TITLE: Analog of "Spin" Chern-Simons Theory
QUESTION [9 upvotes]: 3-dimensional Chern-Simons theories, with compact gauge group $G$, are determined by $H^4(BG)$. Looking at $U(1)$, with generator $c_1^2\in H^4(BU(1))=\mathbb{Z}$ for 1st Chern class $c_1$, there are infinitely many topological theories indexed by the level $k\in\mathbb{Z}$ (for class $k\cdot c_1^2$).  However, we can define other topological theories when an additional structure is put on the manifold. For a spin structure, this new "spin" U(1) Chern-Simons theory (in 3-dimensions) happens to exist for each half-integer $k\in\frac{1}{2}\mathbb{Z}$. Main reason behind this is that for your 3-manifold $Y$, a spin 4-manifold $X$ which bounds it will have even intersection-pairing.
[[Edit]]: To expand on this, my source is Dijkgraaf/Witten. The "spin" topological theory is just taking your topological theory but requiring a choice of spin-structure. Given our functional $S=\frac{k}{4\pi^2}\int_XF^2$ modulo 1, the spin structure makes this $=k\cdot(\text{even number})$, and so we can now enlarge the levels to $\frac{1}{2}\mathbb{Z}$.
My questions are (if they make sense): Is there a structure on your manifold to have levels in $\frac{1}{n}\mathbb{Z}$ for some $n\ge 3$? Can you formulate "spin-c" Chern Simons?
From a physical-perspective I would guess no to the first one, because at least in the case of half-integers it corresponds to fermions in nature (and fermions/bosons are what we live with). But then again there is the Fractional Quantum Hall Effect...

REPLY [8 votes]: As David Roberts mentions in the comments above, indeed as one climbs up the Whitehead tower of the orthogonal group the higher Pontraygin classes become divisible by higher factors.
Notably in the next step, if you demand that also $\tfrac{1}{2}p_1$ vanishes, hence that you have not only a Spin-structure but even a String structure, then the second Pontryagian class becomes divisble by 6.
In this case one can also divide the Lagrangian for 7-dimensional Chern-Simons theory by 6. This is discussed in
Fiorenza, Sati, Schreiber, Multiple M5-branes, String 2-connections, and 7d nonabelian Chern-Simons theory.


TITLE: Construct generator polynomial with cyclotomic sets
QUESTION [0 upvotes]: Suppose that $q=2$ and $n=31$.
$C_0=\{0\}$
$C_1=\{1,2,4,8,16\}=C_2=C_4=C_8=C_{16}$
$C_3=\{3,6,12,24,17\}=C_6=C_{12}$
$C_5=\{5,10,20,9,18\}=C_9=C_{10}$
$C_7=\{7,14,28,25,19\}=C_{14}$
$C_{11}=\{11,22,13,26,21\}=C_{13}$
$C_{15}=\{15,30,29,27,23\}$
$ord_{31}(3)=5$, $x^{31}-1$ has roots in $\mathbb{F}_{2^5}$. Let $\alpha$ be the primitive element in $\mathbb{F}_{2^5}$.
Monic polynomial of $C_1$ will be $(x-1)(x-2)(x-4)(x-8)(x-16)=-1024+1984 x-1240 x^2+310 x^3-31 x^4+x^5\equiv30-30x^2+x^5\mod{13}$
But this monic polynomial isn't right, can someone give me a hint or suggestion to get a right a one so that I can construct the generator polynomial? Thanks

REPLY [2 votes]: If by generator polynomial of $C_1$ you mean the minimal polynomial of
$\alpha$ over $\mathbb F_2$, that is, the binary polynomial of least
degree that has $\alpha$ as a root, then the polynomial
is
$$(x-\alpha)(x-\alpha^2)(x-\alpha^4)(x-\alpha^8)(x-\alpha^{16}).\tag{1}$$
It is also the minimal polynomial of $\alpha^2$ and of $\alpha^4$
and of $\alpha^8$ and of $\alpha^{16}$. The polynomial is
irreducible.
In multiplying out the factors, you need use arithmetic in
$\mathbb F_{2^5}$ (and not modulo $13$ or $31$ as you have it)
but to set up the arithmetic you need to identify which of the
$30$ primitive elements of $\mathbb F_{2^5}$ you are choosing to
call $\alpha$.  The only way to make this identification is to
choose one of the six known irreducible
binary quintic polynomials and say
that $\alpha$ is one of its roots. Of course, once you have chosen,
say, $x^5+x^2+1$ as the polynomial, the immediate need of
setting up arithmetic to verify that $(1)$ multiplies out to
$x^5+x^2+1$ disappears. You will need the arithmetic to figure
out the generator polynomials of $C_3$,$C_5$ etc.

So, how do I know that there are 6 quintics and $x^5+x^2+1$ is one
of them? I worked it out in this answer.


TITLE: Why doesn't the light from a star overwhelm our field of view?
QUESTION [2 upvotes]: This is probably a naive question :)
I was looking at a far off street light (also applies to a star) and thinking about the spherical EM wave it was emitting. From afar, the wave-front will be pretty much planar, right? Shouldn't that mean that the EM field should fill up my eyes? In other words, why do we see the object with a specific angular size while the wave we perceive has the same intensity over the area of our eyes?
I hope my question makes sense :)
Best
M.

REPLY [1 votes]: We'll assume the star/lamp is far enough away that the light rays coming from it are effectively parallel. start by replacing your eye with a sheet of photographic film, or a CCD or something similar.

As you say, the light intensity will be constant over the whole area of the film.
In your eye the equivalent of the sheet of film is the retina, but your eye contains a lens that focusses the light. So the light rays now behave like this:

A lens bends all parallel rays so they converge in  spot at the focal length $f$, and this is where your retina is. So the distant lamp/star does produce a uniform illumination over the whole of the lens, but the lens brings all this light to a single spot on the retina. That's why you see the star/lamp as a bright spot not a uniform wash of light.

REPLY [0 votes]: If our eyes were just big photosensitive plates, then yeah. You'd be facing directly towards the light, the whole wavefront hits/excites the whole photosensitive plate at the same time, and the whole thing is excited. This would make everything look like a blur, because all light from all directions would be present everywhere on your eye.
Instead, our eye uses lenses. Ideally, you want a lens to behave as follows: If you have an infinitely far away object, the lens focuses its parallel rays (or, alternatively, flat wavefront) into a point behind the lens - the focus. In this way the lens turns a direction into a point. Parallel rays coming from a single direction all get focused onto that point.
This sounds magical, that everything in a flat wavefront coming from a given direction can be mapped onto a point, using as dumb an object as a piece of glass, but that's how it works! Indeed it's a bit too magical. In physics it comes about from the "paraxial ray approximation" - that the light rays are close to the axis of your eye or microscope or whatever, and that they move for the most part straight down the axis. You get lots of complicated effects when you try to account for the fact that light is a wave and that things aren't always paraxial.


text:
\begin{document}

\maketitle

\begin{abstract} In this paper we consider the problem of prescribing the Gaussian and geodesic curvature on a disk and its boundary, respectively, via a conformal change of the metric. This leads us to a Liouville-type equation with a nonlinear Neumann boundary condition. We address the question of existence by setting the problem in a variational framework which seems to be completely new in the literature. We are able to find minimizers under symmetry assumptions.
\end{abstract}

\section{Introduction}

The problem of prescribing the Gaussian curvature on a compact surface $\Sigma$ under a conformal change of the metric is a classical one, and dates back to \cite{Ber, KazWar}. Let us denote by $g$ the original metric, $\tilde{g}$ the new one and $e^u$ the conformal factor (that is, $\tilde{g} = e^u g$). This problem reduces to solving the problem
$$ - \Delta_g u + 2 K_g = 2 K_{\tilde{g}}e^{u},$$
where $K_g$, $K_{\tilde{g}}$ denote the curvature with respect to $g$ and $\tilde{g}$, respectively. The solvability of this equation has been studied for a long time, and it is not possible to give here a comprehensive list of references.

If $\Sigma$ has a boundary, then boundary conditions are in order. Homogeneous Dirichlet and Neumann boundary conditions have already been considered in the literature. In this paper our aim is to prescribe not only the Gaussian curvature in $\S$, but also the geodesic curvature on $\partial \S$. In this case we are led with the boundary value problem:
\begin{equation} \label{gg}
\left\{\begin{array}{ll}
-\Delta_{g} u +2K_g = 2 K_{\tilde{g}} e^u  \qquad & \text{in $\S$,}\\
\frac{\partial u}{\partial n} +2h_g = 2h_{\tilde{g}}e^{u/2} \qquad  &\text{on $\partial\S $,}
\end{array}\right.
\end{equation}
 where $h_g, h_{\tilde{g}}$ are the geodesic curvatures of $\partial \S$ relative to $g$, $\tilde{g}$, respectively.

Some versions of this problem have been studied in the literature. The case $h_{\tilde{g}}=0$ has been treated  by A. Chang and P. Yang in \cite{chang1}. Moreover, the case $K_{\tilde{g}}=0$ has been treated in \cite{chang2, li-liu, liuwang}. There is also some progress in the blow-up analysis, see \cite{bao, francesca}, although a complete description of the phenomenon is still missing.

The case of constants $K_{\tilde{g}}$, $h_{\tilde{g}}$ has also been considered. For instance, Brendle (\cite{brendle}) uses a parabolic flow to show that this problem admits always a solution for some constant curvatures. By using complex analysis techniques, explicit expressions for the solutions and the exact values of the constants are determined if $\Sigma$ is a disk or an annulus, see \cite{otro, Asun}. The case of the half-plane has also been studied, see \cite{li-zhu, mira-galvez, zhang}. However, the case in which both curvatures are not constant has not been much considered. In \cite{cherrier}, some partial existence results are given, but they include a Lagrange multiplier which is out of control. Moreover, a Kazdan-Warner type of obstruction to existence has been found in \cite{hamza}. In a forthcoming work, the case of $K<0$ in domains different from the disk is treated, and also a blow-up analysis is performed, see \cite{LS-M-R}. At present, as far as we know, those are the only works considering non-constant curvatures.

The higher dimensional analogue of this question (that is, prescribing scalar curvature of a manifold and mean curvature of the boundary) has been more studied. The case of zero scalar curvature and constant mean curvature is known as the Escobar problem, in strong analogy with the Yamabe problem. In this regard, see \cite{amb, dja, escobar Annals, escobar Indiana, felli, HanLi1, HanLi2, marques}, and the references therein.

Integrating \eqref{gg} and applying the Gauss-Bonnet Theorem, one obtains
\begin{equation} \label{GB}
\int_{\Sigma}K_{\tilde{g}} e^u +  \int_{\partial \Sigma}h_{\tilde{g}} e^{u/2}=  2\pi\chi(\Sigma).
\end{equation}

In this paper we shall consider the case in which $\chi(\Sigma)=1$. By the Uniformization Theorem, we can pass via a conformal map to a disk, obtaining $K_{\tilde{g}}=0$, $h_{\tilde{g}}=1$. Taking this into account we can consider the problem:

\begin{align}
\left\lbrace \begin{array}{ll}
-\Delta u = 2Ke^u & \mbox{in } \D, \\
\frac{\partial u}{\partial \eta} + 2 = 2he^{u/2} & \mbox{on } \Sph,
\end{array}\right. \label{Probl}
\end{align}
where now $K$, $h$ are the curvatures to be prescribed.

Generally speaking, the case of a disk is specially challenging because of the non-compact action of the group of conformal maps of the disk, as happens in the Nirenberg problem for $\S = \mathbb{S}^2$. This issue has been only treated in \cite{chang2} for $K=0$ (see also \cite{francesca}). A blow-up analysis in this case for non-constant $K$, $h$ is yet to be done, and will be the target of further research. In this paper, as a first step in the understanding of the problem, we shall impose symmetry conditions on $K$, $h$ in order to rule out this phenomenon. This idea goes back to Moser (\cite{moser2}) for the Nirenberg problem.

Let $G$ be a symmetry group of $\D$ without fixed points on $\Sph$, that is, for each $x\in\Sph$ there exists $g\in G$ such that $g(x)\neq x$. We say that a function $f$ is $G-$symmetric if $f(x)=f(g(x))$ for all $g \in G$ and for all $x$ in the domain of $f$.

Our main results is the following.

\begin{theorem} \label{TheoremNonNegative} Let $K: \D \to \R$, $h: \Sph \to \R$ be H\"{o}lder continuous, nonnegative and $G-$symmetric functions, not both of them identically equal to $0$. Then problem \eqref{Probl} admits a solution.
\end{theorem}

We can also deal with changing sign curvatures $K$, $h$, as long as their negative part is small:

\begin{theorem} \label{TheoremPerturbed}Let $K_0: \D \to \R$, $h_0: \Sph \to \R$ be H\"{o}lder continuous, nonnegative and $G-$symmetric functions, none of them identically equal to $0$. Then there exists $\e >0$ such that problem \eqref{Probl} admits a solution for any H\"{o}lder continuous and $G-$symmetric functions $K$, $h$ with $\| K- K_0\|_{L^{\infty}} + \| h - h_0\|_{L^{\infty}} < \e$.
\end{theorem}

One of the main goals of this paper is to find an original variational setting to this problem, which we think is natural and could be of use in future research on the topic. Let us be more specific. We define the parameter $\rho := \int_\D K e^u = 2\pi - \int_{\Sph} he^{u/2}$. In order to fix ideas, let us assume that both $K$, $h$ are nonnegative functions; by \eqref{GB},  $0<\rho<2\pi$.

