text:
\begin{document}

	\title{Complex Links and Hilbert-Samuel Multiplicities}
	\author{Martin Helmer}
	\address[MH]{
		Mathematical Sciences Institute,
		The Australian National University,
		Canberra, Australia}\email{martin.helmer@anu.edu.au}
	\author{Vidit Nanda}\address[VN]{Mathematical Institute,
		University of Oxford, Oxford, United Kingdom}\email{nanda@maths.ox.ac.uk}

	\begin{abstract}
		We describe a framework for estimating Hilbert-Samuel multiplicities $\e_XY$ for pairs of projective varieties $X \subset Y$ from finite point samples rather than defining equations. The first step involves proving that this multiplicity remains invariant under certain hyperplane sections which reduce $X$ to a point $p$ and $Y$ to a curve $C$. Next, we establish that $\e_pC$ equals the Euler characteristic (and hence, the cardinality) of the complex link of $p$ in $C$. Finally, we provide explicit bounds on the number of uniform point samples needed (in an annular neighborhood of $p$ in $C$) to determine this Euler characteristic with high confidence.
	\end{abstract}
	\maketitle

	\section{Introduction}

	One of the most fundamental quantities of interest in intersection theory is the {\bf Hilbert-Samuel multiplicity}, which associates an integer $\e_XY \geq 0$ to each pair consisting of an irreducible subvariety $X$ inside a pure-dimensional scheme $Y$. This integer serves -- among other things -- as a coarse measurement of the singularity type of $X$ inside $Y$. When $Y$ is reduced, $\e_XY = 1$ holds if and only if $X$ is nonempty and smoothly embedded in $Y$. The importance of Hilbert-Samuel multiplicities stems from their wide-ranging connections with several other intersection-theoretic invariants. For instance, $\e_XY$ appears as the coefficient of $[X]$ in the {Segre class} $s(X,Y)$ \cite[Chapter 4.3]{fulton2013intersection}, in  Fulton and MacPherson's intersection product \cite[Chapter 12.3]{fulton2013intersection}, and in Serre's Tor formula \cite[Theorem 1, pg 112]{serre2012local}. Computing $\e_XY$, either directly from its definition or as a consequence of these connections,  requires serious algebraic manipulations of the defining equations for $X$ and $Y$.

	Our goal in this paper is to describe a new framework for estimating $\e_XY$ from  finite local point samples without recourse to any such equations.  In this setting, we have no means to capture the scheme structure of $Y$, and will therefore restrict to the case where $Y$ is reduced, i.e.,~a pure dimensional variety in some $n$-dimensional projective space $\pp^n$. The methods developed here could also be applied to any data set which we would expect to have the structure of a complex variety, even if the variety is not known. Here is an informal version of our main result for estimating $\e_XY$ for such pairs $X \subset Y$.

	\begin{theorem*}
		Let $L \subset \pp^n$ be a linear space obtained by intersecting $(\dim Y - 1)$ hyperplanes which are general except for the requirement that they all pass through a generic point $p$ of $X$. The Hilbert-Samuel multiplicity $\e_XY$ can be determined with high confidence from a sufficiently large (but finite) uniform point sample $S$ lying on the curve $Y \cap L$ in a local annular neighbourhood around $p$.
	\end{theorem*}

	\noindent We provide explicit bounds on how large $S$ must be in terms of the local geometry of $Y \cap L$ near $p$ and the desired probability of successful estimation. Our proof has three basic steps, each involving a different key ingredient and producing an intermediate result. These steps are summarized below.

	\subsection*{Step 1: Algebra} We first establish that $\e_XY$ is invariant under the operation of slicing both $X$ and $Y$ by certain hyperplanes. The key ingredient here is a new {\em degree formula} for Hilbert-Samuel multiplicities \cite[Theorem 5.3]{HH19}. Using this formula, we prove the following result.

	\begin{theorem*} [\textbf{A}]
		Given ${X} \subset {Y}$ as above, let $L$ be the intersection of $k$ general hyperplanes which all pass through some general point $p$ of $X$; then,
		\begin{enumerate}
			\item if $k \leq \dim {X}$, then $\e_{{X}}{Y} = \e_{{X} \cap L}({Y} \cap L)$; moreover,
			\item if $k = \dim {X}$, then $\e_{{X}}{Y} = \e_p({Y} \cap L)$; and finally,
			\item if $\dim X < k \leq \dim Y - 1$, then $\e_{{X}}{Y} = \e_p({Y} \cap L)$.
		\end{enumerate}
	\end{theorem*}

	In fact, the first two assertions follow readily from basic properties of Segre classes whereas the last one is new and makes essential use of the aforementioned degree formula from \cite{HH19}. As a consequence of this third assertion (for $k = \dim Y - 1$), every $\e_XY$ calculation can be reduced to the case where $X = p$ is a point and $Y = C$ is a curve in $\pp^n$ containing $p$. In this special case, the degree formula for Hilbert-Samuel multiplicity simplifies to
	\[
	\e_pC = \deg(C) - \deg((C \cap H) - p),
	\]
	where $H$ is a general hyperplane passing through $p$. While this is a convenient reformulation for algebraic computation of $\e_pC$, both degrees appearing on the right side are global computations in the sense that they require checking for intersections far away from $p$. The purpose of the next step is to replace these with a local computation near $p$.

	\subsection*{Step 2: Topology}

	The starting point for our second step is the observation that $\deg(C)$ equals the cardinality of $C \cap H'$, where $H'$ is a general hyperplane in $\pp^n$. Crucially, we let $H'$ be parallel to the plane $H$ when restricted to an affine chart of $\pp^n$ containing $p$. The special ingredient here is {\em Thom's first isotopy lemma} \cite[Chapter I.1.5]{SMTbook}, which allows us to relate $\e_pC$ to the Euler characteristic (and hence, cardinality) of the zero-dimensional space
	\[
	\Clk_p := C \cap \BB_\epsilon(p) \cap H'.
	\]
	Here $\BB_\epsilon(p)$ denotes a small closed ball around $p$ in a chart of $\pp^n$. In particular, we employ a homological argument to show the following result.

	\begin{theorem*}[\textbf{B}] If $p$ is any (possibly singular) point on a curve $C \subset \pp^n$, then its Hilbert-Samuel multiplicity satisfies $\e_pC = \chi(\Clk_p)$, where $\chi$ denotes Euler characteristic.
	\end{theorem*}

	The space $\Clk_p$ plays a fundamental role in (complex) stratified Morse theory --- it provides {\em normal Morse data} for $p$ with respect to a stratified Morse function defined on $C$, and is called the {\bf complex link} of $p$ in $C$ \cite[Ch II.2]{SMTbook}. Variants of Theorem ({\bf B}) have been assigned as exercises to the reader on several occasions, including the Introduction to \cite{SMTbook} and \cite[Ex 4.6]{masseychar}. However, we were unable to locate a proof in the literature; since it forms an essential part of our overall argument, we have included a proof here.

	\subsection*{Step 3: Geometry} It remains to estimate the cardinality of $\Clk_p$ using a uniform finite point sample $S$ chosen from $B := C \cap \BB_\epsilon(p)$. The main difficulty here is that generically the intersection $S \cap H'$ will be empty even when the sample size is enormous. As such, we are compelled to search for points of $S$ which lie within some small distance $\eta > 0$ of $H'$, and hope that these points naturally organize into $\e_XY$ many clusters. The key ingredient here is a suite of {\em geometric inference results}, which date back to the work of Niyogi, Smale and Weinberger from \cite{niyogietal}. Given a compact Riemannian submanifold $M \subset \RR^d$ and a probability parameter $\gamma \in (0,1)$, these results give explicit bounds on the cardinality of a finite point sample $P \subset M$ required to estimate the homology of $M$ with probability exceeding $(1-\gamma)$.

	Recently, Wang and Wang have extended results of \cite{niyogietal} to the case where $M \subset \RR^d$ is a smooth submanifold with boundary \cite{wang2}; they provide an explicit lower bound $N_M(\alpha,\gamma)$ on the size of a uniform point sample $P \subset M$ required to ensure, again with probability at least $(1-\gamma)$, that $P$ is $\alpha/2$-dense in $M$. These results require $\alpha$ to be sufficiently small relative to the injectivity radii of the embeddings $M \inj \RR^d$ and $\partial M \inj \RR^d$ of the manifold and its boundary respectively. Although the space $B$ of interest to us is not a manifold (thanks to the singularity at $p$), it does become a manifold with boundary by removing the interior of a smaller ball $\BB_{\epsilon_0}(p)$ with $\epsilon_0 < \epsilon$. After this excision, we can safely apply the density results from \cite{wang2} and obtain the following result.

	\begin{theorem*} [\textbf{C}] There exists, for all sufficiently small radius $\alpha > 0$ and probabilities $\gamma \in (0,1)$, an explicit bound $N_M(\alpha,\gamma)$ with the following property. Any uniformly sampled subset
		\[
		S \subset C \cap \left(\BB_\epsilon(p) - \BB_{\epsilon_0}(p)^\circ\right)
		\] of cardinality $\#S > N_M(\alpha,\gamma)$ can be used to correctly estimate the Euler characteristic $\chi(\Clk_p)$ with probability exceeding $(1-\gamma)$.
	\end{theorem*}

	Combining Theorems ({\bf A}), ({\bf B}) and ({\bf C}) gives the promised main result.

	\subsection*{Organisation} In Section \ref{section:complink} we briefly review the definition of the complex link. Section \ref{sec:algmult} focuses on Step 1; here we give a brief overview of the Hilbert-Samuel multiplicity and prove Theorem ({\bf A}). In Section \ref{sec:pointcurve} we implement Step 2 by providing a proof of the folklore Theorem ({\bf B}). And finally, in Section \ref{section:Multiplcity_From_Finite_Samples} we carefully state and establish Theorem ({\bf C}) by describing not only the precise form of the bound $N(\alpha,\gamma)$, but also the precise constraints on $\alpha$ imposed by the local geometry of $C$ near $p$.

	{\footnotesize
		\subsection*{Acknowledgements}

		Mark Goresky kindly shared an advance copy of his survey {\em Morse theory, stratifications and sheaves} \cite{goresky-survey} with us; that paper served as our Polaris while we navigated the formidable waters surrounding these topics. We  are grateful to Heather Harrington for organizing the event in Oxford where our collaboration was conceived, to Kate Turner for hosting VN's visit to Canberra where it reached adolescence, and to Yossi Bokor for writing helpful software during these early days. We thank the Sydney Mathematics Research Institute (SMRI) at the University of Sydney for their generous hospitality. VN's work was supported by the EPSRC grant EP/R018472/1 and by the DSTL grant D015 funded through the Alan Turing Institute.
	}

	\section{Complex Links}\label{section:complink}

	A {\em stratification} of a topological space $\bW$ is a filtration
	\[
	\emptyset = W_{-1}\subset W_0 \subset \cdots \subset W_k=\bW
	\] by closed subspaces so that each consecutive difference {$W_{i}-W_{i-1}$} is a (possibly empty or disconnected) $i$-dimensional manifold called the $i$-{\em stratum}. Throughout this section, $\bW$ will denote a Whitney-stratified complex analytic subspace of $\CC^n$. We assume that each stratum $X \subset \bW$ is a connected complex analytic manifold, and write $Y > X$ to indicate that the closure of the stratum $Y$ contains the stratum $X$. We further require all pairs of strata $X < Y$ to satisfy Whitney's {Condition (B)} --- see \cite[Section 19]{whitney1965tangents}, \cite[Section 2]{Mather2012}, or \cite[Chapter I.1.2]{SMTbook}.  Let $T_pX$ denote the $(\dim X)$-dimensional linear subspace of $\CC^n$ which corresponds to the tangent space of a stratum $X$ at a point $p$ in $X$.

	Fix a connected component of a stratum $X \subset \bW$ and consider an arbitrary point $p$ in $X$. Since it remains difficult to illustrate even $2$-dimensional complex varieties, the following real picture (where $n=3$ and $\dim \bW = 2$ while $\dim X=1$) will serve as a proxy for the local structure of $\bW$ near $p$.

	\begin{figure}[h!]
		\centering
		\includegraphics[scale=.35]{stratspace.png}
	\end{figure}

	\noindent The stratum $X$ is represented by the horizontal line along which the four sheets intersect, and the chosen point $p$ is located near the center of $X$. We say that an affine subspace $A \subset \CC^n$ containing $p$ is {\em transverse} to $X$ at $p$ if the sum of tangent subspaces given by \[T_pX + T_pA  = \set{v+w \mid v\in T_pX \text{ and }w \in T_pA}\] equals $T_p\CC^n = \CC^n$.

	\begin{definition} \label{def:normslice} A subset $\bN \subset \bW$ is called a {\bf normal slice} to $X$ at $p$ if it equals the intersection $\bW \cap A$ for some $(n-\dim X)$-dimensional affine subspace $A \subset \CC^n$ which intersects $X$ transversely at $p$.
	\end{definition}
	One possible choice of $\bN$ for our example is shown below:

	\begin{center}
		\includegraphics[scale=.35]{normslice.png}
	\end{center}

	\noindent Here $A$ is the plane which crosses $X$ at $p$, while $\bN$ is the union of four half-open arcs, all of which intersect at $p$. Evidently, $\bN$ will not be a manifold in general; on the other hand, it follows from the definition of a Whitney stratification that $A$ will remain transverse, at least in a small neighbourhood around $p$, to all higher strata $Y > X$. Thus, $\bN$ inherits a Whitney stratification from $\bW$ near $p$ as follows. Each $(\dim Y)$-dimensional stratum $Y > X$ carves out a (possibly disconnected, $(\dim Y-\dim X)$-dimensional) stratum $Y \cap A$ of $\bN$.

