big)(X Y)\big) = \mathfrak{t} \big(F\big(\mo{1},1\big)(X) Y\big).
\end{gather*}
and in case of one variable
\begin{gather*}
\mathfrak{t} \big(  F\big(\mo{1}\big)(X) \big) = \mathfrak{t}  (  F(1) X  ).
\end{gather*}
\end{lem}

\begin{proof}
We have
\begin{gather*}F(s,t) = \sum_{n,m \geq 0} f_{nm} s^n t^m,
\end{gather*}
so
\begin{gather*}
\mathfrak{t} \big(  F\big(\mo{1},\mo{1} \mo{2}\big)(X \cdot Y) \big)   = \sum_{n,m \geq 0} f_{nm} \mathfrak{t} \big(
\Delta^{n+m}(X) \Delta^m (Y) \big) \\
\qquad = \sum_{n,m \geq 0} f_{nm} \mathfrak{t} \big( \Delta^m \big( \Delta^n(X) Y \big) \big)
  = \sum_{n,m \geq 0} f_{nm} \mathfrak{t} \big(\Delta^n(X) Y  \big)
  = \mathfrak{t} \big(  F\big(\mo{1}, 1\big)(X \cdot Y) \big).
\end{gather*}
The other identity is a simple consequence of the above one.
\end{proof}

\subsection{Curvature and chiral curvature}

We can present the f\/irst main result on the dressed scalar curvature of the asymmetric noncommutative torus:

\begin{pro}\label{main}
The dressed scalar curvature for the asymmetric torus is
\begin{gather*}
\widetilde{R} = F_{11}\big(\mo{1},\mo{1} \mo{2}\big)(\delta_1(k) \cdot\delta_1(k)) +  F_{22}\big(\mo{1},\mo{1} \mo{2}\big)(\delta_2(k), \delta_2(k)) \\
\hphantom{\widetilde{R} =}{}
+
F_{11}'\big(\mo{1}\big)(\delta_1(k)^2)  + F_{22}'\big(\mo{1}\big)(\delta_2(k)^2)
+ F_{1}\big(\mo{1}\big)(\delta_{11}(k)) + F_2\big(\mo{1}\big)(\delta_{22}(k)),
\end{gather*}
where
\begin{gather*} F_{11}(s,t) =  - \frac{2\pi }{3k^3} \frac{(2 s^2 + 4st + 4s +3 + 8t + 3t^2)}{(t+1)^3 (s+1) (s+t) },\qquad
 F_{22}(s,t) = \frac{\pi }{2k} \frac{(t^2 - 6t +1)}{(t+1)^3 },\\
 F_{11}'(s) =   \frac{4\pi }{3k^3} \frac{1}{(s+1)^3 }, \qquad
 F_{22}'(s) = - \frac{\pi }{2k} \frac{(s^2 - 6s + 1)}{(s+1)^3 },
\end{gather*}
and
\begin{gather*} F_1(s) =  \frac{2\pi }{3k^2} \frac{1}{(s+1)^2 }, \qquad
  F_2(s) = 0. \end{gather*}
The trace of $\widetilde{R}$ vanishes.
\end{pro}

\begin{proof}
First of all, observe that
\begin{gather*}
F_{22}(s,1) + F_{22}'(1) = 0, \qquad F_2(1) = 0,
\end{gather*}
so all terms containing $\delta_2(k)$ and $ \delta_{22}(k)$ vanish.

For the terms containing $\delta_1(k)$ we have
\begin{gather*}
F_{11}(s,1) + F_{11}'(1) =  - \frac{\pi}{3 k^3} \frac{s+3}{(s+1)^2 }.
\end{gather*}
Then using the identity
\begin{gather*}
\mathfrak{t}\left( k^{-2} \delta_{11}(k) \right) = 2 \mathfrak{t}\left(  k^{-2} \delta_1(k) k^{-1} \delta_1(k) \right) =
2 \mathfrak{t}\left(  k^{-3} \Delta^{-1}(\delta_1(k)) \delta_1(k) \right),
\end{gather*}
which follows directly from the Leibniz rule and the fact that the trace is closed, we can rewrite the trace of the sum of the remaining  terms $F_{11}$, $F_{11}'$ and $F_{1}$ as
\begin{gather*}
 \mathfrak{t} \left( k^{-3} H(\Delta) (\delta_1(k) ) \delta_1(k) \right),
\end{gather*}
where
\begin{gather*}
H(s) =  \frac{\pi}{3 k^3} \frac{1-s}{s (s+1)^2 }.
\end{gather*}

Next, we observe that for any $A$ and $B$ and an entire function $H$
\begin{gather*}
\mathfrak{t} \left( k^{-3} H(\Delta)(A) B \right) = \mathfrak{t} \left(  H(\Delta)\big(\Delta^3(A) \big) k^{-3} B \right) =  \mathfrak{t} \left(  k^{-3} B H(\Delta)\big(\Delta^3(A) \big) \right),
\end{gather*}
and
\begin{gather*}
\mathfrak{t} \left( k^{-3} H(\Delta)(A) B \right) = \mathfrak{t} \left(  k^{-3} A H\big(\Delta^{-1}\big)(B) \right).
\end{gather*}

Now if $A=B$ then both expressions on the right-hand side are identical. In our case, however
\begin{gather*}
H(s) s^3 = \frac{\pi}{3 k^3} \frac{s^2 (1-s)}{(s+1)^2 }
\qquad \text{and}\qquad
H\big(s^{-1}\big) = \frac{\pi}{3 k^3} \frac{s^2 (s-1)}{(s+1)^2 },
\end{gather*}
and therefore since
\begin{gather*}
H(s) s^3 = - H\big(s^{-1}\big),
\end{gather*}
the trace of the above expression must vanish, hence, the Gauss--Bonnet theorem holds.
\end{proof}