We shall show that \eqref{Probl} is equivalent to:

\begin{align}
\left\lbrace \begin{array}{ll}
-\Delta u = 2\rho \frac{Ke^u}{\int_\D Ke^u} & \mbox{in } \D, \\
\frac{\partial u}{\partial \eta} + 2 = 2(2\pi-\rho)\frac{he^{u/2}}{\int_\Sph he^{u/2}} & \mbox{on } \Sph, \\ \frac{(2\pi-\rho)^2}{\rho}=\frac{\left(\int_\Sph he^{u/2}\right)^2}{\int_\D Ke^u} & \mbox{for } 0<\rho<2\pi.
\end{array}\right. \label{Probl3}
\end{align}

Observe that problem \eqref{Probl3} is now invariant under addition of constants to $u$, and $\rho$ is here an unknown. This formulation may seem rather artificial but it has the advantage of being related to the critical points of the energy functional:

\begin{align}
I(u,\rho) &= \frac{1}{2} \int_\D \vert \nabla u\vert^2 - 2\rho \log \int_\D Ke^u + 2\int_{\Sph} u - 4(2\pi-\rho)\log\int_{\Sph} he^{u/2} \nonumber\\ &+4(2\pi-\rho)\log(2\pi-\rho)+2\rho +2\rho\log\rho. \label{DefFunctional}
\end{align}

We highlight the fact that the functional above depends on the couple $(u, \rho)$, where $u \in H^1(\D)$ and $\rho \in (0, 2\pi)$. The form of this energy functional seems to be completely new in the related literature.

If we freeze the variable $\rho$, the form of this functional is adequate for the use of Moser-Trudinger type inequalities (or Onofri-type inequalitites) which are already available also for boundary terms. Indeed, by interpolating these inequalities we will show that $I$ is bounded from below. We will gain coercivity in the $u$ variable by imposing symmetry, as first done by Moser in \cite{moser2}. Finally, we will need to exclude the possibility of obtaining minima at the endpoints $\rho=0$ or $\rho = 2 \pi$. Those limit cases correspond to the problem in which $K=0$ or $h=0$, respectively, so some study of these cases is needed. By energy estimates we can assure that the minimum is attained at $\rho \in (0,2\pi)$, concluding the proof.

If either $K$ or $h$ changes sign  the above approach fails. We shall also give in Theorem \ref{TheoremGeneral} a more general result; as a corollary, and making use of a compactness result for minima of the functional $I$, we will obtain the perturbation result stated in Theorem \ref{TheoremPerturbed} .

The rest of the paper is organized as follows. In Section 2 we set the notation and the variational formulation of the problem. After that, an analysis of the properties of the energy functional is performed by means of Moser-Trudinger type inequalities. Section 3 is devoted to the proof of Theorem \ref{TheoremNonNegative}, for which we first need to address the limiting cases $\rho=0$ and $\rho=2\pi$. A more general version is also given. Finally, the proof of Theorem \ref{TheoremPerturbed} is completed in Section ~\ref{sect4}.

\section{Variational Setting}
\setcounter{equation}{0}
\label{sect2}

\subsection{Notations}

Let us first set some notations. Given a set $A \subset X$ in a metric space, we denote:

$$(A)^r = \{x \in X: \mbox{dist}(x,A)<r\}.$$
Regarding the integrals, in this paper we shall consider only the Lebesgue measure and we drop the element of area or length, that is, we shall only write $\int_\D K e^u$ or $\int_{\Sph} h e^{u/2}$. We also use the symbol $\fint f$ to denote the mean value of $f$, that is, $$\fint_{\Sigma} f = \frac{1}{\vert \Sigma \vert}\int_\S f.$$

In our estimates we sometimes write $C$ to denote a positive constant, independent of the variables considered, that may change from line to line.

\subsection{Variational formulation}

As commented in the introduction, we will consider the functional $I$ given by \eqref{DefFunctional} and defined on the space
$$\X\times (0,2\pi) = \left\lbrace u\in H^1(\D): \int_\D Ke^u >0, \int_{\Sph} he^{u/2} >0 \right\rbrace\times (0,2\pi).$$
With the purpose of clarifying the notation, for a fixed $\rho \in (0,2\pi)$ we call $I_{\rho}$ to the functional $u\to I(u,\rho)$ defined for every $u\in\X$. We should notice that the functionals $I_\rho$ are invariant under the addition of constants.

\begin{lemma} \label{notempty} $\X$ is nonempty if and only if $K$ and $h$ are positive somewhere.
\end{lemma}

\begin{proof}
We reduce ourselves to prove that if $K$ and $h$ are positive somewhere then $\X$ is nonempty, as the reciprocal is immediate. As $K$ is continuous and there exists $x_0\in\mbox{Int}(\D)$ such that $K(x_0)>0$, then there exists $r>0$ such that $(\{x_0\})^r\cap \Sph = \emptyset$ and $K(x)>0$ for all $x\in (\{x_0\})^r$.

Moreover, we know that there exists $x_1\in\Sph$ satisfying $h(x_1)>0$, and again by continuity we get $s>0$ such that $h(x)>0$ for all $x\in (\{x_1\})^s\cap\Sph$. It is not restrictive to assume $(\{x_0\})^r\cap (\{x_1\})^s=\emptyset$. We call $\Omega_0^r := (\{x_0\})^r$ and $\Omega_1^s := (\{x_1\})^s$ and consider a cutoff function $\varphi \in H^1(\D)$ satisfying
\begin{align*}
\varphi(x)=\left\lbrace\begin{array}{ll} a & \mbox{if } x\in\Omega_0^{r/2}, \\ b & \mbox{if } x\in\Omega_1^{s/2}, \\ 0 & \mbox{if } \D\backslash \left(\Omega_0^r \cup \Omega_1^s\right),\end{array} \right.
\end{align*}
where $a$ and $b$ are real constants to determine. We see that:
\begin{align*}
\int_\Sph he^{\varphi/2} &= \int_{\Omega_1^{s/2}\cap\partial\D}he^{\varphi/2}+\int_{\left(\Omega_1^s\backslash\Omega_1^{s/2}\right)\cap\partial\D}he^{\varphi/2}+ \int_{\partial\D\backslash \Omega_1^s}he^{\varphi/2} \\
&\geq e^{b/2}\int_{\Omega_1^{s/2}\cap\partial\D}h+ \int_{\partial\D\backslash \Omega_1^{s/2}}h = C_1e^{b/2} + C,
\end{align*}
being $C_1>0$ and $C\in\R$. We can choose $b$ large enough so that $$\int_\Sph he^{\varphi/2}>0.$$
Furthermore,
\begin{align*}
\int_\D Ke^\varphi &= \int_{\Omega_0^{r/2}}Ke^\varphi + \int_{\Omega_1^{s/2}}Ke^\varphi + \int_{\D\backslash\left(\Omega_0^r\cup\Omega_1^r\right)}Ke^\varphi + \int_{\Omega_1^s\backslash \Omega_1^{s/2}}Ke^\varphi \\ &+ \int_{\Omega_0^r\backslash \Omega_0^{r/2}}Ke^\varphi \geq e^a \int_{\Omega_0^{r/2}}K - e^b C_2 \Vert K \Vert_\infty + C = C_1'e^a - C_2'e^b + C.
\end{align*}
So we can also set $a$ big enough so that $$\int_\D Ke^\varphi >0.$$
\end{proof}

Let us point out that the Euler-Lagrange equation of $I$ is given by \eqref{Probl3}, which is a reformulation of \eqref{Probl}, in view of next Lemma:
\begin{lemma} Problems \eqref{Probl} and \eqref{Probl3} are equivalent.
\end{lemma}
\begin{proof}
In order to check that every solution of \eqref{Probl} is a solution of \eqref{Probl3} we just need to take $\rho = \int_\D Ke^u = 2\pi-\int_\Sph he^{u/2} >0.$ Reciprocally, if $u\in \X$ solves (\ref{Probl3}), applying the invariance under addition of constant of that problem we have, for any $C\in\R:$
\begin{align*}
-\Delta (u+C) &= 2\rho \frac{Ke^{u+C}}{e^C\int_\D Ke^u}\, , \\
\frac{\partial(u+C)}{\partial \eta}+2 &= 2(2\pi-\rho)\frac{he^{u/2}}{e^{\frac{C}{2}}\int_\Sph he^{u/2}}\, .
\end{align*}
If we want $u+C$ to solve \eqref{Probl}, we need $C\in\R$ such that $$e^C = \frac{\rho}{\int_\D K e^u}\, ,\ \ e^{\frac{C}{2}}=\frac{(2\pi-\rho)}{\int_\Sph he^{u/2}}\,.$$
The third equation of \eqref{Probl3} tells us that both conditions are actually the same. Thus, it is enough to choose $C=\log\rho - \log \int_\D Ke^u$.
\end{proof}

\subsection{Moser-Trudinger inequalities}
The Moser-Trudinger inequalities (see \cite{chang1,moser1,moser2,trudinger1}) and their variations are useful tools to deal with the non-linear terms of exponential type which appear in our functional. In particular we are interested in weaker versions of these inequalities, also called Onofri type inequalities.

\begin{theorem}\label{MosTrudWeak} Let $\S$ be a compact surface with $C^1$ boundary. Then there exists a constant $C\in\R$, depending only on $\S$, such that
\begin{equation}
\log \int_\S e^u \leq \frac{1}{16\pi} \int_\S \vert \nabla u \vert ^2 + C \hsp \hsp \forall u\in H^1_0(\S), \label{equa3}
\end{equation}
and
\begin{equation}
\log \int_\S e^u \leq \frac{1}{8\pi} \int_\S \vert \nabla u\vert^2 + \fint_{\S}u + C \hsp \hsp \forall u\in H^1(\S). \label{equa4}
\end{equation}
\end{theorem}

The first inequality is classical, whereas the second is given in \cite[Proposition 2.3 and subsequent corollary]{chang1}. In both cases the constant is optimal.

In order to address the non-linear boundary terms of the functional $I$, we will use an analogous version of Theorem \ref{MosTrudWeak} for the boundary of a compact surface that can be found in \cite{li-liu}, for instance.

\begin{proposition}\label{MosTrudBoundWeak} Let $\S$ be a compact surface with $C^1$ boundary. Then there exists a constant $C>0$, depending only on $\S$, such that
	$$\log \int_{\de\S} e^u \leq \frac{1}{4\pi} \int_\S \vert \nabla u\vert^2 + \fint_{\de\S} u + C, \hsp \hsp \forall u\in H^1(\Sigma).$$
\end{proposition}

In the case of the disk, the above inequality is the so-called Lebedev-Milin inequality (with $C=0$, see for instance \cite[equation (4')]{osgood}).

By interpolating the previous inequalities we will obtain a lower bound for the functional $I$. First, we notice that inequality \eqref{equa4} can be manipulated so that the mean value of $u$ in $\de\S$ replaces the mean in $\S$.
\begin{corollary}\label{WeakModified}Let $\S$ be a compact surface with $C^1$ boundary. There exists a constant $C\in\R$, only depending on $\S$, such that $$\log \int_\S e^v \leq \frac{1}{8\pi} \int_\S \vert \nabla v \vert ^2 +\fint_\deS v + C\hsp \hsp \forall v\in H^1(\Sigma).$$
\end{corollary}

\begin{proof} We consider the problem
	\begin{equation}
	\left\lbrace \begin{array}{ll} -\Delta w = \frac{-4\pi}{\vert \Sigma \vert} & \mbox{in } \Sigma, \\[0.15cm]
	\frac{\partial w}{\partial \eta} = \frac{4\pi}{\vert \deS\vert} & \mbox{on } \partial \Sigma. \end{array}\right. \label{equa5}
	\end{equation}
	Let us point out that \eqref{equa5} is solvable in $H^1(\Sigma)$ because $\int_\deS \frac{4\pi}{\vert \deS\vert} = -\int_\S \frac{4\pi}{\vert \S\vert} = 4\pi$. We fix a solution $w$ of \eqref{equa5} and apply \eqref{equa4} to $v+w$, obtaining:
	$$\log \int_\S e^v \leq \frac{1}{8\pi}\int_\S \vert \nabla v \vert ^2 + \frac{1}{4\pi} \int_\deS \frac{\partial w}{\partial \eta} v - \frac{1}{4\pi}\int_\S (\Delta w)v + \fint_{\Sigma}v + C.$$
	Finally, we use that $w$ solves \eqref{equa5}:
	\begin{align*}
	\log \int_\S e^v &\leq \frac{1}{8\pi} \int_\S \vert \nabla v \vert ^2 + \int_\deS v-\fint_{\Sigma} v + \fint_{\Sigma} v + C \\ =& \frac{1}{8\pi} \int_\S \vert \nabla v \vert ^2 + \int_\deS v+ C.
	\end{align*}
\end{proof}
In a similar way one can obtain a modified version of Proposition \ref{MosTrudBoundWeak} in which the mean value of $u$ on $\S$ substitutes the mean on $\partial\S$.
\begin{corollary}\label{WeakModified2}Let $\Sigma$ be a compact surface with $C^1$ boundary. There exists $C\in\R$, only depending on $\S$, such that
\begin{align*}
\log \int_\deS e^u \leq \frac{1}{4\pi} \int_\S \modgrad{u} + \fint_{\Sigma} u + C \hsp \forall u\in H^1(\Sigma) .
\end{align*}
\end{corollary}