	Fix a radius $\epsilon > 0$ so that the intersection of $\bN$ with the open ball $\BB_\epsilon(p)$ of radius $\epsilon$ around $p$ inherits a Whitney stratification from $\bW$ in the manner described above\footnote{\noindent More precisely, two natural transversality constraints must hold for every radius $e \leq \epsilon$ and for every stratum $Y$ of $\bW$. First, the boundary of $\BB_e(p)$ must be transverse to $Y$ in $\CC^n$, and second, the boundary of $\BB_e(p) \cap A$ must be transverse to $Y \cap A$ in $A$.}. We write $\bN_\epsilon(p) = \bN \cap \BB_\epsilon(p)$ to indicate this {\em restricted} normal slice. The next definition will make use of our chosen $\bN$ and $\epsilon$, and also of the usual inner product $\ip{\bullet,\bullet}$ defined on the ambient space $\CC^n$.

	\begin{definition}\label{def:xi}
		A vector $\xi$ in the affine space $A$ is called {\bf nondegenerate} for the pair $(\bN,\epsilon)$ if the following property holds for all strata $Y > X$. Given any sequence $\{(q_i,v_i)\}$ in the tangent bundle of $Y' = Y \cap \bN_\epsilon(p)$ where $q_i$ limits to $p$, if the $v_i$ limit to some nonzero vector $v$ then $\ip{\xi,v} \neq 0$.
	\end{definition}

	If we restrict our pictorial example to the affine plane $A$, then the set of degenerate vectors will span the vertical line through $p$ because the orthogonal complement of this vertical line (in $A$ through $p$, as drawn below) shares a limiting tangent with all four arcs of $\bN$. For any vertically-aligned $\xi$, one can find a sequence $(q_i,v_i)$ in the tangent bundle of each arc with $q_i \to p$ and $v_i \to v \neq 0$ lying along the horizontal line, which in turn forces $\ip{\xi,v} = 0$. Any $\xi$ off the vertical line will be nondegenerate.
	\begin{center}
		\includegraphics[scale=.4]{degen.png}
	\end{center}

	\noindent Fix a nondegenerate vector $\xi$ for $(\bN,\epsilon)$, and consider the map
	\[
	\pi_\xi:A \to \RR
	\] given by taking the real part of the affine-linear complex functional $z \mapsto \ip{z-p,\xi}$. By nondegeneracy, there exists a  $\delta > 0$ so that if the differential \[(d\pi_\xi)_q:T_qY' \to \RR\] is identically zero at some point $q \neq p$ in a stratum $Y'$ of $\bN_\epsilon(p)$, then $|\pi_\xi(q)| > \delta$ . In other words, the preceding definitions and choices have been concocted in order to ensure that $\pi_\xi$ restricts to a {\bf stratified Morse function} on  $\bN_\epsilon(p)$ as in \cite[Chapter I.2]{SMTbook}; and moreover, $p$ is its unique critical point valued in the interval $[-\delta,\delta]$. Here is a summary of all these choices that have been made for the stratum $X \subset \bW$ of dimension $\dim X$:
	\begin{enumerate}
		\item a point $p \in X$,
		\item an $(n-\dim X)$-dimensional affine subspace $A \subset \CC^n$ transverse to $X$ at $p$,
		\item a radius $\epsilon > 0$ so that $\bN_\epsilon(p) = \bW \cap A \cap \BB_\epsilon(p)$ inherits a  stratification from $\bW$,
		\item a nondegenerate vector $\xi \in A$, and finally,
		\item another radius $\delta \in (0,\epsilon)$ so that $\pi_\xi:A \to \RR$ has no critical points $q \neq p$ in $\bN_\epsilon(p)$ with $\pi_\xi(q)$ in $[-\delta,\delta]$.
	\end{enumerate}
	The following definition makes provisional use of this tuple $(p,A,\epsilon,\xi,\delta)$.

	\begin{definition}\label{def:clink}
		The {\bf complex link} of the stratum $X \subset \bW$ with respect to the choices $(p,A,\epsilon,\xi,\delta)$ is the intersection
		\[
		\Clk_X = \bN_\epsilon(p) \cap \pi_\xi^{-1}(\delta).
		\]
	\end{definition}

	Returning to our example one final time: the hyperplane $\pi_\xi^{-1}(\delta)$ is a non-horizontal line in the plane $A$ which passes near, but not through, the central point $p$. In a small $\epsilon$-ball around $p$, this line generically intersects the arcs which form $\bN_\epsilon(p)$ in two points, so the complex link $\Clk_X$ in this case is just the two-point space:

	\begin{figure}[h!]
		\includegraphics[scale=.4]{intersect.png}
	\end{figure}

	\noindent The (stratified homeomorphism type of the) complex link $\Clk_X$ depends only on the stratum $X$, and not on the auxiliary choices $(p,A,\epsilon,\xi,\delta)$ described above \cite[Chapter II.2]{SMTbook}. It is also interesting to note that the invariance of $\Clk_X$ to the chosen direction $\xi$ is entirely a feature of complex analytic geometry --- for real analytic Whitney stratified spaces, the intersection $\bN_\epsilon(p) \cap \pi_\xi^{-1}(\delta)$ is liable to change as $\xi$ is varied.

	\begin{remark}
		In this paper we will be exclusively interested in the complex link of a point (i.e., a zero-dimensional stratum) within a complex projective variety. In this  special case, one is not required to construct a normal slice, so the formula from Definition \ref{def:clink} reduces to $\Clk_X = \bW \cap \BB_\epsilon(p) \cap \pi_\xi^{-1}(\delta)$.
	\end{remark}

	\section{The Hilbert-Samuel Multiplicity}\label{sec:algmult}

	Let $R = \CC[x_0,\ldots,x_n]$ denote either the coordinate ring of projective space $\pp^n$ or affine space $\CC^{n+1}$. In the projective case we will implicitly assume that $R$ is graded and all of its ideals considered below are homogeneous. Let $X$ be an irreducible complex (affine or projective) algebraic variety given by a prime ideal $I \lhd R$ and let $Y$ be a scheme corresponding to a primary ideal $J \subset I$. The {\em local ring of $Y$ along $X$}, usually written $\cO_{X,Y}$, is the localization of $(R/J)$ at $I$. The following notion is due to Samuel \cite{samuel1955methodes}.

	\begin{definition} \label{def:algmult}
		Let $\cM$ be the maximal ideal of $\cO_{X,Y}$ and let $c$ be the codimension $\dim Y - \dim X$. The {\bf Hilbert-Samuel function} of $Y$ along $X$ is \[
		\textbf{HS}(t) = \text{length}\left(\cO_{X,Y}/\cM^t\right).
		\]
		For all $t \gg 0$, this function is a polynomial in $t$ of degree $c$ whose leading coefficient is a strictly positive integer divisible by $c!$ --- and the {\bf Hilbert-Samuel multiplicity of $Y$ along $X$}, written $\e_XY$, is the leading coefficient of the normalized polynomial $(1/c!)\cdot {\bf HS}(t)$.
	\end{definition}

	It is shown in \cite[Section 4.3]{fulton2013intersection} that $\e_XY$ is also equal to the coefficient of $[X]$ in the {\bf Segre class} $s(X,Y)$, which naturally lives in the {\em Chow group} of $X$ (or in the {\em Chow ring} of an ambient smooth variety $M$ via push-forward, $X\subset Y\subset M$; we will often work in the $M=\pp^n$ setting).

	\subsection{Multiplicities from Degrees}

	When $X \subset Y \subset \pp^n$ are projective varieties, it is often algorithmically convenient to extract $\e_XY$ from a choice $B_X = \set{f_0,\dots, f_r}$ of homogeneous polynomials that generate the defining ideal $I \lhd R$ of $X$. We will assume here that all the $f_i$ have the same degree $d$, which is always possible to arrange without loss of generality \cite[Section 2.1.4]{HH19}. Let $L_i$ be a generic $(n-i)$-dimensional linear subspace of $\pp^n$, and let $V_i \subset \pp^n$ be the varieties given by
	\begin{align} \label{eq:V}
	V_i = \set{x \in \pp^n \mid F_1(x) = F_2(x) = \cdots = F_{\dim Y-i}(x) = 0},
	\end{align} where the $F_j$ are homogeneous polynomials of degree $d$ that have the form
	\[
	F_j = \sum_{k=0}^r \lambda_k^j f_k
	\]
	for general choices of $\lambda_k^j \in \CC$. (Note $V_i$ contains $X$ for all $i$ by design).

	In \cite{HH19} it is shown that the Segre class $s(X,Y)$, and hence the  multiplicity $\e_XY$, is determined by the numbers
	\begin{align}\label{eq:lambdadeg}
	\LC^i_XY = \deg(Y) \cdot d^{\dim Y - i}- \deg((Y \cap V_i \cap L_i) - X),
	\end{align}
	for each $i$ between $0$ and $\dim X$. In particular, \cite[Theorem 5.3]{HH19} establishes that
	\begin{align}\label{eq:multlambda}
	\e_XY = \frac{\LC^{\dim X}_XY}{\deg X},
	\end{align}
	or more explicitly,
	\begin{align}\label{eq:multlambdalong}
	\e_XY = \frac{\deg(Y) \cdot d^{\dim Y - \dim X}- \deg((Y \cap V_{\dim X} \cap L_{\dim X}) - X)}{\deg X}.
	\end{align}
	In order to treat affine varieties on an equal footing with projective ones when it comes to using (\ref{eq:multlambda}) and related formulas, we will appeal to the following result. The idea is to replace the affine varieties $X\subset Y$ in $\CC^n$ by their {\em projective closures} $PX \subset PY$ in $\pp^n$ --- see \cite[Exercise I.2.9]{Hartshorne1974} for a definition.
	\begin{proposition}\label{prop:affine_mult}
		Let $X\subset Y$ be closed subvarieties of affine space $\CC^n$ and let $PX \subset PY$ denote their projective closures in $\pp^n$. Then, we have
		\[
		\e_XY=\frac{\LC_{PX}^{\dim PX} PY}{\deg PX},
		\]
		i.e., the multiplicity of $Y$ along $X$ can be computed from (\ref{eq:multlambda}) applied to the projective closures of $X$ and $Y$.
	\end{proposition}
	\begin{proof}
		Select an affine chart of $\pp^n$ specified by a general hyperplane $H$, i.e. $\CC^n\cong \pp^n-H$ and an open set $U \subset \pp^n$ so that ({\bf a}) $PY \cap U$ is open and dense in $PY$,  ({\bf b}) if $V=U-H$ then $Y \cap V$ is open and dense in $Y\subset \CC^n$, and ({\bf c}) $PX \cap U$ is open and dense in $PX$ (i.e. $PX$ is not contained in $\pp^n-U$). By ({\bf a}) and ({\bf c}) we obtain an isomorphism of local rings \[\cO_{PX,PY} \simeq \cO_{PX \cap U,PY \cap U}.\] Next, since $X$ is a subvariety of $Y$ with $Y \cap V$ dense in $Y$ by ({\bf b}), we have a second isomorphism of local rings \[
		\cO_{PX \cap U,PY \cap U} \simeq \cO_{X \cap V,Y \cap V}.\] And finally, since $Y\cap V$ is dense in $Y$ and since $PX$ is not contained in $\pp^n-U$, then $X\cap V$ is dense in $X$ and we have a third isomorphism \[\cO_{{X}\cap V,{Y}\cap V} \simeq \cO_{X,Y}.\] Stringing together these three isomorphims, one obtains $\cO_{X,Y} \simeq \cO_{PX,PY}$; since the multiplicities $\e_XY$ and $\e_{PX}{PY}$ are completely determined by the corresponding local rings, they must be equal. The desired conclusion now follows from (\ref{eq:multlambda}).
	\end{proof}

	We expect that some version of the above argument, (i.e., that $\e_XY$ must equal $\e_{PX}{PY}$ because the two associated local rings are isomorphic) already exists in the literature, but we were unable to locate it and have therefore included this proof for completeness. This result facilitates the use of \eqref{eq:multlambdalong} for a pair of affine varieties.