Next we give the second main result on the dressed chiral curvature of the asymmetric noncommutative torus:

\begin{pro}\label{main2}
The dressed chiral curvature for the asymmetric torus is
\begin{gather*} \widetilde{R}_\gamma = G_{12}\big(\mo{1},\mo{1} \mo{2}\big)(\delta_1(k), \delta_2(k)) \\
\hphantom{\widetilde{R}_\gamma =}{}
+ G_{21}\big(\mo{1},\mo{1} \mo{2}\big)(\delta_2(k), \delta_1(k)) + G\big(\mo{1}\big)(\delta_{12}(k)),
\end{gather*}
where
\begin{gather*} G_{12}(s,t) = \frac{\pi }{k^2} \frac{(t-1)}{(t+1)^2 (s+1)}, \qquad
 G_{21}(s,t) = \frac{\pi }{k^2} \frac{(t-1)}{(t+1)^2 (s+t)},
 \qquad
G(s) = - \frac{\pi }{k} \frac{(s-1)}{(s+1)^2 }.
\end{gather*}
The trace of~$R_\gamma$ vanishes.
\end{pro}

\begin{proof}
By computation.
Then the last statement follows from
\begin{gather*}
G_{12}(s,1) = G_{21}(s,1) = G(1) = 0. \tag*{\qed}
\end{gather*}
\renewcommand{\qed}{}
\end{proof}

\section{Final comments}\label{section4}

\looseness=-1
We have introduced a new class of spectral triples that describe a curved geometry of the noncommutative torus.
The associated Dirac operators are obtained
through a partial rescaling of the f\/lat Dirac operator by elements from the opposite coordinate algebra.
It turns out that the dressed scalar curvature does not vanish,
but the Gauss--Bonnet theorem holds for that family.

\looseness=-1
Let us stress that even though in the classical limit they arise from the metric, which is conformally equivalent to the f\/lat one, this is an open problem in the noncommutative situation. An interesting question is if and in which sense this
can be established also in the noncommutative case.

It becomes now more evident that the class of admissible Dirac operators on the noncommutative torus is certainly bigger than the one-parameter family of ``f\/lat metric'' (equivariant) Dirac operators.
It is therefore necessary to study the conditions and the general setup of such construction, and we
conjecture that the class should contain all operators proposed in~\cite{DS} and their f\/luctuations.

Although in this paper we have concentrated on the 2-dimensional case, it is natural to consider
generalisations of the result to $3$ and more dimensions. In particular it is interesting
to study the curvature and minimality of yet another class Dirac operators introduced in
\cite{DS2}, which are similar to the ones encountered here.

Finally, we can comment on the purely algebraic computation by  Rosenberg~\cite{RsNC}
of the curvature and Gauss--Bonnet term for the conformally rescaled metric on the noncommutative torus using the noncommutative notion of Levi-Civita connection.
His method can be extended to the case we consider and a direct computation yields the Gauss--Bonnet functional
\begin{gather*}
2 {\mathfrak t} \left( - \tfrac{9}{4} k^{-2} \delta_1(k) k^{-1} \delta_1(k)
+ \tfrac{1}{4}  k^{-3} \delta_1(k) \delta_1(k) + k^{-2} \delta_{11}(k) \right),
\end{gather*}
which however does not vanish,
and the Gauss--Bonnet theorem does not hold.

Instead there is an even simpler possibility that yields an algebraic expression for a candidate for the scalar curvature.
Namely, we note that the formula~(2.1) in~\cite{DS} for the classical scalar curvature in terms of the orthonormal basis of vector f\/ields ({\em moving frame} or {\em two-bein}) for the partially rescaled metric, makes a perfect sense also for the noncommutative torus
(there is no ordering ambiguity).
Taking the two-bein as $(\delta_1, k \delta_2)$ one obtains
\begin{gather*} {\mathfrak t} \left( - 4  k^{-2} \delta_1(k) k^{-1} \delta_1(k)
+  2k^{-2} \delta_{11}(k) \right),\end{gather*}
which vanishes due to the properties of the trace.

\appendix

\section{Appendix}\label{appendixA}

{\sloppy The unital $*$-algebra ${\mathcal A}_\theta$ of smooth functions on the noncommutative 2-torus
consists of power series of the form
\begin{gather*} a =\sum_{m,n \in{\mathbb Z}^2} a_{mn}  U_1^m U_2^n,
 \end{gather*}
where the (double) sequence of coef\/f\/icients $\{a_{mn} \in {\mathbb C}, \, (m,n) \in{\mathbb Z}^2 \}$
decreases rapidly at `inf\/inity'
\begin{gather*}
\| a\|_k := \sup_{m,n} \big\{\big(1+m^2+n^2\big)^k  |a_{mn}|^2\big\} < \infty ,\qquad\forall\, k\in {\mathbb N} .
\end{gather*}
The $*$-algebra structure is specif\/ied by the commutation relations of the unitary genera\-tors~$U_1$,~$U_2${\samepage
\begin{gather*}

U_1  U_2 = e^{2\pi i \theta} U_2  U_1,
\end{gather*}
where $\theta$ is a real parameter.}

}

On $A_\theta$ there is a
(unique if  $\theta$ is irrational)
normalised faithful positive def\/inite tracial state
$\mathfrak{t} \colon {\mathcal A}_\theta \rightarrow {\mathbb C}$; it is given by
\begin{gather*}
\mathfrak{t}(a) =
\mathfrak{t} \bigg(\sum_{(m,n) \in {\mathbb Z}^2} a_{mn} ~U_1^m U_2^n\bigg) := a_{00}.
\end{gather*}
We call $\mathfrak{t}$ {\em trace}, for brevity.
It is invariant under the action of the commutative torus ${\mathbb T}^2= U(1)\times U(1)$
on ${\mathcal A}_\theta$ by automorphisms,
$ U_1\mapsto z_1U_1$ and $ U_2\mapsto z_2U_2$, where $ |z_1|=1= |z_2|$.
The inf\/initesimal action is determined by two commuting derivations $\delta_1$, $\delta_2$
acting by
\begin{gather*}
\delta_\mu (U_\nu) = 2\pi i  \delta_\mu^\nu U_\nu, \qquad \mu,\nu = 1, 2.
\end{gather*}
The invariance means just that $\mathfrak{t}(\delta_\mu a) = 0$, for $\mu = 1, 2$
and any $a\in{\mathcal A}_\theta$.