The combined use of the inequality \eqref{equa5} and Corollary \ref{WeakModified} allows us to prove that $I$ is bounded from below in $H^1(\D)$.

\begin{proposition}\label{LowerBound}There exists a constant $C\in\R$ such that $I_\rho(u)\geq C$ for every $u\in \X$ and every $\rho\in\:[0,2\pi]$.
\end{proposition}
\begin{proof}
Let us define $f:(0,2\pi)\to\R$ as the correction term in \eqref{DefFunctional}, that is
\begin{align*}
f(\rho)=4(2\pi-\rho)\log(&2\pi-\rho)+2\rho +2\rho\log\rho.
\end{align*}
It is clear that $$\lim_{\rho\to 0} f(\rho) = 8\pi \log(2\pi),\hsp \lim_{\rho\to 2\pi}f(\rho)= 4\pi + 4\pi\log(2\pi).$$
Then, $f$ can be continuously extendended to the compact $[0,2\pi]$. Thus, there exists a constant $M>0$ such that $\vert f(\rho)\vert \leq M$ for all $\rho\in$ $[0,2\pi]$. Moreover, since $K$ and $h$ are continuous, there exist $M_1, M_2 \in \R$ such that $$\log\int_\D K e^u \leq \log \int_\D e^u + C, \hsp \log\int_\Sph h e^{u/2}\leq \int_\Sph e^{u/2}+C.$$
Then, for every $a,b\in \R$:
\begin{align*}
I_\rho(u)&\geq \frac{1}{2}\int_\D\vert\nabla u\vert^2 -2\rho \log \int_\D e^u -4(2\pi-\rho)\log \int_\Sph e^{u/2} + 2\int_\Sph u + C \\
&=\frac{8\pi - 2a -b}{16\pi}\int_\D \vert\nabla u\vert^2 + \frac{a}{8\pi}\int_\D \vert\nabla u\vert^2 +\frac{b}{16\pi}\int_\D\vert \nabla u \vert ^2  \\&- 2\rho \log \int_\D e^u - 4(2\pi-\rho)\log \int_\Sph e^{u/2} + 2\int_\Sph u + C.
\end{align*}
As the functional $I$ is invariant under the addition of constants, we can assume that $\int_\S u = 0$ and apply Corollary \ref{WeakModified} and Proposition \ref{MosTrudBoundWeak} taking $a=2\rho$ and $b=4(2\pi-\rho)$, obtaining:
\begin{align*}
I_\rho(u)\geq -2\rho \fint_\Sph u - 2(2\pi-\rho)\fint_\Sph u + 2 \int_\Sph u + C = C.
\end{align*}
We highlight that the constant $C$ does not depend on $\rho$.
\end{proof}

Proposition \ref{LowerBound} states that the functional $I$ is bounded from above, but we do not have coercivity. The reason for that is the non-compact action of the conformal group of the disk. This effect appears also in the Nirenberg problem in the sphere, for instance, and makes the problem rather difficult.

We will show now that we can gain coercivity by restricting ourselves to spaces of symmetric functions. In order to do that, we introduce local versions of the inequalities above. These results are known as Chen-Li type inequalities (see \cite{chen-li} for more details).

\begin{proposition}\label{Local1} Let $\Sigma$ be a compact surface with $C^1$ boundary, $\Sigma_1\subset\Sigma$ and $\delta>0$ such that $(\Sigma_1)^{\delta}\cap\partial\Sigma = \emptyset.$ Then, for every $\e>0$, there exists a constant $C\in\R$ depending on $\e$ and $\delta$ such that $$16\pi \log \int_{\Sigma_1}e^u \leq \int_{(\Sigma_1)^\delta}\vert \nabla u \vert^2 + \e \int_\S \vert \nabla u\vert^2 + C, \hsp \forall u\in H^1(\Sigma) \mbox{ with } \int_\S u = 0.$$
\end{proposition}

 The details of the proof of this precise statement can be found in \cite[Proposition 2.2]{Ruiz}, for instance, but the idea dates back to \cite{chen-li}. Roughly speaking, one applies \eqref{equa3} to the function $u$ multiplied by a cut-off function in $\Sigma_1$.

If the function $u$ has mass in several separated regions satisfying the hypothesis of the propositions above, the obtained bounds improve by a factor of the number of such regions. This information is collected in the following corollary (see for instance \cite[Lemma 2.4]{Ruiz} for the case $l=2$; the case of general $l$ is analogous).

\begin{corollary} \label{Local2Corollary} Let $\Sigma$ be a compact surface with $C^1$ boundary, $l\in\N$ and $\Sigma_1, \ldots, \Sigma_l \subset \Sigma$ for which there exists a $\delta>0$ such that
	$(\Sigma_i)^\delta\cap(\Sigma_j)^\delta = \emptyset$ if $i\neq j$. Assume that there exists $\gamma \in (0,\frac{1}{l})$ such that $$\frac{\int_{\Sigma_i}e^u}{\int_\S e^u}\geq \gamma,\hsp\forall i=1,\ldots,l.$$Then, for every $\e>0$ there exists a constant $C\in\R$ depending on $\e$, $\delta$ and $\gamma$ such that
	\begin{align*}
	8l\pi \log \int_\S e^u \leq \int_\S \vert \nabla u \vert^2 + \e \int_\S \vert \nabla u\vert^2 + C, \hsp \forall u\in H^1(\Sigma) \mbox{ with } \int_\S u = 0.
	\end{align*}
\end{corollary}

Using the same techniques we can give a localized version of the Proposition \ref{MosTrudBoundWeak}.

\begin{proposition}\label{Local3} Let $\S$ be a compact surface with $C^1$ boundary, and $\Gamma_1\subset \de\S$. Then, for every $\e,\delta>0$ there exists a constant $C\in\R$ depending on $\e$ and $\delta$ such that $$4\pi \log \int_{\Gamma_1}e^u \leq \int_{(\Gamma_1)^\delta}\vert \nabla u \vert ^2 + \e \int_\S \vert \nabla u \vert^2 + C, \hsp \forall u\in H^1(\Sigma) \mbox{ with } \int_\S u = 0$$
\end{proposition}

\begin{proof} Following \cite{chen-li}, we consider a cutoff function $g_\delta: \S \to [0,1]$ satisfying
\begin{equation*}
g_\delta=\left\lbrace \begin{array}{ll} 1 & \mbox{ if } x\in\Gamma_1, \\ 0 & \mbox{ si } x\in \Sigma\backslash (\Gamma_1)^{\delta/2}. \end{array} \right.
\end{equation*}
We have $g_\delta u\in H^1(\Sigma)$, hence we can apply Corollary \ref{WeakModified2}:
\begin{align}
4\pi \log \int_{\Gamma_1} e^u &= 4\pi \log \int_{\Gamma_1} e^{g_\delta u} \leq 4\pi \log \int_\deS e^{g_\delta u} \nonumber \\ &\leq \int_\S \vert \nabla(g_\delta u)\vert^2 + 4\pi\fint_\S g_\delta u + C. \label{equa6}
\end{align}
Then,
\begin{align}
\int_\S \vert \nabla(g_\delta u)\vert^2 &= \int_\S u^2 \vert \nabla g_\delta \vert^2 + 2 \int_\S g_\delta u \langle \nabla u, \nabla g_\delta \rangle + \int_\S (g_\delta)^2 \vert \nabla u \vert^2 \nonumber \\
&\leq C_\delta \int_\S u^2 + 2\int_\S g_\delta u \vert \nabla u \vert \vert \nabla g_\delta \vert + \int_{(\Gamma_1)^\delta} \vert \nabla u \vert^2. \label{equa7}
\end{align}
The central term can be bounded using Cauchy's inequality, obtaining
\begin{align}
\int_\S g_\delta u \vert \nabla u\vert \vert \nabla g_\delta \vert \leq C_\delta \int_\S u\vert \nabla u\vert \leq C_{\e,\delta} \int_\S u^2 + \e \int_\S \vert \nabla u \vert ^2. \label{equa8}
\end{align}
Combining \eqref{equa7} and \eqref{equa8}:
\begin{align}
\int_\S \vert \nabla(g_\delta u)\vert^2 \leq \int_{(\Gamma_1)^\delta}\vert \nabla u\vert^2 + \e \int_\S \vert \nabla u \vert^2 + C_{\e,\delta}\int_\S u^2. \label{equa9}
\end{align}
Also, we have the following bound for the mean value of $g_\delta u$ on $\de\S$:
\begin{equation}
\fint_\S g_\delta u \leq \fint_\S \frac{1}{2}((g_\delta)^2+u^2) \leq \frac{1}{2}\fint_\S (g_\delta)^2 + \frac{1}{2\vert \Sigma\vert}\int_\S u^2 \leq C_\delta + C\int_\S u^2. \label{equa10}
\end{equation}
Now, apply both inequalities \eqref{equa9} and \eqref{equa10} to \eqref{equa6} to get:
\begin{equation}
4\pi \log \int_{\Gamma_1}e^u \leq \int_{(\Gamma_1)^\delta} \vert \nabla u\vert^2 + \e \int_\S \vert \nabla u\vert^2 + C_{\e,\delta}\int_\S u^2 + C. \label{equa11}
\end{equation}
Finally we address the term $\int_\S u^2$.

Let $a\in\R$, $\eta = \vert \{x\in\Sigma: u(x)\geq a\}\vert$ and $(u-a)^+ = \max\:\{0,u-a\}$. Clearly, $u\leq (u-a)^+ + a$. We now apply formula \eqref{equa11} to the function $(u-a)^+$:
\begin{align}
4\pi \log \int_{\Gamma_1} e^u &\leq 4\pi \log \left(e^a \int_{\Gamma_1}e^{(u-a)^+} \right) \leq 4\pi a + \log \int_{\Gamma_1} e ^{(u-a)^+} \nonumber \\ &\leq 4\pi a + \int_{(\Gamma_1)^\delta} \vert \nabla(u-a)^+\vert^2 + \e \int_\S \vert \nabla(u-a)^+\vert^2 + C_{\e,\delta}\int_\S \left((u-a)^+\right)^2 \nonumber\\ &\leq 4\pi a + \int_{(\Gamma_1)^\delta}\vert \nabla u\vert^2 + \e \int_\S\vert \nabla u \vert ^2 + C_{\e,\delta}\int_\S \left((u-a)^+\right)^2. \label{equa12}
\end{align}
By means of Sobolev, H\"{o}lder and Poincar\'{e}-Wirtinger inequalities:
\begin{align} \label{equa13}
\int_\S \left((u-a)^+\right)^2 &= \int_{\{x\in\Sigma:\: u(x)\geq a\}} \left((u-a)^+ \right)^2 \leq \eta^{1/2} \left(\int_\S \left((u-a)^+\right)^4\right)^{1/2} \nonumber \\ &\leq \eta^{1/2} \Vert (u-a)^+\Vert^2_{H^1(\Sigma)} \leq C\eta^{1/2}\int_\S\vert \nabla u\vert^2.
\end{align}
Again by Poincar\'{e}-Wirtinger inequality:
\begin{align} \label{equa14}
a\eta \leq \int_{\{x\in\Sigma:\: u(x)\geq a\}} u \leq \int_\S \vert u \vert \leq C \left(\int_\S \vert u\vert^2\right)^{1/2} \leq C\left(\int_\S\vert \nabla u \vert^2\right)^{1/2}.
\end{align}
From \eqref{equa14}, using Cauchy's inequality:
\begin{align} \label{equa15}
a \leq \theta \int_\S \vert\nabla u\vert^2 + \frac{C^2}{\eta^2 \theta}, \hsp \forall \theta >0.
\end{align}
Mixing \eqref{equa12}, \eqref{equa13} and \eqref{equa15}:
\begin{align*}
4\pi\log\int_{\Gamma_1} e^u \leq 4\pi \theta \int_\S \vert\nabla u\vert^2 + \int_{(\Gamma_1)^\delta} \vert\nabla u\vert^2 + \e \int_\S \vert\nabla u \vert^2 + C_{\e,\delta}\eta^{1/2}\int_\S \vert \nabla u\vert^2 + C,
\end{align*}
and it is enough to take $\theta = \frac{1}{4\pi}$ and $\eta^{1/2}\leq \frac{\e}{C_{\e,\delta}}$ to conclude.
\end{proof}

\begin{corollary} \label{Local3Corollary} Let $\S$ be a compact surface with $C^1$ boundary, $l\in\N$ and $\Gamma_1, \ldots, \Gamma_l \subset \partial\Sigma$ for which there exists a $\delta>0$ such that $(\Gamma_i)^\delta\cap(\Gamma_j)^\delta = \emptyset$ if $i\neq j$. Moreover, assume that there exists $\gamma \in \:(0,\frac{1}{l})$ such that

\begin{equation} \label{nueva} \frac{\int_{\Gamma_i}e^u}{\int_\deS e^u}\geq \gamma,\hsp \forall i=1,\ldots,l .\end{equation}