	\subsection{Multiplicities of Linear Sections}\label{sec:eslice}

	Here we describe the behaviour of the Hilbert-Samuel multiplicity $\e_XY$ for complex projective varieties $X \subset Y$ when both $X$ and $Y$ are replaced by their intersections with (sufficiently generic) linear spaces. The proposition below can be seen as a direct consequence of standard properties of Segre classes (along with the relation between Segre classes and multiplicities).

	\begin{proposition}\label{prop:algmultsection}
		Let $Y$ be a pure dimensional subscheme of the complex projective space $\PP^n$, let $X$ be an irreducible subvariety of $Y$ and let $L \subset \pp^n$ be given by an intersection
		\[
		L = H_1 \cap H_2 \cap \cdots \cap H_\ell,
		\]
		where each $H_i \subset \pp^n$ is a generic hyperplane. If the codimension $\ell = n - \dim L$ is strictly less than $\dim X$, then the multiplicities $\e_XY$ and $\e_{X \cap L}(Y \cap L)$ are equal. Further if $\ell=\dim(X)$ then $\e_XY=\e_p(Y\cap L)$, where $p$ is any of the $\deg(X)$ points in $X\cap L$.
	\end{proposition}
	\begin{proof}In this proof we will work with the pushforward to the Chow ring of $\pp^n$ of the Segre class $s(X,Y)$, in a slight abuse of notation this will also be denoted $s(X,Y)$. Denote this Chow ring as $A^*(\PP^n)\cong \ZZ[h]/\langle h^{n+1} \rangle$ where $h$ is the rational equivalence class of a general hyperplane.
		Since each $H_i$ is a general divisor on $\pp^n$, the coefficient of $h^{\dim X - \ell}$ in the Segre class $s(X \cap L,Y \cap L)$ equals the coefficient of $h^{\dim X}$ in the Segre class $s(X,Y)$, i.e.,
		\[
		\{s(X\cap L,Y\cap L)\}_{\dim X - \ell}  = \{s(X,Y)\}_{\dim X} \cdot h^\ell.
		\]
		A proof of the above property of Segre classes can be found, for example, in \cite[Corollary~3.2]{Harris2017}. First suppose that $\ell<\dim(X)$. Using the fact that $\e_{X \cap L}(Y \cap L)$ is the coefficient of $[X \cap L]$ in $s(X \cap L,Y \cap L)$, one obtains
		\begin{align*}
		\{s(X\cap L,Y\cap L)\}_{\dim X - \ell} & = \e_{X\cap L}(Y\cap L) \cdot [X\cap L] \\
		&=\e_{X\cap L}(Y\cap L) \cdot \deg X \cdot h^{n-\dim X+\ell},
		\end{align*}
		where the second equality follows from the fact that each $H_i$ is a general divisor, so in particular $\deg X = \deg (X \cap L)$. Now take $\ell=\dim(X)$. Then $X\cap L$ consists of $\deg(X)$ reduced points $p_1,\dots, p_{\deg(X)}$; by \cite[Example 4.3.4]{fulton2013intersection} we have that \begin{align*}
		\{s(X\cap L,Y\cap L)\}_{\dim X - \ell} & = \e_{p_1}(Y\cap L)[p_1] +\cdots +\e_{p_{\deg(X)}}(Y\cap L)[p_{\dim(X)}]\\
		&=\e_p(Y\cap L)\deg(X)[p]\\
		&=\e_p(Y\cap L) \cdot \deg X \cdot h^{n-\dim X+\ell}
		\end{align*} where $p$ is any point in $X\cap L$ (all of which are rationally equivalent). 	On the other hand, we also have
		\begin{align*}
		\{s(X,Y)\}_{\dim X} \cdot h^\ell  &=\e_XY\cdot [X] \cdot h^\ell \\
		&= \e_XY\cdot \deg X \cdot h^{n-\dim X+\ell},
		\end{align*} which forces $\e_{X\cap L}(Y\cap L)=\e_XY$ for $\ell<\dim(X)$ and $\e_{p}(Y\cap L)=\e_XY$ when $\ell=\dim(X)$ as desired.
	\end{proof}
	Having cut $X$ down to a single point $p$, we turn our attention to simplifying $Y$. The next result uses \eqref{eq:multlambda} to show that we can always replace $Y$ by a curve, i.e., a one-dimensional projective variety, when computing $\e_pY$.

	\begin{proposition}\label{prop:algmultcurve}
		Let $Y$ be a pure dimensional subscheme of the complex projective space $\PP^n$, let $p$ be any reduced point in $Y$, and consider a linear space $L \subset \pp^n$ given by the intersection of $m \geq 0$ general hyperplanes containing $p$. If $m \leq  \dim Y - 1$, then $\e_p(Y \cap L)$ is well-defined and equals $\e_{p}Y$.
	\end{proposition}
	\begin{proof}
		The generating ideal of $p$ in the polynomial ring $\CC[x_0,\dots,x_n]$ can be chosen to consist of $n$ linear forms $\{\ell_1,\ldots,\ell_n\}$. It follows that $d=1$ and $\deg X=1$ in \eqref{eq:multlambdalong}. Since $\dim p = 0$, we have
		\[
		\e_{p}Y=\deg(Y)-\deg((Y\cap V_{\dim Y})-p),
		\] where $V_{\dim Y}$ is the variety defined by the polynomials $\{P_1,\ldots,P_{\dim Y}\}$, with each $P_j$ being a linear combination of the form \[P_j = \sum_{i=1}^n \lambda^j_i \ell_i \text{ for general } \lambda^j_i\in \CC.\] Without loss of generality, we may take $L$ to be the variety defined by the first $m$ of these, say $\{P_1,\dots, P_m\}$. Thus, $L$ is a linear system with base locus $p$, so it forms a smooth complete intersection outside of $p$. Hence the intersection $Y\cap L$ is transverse in the expected dimension, i.e.~in dimension $\dim (Y\cap L) = \dim Y - m > 0$ and moreover, $\deg(Y\cap L)=\deg(Y)$. Letting $V_{> m}$ be the variety defined by $\{P_{m+1},\ldots P_{\dim Y}\}$, we have
		\begin{align*}
		\e_{p}Y&=\deg(Y)-\deg((Y\cap V_{\dim Y})-p) \\&=\deg(Y\cap L)-\deg((Y\cap L \cap V_{> m})-p)  \\
		&=\e_{p}(Y\cap L).
		\end{align*} This argument fails whenever $m=\dim Y$, since the intersection $Y\cap L$ may not be transverse in this case.
	\end{proof}

	The following theorem serves to summarize the main results in this section by combining Propositions \ref{prop:algmultsection}  and \ref{prop:algmultcurve}.

	\begin{theorem}\label{thm:algmultslice}
		Let $X \subset Y$ be a pair of complex projective subvarieties of $\pp^n$, and let $L \subset \pp^n$ be a linear space given by the intersection of $k \geq 0$ general hyperplanes $H_1, \ldots, H_k$ containing a point $p$ of $X$.
		\begin{enumerate}
			\item If $k \leq \dim X$, then $\e_XY = \e_{X \cap L}(Y \cap L)$.
			\item If $k = \dim X$, then $\e_p(Y \cap L) = \e_XY$.
			\item If $\dim X < k \leq \dim Y - 1$, then $\e_XY = \e_p(Y \cap L)$.
		\end{enumerate}
	\end{theorem}

	\noindent This is Theorem ({\bf A}) from the Introduction. We note that we may allow $Y$ in the statement above to be any pure dimensional subscheme of $\pp^n$, but have restricted the case where $Y$ is a variety as this will be the only case we employ in later sections. Assertion (3) of this result (for $k = \dim Y - 1$) implies that the evaluation of $\e_XY$ for arbitrary projective varieties $X \subset Y$ in $\pp^n$ can be reduced to the computation of $\e_pC$ where $p$ is a point lying on the curve $C = Y \cap L$; this scenario will be the central focus of the next section.

	\section{Point-Curve Multiplicities via Complex Links}\label{sec:pointcurve}

	Our goal here is to provide a stratified Morse-theoretic proof of  Theorem ({\bf B}) from the Introduction.
	\begin{theorem}\label{thm:toplink}
		If $p$ is any (possibly singular) point on a curve $C \subset \pp^n$, then we have
		\[
		\e_pC = \chi(\Clk_p),
		\]
		where $\e_pC$ is the Hilbert-Samuel multiplicity (from Definition \ref{def:algmult}) and $\chi(\Clk_p)$ is the Euler characteristic of $p$'s complex link in $C$ (from Definition \ref{def:clink}).
	\end{theorem}
	Since $p$ can be defined as the zero set of $n$ linear polynomials, by \eqref{eq:lambdadeg} we have
	\[
	\e_pC = \LC^0_pC = \deg(C) - \deg((C \cap H_p) - p),
	\] where $H_p$ is a generic hyperplane in $\pp^n$ passing through $p$. Let $\xi$ be the unit normal to $H_p$, and denote by $\pi_\xi:\pp^n \to \RR$ the projection map $z \mapsto \text{Re}\ip{z-p,\xi}$ so that the level set of $\pi_\xi$ at $0$ is precisely $H_p$. By our genericity assumption on $H_p$, the vector $\xi$ is nondegenerate in the sense of Definition \ref{def:xi}. Thus, there is some small positive $\delta$ so that $\pi_\xi$ is a stratified Morse function on $C$ with no critical point other than $p$ taking values in $[-\delta,\delta]$. In particular, this means that no singular points of $C$ other than $p$ are allowed to lie in $C \cap \pi_\xi^{-1}[0,\delta]$, which is the part of $C$ lying within the shaded rectangular region below\footnote{Here we have resorted to drawing a real picture for simplicity; the one-dimensional complex curve $C$ should in fact have $\RR$-dimension two.}.

	\begin{center}
		\includegraphics[scale=.55]{curvelink.png}
	\end{center}

	The upper levelset $\pi_\xi^{-1}(\delta)$ intersects $C$ in a set of cardinality $\deg C$ since $\pi_\xi^{-1}(\delta)$ is sufficiently generic. On the other hand, since the lower levelset $\pi_\xi^{-1}(0)$ is forced to pass through $p$, it may intersect $C$ in fewer points. Thus, the quantity of interest to us here is
	\begin{align}
	\LC^0_pC = \#\set{C \cap \pi_\xi^{-1}(\delta)} - \#\set{C \cap \pi_\xi^{-1}(0)} + 1, \label{eq:lambda0}
	\end{align}
	where the last $+1$ term comes from the fact that we are required to discard $p$ from the second intersection. The main tool in our argument here is one of Thom's celebrated Isotopy Lemmas --- see \cite[Proposition 11.1]{Mather2012} or \cite[Chapter I.1.5]{SMTbook}.

	\begin{lemma}\label{lem:thom}{\bf [Thom's first isotopy lemma]}
		Let $M$ and $N$ be smooth manifolds and $Z \subset M$ a Whitney stratified subset. If $f:M \to N$ is a smooth proper map whose restriction $f|_X$ to each stratum $X \subset Z$ is a submersion (i.e., the derivative $df_p:T_pX \to T_{f(p)}N$ is surjective for all $p$ in $X$), then $f|_X:X \to f(X)$ is a (locally trivial) fiber bundle.
	\end{lemma}

	By our choice of $\delta$, the function $\pi_\xi$ when restricted to $C \cap \pi_\xi^{-1}(0,\delta)$ satisfies the hypotheses of this lemma. Reducing $\delta$ further if necessary, we are therefore guaranteed the existence of a local trivialization, i.e., a homeomorphism
	\[
	\left[C \cap \pi^{-1}_\xi(0,\delta)\right] \simeq (0,\delta) \times \left[C \cap \pi_\xi^{-1}(\delta)\right].
	\] Our strategy here is to examine the following zigzag diagram of inclusion maps
	\begin{align}
	C \cap \pi_\xi^{-1}(0) \hookrightarrow C \cap \pi_\xi^{-1}[0,\delta] \hookleftarrow C \cap  \pi_\xi^{-1}(\delta).\label{eq:zz}
	\end{align} This next result is concerned with the first inclusion.

	\begin{proposition}\label{prop:zerodelta}
		The inclusion $C \cap \pi_\xi^{-1}(0) \hookrightarrow C \cap \pi_\xi^{-1}[0,\delta]$ is a homotopy equivalence, and in particular it admits a homotopy-inverse $\phi:C \cap \pi_\xi^{-1}[0,\delta] \to C \cap \pi_\xi^{-1}(0)$.
	\end{proposition}
	\begin{proof}
		By Thom's first Isotopy Lemma applied to the restriction of $\pi_\xi$ to $C \cap \pi_\xi^{-1}(0,\delta]$, the constant vector field $-\xi$ on $(0,\delta]$ lifts to a vector field $V$ on $C \cap \pi_\xi^{-1}(0,\delta]$ so that the differential $d\pi_\xi$ sends each vector of $V$ to $-\xi$, as depicted below:

		\begin{center}
			\includegraphics[scale=.55]{curveret.png}
		\end{center}
		\noindent The desired map $\phi$ is obtained by flowing along the integral curves of $V$.
	\end{proof}

	The second map from our zigzag (\ref{eq:zz}) will be described via the corresponding relative homology group, namely
	\[
	\HG_\bullet\left(C \cap \pi_\xi^{-1}[0,\delta],C \cap \pi_\xi^{-1}(\delta)\right),
	\] where we have implicitly assumed rational coefficients throughout. The following result shows that this group only depends on local data pertaining to the fibers of $\phi$ over $p$.

	\begin{lemma}\label{lem:local}
		Let $\phi:C \cap \pi_\xi^{-1}[0,\delta] \to C \cap \pi_\xi^{-1}(0)$ be a homotopy inverse to the inclusion (as from Proposition \ref{prop:zerodelta}). There is an isomorphism of relative homology groups:
		\[
		\HG_\bullet\left(C \cap \pi_\xi^{-1}[0,\delta],C \cap \pi_\xi^{-1}(\delta)\right) \simeq  \HG_\bullet\left(\phi^{-1}(p),\phi^{-1}(p)\cap\pi_\xi^{-1}(\delta)\right).
		\]
		Consequently, the associated Euler characteristics satisfy
		\[
		\chi\left(C \cap \pi_\xi^{-1}[0,\delta]\right) - \chi\left(C \cap \pi_\xi^{-1}(\delta)\right) = 1-\chi\left(\phi^{-1}(p) \cap \pi_\xi^{-1}(\delta)\right).
		\]
	\end{lemma}
	\begin{proof}
		The set $C \cap \pi_\xi^{-1}[0,\delta]$ decomposes as a disjoint union
		\[
		C \cap \pi_\xi^{-1}[0,\delta] = \coprod_q \phi^{-1}(q),
		\]
		where $q$ ranges over the points in $C \cap \pi_\xi^{-1}(0)$. By the additivity of homology, we have a direct sum decomposition
		\[
		\HG_\bullet\left(C \cap \pi_\xi^{-1}[0,\delta],C \cap \pi_\xi^{-1}(\delta)\right) = \bigoplus_q \HG_\bullet\left(\phi^{-1}(q),\phi^{-1}(q)\cap\pi_\xi^{-1}(\delta)\right).
		\]
		It therefore suffices to show that the summands corresponding to $q \neq p$ are all trivial. Since no such $q$ is a critical point of $\pi_\xi$, the vector field $V$ on $C \cap \pi_\xi^{-1}(0,\delta]$ which was used to construct $\phi$ in the proof of Proposition \ref{prop:zerodelta} extends non-trivially through $q$. Now by Thom's Isotopy Lemma \ref{lem:thom} above, the stratified homeomorphism type of $\phi^{-1}(q) \cap \pi_\xi^{-1}(t)$ remains unchanged across all $t \in [0,\delta]$, so in particular there is a homeomorphism of pairs
		\[
		\Big(\phi^{-1}(q), \phi^{-1}(q) \cap \pi_\xi^{-1}(\delta)\Big) \simeq \Big([0,\delta],\delta\Big),
		\] and hence the relative homology is trivial as desired. To extract the statement about the Euler characteristics from the statement about relative homology groups, one uses the observation that $\phi^{-1}(p)$ is homeomorphic to the cone at $p$ over $\phi^{-1}(p) \cap \pi_\xi^{-1}(\delta)$. Since all cones are contractible, we obtain $\chi(\phi^{-1}(p)) = 1$.
	\end{proof}

