TITLE: Constructing the projection of a projective module from its dual basis
QUESTION [2 upvotes]: Let $M$ be a (left) module over a not necessarily commutative algebra $A$.
We can say that $M$ is projective if, for some $n \in \mathbb{N}$,
i)  there exists a projection $P$ (i.e. $P \in M_n(A)$ and $P^2 = P$) such that
$$
P(A^n) = M,
$$
OR EQUIVALENTLY
ii) if we have a dual basis, that is to say, a set $\{m_{i}\in M \mid i = 1, \dots,n\}$, and a set $\{f_{i} \in \mathrm {Hom}(M,A)\mid i = 1, \dots,n\}$ such that for every $x \in M$,
$$
x = \sum_{i = 1}^n f_{i}(x)m_{i}.
$$
What I would like to know is: Given the dual basis, how can one construct the projection $P$?

REPLY [2 votes]: The definitions given in the question are for finitely generated projective modules (although they can both be adapted for arbitrary projectives).
Given the dual basis, there are module homomorphisms $f:M\to A^n$ given by
$$f(x)=(f_1(x),\dots,f_n(x)),$$
and $g:A^n\to M$ given by
$$g(a_1,\dots,a_n)=a_1m_1+\dots+a_nm_n.$$
The definition of a dual basis gives that $g\circ f=\text{id}_M$, so $f$ is a split monomorphism and $f\circ g:A^n\to A^n$ is an idempotent endomorphism of $A^n$ with image isomorphic to $M$.
So the projection you want is just $P=f\circ g$, or in other words
$$P(a_1,\dots,a_n)=(f_1(a_1m_1+\dots+a_nm_n),\dots,f_n(a_1m_1+\dots+a_nm_n)).$$


TITLE: Does the eigenvectors of Hermitian operator constitute a basis
QUESTION [2 upvotes]: Eigenvectors of a Hermitian operator corresponding to different eigenvalues are orthogonal. Even for a degenerate eigenvalue we can produce orthogonal eigenvectors in that eigensubspace.
Does this system of orthogonal vectors necessarily span the whole vector space, i.e., do they constitute a basis?

REPLY [0 votes]: Yes. Not only the eigenvectors of a Hermitian operator constitute a basis, but it is a complete basis, i.e., and function in the space where the operator acts, can be expanded in terms of this operator eigenfunctions. The latter fact is sometimes stated differently, as the resolution of identity, see here.
Note that the above is true even in case of degenerate spectrum, provided that one appropriately orthogonalizes the eigenfunctions corresponding to the degenerate eigenvalues (which can always be done).


text:
\section{Introduction}\label{section:introduction}

\IEEEPARstart{N}{a}tural signals have an underlying structure, an insight that has triggered a paradigm shift in the last fifteen years, and spurred fundamental  progress in estimation and inference. Compressive sensing \cite{CandesRombergTao_2006,Donoho_CompressedSensing2006} takes sparsity as the model of structure when a signal $\bX\in\mathbb{R}^n$ has a sparse representation in an appropriate basis, that is, $\bX = \Psi \bZ$ with $\Psi$ an $n\times n$ change of basis matrix and $\bZ\in \mathbb{R}^n$ a sparse vector with $p \ll n$ non-zero components.
For example, $\bX$ can represent a natural image and $\Psi$ a wavelet basis \cite{Mallat_book_1999}.
Despite its success, this model of structure is often too constrained because the appropriate basis  may be unknown and, more generally, the linearity of the transformation may be a severe limitation.
Deep networks have been proposed as an alternative \cite{MousaviPatel_2015} and,
with the advent of generative adversarial networks (GAN) \cite{GoodfellowGAN_2014} and variational auto-encoders (VAE) \cite{Hinton504},
such flexible and non-linear ``generative models'' of structure have been the object of intense interest. Roughly speaking, a generative model can be viewed as a mapping $G: \bS\in \mathbb{R}^p \mapsto \bX = G(\bS)\in \mathbb{R}^n$ with $p\ll n$ and satisfying certain general regularity assumptions \cite{Bora_2017}.
In other words, the signal $\bX$ lies on a low $p$-dimensional ``manifold'' parametrized by $\bS$.
Such models have been studied in the framework of classical denoising problems from observations $\bY=A \bX + \bZ$ where $A$ is a sensing matrix and $\bZ$ some Gaussian noise.
In particular, \cite{Bora_2017} studies fundamental limits under minimal Lipshitz conditions on $G$ and empirically investigates the problem with learned mappings coming from GAN and VAE
Another kind of generative model takes $G$ equal to a one-layer or multi-layer neural network with fixed weights (i.e., frozen and not learned) drawn from a random matrix ensemble \cite{ManoelKrzakala_2017, DBLP:journals/tit/HandV20, heckel2018rateoptimal, HandLeongVoroninski_2018, DBLP:journals/corr/abs-1803-09319}.
Such mappings $G$ are often referred to as \textit{generalized linear models} and this is the terminology that we adopt here.
The simplification of fixed random weights has the virtue of being much more amenable to mathematical (or at least analytical) analysis.
Especially, the mutual information as well as the message passing algorithmic behaviour for classical denoising have been discussed in depth in a Bayesian setting at various levels of rigor \cite{ManoelKrzakala_2017, Gabrie_TwoLayerGLM_JSTAT_2019}.

In this work we investigate generalized models of structure in the context of non-linear estimation (or factorization) of noisy tensors. Tensors representing data have found many modern applications in signal processing, graph analysis, data mining and machine learning \cite{sidiropoulos2016, cichoki2015, kolda2009}, with a large part of the literature focusing on tensor decompositions, either in deterministic settings, or in random settings with independent structureless components.
Here we focus on a simple statistical model of noisy symmetric rank-one tensors. A {\it structured} signal
$\bX=(X_1, \cdots, X_n) \in \mathbb{R}^n$ is generated by a one-layer GLM $X_i = \varphi( (\bW \bS)_i/\sqrt p)$ where the latent vector $\bS\in \mathbb{R}^p$ has independent and identically distributed (i.i.d.) entries and $\bW$ is a \textit{known} random matrix with independent standard Gaussian entries. We only observe a noisy version of the rank-one tensor $\bX^{\otimes r}$ ($r\geq 2$) through an additive white Gaussian noise channel, i.e., $\bY = \frac{\sqrt{\lambda}}{n^{(r-1)/2}} \bX^{\otimes r} + \bZ$ where the noise $\bZ$ is a symmetric tensor with independent standard Gaussians entries and $\lambda > 0$ is the signal-to-noise ratio.
We study the high dimensional limit $n, p\to \infty$ such that $n/p\to \alpha = \Theta(1)$ and show that, quite remarkably,
the asymptotic mutual information $\lim_{n\to +\infty}I(\bX;\bY\vert \bW)/n$ is given by a finite-dimensional variational problem (see Theorem~\ref{theorem:limit_mutual_information} in Section~\ref{subsec:results}).
We also rigorously deduce the corresponding asymptotic \textit{minimum mean square error} (MMSE), which is given by a simple function of the solution to the variational problem (see Theorem ~\ref{theorem:tensor_MMSE} in Section~\ref{subsec:results}).
For concreteness, and to keep the analysis as simple as possible, we focus on the case $r=3$ and one-layer GLM.
However, extensions to any order $r > 3$, multi-layer GLM and asymmetric tensors are possible with the techniques used here.
An extensive recent study of the matrix case $r=2$ can be found in \cite{aubin2019spiked}.