There is a sesquilinear form on ${\mathcal A}_\theta$
\begin{gather*}
(a,b)= \mathfrak{t}(a^*b)
\end{gather*}
and a norm
\begin{gather*}
\|a\| := \sqrt{\mathfrak{t}(a^*a)}.
\end{gather*}
The completion of ${\mathcal A}_\theta$ in this norm yields a Hilbert space
$H_{\rm o}:= L^2({\mathcal A}_\theta ,\mathfrak{t} )$,
which carries a~$*$-representation of ${\mathcal A}_\theta$ by left multiplication operators
\begin{gather*}
\pi(a)\colon \  b \mapsto ab.
\end{gather*}
Moreover $\pi$ is the GNS representation w.r.t.\ the state~$\mathfrak{t}$,
 the cyclic and separating vector is $1\in{\mathcal A}_\theta\subset H_{\rm o}$.
The operator norm of~$\pi(a)$ is~$\|a\|$.
The associated {\em Tomita conjugation} provides an antiunitary operator $J_{\rm o}$ on~$H_{\rm o}$,
\begin{gather*}
J_{\rm o}({b}) := {b^*}.
\end{gather*}
Clearly $J_{\rm o}^2 = 1$ and
\begin{gather*}
\pi^{\rm op}(a) := J_{\rm o} \pi(a^*) J_{\rm o} \colon \  b \mapsto ba
\end{gather*}
(right multiplication by~$a$) is a representation of the opposite algebra of~${\mathcal A}_\theta$.
Since the left and right multiplications commute, we have
\begin{gather}
[\pi(a), J_{\rm o} \pi(b^*) J_{\rm o}] = 0, \qquad  \forall\, a,b.
\label{eq:Jzero-LRcomm}
\end{gather}
(In the paper we are omitting the symbol~$\pi$.)

For Dirac spinors (Hilbert space)
we take the trivial module over $A_\theta$ of rank two
\begin{gather*}
H=H_{\rm o}\otimes {\mathbb C}^2 \simeq H_{\rm o} \oplus H_{\rm o} =: H_+ \oplus H_-  .
\end{gather*}
The inner product is $\langle (a,a^\prime),(b,b^\prime)\rangle=\mathfrak{t}(a^*b+a^\prime b^{\prime*})$.
The \textit{charge conjugation} operator on~$H$ is given by
\begin{gather*}
J := -iJ_{\rm o} \otimes (\sigma_2\circ {\rm c.c.}).
\end{gather*}

\subsection*{Acknowledgements}

L.D.~gratefully acknowledges the hospitality of the Institute of Physics, Jagiellonian University in Krak\'ow.
L.D.~partially supported by PRIN 2010 grant ``Operator Algebras,
Noncommutative Geometry and Applications'', A.S.~partially supported by NCN grant 2012/06/M/ST1/00169.
 The authors express their gratitude to the referees for valuable comments.

\pdfbookmark[1]{References}{ref}


TITLE: Is every almost complex structure tame up to sign?
QUESTION [1 upvotes]: Let $V$ be a real vector space equipped with a symplectic form $\omega : V\times V \to \mathbb{R}$. An almost complex structure $J$ is compatible with $\omega$ if $\omega(J(u), J(v)) = \omega(u, v)$ for all $u, v \in V$. In addition, $J$ is said to tame $\omega$ if $J$ is compatible with $\omega$ and $\omega(u, J(u)) > 0$ for all $u \in V\setminus\{0\}$.
Given a compatible almost complex structure $J$ which tames $\omega$, we have another compatible almost complex structure given by $-J$ and satisfies $\omega(u, -J(u)) < 0$ for all $u \in V\setminus\{0\}$.
In both of these cases, the resulting bilinear form $\omega(\cdot, J\cdot)$ is definite; is this always the case? Alternatively phrased,

Given an almost complex structure $J$ which is compatible with the symplectic form $\omega$, must one of $J$ or $-J$ tame $\omega$?

REPLY [2 votes]: What if $J(e_1)=e_2$, $J(e_3)=e_4$, and $\omega=dx_1\wedge dx_3+dx_2\wedge dx_4$?


TITLE: measure of a probability measure?
QUESTION [1 upvotes]: Recall that the Lebesgue measure $\mu$ is often used to show how ''insignificant'' a subset is compared to $\mathbb{R}$. For example, $\mu(\mathbb{Q})=0$ even though $\mathbb{Q}$ has an infinite number of members.
I would like to show a similar thing for probability measures.
Specifically, consider two sets $A$ and $B$, their Cartesian product $A \times B$, and a sigma algebra over the Cartesian product $\sigma(A \times B)$. Let $U$ denote the set of all probability measures on $A \times B$. Using some ''measure of probability measures'' $\delta$, I would like to show that the subset $V$ (of $U$) consisting of all independent probability measures on $A \times B$ is ''insignificant'' compared to $U$. That is, $\delta(V) = 0$. Here, $V$ is composed of the probability measures which satisfy $P(C_A, C_B)=P(C_A)P(C_B)$ for any $C_A \in \sigma(A), C_B \in \sigma(B)$ and $ C_A, C_B \in \sigma(A \times B)$.
Does anyone have any ideas on how such a ''measure of probability measures''  $\delta$ would be defined?

REPLY [1 votes]: One way would be using random measures: https://en.wikipedia.org/wiki/Random_measure. Then an insignificant set of probability measures is a set with probability zero (this does not have to be the empty set).
Obviously this notion of insignificance depends on the chosen distribution of the random measure (i.e. the probability measure on the space of probability measures), although the notion will coincide for distributions (probability measures on the space of probability measures) which are absolutely continuous with respect to each other, although that is more or less by definition. https://en.wikipedia.org/wiki/Absolute_continuity#Absolute_continuity_of_measures
Needless to say any way to do so requires a lot of measure theory machinery.
You say you want to show that the space of probability measures on the product space which satisfy the independence property is "trivial". If we have any measure of probability measure which does not assign non-zero mass to any single individual probability measure (i.e. to any single point of the space of probability measures), then your desired conclusion will follow from the fact that the "product measure", which corresponds to independent probability measures is a single point (when we fix the probability measures on the factor spaces $A$ and $B$).
This is because the product measure is unique https://en.wikipedia.org/wiki/Product_measure, since the probability measures on $A$ and $B$, being finite measures, are in particular $\sigma-$finite.
I think a better argument that such measures are "insignificant", however, is whenever we fix two marginal distributions on $A$ and $B$, there are uncountably many possible probability measures on the product space $A \times B$ (provided of course that $A$ and $B$ are reasonably complicated, which is waht I am assuming), whereas the product distribution corresponding to independence is a single point. And clearly any reasonable notion of "significance" will designate a single point in a set of uncountable cardinality as insignificant.
Another way to show that there are uncountably many possible probability measures on the product space is to think in terms of correlation coefficients for the two marginal distributions depending on the joint distribution. One can construct at least one joint distribution on $A \times B$ such that the marginal distributions have correlation coefficient $\rho \in [-1,0) \cup (0,1]$, and then even of those joint distributions such that the two marginals are uncorrelated, not even all of them will be independent (in fact, only one of them will be). Again, all of this is assuming that $A$ and $B$ as well as the probability measures on them are reasonably complicated enough; as an example I am thinking of probability measures absolutely continuous with respect to Lebesgue measure on $\mathbb{R}$.
Without more details on your part (which ultimately might be more a matter of taste or convenience than true mathematical insight), I don't think I am able to say anything more specific.