	To conclude our proof of Theorem \ref{thm:toplink}, we observe that
	\begin{align*}
	\e_pC &=  \#\set{C \cap \pi_\xi^{-1}(\delta)} - \#\set{C \cap \pi_\xi^{-1}(0)} + 1, & \text{by }(\ref{eq:lambda0}) \\
	&= \chi\left(C \cap \pi_\xi^{-1}(\delta)\right) - \chi\left(C \cap \pi_\xi^{-1}(0)\right) + 1 & \text{since $\dim C = 1$} \\
	&= \chi\left(C \cap \pi_\xi^{-1}(\delta)\right) - \chi\left(C \cap \pi_\xi^{-1}[0,\delta]\right) + 1 & \text{by Proposition } \ref{prop:zerodelta} \\
	&= \chi\left(\phi^{-1}(p) \cap \pi_\xi^{-1}(\delta)\right) & \text{by Lemma } \ref{lem:local} \\
	&= \chi(\Clk_pC) & \text{by Definition }\ref{def:clink}
	\end{align*}
	as desired.

	\begin{remark} \label{rem:topargcodim}
		Neither Proposition \ref{prop:zerodelta} nor Lemma \ref{lem:local} require any constraint on the dimension of $C$, and both would work just as well when $\dim C > 1$. On the other hand, it is only when $\dim C = 1$ that one obtains $\dim (C \cap \pi_\xi^{-1}(t)) = 0$ for $t$ in $[0,\delta]$, and it is a miracle of zero-dimensionality that degree and Euler characteristic coincide. This accident is exploited only once in our argument, i.e., when transitioning from the first line to the second one in the string of equalities above.
	\end{remark}

	Unfortunately, we do not anticipate any direct relationship between degrees and Euler characteristics of higher-dimensional projective  varieties. Thus, this argument does not extend directly to the scenario where our curve $C$ is replaced by a variety $Y$ of dimension $> 1$. In any event, Theorems \ref{thm:algmultslice} and \ref{thm:toplink} guarantee that all Hilbert-Samuel multiplicity computations can be reduced to Euler characteristic estimation for a finite collection of points. We now turn our attention to inferring such multiplicities from point samples.

	\section{Estimating Multiplicities from Finite Samples}
	\label{section:Multiplcity_From_Finite_Samples}
	The {\em reach} $\tau_M > 0$ of a compact submanifold $M \subset \RR^n$ is the smallest radius $r > 0$ for which the radius-$r$ normal bundle around $M$ self-intersects. This notion was first introduced by Federer in \cite{federer}, and it serves as an important measure of the regularity of the embedding $M \hookrightarrow \RR^n$. The reciprocal $1/\tau_M$, called the {\em condition number} of $M$, features prominently in the homological inference results of Niyogi, Smale and Weinberger from \cite{niyogietal}. These results have been extended to the case where $M$ has a smooth boundary $\partial M$ by Wang and Wang \cite{wang2}. In this setting, the role of the reach is played by a new parameter
	\begin{align}\label{eq:Dwang}
	\Delta_M := \min\set{\tau_M, \tau_{\partial M}, \rho_M},
	\end{align}
	where $\rho_M$ is the largest radius $r > 0$ so that at each point $x$ in $X$, the exponential map $M \to T_xM$ is a diffeomorphism onto its image when restricted to the open ball $\BB_r(x)^\circ \cap M$. The following result is \cite[Theorem 3.3]{wang2}.

	\begin{theorem} \label{thm:wangdense}
		Let $M$ be a smooth, nonempty $k$-dimensional submanifold with boundary of $\RR^n$. For any radius $r \in (0,\Delta_M/2)$ and probability $\gamma \in (0,1)$, there exists an explicit bound $N_M(r,\gamma)$ satisfying the following property. Any uniformly sampled\footnote{i.e., independent and identically distributed with respect to the uniform measure on $M$.} finite set $S \subset M$ of cardinality larger than $N_M(r,\gamma)$ is $(r/2)$-dense in $M$ with probability exceeding $(1-\gamma)$.
	\end{theorem}

	\noindent In other words, we can guarantee with high confidence that every point of $M$ is no more than $r/2$ away from some point of $S$ whenever $\#S > N_M(r,\gamma)$.

	\begin{remark} This bound $N_M(r,\delta)$ has the form
		\begin{align}\label{eq:N}
		N_M(r,\gamma) = \beta_M(r) \cdot \left[\beta_M\left(\frac{r}{2}\right) + \ln\left(\frac{1}{\gamma}\right)\right],
		\end{align}
		where $\beta_M:\RR \to \RR_{> 0}$ is the function
		\[
		\beta_M(x) := \frac{\text{Vol}(M)}{\frac{\cos^k(\theta)}{2^{(k+1)}}\cdot I_{y}\left(\frac{k+1}{2},\frac{1}{2}\right) \cdot \text{Vol}(\BB_x^k)}.
		\]
		Here $\text{Vol}(\bullet)$ is standard $k$-dimensional Lebesgue volume and the auxiliary variables are
		\[
		\theta := \arcsin\left(\frac{x}{4\Delta_M}\right) \quad \text{ and } \quad y := 1-\frac{x^2 \cdot \cos^2(\theta)}{16\Delta_M^2};
		\] moreover, $I_y(a,b)$ denotes the {\em regularized incomplete beta function}
		\begin{align*}
		I_y(a,b) := \frac{B_y(a,b)}{B_1(a,b)}, \quad \text{ with } \quad
		B_y(a,b) := \int_0^y t^{a-1}(1-t)^{b-1}~dt.
		\end{align*}
		And finally, $\BB^k_x$ is the ball of radius $x$ in $k$-dimensional Euclidean space.
	\end{remark}

	\subsection{Setup and Parameter Choices}

	Let $p$ be any (not necessarily singular) point on a curve $C \subset \PP^n$. Passing to an affine chart of $\pp^n$ containing $p$, we may as well work within $\CC^{n+1} \simeq \RR^{2n+2}$. In light of this identification, all dimensions of spaces mentioned henceforth are to be understood as dimensions over $\RR$ rather than $\CC$. Consider any choice of positive radii $\epsilon \gg \delta$ and (unit length) direction $\xi$ so that the complex link of $p$ in $C$ is given by the intersection
	\begin{align}\label{eq:embclk}
	\Clk_p = C \cap \BB_\epsilon(p) \cap \pi_\xi^{-1}(\delta).
	\end{align}
	By Theorem \ref{thm:toplink}, this is a set of (finite) cardinality $\ell := \e_pC$, so we may enumerate its points as $\set{x_1,x_2,\ldots,x_\ell}$. Let $\mu > 0$ be the smallest pairwise Euclidean distance between these points
	\begin{align}\label{eq:mu}
	\mu := \min\set{\|x_i-x_j\|, \text{ where } 1 \leq i \neq j \leq \ell},
	\end{align} and let $\kappa > 0$ be the distance between $\Clk_p$ and the boundary of the closed ball $\BB_\epsilon(p)$:
	\begin{align}\label{eq:kappa}
	\kappa := \min\set{\epsilon - \|p-x_i\|, \text{ where } 1 \leq i \leq \ell}.
	\end{align}
	These new distances $\kappa$ and $\mu$ are determined by the initial choices of $\epsilon, \delta$ and $\xi$. We also select a new parameter $\epsilon_0 \in (0,\delta)$, called the {\em inner radius}. Since $\epsilon_0$ is smaller than $\delta$, the open ball $\BB_{\epsilon_0}(p)^\circ$ does not intersect the offset hyperplane $\pi_\xi^{-1}(\delta)$, and hence does not contain any of the points $\set{x_1,\ldots,x_\ell}$.

	\begin{proposition}\label{prop:annmwb}
		The intersection $C' := C \cap \left[\BB_\epsilon(p) - \BB_{\epsilon_0}(p)^\circ\right]$ forms a two-dimensional manifold with boundary embedded within $C_\text{\rm reg} \subset \RR^{2n+2}$.
	\end{proposition}
	\begin{proof}
		We recall that the radius $\epsilon$ from \eqref{eq:embclk} satisfies the property that the boundary sphere $\partial\BB_e(p)$ is transverse to $C_\text{reg}$ for all $e \in (0,\epsilon]$. Thus, the smooth map ${\bf d}_p:\RR^{2n+2} \to \RR$ given by
		\[
		x \mapsto \|x-p\|^2
		\] has no critical points on $C_\text{reg}$ valued in $(0,\epsilon]$. By Lemma \ref{lem:thom}, the restriction ${\bf d}_p:C' \to [\epsilon_0,\epsilon]$ forms a trivial fiber bundle. Since all such $e$ are regular values of ${\bf d}$, the implicit function theorem guarantees that each fiber $F_e := C_\text{reg} \cap {\bf d}_p^{-1}(e)$ is a smooth one-dimensional submanifold of $C'$.  Therefore, the desired result now follows from the fact that $C'$ is diffeomorphic to the product $F \times [\epsilon_0,\epsilon]$ where $F$ is a smooth one-dimensional manifold.
	\end{proof}

	Recalling the fact that $\set{x_1,\ldots,x_\ell} \cap \BB_{\epsilon_0}(p)$ is empty since $\epsilon_0 < \delta$, we have
	\[
	C' \cap \pi_\xi^{-1}(\delta) = \set{x_1,\ldots,x_\ell}.
	\]
	An immediate side-effect of replacing $C'$ by a dense uniform sample is that none of the sample points will lie exactly on $\pi_\xi^{-1}(\delta)$. Therefore, we require a final {\em thickness} parameter $\alpha > 0$. This is a sufficiently small positive radius for which the following property holds: the set of all points in $C$ that lie within distance $\alpha$ of the offset hyperplane $\pi_\xi^{-1}(\delta)$ is entirely contained within the union of open radius-$\alpha$ balls around points of the complex link:
	\begin{align}\label{eq:alphasep}
	\set{y \in C \mid \text{dist}[y,\pi_\xi^{-1}(\delta)] < \alpha} \subset \bigcup_{i=1}^\ell \BB_\alpha(x_i).
	\end{align}
	By sufficiently small here we mean that $\alpha$ must be bounded above by the minimum
	\begin{align}\label{eq:alphabound}
	\alpha < \min\set{(\epsilon-\delta), (\delta-\epsilon_0),{\mu}/{4},\kappa,{\Delta_{C'}}/{2}}.
	\end{align}
	This inequality encodes all of the geometric constraints required for our inference result. As atonement for introducing this deluge of parameters, we remind the reader that $\epsilon$ and $\delta$ were fixed in \eqref{eq:embclk}, while $\epsilon_0$ is the inner radius used in Proposition \ref{prop:annmwb}; the quantities $\mu$ and $\kappa$ are described in \eqref{eq:mu} and \eqref{eq:kappa} respectively, and $\Delta_{C'}$ is from \eqref{eq:Dwang}. Below we have illustrated the typical local picture of $C$ near $p$ in the case where $\e_pC = 3$.

	\begin{center}
		\includegraphics[scale=.4]{annulus.png}
	\end{center}

	\noindent The first four terms within the minimum on the right side of \eqref{eq:alphabound} are designed simply to ensure that balls of radius $\alpha$ around each of the $x_i$ are well-separated and fully contained within the annulus $[\BB_{\epsilon}(p) - \BB_{\epsilon_0}(p)^\circ]$; the final term is required for applying Theorem \ref{thm:wangdense}.

	\subsection{Inferring Multiplicities with High Confidence}

	Here we prove Theorem {({\bf C}) from the Introduction. The parameters encountered in its statement below were chosen in the previous subsection.
		\begin{theorem}\label{thm:mainfull}
			Let $S \subset \RR^{2n+2}$ be a finite set of points sampled uniformly from the intersection $C' = C \cap [\BB_\epsilon(p)-\BB_{\epsilon_0}(p)^\circ]$. For any $\gamma \in (0,1)$, if the cardinality $\# S$ exceeds the bound $N_{C'}(\alpha,\gamma)$ from \eqref{eq:N}, then the following holds with probability exceeding $(1-\gamma)$: the set
			\[
			S' := \set{y \in S \mid \text{\em dist}[y,\pi_\xi^{-1}(\delta)] < \alpha}
			\] consists of exactly $\ell$ nonempty point-clusters, each of diameter at most $2\alpha$, with the distance between distinct clusters exceeding $\mu - 2\alpha$.
		\end{theorem}
		\begin{proof}
			The intersection $C'$ is an embedded two-dimensional submanifold with boundary of $\RR^{2n+2}$ by Proposition \ref{prop:annmwb}, and $\alpha < \Delta_{C'}/2$ holds by \eqref{eq:alphabound}. Thus, we may safely apply Theorem \ref{thm:wangdense} to conclude that the set $S$ is $\alpha$-dense in $C'$ with probability exceeding $(1-\gamma)$. We will assume throughout the remainder of the argument that this density holds.

			Since the inner radius $\epsilon_0$ is smaller than $\delta$, all points of the complex link $\set{x_1,\ldots,x_\ell}$ lie in $C'$. Let $B_i$ denote the open ball of radius $\alpha$ around each $x_i$. The inequalities which involve $\epsilon, \epsilon_0, \delta$ and $\kappa$ in  $\eqref{eq:alphabound}$ guarantee that  each $B_i$ is entirely contained within the annulus $[\BB_\epsilon(p)-\BB_{\epsilon_0}(p)]$. It follows from the $\alpha$-density of $S$ in $C'$ that the intersections $S_i := B_i \cap S$ are all nonempty; and moreover, the diameter of each $S_i$ is no larger than the diameter $2\alpha$ of $B_i$. We claim that these $S_i$ form the $\ell$ desired point clusters of $S'$.

