text:
\section{Introduction}
\label{sec:intro}

Critical to many statistical inference tasks in complex real-world systems, such as prediction and detection, is the ability to extract and estimate distributional characteristics from the observations. Unfortunately, in the high-dimensional regime such model estimation often leads to ill-posed problems, particularly when the number of observations $n$ (or sample size) is comparable to or fewer than the ambient dimensionality $p$ of the model ({\em i.e.,} the ``large $p$, small $n$'' problem). This challenge arises in many modern real-world applications ranging from recommender systems, gene microarray data, and financial data, to name a few. To perform accurate model parameter estimation and subsequent statistical inference, low dimensional structure is often imposed for regularization~\citep{negahban2012unified}.

For Gaussian-distributed data, the central problem is often to estimate the inverse covariance matrix (alternatively known as the precision, concentration or information matrix).  Gaussian graphical models (GGM)

provide an efficient representation of the precision matrix through a graph that represents non-zeros in the matrix~\citep{lauritzen1996graphical}.  In high-dimensional regimes, this graph can be forced to be sparse, imposing a low-dimensional structure on the GGM.  For sufficiently sparse GGM, statistically consistent estimates of the model structure (\emph{i.e.,} sparsistency) can be achieved ({\em e.g.,}~\citet{ravikumar2011high}).  On the computational side, sparsity also leads to reduced complexity of the estimator~\citep{hsieh2013sparse}.
However, when the true distribution can not be well-approximated by a sparse GGM, the standard learning paradigm suffers from either large estimation bias due to enforcing a overly sparse model, or degraded computation time for a dense model. Both result in suboptimal performance in the subsequent inference tasks.

In this paper, we consider a new class of high-dimensional GGM for extending the standard sparse GGM. The proposed model is motivated by many real-world applications, where there exist certain exogenous and often latent factors affecting a large portion of the variables.
Examples are the price of oil on the airlines' stock price variables~\citep{choi2010gaussian}, and the genres on movie rating variables.
Conditioning on these \emph{global} effects, the variables are assumed to
have highly \emph{localized} interactions, which can be well-fitted by a sparse GGM. However, due to the marginalization over global effects, the observed (marginal) GGM, and its corresponding precision matrix, is not sparse.
Unfortunately, in this regime, existing theoretical results and computational tools for sparse GGM are not applicable.

To address this problem, we propose to use latent variable Gaussian graphical models (LVGGM) for modeling and statistical inference. LVGGM introduce latent variables to capture the correlations due to the global effects, and the remaining effects are captured by a conditionally sparse graphical model.
The resulting marginal precision matrix of the LVGGM has a sparse plus low-rank structure, therefore we consider a regularized maximum likelihood (ML) approach for parameter estimation (previously considered by~\citet{chandrasekaran2012latent}).  By utilizing the \emph{almost strong convexity}~\citep{kakade2009learning} of the log-likelihood, we derive a non-asymptotic parameter error bound for the regularized ML estimator. Our derived bounds apply to the high-dimensional setting of $p \gg n$ due to restricted strong convexity~\citep{negahban2012unified} and certain structural incoherence between the sparse and low-rank components of the precision matrix~\citep{yang2013dirty}.

We show that for sufficiently large $n$, the Frobenius norm error of the precision matrix of LVGGM converges at the rate $\mcal{O}( \sqrt{\frac{(s + \reff \cdot r) \log p}{n}})$, where $s$ is the number of non-zeros in the conditionally sparse precision matrix, $\reff$ is the effective rank of the covariance matrix and $r$ is the number of latent variables.  This rate is in general significantly faster than the standard convergence rate of $\mcal{O}(\sqrt{\frac{p^2 \log{p}}{n}})$ for an unstructured dense GGM.
This result offers a compelling argument for using LVGGM over sparse GGM for many inference problems.

The paper is structured as follows. In Section~\ref{sec:relWork} we review the relevant prior literature.  In Section~\ref{sec:setup} we formulate the LVGGM estimation problem. In Section~\ref{sec:theory} the main theoretical results are presented.  Experimental results are shown in Section~\ref{sec:exp} and we conclude in Section~\ref{sec:conclude}.
We use boldface letters to denote vectors and matrices. ${\| \cdot \|_1}$, $\| \cdot \|_2$, $\| \cdot \|_F$, $\| \cdot \|_*$ denote the elementwise $\ell_1$, spectral, Frobenius, and nuclear matrix norms, respectively.


TITLE: Prove that $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}} \leq \frac{3}{2}.$
QUESTION [2 upvotes]: Let $a,b,$ and $c$ be positive real numbers such that $a+b+c = abc$. Prove that $$\dfrac{1}{\sqrt{1+a^2}}+\dfrac{1}{\sqrt{1+b^2}}+\dfrac{1}{\sqrt{1+c^2}} \leq \dfrac{3}{2}.$$

This question seems tricky because how do we incorporate the $a+b+c = abc$ condition to solve this? I was thinking of solving for the variables and substituting like $a = \dfrac{b+c}{1-bc}$ etc., but I am not sure how that will help.

REPLY [3 votes]: Change the notation using  $x=\frac{1}{a}$ , $y=\frac{1}{b}$ and $z=\frac{1}{c}$
Now the condition is that $xy+yz+zx=1$.
And the inequality to prove is :
$$\frac{x}{\sqrt{x^2+1}}+\frac{y}{\sqrt{y^2+1}}+\frac{z}{\sqrt{z^2+1}} \leq \frac{3}{2}$$
Now use the condition and notice that :
$$x^2+1=x^2+xy+yz+zx=(x+y)(x+z)$$
So now using the AM-GM inequality :
$$\frac{x}{\sqrt{x^2+1}}=\frac{x}{\sqrt{(x+y)(x+z)}} =\sqrt{\frac{x^2}{(x+y)(x+z)}} \leq \frac{1}{2} \left (\frac{x}{x+y}+\frac{x}{x+z} \right )$$
Do the same thing for the other terms and add them to get the conclusion .


TITLE: Doubt in the definition of Centralizer
QUESTION [2 upvotes]: Centralizer
Given : $G, H \le \text{Sym}(\Omega)$ and $G$ normalises $H$ means $\{g \in G \mid gH = Hg\}$
Find : $C_H(G) = \{g \in G \mid hg = gh, \forall h \in H \}$
I am not getting the difference $C_H(G)$ and normaliser of $H$, which is already given. What I am missing here?

REPLY [2 votes]: That' very simple: $G$ normalises $H$ if for all $g\in G$,and every $h\in H$, there is an $h'\in H$ such that
$$gh=h'g.$$
For the centraliser, $h'=h$ for any $h\in H$.
In other words, $G$ normalises $H$ if you have a homomorphism
\begin{align}
G&\longrightarrow \operatorname{Aut} H\\
g&\longmapsto(h\mapsto ghg^{-1})
\end{align}
and if $G$ centralises $H$, this homomorphism is trivial.


TITLE: The coordinate ring of the circle is not UFD
QUESTION [2 upvotes]: $$\mathbb{K}$$ is a field of characteristic different of 2 and $$\sqrt{-1} \not \in \mathbb{K}$$
I have some trouble to show that $$\mathbb{K}[X,Y]/(X^2+Y^2-1)$$
is not UFD. I show that $(\overline{X})$ is not prime (by contradiction) but I can't find a contradiction that $X$ is irreducible.
Cordialy,
doeup

REPLY [4 votes]: Lets assume that the factor ring $A:=\mathbb{K}[X,Y]/(X^2+Y^2-1)$ is a UFD. Since $A$ has dimension $1$ and is noetherian, it is then a principal ideal domain.
The ideals $(X-1,Y)$ and $(X+1,Y)$ are different maximal ideals of $\mathbb{K}[X,Y]$ containing $X^2+Y^2-1$. Hence the images $(\overline{X}-1,\overline{Y})$ and $(\overline{X}+1,\overline{Y})$ are different non-zero prime ideals of $A$. Let $p$ and $q$ be prime elements of $A$ generating these ideals. Then $p$ divides $\overline{X}-1$ and $\overline{Y}$ as well as $q$ divides $\overline{X}+1$ and $\overline{Y}$.
Now the equation $\overline{Y}^2=(\overline{X}-1)(\overline{X}+1)$ in combination with the assumption about $A$ shows that $p^2$ must divide $\overline{X}-1$, since otherwise $p$ must divide $\overline{X}+1$, which implies the contradiction $p=q$.
The ring extension $A|\mathbb{K}[\overline{X}]$ is free of rank $2$ with basis $1$, $\overline{Y}$, hence every prime element of $\mathbb{K}[\overline{X}]$ splits into at most two prime factors in $A$. Consequently $\overline{X}-1=p^2$. Lifting to the polynomial ring yields an equation of the form
$
X-1=(f(X)+g(X)Y)^2+h(X,Y)(X^2+Y^2-1).
$
Note that obviously $g(X)\neq 0$ and moreover $f(X)\neq 0$, since otherwise $\overline{X}-1$ were divisible by $\overline{Y}$.
The first summand of the right hand side has degree $2$ in $Y$. Due to the term $2f(X)g(X)Y$ appearing in this summand, the polynomial $h(X,Y)$ must contain at least one monomial of the form $X^kY$, hence the second summand on the right hand side has degree at least $3$ in $Y$, a contradiction.


text:
\section{Localizations of exact categories}\label{section:Preliminaries}

This section is preliminary in nature.  We summarize the results of \cite{HenrardVanRoosmalen19b,HenrardVanRoosmalen19a} about localizations of exact categories.  One salient feature of the localizations we consider is that the resulting category need not be exact, but will be one-sided exact (in the sense of \cite{BazzoniCrivei13, Rump11}).

\subsection{One-sided exact categories}

\begin{definition}
	A \emph{conflation category} is an additive category $\CC$ together with a chosen class of kernel-cokernel pairs (closed under isomorphisms), called \emph{conflations}. The kernel part of a conflation is called an \emph{inflation} and the cokernel part of a conflation is called a \emph{deflation}. We depict inflations by $\inflation$ and deflations by $\deflation$.

	An additive functor $F\colon \CC\to \DD$ between conflation categories is called \emph{conflation-exact} if conflations are mapped to conflations.
\end{definition}

\begin{definition}\label{Definition:OneSidedExactCategory}
	A conflation category $\EE$ is called an \emph{inflation-exact} or \emph{left exact category} if $\EE$ satisfies the following axioms:
	\begin{enumerate}[label=\textbf{L\arabic*},start=0]
		\item\label{L0} For each $X\in \EE$, the map $0\to X$ is an inflation.
		\item\label{L1} The composition of two inflations is again an inflation.
		\item\label{L2} The pushout of any morphism along an inflation exists, moreover, inflations are stable under pushouts.
	\end{enumerate}
	Dualizing the above axioms yields the notion of a \emph{deflation-exact} category. A \emph{Quillen exact} category is simply a conflation category which is both inflation-exact and deflation-exact by \cite[Appendix~A]{Keller90}.
\end{definition}

Let $\EE$ be one-sided exact category. Analogous to exact categories, one can define the bounded derived category $\Db(\EE)$ as the Verdier localization $\Kb(\EE)/\left\langle\Ac(\EE)\right\rangle_{\text{thick}}$ of the bounded homotopy category by the thick closure of the triangulated subcategory of acyclic complexes (see \cite[Corollary~7.3]{BazzoniCrivei13}). The canonical embedding $i\colon\EE\to\Db(\EE)$, mapping objects to stalk complexes in degree zero, is a fully faithful embedding mapping conflations to triangles.
We write $\Dbinf(\EE)$ for the corresponding stable $\infty$-category; in particular, the homotopy category of $\Dbinf(\EE)$ recovers $\Db(\EE)$.