The analysis and results presented here go beyond many recent works dealing with i.i.d.\ components for $\bX$, for matrices $r=2$ \cite{XXT, Lelarge_fundamental_2019, miolane2017fundamental}, and tensors $r\geq 3$ \cite{LesieurMiolane_2017, barbier2017layered}.
There is a rich phenomenology of phase transitions already for the i.i.d.\ case which stems from the (simpler) variational formula for the mutual information.
In Section~\ref{subsec:examples} we discuss the (numerical) solutions to the new variational problem obtained for structured signals for various examples of priors and activation functions, and we illustrate properties of phase transitions. Furthermore we discuss the similarities and differences between the genuine tensor and matrix cases.

Let us say a few words about the techniques used in this work.
There is a long history in the literature connecting Bayesian inference problems with spin-glass models of statistical mechanics \cite{nishimori01, mezard2009information} and it has been conjectured for some time
that the true variational expressions for the mutual information should coincide with the so-called ``replica-symmetric''
formula for the free energy derived by analytical non-rigorous methods.
The veracity of these conjectures has now been established by a variety of methods for various problems, e.g.,
coding theory \cite{Giurgiu_SCproof}, random linear estimation \cite{8606971, 9079920}, matrix and tensor estimation \cite{koradamacris, XXT, Lelarge_fundamental_2019, miolane2017fundamental, barbier2017layered}.
In all these cases the signal has i.i.d.\ components.
For structured signals, rigorous proofs of the low-dimensional variational expression for the asymptotic mutual information are virtually non-existent.
To the best of our knowledge, besides the case where $\bX$ is uniformly distributed on the sphere \cite{luneau2020highdimensional} (which turns out to be equivalent to an i.i.d.\ Gaussian prior), there are two recent exceptions:
\cite{Gabrie_TwoLayerGLM_JSTAT_2019} which includes the rigorous calculation of a mutual information for a GLM with input generated by another GLM, and \cite{aubin2019spiked} which treats the rank-one matrix case with input coming from a GLM.
The later work uses two different flavors of the interpolation method \cite{Guerra-Toninelli-2002, Alaoui2018} which do not extend to odd-order tensors nor asymmetric ones.
Moreover, certain (reasonable) assumptions are required.
In this work we leverage on recent progress on the proofs of replica-symmetric formulas by the \textit{adaptive interpolation method} \cite{barbier_adaptive_2019, Barbier_Macris_jphysA_2019} which is a powerful evolution of the celebrated Guerra-Toninelli interpolation scheme \cite{Guerra-Toninelli-2002}.
Our treatment is completely self-contained, leverages on only one method, and can also deal with asymmetric matrices and tensors.

In Section~\ref{section:main_results} we formulate the model, present the main theorems for the asymptotic mutual information and MMSE along with examples and illustrations of phase transitions, and explain key ideas behind the proofs.
In Sections~\ref{sec:proofthm1} and \ref{sec:proofthm2} we go through the proofs and
in Section~\ref{section:alphatendstozero} we give an analysis of the limit $\alpha \to 0$.
The appendices contain technical derivations.

config: arxiv


TITLE: How do I solve $\lim \limits_{x \to \frac{π}{3}} \frac{2 \sin x - \sqrt{3}}{\cos \frac{3x}{2}}$
QUESTION [3 upvotes]: Alright, I know, there are easier ways to solve this, like L'hopitals Rule etc.
But I'm not solving it for the answer, just doing it for the fun so I tried using substitution method.
Put $t= x- \dfrac{π}{3}$
$\lim \limits_{x \to \frac{π}{3}} \dfrac{2 \sin x - \sqrt{3}}{\cos \frac{3x}{2}}$
$= \lim\limits_{t \to 0} \dfrac{2 \sin \left(t+\frac{π}{3} \right) - \sqrt{3}}{\cos \left( \frac{3t}{2} + \frac{π}{2}\right)}$
$= \lim\limits_{t \to 0} \dfrac{\sqrt{3}- \sin t - \sqrt{3} \cos t}{ \sin \frac{3t}{2}}$
Where do I go from here, I'm not able to eliminate the $t$ fully from the Nr and Dr, any help?
Or any other alternative way that uses only the fact that $\lim\limits_{x \to 0} \dfrac{ \sin x}{x} = 1$?
Thanks :)

REPLY [3 votes]: Just to give another approach, let's let $f(x)=2\sin x$ and $g(x)=\cos(3x/2)$ and use the fact(s) that $f'(x)=2\cos x$, $g'(x)=-{3\over2}\sin(3x/2)$, $f(\pi/3)=2\sin(\pi/3)=\sqrt3$, and $g(\pi/3)=\cos(\pi/2)=0$.  From the definition of the derivative, it follows that
$$\lim_{x\to\pi/3}{2\sin x-\sqrt3\over x-\pi/3}=\lim_{\pi/3}{f(x)-f(\pi/3)\over x-\pi/3}=f'(\pi/3)=2\cos(\pi/3)=1$$
and
$$\lim_{x\to\pi/3}{\cos(3x/2)\over x-\pi/3}=\lim_{x\to\pi/3}{g(x)-g(\pi/3)\over x-\pi/3 }=g'(\pi/3)=-{3\over2}\sin(\pi/2)=-{3\over2}$$
Thus
$$\lim_{x\to\pi/3}{2\sin x-\sqrt2\over\cos(3x/2)}={\displaystyle\lim_{x\to\pi/3}{2\sin x-\sqrt3\over x-\pi/3}\over\displaystyle\lim_{x\to\pi/3}{\cos(3x/2)\over x-\pi/3}}={1\over-{3\over2}}=-{2\over3}$$


TITLE: Circle is a straight line
QUESTION [3 upvotes]: I was thinking about regular polygons when I tried this out.
The sum of the internal angles of an $n$ - sided polygon is $\frac{180(n-2)}{n}$. The limit of this function as n approaches infinity is $180$. Since a circle has infinite sides, the internal angles of the circle are $180$ degrees, which is a straight line.
Is there a flaw in my reasoning? If so, what is it?