Then, for every $\e>0$ there exists a constant $C\in\R$, depending on $\e, \delta$ and $\gamma$, such that
\begin{align*}
4l\pi \log \int_\deS e^u \leq \int_\S \vert \nabla u \vert^2 + \e \int_\S \vert \nabla u \vert^2 + C, \hsp \forall u\in H^1(\Sigma) \mbox{ with } \int_\S u = 0.
\end{align*}
\end{corollary}
\begin{proof}
First, we apply to each $\Gamma_i$ the previous result, obtaining
$$4\pi \log \int_{\Gamma_i} e^u \leq \int_{(\Gamma_i)^\delta}\vert \nabla u \vert^2 + \e \int_\S \vert \nabla u\vert^2 + C.$$
Using \eqref{nueva}, we obtain:
\begin{align*}
4\pi \log \int_{\Gamma_i} e^u \geq 4\pi \log \int_\deS e^u + C.
\end{align*}
Then,
\begin{align*}
4\pi \log \int_\deS e^u \leq \int_{(\Gamma_i)^\delta} \vert \nabla u\vert^2 + \e \int_\S\vert\nabla u\vert^2 + C.
\end{align*}
Finally, summing on $i\in\{1,\ldots,l \}$:
\begin{align*}
4l\pi \log \int_\deS e^u &\leq \int_{\bigsqcup_{i} (\Gamma_i)^\delta} \vert\nabla u\vert^2 + \e l \int_\S \vert\nabla u\vert^2 + C \\ &\leq \int_\S \vert \nabla u \vert^2 + \e l \int_\S \vert\nabla u\vert^2 + C.
\end{align*}
\end{proof}
We have just seen that the more regions the mass of a function is separated in, the better bounds we obtain using the local versions of the Moser-Trudinger inequalities. If a $H^1(\D)$ function is concentrated in an interior point of the disk, Proposition \ref{Local1} gives us a lower bound which is sufficient to achieve coercivity, but that is not the case when a function concentrates around a boundary point. To avoid this we will restrict ourselves to consider functions satisfying a symmetry condition guaranteeing that a function cannot concentrate around a single point of the boundary. Hence we will obtain coercivity by interpolating \ref{Local2Corollary} and \ref{Local3Corollary} with $l=2$.
\begin{figure}[h]
\includegraphics[scale=0.3]{symm.pdf}
\caption{If a symmetric function concentrates around $x\in\Sph$, then it also concentrates around $g(x)$ for all $g\in G$.}
\end{figure}

We let $G$ be a subgroup of the orthogonal transformation group of $\D$ such that the set of fixed points on $\Sph$ under the action of $G$ is empty, in other words,
\begin{equation*}
\{x\in\Sph: g(x)=x \hsp \forall g\in G\} = \emptyset.
\end{equation*}
For instance, we can take $G$ as the group of rotations generated by $g(z)= e^{\frac{2 \pi i}{k}}z$,  as well as the dihedral groups $\mathbb{D}_k$ ($k \in \mathbb{N}$, $k>1$).

In the sequel, $K$ and $h$ will be assumed to be $G-$symmetric functions, and we denote $H^1_G(\D)= \{ u \in H^1(\D):\ u \circ g = u \ \forall \ u \in G \}$, and:
\begin{equation*}
\X_G = \{u\in \X: u\circ g = u \hsp \forall g\in G\}.
\end{equation*}

As in Lemma \ref{notempty} we observe that if $K$ and $h$ are G-symmetric functions somewhere positive, then $\X_G$ is not empty.

\begin{proposition}\label{Coercivity} Given $\rho\in \: [0,2\pi]$, the functional $I_\rho$ is coercive on $\X_G$, that is, $$I_\rho(u)\to +\infty \hsp (\Vert u\Vert_{H^1(\D)}\to +\infty, u\in \X_G).$$
\end{proposition}

\begin{proof} Take a sequence $(u_n)$ in $\X_G$. We know that $I_{\rho}$ is invariant under the addition of constants, so we can assume that $\int_\D u_n = 0$ for every $n\in\N$. We have
\begin{align*}
I_\rho(u_n)&\geq \frac{1}{2}\int_\D\vert \nabla u_n\vert^2 -2\rho \log \int_\D e^{u_n} -4(2\pi-\rho)\log \int_\Sph e^{u_n/2} + 2\int_\Sph u_n + C.
\end{align*}
Then, for any $a,b\in\R$ one has:
\begin{align*}
I_\rho(u_n) &\geq \frac{16\pi - 2a -b}{32\pi} \int_\D\vert \nabla u_n\vert^2 + \frac{a}{16\pi}\int_\D \vert \nabla u_n\vert^2 + \frac{b}{32\pi}\int_\D \vert \nabla u_n\vert^2 + 2\int_\Sph u_n \\ &-2\rho \log \int_\D e^{u_n} -4(2\pi -\rho)\log\int_\Sph e^{u_n/2}.
\end{align*}
We can now apply Corollaries \ref{Local2Corollary} and \ref{Local3Corollary} with $l=2$:
\begin{align*}
I_\rho(u_n)&\geq \frac{16\pi-2a-b}{32\pi}\int_\D \vert \nabla u_n\vert^2 + a\log\int_\D e^{u_n} - a \e \int_\D \vert \nabla u_n\vert^2 + b \log \int_\Sph e^{u_n/2} \\ &-b \e \int_\D \vert \nabla u_n\vert^2 - 2\rho \log \int_\D e^{u_n} - 4(2\pi-\rho)\log \int_\Sph e^{u_n/2} + 2 \int_\Sph u_n + C.
\end{align*}
Choosing $a=2\rho$ and $b=4(2\pi-\rho)$ and applying the trace inequality:
\begin{align*}
I_\rho(u_n) \geq \left(\frac{1}{4}-\e\right) \int_\D \vert \nabla u_n\vert^2 - 2C_2 \Vert u_n \Vert_{H^1(\D)} + C, \hsp C_2 >0.
\end{align*}
Finally, taking $\e$ small enough and using the Poincar\'{e}-Wirtinger inequality we obtain
\begin{equation}
I_\rho(u_n) \geq C_1 \Vert u_n \Vert_{H^1(\D)} ^2 - C_2 \Vert u_n \Vert_{H^1(\D)} + C, \hsp C_1, C_2 >0. \label{equa16}
\end{equation}
Again, we remark that the constant $C_1$ is independent of $\rho$.
\end{proof}

\section{Proof of Theorem \ref{TheoremNonNegative} and its generalization}
\label{sect3}
\setcounter{equation}{0}

We begin this section considering the limiting cases $\rho=0$ and $\rho=2\pi$. These cases have their own interest, as will be shown, but their study will be useful also for the proof of Theorems \ref{TheoremNonNegative}, \ref{TheoremPerturbed} and \ref{TheoremGeneral}.

\medskip Observe that:

\begin{equation} \label{I(u,0)} I(u,0) = \frac{1}{2} \int_\D \vert \nabla u\vert^2  + 2\int_{\Sph} u - 8 \pi \log\int_{\Sph} he^{u/2} + 8\pi \log(2\pi),\end{equation}
and, as $K$ does not play any role, it can be defined on the bigger space
$$\X^1=\left\lbrace u\in H^1(\D): \int_\Sph he^{u/2} > 0 \right\rbrace \supset \X.$$
The critical points of $I_0$ on $\X^1$ are weak solutions of the problem
\begin{align*}
\left\lbrace \begin{array}{cc} -\Delta u = 0 & \mbox{ in } \D, \\[0.15cm] \frac{\partial u}{\partial \eta} + 2 = 4\pi \frac{he^{u/2}}{\int_\Sph he^{u/2}} & \mbox{ on } \Sph,
\end{array} \right.
\end{align*}
which is clearly equivalent to the problem of prescribing Gaussian curvature $K=0$ and geodesic curvature $h$, that is,

\begin{align}
\left\lbrace \begin{array}{cc} -\Delta u = 0 & \mbox{ in } \D, \\[0.15cm] \frac{\partial u}{\partial \eta} + 2 = 2h e ^{u/2} & \mbox{ on } \Sph.
\end{array} \right.\label{Probl4}
\end{align}

Under the hypothesis that $h$ is $G$-symmetric, we can seek a minimizer of $I_0$ on the space of symmetric functions $$\X^1_G = \left\lbrace u\in \X^1:u\circ g = u \hsp \forall g\in G \right\rbrace.$$

\begin{theorem} \label{TheoremGeodesic} Let $h: \Sph \to \R$ be a H\"{o}lder continuous, somewhere positive $G-$symmetric function. Then Problem \eqref{Probl4} admits a solution as a minimum of $I_0$ on $\X^1_G$.
\end{theorem}

\begin{proof} The functional is bounded from below as seen in Proposition \ref{LowerBound}, so there exists $$\alpha = \inf_{u\in\X^1_G}I_0(u).$$
Let $(u_n)$ be a minimizing sequence in $\X^1_G$, that is, $I_0(u_n)\to \alpha$. By Proposition \ref{Coercivity} we know that $I_0$ is coercive so $u_n$ is bounded in the $H^1(\D)$ norm and we can assume that, there exists $u_0$ in $H^1(\D)$ such that, up to a subsequence, $u_n\weakto u_0$. Then, we also have $$\int_\Sph u_n \to \int_\Sph u_0,\hsp \int_\Sph he^{u_n/2} \to \int_\Sph he^{u_0/2}.$$
Combining this information with the fact that the function $u\to \int_\D \modgrad{u}$ is weakly lower semicontinuous, we have $I_0(u_0)\leq \alpha$. It is easy to check that $\int_\Sph he^{\frac{u_0}{2}}>0$, because if we had $\int_\Sph he^{u_n/2}\to 0$ then $I_0(u_n)\to +\infty$, which contradicts that $u_n$ is minimizing. Also, notice that weak convergence respect symmetry, so $u_0$ is a $G-$symmetric function.
\end{proof}

Analogously, we can consider the functional related to the limiting case $\rho= 2\pi$,

 \begin{equation} \label{I(u, 2pi)} I(u, 2\pi)= \frac{1}{2} \int_\S \modgrad{u} + 2\int_\Sph u -4\pi \log \int_\D K e^u + 4\pi +4\pi\log(2\pi). \end{equation}
defined on the space
$$\X_G^2 = \left\{u\in H^1_G(\D): \int_\D Ke^u >0 \right\} \supset \X_G.$$
One can check that its variation with respect to $u$ produces weak solutions of the problem
\begin{align*}
\left\lbrace \begin{array}{cc} -\Delta u = 4\pi\frac{Ke^u}{\int_\D Ke^u} & \mbox{ in } \D, \\[0.15cm] \frac{\partial u}{\partial \eta} + 2 = 0 & \mbox{ on } \Sph,
\end{array} \right.
\end{align*}
which is equivalent to the problem of prescribing geodesic curvature $h=0$ and Gaussian curvature $K$:
\begin{align}
\left\lbrace \begin{array}{cc} -\Delta u = 2Ke^u & \mbox{ in } \D, \\[0.15cm] \frac{\partial u}{\partial \eta} + 2 = 0 & \mbox{ on } \Sph.
\end{array} \right. \label{Probl5}
\end{align}

A trivial adaptation of the proof of Theorem \ref{TheoremGeodesic} gives the following:

\begin{theorem} \label{TheoremGaussian} Let $K: \D \to \R$ be a H\"{o}lder continuous, somewhere positive $G-$symmetric function. Then Problem \eqref{Probl4} admits a solution as a minimum of $I_{2\pi}$ on $\X^2_G$.
\end{theorem}

\begin{remark} The existence result of Theorem \ref{TheoremGeodesic} is known, see for instance \cite{liuwang}. We have not found a explicit statement of the existence result of Theorem \ref{TheoremGaussian}, but we guess that it must be also known. However in this section we have reinterpreted those solutions as minimizers of $I_0$ and $I_{2\pi}$, respectively. This will be of use in what follows.

\end{remark}

Let us now conclude the proof of Theorem \ref{TheoremNonNegative}.

\begin{proof}[Proof of Theorem \ref{TheoremNonNegative}.]

If $K =0$ or $h = 0$, then we are under the assumptions of  Theorem \ref{TheoremGeodesic} or Theorem \ref{TheoremGaussian}. Then, we can assume that both $K$ and $h$ are positive in some point and non-negative. In this case, $\X_G = \X^1_G =\X^2_G = H^1_G(\D)$.

By Proposition \ref{Coercivity}, there exists $(\hat{u}, \hat{\rho}) \in H^1_G(\D) \times [0,2\pi]$ a minimizer for $I$. We conclude if we exclude the possibilities $\hat{\rho}=0$ or $\hat{\rho}=2\pi$.

Assume that $\hat{\rho}=0$. Observe that in this case, $\hat{u}$ is a minimizer for $I(\cdot, 0)$. Then,

\begin{align*} I(\hat{u},0) \leq I(\hat{u}, \rho) = I(\hat{u},0)  -2 \rho \log \Big (\int_{\D} K e^{\hat{u}} \Big ) + 4 \rho \log \Big (\int_{\Sph} h e^{\hat{u}/2} \Big ) \\ + 8 \pi \log \Big (\frac{2\pi-\rho}{2\pi} \Big)- 4 \rho \log (2 \pi- \rho) + 2 \rho + 2 \rho \log \rho.\end{align*}
But observe that, as $\rho \to 0$, the main term above is $2 \rho \log \rho$, which is negative. This gives a contradiction that excludes the case $\hat{\rho}=0$. One can exclude the case $\hat{\rho} = 2 \pi$ in an analogous way.