The derived category allows a construction of the \emph{exact hull} $\ex{\EE}$ of $\EE$ (see \cite{HenrardVanRoosmalen19b}): the exact hull is given by the extension closure of $\EE$ in the (bounded) derived category $\Db(\EE)$.  A sequence $X\to Y \to Z$ in $\ex{\EE}$ is a conflation if and only if it fits in a triangle $X \to Y \to Z\to \Sigma X.$  The following proposition is shown in \cite{HenrardVanRoosmalen19b}.

\begin{proposition}\label{proposition:ExactHull}
Let $j\colon \EE \to \ex{\EE}$ be the embedding of an inflation-exact category in its exact hull.
\begin{enumerate}
	\item The embedding $j\colon \EE\to \ex{\EE}$ is $2$-universal among conflation-exact functors to exact categories.
	\item The embedding $j$ lifts to an equivalence $\Dbinf(\EE)\stackrel{\simeq}{\rightarrow}\Dbinf(\ex{\EE})$ of stable $\infty$-categories.
\end{enumerate}
\end{proposition}

\subsection{Strictly percolating subcategories}

\begin{definition}\label{Definition:InflationPercolating}
	Let $\EE$ be an exact category. A full subcategory $\AA\subseteq \EE$ is called a \emph{strictly inflation-percolating subcategory} if the following properties are satisfied:
\begin{enumerate}[label=\textbf{A\arabic*},start=1]
		\item\label{A1} The category $\AA$ is a \emph{Serre subcategory} of $\EE$, i.e.~for any conflation $X\inflation Y\deflation Z$ in $\EE$, we have $Y\in \AA\Leftrightarrow X,Z\in \AA$.
		\item\label{A2} Every morphism $f\colon A\to X$ with $A\in \AA$ is \emph{strict}, i.e.~factors as $A\deflation \im(f)\inflation X$, and $\im(f)\in \AA$.
	\end{enumerate}
\end{definition}

\begin{remark}
If $\AA\subseteq \EE$ is a strictly inflation-percolating subcategory, then $\AA$ is a fully exact abelian subcategory of $\EE$.
\end{remark}

The following observation will be of use later.

\begin{proposition}\label{proposition:A1A2+InjectiveLiftsToHull}
Let $\EE$ be an inflation-exact category and let $\VV\subseteq \EE$ be a full additive subcategory satisfying axioms \ref{A1} and \ref{A2}. If every object of $\VV$ is injective in $\EE$, then $\VV$ is a strictly inflation-percolating subcategory of the exact hull $\ex{\EE}$.
\end{proposition}

\begin{proof}
As each $V\in \VV\subseteq \EE$ is injective, $\Hom_{\EE}(-,V)$ is exact and hence $\Ext_\EE^1(-,V)= 0.$  As $\ex{\EE}$ is the extension-closure of $\EE$ in $\ex{\EE}$, it follows that $V$ is injective in $\ex{\EE}$ as well.

	Note that $\ex{\EE}=\bigcup_{n\geq 0}\EE_n$ where $\EE_0=\EE$ and $\EE_n$ for $n\geq 1$ is defined recursively as extensions of objects in $\EE_{n-1}$. As $\VV\subseteq \ex{\EE}$ consists of injective objects, it follows that $\VV\subseteq \ex{\EE}$ is an extension-closed subcategory.

	We now show axiom \ref{A2}. Let $f\colon V\to Y$ be a map in $\ex{\EE}$ with $V\in \VV$. By definition, there is an $n$ such that $Y\in \EE_{n}$. We proceed by induction on $n\geq 0$. If $n=0$, the result follows as $\VV\subseteq \EE$ satisfies axiom \ref{A2}. If $n\geq 1$, then there is a conflation $X\stackrel{\iota}{\inflation} Y\stackrel{\rho}{\deflation} Z$ in $\ex{\EE}$ with $X,Z\in \EE_{n-1}$. By the induction hypothesis, the composition $\rho\circ f$ factors as $V\stackrel{p}{\deflation}V''\stackrel{h}{\inflation}Z$ with $V''\in \VV$. As $\VV\subseteq\ex{\EE}$ is an abelian subcategory, $(V' \coloneqq) \ker(p)\in \VV$. Note that there is an induced map $g\colon V'\to X$ such that $fi=\iota g$. Again the induction hypothesis yields that $g$ factors as $V'\stackrel{g'}{\deflation}U\stackrel{g''}{\inflation}X$ with $U\in \VV$. Taking the pushout of $g'$ along $i$ in $\ex{\EE}$ yields the following commutative diagram (where the rows are conflations):
	\[\xymatrix{
		V'\ar@{>->}[r]^i\ar@{->>}[d]^{g'} & V\ar@{->>}[r]^{p}\ar@{->>}[d]^{f'} & V''\ar@{=}[d]\\
		U\ar@{>->}[r]^{i'}\ar@{>->}[d]^{g''} & W\ar@{->>}[r]^{p'}\ar[d]^{f''} & V''\ar@{>->}[d]^h\\
		X\ar@{>->}[r]^{\iota} & Y\ar@{->>}[r]^{\rho} & Z
	}\] Here the upper-left square is bicartesian, $f''f'=f$ and $W\in \VV$ as $\VV\subseteq\ex{\EE}$ is extension-closed. It follows from \cite[Corollary~3.2]{Buhler10} that $f''$ is an inflation in $\ex{\EE}$. This shows axiom \ref{A2}.

	To show axiom \ref{A1}, it remains to show that given a conflation $X\stackrel{\iota}{\inflation}V \stackrel{\rho}{\deflation} Z$ in $\ex{\EE}$ with $V\in\VV$, $X,Z$ belong to $\VV$ as well. By axiom \ref{A2}, $\rho$ is admissible with image in $\VV$.  It follows that $Z\in \VV$. As $X$ is the kernel of a morphism in $\VV$ and $\VV$ is an abelian subcategory of $\EE$, we know that $X$ belongs to $\VV$ as well. This concludes the proof.
\end{proof}

\subsection{Quotients by strictly percolating subcategories}

The next definition is based on \cite[Definition~1.12]{Schlichting04}.

\begin{definition}
	Let $\EE$ be an inflation-exact category and let $\AA\subseteq\EE$ be a strictly inflation-percolating subcategory. A morphism $f\colon X\to Y$ in $\EE$ is called a \emph{weak $\AA$-isomorphism} (or simply a \emph{weak isomorphism}) if $f$ is strict and $\ker(f),\coker(f)\in \AA$. The set of weak $\AA$ -isomorphisms is denoted by $S_{\AA}$.
\end{definition}

The following theorem summarizes the main results of \cite{HenrardVanRoosmalen19b,HenrardVanRoosmalen19a}.

\begin{theorem}\label{theorem:Quotient/LocalizationTheorem}
	Let $\AA$ be a strictly inflation-percolating subcategory of an exact category $\EE$.
	\begin{enumerate}
		\item The set $S_{\AA}$ of weak $\AA$-isomorphisms is a left multiplicative system.
		\item The natural localization functor $Q\colon\EE\to \EE[S^{-1}_\AA]$ endows the localization $\EE[S^{-1}_\AA]$ with an inflation-exact structure such that $Q$ preserves and reflects conflations.
		\item  The localization functor $Q$ is also a quotient in the category of conflation categories, i.e.~it satisfies the following universal property: if $F\colon \EE\to \FF$ is a conflation-exact functor between conflation categories such that $F(\AA)=0$, then $F$ factors uniquely through $Q$ via a conflation-exact functor $F'\colon \EE/\AA=\EE[S_\AA^{-1}]\to \FF$.
	\end{enumerate}

	Moreover, the localization sequence $\AA\hookrightarrow\EE \stackrel{Q}{\rightarrow}\EE/\AA$ induces a Verdier localization sequence on the bounded derived categories
	\[\DAb(\EE)\to \Db(\EE)\to \Db(\EE/\AA)\]
	where $\DAb(\EE)$ is the thick subcategory of $\Db(\EE)$ generated by $\AA$ under the canonical embedding $\EE\hookrightarrow \Db(\EE)$.

	If $\AA$ has enough $\EE$-injectives, then the natural embedding $\Db(\AA)\hookrightarrow \DAb(\EE)$ is a triangle equivalence, and there is an exact sequence in $\StabInfCat$:
	\[\Dbinf(\AA)\to \Dbinf(\EE)\to \Dbinf(\EE/\AA).\]
\end{theorem}

config: arxiv


TITLE: Testing whether two vertices are neighbours
QUESTION [1 upvotes]: I face the following problem: I am given a high-dimensional, convex, bounded polyhedron in both vertex description: $X = \mathrm{conv} \, \{ v_1, \ldots, v_K \}$ and halfspace description: $X = \{ x \in \mathbb{R}^k \, : \, A x \leq b \}$. I can assume that the halfspace description does not contain any redundant inequalities, but in general some of the vertices will be highly degenerate.
I would like to know whether two specified vertices $v_i$ and $v_j$ are neighbours, that is, they are connected by an edge. This problem is fairly simple in dimensions 2 and 3, but I'm interested in the high(er)-dimensional case.
Could anybody please point me into the right direction?
Many thanks

REPLY [1 votes]: You should first check that $v_i$ and $v_j$ are both vertices of $X$.  Given the description $X = \{ x \in \mathbb{R}^k \, : \, A x \leq b \}$, this is easy.
Suppose that $A$ is an $m \times n$ matrix.  Then $v \in X$ is a vertex if and only if $A'v=b'$ for some subsystem $A'x \leq b'$ of $Ax \leq b$, where $A'$ has rank $n$.  Such a subsystem is called a basis for $v$.
Now, two vertices $v_i$ and $v_j$ of $X$ are adjacent if and only if there is a rank-$(n-1)$ subsystem $A^*x \leq b^*$ of $Ax \leq b$ such that $A^*L=b^*$, where $L$ is the line through $v_i$ and $v_j$.
Thanks to Brendan McKay for this last step.


TITLE: 90 days to finish a task with every day probability to complete is 1/(i+1), what is a probability to finish within given time?
QUESTION [0 upvotes]: For each day i (staring from 1) probability to finish a task at that i day is $\frac{1}{(1+i)}$. What is a probability to finish the task within 90 days?
I solved it as $\sum_{\substack{0<i<91}}$ P(to finish the task at day i)
and
P(to finish the task at day i) = $\frac{1}{(1+i)}$*$\sum_{\substack{0<j<i}}$ P(not finishing at day j)
I just want to check whether this answer is right.

REPLY [1 votes]: Presumably the probability to finish on day $i$ assumes you did not finish earlier.  Otherwise the probability to finish by day $3$ is $\frac 12 + \frac 13 + \frac 14 \gt 1$.  On that assumption, the chance you do not finish by day $i$ is $$\prod_{k=1}^i \frac k{k+1}=\frac {i!}{(i+1)!}=\frac 1{i+1}$$
and the chance you finish by day $90$ is $\frac {90}{91}$


TITLE: Elementary Truth Functional Logic question
QUESTION [1 upvotes]: So I am currently attacking a question from the first chapter of my logic book. I know that the question is true, but I am having a hard time actually proving it. The question is as follows.
If a argument is valid, then it cannot be made invalid by adding premises.
If someone could please give me some guidance on how I might approach this proof. This is as far as I could get. If V is an argument then there must not exist a premises P that could be added to the a valid argument V such that when P is added to V V' becomes invalid.
All help would be much appreciated. I am of course not looking for a full solution but perhaps some insight or guidance.
Thank You
EDIT: Also, I suppose the reason I'm having trouble is because I cannot show why such a P cannot exist. I don't know what kind of contradiction it may offer if it did either.