TITLE: Why is 1+$\sqrt{r}*e^{ia}$ already in phasor/polar form?
QUESTION [0 upvotes]: I am new to phasors/polar numbers. so I am confused as to why when I try to convert $$1-\sqrt{r}*e^{ia} $$ to a polar form  on my calculator It comes back with the same thing, indicating this is already in polar form.
The form I want to get it in is $$M*e^{i\phi}$$ (M= magnitude (a real number) , $\phi$ is the phase (also a real number)
I don't understand how this can be if the complex exponential $e^{ia}$ is only multiplied with $\sqrt{r}$ but not $1$.
I intend to multiply this expression with another polar number later and add their phase angles later. I would appreciate any help, and thanks in advance.

REPLY [1 votes]: No $f=1+ t e^{ia}$ is not in a polar form but we can write it the polar form as $f=R e^{i\theta}$:
$$f=1+t\cos a+i t \sin a= R \cos \theta + i R \sin \theta \implies R\cos \theta=1+t\cos a, ~~R\sin \theta= t \sin a $$ $$ \implies R=\sqrt{1+t^2+2t \cos a},~~~ \theta= \tan^{-1}\frac{1+t \cos a}{t\sin a}.$$


TITLE: Bézout's theorem for arcs in the plane
QUESTION [4 upvotes]: Consider two polynomials $p,q \in {\mathbb R}[x,y]$, both of degree $d$. Let $\gamma_p$ and $\gamma_q$ be the two curves in ${\mathbb R}^2$ that are defined by these polynomials, and assume that these curves have no common components. As is well known, by Bézout's theorem the two curves intersect in at most $d^2$ points.
We cut $\gamma_p$ at each point where $\frac{\partial p}{\partial y} =0$, and similarly for $\gamma_q$. We obtain a set of arcs that behave like functions in the sense that each arc has at most one point for every $x$-coordinate.
Consider one such arc of $\gamma_p$ and one such arc of $\gamma_q$. My question is: Is the number of intersection points between two such arcs at most $d$? Or at least significantly smaller than $d^2$?
At first glance this may seem false. However, when $d=2$ it is clear that the two arcs have at most two intersection points. I played with the case of $d=3$ for a while and could not get more than three intersection points.
Many thanks!
Adam

REPLY [3 votes]: Let $f(x)$ be a cubic polymomial such that $df/dx$ has no real roots. Let $g(x)$ be a cubic polynomial with $3$ real roots on the half-line where $f(x)>0$.
For $\epsilon$ small enough, the derivative of $f(x)+\epsilon g(x)$ also has no real roots. Hence $x^2=f(y)$ and $x^2=f(y)+\epsilon g(y)$ have no critical points. But they intersect at six points, the three roots of $g(y)$ with two values of $x$ each.


TITLE: Proving that a sequence converges to $\sqrt{2}$
QUESTION [1 upvotes]: I'm trying to prove from the that the following sequence converges to $\sqrt{2}$, even though it seems obvious.
We define $x_n = \frac{m_n}{10^n}$ where $m_n$ is defined as the largest natural number such that $\left(\frac{m_n}{10^n}\right)^2 < 2$. After plugging in some values, I find
$$
(x_n) = \{1.4, 1.41, 1.414, 1.4142, \ldots \}
$$
and the idea is that $x_n$ agrees with $\sqrt{2}$ up to the $n$th decimal position. The problem is that I don't "know" that by the construction. If I did, I could write out the decimal expansion of $\sqrt{2}$ and find $N$ such that $\frac{1}{10^N} < \epsilon$ by the Archimedean property. Then taking $n \geq N + 1$, we have $|x_n - \sqrt{2}| < \frac{1}{10^N} < \epsilon$. But I don't believe I know at the start of the problem that this actually  is the decimal expansion of $\sqrt{2}$, so this proof assumes the conclusion. I can prove that the sequence is Cauchy and hence convergent, but I can't prove that the limit is actually $\sqrt{2}$.

REPLY [6 votes]: If $m_n$ is the largest integer such that $\left(\frac{m_n}{10^n}\right)^2 < 2$ then
$$
\left(\frac{m_n}{10^n}\right)^2 < 2 \le \left(\frac{m_n+1}{10^n}\right)^2
$$
which implies
$$
x_n < \sqrt 2 \le x_n + 10^{-n}
$$
or
$$
 \sqrt 2 - 10^{-n} \le x_n < \sqrt 2 \, .
$$
$x_n \to \sqrt 2$ now follows, e.g. by the  squeeze theorem.
Actually all inequalities are strict since $x_n$ is rational and $\sqrt 2$ is irrational, but that is not needed in order to prove the convergence.


TITLE: Reference request modules over a small category
QUESTION [1 upvotes]: I am currently writing up one of my results where I am using modules over small $\mathbb{k}$-linear categories. I so far sadly did not find a proper introduction of them to refer the reader to and was hoping one of you could help.
Best wishes,
Felix

REPLY [2 votes]: The classic reference is Mitchell's long paper Rings with Several Objects.