\end{proof}

The proof of Theorem \ref{TheoremNonNegative} can be adapted to a more general setting as follows:
\begin{theorem}\label{TheoremGeneral} Let $K$ and $h$ be H\"{o}lder continuous $G-$symmetric functions that are positive somewhere. We define
\begin{align*}
S_0 = \left\lbrace u\in \X^1_G:\: I_0(u)=\min_{\X^1_G} I_0  \right\rbrace, \hsp S_{2\pi} = \left\lbrace u\in \X^2_G:\: I_{2\pi}(u)=\min_{\X^2_G} I_{2\pi} \right\rbrace.
\end{align*}
If $S_0\cap\X_G$ and $S_{2\pi}\cap\X_G$ are nonempty, then \eqref{Probl} admits a solution.
\end{theorem}

Clearly, Theorem \ref{TheoremNonNegative} is an immediate consequence of Theorem \ref{TheoremGeneral}. Notice also that the sets $S_0$ and $S_{2\pi}$ of the hypotheses are nonempty because of Theorems \ref{TheoremGeodesic} and \ref{TheoremGaussian}.

\begin{proof}
The proof follows the same energy comparison argument than above, but a couple of details are worth to be written down. First, the existence of a minimizer is not clear a priori. Then let $(u_n, \rho_n ) \in \X_G \times (0,2\pi)$ be a minimizing sequence, that is, $I(u_n, \rho_n) \to \inf I$. Clearly $u_n$ is bounded in $H^1_G(\D)$ by Proposition \ref{Coercivity}, but its weak limit $\hat{u}$ could fall outside $\X_G$.

If $\rho_n \to \hat{\rho} \in (0,2\pi)$, from the fact that $I(u_n, \rho_n)$ is bounded we obtain:

$$ 0< \e < \int_{\D} K e^{u_n} < C, \ \ 0< \e < \int_{\Sph} h e^{u_n/2}< C,$$
for some $\e>0$, $C>0$. As a consequence $u_n \rightharpoonup \hat{u} \in \X_G$ and we are done.

\medskip Assume now that $\rho_n \to 0$. If $n$ is sufficiently large we have the estimate:

$$I(u_n,\rho_n)\geq -2\rho_n \log\left(\int_\D Ke^{u_n}\right)-4(2\pi-\rho_n)\log\left(\int_\Sph he^{u_n/2}\right)+C.$$
Notice that $$\liminf\limits_{n\to\infty}-2\rho_n \log\left(\int_\D Ke^{u_n} \right)\geq 0.$$ Thus, $-\log\left(\int_\Sph h e^{u_n/2}\right)$ must be bounded from above, which means that $$0<\e<\int_\Sph he^{u_n/2}.$$

Now, we write:
\begin{align*}I(u_n,\rho_n) = I(u_n,0)  -2 \rho_n \log \Big (\int_{\D} K e^{u_n} \Big ) + 4 \rho_n \log \Big (\int_{\Sph} h e^{u_n/2} \Big ) \\+ 8 \pi \log \Big (\frac{2\pi-\rho_n}{2\pi} \Big)- 4 \rho_n \log (2 \pi- \rho_n) + 2 \rho_n + 2 \rho_n \log \rho_n.\end{align*}
From this we deduce that:

$$ \inf I = \lim_{n \to \infty} I(u_n,\rho_n) \geq \liminf_{n \to \infty}  I(u_n,0) \geq I(u_0,0),$$
where $u_0 \in S_0 \cap \X_G$. But, as in the proof of Theorem \ref{TheoremNonNegative},

$$I(u_0,0) > I(u_0, \rho),$$
for small values of $\rho$. This contradiction shows that $\rho_n $ cannot converge to $0$. In an analogous way we can exclude its convergence to $2 \pi$.

\end{proof}

\section{A perturbation result}
\label{sect4}
\setcounter{equation}{0}

In this section it is necessary to specify the dependence of $I$ on the curvature functions $K$ and $h$, so we are writting $I(u, \rho) = I[K,h](u,\rho)$. We begin with a compactness result:

\begin{lemma}\label{Compactness} Let $(K_n)$ and $(h_n)$ be sequences of  H\"{o}lder continuous $G-$symmetric functions, defined on $\D$ and $\Sph$ respectively, such that
\begin{align*}
K_n &\to K \hsp \mbox{uniformly in } \D \mbox{ and } K\in C^{0,\alpha}(\D), \\
h_n &\to h \hsp \mbox{uniformly on } \Sph \mbox{ and } h\in C^{0,\alpha}(\Sph).
\end{align*}
Let us consider a sequence $(u_n)$, where each $u_n$ is a solution of the problem
\begin{align} \label{Probl6}
\left\lbrace \begin{array}{ll} -\Delta u = 2K_n e^u & \mbox{in } \D, \\ \frac{\partial u}{\partial n}+2 = 2h_ne^{u/2} & \mbox{on } \Sph, \end{array}\right.
\end{align}
satisfying
\begin{align}
\rho_n = \int_\D K_n e^{u_n} > 0,\hsp \int_\Sph h_n e^{u_n/2} > 0, \hsp \forall n\in\N. \label{equa17}
\end{align}
Assume that $I[K_n,h_n](u_n, \rho_n)$ is uniformly bounded from above. Then $u_n\weakto u_\infty$ on $H^1(\D),$ being $u_\infty$ a solution of the problem
\begin{align}
\left\lbrace \begin{array}{ll} -\Delta u = 2K e^u & \mbox{in } \D, \\ \frac{\partial u}{\partial n}+2 = 2he^{u/2} & \mbox{on } \Sph. \end{array}\right. \label{Probl7}
\end{align}
\end{lemma}
\begin{proof} First, we notice that  $\Vert K_n - K\Vert_\infty \to 0$ and $\Vert h_n - h\Vert_\infty \to 0$ imply that for every $\e > 0$ there exists $n_0\in\N$ such that, for $n\geq n_0$: $$\Vert K_n \Vert_\infty < \Vert K \Vert_\infty+\e,\hsp \Vert h_n \Vert_\infty < \Vert h \Vert_\infty+\e.$$
Hypothesis \eqref{equa17} gives us $0< \rho_n < 2\pi$ for all $n\in\N$. Then, for $n\geq n_0$ we have the following bound:
\begin{equation*}
I[K_n, h_n](u_n,\rho_n) \geq I(\Vert K\Vert_\infty + \e, \Vert h\Vert_\infty + \e)(u_n,\rho_n) .
\end{equation*}
And then, by Proposition \ref{Coercivity}, there exist constants $C_1, C_2 >0$, independent of $n$, such that
\begin{align*}
I[K_n, h_n](u_n,\rho_n)\geq C_1\Vert u_n\Vert_{H^1}^2-C_2\Vert u_n\Vert_{H^1}+C_\e.
\end{align*}
Taking into account the hypothesis that $I[K_n, h_n](u_n,\rho_n)$ is uniformly bounded from above we have immediately that $u_n$ is bounded in the $H^1(\D)$ norm. Hence, up to a subsequence we can assume that there exists $u_\infty\in H^1(\D)$ such that $u_n\weakto u_\infty$.

Then, it is known that $2K_ne^{u_n}\to 2Ke^{u_\infty}$ and $2h_n e^{u_n/2} \to 2he^{\frac{u_\infty}{2}}$ on $L^p$ for $1\leq p < +\infty$, and that $\langle \nabla u_n, w \rangle\to \langle \nabla u_\infty, w\rangle$ for all $w\in H^1(\D)$. In particular $$ u_n|_{\Sph} \to  u_\infty|_{\Sph} \ \mbox{ in } L^2(\Sph).$$
We now pass to the limit in the weak formulation of \eqref{Probl6}:
\begin{align}
\int_\D \langle\nabla u_n ,\nabla v\rangle -2  \int_\D K_n e^{u_n} v +2 \int_\Sph v -\int_\Sph h_ne^{u_n/2} v = 0. \label{WeakSol}
\end{align}
for all $v\in H^1(\D)$. As a consequence $u_{\infty}$ is a weak solution of \eqref{Probl7}. By standard regularity estimates $u_\infty$ is indeed a classical solution.

\end{proof}

The next step is to check that, when considering a sequence of minimum type solutions, the hypothesis of Lemma \ref{Compactness} are automatically satisfied. Observe that under our hypotheses $I[K_n,h_n](\cdot,\cdot)\to I[K,h](\cdot,\cdot)$ pointwise in $\X\times (0,2\pi)$.

Then, if $(u_n,\rho_n)$ is a sequence of minimum type solutions of \eqref{Probl6}, $$\limsup_{n\to+\infty}I[K_n,h_n](u_n,\rho_n)= \limsup_{n\to+\infty}\min_{\X\times(0,2\pi)}I[K_n,h_n](\cdot,\cdot)\leq \min_{\X\times(0,2\pi)}I.$$
Where the previous inequality is due to the fact that $(f_n)$ converging pointwise to $f$ implies
$\lim_{n\to+\infty}\inf f_n(y)\leq \inf f(y)$.
\begin{proof}[Proof of Theorem \ref{TheoremPerturbed}] We apply Theorem \ref{TheoremGeneral} to the problems
\begin{align*}
\left\lbrace \begin{array}{ll} -\Delta u = 2K e^u & \mbox{in } \D, \\[0.15cm] \frac{\partial u}{\partial \eta}+2 = 2he^{u/2} & \mbox{on } \Sph \end{array}\right.
\end{align*}
for which we need that the limiting problems
\begin{align*}
(P^1_{K})\left\lbrace \begin{array}{ll} -\Delta u = 2K e^u & \mbox{in } \D \\ \frac{\partial u}{\partial n}+2 = 0 & \mbox{on } \Sph \end{array}\right.,\hsp (P^2_{h})\left\lbrace \begin{array}{ll} -\Delta u = 0 & \mbox{in } \D \\ \frac{\partial u}{\partial \eta}+2 = 2he^{u/2} & \mbox{on } \Sph \end{array}\right.
\end{align*}
admit minimum type solutions, $u_1$ and $u_2$ respectively, verifying
$$\int_\D Ke^{u_2}>0,\hsp \int_\Sph he^{u_1/2}>0.$$

By contradiction, take $K_n$ and $h_n$  H\"{o}lder continuous functions converging uniformly to $K_0$ and $h_0$. We can assume that $n$ is large enough so that $K_n$ and $h_n$ are somewhere positive, so that solutions for the limiting problems in the form of minimizers can be found via Theorems \ref{TheoremGeodesic} and \ref{TheoremGaussian}. Now, take $(\tilde{u}_n)$ a sequence of minimum type solutions of the problems $(P^1_{K_n})$ and $(\hat{u}_n)$ a sequence of minimum type solutions of the problems $(P^2_{h_n})$ such that
\begin{equation}\label{equa18}
\mbox{either } \int_\D K_n e^{\hat{u}_n}\leq 0, \ \mbox{ or }\ \int_\Sph h_ne^{\tilde{u}_{n}/2}\leq 0, \hsp \forall n\in \N.
\end{equation}
By Lemma \ref{Compactness} we know that $\hat{u}_n\weakto\hat{u}$ and $\tilde{u}_n\weakto\tilde{u}$, solutions for the limiting problems $(P^1_{K_0})$ and $(P^2_{h_0})$. Taking limit when $n\to +\infty$ in \eqref{equa18} we obtain:

$$ \mbox{either }  \int_\D K_0\,e^{\tilde{u}}\leq 0, \ \mbox{ or }\ \int_\Sph h_0\,e^{\hat{u}/2}\leq 0,$$ which is a contradiction since both $K_0$ and $h_0$ are nonnegative functions somewhere positive.
\end{proof}

config: arxiv


TITLE: Permutations of Intersections of Parabola and Circles
QUESTION [3 upvotes]: What is the maximum number of points of intersection for four distinct parabolas and three distinct circles drawn on a sheet of paper?
Anyone good enough to solve this.?
Well considering the two different parabolas can intersect at 4 different points. WHen you bring a third one in, its even more.
Two circles intersecting bring 2 points. A circle and a parabola brings 4 I guess. So how many in total?

REPLY [5 votes]: 6 times 4 parabola-parabola intersections, 12 times 4 parabola-circle intersections, 3 times 2 circle-circle intersections.


TITLE: $\mathbb{F}_{3}[x]/(x^3-x^2+1) \cong \mathbb{F}_{3}[x]/(x^3-x^2+x+1)$.
QUESTION [1 upvotes]: I want to prove that $\mathbb{F}_{3}[x]/(x^3-x^2+1) \cong \mathbb{F}_{3}[x]/(x^3-x^2+x+1)$.
I have by Proposition 3.117 from Rotman´s Advanced Algebra that both are fields and we have that:
$$\mathbb{F}_{3}[x]/(x^3-x^2+1)= \lbrace a+ b\alpha / \alpha^3-\alpha^2+1=0 \rbrace$$
And;
$$\mathbb{F}_{3}[x]/(x^3-x^2+x+1)= \lbrace a+b \beta / \beta^3-\beta^2+\beta+1  \rbrace$$.
So how can I define a properly isomorphism between both fields? Thanks

REPLY [1 votes]: Both $x^3-x^2+1$ and $x^3-x^2+x+1$ are irreducible over $\Bbb F_3$ (since  neither has a root).  Thus both quotients are the field $\Bbb F_{27}=\{\alpha+\beta x+\gamma x^2\mid \alpha,\beta,\gamma \in\Bbb F_3\}$.