REPLY [2 votes]: Consider these two claims:

(1) There is no possible situation in which $A$ is true and $C$ false.
(2) There is no possible situation in which $A$ is true and $B$ is true and $C$ false.

Question: do you see that if (1) is true then (2) is true too?
Question: what does that tell us about the relation between the validity of the argument $A \therefore C$ and the validity of $A, B \therefore C$?
Question: how do we generalize the answer to the last question to answer your original question about adding more than one premiss to a valid argument?


TITLE: Bijective proof check
QUESTION [0 upvotes]: I have been given the problem: "Say $$f:X→Y$$ and $$g:Y→Z$$are functions so that $$g◦f:X→Z$$ is bijective and $f$ is surjective. Can you deduce that $g$ is bijective?
My proof went as follows, I am not sure if this is valid?
$$ $$
Since $f$ is surjective then for any $y\in Y$ there exists and $x \in X$ such that $$f(x)=y$$
and $gf$ is bijective so it is surjective and injective so holds for:
$$g(f(x))=z$$for any $x$ and $$gf(x_1)=g(f(x_2)) \Rightarrow x_1=x_2$$
Then since we know $gf$ is surjective so $gf(x)=z$, since $f(x)=y$, then $g(y)=z$, so for any $z \in Z$ there exists $y \in Y$ such that $g(y)=z$, so $g$ is surjective.
Finally we say $$x_1=x_2$$ and call $h=gf$. Since $h$ is injective and surjective $$h(x_1)=h(x_2), g(f(x_1))=g(f(x_2))$$. Since we know that $f$ is surjective so $f(x)=y$ then $g(y_1)=g(y_2)$ so it is injective. Then since we have proven $g$ is both injective and surjective then it is bijective.

REPLY [0 votes]: You can prove it more easily(I think much more) by using RAA argument. That is, first suppose that gf is not bijective and lead a contradiction from the supposition and the fact that f is a "surjective" "function".


TITLE: Why is an adjunction a “weak form of equivalence”?
QUESTION [2 upvotes]: Wikipedia says:

An adjunction between categories $C$ and $D$ is somewhat akin to a "weak form" of an equivalence between $C$ and $D$.

I have heard this idea before, e.g. from Qiaochu.
Can you give an extremely simple example or argument that clarifies the intuitive idea that an adjunction is a “weak form of equivalence”? I’ve only read some explanations that are technical enough that my intuition doesn’t get that idea.

REPLY [0 votes]: As the comments have mentioned, there are a few ways to think of this:

Equivalences are Adjunctions, but not conversely.
The unit/counit are natural isomorphisms if the adjunction is an equivalence.
An adjunction is a specific type of bijection between hom sets, or one could say they have the same number of morphisms/size of morphism sets. An equivalence already has this, but the adjunction is weaker but is quite ambient in Mathematics.

Helpful examples would include the forgetful functor $F:\textbf{Grp}\rightarrow \textbf{Set}$, and the free group functor $G:\textbf{Set}\rightarrow \textbf{Grp}$.
Grp and Set are definitely not equivalent, but there is a one-to-one correspondence between morphism sets defined by the adjunction between $F$ and $G$.


TITLE: How do I find the range of $\sqrt{x^2-4}$?
QUESTION [0 upvotes]: I tried to find the range of $\sqrt{x^2-4}$ by the method of finding the domain of its inverse function. The inverse function will be $\sqrt{x^2+4}$. It's domain will be $(-\infty,\infty)$ as ${x^2}$ will always be greater than ${-4}$. But the range of the original function is $[0,\infty)$. How is that?

REPLY [2 votes]: The domain of a function $y=f(x)$ is set of all real values  of $x$ where the function is real finite and unique.  So in this case the domain is $(-\infty, -2] \cup [2, \infty)$. The range is set of all real values of $y$ taken by the function over the domain. So in this case the
range is $[0, \infty)$. This is the direct method.
Otherwise, you have to realize the your function $f(x)$ is positive definite so its range is only $[0, \infty)$. In fact, $(-\infty,0]$ is the range of another function $g(x)=-\sqrt{x^2-4}$ which is the other branch of the curve: $x^2-y^2=4$ (hyperbola). So this curve defines two functions: $f(x)$ and $g(x)$. Here the question chooses  $f(x)$ hence its range is $[0,\infty)$.


TITLE: Proving part of the theorem that a continuous function on the bounded interval is uniformly continuous
QUESTION [0 upvotes]: I am asked to prove the following.  Suppose $D$ is a closed and bounded subset of $R$ and suppose $f:D\to R$ is continuous on $D$.  Then $f$ is uniformly continuous:
$
$Proof: Suppose by contradiction that $\{x_n\}$ and $\{y_n\}$ are sequences in $D$ s.t.
$$ \lim_{n\to\infty} [x_n-y_n] =0$$
and the sequence $[f(x_n)-f(y_n)]$ does not converge to $0$.
We want to negate the statement of convergence for the above s.t. we have: "There exists an $\epsilon>0$ s.t. for all natural number k there exists an $n_k>k$ s.t. |$f(x_{n_k})-f(v_{n_k})$|$\geq \epsilon$.
Now apply the sequential compactness theorem to $\{x_{n_k}\}$ which is a subset of $D$.  There is a limit point $x_0$ s.t. there is a further subsequence $\{x_{n_{k_j}}\}$ s.t. the limit as j tends to infinity is $x_0$, which is in $D$.
Now here's what I am supposed to prove: By supposing that $\{x_n\}$ and $\{y_n\}$ are sequences satisfying the fact that the image of a continuous function on a closed bounded interval is bounded above and assuming $\{x_n\}$ converges to $x$ I want to show that the limit as $j$ tends to infinity of $y_{n_{k_j}}$ is $x_0$.  How would you do this part?
So basically I am just proving this one last part of the proof.

REPLY [0 votes]: Here is other proof maybe this would help you
If not, there is a $\epsilon>0$ and points $x_n,y_n\in D$, s.t. $d(x_n,y_n)<1/n$ and $d(fx_n,fy_n)\ge \epsilon$ for all $n$. Since any sequence in $D$ has a convergent subsequence, we may assume $x_n\to x$, for $x\in D$, so $y_n\to x$. By continuity at $x$,  for $n$ large  enough $fx_n,fy_n\in  B(fx,\epsilon/2)$. Thus $d(fx_n,fy_n)<\epsilon$, contradiction.


TITLE: Change of order of integration of a triple integral
QUESTION [1 upvotes]: Consider $$ I = \int_0^{\omega}\int_0^{\alpha}\int_0^{\alpha}F(\beta){\tilde{F}(\gamma)}e^{i\beta t}e^{-i\gamma t}R(\alpha)d\beta d\gamma d\alpha$$
In this triple integral,I want to bring about, a change of order of integration, where in I take integration with respect to $\alpha$ inside most and integration with respect to other variables outside. Appreciate some help in this regard.
Here $\tilde{F}(\gamma)$ is the complex conjugate of $F(\gamma)$.
My approach yielded me this result, which I am not sure about :
$$I = \int_0^{\omega}\int_0^{\gamma}F(\beta){\tilde{F}(\gamma)}e^{i(\beta-\gamma )t}\int_0^{2(\omega-\gamma)}R(\alpha)d\alpha d\beta d\gamma $$

REPLY [2 votes]: What you have is wrong.
Hint. The original integral is over the domain consisting of all triples $(\alpha,\beta,\gamma)$ for which
$$0 \leq \beta, \gamma \leq \alpha \leq \omega.$$
To change the order, note that this is equivalent to picking $\beta$ and $\gamma$ anywhere in $[0,\omega]$, and then picking $\alpha$ in the interval $[\max(\beta,\gamma),\omega]$.
Edit: It might be convenient to divide the domain into two parts, one where $\beta \leq \gamma$, and one where $\gamma \leq \beta$.


TITLE: Number recursion
QUESTION [0 upvotes]: I think it is not called as in the title, but anyway, that is it below. Can anyone please explain the magics?
1/99 = .01010101010101010101
1/999 = .00100100100100100100
1/9999 = .00010001000100010001
1/99999 = .00001000010000100001
1/999999 = .00000100000100000100
1/9999999 = .00000010000001000000
1/999999999 = .00000000100000000100
1/9999999999 = .00000000010000000001

REPLY [0 votes]: There are all special cases of a more general pattern.  Whenever $a>1$, we have
$$\frac1a + \frac1{a^2} + \frac1{a^3} + \frac1{a^4} + \ldots = \frac1{a-1}.$$
For example, you may be familiar with the fact $\frac12 + \frac14 + \frac18 + \ldots = 1$, which is the case of the formula above when $a=2$.
Now $0.01010101\ldots = \frac1{100} + \frac1{10,000} + \frac1{1,000,000}+\ldots$, so this is an instance of the formula for $a={100}$. In this case the right-hand side $\frac{1}{a-1} = \frac1{99}$.  Your other examples are similar.

REPLY [0 votes]: Hint $\ \dfrac{1}{10^n-1}\, =\, \dfrac{1}{10^n}\, \dfrac{1}{1-10^{-n}}\, =\, 10^{-n} (1 + 10^{-n} + 10^{-2n}+\,\cdots\,)\ $ via geometric power series.


TITLE: character formula for demazure modules
QUESTION [2 upvotes]: Is there any character formula for demazure modules in arbitary kac moody settings which does not use demazure operators?

REPLY [2 votes]: Ryom-Hansen, Steen(DK-CPNH)
Littelmann's refined Demazure character formula revisited. (English summary)
Sém. Lothar. Combin. 49 (2002/04), Art. B49d, 10 pp.
The review:
"The author provides a purely combinatorial proof the Demazure character formula, a generalisation of Weyl's character formula. This is done using only the combinatorial properties of crystals, namely Kashiwara's crystal operators and the ∗-operation. Prior proofs required an appeal to either representation theory or Littelmann's path models."