TITLE: Why we never use the product between vectors like between elements of direct groups product?
QUESTION [2 upvotes]: We can say, that any field $\mathbb{K}$ -- $1$-dim vector space on itself: $\mathbb{K}_{\mathbb{K}}$. So any vector of one another finite-dimensional vector space $V_{\mathbb{K}}$, after choosing the some basis can be represented as the element of isomorphic space $\mathbb{K}_{\mathbb{K}}^{n} = \prod_{i=1}^n\mathbb{K}_i$, where $n$ -- dimension of $V_{\mathbb{K}}$. But we can determine operations on elements of $\mathbb{K}_{\mathbb{K}}^{n}$ like on the direct group product: $(x_1,\dots,x_n) + (x'_1, \dots, x'_n) = (x_1 + x'_1, \dots ,x_n + x'_n)$ and similar for second field operation: $(x_1,\dots,x_n) \times (x'_1, \dots, x'_n) = (x_1 \times x'_1, \dots ,x_n \times x'_n)$.
But usually we doing it only with one field operation $+$. Why?

REPLY [0 votes]: The category of fields has no product. If it had, then the projection on one factor would be injective, which is clearly false!


TITLE: Closed form of a sum with a falling factorial
QUESTION [0 upvotes]: I am having trouble finding the closed form for this:
$\sum_{k=1}^n  k^\underline{-2} + 2^k$
If it were just
$\sum_{k=1}^n 2^k + 2k$
I would just use the geometric series to solve for the closed form, but I do not know how to apply the geometric sum to the falling factorial. Can someone help me out?

REPLY [1 votes]: Using the definition of Falling Factorial, rewrite the sum as
$$\begin{align} \sum_{k=1}^n  k^\underline{-2} + 2^k &= \sum_{k=1}^n k(k-1) + 2^k \\&= \sum_{k=1}^n k^2-k + 2^k \\&= \sum_{k=1}^nk^2 - \sum_{k=1}^nk + \sum_{k=1}^n 2^k \end{align}$$ Then, use the known formulas.


TITLE: Atom interferometry,gravity and inertia: What can it measure that light interferometry can't?
QUESTION [2 upvotes]: What previously unexplored effects in gravity and inertia can be examined with atom interferometry in ways that hasn't already been done through light interferometry?
Can atom interferometry be used to measure gravitational waves like LIGO has done with light interferometry?
Can atom interferometry be used to measure effects like frame dragging?
Can atom interferometry be used to measure machian relationships and gravinertial field?

REPLY [1 votes]: Well i don't know much about an atom interferometer but according to Wikipedia: "Atom interferometers have many uses in fundamental physics including measurements of the gravitational constant, the fine-structure constant, the universality of free fall, and have been proposed as a method to detect gravitational waves.[1] They also have applied uses as accelerometers, rotation sensors, and gravity gradiometers."  Sources are available at Wikipedia.


TITLE: As resolved the identity of this series?
QUESTION [0 upvotes]: Use the identity $\cos((k-\frac{1}{2})x) - \cos((k+\frac{1}{2})x) = 2\sin kx \sin \frac{x}{2}$  to show that
$S_n:=\sum_{k=1}^{n}\sin kx=\frac{1}{2\sin \frac{x}{2}}(\cos\frac{x}{2}-\cos((nx+\frac{x}{2}))$
for $\sin \frac{x}{2}\neq0$, and conclude that $(S_n)$ is bounded. Get a similar result for
 $$T_n:=\sum_{k=1}^{n}\cos kx
$$

REPLY [0 votes]: Hint You have a telescopic sum in $k$.$$\cos\left(k-\frac{1}{2}\right) x- \cos\left(k+\frac{1}{2}\right) x=a_k-a_{k+1}  $$


text:
To detect the change points in \Cref{model:sepp change point}, we propose the penalized dynamic programming (\pdp) algorithm, which is stated in \eqref{eq:SEPP DP} with necessary notation in \eqref{eq:definition of A hat}, \eqref{eq:penalized likelihood} and \eqref{eq:constrain set}.  The \pdp consists of two layers: estimation of the coefficient matrices $A^*(t)$, $t \in [1, T]$, and estimation of the change points.

For the coefficient matrix estimation, we let $\widehat{A}(I)$ be the penalized log-likelihood estimator of the coefficient matrix over an integer interval $I \subset [1, T]$, i.e.
	\begin{align} \label{eq:definition of A hat}
		\widehat A(I) = \argmin_{A \in \mathcal C} H(A, I),
	\end{align}
	where $H(A, I)$ and $\mathcal C$ are the penalized log-likelihood function and the constrained domain of the coefficient matrices, respectively.  To be specific, with a pre-specified tuning parameter $\lambda > 0$ and $I = [s, e]$, let
	\begin{align}\label{eq:penalized likelihood}
		H(A, I) = \sum_{t = s}^{e - 1} \sum_{m=1}^M \left(\exp\left[v + A_m g_t\{\mathcal X (t)\}\right] - X_m (t+1) [v + A_m g_t\{\mathcal X (t)\}] + \lambda |I |^{1/2} \|A\|_1\right)
	\end{align}
	and
	\begin{align} \label{eq:constrain set}
		\mathcal C = \left \{A \in \mathbb R^{M \times M }:\, \max_{m = 1, \ldots, M} \|A_{m}\|_1 \le 1 \right \}.
	\end{align}

The loss function $H(\cdot, \cdot)$ is a penalized logarithmic conditional likelihood function, recalling that $X_m(t+1)$ given $\mathcal{X}(t)$ follows a Poisson distribution with intensity $\exp\left[v + A_m g_t\{\mathcal X (t)\}\right]$.  The penalty term $\lambda |I|^{1/2}$ in \eqref{eq:penalized likelihood} is introduced in a way such that the tuning parameter $\lambda$ is independent of the interval length.  The term $|I|^{1/2}$ reflects the order of the standard error of the sum of $|I|$ marginal log-likelihood functions.  We elaborate on this scaling factor and its derivation in \Cref{lemma:Azuma Hoeffding} and its proof.

The constraint on $\mathcal{C}$ is to ensure that the \sepp process as vector-valued time series is stable \citep[see e.g.][]{lutkepohl2005new}.  As for stationary \sepp estimation, \cite{mark2018network} proposed a constraint similar to \eqref{eq:constrain set}.

