TITLE: A scheme that can be covered by finitely many affine open subschemes with noetherian topology is not necessarily a noetherian scheme?
QUESTION [0 upvotes]: Definition: A noetherian scheme is a scheme that admits a finite covering by open affine subsets $\operatorname {Spec} A_{i}$, where $A_{i}$ are noetherian rings.
I wonder if there is a non-noetherian scheme that satisfies the following slightly weakened conditions

A scheme that admits a finite covering by open affine subsets $\operatorname {Spec} A_{i}$, where  $\operatorname {Spec} A_{i}$ are noetherian topological spaces.

REPLY [1 votes]: The following reassuring result simplifies Pavel's and Eric's answers: given a ring $A$ we have

$\operatorname {Spec} A$ is a noetherian scheme $\iff$ $A$  is a noetherian ring

[EGA I, Proposition (6.1.3), page 141]


TITLE: Proof that this equation is correct
QUESTION [1 upvotes]: Using a truth table I had no problems to proof, that this equation is correct.
But how can I transform the first part to get to the second?
I tried using de morgan but I never made it.
Can anyone give me advice?

REPLY [2 votes]: $$\neg p \wedge q \rightarrow r =$$
First we know that $A\rightarrow B$ is equivalent with $\neg A\vee B$ thus.
$$\neg(\neg p \wedge q) \vee r  =$$
Now use de morgan:
$$ (p \vee \neg q) \vee r =$$
Switch place between $r$ and $(p \vee \neg q)$ and notice that $r$ is equivalent with $\neg \neg r$.
$$\neg \neg r \vee (p \vee \neg q) =$$
Now again use that $A\rightarrow B$ is equivalent with $\neg A\vee B$ and we reach our conclusion.
$$\neg r \rightarrow (p \vee \neg q)$$


TITLE: Arbitrary 2-forms on $\mathbb{R}^4$
QUESTION [0 upvotes]: If I want to compute with arbitrary 2-forms on $\mathbb{R}^4$, I wonder if each form must contain all 4 variables e.g $x, y, z, t$.
If so, I figure that two-forms in 4 dimension will correspond to the multi-index $$I = \{(1,2), (1,3), (1,4), (2,3), (2,4), (3,4) \}.$$ That is a set of $$\binom{4}{2}$$  strictly increasing 2-tuples if I have understood correctly. Given that the definition of $k$-forms is $$\omega = \sum a_I dx_I,$$ I pick arbitrary elements from $I$ to form 2-forms. Example $$\omega = dx_1 \wedge dx_2 + dx_1 \wedge dx_3$$ and $$\tau = dx_2 \wedge dx_3 + dx_3 \wedge dx_4.$$
The wedge product $\omega \wedge \tau$ is trivial because each term contain $dx_i \wedge dx_i = 0$.
EDIT: Can I choose how many basis elements to include in the linear combinations i.e. the $k$-forms?

REPLY [1 votes]: Well, you can certainly compute products like $\omega \wedge \tau = 0$ without all four indices appearing.
If you want $\alpha \wedge \beta \ne 0$ for 2-forms $\alpha, \beta$ on $4$-space, then you do, indeed, need to have all four indices appear somewhere -- two of them in $\alpha$ (and possibly in $\beta$ as well), and the complementary pair in $\beta$ (and possible in $\alpha$ as well). Even so, that won't guarantee that $\alpha \wedge \beta \ne 0$.
For instance, if
$$
\alpha = dx_1 \wedge dx_2 + dx_3 \wedge dx_4\\
\beta = dx_1 \wedge dx_2 - dx_3 \wedge dx_4
$$
then we have $\alpha \wedge \beta = 0$, despite all four indices appearing.


TITLE: Point-wise Convergence Timing
QUESTION [0 upvotes]: Consider a sequence of non-negative simply functions $\{\phi_k(x)\}_{k=1}^\infty$ that converges point-wise to a non-negative measurable function $f$.
$$\lim_{k\rightarrow\infty}\phi_k(x)=f(x)\space\space\forall x.$$
When we actually check this with a specific example, is the following timing of steps correct?