REPLY [1 votes]: Littelmann's, which gives a positive formula (counting Littelmann paths). His ICM address is here: http://www.mathunion.org/ICM/ICM1994.1/Main/icm1994.1.0298.0308.ocr.pdf
He proves its validity using Demazure operators -- I hope that doesn't disqualify it!


text:
\begin{document}

\begin{abstract}
We compute the primary decomposition of certain ideals generated by
subsets of minors in a generic matrix or in a generic symmetric matrix,
or subsets of Pfaffians in a generic skew-symmetric matrix.
Specifically, the ideals we consider are generated by minors
that have at least some given number of rows and columns in certain submatrices.
\end{abstract}

\maketitle

\section{Introduction}

The paper \cite{MR853243} concerns ideals of minors fixing a submatrix,
meaning the set of minors in an $m \times n$ matrix that involve all $r$ of the first $r$ columns of the matrix.
One of the main results of that paper, Theorem~A, gives the primary decomposition of the ideal
generated by this set of minors.
We generalize this to consider minors that involve at least $r$ of the first $a$ columns:
\begin{thm}\label{thm: two block minors}
Let $k$ be a field, let $X$ be a \emph{generic} $m \times n$ matrix,
that is $X = (x_{i,j})$ for $1 \leq i \leq m$, $1 \leq j \leq n$,
and let $R = k[X] = k[x_{i,j}]$.
Regard $X$ as a block matrix, $X = (AB)$, where $A$ has size $m \times a$ and $B$ has size $m \times (n-a)$.
Let $J$ be the ideal generated by the set of $t$-minors of $X$ that involve at least $r$ columns of $A$,
let $I_t(X)$ be the ideal generated by the $t$-minors of $X$,
and similarly let $I_r(A)$ be the ideal generated by the $r$-minors of $A$.
Then $J = I_t(X) \cap I_r(A)$.
\end{thm}
We generalize further than this, to allow several blocks
as well as restrictions on both rows and columns.
We also give similar statements for ideals generated by
sets of minors of a generic symmetric matrix,
requiring some number of rows or columns in certain submatrices.
Before we give these statements,
we consider one possible application in the setting of the two-block theorem above.

It is sometimes useful to consider, for a homogeneous ideal $I$,
the ideal $I_{\leq d}$ generated by the forms in $I$ of degree $\leq d$.
For example, in resolving the singularities of the affine cone $V(I) \subset \mathbb{A}^n$,
upon blowing up the origin, the total transform of $I$ may have embedded components
supported along the projective variety $V(I_{\leq d})$, lying in the exceptional divisor $\cong \mathbb{P}^{n-1}$,
for various $d$; see \cite{zct:mla}.
When $I$ is a determinantal ideal, generated by minors of a matrix whose entries are homogeneous forms,
then $I_{\leq d}$ is generated by just some of the minors of the matrix.

\begin{cor}
Let $X = (x_{i,j})$ be a generic $m \times n$ matrix,
regarded as consisting of two blocks, $X = (AB)$,
where $A$ has size $m \times a$ and $B$ has size $m \times (n-a)$.
Fix the ring $R = k[X] = k[x_{i,j}]$
where every entry in $A$ has degree $p$ and every entry in $B$ has degree $q > p$;
that is, $\deg(x_{i,j}) = p$ if $1 \leq j \leq a$, $\deg(x_{i,j}) = q$ otherwise.
Fix $t$ and $d$.
Let $I_t(X)_{\leq d}$ be the ideal generated by those $t$-minors of $X$
of degree less than or equal to $d$,
and for each $r$ let $I_r(A)$ be the ideal generated by the $r$-minors of $A$.
Then $I_t(X)_{\leq d} = I_t(X) \cap I_r(A)$ for $r = \lceil \frac{tq-d}{q-p} \rceil$.
\end{cor}
Indeed, a $t \times t$ minor $M$ with $r$ columns in $A$ and $t-r$ columns in $B$ will have degree
$\deg M = pr+q(t-r)$; the value of $r$ in the statement is the least integral solution to $\deg M \leq d$.

\begin{cor}
Consider the vector bundles
$F = \mathcal{O}_{\PP^N}^m$, $G = \mathcal{O}_{\PP^N}(p)^{\oplus a} \oplus \mathcal{O}_{\PP^N}(q)^{\oplus n-a}$.
Let $f : F \to G$ be a general map,
let $f' : F \to \mathcal{O}_{\PP^N}(p)^{\oplus a}$ be the induced map,
let $\Delta_t(f)$ be the degeneracy locus, $\Delta_t(f) = \{ x \in \PP^N \mid \rank f_x < t \}$,
and let $\Delta_t(f)_{\leq d}$ be the locus defined by the ideal $I(\Delta_t(f))_{\leq d}$.
Then $\Delta_t(f)_{\leq d} = \Delta_t(f) \cup \Delta_r(f')$
for $r = \lceil \frac{tq-d}{q-p} \rceil$.
\end{cor}

This is similar to \cite{zct:smoothness-of-envelopes},
which dealt with $\PP^2$ and had $n = m+1$ in order to obtain general Hilbert-Burch matrices of a given type.
(In particular \cite[Prop.~3.4]{zct:smoothness-of-envelopes} simply recreated
a special case of \cite[Thm.~A]{MR853243}.)

Sections~\ref{section: orders and straightening} and \ref{section: order ideals}
review some background of posets, dosets,
algebras with straightening law, and doset algebras with straightening law.
Then we give our results for minors in generic matrices (Section~\ref{section: minors}),
minors in generic symmetric matrices (Section~\ref{section: symmetric minors}),
and Pfaffians in generic skew-symmetric matrices (Section~\ref{section: pfaffians}).

Throughout, all rings are commutative with unity.

\section{Orders and straightening}\label{section: orders and straightening}

A \defining{poset} (partially ordered set)
is a set together with a transitive, reflexive, antisymmetric relation $\leq$.

\begin{definition}[{\cite[Definition~1.0.3]{MR2298637}}]
A \defining{doset} of a poset $P$ is a subset $D \subset P \times P$ such that
\begin{enumerate}
\item $(a,a) \in D$ for all $a \in P$,
\item if $(a,b) \in D$ then $a \leq b$, and
\item if $a \leq b \leq c \in P$, then $(a,c) \in D$ if and only if $(a,b) \in D$ and $(b,c) \in D$.
\end{enumerate}
\end{definition}

\begin{example}
\begin{enumerate}
\item Let $[n] = \{1,\dotsc,n\}$
and let $P_n = 2^{[n]}$, the power set of $[n]$.
We order $P_n$ as follows.
For $A = \{a_1 < \dotsb < a_s\} \subset [n]$
and $B = \{b_1 < \dotsb < b_t\} \subset [n]$,
$A \leq B$ if and only if $s \geq t$ and $a_i \leq b_i$ for $i = 1,\dotsc,t$.
This makes $P_n$ a poset.

Note, $A \leq B$ if and only if in the diagram
\[
\begin{array}{|c|c|c|c|c|c|}
  \hline
  a_1 & a_2 & \cdots & a_t & \cdots & a_s \\
  \hline
  b_1 & b_2 & \cdots & b_t \\
  \cline{1-4}
\end{array}
\]
the first row is at least as long as the second
and the entries are weakly increasing down each column.

\item Fix $m$ and $n$.
Let $P_{m,n} \subset P_m \times P_n$
consist of pairs of subsets $(A,B)$ such that $|A|=|B|$,
with $(A,B) \leq (A',B')$ if and only if $A \leq A'$ and $B \leq B'$.

\item Let $D_n \subset P_n \times P_n$
consist of pairs $(A,B)$ such that $|A|=|B|$ and $A \leq B$.
Then $D_n$ is a doset.
\end{enumerate}
\end{example}

\begin{example}\label{example: posets of minors}
Here are some key examples of posets and dosets of minors in matrices.
\begin{enumerate}
\item $P_{m,n}$ is the \defining{poset of minors} (of an $m \times n$ matrix):
$(A,B) \in P_{m,n}$ corresponds to the minor with rows indexed by $A$ and columns indexed by $B$.
This poset is usually denoted $\Delta(X)$, where $X$ is an $m \times n$ matrix.

\item $D_n$ is the \defining{doset of minors} of a symmetric $n \times n$ matrix:
$(A,B) \in D_n$ corresponds to the minor with rows indexed by $A$ and columns indexed by $B$.
An element of $D_n$ is called a \defining{doset minor}.
We denote this poset $\Delta^s(Y)$, where $Y$ is a symmetric $n \times n$ matrix.
(It is denoted $\Delta(Y)$ in \cite{MR2298637};
we adjoin the $s$ for ``symmetric'' in order to avoid ambiguity.)

The condition $A \leq B$ means that a minor is a doset minor if and only if
the main diagonal of the minor lies in the upper triangle of the matrix (including the diagonal).

\item Let $P_n(2)$ be the subset of $A \in P_n$ such that $|A|$ is even.
Then $P_n(2)$ is the \defining{poset of Pfaffians} of a skew-symmetric $n \times n$ matrix:
$A \in P_n(2)$ corresponds to the Pfaffian of the submatrix with rows and columns indexed by $A$.
Following \cite{MR2298637} we denote this poset $\Pi(Z)$, where $Z$ is a skew-symmetric $n \times n$ matrix.
\end{enumerate}
\end{example}

\begin{definition}[{\cite[\textsection4.A]{MR953963},\cite[Definition~1.0.1]{MR2298637}}]
Let $A$ be a $B$-algebra and $P \subset A$ a subset with a partial order $\leq$.
Then $A$ is a \defining{graded algebra with straightening law} (abbreviated ASL) on $P$ over $B$ if
\begin{enumerate}
\item $A = \bigoplus_{i \geq 0} A_i$ is a graded $B$-algebra such that $A_0 = B$,
$P$ consists of homogeneous elements of positive degree, and $P$ generates $A$ as a $B$-algebra.
\item $A$ is a free $B$-module with a basis given by products
$\xi_1 \dotsm \xi_m$, $m \geq 0$, $\xi_i \in P$, such that $\xi_1 \leq \dotsb \leq \xi_m$.
These products are called \defining{standard monomials}.
\item For all incomparable $\xi, \nu \in P$, the product $\xi \nu$ can be written as a combination
of standard monomials
\[
  \xi \nu = \sum a_\mu \mu, \qquad a_\mu \in B, a_\mu \neq 0, \qquad \text{$\mu$ standard monomial},
\]
in which every $\mu$ contains a factor $\zeta \in P$ such that $\zeta \leq \xi$ and $\zeta \leq \nu$.
These are called \defining{straightening relations}.
\end{enumerate}
\end{definition}

\begin{definition}[{\cite[Definition~1.0.4]{MR2298637}}]
Let $A$ be a $B$-algebra and $D \subset A$ a subset such that $D$ is a doset of a poset $P$.
Then $A$ is a \defining{graded doset algebra with straightening law} (abbreviated DASL)
on $D$ over $B$ if
\begin{enumerate}
\item $A = \bigoplus_{i \geq 0} A_i$ is a graded $B$-algebra such that $A_0 = B$,
$D$ consists of homogeneous elements of positive degree, and $D$ generates $A$ as a $B$-algebra.
\item $A$ is a free $B$-module with a basis given by products
$(\alpha_1,\alpha_2)\dotsm(\alpha_{2k-1},\alpha_{2k})$, $k \geq 1$, $(\alpha_{2i-1},\alpha_{2i}) \in D$,
$\alpha_1 \leq \dotsb \leq \alpha_{2k}$.
These products are called \defining{standard monomials}.
\item Suppose $M = (\alpha_1,\alpha_2)\dotsm(\alpha_{2k-1},\alpha_{2k})$,
with \defining{standard representation} $M = \sum \lambda_N N$, $0 \neq \lambda_N \in B$,
each $N$ a standard monomial.
Let $N = (\beta_1,\beta_2)\dotsm(\beta_{2\ell-1},\beta_{2\ell})$ be one of the standard monomials
appearing in the standard representation of $M$.
Then for every permutation $\sigma$ of $\{1,\dotsc,2k\}$,
the sequence $\{\alpha_{\sigma(1)},\dotsc,\alpha_{\sigma(2k)}\}$
is lexicographically greater than or equal to the sequence $(\beta_1,\dotsc,\beta_{2\ell})$.
\item In the notation above, if there is a permutation $\sigma$ such that
$\alpha_{\sigma(1)} \leq \dotsb \leq \alpha_{\sigma(2k)}$
then the standard monomial $(\alpha_{\sigma(1)},\alpha_{\sigma(2)})\dotsm(\alpha_{\sigma(2k-1)},\alpha_{\sigma(2k)})$
must appear in the standard representation of $M$ with coefficient $\pm1$.
\end{enumerate}
\end{definition}

\begin{example}
Fix an arbitrary commutative ring $B$ with unity.
\begin{enumerate}
\item Let $X$ be an $m \times n$ \defining{generic matrix}, that is $X = (x_{i,j})$,
$1 \leq i \leq m$, $1 \leq j \leq n$, the $x_{i,j}$ variables over $B$.
Then $A = B[X] = B[x_{i,j}]$ is a graded ASL on $\Delta(X)$ over $B$
\cite[Chap.~4]{MR953963},\cite[Thm.~1.0.5]{MR2298637}.