\medskip
Given the above framework, we can now consider estimating change points by setting
	\begin{align} \label{eq:SEPP DP}
		\widehat { \mathcal P }  = \argmin_{\mathcal P} \left\{\sum_{ I \in \mathcal P }   H( \widehat A(I), I) + \gamma | \mathcal P|\right\},
	\end{align}
where $\gamma > 0$ is a tuning parameter, the minimization is over all possible interval partitions of $[1, T]$  and $\mathcal P$ denotes one such partition.  To be specific, an interval partition has the form $\mathcal P = \{I_k, \, k=1, \ldots, K_{\mathcal{P}}\}$ and satisfies $I_{k'} \cap I_{k} = \emptyset$ and $\bigcup_{k=1}^{K_{\mathcal{P}}} I_k = [1, T]$.  Once $\widehat{\mathcal{P}}$ is at hand, we let $\widehat K = |\widehat {\cal P}| - 1 \geq 0$, $\eta_{\widehat{K} +  1} = T+1$ and
	\[
		\widehat{\mathcal{P}} = \left\{\{1, \ldots, \widehat{\eta}_1 - 1\},\, \{\widehat{\eta}_k, \ldots, \widehat{\eta}_{k+1} - 1\}_{k = 1}^{\widehat{K}}\right\}.
	\]
	We call $\{\widehat{\eta}_k\}_{k = 1}^{\widehat{K}}$ the change point estimators induced by  $\widehat{\mathcal{P}}$.

The optimization problem in \eqref{eq:SEPP DP} is known as the minimal partition problem on a linear chain graph and can be solved using dynamic programming \citep[e.g.][]{friedrich2008complexity} with the worst case computational cost of order $O\{T^2 \mathrm{Cost}(T)\}$, where $\mathrm{Cost}(t)$ denotes, in our case, the computational cost of computing $\widehat A(I)$ in the interval $I$ with $|I| = t$.  We remark that there has been a line of attack on the computational aspect of optimizing the minimal partition problem, including \cite{killick2012optimal} and \cite{maidstone2017optimal}, among others.

For completeness, we summarize the \pdp procedure in \Cref{algorithm:PDP} below.  The quantities and functions involved there are defined in \eqref{eq:definition of A hat}, \eqref{eq:penalized likelihood} and \eqref{eq:constrain set}.
\begin{algorithm}[htbp]
\begin{algorithmic}
	\INPUT Data $\{X(t)\}_{t=1}^{T}$, tuning parameters $\lambda, \gamma > 0$.
	\State $(\mathcal{B}, s, t, \mathrm{FLAG}) \leftarrow (\emptyset, 0, 2, 0)$
	\While{$s < T-1$}
		\State $s \leftarrow s + 1$
		\While{$t < n$ and $\mathrm{FLAG} = 0$}
			\State $t \leftarrow t+1$
			\If{$\min_{l = s+1, \ldots, t-1}\{H(\widehat{A}([s, l]), [s, l]) + H(\widehat{A}([l+1, t]), [l+1, t]) + \gamma < H(\widehat{A}([s, t]), [s, t])\}$}
				\State
				\begin{align*}
					& s \leftarrow \min \argmin_{l = s+1, \ldots, t-1}\{H(\widehat{A}([s, l]), [s, l]) + H(\widehat{A}([l+1, t]), \\
					& \hspace{5cm} [l+1, t]) + \gamma < H(\widehat{A}([s, t]), [s, t])\}
				\end{align*}
				\State $\mathcal{B} \leftarrow \mathcal{B} \cup \{s\}$
				\State $\mathrm{FLAG} \leftarrow 1$
			\EndIf
		\EndWhile
	\EndWhile
	\OUTPUT The set of estimated change points $\mathcal{B}$.
\caption{Penalized dynamic programming. PDP$(\{X(t)\}_{t=1}^{n}, \lambda, \gamma)$}
\label{algorithm:PDP}
\end{algorithmic}
\end{algorithm}

\subsection{The localization rate of the Penalized Dynamic Programming estimators}

In order to establish the consistency of the change point estimators resulting from the \pdp procedure detailed in \Cref{algorithm:PDP}, we first impose \Cref{assume:parameters}.

\begin{assume}\label{assume:parameters}
Let $\{X(t)\}_{t=1}^T \subset \mathbb{Z}^M$ be a discrete-time \sepp generated according to \Cref{model:sepp change point} and satisfying the following.
\begin{itemize}
\item [{\bf A1.}] There exists a subset $S \subset \{(m, m'): m, m' = 1, \ldots, M\}$ such that for all $t \in [1, T]$, $A_{m, m'}^*(t) = 0$, if $(m, m') \notin S$.  Let $d = |S|$.
\item [{\bf A2.}] It holds that
	\[
		\max_{t = 1, \ldots, T} \max_{m =1, \ldots, M} \|A^*_{m}(t)\|_1 \le 1.
	\]
\item [{\bf A3.}] For any $\xi > 0$, there exist absolute constants $C_{\Delta, 1}, C_{\Delta, 2} > 0$ such that
	\[
		\Delta \ge C_{\Delta, 1} T \quad \mbox{and} \quad \Delta \ge C_{\Delta, 2} \log^{2+\xi}(TM) d ^2\max\{\kappa^{-2},\, \kappa^{-4}\}.
	\]
\item [{\bf A4.}] There exist absolute constants $p \in \mathbb Z^+$ and $\omega > 0$ such that for any $t$, the matrix
	\[
		\mathbb{E}[g_t\{\mathcal X(t)\}g_t\{\mathcal X(t)\}^\top |\mathcal X (t - p)] -\omega I_{M}
	\]
	is positive definite, where $I_M \in \mathbb{R}^{M \times M}$ is an identity matrix.  In addition, $v$ and $\|g_t(\cdot)\|_{\infty}$, for all $t$, are uniformly upper bounded by an absolute constant $C_g > 0$.
\end{itemize}
\end{assume}

\Cref{model:sepp change point} and \Cref{assume:parameters} completely charaterize the problem with model parameters $M$ (the dimensionality of the time series), $d$ (the sparsity parameter indicating an upper bound of the number of nonzero entries in all the coefficient matrices), $\Delta$ (the minimal spacing between change points), and $\kappa$ (the minimal jump size), along with the sample size $T$.  The consistency we are to establish is based on allowing $M$ and $d$ to diverge and $\kappa$ to vanish as the sample size $T$ diverges unbounded.