Fix $x$.
Let $k$ go to infinity and see if $|\phi_k(x)-f(x)|<\varepsilon$ for arbitrarily small $\varepsilon$.
Repeat step #1 and 2 for the entire $x$ in the domain of the function $f$.

REPLY [0 votes]: Yes, if you remove step $1$ it checks the uniform convergence.


TITLE: Completion of a scalar extension in the metric topology
QUESTION [0 upvotes]: Let $V$ be a complex vector space (not necessarily finite dimensional). Let $V[[t]]\cong V\hat{\otimes}_{\mathbb C}\mathbb C[[t]]$ be the completion of $V\otimes_{\mathbb C}\mathbb C[[t]]$ with respect to the $(t)$-adic topology on $\mathbb C[[t]]$. Let $K:=\mathbb C((t))$ be the field of formal Laurent series. I would like to know what the completion of the tensor product $V[[t]]\otimes_{\mathbb C[[t]]}K$ with respect to the $(t)$-adic topology on $V[[t]]$ and the topology $\{t^i\mathbb C[[t]]\}$ on $K$ is. I come up with the result zero but I do not believe that zero is what should come out. So maybe I have made a mistake.
Here is my computation:
$$V[[t]]\hat{\otimes}_{\mathbb C[[t]]}K=\lim_{i, j}(V[[t]]/(t^i)\otimes_{\mathbb C[[h]]}K/t^j\mathbb C[[t]])\\
=\lim_{i, j}\frac{V[[t]]\otimes_{\mathbb C[[t]]} K}{t^iV[[t]]\otimes_{\mathbb C[[t]]}K+ V[[t]]\otimes_{\mathbb C[[t]]}t^j\mathbb C[[t]]}\\
=\lim_{i}\frac{V[[t]]\otimes_{\mathbb C[[t]]}K}{t^iV[[t]]\otimes_{\mathbb C[[t]]}K}\\
=\lim_i\frac{V[[t]]}{(t^i)}\otimes_{\mathbb C[[t]]} K\\
=0.$$

REPLY [1 votes]: You are correct that with your definition you get $0$.  If you see people talking about getting a completion of $V((t))$ they are presumably using a different topology on the tensor product from yours, probably the one generated by neighborhoods of $0$ of the form $t^iV[[t]]\otimes t^j\mathbb{C}[[t]]$.
To give some intuition for why this should be sensible, your topology corresponds to saying that $x\otimes y$ is small if either $x$ or $y$ is small, whereas that topology would correspond to saying $x\otimes y$ is small if both $x$ and $y$ are small.  But if the potential size of $y$ is unbounded (as is the case for $y\in\mathbb{C}((t))$ when $y$ could be $t^{-n}$ for arbitrarily large $n$), you shouldn't expect $x\otimes y$ to be small just because $y$ is small.  Saying that $x\otimes y$ is small when $x$ is small just ends up meaning that everything is arbitrarily small, and so becomes $0$ in the completion.


text:
\begin{document}
\maketitle

\begin{abstract}
We prove that the Iitaka conjecture $C_{n,m}$  for algebraic fibre spaces holds up to dimension $6$, that is, when $n\le 6$.
\end{abstract}

\section{Introduction}

We work over an algebraically closed field $k$ of
characteristic zero. Let $X$ be a normal variety. The canonical divisor $K_X$ is one of
the most important objects associated with $X$ especially in birational geometry.
If another normal variety $Z$ is in some way related to $X$, it is often crucial to find a relation
between $K_X$ and $K_Z$. A classical example is when $Z$ is a
smooth prime divisor on a smooth $X$ in which case we have $(K_X+Z)|_Z=K_Z$.