\item Let $Y$ be an $n \times n$ \defining{generic symmetric matrix},
that is $Y = (y_{i,j})$, $1 \leq i, j \leq n$, $y_{i,j} = y_{j,i}$.
Then $A = B[Y]$ is a graded DASL on $\Delta^s(Y)$ over $B$
\cite[Thm.~1.0.10]{MR2298637}.

\item Let $Z$ be an $n \times n$ \defining{generic skew-symmetric matrix},
that is $Z = (z_{i,j})$, $1 \leq i,j \leq n$, $z_{i,j} = -z_{j,i}$, $z_{i,i} = 0$.
Then $A = B[Z]$ is a graded ASL on $\Pi(Z)$ over $B$
\cite[Thm.~1.0.14]{MR2298637}.
\end{enumerate}
\end{example}

\section{Order ideals}\label{section: order ideals}
We use ASLs and DASLs entirely for the following properties.

\begin{definition}
Let $P$ be a poset.
An \defining{order ideal} is a subset $I \subset P$
such that if $\alpha \in I$ and $\beta \leq \alpha$
then $\beta \in I$.
The order ideal \defining{generated by} $S \subset P$ is the smallest order ideal containing $S$,
that is, $\{ \alpha \in P \mid \alpha \leq s \text{ for some $s \in S$} \}$.
The order ideal \defining{cogenerated by} $S \subset P$ is the largest order ideal disjoint from $S$,
that is, $\{ \alpha \in P \mid \alpha \not\leq s \text{ for all $s \in S$} \}$.
\end{definition}

When $A$ is an ASL on $P$ and $I \subset P$, we write $AI$ for the (ring) ideal generated by $I$.

\begin{lemma}[{\cite[Prop.~5.2]{MR953963}}]\label{lemma: ASL order ideal intersection}
Let $A$ be an ASL on $P$ and let $I, J \subset P$ be ideals.
Then $AI \cap AJ = A(I \cap J)$.
\end{lemma}

We will prove a similar lemma for DASLs.
First, we introduce a partial order for dosets.

\begin{definition}
Let $D$ be a doset of $P$.
Then $D$ is a poset with the partial order $(a,b) \leq_1 (c,d)$ if and only if $a \leq c$ in $P$.

A \defining{doset order ideal} is an order ideal in the poset $(D,\leq_1)$.
As before, the ideal generated by $S \subset D$ is the smallest ideal containing $S$
and the ideal cogenerated by $S \subset D$ is the largest ideal disjoint from $S$.
\end{definition}

A DASL on $D$ is not necessarily an ASL on $(D,\leq_1)$.
Again when $A$ is a DASL on $D$ and $I \subset D$ is a doset order ideal,
we write $AI$ for the ring ideal generated by $I$.

\begin{lemma}
Let $A$ be a DASL on $D$ over $B$ and let $I \subset D$ be a doset order ideal.
Then $AI$ is spanned over $B$ by the standard monomials
$N = (\beta_1,\beta_2)\dotsm(\beta_{2\ell-1},\beta_{2\ell})$
such that $(\beta_1,\beta_2) \in I$.
\end{lemma}
\begin{proof}
Let $(\alpha_1,\alpha_2) \in I$, $f \in A$, and let
$N = (\beta_1,\beta_2)\dotsm(\beta_{2\ell-1},\beta_{2\ell})$
be one of the standard monomials appearing in the standard representation of $(\alpha_1,\alpha_2) f$.
The sequence $(\beta_1,\beta_2,\dotsc,\beta_{2\ell})$ is lexicographically less than or equal
to $(\alpha_1,\alpha_2)$,
so in particular $\beta_1 \leq \alpha_1$.
Hence $(\beta_1,\beta_2) \leq_1 (\alpha_1,\alpha_2)$ and hence $(\beta_1,\beta_2) \in I$.
Thus every standard monomial appearing in every element of $AI$ has a factor in $I$.
\end{proof}

\begin{lemma}\label{lemma: DASL order ideal intersection}
Let $A$ be a DASL on $D$ and let $I, J \subset D$ be ideals.
Then $AI \cap AJ = A(I \cap J)$.
\end{lemma}
\begin{proof}
A standard monomial $N = (\beta_1,\beta_2)\dotsm(\beta_{2\ell-1},\beta_{2\ell})$
appearing in the standard representation of an element of $AI \cap AJ$
has $(\beta_1,\beta_2) \in I$ and $\in J$, hence in $I \cap J$.
This shows $AI \cap AJ \subset A(I \cap J)$ and the reverse inclusion is obvious.
\end{proof}

Finally we recall the following results.

\begin{prop}[{\cite[Thm.~6.3]{MR953963}}]
Let $B$ be a domain, $X$ a generic matrix, $A = B[X]$, and $\delta \in \Delta(X)$,
the poset of minors (see Example~\ref{example: posets of minors}(1)).
Let $I(X,\delta)$ be the ideal in $A$ generated by the order ideal cogenerated by $\delta$.
Then $I(X,\delta)$ is a prime ideal.
\end{prop}

\begin{prop}[{\cite[Theorem 1]{MR0352082}, \cite[Remark~2.5(a)]{MR1279266}}]
\label{prop: doset 1-cogenerated prime}
Let $B$ be a domain, $Y$ a generic symmetric matrix, $A = B[Y]$,
and $\delta \in \Delta^s(Y)$,
the doset of minors (see Example~\ref{example: posets of minors}(2)).
Let $I(Y,\delta)$ be the ideal in $A$ generated by the doset order ideal cogenerated by $\delta$.
Then $I(Y,\delta)$ is a prime ideal.
\end{prop}

\begin{prop}[{\cite[Thm.~2.1.12]{MR2298637}}]
Let	$B$ be a domain, $Z$ a generic skew-symmetric matrix, $A = B[Z]$,
and $\delta \in \Pi(Z)$,
the poset of Pfaffians (see Example~\ref{example: posets of minors}(3)).
Let $I(Z,\delta)$ be the ideal in $A$ generated by the order ideal cogenerated by $\delta$.
Then $I(Z,\delta)$ is a prime ideal.
\end{prop}

\section{Minors}\label{section: minors}

We are interested in ideals generated by certain sets of $t$-minors in a generic matrix $X$.
Specifically, we will require the generating minors to have at least $r_1$ rows in the first $R_1$ rows of $X$,
at least $r_2$ rows contained in the first $R_2$ rows of $X$, and so on;
and similarly for columns.

Let $X$ be a generic $m \times n$ matrix, $X = (x_{i,j})$ for $1 \leq i \leq m$, $1 \leq j \leq n$,
and fix $A = B[X] = B[\{x_{i,j}\}]$ for a commutative ring $B$ with unity.

For $1 \leq t \leq \min(m,n)$, a $t$-minor may be specified by listing its rows and columns;
we write $[a_1,\dotsc,a_t \mid b_1,\dotsc,b_t]$,
where $1 \leq a_1 < \dotsb < a_t \leq m$ and $1 \leq b_1 < \dotsb < b_t \leq n$,
for the minor with rows $a_1,\dotsc,a_t$ and columns $b_1,\dotsc,b_t$.

Fix sequences $1 \leq R_1 \leq \dotsb \leq R_p \leq m$
and $1 \leq C_1 \leq \dotsb \leq C_q \leq n$
where $p, q \geq 0$.
The sequences $R = (R_1,\dotsc,R_p)$ and $C = (C_1,\dotsc,C_q)$
(possibly empty if $p=0$ or $q=0$)
describe the division of $X$ into row and column blocks, respectively.
Specifically, let $X_{R_i}$ be the submatrix of $X$ consisting of the first $R_i$ rows
and let $X^{C_j}$ be the submatrix consisting of the first $C_j$ columns.
Fix also sequences $r = (r_1,\dotsc,r_p)$ and $c = (c_1,\dotsc,c_q)$.

We are interested in the $t$-minors that have at least $r_i$ rows contained in $X_{R_i}$
and at least $c_j$ columns contained in $X^{C_j}$, for each $i, j$
(with no restriction if $p=0$ or $q=0$).
\begin{thm}\label{thm: minors}
Let $B$ be a ring and $A = B[X]$.
Let $J = J(X,t,R,C,r,c)$ be the ideal generated by $t$-minors of $X$
that have at least $r_i$ rows contained in $X_{R_i}$ for each $1 \leq i \leq p$ (no restriction if $p=0$)
and at least $c_j$ columns contained in $X^{C_j}$ for each $1 \leq j \leq q$ (no restriction if $q=0$).
Then
\begin{equation}\label{eq: minor decomposition}
  J = I_t(X) \cap I_{r_1}(X_{R_1}) \cap \dotsb \cap I_{r_p}(X_{R_p})
             \cap I_{c_1}(X^{C_1}) \cap \dotsb \cap I_{c_q}(X^{C_q}) .
\end{equation}
\end{thm}

\begin{example}
When $p=q=0$, $J = I_t(X)$.

When $p=0$ and $q=1$, we are in the two-block setting of the Introduction.
If also $c_1=C_1$, we recover \cite[Thm.~A]{MR853243}.
\end{example}

\begin{remark}
We are essentially working with the special case of Mohammadi's block adjacent simplicial complexes \cite{Mohammadi:2012fk}
in which each block is contained in the previous one and they all have the last column of the matrix
as a common endpoint (in Mohammadi's indexing; for us, we take blocks to start at the first column or row).
Unlike Mohammadi, we allow non-maximal minors,
we allow restrictions on both the rows and columns appearing in the minor,
and we allow the overlaps between ``consecutive'' blocks to be arbitrarily large.