REPLY [4 votes]: To summarise, the flaw is that you're treating infinity as if it's something finite. You can't throw around infinities and infinitesimals too casually. My answer will have a few hand-wavy arguments, but it should be enough for you to realise why your argument was flawed.
In your argument you're taking the limit as side-lengths ($l$) go to zero and the limit as the number of sides ($n$) tend to infinity simultaneously. The rule for "simultaneous limits" is that there must be no dependence on order. This means you want:
\begin{align*}
\lim_{n\rightarrow\infty}\lim_{l\rightarrow0}P(n, l) = \lim_{l\rightarrow 0}\lim_{n\rightarrow\infty}P(n, l) = \mathrm{Circle}
\end{align*}
Where $P(n, l)$ is a regular polygon with side-lengths $l$, and $n$ sides.
Imagine you have an infinite number of unit-length line segments. If you use $3$ of them, you have an equilateral triangle with side lengths of $1$. If you use $n$ of them, you get the regular $n$ sided polygon with side lengths of $1$. Suppose you were to measure the interior angle of such a polygon. If this angle is less than $180^{\circ}$, then you can work out how many sides were used to construct it. In other words, it's still only a finite number of sides. Thus $\lim_{n\rightarrow\infty}P(n, l)$ must be equal to an infinite straight line. This no longer has any dependence on $l$, because $\infty/l = \infty$ for any $l\neq 0$. Therefore,
\begin{align*}
\lim_{l\rightarrow 0}\lim_{n\rightarrow\infty}P(n, l) = \lim_{l\rightarrow 0}(\lim_{n\rightarrow\infty}P(n, l)) = \lim_{l\rightarrow 0}(\mathrm{Infinite\ straight\ line}) = \mathrm{Infinite\ straight\ line}
\end{align*}
How about the other limit? Suppose you have chosen some number of sides, and their lengths can vary. If you have $3$ sides and they have lengths of $1$, you have an equilateral triangle with side lengths of $1$. For any finite $l$, you will have equilateral triangle with side length $l$. Now if you suppose you have a tiny tiny polygon and it's still a triangle, the side length is clearly some finite $l > 0$. Hence $\lim_{l\rightarrow0}P(n, l)$ must be a single point. No matter how many "sides" you give to a single point it will always just be a single point. Therefore,
\begin{align*}
\lim_{n\rightarrow\infty}\lim_{l\rightarrow0}P(n, l) = \lim_{n\rightarrow\infty}(\lim_{l\rightarrow0}P(n, l)) = \lim_{n\rightarrow\infty}(\mathrm{Single\ point}) = \mathrm{Single\ point}
\end{align*}
Clearly a single point is not an infinite line, and they're both certainly not circles. The flaw in your argument is that "the polygon formed when side lengths go to zero and the number of lines goes to infinity" cannot be uniquely described as a circle.
Also I apologise to everybody for my loose use of the word "polygon"


TITLE: Composing two discontinuous functions into a continuous one
QUESTION [6 upvotes]: Please help me think of an example of two discontinuous functions on $\mathbb R$ whose composition gives a continuous function on $\mathbb R$.

REPLY [6 votes]: Hints for a general process of producing such functions:

The identity function $\mathrm{id}:\mathbb{R}\to\mathbb{R}$, defined by $\mathrm{id}(x)=x$, is continuous.
Think of two non-trivial functions on $\mathbb{R}$ that, when composed, give the identity.
Define $f$ and $g$ to be those functions, except only applied to some subset $S\subset\mathbb{R}$ (for example, the irrationals), defining them to be the identity on $\mathbb{R}\setminus S$.

As long as the functions in step 2 map $S$ to itself, this (usually) produces discontinuous functions with the desired properties.
Some examples:

Choose the function "multiply by $-1$", apply it to any subset $S$ of $\mathbb{R}$ symmetric about the origin other than $\mathbb{R}$ itself.
Choose the function "take reciprocals", apply it to any subset $S$ of $\mathbb{R}\setminus\{0\}$ which is closed under taking reciprocals.


TITLE: Prove: If $f$ and $g$ are bijective, then $g\circ f$ is bijective.
QUESTION [0 upvotes]: I want to prove this theorem:

If $f: X \to Y$ and $g: Y \to Z$ are both bijective, then $g \circ f: X \to Z$ is bijective, as well.

Up until now, I've only used induction for proofs, but that isn't applicable here. Any ideas on how to proceed?

REPLY [3 votes]: If $f, g$ are both bijective, then there exist inverses $f^{-1} : Y \to X$ and $g^{-1} :  Z \to Y$.
Then we have $f^{-1} \circ g^{-1}$ is an inverse to $g \circ f$ because they compose to the identities of $X$ and $Z$:
$$ (f^{-1} \circ g^{-1}) \circ (g \circ f)(x) = (f^{-1} \circ f)(x) = x $$
and
$$ (g \circ f) \circ (f^{-1} \circ g^{-1})(z) = (g \circ g^{-1})(z) =z$$
So $g \circ f$ is a bjiection.


TITLE: Density process/Radon-Nikodym derivative problems
QUESTION [1 upvotes]: I have two (related), ungraded homework problems. I am hoping to receive advice on how to proceed, or to be alerted if I'm on the wrong path (or, alternatively, to have the problems solved for me).

Let $P\sim Q$ be equivalent probability measures.
QUESTION 1
Prove that $dQ/dP > 0 \,\,\,\,\,\text{ P-a.s. (and hence Q-a.s.)}$
Attempt:
$\displaystyle \ \ \int_\Omega dQ = 1 $
If $P\sim Q$ and $dQ/dP < 0$ then:
$1 = \int_\Omega \frac{dQ}{dP}dP = E^P[\frac{dQ}{dP}]\leq 0$
Which is a contradiction, therefore $\frac{dQ}{dP} > 0 \,\,\,\text{P-a.s. (and hence Q-a.s.)}$.
QUESTION 2
Prove that $dQ/dP = (dP/dQ)^{-1} \,\,\,\text{P-a.s. (and hence Q-a.s.)}$
Attempt:
By the definition of equivalent probability measures, we have that:
$\displaystyle \ \ \int_\Omega dP = \int_\Omega \frac{dP}{dQ}\frac{dQ}{dP}dP = \int_\Omega \frac{dP}{dQ}dQ$
Therefore the equality is true.

REPLY [1 votes]: Let $f:=\frac{dQ}{dP}$, and assume that "$f>0$ $P$-a.e. is not true". That means that we can find $S\subset\Omega$ such that $P(S)>0$ and $f(x)\leqslant 0$ for all $x\in S$. This gives
$$Q(S)=\int_{S}f(x)dP(x)\leqslant 0,$$
so $Q(S)=0$. It's a contradiction as $P\ll Q$.
What you did would be correct if you replace $\Omega$ by an arbitrary measurable set $A$.


TITLE: Average measure of intersection of a convex region with its translate
QUESTION [24 upvotes]: Let $\lambda$ denote the Lebesgue-measure on $\mathbb{R}^n$, and let $C\subset\mathbb{R}^n$ be a convex region.
My question is about
$$f(C):=\int_{C} \lambda(C \cap (x + C) ) \mathrm{d} x.$$
How large can $f(C)$ be? Of course, there is the trivial bound $f(C)\leq \lambda(C)^2$ but I would expect more something like $f(C) = O( \lambda(C) )$ at least. This question has probably been answered somewhere, and I suspect that the following might be very well known: for regions $C$ of constant measure, the value of $f(C)$ is maximal if $C$ is an $n$-ball (or something similar). I couldn't find any useful references regarding this question. If someone could help me out with some reference, or even a simple solution, I would be very grateful.