An algebraic fibre space is a surjective morphism $f\colon X\to Z$
of normal projective varieties, with connected fibres. A central problem in birational geometry
is the following conjecture which relates the
Kodaira dimensions of $X$ and $Z$. In fact, it is an attempt to relate $K_X$ and $K_Z$.

\begin{conj}[Iitaka]\label{c-iitaka}
Let $f\colon X\to Z$ be an algebraic fibre space where $X$ and $Z$ are smooth projective varieties of dimension $n$ and $m$,
respectively, and let $F$ be a general fibre of $f$. Then,
$$
\kappa(X)\ge \kappa(F)+\kappa(Z)
$$
\end{conj}

This conjecture is usually denoted by $C_{n,m}$. A strengthend version was proposed by
Viehweg (cf. [\ref{Viehweg}]) as follows which is denoted by $C_{n,m}^+$.

\begin{conj}[Iitaka-Viehweg]\label{c-iitaka-viehweg}
Under the assumptions of \ref{c-iitaka},
$$
\kappa(X)\ge \kappa(F)+\max\{\kappa(Z),\var(f)\}
$$
when $\kappa(Z)\ge 0$.
\end{conj}

 Kawamata [\ref{Kaw3}] showed that these conjectures  hold if the general fibre
$F$ has a good minimal model, in particular, if the minimal model and the
abundance conjectures hold in dimension $n-m$ for varieties of nonnegative Kodaira dimension.
However, at the moment the minimal model conjecture for such varieties is known only up to dimension $5$
[\ref{B2}] and the abundance conjecture up to dimension $3$ [\ref{Miyaoka}][\ref{Kaw-abundance}] and some cases
in higher dimensions which will be discussed below. Viehweg [\ref{Viehweg}] proved $C_{n,m}^+$ when $Z$ is of general type. When $Z$ is a curve $C_{n,m}$ was setteld by Kawamata [\ref{kaw2}]. Koll\'ar [\ref{kollar}] proved $C_{n,m}^+$ when $F$ is of general type. The latter also follows from Kawamata [\ref{Kaw3}] and the  existence of good minimal models for varieties of general type by Birkar-Cascini-Hacon-M$^c$Kernan [\ref{BCHM}].  We refer the reader to Mori [\ref{Mori}] for a detailed survey of the above conjectures and related problems. In this paper, we prove the following

\begin{thm}\label{6-iitaka}
Iitaka conjecture $C_{n,m}$ holds when $n\le 6$.
\end{thm}

\begin{thm}\label{p-surface}
Iitaka conjecture $C_{n,m}$ holds when $m=2$ and  $\kappa(F)=0$.
\end{thm}

When $n\le 5$ or when $n=6$ and $m\neq 2$, $C_{n,m}$ follows immediately from theorems of Kawamata and deep results of the
minimal model program.

Iitaka conjecture is closely related to the following

\begin{conj}[Ueno]\label{c-ueno}
Let $X$ be a smooth projective variety with $\kappa(X)=0$. Then, the Albanese map $\alpha\colon X\to A$
satisfies the following\\\\
$(1)$ $\kappa(F)=0$ for the general fibre $F$,\\
$(2)$ there is an etale cover $A'\to A$ such that $X\times_A A'$ is birational to $F\times A'$ over $A$.
\end{conj}

Ueno conjecture is often referred to as Conjecture K. Kawamata [\ref{kaw-albanese}] showed that $\alpha$ is an algebraic fibre space. See Mori [\ref{Mori}, \S 10] for a discussion of this conjecture.

\begin{cor}\label{6-ueno}
Part $(1)$ of Ueno conjecture holds when $\dim X\le 6$.
\end{cor}
\begin{proof}
Immediate by Theorem \ref{6-iitaka}.
\end{proof}

Concerning part $(1)$ of Ueno conjecture, recently Chen and Hacon [\ref{CH}] showed that $\kappa(F) \le \dim A$.

\section*{Acknowledgements}

I would like to thank Burt Totaro for many helpful conversations and comments. I am grateful to Fr\'ed\'eric Campana for reminding me
of a beautiful theorem of him and Thomas Peternell which considerably simplified my arguments.