\end{remark}

\begin{proof}
Each of the following sets of minors is an order ideal in $\Delta(X)$:
\begin{enumerate}
\item The set of minors of size $\geq t$, the generating set of $I_t(X)$, is the order ideal generated
by $[m-t+1,\dotsc,m \mid n-t+1,\dotsc,n]$,
or cogenerated by $[1,\dotsc,t-1 \mid 1,\dotsc,t-1]$.
\item The set of $(\geq r_i)$-minors of $X_{R_i}$ is the order ideal generated by
$[R_i-r_i+1,\dotsc,R_i \mid n-r_i+1,\dotsc,n]$,
or cogenerated by $[1,\dotsc,r_i-1,R_i+1,\dotsc,n \mid 1,\dotsc,n-R_i+r_i-1]$.
\item Similarly, the set of $(\geq c_j)$-minors of $X^{C_j}$ is the order ideal generated by
$[m-c_j+1,\dotsc,m \mid C_j-c_j+1,\dotsc,C_j]$,
or cogenerated by $[1,\dotsc,n-C_j+c_j-1 \mid 1,\dotsc,c_j-1,C_j+1,\dotsc,n]$.
\end{enumerate}
By Lemma~\ref{lemma: ASL order ideal intersection},
the intersection of the ideals generated by these sets
is equal to the ideal generated by the intersection of the sets.
\end{proof}

Note, if $B$ is a domain this gives $J$ as an intersection of prime ideals.
However it may fail to be a primary decomposition of $J$,
as redundancies may arise in the following ways.
For example, if $r_j > R_j - R_i$ then every minor containing
at least $r_j$ rows of $X_{R_j}$ must contain at least $r_j - (R_j-R_i)$ rows of $X_{R_i}$;
now if $r_i \leq r_j - R_j + R_i$ then the condition imposed by $r_i$ is implied by the $r_j$ condition
and the prime ideal $I_{r_i}(X_{R_i})$ is redundant.
Or if $t - (m-R_i) \geq r_i$ then every $t$-minor has at least $r_i$ rows in $X_{R_i}$.
Finally there are a few trivial situations: if $r_i > R_i$ the whole thing is zero;
if $R_i = R_j$ or $r_i = r_j$ then one condition is obviously redundant.
These are the only possible redundancies as the following proposition shows.
\begin{prop}
Suppose
\begin{enumerate}
\item $R_1 < \dotsb < R_p$ and $C_1 < \dotsb < C_q$,
\item $r_1 < \dotsb < r_p < t$ and $c_1 < \dotsb < c_q < t$,
\item $0 \leq r_i \leq R_i$ for each $i$ and $0 \leq c_j \leq C_j$ for each $j$,
\item $R_1 - r_1 < \dotsb < R_p - r_p < m-t$ and $C_1 - c_1 < \dotsb < C_q - c_q < n-t$.
\end{enumerate}
Then the intersection \eqref{eq: minor decomposition} is irredundant.
\end{prop}
\begin{proof}
First, fix $1 \leq i \leq p$.
Consider the $t$-minor
\[
  m = [1,\dotsc,r_i-1,R_i+1,\dotsc,t+R_i-r_i+1 \mid 1,\dotsc,t] .
\]
We use $R_i - r_i < m-t$ to verify $t+R_i-r_i+1 \leq m$, so this is a permissible $t$-minor in an $m \times n$ matrix.
For each $j < i$, $m$ has exactly $\min(r_i-1,R_j)$ rows in $X_{R_j}$, and this is $\geq r_j$,
so $m \in I_{r_j}(X_{R_j})$.
For each $j > i$, the number of rows of $m$ in $X_{R_j}$ is either $t$, if $t+R_i-r_i+1 \leq R_j$,
or else $R_j-R_i+r_i-1$, if $R_i+1 \leq R_j \leq t+R_i-r_i+1$.
In the first case $t \geq r_j$ and in the second case $R_j - r_j > R_i - r_i$, so $R_j - R_i + r_i - 1 \geq r_j$;
therefore $m \in I_{r_j}(X_{R_j})$.
And clearly $m$ has only $r_i-1$ rows in $X_{R_i}$.
This shows that
\[
  m \in I_t(X) \cap I_{r_1}(X_{R_1}) \cap \dotsb \cap I_{r_{i-1}}(X_{R_{i-1}})
    \cap I_{r_{i+1}}(X_{R_{i+1}}) \cap \dotsb \cap I_{r_p}(X_{R_p})
\]
but $m \notin I_{r_i}(X_{R_i})$.
Clearly $m \in \bigcap I_{c_j}(X^{C_j})$.
So the term $I_{r_i}(X_{R_i})$ is irredundant for each $i$.
The same argument shows that each $I_{c_j}(X^{C_j})$ is irredundant.

Finally consider the $(t-1)$-minor
\[
  m' = [1,\dotsc,t-1 \mid 1,\dotsc,t-1] .
\]
Since each $r_i < t$, $m'$ has at least $r_i$ rows in each $X_{R_i}$ and similarly at least $c_j$ columns in each $X^{C_j}$.
This shows that the term $I_t(X)$ is irredundant.
\end{proof}

\section{Minors of symmetric matrices}\label{section: symmetric minors}

Now let $Y = (y_{i,j})$ be a generic symmetric $n \times n$ matrix, $y_{i,j} = y_{j,i}$.
Fix sequences $R = (R_1,\dotsc,R_p)$ with $1 \leq R_1 \leq \dotsb \leq R_p \leq n$
and $r = (r_1,\dotsc,r_p)$.
Let $Y_{R_i}$ be the submatrix consisting of the first $R_i$ rows of $Y$.
We are interested in the $t$-minors that have at least $r_i$ rows in $Y_{R_i}$ for each $i$.
Note, at this point we allow all minors, not only doset minors.

\begin{thm}\label{thm: symmetric minors}
Let $B$ be a ring and $A = B[Y]$.
Let $J = J(Y,t,R,r)$ be the ideal generated by $t$-minors of $Y$ that have
at least $r_i$ rows contained in $Y_{R_i}$ for each $1 \leq i \leq p$.
Then
\begin{equation}\label{eq: symmetric minor decomposition}
  J = I_t(Y) \cap I_{r_1}(Y_{R_1}) \cap \dotsb \cap I_{r_p}(Y_{R_p}) .
\end{equation}
If $B$ is a domain then each $I_{r_i}(Y_{R_i})$ is a prime ideal.
\end{thm}
\begin{proof}
First, by \cite[Lemma~2.3]{MR1279266}, every $t$-minor $[a \mid b]$
is a linear combination of doset $t$-minors $[c \mid d]$ with $c \leq a$.
Thus we can take $J$ to be generated by the doset $t$-minors meeting the row conditions.

Next, each of the following sets is a doset order ideal in $\Delta^s(Y)$:
\begin{enumerate}
\item The set of doset minors of size $\geq t$ is the doset order ideal
generated by $[n-t+1,\dotsc,n \mid n-t+1,\dotsc,n]$.

\item The set of doset $(\geq r_i)$-minors of $Y_{R_i}$ is the doset order ideal
generated by $[R_i-r_i+1,\dotsc,R_i \mid n-r_i+1,\dotsc,n]$.
If $[a \mid b] \leq_1 [R_i-r_i+1,\dotsc,R_i \mid n-r_i+1,\dotsc,n]$
then $[a \mid b]$ involves at least $r_i$ rows of $Y_{R_i}$;
by Laplace expansion and \cite[Lemma~2.3]{MR1279266},
$[a \mid b]$ is a linear combination of doset $r_i$-minors of $Y_{R_i}$.
\end{enumerate}
This shows that $J$ is the indicated intersection.

The set of doset $(\geq t)$-minors of $Y$ is
cogenerated by $[1,\dotsc,t-1 \mid 1,\dotsc,t-1]$.
The set of doset $(\geq r_i)$-minors of $Y_{R_i}$
is cogenerated by $m = [1,\dotsc,r_i-1,R_i+1,\dotsc,n \mid 1,\dotsc,r_i-1,R_i+1,\dotsc,n]$
\cite[Remark~2.5(c)]{MR1279266}.
Indeed, $[a \mid b] \not \geq_1 m$ if and only if
$a \not \geq (1,\dotsc,r_i-1,R_i+1,\dotsc,n)$,
if and only if $|a| \geq r_i$ and $a_{r_i} \leq R_i$;
so $[a \mid b]$ involves at most $r_i$ rows of $Y_{R_i}$.
This shows that each of the ideals being intersected is cogenerated by a single doset element.
Therefore if $B$ is a domain then each of them is a prime ideal by Proposition~\ref{prop: doset 1-cogenerated prime}.
\end{proof}

If $B$ is a domain then once again this writes $J$ as an intersection of prime ideals,
but as before it may fail to be a primary decomposition because of redundancy.

\begin{prop}
Suppose
\begin{enumerate}
\item $R_1 < \dotsb < R_p$,
\item $r_1 < \dotsb < r_p < t$,
\item $0 \leq r_i \leq R_i$ for each $i$,
\item $R_1 - r_1 < \dotsb < R_p - r_p < n-t$.
\end{enumerate}
Then the intersection \eqref{eq: symmetric minor decomposition} is irredundant.
\end{prop}
The proof is the same as before.

\section{Pfaffians}\label{section: pfaffians}

Let $Z = (z_{i,j})$ be an $n \times n$ generic skew-symmetric matrix,
so that $z_{i,j} = -z_{j,i}$ and $z_{i,i}=0$,
and let $A = B[Z] = B[\{z_{i,j}\}]$.
The Pfaffian of $Z$, denoted $\Pf(Z)$, is a certain polynomial in the entries of $Z$,
with the property that $\Pf(Z)^2 = \det(Z)$.
When $n$ is odd, $\Pf(Z) = \det(Z) = 0$; for $n = 2,4$ we have
\[
  \Pf\begin{pmatrix} 0 & a \\ -a & 0\end{pmatrix} = a,
  \qquad
  \Pf\begin{pmatrix} 0 & a & b & c \\ -a & 0 & d & e \\ -b & -d & 0 & f \\ -c & -e & -f & 0\end{pmatrix} = af-be+cd .
\]
In general, for $n$ even,
\[
  \Pf(Z) = \sum \operatorname{sgn}(\sigma) z_{\sigma(1),\sigma(2)} \dotsm z_{\sigma(n-1),\sigma(n)} ,
\]
where the sum is over all permutations $\sigma \in S_n$ such that $\sigma(2i-1) < \sigma(2i)$ for all $i$
and $\sigma(1) < \sigma(3) < \dotsb < \sigma(2n-1)$.
Equivalently, the sum is over all unordered partitions of $\{1,\dotsc,2n\}$ into pairs;
the restrictions on $\sigma$ simply amount to choosing one representative ordering for each partition.
There is a Laplace-like expansion: for each $j$, $1 \leq j \leq n$,
\[
  \Pf(Z) = \sum_{i<j} (-1)^{i+j+1} z_{i,j} \Pf(Z^{i,j}) + \sum_{i > j} (-1)^{i+j} z_{i,j} \Pf(Z^{i,j}) ,
\]
where $Z^{i,j}$ is the matrix obtained by deleting the $i$th and $j$th rows and columns of $Z$.
See \cite{MR0453723,MR0257105,MR554859}.

A $t$-Pfaffian of $Z$ is given by a list of $t$ rows and the same columns;
we write briefly $[a_1,\dotsc,a_t]$, where $1 \leq a_1 < \dotsb < a_t \leq n$,
for the Pfaffian of the skew-symmetric submatrix
given by the rows $a_1,\dotsc,a_t$ and the same columns.
Of course this is zero if $t$ is odd.

The ideal generated by the size $t$ Pfaffians of $Z$ is denoted $P_t(Z)$.
If $n$ is odd then $P_{n-1}(Z)$ is a prime ideal of height $3$.
More generally, $P_{2p}(Z)$ is a prime ideal of height $\mu(p,n) = (n-2p+1)(n-2p+2)/2$, see \cite{MR554859}.

We are interested in the ideal generated by the subset of Pfaffians with at least $r_1$ rows
in the first $R_1$ rows of $Z$, at least $r_2$ rows in the first $R_2$ rows of $Z$, and so on;
the row condition implies that these Pfaffians meet the corresponding column conditions as well,
i.e., at least $r_1$ columns in the first $R_1$ columns of $Z$, and so on.