The number of parameters at each time point is of order $M^2$, which is allowed to well exceed the sample size.  A sparsity constraint therefore comes into force in Assumption {\bf A1}, which is a standard assumption in the high-dimensional statistics literature.  Note that the set $S$ is the union of all $(m,m')$ pairs with a nonzero entry in any coefficient matrix.  Assumption {\bf A2} echoes the imposition of the constraint domain $\mathcal{C}$ \eqref{eq:constrain set} in the optimization \eqref{eq:definition of A hat}, to ensure the stationarity of the \sepp.  In fact, the constant one in the upper bound can be relaxed to any absolute constant and is set to be one in this paper for identification issue.  To be specific, what goes into the model is the product of $A_m(t)$ and $g_t\{\mathcal{X}(t)\}$, and the latter is assumed to be upper bounded in sup-norm in  Assumption {\bf A4}.

Assumption {\bf A3} can be regarded as a signal-to-noise assumption.  It is required that the minimal spacing $\Delta$ is at least of a constant fraction of the total sample size, which implies the number of change points is of order $O(1)$.  This might appear to be strong compared to other change point detection literature, however, the problem we are facing here is challenging due to the nonlinearity of the \sepp model.  In fact, Assumption {\bf A3} is a mild condition and covers some challenging scenarios.  For instance, Assumption {\bf A3} holds if $M \asymp \exp(T^{1/2})$, $d \asymp T^{1/4}$ and $\kappa \asymp \log(T)$.  The quantity $\xi$ can be set arbitrarily small and it ensures the consistency of the estimator which will be explained after \Cref{theorem:SEPP change point}.

Assumption {\bf A4} can be interpreted as the restricted eigenvalue condition for \sepp processes.  We refer readers to Section~4 of \cite{mark2018network} for a number of common self-excited point process models satisfying Assumption {\bf A4}.

\medskip

In what follows, we show the consistency of \pdp in \Cref{theorem:SEPP change point}.

\begin{theorem} \label{theorem:SEPP change point}
Let $\{X(t)\}_{t=1}^T \subset \mathbb{Z}^M$ be an \sepps generated from \Cref{model:sepp change point} and satisfying \Cref{assume:parameters}.  Let $\{\widehat \eta_k\}_{k=1}^{\widehat K}$ be the change point estimators from the \pdp algorithm detailed in \Cref{algorithm:PDP} with tuning parameters
	\begin{equation}\label{eq-tuning-order}
		\lambda =  C_\lambda \log(TM) \quad \text{and} \quad \gamma =  C_\gamma \log^2(TM) d\left(1 +  d\kappa^{-2}\right),
	\end{equation}
	where $C_\lambda, C_\gamma > 0$ are absolute constants, depending only on $p$, $\omega$, $C_{\Delta, 1}$, $C_{\Delta, 2}$ and $C_g$.  We have that
	\[
		\mathbb{P}\left\{\widehat{K} = K \quad \mbox{and} \quad \max_{k = 1, \ldots, K}|\widehat{\eta}_k - \eta_k| \le C_{\epsilon}d^2 \log^2(TM) \max\{\kappa^{-2}, \kappa^{-4}\}\right\} \geq 1- 2 (TM)^{-1},
	\]
	where $C_\epsilon > 0$ is an absolute constant only depending on $p$, $\omega$, $C_{\Delta, 1}$, $C_{\Delta, 2}$ and $C_g$.
\end{theorem}

The proof of \Cref{theorem:SEPP change point} is deferred to \Cref{sec-proof-thm}, where it can be seen that the order of the estimation error is of the form
	\[
		\frac{\lambda^2 d}{\kappa^2} + \frac{\lambda^2 d^2}{\kappa^4} +  \frac{\gamma}{\kappa^2}.
	\]
	Due to the signal-to-noise ratio condition in Assumption~\textbf{A3}, we have that
	\begin{align*}
		\frac{\max_{k = 1, \ldots, K}|\widehat{\eta}_k - \eta_k|}{\Delta} \lesssim \frac{d^2 \log^2(TM) \max\{\kappa^{-2}, \kappa^{-4}\}}{\Delta} \lesssim \frac{d^2 \log^2(TM) \max\{\kappa^{-2}, \kappa^{-4}\}}{d^2 \log^{2 + \xi}(TM) \max\{\kappa^{-2}, \kappa^{-4}\}} \to 0,
	\end{align*}
	as $T \to \infty$.  This explains the role of the quantity $\xi$ in Assumption~\textbf{A3} and shows the consistency of the \pdp algorithm.  In fact, if we let $d = 1$ and assume $\kappa > 1$, then the localization error we derived here coincides with the optimal localization error in the univariate mean change point detection problem \citep[e.g.][]{wang2020univariate}.

Two tuning parameters are involved, where $\lambda$ is used in the optimization \eqref{eq:penalized likelihood} to recover the sparsity in estimating high-dimensional coefficient matrices, and $\gamma$ is involved in optimizing \eqref{eq:SEPP DP} to penalize the over-partitioning.  The order of $\lambda$ required in \eqref{eq-tuning-order} is a logarithmic quantity in $T$ and $M$, which is resulted from a union bound argument applied to a sub-exponential concentration bound.  The requirement on $\gamma$ is essentially that $\gamma \asymp \lambda^2\left(d + d^2\kappa^{-2}\right)$, which can be intuitively explained as an upper bound on the difference between $H(\widehat{A}(I_1), I_1) + H(\widehat{A}(I_2), I_2)$ and $H(\widehat{A}(I_1 \cup I_2), I_1 \cup I_2)$, where $I_1$ and $I_2$ are two relatively long, non-overlapping and adjacent intervals, and there is no true change point near the shared endpoint of $I_1$ and $I_2$.  In this case, one would not wish to partition $I_1 \cup I_2$ into $I_1$ and $I_2$.  If we only focus on the log-likelihood functions, over-estimating will result in that
	\[
		H(\widehat{A}(I_1), I_1) + H(\widehat{A}(I_2), I_2) < H(\widehat{A}(I_1 \cup I_2), I_1 \cup I_2).
	\]
	The penalty we impose through $\gamma$ will therefore avoid this over-partitioning.