REPLY [5 votes]: How about a short solution to a much general problem, and without convexity assumption (which turns out to  be really just a distraction here) ?

So, more generally, for nonempty compact subsets $A$, $B$, and $C$ of $\mathbb R^n$, define $F(A,B,C)$ by
$$
F(A,B,C) := \int_{A}\lambda(B \cap (x + C))dx.
$$

In particular, the OP's question is concerned with $f(C) := F(C,C,C)$.

We shall proceed in 3 short steps.

Step 1. The first step in analyzing $F$ is to realize that we can rewrite it as $F(A,B,C) = G(1_A,1_B,1_C)$, where

\begin{align}
G(u,v,w) = \int_{\mathbb R^n}\int_{\mathbb R^n} u(x)w(x-y)v(y)\,dxdy.
\end{align}

Step 2. Now, by the Riez rearrangement inequality, we know that
$$
G(u,v,w) \le G(u^\star,v^\star,w^\star), $$
where $u^\star$ is the Symmetric decreasing rearrangement of the function $u$. On the other hand, it is clear that
$$
(1_A)^{\star} = 1_{K(A)},
$$
where $K(A)$ is the centered ball of same volume as $A$.
We deduce that
$$
F(A,B,C) \le F(K(A),K(B),K(C)),
$$

and in particular, $f(C) \le f(K(C))$. The solution to the OP's question then follows from computation for the ball, in the begining of the accepted answer https://mathoverflow.net/a/282941/78539.


TITLE: Defining $D\in \mathcal L(V)$ by $Df=f'$, show $D^*=-D$
QUESTION [0 upvotes]: Let $V = \text{span}(1, \cos x,..., \cos nx, \sin x,...,\sin nx)$ for a fixed positive $n$.
In the inner product space of real-valued, continuous functions on $[-\pi,\pi]$ with inner-product
$$\langle f,g\rangle= \int_{-\pi}^{\pi}f(x)g(x)dx$$
Define $D\in \mathcal L(V)$ by $Df=f'$
Show $D^*=-D$
In this case:
$$\langle Tf(x),g(x)\rangle=\langle f(x),T^*g(x)\rangle$$ and specifically $$\langle Df(x),g(x)\rangle=\langle f(x),-D g(x)\rangle$$
What I have done so far: I took $f(x)= a_0+a_1\sin x+a_2\cos x$ and $g(x)=b_0+b_1\sin x+b_2\cos x$ and applied the above-mentioned inner-products to see an example of how things worked.
My question: I would please appreciate help with a general proof of the assertion. (Also, while I have no real knowledge of Fourier series, is there a connection with this problem?)
Thanks

REPLY [3 votes]: Integration by parts gives
$$ \langle Df,g \rangle = \int_{-\pi}^{\pi} f'(x)g(x) \, dx = [f(x)g(x)]_{-\pi}^{\pi} + \int_{-\pi}^{\pi} -f(x)g'(x) \, dx. $$
All of the functions in $V$ have period $2\pi$, so the boundary terms vanish and the right-hand side becomes $ \langle f, -Dg \rangle $.


TITLE: Four Color Theorem
QUESTION [6 upvotes]: My wife is making a quilt. She has a whole bunch of colors and is making a very simple pattern. I enjoy watching the whole process because it's very mathematical, but it has made me question the four color theorem because I've come up with a counterexample that indicates that I either:

Do not understand the problem.
Am missing a solution for this example.

The "counterexample" is this:
++++
+wx+
+yz+
++++

Unless w andz or x and y are allowed to be the same color you need 5 colors to make this work.
Does adjacency in this problem not include single-point adjacency?

REPLY [5 votes]: That is correct: it does not include single-point adjacency. Otherwise the Four Corners states of Utah, Colorado, Arizona, and New Mexico would provide a familiar core for a potential counterexample, one that would be hard for any cartographer to overlook. It’s also required that each region be a contiguous territory. A full informal statement of the theorem, adapted from Wikipedia:

Given any separation of a plane into contiguous regions, producing a figure called a map, no more than four colors are required to color the regions of the map so that no two adjacent regions have the same color. Two regions are said to be adjacent if they share a common boundary that is not a corner, where a corner is a point shared by three or more regions.


TITLE: Defining an unusual norm on $\mathbb{R}^3$
QUESTION [2 upvotes]: For vectors $\vec v = (v_1,v_2,v_3) \in \mathbb R^3$, does $||\vec v|| = |v_1| + \max\{ |v_2|,|v_3|\}$ define a norm on $\mathbb{R}^3$?
I know I need to show positivity, homogeneity and the triangle inequality.
I can do positivity, I'm lost for the other two conditions though.
Let $c \in \mathbb R$ and $v,w \in \mathbb R^3$.
\begin{equation}\begin{split}
||c\vec v|| &= |cv_1| + \max\{|cv_2|,|cv_3|\} \\
&= |c||v_1| + \max\{|c||v_2|,|c||v_3|\} \\
&= |c||v_1| + |c|\max\{|v_2|,|v_3|\} \\
&= |c|(|v_1| + \max\{|v_2|,|v_3|\}) \\
&= |c|||\vec v||
\end{split}\end{equation}
And
\begin{equation}\begin{split}
||\vec v+ \vec w|| &= |v_1 + w_1| + \max\{|v_2+w_2|,|v_3+w_3|\} \\
&\leq |v_1| + |w_1| + \max\{|v_2|+|w_2|,|v_3|+|w_3|\} \\
&= (|v_1| + \max\{|v_2|,|v_3|\}) + (|w_1| + \max\{|w_2|,|w_3|\}) \\
&= ||\vec v||+||\vec w||
\end{split}\end{equation}
Thus
$||\vec v+ \vec w|| ≤ ||\vec v|| + ||\vec w|| $
That's what I've tried so far.

REPLY [0 votes]: For the Triangle Inequality, we need to show the following
$|v_1+u_1|+max{\{(v_2+u_2),(v_3+u_3)\}}\leq (|v_1|+max{\{v_2,v_3}\})+(|u_1|+max{\{u_2,u_3\}})$.
note you can use the Triangle inequality for each component to arrive at the Triangle Inequality for the norm.
we have $|v_1+u_1|<|v_1|+|u_1|$ for the first component, and also we have
$max{\{|v_2+u_2|,|v_3+u_3|\}}$. Note that you can split this into two cases where if $|v_2+u_2|>|v_3+u_3|$ and $|v_2+u_2|<|v_3+u_3|$  the Triangle Inequality (for the norm) naturally follows with a little bit of computation. So  the proposed norm is indeed a norm.


TITLE: Proof by contradiction proving both numbers are not odd.
QUESTION [1 upvotes]: I have to do a proof by contradiction:
Suppose $a,b,\in\mathbb{Z}$. If $4| (a^2 + b^2)$ then a and b are not both odd.
So far I know that I need to prove that if $4|(a^2+b^2)$ then a and b are both odd.
I would use the definition of an odd number $(2k+1), k\in\mathbb{Z}$
I'm a little but unsure of where to go from here. Any help would be appreciated.