TITLE: Does the Grothendieck group of finitely generated modules form a commutative ring where the multiplication structure is induced from tensor product?
QUESTION [7 upvotes]: For a commutative ring $R$, let $\mathrm{mod-}R$ denote the category of finitely generated $R$-modules. Let $\mathcal C$ be an abelian, full, isomorphism closed (i.e. $M\cong N$ in $\mathrm{mod-}R$ and $M\in Ob(\mathcal C)\implies N\in Ob(\mathcal C)$) subcategory of $\mathrm{mod-}R$.
The Grothendieck group of $\mathcal C$, usually denoted by $G(\mathcal C)$, is defined as
$$\dfrac{\bigoplus_{X\in \operatorname{Ob}(\mathcal C)}\mathbb Z[X]}{\langle [A]-[B]+[C] \mid 0\to A \to B \to C\to 0  \text{ is exact}   \rangle }$$
where $[X]$ denotes the isomorphism class of $X$.
Now if for every $M,N\in Ob(\mathcal C)$, we also have $M\otimes_R N\in Ob(\mathcal C)$, then the $\mathbb Z$-module
$$\bigoplus_{X\in \operatorname{Ob}(\mathcal C)}\mathbb Z[X]$$
can be given a $\mathbb Z$-algebra structure where the ring multiplication is induced by the tensor product. If every module in $\mathcal C$ is flat, then the $\mathbb Z$-submodule $${\langle [A]-[B]+[C] \mid 0\to A \to B \to C\to 0  \text{ is exact}   \rangle }$$ of $\bigoplus_{X\in \operatorname{Ob}(\mathcal C)}\mathbb Z[X]$ is also an ideal, so the Grothendieck group actually becomes a commutative ring where the product structure is induced from tensor product.
My question is: What happens in case not every module of $\mathcal  C$ necessarily flat? Does the Grothendieck group necessarily become a commutative ring where the product structure is induced from tensor product? I'm particularly interested in the case of $G(\mathrm{mod-}R)$: is it necessarily a commutative ring?

REPLY [5 votes]: Suppose $R$ is a ring such that for every $R$-module $M$ there is a projective resolution $P_\bullet \to M$ of finite length (actually, this condition can be loosened quite a bit as I will note at the end). Then we can define a ring structure on $G(R)$ the Grothendieck group of finitely generated $R$-modules by defining the following product
$$
[A] \cdot [B] = \sum_i (-1)^i \text{Tor}_i(A,B).
$$
Note that if $P$ is a projective $R$-module, then $[A]\cdot[P] = [A \otimes P]$ since the $\text{Tor}_i(A,P)$ vanish for $i >0$.
The way to show this is a little painful. We need to show that

This product is independent of choice of resolution,
this product is well-defined,
this product is symmetric,
this product satisfies the ring axioms.

It is in my opinion easier to define this product as a map on $G(R)$ first, though tastes may vary. Fix a finitely generated $R$-module $M$ and let
$$
- \cdot M : G(R) \to G(R),\; [A] \mapsto [A] \cdot M := \sum_i (-1)^i [\text{Tor}_i(A,M)].
$$
Then one can show that this map is independent of the choice of resolution by showing $\text{Tor}_i(-,M)$ is. The well-definedness of the map follows since for any short exact sequende $0 \to A \to B \to C \to 0$ we have a long exact sequence of Tor, which shows $[A] \cdot M - [B] \cdot M + [C] \cdot M = 0$. The symmetry follows from the symmetry of Tor, so $[A] \cdot B = [B] \cdot A$.
From this point it is therefore acceptable to write $[A] \cdot [B]$, and we can prove the ring axioms one by one. For this it is good to note that for an $R$-module $M$ with projective resolution $P_\bullet \to M$ we have
$$
[M] \cdot [N] = \sum_i (-1)^i [\text{Tor}_i(M,N)] = \sum_i(-1)^i H_i(P_\bullet \otimes N) = \sum_i (-1)^i [P_i \otimes N].
$$
This computation involves the nontrivial lemma that in the Grothendieck group we have $\sum_i(-1)^i [C_i] = \sum_i (-1)^i [H_i(C_\bullet)]$ for any finite complex $C_\bullet$ in the category.
Then the ring axioms can easily be checked by using properties of the tensor product.
The requirement that every object in the category have a finite projective resolution is actually too strong. If every object has a finite resolution that is $F$-acyclic for a functor $F$ on the category, then it induces a derived morphism on Grothendieck groups. Actually, the resolution need not even be in the category itself. For example for categories of coherent sheaves we almost never have projective resolutions, but in some situations (details omitted on purpose) we may have locally free resolutions, which are $\text{Tor}_i(-, *)$ acyclic, so that we can define the product.


text:
{\bf Problem.} Two positive integers differ by 6 and their product is 135. What is the larger integer?
{\bf Level.} Level 1
{\bf Type.} Algebra
{\bf Solution.} 135 factors into four possible pairs: $(1,135)$, $(3,45)$, $(5,27)$, and $(9,15)$. The only one of these with difference 6 is $(9,15)$, which has a larger integer of $\boxed{15}$.

set_name: MATH


TITLE: Long exact sequence of cohomology group "without" Snake lemma
QUESTION [5 upvotes]: Let a short exact sequence
$$
0 \to L \to M \to N \to 0
$$
is a short exact sequence of $G$-modules, then a long exact sequence is induced:
$$
0\longrightarrow L^G \longrightarrow M^G \longrightarrow N^G \overset{\delta^0}{\longrightarrow} H^1(G,L) \longrightarrow H^1(G,M) \longrightarrow H^1(G,N) \overset{\delta^1}{\longrightarrow} H^2(G,L)\longrightarrow \cdot
$$
The connecting homomorphism is,
$$
\delta^n : H^n (G,N) \to H^{n+1}(G, L)
$$

Question 1: How to prove the above long exact sequence is true? Is this simply based on the Snake Lemma [It's My Turn (1980)]? Or is there other simpler way to think about it, without using the Snake Lemma?

REPLY [6 votes]: Let's see that the long exact sequence theorem is equivalent to the snake lemma.
From the short exact sequence we have an induced diagram
$$\require{AMScd}\begin{CD}&&L^n/B^n(G,L)@>>>L^n/B^n(G,M)@>>>L^n/B^n(G,N)@>>>0\\&@VVV@VVV@VVV\\0@>>>Z^{n+1}(G,L)@>{}>>Z^{n+1}(G,M)@>>>Z^{n+1}(G,N)\end{CD}$$
Applying snake lemma we have an exact sequence
$$H^n(G,L)\rightarrow H^n(G,M)\rightarrow H^n(G,N)\xrightarrow{\delta^n}H^{n+1}(G,L)\rightarrow H^{n+1}(G,M)\rightarrow H^{n+1}(G,N).$$
For other hand, a commutative diagram of $G$-modules
$$\require{AMScd}\begin{CD}0@>>>A^0@>>>B^0@>>>C^0@>>>0\\&@V{f}VV@V{g}VV@V{h}VV\\0@>>>A^1@>>>B^1@>>>C^1@>>>0\end{CD}$$
can be viewed as a short exact sequence between complexes $0\rightarrow A\rightarrow B\rightarrow C\rightarrow0$. Then there is an exact sequence between its cohomologies
$$0\rightarrow\ker(f)\rightarrow\ker(g)\rightarrow\ker(h)\xrightarrow{\delta^0}coker(f)\rightarrow coker(g)\rightarrow coker(h)\rightarrow0$$


TITLE: What are the advantages of the more abstract approaches to nonstandard analysis?
QUESTION [36 upvotes]: This question does not concern the comparative merits of standard (SA) and nonstandard (NSA) analysis but rather a comparison of different approaches to NSA. What are the concrete advantages of the abstract approaches to NSA (e.g., via the compactness theorem), as compared to the more concrete approach using ultrapowers? One can name generic reasons such as naturality, functoriality, categoricity, etc., but I am hoping for a concrete illustration of why a more abstract approach may be advantageous for understanding NSA concepts and/or proving theorems.
Note 1. One of the existing answers provided a bit of information about advantages of the more abstract approach in terms of saturation. I would appreciate an elaboration of this if possible, in terms of a concrete application of saturation.
Note 2. These issues are explored in more detail in this 2017 publication in Real Analysis Exchange.

REPLY [7 votes]: You can quote me on this if you like.
This is such an old issue that I am surprised it is still up for discussion.   I know you like ultra-powers, etc,  but I have always thought they were wrong-headed;  it was Luxemburg who made them popular for NSA originally.
The point is that you actually really NEED the transfer theorems; so you essentially need the logical apparatus; in most cases the compactness theorem and some form of saturation.    Occasionally a type omitting argument could be used but that is rare.
The reason is that you are postulating that "all of mathematics" carries over to the
non-standard model and in fact, that is the underlying intuition in the applications.
I also think that the "more concreteness" of the ultraproduct construction is just an illusion.     You can not answer if the integer produced by (1,2,3,4,5, ...) is even or odd
since it depends on the ultrafilter completion of e.g. any non-principal filter.
Moreover, the basic tools in almost all the proofs is dealing with the seam between
"standard" and "non-standard" elements; which really depends on a feel for expressibility in the language being used.  Nelson's approach was to try to help people manage without the "feel".  That's a matter of taste.
Furthermore, I would differ with your point that NSA is to formalize classical mathematicians arguments.  That is "cool" and Robinson greatly enjoyed it, but really he also was sure that it is a great way to prove new theorems.    Since people sort of like to
hear about the theorems consequence in "classical" settings; then one often had to find
equivalence theorems (this was clear in the work on brownian motion etc) and in my work on inverse limits of finite groups  (by the uniqueness of inverse limits, the non-standard finite groups are essentially equivalent to inverse limits of systems of finite groups and a lot of the very early work of Lubotzky on profinite groups can be easily carried out in this setting)  but actually
a more extreme position can be seen when you just take the position that the "standard"
results are just one specific implementation and one could develop mathematics happily without them.    The argument that the non-standard models are not "unique" is just a habit and not important.
Robinson himself put this viewpoint forward very nicely in his Brouwer medal address  "Standard and NonStandard Number Systems"
Best regards
Larry Manevitz
manevitz@cs.haifa.ac.il
Regarding the use of limit ultrapowers etc; one needs to be careful there.  I once proved there was no measureable cardinal by taking such a large ultrapower; but actually all the
standard sets dont grow in that context :).


text:
\paragraph{Market equilibrium.} In this paper we focus on computing Fisher market equilibrium in the divisible setting.
Let the market consists of $n$ buyers and $m$ goods. Buyer $i$ has budget $B_i > 0$ and utility function $u_i: \RR^m_+ \rightarrow \RR_+$. As mentioned above, $B_i = 1$ for all $i$ corresponds to the CEEI mechanism. Without loss of generality, assume each good $j$ has \textit{unit} supply.
An (aggregate) \textit{allocation} is a matrix $x = [x_1^\top; \dots; x_n^\top]\in \RR^{n\times m}_+$, where $x_{ij}$ is the amount of item $j$ purchased by buyer $i$. Given prices $p\in \RR^m_+$ of the goods, the \emph{demand set} of buyer $i$ is defined as the set of utility-maximizing allocations:
\[ D_i(p) = \argmax \left\{u_i(x_i)\mid  x_i\in \RR^n_+,\, \langle p, x_i\rangle \leq B_i \right\}. \]
A \textit{competitive equilibrium} is a pair of prices and allocations $(p, x)$, $p\in \RR_+^m$ that satisfy the following conditions \cite{eisenberg1959consensus, jain2010eisenberg}:
\begin{itemize}
	\item \textit{Buyer optimality}: $x_i \in D_i(p)$ for all $i\in [n]$,
	\item \textit{Market clearance}: $\sum_{i=1}^n x_{ij} \leq 1$ for all $j\in [m]$, and equality must hold if $p_j > 0$.
\end{itemize}

We say that $u_i$ is \textit{homogeneous} (with degree $1$) if it satisfies $u_i(\alpha x_i) = \alpha u_i(x_i)$ for any $x_i \geq 0$ and $\alpha>0$ \cite[\S 6.2]{nisan2007algorithmic}. We assume that $u_i$ are concave, continuous, nonnegative, and homogeneous utilities (CCNH). This captures many widely used utilities, such as linear, Leontief, Cobb-Douglas, and general \textit{Constant Elasticity of Substitution} (CES) utilities (see, e.g., \cite[\S 6.1.5]{nisan2007algorithmic} and \cite{branzei2014fisher}).