\subsection{Comparisons with related work}
\label{sec:prior}
In a broad sense, as we have mentioned, there have been numerous  existing papers on different aspects of \sepps.  In fact, another related area is the analysis of piecewise-stationary time series models, where we also see a vast volume of existing papers.  The two most related papers are \cite{mark2018network}, which is concerned with a stationary, high-dimensional \sepp process, and \cite{wang2019localizing}, which studies a piecewise-stationary high-dimensional linear process.

\cite{mark2018network} studied a stationary version of \Cref{model:sepp change point} with $K = 0$.  The penalized estimator of the coefficient matrix developed there is almost identical to the ones summoned in our problem in \eqref{eq:definition of A hat}.  There are a few fundamental differences between this paper and \cite{mark2018network}.  (1) Due to the piecewise-stationarity assumed in \Cref{model:sepp change point}, when estimating the coefficient matrices in \eqref{eq:definition of A hat} and \eqref{eq:penalized likelihood}, it is possible that there exists a true change point in the interval of interest and the estimator we seek is an estimator of a mixture of different true coefficient matrices.  (2) We provide a more refined analysis as an improved version of \cite{mark2018network}, for instance, the optimization constrain domain $\mathcal{C}$ defined in \eqref{eq:constrain set} is a cleaner version of its counterpart in \cite{mark2018network}; a subspace compatibility condition is required in \cite{mark2018network} to control the ratio of different norms of the coefficient matrix, and this assumption is shown to be redundant in our new analysis.

The other closest-related work is \cite{wang2019localizing}, where the change point localizing problem in the piecewise-stationary vector autoregressive models is investigated and a penalized dynamic programming approach was deployed there.  The main differences between this paper and \cite{wang2019localizing} comes from the underlying model.  The vector autoregressive model is a linear model in the sense that given $\mathcal{X}(t)$ the history data up till time point $t$, the conditional expectation of $X(t+1)$ is a linear combination of the columns of $\mathcal{X}(t)$, which is not the case here.  The self-exciting point process is a nonlinear model, and as we have mentioned, the logarithm of the conditional intensity is a linear function of the history.  Another key difference is that  \cite{wang2019localizing} are concerned with sub-Gaussian innovation sequences, while the counting processes we study here determine the heavy-tail properties of the data.

config: arxiv


TITLE: Determinant of this coherent sheaf on a surface $S$
QUESTION [0 upvotes]: If $C$ is a curve on a surface $S$, i.e. $i:C\subset S$, and $G$ is a line bundle on $C$, then $G|_U\cong \mathcal{O}_C$ where $U$ is an open subset of $S$, that is, $G$ is trivial on the complement of finitely many points. Then is $\textrm{det}(i_*G)\cong\textrm{det}(i_*\mathcal{O}_C)$? Why is it so?
If $\textrm{det}(i_*G)\cong\textrm{det}(i_*\mathcal{O}_C)$, then from the short exact sequence $0\longrightarrow\mathcal{O}_S(-C)\longrightarrow\mathcal{O}_S\longrightarrow\mathcal{O}_C\longrightarrow 0$, we get that $\textrm{det}(i_*G)=\textrm{det}(i_* \mathcal{O}_C)=\mathcal{O}_S(-C)$.
So is $\textrm{det}(i_*G)\cong\textrm{det}(i_*\mathcal{O}_C)$? Thank you!

REPLY [1 votes]: The determinant is a line bundle hence locally free. But it is easy to see that two locally free sheaves are isomorphic, if they are isomorphic in codimension one. (Because they are reflexive.) And i think it should be $det(i_{*}G)=\mathcal{O}_S(C)$.


TITLE: Sample size to find a death rate
QUESTION [0 upvotes]: I need help, please, in working up a sample size.  I’m working on a real project to track the incipient famine in Afghanistan by estimating crude death rates over time using survey data gathered from a sample of mosques.  Enumerators will visit mosques on a monthly basis and ask the mullah of each how many of his congregants have died since the enumerator’s last visit, and how many congregants there are.  The sum of the deaths divided by the sum of the congregation sizes will give me my estimate of the overall population’s crude death rate.  I don’t have enough information to segment my sample, or even to make my selection of mosques dependably random.  (And there are a thousand other things that could fowl me up, I know, but I’m just looking at the sample size question for now.)
The metro area I’m looking at has an overall population over a million.   The crude death rate for Afghanistan was estimated at 13.6 deaths per thousand per year for 2016.  To identify a famine,  the IPC, a UN affiliated group, uses a threshold rate of 2 deaths per 10,000 per day (=70 deaths per thousand per year).
I’ve tried to use Cochran’s formula to come up with a sample size:
Sample size = $ Z^2* p * (1-p) / M^2 $
Where:
Z is the z score of 1.96 for a 95% confidence level,
p is the population proportion.  I’m setting it at .001178.  I got that by converting the annual crude death rate for Afghanistan in 2016 of 13.6 per thousand to a probability figure for 30 days $(0.001178  = 13.6 / 1000 / 360 * 30)$ .
M is the ‘confidence interval’ or  'margin of error', within which I want my estimate to fall 95% of the time.  I’m putting it at .10 for 10%.
So I do the math, but my answer seems a little low, even when I round it up to 1.
$1.962^2 * 0.001178 * (1-0.001178) /  0.1^2= 0.428934049$
I have tried an alternative approach of writing a hundred thousand rows on a spreadsheet and assigning “life” or “death” randomly to each, based on that tiny daily crude death rate of 0.001178, and then randomly sampling 10,000 rows.  I run the sample repeatedly and get terribly erratic results despite the huge sample size.
I realize that friend Cochran had a better grasp of the subject than I, but I can’t see where I went wrong.  Can anyone think what I’m misunderstanding, please, or steer me to a better approach?

REPLY [1 votes]: In this formula $M$ is the absolute error and not the relative error. Your crude estimate of $p = 0.001178$ means that an error margin of $0.1$ is near meaningless. $M= 0.0001$ is a better choice which gets you a sample size of 452931. Obviously this can be made higher or lower depending on what data you have available.