\section{Preliminaries}

\emph{Nef divisors.}
A Cartier divisor $L$ on a projective variety $X$ is called nef if $L\cdot C\ge 0$
for any curve $C\subseteq X$. If $L$ is a $\Q$-divisor, we say that it is nef if
$lL$ is Cartier and nef for some $l\in \N$. We need a theorem about nef $\Q$-divisors
due to Tsuji [\ref{Tsuji}] and Bauer et al. [\ref{nef}].

\begin{thm}
Let $L$ be a nef $\Q$-divisor on a normal projective variety $X$. Then, there is a dominant almost regular rational map $\pi\colon X\bir Z$
 with connected fibres to a normal projective variety, called the reduction map of $L$, such that\\\\
(1) if a fibre $F$ of $\pi$ is projective and $\dim F=\dim X-\dim Z$, then $L|_F\equiv 0$,\\
(2) if $C$ is a curve on $X$ passing  through a very general point $x\in X$ with $\dim \pi(C)>0$, then $L\cdot C>0$.
\end{thm}

Here by almost regular we mean that some of the fibres of $\pi$ are projective and away from the indeterminacy locus of $\pi$.
Using the previous theorem, one can define the nef dimension $n(L)$ of the nef $\Q$-divisor $L$ to be $n(L):=\dim Z$.
In particular, if $n(L)=0$, the theorem says that $L\equiv 0$.\\

\emph{Minimal models.}
Let $X$ be a smooth projective variety. A projective variety $Y$ with terminal singularities is called a minimal model of $X$
if there is a birational map $\phi\colon X\bir Y$, such that $\phi^{-1}$ does not contract divisors, $K_Y$ is nef, and finally there is a common
resolution of singularities $f\colon W\to X$ and $g\colon W\to Y$ such that $f^*K_X-g^*K_Y$ is effective and its support contains the birational transform of any prime divisor
on $X$ which is exceptional over $Y$. If in addition $lK_Y$ is base point free
for some $l\in \N$, we call $Y$ a good minimal model.

The minimal model conjecture asserts
that every smooth projective variety has a minimal model or a Mori fibre space, in particular, if the variety has
nonnegative Kodaira dimension then it should have a minimal model. The abundance conjecture states that every minimal model is a good one.\\

\emph{Kodaira dimension}. Campana and Peternell [\ref{CP}] made the following interesting conjecture.

\begin{conj}
Let $X$ be a smooth projective variety and suppose that $K_X\equiv A+M$ where $A$ and $M$ are effective and pseudo-effective
$\Q$-divisors respectively. Then, $\kappa(X)\ge \kappa(A)$.
\end{conj}

They proved the conjecture in case $M\equiv 0$ [\ref{CP}, Theorem 3.1]. This result is an important ingredient of the
proofs below.\\

\section{Proofs}

\begin{proof}(of Theorem \ref{p-surface})
We are given that the base variety $Z$ has dimension $2$ and that $\kappa(F)=0$.
We may assume that $\kappa(Z)\ge 0$ otherwise the theorem is trivial.
Let $p\in \N$ be the smallest number such that $f_*\mathcal{O}_X(pK_X)\neq 0$. By Fujino-Mori
[\ref{FM}, Theorem 4.5], there is a diagram
$$
\xymatrix{
X' \ar[d]^g\ar[r]^\sigma & X \ar[d]^f\\
Z' \ar[r]^\tau & Z
}
$$
in which $g$ is an algebraic fibre space of smooth projective varieties, $\sigma$ and $\tau$ are birational, and there are $\Q$-divisors $B$
and $L$ on $Z'$ and a $\Q$-divisor $R=R^+-R^-$ on $X'$ decomposed into its positive and
negative parts satisfying the following:\\

(1) $B\ge 0$,\\

(2) $L$ is nef,\\

(3) $pK_{X'}=pg^*(K_{Z'}+B+L)+R$,\\

(4) $g_*\mathcal{O}_{X'}(iR^+)=\mathcal{O}_{Z'}$ for any $i\in\N$,\\

(5) $R^-$ is exceptional$/X$ and codimension
of $g(\Supp R^-)$ in $Z'$ is $\ge 2$.