For CCNH utilities, a market equilibrium can be computed using the following \emph{Eisenberg-Gale convex program} (EG): \begin{align}
\max \sum_{i=1}^n B_i \log u_i(x_i)\quad {\rm s.t.} \ \sum_{i=1}^n x_{ij} \leq 1, \ x\geq 0.
\label{eq:eisenberg-gale-primal}
\end{align}
More precisely, we state the following theorem, which is well-known in various forms in the literature: see, e.g., \cite[\S 6.2]{nisan2007algorithmic} for the case of differentiable $u_i$ and Eisenberg's original work \cite[Theorem 4]{eisenberg1961aggregation} for the convex program with a maximization objective of the Nash social welfare).
For completeness, we present an elementary, self-contained proof (all proofs are in the Appendix). Note that the assumption essentially means that everyone can at least get positive utility.

\begin{theorem}
	Let $u_i$ be concave, continuous, nonnegative, and homogeneous (CCNH). Assume $u_i(\ones) >0$ for all $i$. Then, (i) \eqref{eq:eisenberg-gale-primal} has an optimal solution and (ii) any optimal solution $x^*$ to \eqref{eq:eisenberg-gale-primal} together with its optimal Lagrangian multipliers $p^*\in \RR_+^m$ constitute a market equilibrium, up to arbitrary assignment of zero-price items. Furthermore, $\langle p^*, x^*_i \rangle = B_i$ for all $i$.
	\label{thm:eg-gives-me-for-certain-ui}
\end{theorem}

We focus on linear, quasilinear (QL) and Leontief utilities. Under linear utilities, the utility of buyer $i$ is $u_i(x_i) = \langle v_i, x_i \rangle$. As mentioned above, linear utilities are used in the CEEI mechanism in fair division and fair recommender systems. QL utilities have the form $u_i(x_i) = \sum_j (v_{ij} - p_j) x_{ij}$, which depends on the prices \cite{chen2007note, cole2017convex}.  It captures budget-smoothing problems in first- and second-price auction markets~\cite{conitzer2018multiplicative,conitzer2019pacing}. Leontief utilities have the form $u_i(x_i) = \min_{j\in J_i} \frac{x_{ij}}{a_{ij}}$ (see \S \ref{sec:leontief-utilities}). They model perfectly complementary goods and are suitable in resource sharing where an agent's utility is capped by its \emph{dominant resource}, such as allocating different types of compute resources to computing tasks \cite{ghodsi2011dominant, kash2014no}.

Another notable convex program that captures market equilibrium under linear utilities is Shmyrev's \cite{shmyrev2009algorithm} (see Appendix \ref{app:shmyrev-cp} and \eqref{eq:shmyrev} there for more details). \cite{birnbaum2011distributed} shows that the well-known Proportional Response dynamics (PR) \cite{zhang2011proportional} for computing market equilibrium under linear utilities is in fact applying Mirror Descent to Shmyrev's convex program \eqref{eq:shmyrev}. In \S \ref{sec:ql-utilities}, we will show that both Shmyrev's convex program and PR generalize elegantly to QL utilities.

In principle, \eqref{eq:eisenberg-gale-primal} can be solved using an interior-point method (IPM) that handles exponential cones \cite{skajaa2015homogeneous, dahl2019primal}. However, it does not scale to large markets since IPM requires solving a very large linear system per iteration. This linear system can often be ill-conditioned and relatively dense, even when the problem itself is sparse. This makes IPM impractical for applications such as fair recommender systems and ad markets, where the numbers of buyers and items are typically extremely large. In this paper, we investigate iterative (gradient-based) first-order methods (FOMs) for computing equilibria of Fisher markets. Because each FOM iteration has low cost, which usually scales only with the number of nonzeros when the problem is sparse, these methods are suitable for scaling up to very large markets.
\highlight{Furthermore, there have been variants of FOMs that allow parallel and distributed updates, making them even more scalable \cite{chen2012fast, shi2015proximal, wang2016parallel, lin2014accelerated}.}
This is analogous to other equilibrium-computation settings such as zero-sum Nash equilibrium, where iterative first-order methods are also the state-of-the-art~\cite{brown2018superhuman,moravvcik2017deepstack,kroer2020faster}. We will focus on the important classes of linear, quasi-linear and Leontief utilities, which are discussed in \S \ref{sec:linear-utilities}, \ref{sec:ql-utilities} and \ref{sec:leontief-utilities}, respectively.
\highlight{We also note that under CES utilities, PR yields linearly convergent prices and utilities \cite[Theorem 4]{zhang2011proportional}. Meanwhile, market equilibrium under Cobb-Douglas utilities can be computed explicitly. See Appendix \ref{app:CD-and-CES} for more details.}
We note that there have been highly nontrivial algorithms that find market equilibria in time polynomial in $m, n,\log \frac{1}{\epsilon}$ in theory (where $\epsilon$ is a desired upper bound on error in prices) \cite{devanur2002market,vazirani2007combinatorial,bei2019ascending}.
However, none of these are as easily implementable as the FOMs we consider here, which also achieves $\epsilon$-equilibrium prices in $O\left(\log \frac{1}{\epsilon}\right)$ time thanks to the linear convergence guarantees. Secondly, these methods all have prohibitively-expensive per-iteration costs from a large-scale perspective.


TITLE: Help with an exponential/generating functions problem.
QUESTION [1 upvotes]: Is there anyone that would be willing to help me solve this problem?
While I am comfortable with using ordinary generating functions, I do not have a lot of experience or intuition when it comes to exponential generating functions. Any tips or a push in the right direction would be much appreciated!
Let $\mathit{A(t)}$ be an ordinary generating function and let $\mathit{E(t)}$ be an exponential generating function for a sequence {$\mathit{a_n}$}.
Show that $\mathit{A(t)}$ = $\int^\infty_0$$e^{-x}$$\mathit{E(xt)}$$\mathit{dx}$.
Hint: Use the equality $\mathit{n}$! = $\int^\infty_0$$e^{-x}$$\mathit{x}^n$$\mathit{dx}$
Thanks in advance!

REPLY [1 votes]: Let $$A(t) = \sum_{n\ge 0} a_n t^n$$ be the o.g.f. and $$E(t) = \sum_{n\ge 0}a_n \frac{t^n}{n!}$$ be the e.g.f.
Then we have
\begin{align}
\int_0^\infty e^{-x}E(xt) dx &= \int_0^\infty e^{-x} \sum_{n\ge 0}\frac{a_nt^nx^n}{n!} dx\\
&=  \sum_{n\ge 0}\int_0^\infty e^{-x}\frac{a_nt^nx^n}{n!} dx\\
&=  \sum_{n\ge 0}a_nt^n\frac{1}{n!}\int_0^\infty e^{-x} x^n dx\\
&=  \sum_{n\ge 0}a_nt^n\frac{1}{n!} n!\\
&= \sum_{n\ge 0}a_nt^n\\
&= A(t)
\end{align}
Note we use the fact that $\int (f_1(x) + f_2(x) + \cdots) dx = \int f_1(x) dx + \int f_2(x)dx + \cdots$. Then since we are integrating with respect to $x$ we have $\frac{a_nt^n}{n!}$ as a constant we can pull out, and we use the hint.


TITLE: Derivation of Newton's Second Law of Motion
QUESTION [2 upvotes]: How does one derive the equation that $F=Δp/Δt$ without using $F=ma$?
I reviewed Deriving $F = ma$ - Newton's Second Law of Motion, but it just assumed $F=Δp/Δt$.

REPLY [1 votes]: You haven't specified what we are supposed to take as axioms in deriving this equation. One possibility is the following. Suppose a particle is moving in a one-dimensional potential $V(x)$. We assume that the particle obeys the conservation of energy, given as
$$
\frac{d}{dt}\left(\frac{p^2}{2m}+V(x)\right)=0,
$$
where we define momentum as
$$
p = m\frac{dx}{dt}.
$$
We also define the force as
$$
F=-\frac{dV}{dx}.
$$
Then differentiating the energy conservation equation gives
$$
\frac{d}{dt}\left(\frac{p^2}{2m}+V(x)\right)=\frac{p}{m}\frac{dp}{dt} + \frac{dx}{dt}\frac{dV}{dx}=0
$$
Then using the definition of momentum and the definition of force, we have
$$
\frac{dx}{dt}\frac{dp}{dt} - \frac{dx}{dt}F=0.
$$
Whenever $dx/dt\neq 0$, it follows that
$$
\frac{dp}{dt}=F.
$$
However, conservation of energy alone is not sufficient to prove the statement when $dx/dt=0$.


TITLE: Integration Technique Strategy
QUESTION [0 upvotes]: There are many types of integration techniques. Are there any good ways to figure out which one to use or is that just a developed intuition that comes with experience?

REPLY [0 votes]: Mainly there are three techniques
1) Integration by substituion
Here you take a small expression of x from whole equation that is goin to be integrated and put that equal to a new variable. By this you make the integral in terms of a new variable and much simpler to integrate.
2)Integration by parts
It is integration technique like product rule of diffrentiation.
3)Integration by trigonometric substituion
It is same as Integration by substituion except for that in this x becomes any trigonometric expression (sin,cos, and tan) and we have to take care of the range of x.
Trigonometric substituion is simple to indentify whereas Integration by substituion and Integration by parts are  difficult to indentify.So for Integration by substituion and by parts, you have to practice a so that you are able to identify which technique to use in first glance.


TITLE: how to calculate the value of $\int_{-\infty}^\infty \frac{e^{ax}}{\cosh x}\,dx$
QUESTION [6 upvotes]: So I have to evaluate $$\int_{-\infty}^\infty\frac{e^{ax}}{\cosh x} \, dx$$
I tried takeing the analytic expansion, and integrating over the real axis. I took this as being a half circle from $-\infty$ to $\infty$, minus the arc of the circle. over the arc I proved that the integral is zero, and I have left only with a sum of the residues of the function.
With the help of a previous question I calculated the values of the residues, but I could not manage to converge the sum.
Help will be highly appreciated

REPLY [0 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\left.\int_{-\infty}^{\infty}{\expo{ax} \over \cosh\pars{x}}\,\dd x\,\right\vert_{\ -1\ <\ \Re\pars{a}\ <\ 1}} =
\int_{0}^{\infty}{2\cosh\pars{ax} \over \cosh\pars{x}}\,\dd x
\\[5mm] & =
2\int_{0}^{\infty}{\cosh\pars{ax}\sinh\pars{x} \over \cosh\pars{x}\sinh\pars{x}}\,\dd x
\\[5mm] & =
2\int_{0}^{\infty}{\bracks{\sinh\pars{x - ax} + \sinh\pars{x + ax}}/2 \over \sinh\pars{2x}/2}\,\dd x
\\[5mm] \stackrel{x\ \mapsto\ x/4}{=}\,\,\, &\
{1 \over 2}\int_{0}^{\infty}{\bracks{\sinh\pars{\bracks{1 - a}x/4} + \sinh\pars{\bracks{1 + a}x/4}} \over \sinh\pars{x/2}}\,\,\dd x
\\[5mm] = &\
\mrm{f}\pars{1 - a \over 2} + \mrm{f}\pars{1 + a \over 2}\label{1}\tag{1}
\quad\mbox{where}\quad
\mrm{f}\pars{\xi} \equiv
{1 \over 2}\int_{0}^{\infty}{\sinh\pars{\xi x/2} \over \sinh\pars{x/2}}\,\dd x
\end{align}

Moreover,

$$
\mrm{f}\pars{\xi} =
{1 \over 2}\int_{0}^{\infty}{\expo{\xi x/2} - \expo{-\xi x/2} \over \expo{x/2} - \expo{-x/2}}\,\dd x =
2\int_{0}^{\infty}{\expo{-\pars{1 - \xi}x/2} - \expo{-\pars{1 + \xi}x/2} \over 1 - \expo{-x}}\,\dd x
$$

With $\ds{\pars{~\expo{-x} \equiv t \implies x = -\ln\pars{t}~}}$:

\begin{align}
\mrm{f}\pars{\xi} & =
{1 \over 2}\int_{0}^{1}{t^{-1/2 - \xi/2} - t^{-1/2 + \xi/2} \over 1 - t^{}}\,\dd x
\\[5mm] & =
{1 \over 2}\pars{%
\int_{0}^{1}{1 - t^{-1/2 + \xi/2} \over 1 - t}\,\dd x -
\int_{0}^{1}{1 - t^{-1/2 - \xi/2} \over 1 - t}\,\dd x}
\\[5mm] & =
{1 \over 2}\bracks{%
\Psi\pars{{1 \over 2} + {\xi \over 2}} -
\Psi\pars{{1 \over 2} - {\xi \over 2}}}
\end{align}
where $\ds{\Psi}$ is the
Digamma Function. With
Euler Reflection Formula:
\begin{align}
\mrm{f}\pars{\xi} & =
{1 \over 2}\braces{\pi\cot\pars{\pi\bracks{{1 \over 2} - {\xi \over 2}}}} =
{1 \over 2}\,\pi\tan\pars{{\pi \over 2}\,\xi}\label{2}\tag{2}
\end{align}

With \eqref{1} and \eqref{2}

$$
\bbx{\left.\int_{-\infty}^{\infty}{\expo{ax} \over \cosh\pars{x}}\,\dd x\,\right\vert_{\ -1\ <\ \Re\pars{a}\ <\ 1} =
\pi\sec\pars{\pi a \over 2}}
$$


TITLE: Prove $f$ is surjective or not
QUESTION [2 upvotes]: For $\mathbb{R} \to \mathbb{R}$, prove/disprove $f(x) = x^2 + 4x + 9$ is surjective.