Fix a sequence $1 \leq R_1 \leq \dotsb \leq R_p \leq n$, $R = (R_1,\dotsc,R_p)$,
and another sequence $r = (r_1,\dotsc,r_p)$ of the same length.
Let $Z_{R_i}$ be the submatrix of $Z$ consisting of the first $R_i$ rows
and let $Z_{R_i}^{R_i}$ be the $R_i \times R_i$ submatrix of $Z$ in the upper left corner,
consisting of the first $R_i$ rows and the first $R_i$ columns.
We will also need, for each $R_i+1 \leq k \leq n$,
the $(R_i+1) \times (R_i+1)$ submatrix given by the
first $R_i$ rows and columns plus the $k$th row and column,
that is, the set of rows (and columns) corresponding to the set $\{1,\dotsc,R_i,k\}$.
Recall the common notation $[R_i] = \{1,\dotsc,R_i\}$, so we may write $[R_i] \cup \{k\}$ for the set we want.
To simplify notation, we write $Z([R_i])$ for $Z_{R_i}^{R_i}$
and we write $Z([R_i] \cup \{k\})$ for the $(R_i+1) \times (R_i+1)$ skew-symmetric submatrix of $Z$
given by the rows (and columns) corresponding to the set $\{1,\dotsc,R_i,k\}$.
Since confusion seems unlikely we will drop the brackets and braces
and simply write $Z(R_i)$ and $Z(R_i \cup k)$.
Thus for example
\[
  Z(3\cup5) =
    \begin{pmatrix}
      0 & z_{1,2} & z_{1,3} & z_{1,5} \\
      -z_{1,2} & 0 & z_{2,3} & z_{2,5} \\
      -z_{1,3} & -z_{2,3} & 0 & z_{3,5} \\
      -z_{1,5} & -z_{2,5} & -z_{3,5} & 0
    \end{pmatrix} ,
\]
with rows and columns given by the set $3 \cup 5 = [3] \cup \{5\} = \{1,2,3,5\}$.
\begin{thm}\label{thm: pfaffians}
Let $B$ be a ring and $A = B[Z]$.
Let $J = J(Z,2t,R,r)$ be the ideal generated by $2t$-Pfaffians of $Z$
that have at least $r_i$ rows in $Z_{R_i}$ for $1 \leq i \leq p$.
For each $i$, if $r_i$ is even, let $J_i = P_{r_i}(Z(R_i))$,
and if $r_i$ is odd, let $J_i = \sum_{k=R_i+1}^n P_{r_i+1}(Z(R_i \cup k))$.
Then
\begin{equation}\label{eq: pfaffian decomposition}
  J = P_{2t}(Z) \cap J_1 \cap \dotsb \cap J_p .
\end{equation}
If $B$ is a domain then $P_{2t}(Z)$ is prime and each $J_i$ is a prime ideal.
\end{thm}

\begin{proof}
Each of the following is an order ideal in $\Pi(Z)$:
\begin{enumerate}
\item The set of $2t$-Pfaffians is the order ideal generated by $[n-2t+1,\dotsc,n]$.
\item The set of Pfaffians (of all sizes) with at least $r_i$ rows contained in $Z_{R_i}$.
If $r_i$ is even, this is the order ideal generated by $[R_i-r_i+1,\dotsc,R_i]$.
If $r_i$ is odd, this is the order ideal generated by $[R_i-r_i+1,\dotsc,R_i,n]$.
\end{enumerate}
So, by Lemma~\ref{lemma: ASL order ideal intersection}, $J$ is equal to the intersection
of the ideals $P_{2t}(Z)$ and, for each $i$, the ideal generated by the Pfaffians (of any size)
having at least $r_i$ rows in $Z_{R_i}$.

If $r_i$ is even then the ideal generated by Pfaffians
with at least $r_i$ rows in $Z_{R_i}$ is $P_{r_i}(Z(R_i))$.
Indeed, if $P$ is any Pfaffian with at least $r_i$ rows in $Z_{R_i}$ then
$P$ can be expanded as a combination of $r_i$-Pfaffians
involving those rows.

If $r_i$ is odd and $P$ is any Pfaffian with at least $r_i$ rows in $Z_{R_i}$,
then either $P$ actually has at least $r_i+1$ rows in $Z_{R_i}$
or else $P$ involves at least one more row, say the $k$th row, with $k > R_i$.
Either way, $P$ can be expanded as a combination of $(r_i+1)$-Pfaffians in $Z(R_i \cup k)$.
So $P$ lies in the sum given in the statement.
Conversely, every Pfaffian generator of the sum in the statement must have at least $r_i$ rows in $Z_{R_i}$.

Now suppose $B$ is a domain.
The set of $(\geq 2t)$-Pfaffians of $Z$ is cogenerated by $[1,\dotsc,2t-2]$.
This shows $P_{2t}(Z)$ is prime.
(Of course $P_{2t}(Z)$ is already well-known to be prime.)

To see that each $J_i$ is prime, note that the order ideal of Pfaffians generating $J_i$ is cogenerated by
either $m = [1,\dotsc,r_i-1,R_i+1,\dotsc,n]$ or $m' = [1,\dotsc,r_i-1,R_i+1,\dotsc,n-1]$,
whichever has even length (regardless of whether $r_i$ is even or odd).
Let us verify this.
For simplicity, suppose that $m$ has even length.
We must show that $\alpha \not \geq m$ if and only if $\alpha$ has at least $r_i$ rows in $Z_{R_i}$,
equivalently $\alpha \geq m$ if and only if $\alpha$ has $r_i-1$ or fewer rows in $Z_{R_i}$;
note that this is the criterion whether $r_i$ is even or odd.
Now $\alpha \geq m$ if and only if $|\alpha| \leq r_i-1$ or $|\alpha| \geq r_i$ and $\alpha_{r_i} \geq R_i + 1$.
The forward direction is obvious; conversely, under these conditions,
$\alpha_{r_i+t} \geq \alpha_{r_i} + t \geq R_i + 1 + t = m_{r_i + t}$ for all $0 \leq t \leq |\alpha|-r_i$,
so each entry of $\alpha$ is at least as great
as the corresponding entry of $m$; and in particular since every entry of $\alpha$ is at most $n$, $|\alpha| \leq |m|$.
This shows that $\alpha \geq m$.
So indeed $\alpha \geq m$ if and only if $\alpha$ has $r_i-1$ or fewer rows in $Z_{R_i}$.

The argument in case $|m'|$ is even is similar.
Note that $m'$ is as long as possible for a member of $\Pi(Z)$ with $R_i+1$ in the $r_i$ position;
so if $\alpha_{r_i} \geq R_i+1$ then $|\alpha| \leq |m'|$.
\end{proof}

Once again this writes $J$ as a possibly redundant intersection of prime ideals, if $B$ is a domain.
\begin{prop}
Suppose
\begin{enumerate}
\item $R_1 < \dotsb < R_p$,
\item $r_1 < \dotsb < r_p < 2t$,
\item $0 \leq r_i \leq R_i$ for each $i$,
\item $R_1 - r_1 < \dotsb < R_p - r_p < n-2t$.
\end{enumerate}
Then the intersection \eqref{eq: pfaffian decomposition} is irredundant.
\end{prop}
The proof is the same as before.

\begin{cor}
Let $Z = (z_{i,j})$ be a generic skew-symmetric matrix,
let $0 < p < q$,
and fix $A = B[Z]$ with degree $\deg z_{i,j} = 2p$ if $i,j \leq R$,
$\deg z_{i,j} = p+q$ if $i \leq R < j$,
and $\deg z_{i,j} = 2q$ if $i,j > R$.
Fix $t$ and $d$.
Let $r = \left\lceil \frac{2tq - d}{q-p} \right\rceil$.
Then $P_{2t}(Z)_{\leq d} = P_{2t}(Z) \cap I_r$
where $I_r$ is the ideal generated by Pfaffians with at least $r$ rows in $Z_R$.
If $r$ is even, $I_r = P_r(Z(R))$.
If $r$ is odd, $I_r = \sum_{k=R+1}^n P_{r+1}(Z(R \cup k))$
where $Z(R \cup k)$ is the $(R+1) \times R+1$ skew-symmetric submatrix of $Z$ given by the rows (and columns)
corresponding to the set $\{1,\dotsc,R,k\}$.
If $B$ is a domain then $P_{2t}(Z)$ and $I_r$ are prime.
\end{cor}
Indeed, a $2t$-Pfaffian $P$ with $r$ rows in $Z_R$ has degree $pr+q(2t-r)$;
the value of $r$ in the statement is the least integral solution to $\deg P \leq d$.

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TITLE: example of a open function such that the restriction is not open
QUESTION [4 upvotes]: Give an example of a function $f:X \to Y$ and  a subset $A \subset X$ such that f is open but $f_A$, the restriction of $f$ to $A$ is not open.
Can someone help me please? Thanks

REPLY [13 votes]: HINT: Let $X=Y=\Bbb R$, take $f$ to be the identity function, and take $A$ to be a set that isn’t open in $X$.


TITLE: Show no open ball in $C[0,1]$ is contained in the space of all Lipschitz-continuous functions on $[0,1]$
QUESTION [2 upvotes]: This is a question I encountered in a textbook exercise. Here $C[0,1]$ is the space of all real-valued continuous functions on $[0,1]$ endowed with the sup metric, $d(f,g):=\sup_{x\in[0,1]}|f(x)-g(x)|$ for $f,g\in C[0,1]$.
Let $f_0$ be an arbitrary element of $C[0,1]$; for an $\varepsilon>0$, let $B(f_0, \varepsilon):=\{f\in C[0,1]:d(f,g)<\varepsilon\}$ be an open ball in $C[0,1]$. When $f_0$ is not Lipschitz-continuous, the claim obviously follows. Suppose $f_0$ is Lipschitz-continuous. For me, the claim seems to require finding a continuous function that is 'close' to $f_0$ (contained in $B(f_0,\varepsilon)$) but not Lipschiz-continuous. I don't have any clue now.

REPLY [1 votes]: Choose some $f_0 \in C[0,1]$, and $\epsilon>0$. Let $f_1$ be a piecewise affine function such that $\|f_0-f_1\|_\infty < {\epsilon \over 2}$. (Actually, $f_1$ just needs to be smooth in the neighbourhood of $0$ for this proof, a nearby polynomial would work fine too.)
Now let $f_2(x) = f_1(x)+ {\epsilon \over 2} \sqrt{x}$. Clearly $\|f_1-f_2\|_\infty \le {\epsilon \over 2}$, and $f_2$ is not Lipschitz at $x=0$.
Finally, we have $\|f_2-f_0\|_\infty \le \|f_2-f_1\|_\infty  + \|f_1-f_0\|_\infty  < \epsilon$.


TITLE: How many different two-scoop cones are there from 10 flavors?
QUESTION [0 upvotes]: In the local ice cream shop, there are 10 different flavors. How many
  different two-scoop cones are there? Here, the order does not matter.
  A cone with a vanilla scoop on top of a chocolate scoop is considered
  the same as a cone with a chocolate scoop on top of a vanilla scoop.

Here's my approach so far:
I'm choosing 2 flavors from 10, so I use ${10\choose 2} = 45$ different two-scoop cones.

Now suppose that the order of the scoops does matter. How many
  different possible two-scoop cones are there?