\vspace{0.5cm}

Thus for any sufficiently divisible $i\in\N$ we have\\

(6) $g_*\mathcal{O}_{X'}(ipK_{X'}+iR^-)=\mathcal{O}_{Z'}(ip(K_{Z'}+B+L))$\\

 If the nef dimension $n(L)=2$ or if $\kappa(Z)=\kappa(Z')=2$, then $ip(K_{Z'}+L)$ is big for some $i$
by Ambro  [\ref{Ambro}, Theorem 0.3]. So, $ip(K_{Z'}+B+L)$ is also big and by (6)
and by the fact that $\sigma$ is birational and $R^-\ge 0$ is exceptional$/X$ we have
$$
H^0(ipK_X)=H^0(ipK_{X'}+iR^-)=H^0(ip(K_{Z'}+B+L))
$$
for sufficiently divisible $i\in \N$. Therefore, in this case $\kappa(X)=2\ge \kappa(Z)$.

 If  $n(L)=1$, then the nef reduction map $\pi\colon Z'\to C$ is regular where $C$ is a smooth projective curve, and there is a
$\Q$-divisor $D'$ on $C$ such that $L\equiv \pi^*D'$ and $\deg D'>0$ by [\ref{nef}, Proposition 2.11]. On the
other hand, if $n(L)=0$ then $L\equiv 0$. So, when $n(L)=1$ or $n(L)=0$, there is a $\Q$-divisor $D\ge 0$ such that
$L\equiv D$.  Now letting $M:=\sigma_*g^*(D-L)$, for sufficiently divisible $i\in \N$, we have
$$
H^0(ip(K_{X}+M))=H^0(ip(K_{X'}+g^*D-g^*L)+iR^-)
$$
$$
\hspace{1.6cm} =H^0(ip(K_{Z'}+B+D))
$$
and by Campana-Peternell [\ref{CP}, Theorem 3.1]
$$
\kappa(X)\ge \kappa(K_{X}+M)=\kappa(K_{Z'}+B+D)\ge \kappa(Z)
$$
\end{proof}

\begin{proof}(of Theorem \ref{6-iitaka})
We assume that $\kappa(Z)\ge 0$ and $\kappa(F)\ge 0$ otherwise the theorem is trivial.

If $m=1$, then the theorem follows from  Kawamata [\ref{kaw2}]. On the other hand, if $n-m\le 3$, then
the theorem follows from Kawamata [\ref{Kaw3}] and the existence of good minimal models in dimension $\le 3$. So,
from now on we assume that $n=6$ and $m=2$ hence $\dim F=4$.  By the flip theorem of Shokurov [\ref{shokurov}] and the termination theorem of Kawamata-Matsuda-Matsuki [\ref{KMM}, 5-1-15] $F$ has a minimal model (see also [\ref{B2}]). If $\kappa(F)> 0$, by Kawamata [\ref{Kaw}, Theorem 7.3] such a minimal model is good, so we can apply [\ref{Kaw3}] again. Another possible argument would be to apply Koll\'ar [\ref{kollar}] when $F$ is of general type and to use
the relative Iitaka fibration otherwise.

Now assume that $\kappa(F)=0$. In this case, though we know that $F$ has a minimal model, abundance is not yet known. Instead, we use Theorem \ref{p-surface}.
\end{proof}

config: arxiv


TITLE: Thermal imaging camera always blurry. Expected?
QUESTION [1 upvotes]: I bought a cheap thermal imaging camera for establishing hotspots in a circuit to see if any components get near or above maximum rated temperature.
But...

even though some hot spots should be very clear and distinct, they appear blurry.
Those small hot spots' temperatures seem to be hotter than the camera indicates.