REPLY [1 votes]: I'm afraid you aren't correctly stating the goal of proof by contradiction. To prove "If $P$ then $Q$" by contradiction (where $P$ and $Q$ are some statements), we must assume that $P$ is true and $Q$ is false (which is not the same as proving "If $P$ then not $Q$"), and then try to derive a nonsensical conclusion.

In your particular case, you must assume that $4\mid(a^2+b^2)$ and that $a$ and $b$ are both odd. Since $a$ and $b$ are odd integers, then there are some integers $j,k$ such that $a=2j+1$ and $b=2k+1.$ Then $a^2=4j^2+4j+1$ and $b^2=4k^2+4k+1,$ so $$a^2+b^2=4j^2+4j+4k^2+4k+2=4(j^2+j+k^2+k)+2.$$ Since $j,k$ are integers, then so are $j^2,k^2,$ and so $j^2+j+k^2+k$ is an integer--let's call it $m$--and so since $4\mid(a^2+b^2),$ then $4\mid(4m+2).$ Can you take it from there to find the contradiction, and justify all the claims I made above?


TITLE: Palindrome Remove one Character
QUESTION [2 upvotes]: We are given a string $S$. If we remove just one character from the string $S$, we can get a palindrome.
Suppose $S$ $=$ $aBa$, where $B$ is the middle string, then I need to prove that I can remove one character from the string B and get a palindrome.
How do I prove this?

REPLY [5 votes]: Looks like a proof by contradiction.  Suppose $S=aBa$ and that it is not possible to remove a character from $B$ and turn $S$ into a palindrome.  Then since $S$ can be turned into a palindrome by removal of one character, that one character must be either the first or last $a$.
Suppose it's the first $a$.  Then $Ba$ must be a palindrome, which means that $B$ must begin with $a$.  But then I could have removed that $a$ from $B$ to obtain a palindrome, which contradicts my assumptions.  So it must be the last $a$ that I can remove... but then $B$ must end in $a$ which means I could have removed that $a$ from the $B$ in the first place instead.
So the assumption that it wasn't possible to remove a character from $B$ to turn $S$ into a palindrome was wrong, and I'm done.


TITLE: Fixed point in a metric space with distance at most 1
QUESTION [3 upvotes]: The question is:
Suppose that $X$ is a complete metric space such that the distance function is at most 1, and $f:X\rightarrow X$ is such that $d(f(x),f(y))\le d(x,y)−1/2(d(f(x),f(y)))^2$.
Prove that $f$ has a unique fixed point $x_0\in X$ and $\lim_{n\to\infty} f^{n}(x)=x_0$ for any $x\in X$.
I assume I must use that Cauchy sequences converge in $X$ but when I set up the inequality, I get:
$d(f^m(x),f^n(x))=\sum_{k=0}^{m-n-1} d(f^{n+k+1},f^{n+k})\le (m-n)d(f(x),x)-(1/2)(\sum_{k=0}^{m-n-1} d(f^{n+k+1},f^{n+k})^2)$
attempting to mirror the proof of the contraction principle, but we have no guarantee that the right hand side goes to zero.

REPLY [1 votes]: There must be a quicker proof than this, but the first argument I came up with  is the following (this is just a sketch, I'm leaving the details to you).
Step 1: Solving the inequality $d(fx, fy) \leq d(x,y) - \frac{1}{2}d(fx, fy)^2$ for $d(fx, fy)$ gives the inequality $$d(fx, fy)\leq -1 + \sqrt{1 + 2d(x,y)} = \left(\frac{-1 + \sqrt{1 + 2d(x,y)}}{d(x,y)}\right)d(x,y).$$
Step 2: Note that you have the inequality $$\frac{-1 + \sqrt{1 + 2d(x,y)}}{d(x,y)} < 1$$ when $d(x,y)\neq 0$. From this you can conclude two things:

$d(fx, fy) \leq d(x,y)$ for all $x,y$, and
For each $\delta>0$ there is a $\theta$ with $0<\theta<1$ such that if $d(x,y)\geq \delta$, then $d(fx, fy) \leq \theta d(x,y)$.

Step 3: Fix a point $x\in X$, and consider the sets $Y_n = \{f^nx, f^{n+1}x, \ldots\}$ for $n\geq 0$. Note $Y_{n+1}\subseteq Y_n$, so in particular the sequence $\Delta_n = \mathrm{diam}(Y_n)$ is monotonically decreasing. Let $\delta = \lim \Delta_n$. I claim that $\delta = 0$.
Why is this? Suppose for contradiction that $\delta>0$. By step 2, we know that there is a $\theta\in (0,1)$ such that if $d(x,y)\geq \theta$, then $d(fx, fy)\leq \theta d(x,y)$. Let $\epsilon>0$ be such that $$\epsilon < \delta\left(\frac{1-\theta}{\theta}\right).$$ Since $\Delta_n$ monotonically decreases to $\delta$, there is an $N$ such that $\delta\leq\Delta_n < \delta + \epsilon$ for all $n\geq N$. Applying this to $n = N+1$ yields that there exist integers $i,j\geq N+1$ such that $$\delta \leq d(f^ix, f^jx)< \delta + \epsilon.$$ Since by step 2 we know that $f$ does not increase distances, $d(f^ix, f^jx)\leq d(f^{i-1}x, f^{j-1}x)$. On the other hand, $\Delta_N< \delta + \epsilon$, so we have $\delta \leq d(f^{i-1}x, f^{j-1}x) < \delta + \epsilon$. But here's where we reach a contradiction, because now $$\delta \leq d(f^ix, f^jx) \leq \theta d(f^{i-1}x, f^{j-1}x) <\theta(\delta + \epsilon),$$ which implies $\epsilon > \delta(1-\theta)/\theta$. This proves that in fact $\delta = 0$.
Step 4: The closures $\overline{Y}_n$ have the same diameters $\Delta_n$ as the sets $Y_n$, so $\overline{Y}_n$ is a nested sequence of closed sets with diameters tending to $0$. By completeness, it follows that $\bigcap\overline{Y}_n = \{x_*\}$ for some point $x_*\in X$. It is now easy to check (and you should!) that $f^nx\to x_*$ as $n\to \infty$, and that $x_*$ is a fixed point of $f$. Thus we have shown:

For every $x\in X$, there is a fixed point $x_*\in X$ such that $f^nx\to x_*$ as $n\to \infty$.