This time, there are even more cones that have been excluded previously, and I use permutation:
$\frac{10!}{8!} = 90$ different two-scoop cones.
My issue is that I was told the answers were 100 and 55, respectively.
The only way I could assume 100 is correct for the first one is if you could repeat flavors. That way the number of different cones would be $10*10 = 100$. But if that's the case, then how would I calculate 55 for the second problem? My method seems to fall apart.
Any guidance would be appreciated, thank you.

REPLY [0 votes]: I think it's $10!/6$, it's divided by six for three reasons,

prevent doubles

reverse of same combinations

pair with 2. so the answer is $604800$.


TITLE: Show that if $x^2 + y^2 = z^2$, then at leeast one of x and y is a multiple of 3.
QUESTION [3 upvotes]: Show that if $x^2 + y^2 = z^2$, then at least one of $x$ and $y$ is a multiple of 3.
Attempt:
Given $x,y,z$ is pyhagorean triple,
$$
x^2+y^2=z^2.
$$
Assume neither $x$ nor $y$ is divisible by $3$. Since $x^2+y^2=z^2$, $x^2+y^2 \equiv z^2 \pmod 3$. But neither $x$ nor $y$ is congruent to $0 \pmod 3$. Therefore,
\begin{align}
x \equiv \pm 1 &\pmod 3 \\
y \equiv \pm 1 &\pmod 3.
\end{align}
Hence,
\begin{align}
x^2 \equiv 1 &\pmod 3 \\
y^2 \equiv 1 &\pmod 3,
\end{align}
and so
$$
x^2+y^2 \equiv 2 \pmod 3.
$$
But $2$ is not a square modulo $3$. Therefore,
$$
2 \not\equiv z^2 \pmod 3,
$$
which is a contradiction.

REPLY [3 votes]: Suppose none of $x,y,z$ is divisible by $3$.
then
$$x\equiv \pm 1\mod 3,$$
$$y\equiv \pm 1 \mod 3,$$
and
$$z\equiv \pm 1 \mod 3$$
$\implies$
$$x^2+y^2-z^2\equiv 1+1-1\mod 3$$
$$\implies x^2+y^2\neq z^2.$$


TITLE: Conectionless parallel transport
QUESTION [2 upvotes]: In books I read, the notion of parallel transport along a curve on manifold is defined in terms of connection. Then we have a theorem, which states, that we can recover connection given parallel transport.
So, the question is whether there is a simple enough definition of parallel transport, that does not depend on connection. Especially interesting is the case of riemannian manifold.
I think the question is important, because parallel transport is fundamental concept, which can provide helpful insight towards understanding connections. Of course, that is if we can properly define it.
Thanks.

REPLY [2 votes]: There is a definition of parallel transport, without reference to an infinitesimal notions such as covariant derivative.
Let $M$ be a manifold. Consider the $\operatorname{Path}(M)$ the collection of smooth paths in $M,$ modulo some reparametrization invariance. A parallel transport is a function $\Gamma$ from paths $x\to y$ to linear maps $T_xM\to T_yM,$ such that

If $\alpha$ is a constant path with $\alpha(t)=x$, then $\Gamma(\alpha)=1_{T_xM}.$
If $\alpha$ is a path from $x$ to $y$, and $\beta$ is a path from $y$ to $z$, and $\alpha\cdot\beta$ is the concatenation path from $x$ to $z$, then $\Gamma(\alpha\cdot\beta)=\Gamma(\alpha)\circ \Gamma(\beta).$
$\Gamma(\alpha)$ depends smoothly on $\alpha.$

The right language to formulate this definition is that of groupoids. The paths on $M$ form a groupoid, and the linear isomorphisms between tangent spaces form another groupoid, and a parallel transport is a groupoid homomorphism from the first groupoid to the second.
And with appropriate smoothness conditions, they are both Lie groupoids, and the parallel transport is a smooth groupoid homomorphism, whose derivative is a covariant derivative.
The article at nlab has more.


TITLE: Does there exist a unital ring whose underlying abelian group is $\mathbb{Q}^*$?
QUESTION [5 upvotes]: Let $\mathbb{Q}^*$ be the group of units of the rational numbers.
Does there exist a unital ring whose underlying additive group is $\mathbb{Q}^*$?
I don't really have a gut feeling yea or nea.
Hope I'm not overlooking something obvious.
Thanks!

REPLY [9 votes]: By unique prime factorization of integers, the multiplicative group $\newcommand{\Q}{\mathbb{Q}}\Q^\times$ is isomorphic to the additive group of $\newcommand{\Z}{\mathbb{Z}}(\Z/2\Z) \times \bigoplus_{p \text{ prime}} \Z$. The isomorphism comes from the fact that every rational number can be written uniquely as $\pm \prod_{p \text{ prime}} p^{a_p}$ for some integers $a_p$, only finite many of which are nonzero; and conversely, every such expression corresponds to a rational number. But now we run into a difficulty: $(\Z/2\Z) \times \bigoplus_{p \text{ prime}} \Z$ with the obvious multiplication doesn't have a multiplicative identity, so while it's an associative $\Z$-algebra, it's not a ring.
But, there is a non-obvious multiplication that works! The additive group of the polynomial ring $\Z[x]$ is isomorphic to a countable direct sum of copies of $\Z$, and so (upon choosing a bijection between the natural numbers and the set of prime numbers), the additive group of $(\Z/2\Z) \times \Z[x]$ is isomorphic to the additive group $(\Z/2\Z) \times \bigoplus_{p \text{ prime}} \Z$.
Composing these isomorphisms, we see that the group of units $\Q^\times$ is isomorphic to the additive group of the ring $(\Z/2\Z) \times \Z[x]$.


TITLE: Why do you draw the triangle as in the picture below and not in any other way?
QUESTION [0 upvotes]: In the closed area limited by the graph $y = 4-x^2$, an isosceles triangle is inscribed. The triangle has its top angle in the origin and its base is parallel to the x-axis. Decide the triangle's maximal area.

My question is the following: Why do you draw the triangle as in the picture below and not in any other way?

REPLY [1 votes]: Set the top vertex in the origin. "In the closed area limited by the graph" sounds a little bit sloppy formulated, because the graph itself does not limit any closed area. I guess they mean "limited by the graph and the $x$-axis". Then the base should be above the $x$-axis and below the line $y=4$. "Parallel to $x$-axis" - as it says. "Inscribed" means two other vertices on the parabola. Otherwise, no more restrictions. The base on the picture goes through $y=3$, but it is just a coincidence. You can draw it any other way, it is simply an example of a possible triangle (not likely the one that maximizes the area). When you vary the base all the possible ways between the $x$-axis and the line $y=4$ you'll get different triangles, and you are to find the one with the largest area.


TITLE: What is an extragradient method?
QUESTION [6 upvotes]: I've searched Google, but it seems that only research journal papers appear in search results, where some new, improved, or specialized extragradient method is discussed. I've also searched Wikipedia and Wolfram MathWorld.
I would like to perhaps know the straightforward definition of that, instead of deducing it from a number of articles, where the basic/original method is not defined.

REPLY [2 votes]: A good discussion is in this recent preprint: Proximal Reinforcement Learning. They use the same definition as in dragonxlwang's answer, and discuss its relation to the mirror-prox method.


TITLE: Automorphism group of $UT(n,p)$, the group of unitriangular matrices over the field $\mathbb{F}_p$
QUESTION [6 upvotes]: I am interested in understanding (at least roughly, if no such a description exists) the group of automorphisms for the group $UT(n,p)$, of unitriangular matrices over the field $\mathbb{F}_p$ on $p$-elements. Unfortunately the online searching I've carried out wasn't quite helpful. If no description exists, maybe some information how the automorphisms behave on the set of subgroups of it may be helpful too.
For instance when $n=3$, I know we have that any $\phi \in Aut(UT(3,p))$, permutes the pairs of non-commuting elements of $UT(3,p)$ (which in that case happens always to be a generating set) and the subgroups of order $p^2$.
So am wondering if something similar exists (of course modified somehow) in the general case too.
If the above sounds quite general, the case where $n=4$, is of great importance for me either. Any comment might be useful, and of course references too.
Thanks!

REPLY [3 votes]: for size = 3
https://groupprops.subwiki.org/wiki/Unitriangular_matrix_group:UT(3,p)#Automorphisms
Automorphisms
The automorphisms essentially permute the subgroups of order
containing the center, while leaving the center itself unmoved.
over $F_2$, any size: Maginnis J. S., (1993/11)."Outer Automorphisms of Upper Triangular Matrices." Journal of Algebra 161(2): 267-270.
Abstract: The outer automorphism group of the upper triangular matrices over the field of two elements is calculated. A. J. Weir (Proc. Amer. Math. Soc.6 (1955), 454-464) performed a similar calculation for Fields of odd characteristic, and we borrow the term extremal auto ....
Any field(may be even ring) International Journal of Algebra, Vol. 7, 2013, no. 15, 723 - 733
HIKARI Ltd,
The Automorphism Group of the Group of
Unitriangular Matrices over a Field1
Ayan Mahalanobis
Abstract. This paper finds a set of generators for the automorphism group
of the group of unitriangular matrices over a field. Most of this paper is an
exposition of the work of V.M. Lev˘chuk, part of which is in Russian. Some
proofs are of my own.
From the paper:
The automorphism group of the group of unitriangular matrices over a field
was studied by many [2–4]. In this direction, the first paper was in Russian,
published by Pavlov in 1953. Pavlov studies the automorphism group of unitriangular
matrices over a finite field of odd prime order. Weir [4] describes
the automorphism group of the group of unitriangular matrices over a finite
field of odd characteristic. Maginnis 3 describes it for the field of two elements
and finally Lev˘chuk 2 describes the automorphism group of the group
of unitriangular matrices over an arbitrary ring.
In this expository article, we shall study the automorphism group of the
group of unitriangular matrices over an arbitrary field F. There are two most
commonly used non-abelian finite p-groups in the literature. One is the group
1Research supported by a NBHM research grant.
724 Ayan Mahalanobis
of unitriangular matrices over a field and the other is the extra-special pgroups.
Weir [4] and Lev˘chuk 2 worked on the automorphism group of the
unitriangular matrices.

PS
I remember I have seen some papers with quite explicit description of automorphism, and I thought I collected such links in comments to my old MO-question: Representation theory of p-groups in particular upper tringular matrices over F_p, but it cannot find it now again, hm-m... I'll try to find more...


TITLE: If $B$ satisfies $A B = B A^{-1}$, show that $B^2$ is diagonalisable
QUESTION [2 upvotes]: Let $A$ be a $n \times n$ non-singular matrix having distinct eigenvalues. If $B$ is a matrix satisfying  $A B = B A^{-1}$, show that $B^2$ is diagonalisable.

Answer:
Let $\lambda_i, \ i=1,2,3, \cdots, n$ be the $n$ distinct eigenvalues.
Now, $AB=BA^{-1} \Rightarrow B=ABA $
But then how to proceed?

REPLY [2 votes]: We can show that $AB^2=ABABA=B^2A$. So $B^2$ and $A$ commute. Then the claim follows from these duplicates (replacing $B$ by $B^2$):
$AB=BA$. Prove $B$ is diagonalizable.
If $AB=BA$, show that $B$ is diagonalizable.
Indeed, we have $AB^2=(AB)B=ABABA$ and $B^2A=B(BA)=ABABA$.