			To establish the claim, note from \eqref{eq:alphasep} that every point of $S'$ must lie in one of the $B_i$, whence $S' = \bigcup_{i=1}^\ell S_i$. Thus, it remains to show that the $S_i$ are separated from each other by a distance larger than $\mu-2\alpha$. To this end, note from \eqref{eq:mu} that the points $x_i$ and $x_j$ are separated by distance at least $\mu$ whenever $i \neq j$. Thus, points in distinct $B_i$ and $B_j$ are at least $\mu-2\alpha$ apart from each other, as desired.
		\end{proof}

		We know from \eqref{eq:alphabound} that $2\alpha$ is smaller than $\mu-2\alpha$, so the desired number $\ell = \e_pC$ can be determined with high confidence by clustering together points of $S'$ which lie within $2\alpha$ of each other and then counting the clusters.

		\bibliographystyle{abbrv}
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	\end{document}


TITLE: In the set $Z_n=\left\{0, 1, 2, ..., n-1\right\}$ for $n\geq 1$ why is it $n -1$ for the last term, also why are $1$ and $-1 = n -1$?
QUESTION [0 upvotes]: In the set $Z_n$ = {$0, 1, 2, ..., n-1$}  for  $n$ $\geq$ $1$ why is it $n -1$ for the last term, also why are $1$ and $-1$ = $n -1$?
The full example question is as follows: In the set $Z_n$ = {0, 1, 2, ..., n-1}  for  n $\geq$ 1 is a cyclic group under addition modulo n. $1$ and $-1$ = $n - 1$ are generators.
Similarly, in the statement: If a has finite order say, n, then  = { e, a, $a^2$, ..., $a^{n-1}$} and $a^i$ = $a^j$ if and only if n divides i - j. Why is it n -1 for the last power.
Possibly because we need to subtract e from the set?

REPLY [0 votes]: Really we have $\mathbb{Z}_n = \{[0], [1],[2],\dots,[n-1]\}$ where for any $k \in \mathbb{Z}$ we let $[k]$ denote the class of $k$ in $\mathbb{Z}_n$. These classes are defined so that $[k_1] = [k_2]$ in $\mathbb{Z}_n$ if and only if $n$ divides $k_1 - k_2$. This is why $[-1] = [n-1]$ since $(n-1) - (-1) = n$ and $n$ divides $n$. Often people will just write $\{0,1,2,\dots,n-1\}$ for convenience and it is understood they actually represent classes. Also another notation that is sometimes used is $\bar{k}$ for the class of $k$ modulo $n$.
Thus we can write any of the following
$$\mathbb{Z}_4 = \{[0],[1],[2],[3]\}$$
$$\mathbb{Z}_4 = \{[1],[2],[3],[4]\}$$
$$\mathbb{Z}_4 = \{[-1],[0],[2],[3]\}$$
or infinitely many other choices of representatives of our classes. The first is the most common way, but the others are correct too. Observe this works since $[0] = [4]$ and $[-1] = [3]$ in our example. So, in each of the examples above all 4 classes are present. Note there are exactly 4 classes in $\mathbb{Z}_4$ and will be exactly $n$ classes in $\mathbb{Z}_n$.


TITLE: Does this function define first-order ordinary differential equations?
QUESTION [3 upvotes]: I've read 3 different books but, this condition seems to fit them all:

It's homogeneous if:
  $y'=f(tx,ty)=f(x,y)$

Is this correct?

REPLY [1 votes]: This is one sense in which a differential equation can be homogeneous, but be cautioned that this is not nearly the more common sense, which is very different!

The usual sense of homogeneous in the context of differential equations is the following: A linear ordinary differential equation is one of the form
$$\frac{d^m}{dx^m} y(x) + A_{m - 1}(x) \frac{d^{m - 1}}{dx^{m - 1}} + \cdots + A_1(x) \frac{d}{dx} y(x) + A_0(x) y = F(x).$$
It is furthermore homogeneous iff $F(x)$ (called the source term) is $0$, or equivalently, iff $y(x) = 0$ is a solution. A key feature of homogeneous linear o.d.e.s is that their solutions comprise a vector space under the usual addition and scalar multiplication of functions.
We see immediately that an o.d.e. $y' = f(x, y)$ is homogeneous if $f(x, y)$ is linear in $y$, that is, if $\frac{\partial^2 f}{\partial y^2} = 0$.

Wikipedia describes a second notion of homogeneous first-order equations that I hadn't encountered before. It's closer to the notion you describe but still distinct. A first-order differential equation of the form $$P(x, y) \,dx + Q(x, y) \,dy = 0$$ is of homogeneous type if $P, Q$ are homogeneous functions of the same degree: A function $R(x, y)$ is homogeneous of degree $m$ iff $R(\lambda x, \lambda y) = \lambda^m R(x, y)$. Rearranging, we can rewrite such an equation (ignoring the places where $Q$ takes on the value zero) as $$\frac{d}{dx} y(x) = -\frac{P(x, y)}{Q(x, y)},$$ and the right-hand side satisfies $$-\frac{P(\lambda x, \lambda y)}{Q(\lambda x, \lambda y)} = -\frac{\lambda^m P(x, y)}{\lambda^m Q(x, y)} = -\frac{P(x, y)}{Q(x, y)},$$ so any homogeneous first-order differential equation of homogeneous type can be put in the indicated form.
Conversely, we can rewrite the given equation $y' = f(x, y)$ as $$-f(x, y) \,dx + dy = 0,$$ and both of the coefficients are homogeneous of degree zero, hence so view the given equation as one of homogeneous type, and hence the notion in the question agrees with the notion of homogeneous type, at least up the zero set of $Q$.


TITLE: Is there a machinery describing all the irreducible representations ?
QUESTION [11 upvotes]: Suppose we have a finite dimensional Lie algebra $g$， Is there a machinery to describe all the irreducible representation of $g$.
Consider toy example: $sl_{2}$ or $sl_{3}$, how do we describe all the irreducible representations of them.
Further, consider quantum case, Is there a machinery way(like algorithm)describing all the irreducible representations of $U_{q}(sl_{2})$
EDIT: What I am looking for is an "mechanical" and canonical machinery describing all the irreducible representations(of course, not only finite dimensional representations,not only unitary representations)
EDIT2: What I am looking for is some reference to describe them in explicitly(such as $sl_{3}$）

REPLY [29 votes]: The problem of classifying irreducible $sl_2(\mathbb C)$-representations is essentially
untractable as it contains a wild subproblem. Indeed, the action of the Casimir
element $C$ on any irreducible representation is by a complex scalar (by a
theorem of Quillen I believe). If we consider the case when $C$ acts by zero, by
a result of Beilinson-Bernstein the category of $sl_2$-representations with
$C=0$ is equivalent to the category of quasi-coherent $\mathcal D_{\mathbf
  P^1}$-modules. In this $1$-dimensional case every irreducible $\mathcal
D_{\mathbf P^1}$-module is holonomic. If we restrict ourselves to irreducible
regular holonomic modules we have two possibilities. One case is that they are
supported at a single point and then the point is a complete invariant. In the
other case they are classified by a finite collection of points of $\mathbf P^1$
and equivalence classes of irreducible representation of the fundamental group
of the complement of the points which map the monodromy elements of the points
non-trivially. In particular we can consider the case of three points in which
case the fundamental group is free on two generators (they and the inverse of
their product being the three monodromy elements). The irreducible
representations where one of the monodromy elements act trivially correspond to
removing the corresponding point and thinking of the representation as a
reprentation of the fundamental group of that complement.
Hence, we can embed the category of finite-dimensional representations of the
free group on two elements as a full subcategory closed under kernels and
cokernels of the category of $sl_2(\mathbb C)$-modules. This makes the latter
category wild in the technical sense. However, the irreducible representations
of the free group on two letters are also more or less unclassifiable.
There is no contradiction between this and the result of Block. His result gives
essentially a classification of irreducibles in terms of equivalence classes of
irreducible polynomials in a twisted polynomial ring over $\mathbb C$. So the
consequence is that such polynomials are essentially unclassifiable.
[Added] Intractable depends on your point of view. As an algebraic geometer I agree with Mumford
making (lighthearted) fun of representation theorists that think that wild problems are intractable. After
all we have a perfectly sensible moduli space (in the case of irreducible representations) or moduli stack (in the general case). One should not try to "understand" the points of an algebraic variety but instead try to understand the variety geometrically. Today, I think that this view point has been absorbed to a large degree by representation theory.


text:
\section{Zorn's Lemma Implies Well Ordering Theorem}
Tags: Axiom of Choice

\begin{theorem}
[[Zorn's Lemma]] implies the [[Well-Ordering Theorem]].
\end{theorem}

\begin{proof}
Let $X$ be a set.
If $X = \varnothing$ the theorem holds vacuously, so assume $X$ is not empty.
Let $\mathcal W \in \mathcal P(X)$ be the set of all [[Definition:Well-Orderable Set|well-ordered subsets]] of $X$.
Then [[Subset Relation on Power Set is Partial Ordering|$\mathcal W$ is partially ordered by $\subseteq$]].
Furthermore, there is an upper bound for every chain in $\mathcal W$: the union of all chains in $\mathcal W$.
Thus the hypotheses of Zorn's Lemma hold and we can conclude that $\mathcal W$ has a maximal element.
[[Proof by Contradiction|Seeking a contradiction]] suppose that $\mathcal W \subsetneqq X$.
Let $(E,\subseteq)$ be the maximal well ordering on $X$, as defined above, and assume $x_0 \in X \setminus E$.
Define an ordering $\preceq'$ on $X \cup \{ x_0 \}$ as follows:
:$ \preceq' := \begin{cases} x \preceq y & x,y \in E \\ x_0 \preceq x & x \in E,x_0 \in X \setminus E \end{cases}$
Then $\preceq'$ properly includes $\preceq$, contradicting the maximality of $\preceq$.
Thus the well ordering on $\mathcal W$ is a well ordering on all of $X$.
{{qed}}
\end{proof}

config: wiki


TITLE: Question about limits and logarithms of functions
QUESTION [0 upvotes]: Let $f$ be a smooth function of one-variable.  Suppose that the limit $\lim_{T\rightarrow \infty}\frac{1}{T}\log f(T)$ exists. Now I am trying to see wether or not we will have that if we do a time shift this won't make a difference i.e. if $\lim_{T\rightarrow \infty}\frac{1}{T}\log f(T+c)= \lim_{T\rightarrow \infty}\frac{1}{T}\log f(T)$, where $c$ is a constant.
Intuitively I would think this is true , but I am not being able to prove it.
All that I was able to see is that if $c>0$ then $\lim_{T\rightarrow \infty }\frac{1}{T}\log(f(T+c))=\lim_{T\rightarrow \infty}\frac{1}{T-c}\log(f(t))\geq \lim _{T\rightarrow \infty}\frac{1}{T}\log(f(t))$
Any help is appreciated. Thanks in advance.

REPLY [0 votes]: Let
$$
\phi(T)=\frac{\log f(T)}{T-C} \, .
$$
and define
$$
\phi(\infty)=\lim_{T \to \infty}\phi(T) \, .
$$
Then, $\phi$ is continuous at $\infty$, and so for all $C\in\Bbb{R}$,
$$
\lim_{T\to\infty}\phi(T+C)=\phi\left(\lim_{T\to\infty}T+C\right)=\phi(\infty)=\lim_{T\to\infty}\phi(T) \, .
$$
See this thread for more details.


TITLE: Can a photon survive a collision? If so, is it at rest during the process?
QUESTION [0 upvotes]: Background
Irving Kaplan, in Article 6.7: The Compton Effect of Nuclear Physics (2nd Ed.) explains the Compton effect as follows:

Compton (1923) was able to show
that when a beam of monochromatic x-rays was scattered by a light element
such as carbon, the scattered radiation consisted of two components, one
of the same wavelength as that of the incident beam, the second of slightly
longer wavelength.

The explanation for the observation of unshifted line is that

If the collision is such that the electron is not
detached from the atom, but remains bound, the rest mass $m_0$ of the
electron  must be replaced by the mass of the atom, which is
several thousand times greater. The calculated value of $\Delta \lambda$ then becomes much too small to be detected. A photon which collides with
a bound electron, therefore, does not have its wavelength changed, and
this accounts for the presence of the unshifted spectral line.

Now consider a carbon atom (vibrating in a bulk) moving towards X-ray photon (having frequency $\nu \neq(\delta E_n)/h$, where $\delta E_n$ refers to any possible excitation of electron) and photon is moving towards the carbon atom. photon strikes the electron of carbon atom. Their velocities are such that same energy is given back to photon which is given to carbon atom when photon collides and is thrown back at 180 degrees. The photon has same energy as before hence same frequency. (Everything is taken in a particular laboratory frame of reference.) The energy  of atom and photon are not changed only their direction of motion are changed.

$%Note the visible separation between the background and the questions$
With this in mind, then

Is the  photon ejected after collision the same photon which was thrown in? If not, what happened to the previous photon? What is going on with the energy carried by the previous photon? Since energy cannot be absorbed, where does the energy go?

When this photon collides (perpendicularly), is it at rest at the very moment of collision?

REPLY [0 votes]: Figured it would be interesting to look at some X-ray scattering cross-sections for carbon.  A wide variety of scattering processes are involved.  I suggest the OP have a look at the different scattering mechanisms involved.
Fig. 3-1.    "Total photon cross section $\sigma_{\hbox{tot}}$ in carbon, as a function of energy, showing the contributions of different processes: $\tau$, atomic photo-effect (electron ejection, photon absorption); ,$\sigma_{\hbox{coh}}$ coherent scattering (Rayleigh scattering—atom neither ionized nor excited); $\sigma_{\hbox{incho}}$ incoherent scattering (Compton scattering off an electron);$\kappa_n$ pair production, nuclear field;$\kappa_n$ pair production, electron field; $\sigma_{\hbox{ph}}$ photonuclear absorption (nuclear absorption, usually followed by emission of a neutron or other particle)"  From this Lawrence Berkeley document.