Step 5: To complete the proof, it now suffices to note that $f$ can have at most one fixed point. Indeed, if $x_0$ and $x_1$ were two distinct fixed points, then since $d(x_0, x_1)\neq 0$, we have $d(fx_0, fx_1)<d(x_0, x_1)$, which is absurd, since $fx_0 = x_0$ and $fx_1 = x_1$.


text:
\section{Brill--Noether loci}
\label{Sec_BrillNoetherLoci}

Let us begin by discussing the non-equivariant theory.

\begin{definition}
  \label{Def_FamilyOfEllipticOperators}
  Let $k \in \N_0$.
  A \defined{family of linear elliptic differential operators of order $k$} consists of
  a Banach manifold $\sP$ and
  a smooth map
  \begin{equation*}
    D \co \sP \to \sF(W^{k,2}\Gamma(E),L^2\Gamma(F))
  \end{equation*}
  such that for every $p \in \sP$ the operator $D_p \coloneq D(p)$ is the extension of a linear elliptic differential operator $\Gamma(E) \to \Gamma(F)$ of order $k$.
  \footnote{
    \label{Footnote_BanachManifolds}
    Banach manifolds are assumed to be Hausdorff, paracompact, and separable.
    This is required in \autoref{Sec_CodimensionInBanachManifolds} where the Sard--Smale Theorem is used.
  }
\end{definition}

\begin{definition}
  \label{Def_BrillNoetherLoci}
  Let $(D_p)_{p \in \sP}$ be a family of linear elliptic differential operators.
  For
  $d,e \in \N_0$
  define the \defined{Brill--Noether locus} $\sP_{d,e}$ by
  \begin{equation*}
    \sP_{d,e}
    \coloneq
    \set[\big]{ p \in \sP : \dim \ker D_p = d \textnormal{ and } \dim\coker D_p = e }.
    \qedhere
  \end{equation*}
\end{definition}

\begin{remark}
  \label{Rmk_Index}
  Let $(D_p)_{p \in \sP}$ be a family of linear elliptic operators of index $i \in \Z$.
  If $\sP_{d,e} \neq \emptyset$, then $d-e = i$;
  in particular: $d \geq i$ and $e \geq -i$.
\end{remark}

The following elementary fact from the theory of Fredholm operators reduces the discussion to the finite-dimensional case.

\begin{lemma}
  \label{Lem_LyapunovSchmidtReduction}
  Let $X$ and $Y$ be Banach spaces.
  For every $L \in \sF(X,Y)$ there is an open neighborhood $\sU \subset \sF(X,Y)$ and a smooth map $\sS \co \sU \to \Hom(\ker L,\coker L)$ such that
  for every $T \in \sU$ there are isomorphisms
  \begin{equation*}
    \ker T \iso \ker \sS(T) \qandq
    \coker T \iso \coker \sS(T);
  \end{equation*}
  furthermore, $\rd_L\sS\co T_L\sF(X,Y) \to \Hom(\ker L,\coker L)$ satisfies
  \begin{equation*}
    \rd_L\sS(\hat L)s
    =
    \hat L s \mod \im L.
  \end{equation*}
\end{lemma}

\begin{proof}
  Pick a complement $\coim L$ of $\ker L$ in $X$ and a lift of $\coker L$ to $Y$.
  With respect to the splittings
  $X = \coim L \oplus \ker L$
  and
  $Y = \im L \oplus \coker L$
  every $T \in \sF(X,Y)$ can be written as
  \begin{equation*}
    T
    =
    \begin{pmatrix}
      T_{11} & T_{12} \\
      T_{21} & T_{22}
    \end{pmatrix}.
  \end{equation*}
  Choose an open neighborhood $\sU$ of $L$ in $\sF(X,Y)$ such that for every $T \in \sU$ the operator $T_{11}$ is invertible.
  Define $\sS\co \sU \to \Hom(\ker L,\coker L)$ by
  \begin{equation*}
    \sS(T)
    \coloneq
    T_{22} - T_{21}T_{11}^{-1}T_{12}.
  \end{equation*}
  A brief computation shows that for every $T \in \sU$
  \begin{equation*}
    \Phi
    T
    \Psi
    =
    \begin{pmatrix}
      \one & 0 \\
      0 & \sS(T)
    \end{pmatrix}
    \qwithq
    \Phi
    \coloneq
    \begin{pmatrix}
      T_{11}^{-1} & 0 \\
      -T_{21}T_{11}^{-1} & \one
    \end{pmatrix}
    \qandq
    \Psi
    \coloneq
    \begin{pmatrix}
      \one & -T_{11}^{-1}T_{12} \\
      0 & \one
    \end{pmatrix};
  \end{equation*}
  hence,
  $\ker T \iso \ker \sS(T)$
  and
  $\coker T \iso \coker \sS(T)$.
\end{proof}

\autoref{Lem_LyapunovSchmidtReduction} together with the Regular Value Theorem immediately imply the following.

\begin{theorem}
  \label{Thm_BrillNoetherLoci}
  Let $(D_p)_{p \in \sP}$ be a family of linear elliptic differential operators.
  Let $d,e \in \N_0$.
  If for every $p \in \sP_{d,e}$ the map $\Lambda_p \co T_p\sP \to \Hom(\ker D_p,\coker D_p)$ defined by
  \begin{equation*}
    \Lambda_p(\hat p)s \coloneq \rd_pD(\hat p) s \mod \im D_p
  \end{equation*}
  is surjective,
  then the following hold:
  \begin{enumerate}
  \item
    $\sP_{d,e}$ is a submanifold of codimension
    \begin{equation*}
      \codim \sP_{d,e} = de.
    \end{equation*}
  \item
    If $\sP_{d,e} \neq \emptyset$,
    then $\sP_{\tilde d,\tilde e} \neq \emptyset$ for every $\tilde d,\tilde e \in \N_0$ with $\tilde d \leq d, \tilde e \leq e$, and $\tilde d-\tilde e = d-e$.
    \qed
  \end{enumerate}
\end{theorem}

\begin{remark}
  \label{Rmk_BrillNoetherLoci_Complex}
  If $E$ and $F$ are Hermitian vector bundles and $(D_p)_{p \in \sP}$ is a family of complex linear elliptic differential operators,
  then the map $\Lambda_p$ factors through $\Hom_\C(\ker D_p,\coker D_p)$.
  Therefore,
  the hypothesis of \autoref{Thm_BrillNoetherLoci} cannot be satisfied (unless it holds trivially).
  Of course,
  this issue is rectified by replacing $\R$ with $\C$ throughout the above discussion.
\end{remark}