I say it is not surjective. Let $y = 0$ then the discriminant of $f(x)$ is $16 - 36 < 0$, thus $x \not \in \mathbb{R}$.
Is this the correct way to go about it?

REPLY [2 votes]: $x^2 + 4x + 9 = (x+2)^2 + 5$ thus your function only takes value in $[5, +\infty[$ and is therefore not surjective.


TITLE: Is there any knot showing infinitely many crossings?
QUESTION [3 upvotes]: Let $K$ be any unknot. Is it possible that $K$ shows infinitely many crossings?
And if it is possible: How to get $K$ from the simplest unknot through Reidemeister moves?

REPLY [2 votes]: There are wild knots with infinitely many crossings. A knot is wild if it is not tame, where this means that the knot extends to an embedded $S^1 \times D^2 \subset \mathbb{R}^3$. Every tame knot can be realized as a finite polygonal path.
Note that a smooth or PL knot is necessarily tame.
If you take the idea for the construction of the wild knot in the image in that wikipedia article but do it with different crossings you can make it an unknot.
Knot diagrams for equivalent tame knots have a finite sequence of Reidemeister moves  relating them. This doesn't work for wild knots.

A famous use of wild knots is the Mazur swindle.


TITLE: Number of circles required to reach a length of 100 million units in form of circle
QUESTION [0 upvotes]: I need a formula to find out number of circles required to reach a length of 100 million units in form of circle. All circles are in offset from first circle with +1 unit in radius. all circles have a common centre.
N=X*3.14*2 (x is the value of radius of circle)(N=perimeter of circle)
Y=N+6.28 ((Y is the value of circumference of next circle's radius with x+1unit) (a circle with a radius of 100 units has a perimeter of 628 units,
a circle with 101 unit has a perimeter of 634.28 units, a circle with 102 unit has a perimeter of 640.56 units, the difference is always 6.28 units.))
so, addition of perimeter of first three circles is (628+634.28+640.56) 1902.84) units. what should be the formula to find out number of circles required to reach 100 million units.

REPLY [1 votes]: We consider $n$ circles, where the radius of each circle is equal to $99+n$. The perimeter of the $n$th circle is therefore
$$
p(n) = 2\pi (99+n)
$$
When we add up the perimeters of the circles, we get the sum of an arithmetic progression
$$
\begin{split}
S =& p(1) + p(2) +p(3) +\ldots + p(n)\\ =& 2\pi \left(100+101+102 + \ldots + 99 +n \right)\\
=&2\pi(\underbrace{99+99+99 +\ldots+99}_{\text{n terms}}) + 2\pi \left(1+2+3+\ldots+n \right)
\end{split}
$$
The summing the $99$'s together makes just $99n$, and for the sum $1+2+\dots+n$ we can find the identity
$$
1+2+3+\ldots+n = \frac{n(n+1)}{2}
$$
and therefore
$$
S = 198\pi n + 2\pi \frac{n(n+1)}{2}
$$
We want this to be at least $S= 10^8$, and therefore we get the equation
$$
2\times 10^8 = 396\pi n + 2\pi n^2 + 2\pi n \qquad \Rightarrow \qquad 2\pi n^2 + 398\pi n - 2\times 10^8 = 0
$$
Numerically, we can find $n\approx 5~544$. This is the number of circles required so that the sum of perimeters is at least to 100 million.


TITLE: Exchange of two limits of a function
QUESTION [3 upvotes]: The question is simply given a function $f(x,y)$, when does $\mathop {\lim }\limits_{x \to 0} \mathop {\lim }\limits_{y \to 0} f(x,y) = \mathop {\lim }\limits_{y \to 0} \mathop {\lim }\limits_{x \to 0} f(x,y)$ hold?
I know that if the mixed limit $\mathop {\lim }\limits_{x \to 0,y \to 0} f(x,y)$ exists, then above exchange is valid, but I think there should be more general conditions. Thanks!
Related post about limit of sequence: When can we exchange order of two limits?.

REPLY [5 votes]: The Moore-Osgood Theorem states that if $\lim_{x\to x_0}f(x,y)$ converges uniformly for $y\ne y_0$ and if $\lim_{y\to y_0}f(x,y)$ converges pointwise for $x\ne x_0$, then the iterated limits are equal with
$$\lim_{x\to x_0}\lim_{y\to y_0}f(x,y)=\lim_{y\to y_0}\lim_{x\to x_0}f(x,y)$$


TITLE: The expectation of total number of different states in N time points
QUESTION [3 upvotes]: [Conditions]
(1) An object has K possible states.
(2) This object can have only one state at a single time point.
(3) The probability of each state at any single time point is 1/K, and each time point is independent of another.
[Question] What is the expectation of different states appearing in N time points? (Obviously, this expectation should be less than K. )
Someone told me the solution is $K(1-(1-\frac{1}{K})^N)$, but I could not understand his explanation.
So here I hope you to validate this solution, and provide detailed explanation.

REPLY [1 votes]: To give the answer you quote, you need a stronger version of (3) so that the states at each time point are independent of each other.
Hints:

What is the probability that at any particular time point the object has a partiuclar state $x$?
What is the probability that at any particular time point the object does not have state $x$?
What is the probability that at all $N$ time points the object does not have state $x$?
What is the probability that at at least one of $N$ time points that the object has state $x$?
What is the expectation of an indicator that at at least one of $N$ time points the object has state $x$?
What is the sum over all $K$ states of the expectations of an indicator that at at least one of $N$ time points the object has each state? (Use linearity of expectation)
What is the expectation of the sum over all $K$ states of an indicator that at at least one of $N$ time points the object has each state?
What is the expectation of the number of states the object has over $N$ time points?


TITLE: Vector spaces with unique basis
QUESTION [5 upvotes]: Vector spaces like $\mathbb{R}^n$ can have different bases and we can change the basis with a matrix to get a new one. This made me wonder:
Are there any vector spaces with $dim>1$ that have only one basis?

REPLY [5 votes]: No. Let $b_1, b_2, ..., b_n$ be a basis. Create an element $a=c_1b_1 + c_2b_2+...+c_nb_n$ where not all of the $c_i$ are $0$. WLOG, we may assume $c_n \neq 0$. Then the set $b_1, b_2, ..., b_{n-1}, a$ is a basis for the space.


TITLE: A remark about the map $\partial^*:H^{*}(A;V)\to H^{*+1}(X,A)$ of the l.e.S. in cohomology
QUESTION [0 upvotes]: My question is about a remark from lecture about the connecting-homomorphism of the long exact sequence in homology of a pair $(X,A)$.
Let $(X,A)$ a pair of topological spaces, $V$ be an abelian group and consider the long exact sequence in singular cohomology
$$..H^{n-1}(A;V)\xrightarrow{\partial^{n-1}}H^n(X,A)\to H^n(X;V)\to H^n(A;V)\xrightarrow{\partial^{n}}...$$
Let $\partial^*:H^{*}(A;V)\to H^{*+1}(X,A)$ be the connecting homomorphism in the sequence above.
In lecture we had the remark: The map $\partial^*:H^{*}(A;V)\to H^{*+1}(X,A)$ can be described as follows: Let $\alpha:C_n(A;\mathbb{Z})\to V$ be a cocycle. Extend $\alpha$ on $C_n(X;\mathbb{Z})$ by $\hat{\alpha}(\sigma )=0$ for $\sigma \in C_n(X;\mathbb{Z})$ with $\operatorname{im}(\sigma)\nsubseteq A$. Then we have: $$\partial^n [\alpha]=[d^n(\hat{\alpha})]=[\hat{\alpha} \circ d_{n+1}]\in H^{n+1}(X,A).$$
I don't understand this remark, in my opion there must be a mistake. Because if $d^n$ in the expression $[d^n(\hat{\alpha})]$ is the boundary map $d^n:C^{n}(X;V)\to C^{n+1}(X;V)$, then $[\hat{\alpha} \circ d_{n+1}]$ is not in $ H^{n+1}(X,A)$. If $d^n$ in the expression $[d^n(\hat{\alpha})]$ is the boundary map $d^n:C^{n}(X,A;V)\to C^{n+1}(X,A;V)$, $[d^n(\hat{\alpha})]$ is zero.
Where is the mistake? Is there a mistake on my thoughts or is the remark wrong? If the remark is wrong, what should be the correct version?
Best.

REPLY [0 votes]: You know, since $\alpha$ is a cocycle, that $\alpha \circ d^A_n = 0$. However, you do not know that $\hat\alpha \circ d^X_n = 0$! So $\hat\alpha$ is not even a cocycle. What you know is that, if $i_* : C_*(A) \to C_*(X)$ is the inclusion, then
$$\hat\alpha \circ d^X_n \circ i_{n+1} = \hat\alpha \circ i_n \circ d^A_n = \alpha \circ d^A_n = 0,$$
and so by definition of $C^{n+1}(X,A;V)$, it follows that $\hat\alpha \circ d^X_n \in C^{n+1}(X;V)$ is in fact in $C^{n+1}(X,A;V)$, i.e. it vanishes on $C_{n+1}(A)$, and it is a cocycle.
But it is not a coboundary in $C^*(X,A;V)$, because it would need to be the codifferential of something in $C^n(X,A;V)$; but $\hat\alpha$ does not vanish (in general) on $C_n(A)$, so in general $\hat\alpha$ is in $C^n(X;V)$ but not in $C^n(X,A;V)$. So you cannot conclude from this that $\hat\alpha \circ d^X_n$ is in the image of
$$\delta^n : C^n(X,A;V) \to C^{n+1}(X,A;V),$$
i.e. you cannot conclude that $\hat\alpha \circ d_n^X$ is a coboundary in $C^*(X,A;V)$. The remark is correct.

PS: This is a general construction known as the snake lemma, applied to:
$$\require{AMScd}
\begin{CD}
0 @>>> C^n(X,A) @>>> C^n(X) @>>> C^n(A) @>>> 0 \\
@. @VVV @VVV @VVV \\
0 @>>> C^{n+1}(X,A) @>>> C^{n+1}(X) @>>> C^{n+1}(A) @>>> 0
\end{CD}$$
The snake lemma is a very powerful tool in homological algebra.


TITLE: How to show $y(T) = x(1)$ for $y'=f(y)$ and $x'=Tf(x)$
QUESTION [0 upvotes]: Suppose I have two ODEs,  $$\frac{dy}{dt}=f(y), \ \ \ \frac{dx}{dt}= Tf(x),$$
where $t$ is the time and $T$ is the terminal time. Also $x(0)=y(0)=c$.

How to show that $y(T)=x(1)$? This means the value of $y(T)$ for the LHS ODE is equal to $x(1)$ of the RHS ODE.

My effort:  I only know that for any $t_p\in [0,T]$, LHS ODE $$\frac{dy}{dt}|_{t=t_p}= f(y(t_p))$$however, RHS ODE $$\frac{dx}{dt}|_{t=t_p}= Tf(x(t_p)),$$ i.e., at each point, the slope is $T$-times the previous one.

But I still have no idea about how to show $x(1) = y(T)$?

Consider an example:
Suppose $x(0)=1$, and consider the following MATLAB code
tf = 4;
x0 = 1;
tspan = [0 tf];
sspan = [0 1];
x0 = 1;
[t,x] = ode45(@(t,x) x, tspan, x0);
[s,y] = ode45(@(s,y) tf*y, sspan, x0);

In this case, $x(4)=x(1)$ (blue line is the second system.)
I just edited my question so I am sorry for the first answer from Kavi Rama Murthy.

REPLY [1 votes]: I start with the claim that $y(T \times t) = x(t)$. Here, we can check that $ t = 0$ we obtain $y(0) = x(0) = c$ for some constant $c$. Furthermore, at $t = 1$, we have $y(T) = x(1)$ and we are done.
To verify that, we just have to show that under this claim, the two differential equations are equivalent (ie it satisfy both equations).
From $\frac{dy(\xi)}{d\xi} = f(y(\xi))$, we pick $\xi = Tt$. Hence,
$\frac{dy(\xi)}{d\xi} = f(y(\xi)) \iff \frac{dy(Tt)}{d(Tt)} = f(y(Tt)) \iff \frac{d(x(t))}{d(Tt)} = f(x(t)) \iff \frac{d(x(t))}{d(t)} = Tf(x(t))$
Thus, $y(T \times t) = x(t)$ indeed satisfy both differential equations.


TITLE: How do I finish this summations problem?
QUESTION [0 upvotes]: I have posted a picture since I don't know how to make the summation symbols with the lower and upper summations on keyboard, sorry about that..
$$\sum_{a=1}^9\sum_{b=0}^9(101a+10b)$$
The answer is $49,500$, and I am not sure how I am supposed to solve this. The book does have the steps, but they are all alien to me because it doesn't explain how they got from one step to the other.
To be precise I am confused about how they managed to bring $101a + 10b$ into the middle of the two summations, turning it into $\sum_{a=1}^9 (10(101a)+10)\sum_{b=0}^9 b$
Here is what I am referring to,

REPLY [0 votes]: This kind of problems can't resist solving even though they have been solved by others or answered by others.  Here is what I did.  I am poor at writing math equations on computer so wrote on paper and attached the photo.