TITLE: Why is the trace of this object non-negative?
QUESTION [2 upvotes]: $\phi$ is an arbitrary superposition of powers of a single hermitian matrix, $M^k$:
$$\phi =\sum_{k} \alpha_{k} M^{k}$$
Why is the following statement true?
$$\langle \mathrm{tr}\left(\phi^\dagger \phi \right)\rangle \geq 0$$
I have recently asked a related question, Why is the trace of a hermitian matrix raised to an even power greater than or equal to 0?. I understand that the trace over any $M^{2k}$ would be non-negative. But here the product $\phi^\dagger \phi$ will in general have odd powers, and so there could be negative terms in the trace, according to my understanding.
I am coming across this statement in Eq. (3.9) of https://arxiv.org/abs/2002.08387

REPLY [1 votes]: Actually if $M$ is any matrix we have $\text{tr(M*M)} \geq 0$. This is clear if $M$ is diagonalizable and diagonalizable matrices are dense in all matrices.


TITLE: Validate the proof that $x_n = {\ln n \over n}$ is bounded by $\ln2$ and find $\sup\{x_n\}$
QUESTION [0 upvotes]: Let $n\in \mathbb N$ and:
  $$
x_n = \left\{ \ln n \over n \right\}
$$
  Prove $x_n$ is bounded by $\ln 2$. Find $\sup\{x_n\}$

Consider nominator and denominator, $\ln n$ is growing slower than $n$ therefore:
$$
{\ln n \over n} <1
$$
Suppose:
$$
{\ln n \over n} < \ln2 \iff \ln n < n\ln2 \iff \ln n < \ln{2^n} \iff n < 2^n
$$
So $n < 2^n$ which is true for any $n \in \mathbb N$
It it valid? And also for the second part how can I find supremum $x_n$ without using calculus?

REPLY [3 votes]: if $n\in \mathbb N$
for some $N>0, n>N \implies \frac {\ln n}{n}> \frac {\ln (n+1)}{n+1}$
Or $n^{n+1} > (n+1)^{n}$
It is not true when $n=2$
as $2^3 = 8 < 9 = 3^2$
But it is true for $n > 2$
sup $x_n = \frac {\ln 3}{3}$


TITLE: Is a chain map inducing zero on all homology groups necessarily a boundary?
QUESTION [1 upvotes]: Given the category $\mathbf{Chain}_A$ of chain complexes over an abelian group $A$.
The $n$-th Homology functor is:
$$H_n:\mathbf{Chain}_A\to\mathbf{Ab}$$
Clearly it is additive:
$$H_n(\varphi+\varphi')=H_n(\varphi)+H_n(\varphi')$$
There are special chain maps:
$$\Delta_n:C_n\to D_{n+1}:\quad\vartheta_n:=\partial^D_{n+1}\Delta_n+\Delta_{n-1}\partial^C_n$$
on which the functor vanishes:
$$H_n(\vartheta)=0\quad(\forall n)$$
Does the converse hold? In other words, if a chain map induces zero on all homology groups, is it necessarily of the form $\vartheta$ for some sequence $\Delta_n$?

REPLY [3 votes]: So rephrased in simpler words, you are asking if a chain map which induces zero on homology groups is necessarily a boundary. The answer is no.
Consider the two chain complexes $C_*$ and $D_*$ given respectively by

$C_1 = C_0 = \mathbb{Z}$, $C_i = 0$ for $i \neq 0,1$ and $d_1 : C_1 \to C_0$ is given by $d_1(x) = 2x$;
$D_1 = \mathbb{Z}$, $D_i = 0$ if $i \neq 1$, all the differentials are zero.

Let also $\varphi : C_* \to D_*$ be given by $\varphi_1(x) = x$. In other words we are looking at:
$$\require{AMScd}
\begin{CD}
0 @>>> \mathbb{Z} @>{d_1 = 2 \cdot}>> \mathbb{Z} @>>> 0 \\
@. @V{\varphi_1 = \operatorname{id}}VV @VVV \\
0 @>>> \mathbb{Z} @>>> 0 @>>> 0
\end{CD}$$
Then clearly $H_0(C) = \mathbb{Z}/2\mathbb{Z}$, $H_1(C) = 0$, whereas $H_0(D) = 0$ and $H_1(D) = \mathbb{Z}$; besides $H_i(C) = H_i(D) = 0$ for $i \neq 0,1$. So of course have $H_i(\varphi) = 0$ for all $i$.
However, $\varphi$ is not a boundary. Otherwise there should exists a map $h : C_0 = \mathbb{Z} \to D_1 = \mathbb{Z}$ such that
$$x = \varphi(x) = h(dx) = h(2x) = 2 h(x)$$
 for all $x$. In particular, $\varphi(1) = 1 = 2 h(1)$, which is impossible.


TITLE: Dimension of subsets of nonsigular variety
QUESTION [1 upvotes]: Could  you help me to prove this question? Do you have any idea?
Let $X$ be a nonsingular  variety and $Y \subseteq X$ is  close and nonsingular . Then for any $x \in Y$   which $\dim Y_x=\dim X_x -n$ there is an affine open neighborhood $ U$, $ x\in U\subseteq X$ such that $Y\cap U=V(f_1,\ldots,f_n)$ where   $ (f_1,\ldots,f_n) $ is a regular sequence of $\mathcal{O}(U)  $.

REPLY [1 votes]: Because $X$ and $Y$ are non-sigular, the conormal bundle to $Y$ in $X$
is a locally free sheaf on $Y$ of rank $n$.  Choose an affine n.h. $U = $
Spec $A$ of $x$ in $X$ such that over the intersection $U \cap Y$ the conormal
bundle becomes free.  If we let $I$ be the ideal in $A$ cutting out $Y$
(so $U \cap Y =$ Spec $A/I$), then the conormal bundle is the sheaf associated to the module $I/I^2$.   Thus $I/I^2$ is free of rank $n$ over $A/I$,
and so is generated by $n$ elements.
A Nakayama-type argument will now show that $I$ is generated by $n$ elements,
say $f_1,\ldots,f_n$.  The fact that these form a regular sequence is related
to the fact that they give a free basis for $I/I^2$.


TITLE: Inference from Wronskian determinant to 2nd order linear homogeneous ordinary differential equation.
QUESTION [0 upvotes]: I know that the equation of a homogeneous linear ODE2 is
$$y''+p(x)y'+q(x)y=0$$
I know that the Wronskian of two solutions is a determinant of a matrix.
$$W=\det\pmatrix{y_1 & y_2 \\ y_1' & y_2'} .$$
What is the logic leading to the statement?
$$y''+ \frac1{x} y'- {4\over x^2} y = 0$$
This is discussed in Trench: Elementary Differential Equations, section 5.1.
Thank you. MM

REPLY [1 votes]: Your last ODE is singular in $x=0$. Any statement about solutions is only valid over the interval $(0,\infty)$.
As the ODE is of Euler-Cauchy type, it is easy to identify the basis solutions $x^2,x^{-2}$. Their Wronskian is
$$
W=x^2⋅(-2x^{-3})-2x\cdot x^{-2}=-4x^{-1}.
$$
This, as expected, is a non-zero function over $(0,\infty)$.


TITLE: Calculate aggregate value.
QUESTION [2 upvotes]: I have six scores as:
Score 1 - Large value is better, i.e., 63 is better than 60.
Score 2 - Small value is better, i.e. 5 is better than 6.7.
Score 3 - Small value is better, i.e. 5 is better than 6.7.
Score 4 - Small value is better, i.e. 5 is better than 6.7.
Score 5 - Small value is better, i.e. 5 is better than 6.7.
Score 6 - Small value is better, i.e. 5 is better than 6.7.
Now I want to calculate an aggregate score so that i will be able to compare the score of two persons P1 and P2 and decide which one is better.

REPLY [2 votes]: Just convert all the scores so that higher is better. Say each score x is out of some number y. Thus we can represent a score as $x/y$. Define a function $f$ which maps $x/y$ to $y - (x/y)$. Apply this function to all the scores in which lower is better. You can then either convert all scores to a like denominator, or compute their decimals to easily see which scores are higher.
Let us look at some examples. Say we want to compare 3/5 (higher better) to 1/20 (lower is better). Our function maps 1/20 to 20-1/20 = 19/20. Thus both our scores are now changed to higher is better, and are (3/5,19/20). Converting to like denominator, we have our scores are (12/20,19/20). Clearly 19/20 is greater.
If multiple scores belong to one person, after applying the transformation (ie the function) given above, just compute the average of that person's transformed scores to get an aggregate score.


TITLE: If we have that $X < \infty$, why doesn't it imply $X$ is bounded?
QUESTION [0 upvotes]: Suppose that $X$ is some sequence or random variable, etc. I am trying to understand why if $X < \infty$, then that doesn't necessarily mean it has a bound, i.e., there doesn't exist some value $M$ whereby $X \leq M$ for all possible realizations of $X$. Does anyone have an intuition why? Thanks

REPLY [2 votes]: Assuming it is not a special ad-hoc notaion, but that it means "$X(\omega)<\infty$ for all $\omega$":
For instance $X:\Bbb R\to\Bbb R$, $X(\omega)=\omega$.

REPLY [2 votes]: Say $X$ successively takes the values $1,2,3, ...$. Then $X$ is unbounded, even though each value it takes is finite.


TITLE: Möbius transformation that maps the unit circle to itself
QUESTION [3 upvotes]: I need to find necessary and sufficient conditions on the coefficients of a Möbius transform $T(z)=\frac{\tilde a z+\tilde b}{\tilde c z+ \tilde d}$ so that it maps the unit circle $\{z: |z|=1\}$ into itself.
I initially thought that, since any Möbius transformation can be written as a finite composition of simple transformations (translations ($z+a$), rotations ($e^{i\theta}z$), dilations ($az$) and inversions ($\frac1z$)) and since we do not want to dilate or move the unit circle, we then can write the required transformation as $T(z)=e^{i\alpha}z$ or $T(z)=\frac{e^{i\alpha}}{z}$ for some $\alpha\in (-\pi,\pi]$. However, this does not look like the result I'm supposed to get. Where is my mistake?
I then read the exercise hint, which says I should first write a transformation $R$ that maps the unit circle to $\mathbb{R}_\infty$, and use transformations $S(z)$ that map $\mathbb{R}_\infty$ to $\mathbb{R}_\infty$, which I believe are of the form $S(z)=\frac{az+b}{cz+d}$ where $a,b,c,d\in\mathbb{R}$.
I then chose $R(z)=\frac{z+1}{z-1}i$ and tried composing $T=R^{-1}\circ S\circ R$ to find the answer. However, I'm getting an ugly expression that does not seem to be correct either: $T(z)=\frac{(A+Bi)z-\overline{(A-Bi)}}{(A-Bi)z-\overline{(A+Bi)}}$, where $A=b+ai$ and $B=d+ci$.
Could you help me see how to use the hint? I know there are other solutions for this problem on this site, but they solve it in different ways. Thank you!

REPLY [2 votes]: Your calculations are correct so far. $S(z)=\frac{az+b}{cz+d}$ with $a,b,c,d\in\mathbb{R}$, $ad-bc \ne 0$, is the general Möbius transformation mapping $\Bbb R \cup \{ \infty \}$ onto itself. (It maps the upper halfplane onto itself iff $ad-bc > 0$.)
Your expression for $T$ becomes simpler if we set
$$
C = A+iB = (b-c) + i(a+d) \, ,\\
D = \overline{A-iB} = (b+c) +i (a-d) \, .
$$
Then
$$
 |C|^2 - |D|^2 = 4(ad-bc) \ne 0 \implies |D| \ne |C|
$$
and
$$
 T(z) = \frac{Cz-D}{\overline D z - \overline C} \, .
$$
This is the general form of a Möbius transformation mapping the unit circle onto itself. (It maps the unit disk onto itself iff $|D| < |C|$).
So the conditions on $T(z)=\frac{\alpha z+\beta}{\gamma z+ \delta}$ are that
$$
 \overline \alpha = - \delta, \overline \beta = -\gamma, |\alpha| \ne |\beta| \, .
$$


TITLE: how to prove that $\big \lfloor \frac np\big \rfloor+\big \lfloor \frac n{p^2}\big \rfloor+\big \lfloor \frac n{p^3}\big \rfloor+\cdots≤\frac n{p-1}$
QUESTION [2 upvotes]: Prove:  $$\Big \lfloor \frac np\Big \rfloor+\Big \lfloor \frac n{p^2}\Big \rfloor+\Big \lfloor \frac n{p^3}\Big \rfloor+\cdots≤\frac n{p-1}$$
I tried to calculate both sides with floor function again.when we do that left side of the equation is still same while the right side $\frac n{p-1}$ becomes $\Big \lfloor \frac {n}{p-1}\Big \rfloor$ which is a integer.but i dont know what to do with it

REPLY [2 votes]: An alternative perspective on the proof which works only when $p$ is a prime number and $n$ a nonnegative integer can be seen by Legendre's formula for the power of $p$ dividing $n!$, written in terms of the $p$-adic absolute value as,
$$v_p(n!) = \sum_{k \ge 1} \left\lfloor \frac{n}{p^k}\right\rfloor$$
This has the alternate form in terms of the sum of digits of $n$ written in base $p$ function $s_p(n)$,
$$\sum_{k \ge 1} \left\lfloor \frac{n}{p^k}\right\rfloor = \frac{n-s_p(n)}{p-1}$$
Since for nonnegative integers $n$ we have $s_p(n)\ge 0$:
$$\sum_{k \ge 1} \left\lfloor \frac{n}{p^k}\right\rfloor \le \frac{n}{p-1}$$
This proof is certainly less efficient than the one given by Andreas, but has the nice side effect of giving us a simple upper bound on the power of $p$ dividing $n!$.