\begin{example}[Brill--Noether theory for holomorphic line bundles over a Riemann surface]
  \label{Ex_BrillNoetherHolomorphicLineBundlesRiemannSurface}
  Let $\Sigma$ be a closed, connected Riemann surface of genus $g$.
  Let $L$ be a Hermitian line bundle of degree $d$ over $\Sigma$.
  Denote by $\sA(L)$ the space of unitary connections on $L$.
  Define the family of complex linear elliptic differential operators
  \begin{equation*}
    \delbar\co \sA(L) \to \sF(W^{1,2}\Gamma(L),L^2\Omega^{0,1}(\Sigma,L))
  \end{equation*}
  by assigning to every connection $A$ the Dolbeault operator $\delbar_A \coloneq \nabla_A^{0,1}$.
  Let $A \in \sA(L)$.
  Denote by $\sL$ the holomorphic line bundle associated with $\delbar_A$.
  By Serre duality,
  \begin{equation*}
    \coker \delbar_A = H^1(\Sigma,\sL) \iso H^0(\Sigma,K_\Sigma\otimes_\C \sL^*)^*.
  \end{equation*}
  Since
  $\rd_A\delbar(a) = a^{0,1}$,
  the map
  $\Lambda_A$
  factors through the isomorphism
  $T_A\sA(L) = \Omega^1(\Sigma,i\R) \iso \Omega^{0,1}(\Sigma)$.
  Therefore, the adjoint of $\Lambda_A\co T_A\sA(L) \to \Hom_\C(\ker \delbar_A, \coker \delbar_A)$ is the composition of the Petri map
  \begin{equation}
    \label{Eq_PetriL}
    \varpi_\sL
    \co
    H^0(\Sigma,\sL) \otimes_\C H^0(\Sigma,K_\Sigma\otimes \sL^*) \to H^0(\Sigma,K_\Sigma)
  \end{equation}
  with the inclusion $H^0(\Sigma,K_\Sigma) \incl \Omega^{0,1}(\Sigma)$.
  In particular, $\Lambda_A$ is surjective if and only if $\varpi_\sL$ is injective.
  If $\varpi_\sL$ is injective for every $[\sL] \in \Pic^d(\Sigma)$,
  then
  \begin{equation*}
    \tilde G_d^r
    \coloneq
    \delbar^{-1}(\sF_{r+1,g-d+r})
  \end{equation*}
  is a complex submanifold of codimension $(r+1)(g-d+r)$;
  therefore, so is the classical Brill--Noether locus
  \begin{equation*}
    G_d^r
    \coloneq
    \tilde G_d^r/\sG^\C(L)
    \iso
    \set*{
      \sL \in \Pic^d(\Sigma)
      :
      \begin{array}{l}
        \dim H^0(\Sigma,\sL) = r+1 \textnormal{ and} \\
        \dim H^1(\Sigma,\sL) = g-d+r
      \end{array}
    };
  \end{equation*}
  cf.~\cite[Lemma 1.6, Chapter IV]{Arbarello1985}.
\end{example}

This example motivates the following definitions,
which are particularly appropriate for first order operators appearing in geometric applications.

\begin{definition}
  \label{Def_Flexible}
  Let $U \subset M$ be an open subset.
  A family of linear elliptic differential operators $(D_p)_{p \in \sP}$ is \defined{flexible in $U$} if for every $p \in \sP$ and $A \in \Gamma(\Hom(E,F))$ supported in $U$
  there is a $\hat p \in T_p\sP$ such that
  \begin{equation*}
    \rd_p D(\hat p)s = As \mod \im D_p
  \end{equation*}
  for every $s \in \ker D_p$.
\end{definition}

\begin{definition}
  \label{Def_FormalAdjoint}
  Let $D\co \Gamma(E) \to \Gamma(F)$ be a linear differential operator.
  Set
  \begin{equation*}
    E^\dagger \coloneq E^*\otimes \Wedge^n T^* M \qandq
    F^\dagger \coloneq F^*\otimes \Wedge^n T^* M.
  \end{equation*}
  The \defined{formal adjoint} of $D$ is the linear differential operator $D^\dagger \co \Gamma(F^\dagger) \to \Gamma(E^\dagger)$ characterized by
  \begin{equation*}
    \int_M \Inner{s,D^\dagger t} = \int_M \Inner{Ds,t}.
  \end{equation*}
  Here $\Inner{\cdot,\cdot}$ denotes the canonical pairings $E\otimes E^\dagger \to \Wedge^nT^* M$ and $F\otimes F^\dagger \to \Wedge^nT^* M$.
\end{definition}

\begin{definition}
  \label{Def_PetrisCondition}
  The \defined{Petri map} $\varpi \co \Gamma(E)\otimes\Gamma(F^\dagger) \to \Gamma(E\otimes F^\dagger)$ is defined by
  \begin{equation*}
    \varpi(s\otimes t)(x) \coloneq s(x)\otimes t(x).
  \end{equation*}
  Let $U \subset M$ be an open subset.
  A linear elliptic differential operator $D\co \Gamma(E) \to \Gamma(F)$ satisfies \defined{Petri's condition in $U$} if the map
  \begin{equation*}
    \varpi_{D,U} \co \ker D \otimes \ker D^\dagger \to \Gamma(U,E\otimes F^\dagger)
  \end{equation*}
  induced by the Petri map is injective.
\end{definition}

\begin{prop}
  \label{Prop_Flexible+Petri=>Surjective}
  Let $(D_p)_{p \in \sP}$ be a family of linear elliptic differential operators and let $p \in \sP$.
  Let $U \subset M$ be an open subset.
  If $(D_p)_{p \in \sP}$ is flexible in $U$ and
  $D_p$ satisfies Petri's condition in $U$,
  then the map $\Lambda_p$ defined in \autoref{Thm_BrillNoetherLoci} is surjective.
\end{prop}

\begin{proof}
  Define the map $\ev_p \co \Gamma_c(U,\Hom(E,F)) \to \Hom(\ker D_p,\coker D_p)$ by
  \begin{equation*}
    \ev_p(A)s \coloneq As \mod \im D_p.
  \end{equation*}
  $(D_p)_{p \in \sP}$ is flexible in $U$ if and only $\im \ev_p \subset \im \Lambda_p$.
  $D_p$ satisfies Petri's condition in $U$ if and only if $\ev_p$ is surjective.
  To see this, observe the following.
  The isomorphism $\ker D_p^\dagger \iso (\coker D_p)^*$ induces an isomorphism $\Hom(\ker D_p,\coker D_p)^* \iso \ker D_p \otimes \ker D_p^\dagger$.
  An element $B \in \ker D_p \otimes \ker D_p^\dagger$ annihilates $\im \ev_p$ if and only if
  \begin{equation*}
    \int_M \inner{\varpi(B)}{A} = 0
  \end{equation*}
  for every $A \in \Gamma(\Hom(E,F))$ supported in $U$;
  that is: $\varpi(B) = 0$ in $U$.
  Therefore,
  $(\im \ev_p)^\perp = \ker \varpi$.
\end{proof}

\begin{remark}
  In \autoref{Ex_BrillNoetherHolomorphicLineBundlesRiemannSurface},
  $\im \Lambda_p = \im \ev_p$ (with $U=\Sigma$, and $\R$ replaced with $\C$).
  Therefore, $\Lambda_p$ being surjective is equivalent to Petri's condition.
  Furthermore, tracing through the isomorphisms identifies the restriction of Petri map $\varpi$ to $\ker \delbar_A\otimes_\C \ker \delbar_A^*$ with the Petri map $\varpi_\sL$.
\end{remark}

Flexibility is not a rare condition.
In fact, typically the following condition holds,
which evidently implies flexibility.

\begin{definition}
  \label{Def_StronglyFlexible}
  Let $U \subset M$ be an open subset.
  A family of linear elliptic differential operators $(D_p)_{p \in \sP}$ is \defined{strongly flexible in $U$} if for every $p \in \sP$ there is a $C^0$--dense subset $\sH_p \subset \Gamma_c(U,\Hom(E,F))$ such that for every $A \in \sH_p$ there is a $\hat p \in T_p\sP$ such that
  \begin{equation*}
    \rd_p D(\hat p)s = As
  \end{equation*}
  for every $s \in \Gamma(E)$.
\end{definition}