TITLE: Elementary matrices operations and simplified expression
QUESTION [1 upvotes]: I want to simplify the following expression between matrices. Let $\mathbf{A}$ a $n\times n$ symmetric PSD matrix, $\mathbf{I}$ the identity matrix of order $n$ and $\lambda>0$ a positive constant.
I want to simplify the following expression:
$$\mathbf{A} -\mathbf{A} \left[\mathbf{A} + \lambda \mathbf{I}\right]^{-1} \mathbf{A}$$
My attempt:
I use the SDV of $\mathbf{A}=\mathbf{PDP^T}$. Then as suggested in this answer,
$$\left[\mathbf{A} + \lambda \mathbf{I}\right]^{-1}=\mathbf{P} \left(\mathbf{D} +\lambda \mathbf{I}\right)^{-1}\mathbf{P^T}.$$
Therefore:
$$\mathbf{A} -\mathbf{A} \left[\mathbf{A} + \lambda \mathbf{I}\right]^{-1} \mathbf{A} =\\
= \mathbf{PDP^T} - \mathbf{PDP^T} \left[\mathbf{P} \left(\mathbf{D} +\lambda \mathbf{I}\right)^{-1} \mathbf{P^T}\right] \mathbf{PDP^T}=\\
=\mathbf{PD}\left[\mathbf{I} -\left( \mathbf{D} + \lambda \mathbf{I} \right)^{-1} \mathbf{D}\right] \mathbf{P^T}=\\
=\mathbf{PD}\left[\mathbf{D}^{-1} -\left( \mathbf{D} + \lambda \mathbf{I} \right)^{-1}\right] \left(\mathbf{PD}\right)^{T}.$$
My questions are:

Is it correct?
Can the result be simplified further in term of computational cost? I need to use it in an algorithm where the only thing that changes at each step is the value of $\lambda$.

Thanks a lot!

REPLY [0 votes]: I didn't read your proof, because it looks overly complicated. Anyway, by a change of orthonormal basis, you may assume that $A$ is diagonal and hence the problem reduces to the scalar case. Since $x-\frac{x^2}{x+\lambda}=\frac{\lambda x}{x+\lambda}$, the expression becomes $\lambda A(A+\lambda I)^{-1}$, which in your notations is equal to $P\left[\lambda D(D+\lambda I)^{-1}\right]P^T$.


TITLE: Jacobi Method and Gauss-Seidel Multiple Choice Convergence Answer Verification
QUESTION [1 upvotes]: The book gives the following problems.

A sufficient condition for the Jacobi method to converge for the linear system $Ax=b$ is ...
$a. \quad A-I$ is diagonally dominant
$b. \quad A$ is diagonally dominant
$c. \quad G$ is nonsingular where $G$ is the iteration matrix: $G=I-Q^{-1}A$.
$d.\quad $ The spectral radius of $G$ is less than $1$.
$e.\quad $ none of these.

My answer: $b.$ i.e. the condition that $A$ is diagonally dominant.
Reasoning: The book gives the Jacobi and Gauss-Seidel Convergence Theorem:

If $A$ is diagonally dominant, then the Jacobi and Gauss-Seidel methods converge for any starting vector $x^{(0)}$. (Notice that this is a sufficient, but not a necessary condition.)

The next problem:

A sufficient condition for the Gauss-Seidel Method to work on the linear system $Ax=b$ is ...
$a. \quad A$ is diagonally dominant
$b. \quad A-I$ is diagonally dominant
$c.\quad $ The spectral radius of $A$ is less than $1$.
$d. \quad G$ is nonsingular.
$e.\quad $ none of these.

My answer: $a.$ that $A$ is diagonally dominant.
Reasoning: The book gives the Jacobi and Gauss-Seidel Convergence Theorem, see above.
Concerns: I'm not sure if my answers are correct. It seems a bit too simple/straightforward.  I'm not sure if these are trick questions and I'm missing a key fact or idea.  Also the first question asks for a sufficient condition for Jacobi Method to converge, and the second question asks for a sufficient condition for Gauss-Seidel to work.
I'm not sure if there is a difference between converge and work.  Any help would be much appreciated.  Thank you.

REPLY [0 votes]: For these methods convergence occurs if and only if the spectral radius of the iteration/error propagation matrix $G$ is strictly less then one:
$$\rho(G) \overset{!}<1.$$
Now, for a general matrix-splitting based method like Jacobi or Gauss-Seidel, you do the splitting $$A = Q - N$$ resulting in the iteration matrix $$G = I - Q^{-1} A = Q^{-1}N.$$
For the Jacobi method $Q_J = D, N_J = L+U$. So for question one, a certainly correct answer is d. Concerning diagonal-dominance, it is usually a matter of definition. Jacobi converges guaranteed for strict or irreducible diag. dom. matrices $A$. Question is now, what your definition of diagonal-dominance is, i.e., whether b. is also correct or not.
This matter of definition determines actually also the outcome of the second question - it is either b. or e..


TITLE: Why are $p$-adic characters locally analytic?
QUESTION [4 upvotes]: By definition, a $p$-adic character is a continuous homomorphism $\chi:\mathbb{Z}_p^\times\rightarrow \mathbb{C}_p^\times$. In several books/papers, I have seen it stated that all $p$-adic characters are locally analytic, in the sense that there is a covering $a+p^n\mathbb{Z}_p$ ($a$ prime to $p$) of $\mathbb{Z}_p^\times$ such that $\chi$ is given by a power series $\sum_{n\geq 0} c_n(x-a)^n$ on the open set $a+p^n\mathbb{Z}_p$. I think this is probably easy and I am just overthinking it, but can someone elucidate this for me? In other words:

Why are $p$-adic characters locally analytic?

I think I can see why this is true for Dirichlet characters of $p$-power conductor, but not in general. A Dirichlet character of conductor $p^n$ for some $n$ can be identified (perhaps after fixing an embedding $\mathbb{C}\rightarrow \mathbb{C}_p$) with a $p$-adic character $\chi$ which is constant on $1+p^n\mathbb{Z}_p$. Hence such characters are locally constant, and therefore locally analytic, represented on $a+p^n\mathbb{Z}_p$ by the constant power series $\chi(a)$. I'm not sure how to proceed for a general character though...

REPLY [2 votes]: The topological group $\Bbb Z_p^\times$ is isomorphic to the product of $\Bbb Z_p$ with a finite group. In other words, it is locally isomorphic to the additive group $\Bbb Z_p$. The isomorphism is given by the functions $\log$ and $\exp$, which are locally analytic.
Therefore we may just focus on (additive) characters on $\Bbb Z_p$.
This is then implied by the following exercise:

If $z$ is an element of $\Bbb C_p$ such that $|z - 1| < 1$, then the function $\chi_z: \Bbb Z \rightarrow \Bbb C_p$ sending any $k\in \Bbb Z$ to $z^k$ extends continuously to a character $\chi_z:\Bbb Z_p \rightarrow \Bbb C_p^\times$, which is given by $$\chi_z(x) = \sum_{n \geq 0} \binom x n (z - 1)^n$$ for every $x \in \Bbb Z_p$.
Any continous character $\chi:\Bbb Z_p \rightarrow \Bbb C_p^\times$ is of the form $\chi_z$ for some $z$ as in 1.


TITLE: Proper maps and convergent subnets
QUESTION [2 upvotes]: Let $X$ and $Y$ be Hausdorff topological spaces and let $p: X \rightarrow Y$ be a proper (every compact set has compact preimage) continuous map.

Is the following true?  Given a net $(x_i)$ in $X$ such that $(p(x_i))$ converges, then $(x_i)$ has a convergent subnet.
If so, is there a standard reference?

REPLY [2 votes]: No, this is not true in general.  For instance, let $X$ be an infinite discrete space and let $F$ be a nonprincipal ultrafilter on $X$ (or more generally, a filter that contains the cofinite filter and is not contained in the cofinite filter on any infinite subset of $X$).  Let $Y=X\cup\{\infty\}$, where $X$ is an open subspace of $Y$ and neighborhoods of $\infty$ are sets of the form $\{\infty\}\cup A$ for $A\in F$.  Note that no infinite subset of $Y$ is compact, since if $S\cup\{\infty\}$ is compact then every element of $F$ must be cofinite in $S$.
It follows that the inclusion map $p:X\to Y$ is proper.  But we can choose a net $(x_i)$ in $X$ which converges to $\infty$ in $Y$, and this net has no convergent subnet in $X$.
However, it is true if $Y$ is locally compact.  I don't know of a reference off the top of my head, but here's a proof.  Let $y$ be the limit of $(p(x_i))$ and choose a compact neighborhood $K$ of $y$.  Then $p(x_i)$ is eventually in $K$, so $x_i$ is eventually in $p^{-1}(K)$.  Passing to a subnet, we may assume $x_i\in p^{-1}(K)$ for all $i$.  But $p^{-1}(K)$ is compact since $p$ is proper, so $(x_i)$ must have a convergent subnet.
Actually, it is true more generally as long as $p$ is a closed map (this is automatic if $Y$ is locally compact).  Indeed, suppose $(p(x_i))$ converges to a point $y$.  Note that an accumulation point of $(x_i)$ is just an element of the intersection $\bigcap_j \overline{A_j}$ where $A_j=\{x_i:i\geq j\}$ for each $j$ in the index set of the net.  If $p$ is closed, then $p(\overline{A_j})$ is closed for each $j$, and thus must contain $y$ since it contains $p(x_i)$ for all $i\geq j$.  Thus $B_j=\overline{A_j}\cap p^{-1}(\{y\})$ is nonempty for each $j$.  Each $B_j$ is closed in $p^{-1}(\{y\})$, and they have the finite intersection property since the index set is directed.  Since $p^{-1}(\{y\})$ is compact, $\bigcap B_j$ is nonempty.  Thus $\bigcap\overline{A_j}$ is nonempty, so $(x_i)$ has an accumulation point and hence a convergent subnet.


TITLE: radius of convergence of Taylor expansion of $e^x$
QUESTION [2 upvotes]: The Taylor expansion of $e^x$ at $x=0$ is $1+x+x^2/2!+x^3/3! + ...$
but I get that this expansion is true for all $x$ (not necessarily around $x=0$).
can someone prove this?

REPLY [5 votes]: HINT: Apply the ratio test.
$$\lim_{n\to +\infty} \frac{|a_{n+1}|}{|a_n|}$$
More Details
Applying the Taylor sequence $a_n = x^n/n!$ as requested above, we obtain
$$\lim_{n\to +\infty} \frac{\frac{n^{n+1}}{(n+1)!}}{\frac{x^n}{n!}}$$
Which means the limit of
$$\frac{x}{n+1}$$
Which goes to zero as $n\to +\infty$, hence, being less than $1$ it does converge and the radius of convergence is infinite.
This means the Taylor series holds $\forall x \in \mathbb{R}$
Radius of Convergence
$$R = \displaystyle \lim_{n \to \infty} \dfrac{|a_n|}{|a_{n+1}|}$$


text:
In general, we begin with a collection of $N$ points $\bm{x_k}$ with forces $\bm{f_k}$ in some domain $\Omega$. At each timestep, we compute the velocity of each point $\vec{x_i}$ as
\begin{equation}
\label{eq:totalvel}
\vec{u}(\vec{x_i}) = \bm{u^R}+\sum_{k=1}^N \vec{u}^\epsilon (\vec{x_i},\vec{x_k}),
\end{equation}
where $\bm{u^R}$ is given by Eq.\ \eqref{eq:addedvel} and $\vec{u}^\epsilon (\vec{x_i},\vec{x_k})$ is given by Eq.\ \eqref{eq:regveloc}. Because $\bm{u^R}$ is constant throughout the domain, the calculation in Eq.\ \eqref{eq:totalvel} is $\mathcal{O}(2N+(2N)^2)$ operations. The first $\mathcal{O}(2N)$ operations arise from the computation of the constant additional velocity in Eq.\ \eqref{eq:addedvel}. The second $\mathcal{O}((2N)^2)$ operations come from computing the regularized velocities in Eq.\ \eqref{eq:regveloc} for all of the points.

We compare this operation count to alternative formulations. Suppose that the large circle was discretized with $M$ points and the forces on the large circle solved for explicitly, as in \cite{Vanderlei2011}. This calculation is a $2M \times 2M$ linear solve and requires $\mathcal{O}((2M)^2)$ operations using GMRES or $\mathcal{O}((2M)^3)$ operations if done directly. In addition, the calculation of the added velocities at each point in the domain from the forces on the large circle requires another $\mathcal{O}(2NM)$ flops, and it is unclear how to choose the number and location of the $M$ points. Alternatively, the addition of more Stokeslets in a method similar to the method of images \cite{ainley2008method,blake1971note,cortez2015general} would require $\mathcal{O}(2SN)$ operations to compute the added velocity, where $S$ is the number of added Stokeslets. Our added velocity is computed in $2N$ flops, making it much more efficient than any of these alternatives.

\subsection{Choosing the radius}
Central to our method is the assumption that the velocity computed from Eq.\ \eqref{eq:standvel} on the large circle $\Gamma$ is relatively constant. The validity of this assumption dictates a lower bound on $R$. In order to test the variation of the velocity on the large circle $\Gamma$, we first impose a force of $\bm{f_0}=\begin{pmatrix} 1\\[1 pt] 0 \end{pmatrix}$ at the origin. Next, we use Eq.\ \eqref{eq:standvel} to compute and measure the velocity from the Stokeslet (i.e. the velocity without the addition of $\bm{u^R}$) on the large circle.

Specifically, we discretize the circle $\Gamma$ with $N=100$ points and quantify velocity variability by defining
\begin{equation}
\label{eq:velvar}
\sigma_u(R) = \max_i{\left|\frac{u_x^i - \bar{u}_x}{\bar{u}_x}\right|}.
\end{equation}
Here the index $i$ runs from 1 to $N=100$, $u_x^i$ refers to the velocity in the $x$ direction (the direction of the force) at point $i$, and $\bar{u}_x$ refers to the arithmetic mean of $u_x$ taken over $\Gamma$.

\begin{figure}
\centering
\includegraphics[width = 0.6\textwidth]{velocityvarR_Mean}
\caption{Variation of the velocity on the large circle $\Gamma$ for different values of $R$. Dotted lines indicate 10\% and 5\% variation, $\sigma_u(R)$.}
\label{fig:assumptest}
\end{figure}

Fig.\ \ref{fig:assumptest} shows the values of $\sigma_u(R)$ (as a percentage) for different values of $R$. We observe a variation in the velocity $<10\%$ for $R \geq 10^3$ and a variation $<5\%$ for $R \geq 10^5$.

With this in mind, we specify the lower limit on $R$, $R \geq 10^3$. We can find an upper limit on $R$ based on the Reynolds number. Let $v$ and $L$ be the relevant velocity and length scales. Our systems have values of viscosity $\mu$ and density $\rho$ of the same magnitude as water, so that Re$\displaystyle{=\frac{\rho v L}{\mu}=}$ $10^6vL$ is the relevant Reynolds number. Our model systems are from applications in cell biology, so the relevant velocity scale is in $\mu$m/s. Therefore we take $v=10^{-6}$ m/s and Re $=L$, where $L$ is the relevant length scale. In order for Stokes flow to be valid, we need (Re $\ll 1$), which we define to be Re $\leq 0.1$. Then the relevant length scale cannot exceed 0.1 m = $10^5$ $\mu$m. Because we have confined the domain and effectively introduced forces on the large circle, the radius of the large circle is now the largest relevant length scale, and we have determined an upper bound on $R$, $R \leq 10^5$ $\mu$m. We note that this upper bound may change depending on the characteristic time and length scales used to compute the Reynolds number, but it is straightforward to derive it as we have here. Thus we have determined in general that $10^3 \leq R \leq 10^5$. For the examples in Section \ref{sec:exs}, we choose $R=10^3$ $\mu$m to ensure the validity of Stokes equations for these systems. This value of $R$ results in a velocity variation from Eq.\ \eqref{eq:velvar} less than 10\%. In addition, we did not find any additional stiffness when using $R$ is this range (for larger $R$, e.g. $R=10^{16}$, the velocities in Eq.\ \eqref{eq:addedvel} would increase, thereby increasing the overall problem stiffness.

config: arxiv


TITLE: Infinite product $\Gamma(\tfrac14)=\mathrm{A}^3e^{-\mathrm{G}/\pi}2^{1/6}\sqrt{\pi}\prod_{k\ge1}\left(1-\frac1{2k}\right)^{(-1)^k k}$
QUESTION [1 upvotes]: I saw the following infinite product on Wikipedia:
$$\Gamma\left(\tfrac14\right)=\mathrm{A}^3e^{-\mathrm{G}/\pi}2^{1/6}\sqrt{\pi}\prod_{k\ge1}\left(1-\frac1{2k}\right)^{(-1)^k k}\tag{1}$$
where $\mathrm A$ is the Glaisher constant, and $\mathrm G$ is Catalan's constant. I am looking for a proof of this product.
I haven't gotten very far with this product, other than noting that if
$$\zeta_*(s)=\sum_{k\ge1}\frac1{(1-\frac1{2k})^{(-1)^k ks}}$$
then
$$\zeta_*'(0)=3\ln\mathrm A+\frac16\ln2+\frac12\ln\pi-\frac{\mathrm{G}}{\pi}-\ln\Gamma\left(\tfrac14\right),\tag2$$
of course, assuming that $(1)$ is true. Perhaps $(2)$ is easier to prove. Could I have some help? Thanks.

REPLY [3 votes]: A short proof.
$\displaystyle Q_0(x) :=\Gamma(x+1)=\lim_{n\to\infty}\frac{n^x}{\prod\limits_{k=1}^n\left(1+\frac{x}{k}\right)}~~ , ~~~~ Q_1(x) :=\lim_{n\to\infty}\frac{e^{xn}n^{-x^2/2}}{\prod\limits_{k=1}^n\left(1+\frac{x}{k}\right)^k}$
$\displaystyle \pi^{1/2} = Q_0\left(-\frac{1}{2}\right)~~ , ~~ A^3  = 2^{7/12}Q_1\left(-\frac{1}{2}\right)^2 ~~ , ~~ e^{G/\pi} = 2^{3/4}\left(\frac{ Q_0\left(-\frac{1}{2}\right) Q_1\left(-\frac{3}{4}\right) }{ Q_0\left(-\frac{3}{4}\right) Q_1\left(-\frac{1}{4}\right)  }\right)^2$
It follows:
$\displaystyle \prod\limits_{k=1}^{2n}\left(1-\frac{1}{2k}\right)^{(-1)^k k} =  \prod\limits_{k=1}^n \frac{\left(1-\frac{1}{4k}\right)^{2k}}{\left(1-\frac{1}{4k-2}\right)^{2k-1}} = \prod\limits_{k=1}^n \frac{\left(1-\frac{1}{2k}\right)^{2k-1} \left(1-\frac{1}{4k}\right)^{2k}}{ \left(1-\frac{3}{4k}\right)^{2k-1} } = $
$\displaystyle = \frac{ \prod\limits_{k=1}^n \left(1-\frac{3}{4k}\right) n^{-1/2} }{ n^{-3/4} \prod\limits_{k=1}^n \left(1-\frac{1}{2k}\right) } \left(\frac{ e^{-3n/4} n^{-9/32} \prod\limits_{k=1}^n \left(1-\frac{1}{2k}\right)^k \prod\limits_{k=1}^n \left(1-\frac{1}{4k}\right)^k }{\prod\limits_{k=1}^n \left(1-\frac{3}{4k}\right)^k  e^{-n/2} n^{-1/8} e^{-n/4} n^{-1/32} }\right)^2$
$\displaystyle \to  ~\frac{ Q_0\left(-\frac{1}{2}\right) }{ Q_0\left(-\frac{3}{4}\right) } \left(\frac{ Q_1\left(-\frac{3}{4}\right) }{ Q_1\left(-\frac{1}{2}\right) Q_1\left(-\frac{1}{4}\right) }\right)^2$
$\displaystyle = 2^{-1/6} \frac{ Q_0\left(-\frac{3}{4}\right) }{ Q_0\left(-\frac{1}{2}\right) } \cdot 2^{-7/12} Q_1\left(-\frac{1}{2}\right)^{-2} \cdot 2^{3/4} \left(\frac{ Q_0\left(-\frac{1}{2}\right) Q_1\left(-\frac{3}{4}\right) }{ Q_0\left(-\frac{3}{4}\right) Q_1\left(-\frac{1}{4}\right)}\right)^2$
$\displaystyle = \Gamma\left(\frac{1}{4}\right) 2^{-1/6} \pi^{-1/2} A^{-3} e^{G/\pi}$


text:
\begin{document}
\input{zcom}
\setlength{\arraycolsep}{0.05cm}
\newcommand{\rest}{{T_\text{rest}}}
\newcommand{\zetahat}{\hat{\zeta}}
\newcommand{\tDelta}{{\tilde{\Delta}}}
\newcommand{\tDeltae}{{\tilde{\Delta}_e}}
\newcommand{\tT}{{\tilde{T}}}
\newcommand{\add}{{\cal A}}
\newcommand{\rem}{{\cal R}}
\newtheorem{sigmodel}{Signal Model}

\newcommand{\thr}{{\text{thr}}}
\newcommand{\delthr}{{\text{del-thr}}}
\newcommand{\delbound}{{b}}
\newcommand{\err}{{\text{err}}}
\newcommand{\Q}{{\cal Q}}

\newcommand{\dett}{{\text{add}}}
\newcommand{\CSres}{{\text{CSres}}}
\newcommand{\diff}{{\text{diff}}}

\newcommand{\Section}[1]{ \vspace{-0.13in}  \section{#1} \vspace{-0.12in} }
\newcommand{\Subsection}[1]{  \vspace{-0.12in} \subsection{#1}  \vspace{-0.08in} }
\newcommand{\Subsubsection}[1]{   \subsubsection{#1} }

\date{}
\maketitle

\newcommand{\Aset}{{\cal A}}
\newcommand{\Rset}{{\cal R}}
\newcommand{\Iset}{{\cal I}}
\newcommand{\Dset}{{\cal D}}
\newcommand{\Sset}{{\cal S}}
\newcommand{\Inc}{\text{Inc}}
\newcommand{\Dec}{\text{Dec}}
\newcommand{\Con}{\text{Con}}
\newcommand{\sm}{e}

\input{introduction_long}
\input{Allerton_stability_every_t_main}
\input{addLSdel_lscs}
\input{simulations_long}

\appendix
\input{signalmodel_generative}
\input{proof_simple_modcs}
\input{proof_addLSdel_modcs_d0_2}
\input{proof_addLSdel_lscs}

\bibliographystyle{IEEEbib}
\bibliography{tipnewpfmt_kfcsfullpap}

\end{document}


TITLE: Does this result for two vectors generalize to larger sets?
QUESTION [1 upvotes]: Suppose we have linearly independent two-dimensional column vectors $\mathbf{x}_1=\left[{a}\atop{b}\right],\mathbf{x}_2=\left[{c}\atop{d}\right]$. Now let the "weight" of vector $\mathbf{x}_1$ be $p_1\in(0,1)$, and the "weight" of vector $\mathbf{x}_2$ be $p_2=1-p_1$.  One can show (using straightforward but tedious arithmetic) that the eigenvalues of $\mathbf{A}=p_1\mathbf{x}_1\mathbf{x}_1^T+p_2\mathbf{x}_2\mathbf{x}_2^T$ are the same as the eigenvalues of a $2\times2$ matrix $\mathbf{B}$ whose $(i,j)$-th element $B_{i,j}=\sqrt{p_ip_j}\mathbf{x}_i^T\mathbf{x}_j$.  In fact, the restriction for weights summing to one is unnecessary.
I am wondering if this generalizes to $n$ linearly independent $n$-dimensional vectors $\{\mathbf{x}_1,\ldots\mathbf{x}_n\}$ with corresponding weights $\{p_1,\ldots,p_n\}$, where, say $\sum_{i=1}^np_i=1$.  Thus, $\mathbf{A}=\sum_{i=1}^np_i\mathbf{x}_i\mathbf{x}_i^T$ and $\mathbf{B}$ is defined as above.  I think that their eigenvalues are the same (and confirm this belief with a few numerical experiments).  I also feel that this should be a pretty simple proof, but I can't seem to figure out where to get started.  Any ideas/hints?

REPLY [1 votes]: This can be seen by writing $A$ with respect to the basis $\{x_1,\ldots,x_n\}$, or equivalently conjugating using an invertible matrix. Let $X$ be the matrix whose $i^{\rm th}$ column is $\sqrt{p_i}x_i$. Then
$$
  A(\sqrt{p_i}x_i)=\sqrt{p_i}\sum_j p_jx_jx_j^Tx_i
    =\sum_j(\sqrt{p_ip_j}x_j^Tx_i)(\sqrt{p_j}x_j)
    =\sum_jB_{ji}\sqrt{p_j}x_j.
$$
Thus $AX=XB$. Assuming the $x_i$ are linearly independent and all $p_i>0$, $X$ is invertible so $A=XBX^{-1}$. Thus $A$ and $B$ have the same eigenvalues.


TITLE: Differentiate the Function: y=$\ ln\ tan^2x$
QUESTION [0 upvotes]: y=$(ln\ tan^2x)$
= $2(ln\ tan\ x)\cdot (ln\ tan\ x)'$
=$2(ln\ tan\ x)\cdot \frac{1}{(tan\ x)'}$
=$\frac{2(ln\ tan\ x)}{sec^2x}$
Is this right if not what am I doing wrong?

REPLY [0 votes]: $(\ln(\tan^2(x)))' = (2\ln(\tan(x))' = 2 \frac{\tan'(x)}{\tan(x)} = 2\frac{\cos(x)}{\sin(x)\cos^2(x)} = \frac{2}{\sin(x)\cos(x)} = \frac{4}{\sin(2x)}$


TITLE: Generic singular hypersurface
QUESTION [4 upvotes]: I've heard in informal conversations before the claim that:
"a generic singular hypersurface has a single singularity of type $\mathbf{A}_1$".
What is the precise statement of this result? Where can I find a proof?

REPLY [7 votes]: Consider the space of all hypersurfaces of degree $d$ in $\mathbb P^n$, say, with $n \geq 1$ and $d \geq 2$. Consider the closed subspace of singular hypersurfaces. Then there is a dense open set of this closed subspace over which all hypersurfaces have a single singularity of type $\mathbf A_1$.
The proof is to check that a sufficient condition for a singularity to have type $\mathbf A_1$ is for the Hessian to be nondegenerate, then to check that the locus of singular hypersurfaces has codimension $1$, while the locus of hypersurfaces with two singularities and with a singularity with degenerate Hessian each have codimension at least $2$. These can all be proved by viewing the moduli space of hypersurfaces with a certain type of singularity as a bundle over the space of possible singular points and calculating its dimension as the dimension of the base plus the dimension of the fiber.


TITLE: Notations for groups of order $p^3$
QUESTION [3 upvotes]: Are there any relatively common notations for the two non-isomorphic nonabelian groups of order $p^3$ where $p$ is a prime number? I remember reading some notations like $p_+^{1+2}$ and $p_-^{1+2}$. Where can I find the definition of these notations?

REPLY [3 votes]: These groups are the so-called extra-special groups of order $p^{3}.$ For each prime $p$ there are exactly two isomorphism types of such groups. The case $p=2$ has been discussed in another answer. When $p$ is odd, the two isomorphism types are distinguished by the fact that one has exponent $p,$ the other has exponent $p^{2}.$ I believe an explanation of the notation you refer to may be found in the "Atlas of Finite Groups".