Petri's condition up appears to be more subtle.
(By the uniqueness theorem for ordinary differential equations,
it holds for first operators linear elliptic differential operators on $1$--manifolds.
This is somewhat useful;
see, e.g., \cite{Eftekhary2019}.)
In \autoref{Sec_PetrisConditionRevisited},
we revisit Petri's condition and discuss an algebraic criterion due to \citeauthor{Wendl2016} for Petri's condition to hold away from a subset of infinite codimension.

\begin{remark}
  \label{Rmk_BrillNoetherLoci_EstimateCodimension}
  Assume the situation of \autoref{Thm_BrillNoetherLoci}.
  The following observations are useful in situations where the primary objective is to estimate the codimension of $\sP_{d,e}$.
  \begin{enumerate}
  \item
    \label{Rmk_BrillNoetherLoci_EstimateCodimension_Codimension>=Rank}
    Every $p \in \sP_{d,e}$ has an open neighborhood $\sU$ in $\sP$ such that $\sP_{d,e}\cap \sU$ is contained in a submanifold of codimension $\rk \Lambda_p$.

  \item
    \label{Rmk_BrillNoetherLoci_EstimateCodimension_PetriUpToRankRho}
    Let $\rho \in \N$ and let $U \subset M$ be an open subset.
    A linear elliptic differential operator $D\co \Gamma(E) \to \Gamma(F)$ satisfies \defined{Petri's condition up to rank $\rho$ in $U$}
    if for every non-zero $B \in \ker D\otimes\ker D^\dagger$ of rank at most $\rho$ the section $\varpi(B)$ does not vanish on $U$.
    (A \defined{simple tensor} is tensor of the form $v\otimes w$.
    Every tensor $B$ is a sum of simple tensors.
    The \defined{rank} of $B$ is the minimal number of simple tensors that sum to $B$.)
    If $D_p$ satisfies this condition and $(D_p)_{p\in \sP}$ is flexible,
    then
    \begin{equation*}
      \rk \Lambda_p \geq \min\set{\rho,d,e}\max\set{d,e}.
    \end{equation*}
    \begin{proof}
      Set $\sigma \coloneq \min\set{\rho,d,e}$.
      If $d \leq e$, then choose an injection $\R^\sigma \into \ker D_p$ and set $H \coloneq \Hom(\R^\sigma,\coker D_p)$;
      otherwise, choose a surjection $\coker D_p \onto \R^\sigma$ and set $H \coloneq \Hom(\ker D_p,\R^\sigma)$.
      In either case, composition defines a surjection
      \begin{equation*}
        \pi_p\co \Hom(\ker D_p,\coker D_p) \to H.
      \end{equation*}
      The subspace $\im \pi_p^* \subset \Hom(\ker D_p,\coker D_p)^* \iso \ker D_p \otimes \ker D_p^\dagger$ consists of elements of rank at most $\sigma \leq \rho$.
      The argument of the proof of \autoref{Prop_Flexible+Petri=>Surjective} thus shows that $\pi_p \circ \Lambda_p$ is surjective.
      Therefore, $\rk \Lambda_p \geq \dim H = \min\set{\rho,d,e}\max\set{d,e}$.
      \pushQED{\hfill$\blacksquare$}
    \end{proof}
  \item
    \label{Rmk_BrillNoetherLoci_EstimateCodimension_PetriUpToRankThree}
    (This is due to \citet[Proof of Lemma 4.4]{Eftekhary2016}.)
    Let $U \subset M$ be a non-empty open subset.
    Every first order linear elliptic differential operator $D \co \Gamma(E) \to \Gamma(F)$ satisfies Petri's condition up to rank three in $U$.
    \begin{proof}
      Every $B \in \ker D\otimes \ker D^\dagger$ can be written as $B = s_1\otimes t_1 + \cdots + s_\rho\otimes t_\rho$ with $\rho \coloneq \rk B$, and $s_1,\ldots,s_\rho \in \ker D$ and $t_1,\ldots,t_\rho \in \ker D^\dagger$ linearly independent.
      If $\rho = 1$ and $\varpi(B) = 0$,
      then $s_1$ or $t_1$ vanishes on an open subset;
      hence, by unique continuation, $s_1 = 0$ or $t_1 = 0$: a contradiction.

      Henceforth, assume that $\rho \geq 2$.
      Define $\delta,\epsilon\co U \to \N_0$ by $\delta(x) \coloneq \dim\Span{s_1(x),\ldots,s_\rho(x)}$ and $\epsilon(x) \coloneq \dim\Span{t_1(x),\ldots,t_\rho(x)}$.
      By unique continuation,
      $\delta$ and $\epsilon$ are positive on a dense open subset.
      In fact, $\delta,\epsilon \geq 2$ on a dense open subset.
      To see this,
      observe that if $\delta = 1$ on a non-empty open subset,
      then there is a non-empty open subset $V \subset U$ and a function $f \in C^\infty(V)$ such that $s_1(x) = f(x)s_2(x)$ for every $x \in V$.
      Therefore, $\sigma(\rd f)s_2 = 0$ with $\sigma$ denoting the symbol of $D$.
      Since $D$ is elliptic,
      $f$ must be constant:
      a contradiction to $s_1$ and $s_2$ being linearly independent.
      The same argument applies to $\epsilon$.

      If $\rho = 2$,
      then there exists an $x \in U$ such that $\delta(x) = \epsilon(x) = 2$;
      therefore: $\varpi(B)$ does not vanish at $x$.
      If $\rho = 3$,
      then there is an $x \in U$ such that $\min\set{\delta(x),\epsilon(x)} \geq 2$.
      If $\delta(x) = \epsilon(x) = 3$,
      then $\varpi(B)$ evidently does not vanishing at $x$;
      otherwise,
      without loss of generality,
      $s_1(x)$ and $s_2(x)$ are linearly independent,
      and $s_3(x) = \lambda_1s_1(x) + \lambda_2s_2(x)$.
      In the latter case,
      \begin{equation*}
        \varpi(B)(x) = s_1(x) \otimes \paren*{t_1(x)+\lambda_1t_3(x)} + s_2(x)\otimes \paren*{t_2(x) + \lambda_2t_3(x)}
      \end{equation*}
      which cannot vanish because $\epsilon(x) \geq 2$.
      \pushQED{\hfill$\blacksquare$}
    \end{proof}
    There are examples of first order linear elliptic operators which fail to satisfy Petri's condition up to rank four;
    see \cite[Example 5.5]{Wendl2016} or \autoref{Prop_BInKerHatVarpi}.
    Finally, a brief warning:
    the preceding observation is false when $\R$ is replaced with $\C$ or $\H$.
    The analogue of Petri's condition only holds up to rank one in this case.
    (The issue is that $\sigma(\df)s_2 =0$ does not imply $\rd f = 0$ if $f$ takes values in $\C$ or $\H$.)
    \qedhere
  \end{enumerate}
\end{remark}