Are these expected phenomena? Or is the camera cheap/inappropriate? It has 60x60 pixel, 20 degree FOV and a fixed 50cm focal distance, so really it should be ideal? Cheapest on the market but excellent reviews.
I wonder if conduction, convection and radiation mean the radiation received by the camera will always appear blurry? (Because either the heat actually is distributing itself in the circuitboard) or the radiation diffuses (so to speak) in the air before it reaches the camera? Maybe I'm doing something wrong?
Edit: thanks for responses. Please see images below:

The hot component is the transistor attached to the heatsink.
It looks to me like upscaling is not the issue? Nor the limited resolution? The thermal images in sales material for this and other models look crisp in comparison, so I'm leaning towards getting a refund, something seems wrong.
The model is KKMoon 3600

REPLY [0 votes]: There may be different causes of the problem. You did not answer from what distance you took the images. They say the shortest focal distance is 0.5 m. I don't quite understand that, but suspect this is the shortest distance from which you can take focused images (assuming that the focus is adjustable; but you say the focal distance is fixed?). So you should probably take pictures from this distance. However, I suspect you used a larger distance: the image occupies a small part of your screen. For the distance of 50 cm and FOV of 20 deg, the object covering the entire screen is about 17 cm (20 deg is approximately 1/3 radian), therefore, 1 pixel corresponds to 1/60 of that, i.e., about 0.3 cm. Thus, in principle, the resolution that you can achieve is not bad. On the other hand, maybe the heatsink works too well :-), so the heat is distributed over the heatsink, or the hot spots can be obstructed by something (your visible image is not detailed enough to see).
I don't think the camera is too cheap, but you can get something better in this price range if you choose an add-on for a smartphone, such as seek compact xr (https://www.thermal.com/compact-series.html ; maybe there are newer and better gadgets now). We used it to make pictures of an 11 micron carbon fiber heated by a microwave beam. The minimum distance for the camera is 15 cm, if I am not mistaken, so 1 pixel corresponds to about 250 micron of the object, so the thermal image is heavily averaged. See an example below (the camera probably "sees" not only the fiber but also heated air near the fiber).

So maybe you should try your camera with an object where you have a better idea of what can be achieved in principle, for example, candle flame of heated nichrome wire.


TITLE: Does there exist a notation for the number of terms in a sum?
QUESTION [0 upvotes]: If $S$ were a set comprised of $n$ elements we would say that the cardinality of $S$ is $n$ and denote said cardinality as $|S|=n$. Does there exist any notation alike the mentioned one for the  number of terms in a sum?

REPLY [2 votes]: Mathematical operations deal with properties of mathematical objects. But the number of terms in a sum is a property of the way the sum is written; that is, it is a property of the expression of the sum.  In contrast, the number of elements in a set is a property of the set itself, and is independent of the way it is written.  For example:
$$\{n\in \Bbb Z \mid n^2\lt 4\}\\
\{-1, 0, 1\}\\
\Bbb Z \cap \left(-\frac\pi2, \frac\pi2\right)\\
$$
are three different ways to write the same set, and regardless of how it is written, the set has exactly 3 elements.
A similar issue arises sometimes when one wants to discuss the numerator of a fraction.  One has to be careful to speak only of the numerator of a fraction, which makes sense, and not of the numerator of rational number that might be represented by a fraction, which does not. The rational number $1.25$, for example, could be represented as $\frac54$ or as $\frac{500}{400}$, which have different numerators.    One can of course work around the issue.  All I claim is that the issue exists.
Here you want to discuss the number of terms in a sum, which is a certain type of mathematical expression.  This can be done, but only if you are treating the expression itself as a sort of mathematical object.  This kind of thing is done, but usually only in specialized contexts, and I am not aware of any universal standard notation for it.
One context in which we often discuss the number of terms in a sum is the context of polynomials.  A polynomial is a special sort of mathematical expression, one which comes up often enough that the expression itself is considered to be a kind of mathematical object. One often talks about the number of terms in a polynomial.  The representation issue I mentioned does come up.  For example, is the polynomial $x^2 + 1$ the same as the polynomial $x^2 + 0x + 1$?  What about $0x^3 + x^2 + 0x + 1$? And how many terms are there?  We always agree that these are the same polynomial, and that it has exactly three terms.  The jargon for this is that that polynomial has degree 2, which is often written as something like $\deg P = 2$. In some contexts one might want to talk about the number of nonzero terms, so that a polynomial like $x^{53}+x+1$ has three nonzero terms although its degree is 53. I'm not aware of any special notation for this.
In a more specialized context, dealing with more general expressions, here's the kind of thing we might do:

We might decide to understand an “expression” as a rooted tree whose leaves are numbers and whose interior nodes are labeled with operations, chosen from some suitable set that includes +, ×, ÷, and so forth.  For example, the expression $“2×(9+(5-2))”$ might be identified with this tree:

An expression can then be defined to be a “sum” if each of its interior nodes is labeled with either + or -.  So the tree of the previous illustration is not a sum, but its right-side subtree is a sum.
In this case, the number of terms in the sum can be defined to be the number of leaves in the tree.  The right-side subtree here has three leaves, so it is a sum of three terms.
Now we can define a notation to represent the number of terms in a sum:  If $E$ is a sum expression, then $\operatorname{terms}(E)$ is the number of terms in  $E$.  Or we could use $|E|$ or $|E|_+$ or whatever.  mathematical notation is often fluid and ad hoc.

One could then extend the definition of “expression” to include expressions of the form $\sum_{n\in S} f(n)$ and the like, and the definition of “number of terms” to say that this expression has $|S|$ terms, or $1$, or whatever is desired.
Maybe this was more than you wanted to know, but I hope if so it was at least interesting.


TITLE: Double integral polar coordinates - boundaries for angle $ \phi $
QUESTION [0 upvotes]: Compute
$$ \iint \arctan{\frac{y}{x}}\,\mathrm dx\,\mathrm dy,$$ where $1 \le x^2 + y^2 \le 9$, $y \ge \frac{x}{\sqrt{3}}$, $y \ge x\sqrt{3}$.
Introducing polar coordinates, I get that $1 \le r \le 3$ which is fine. However, I have trouble determining the boundaries for the angle $\phi$
From the conditions:
$y \ge \frac{x}{\sqrt{3}}$ we have:
$$\sin{\phi} \ge \frac{\cos{\phi}}{\sqrt{3}}$$
$$\tan{\phi} \ge \frac{\sqrt{3}}{3}$$
with $\phi \ne \frac{\pi}{2} + 2n\pi$
Applying $\arctan$ to both sides we get
$$\phi \in [\frac{\pi}{6}, \frac{\pi}{2})$$
Similarly, from the second condition we have the boundaries
$$\phi \in [\frac{\pi}{3}, \frac{\pi}{2})$$
Now, the boundaries for the angle are
$$\phi \in [\frac{\pi}{6}, \frac{\pi}{2})$$
But my workbook states that the boundaries are
$$\phi \in [\frac{\pi}{6}, \frac{7\pi}{6})$$
Can anyone help me see where I went wrong? Thanks!

REPLY [2 votes]: In the first quadrant you know that for both $y \geq \dfrac{x}{\sqrt3}$ and $y \geq \sqrt3 x$ to be true, we must have $y \geq x \sqrt3 \implies \theta \in [\dfrac{\pi}{3}, \dfrac{\pi}{2}]$.
In second quadrant, $x \leq 0$ and so the inequality holds in the second quadrant $\theta \in [\dfrac{\pi}{2}, \pi]$.
In the third quadrant, both $x$ and $y$ are negative and the inequality reverses so for both $y \geq \dfrac{x}{\sqrt3}$ and $y \geq \sqrt3 x$ to be true, we must have $y \geq \dfrac{x}{\sqrt3} \implies \theta \in [\pi, \dfrac{7\pi}{6}]$
Here is a diagram that may help explain further.


TITLE: Probability of forming a triangle
QUESTION [0 upvotes]: We are given a line with length $K$.We pick 2 random lines with length less than $K$.What is the probability to form a triangle?
What I have done is : Let the lines be $x $ and $y$ , then we need to find the probability of $x+y > K$ ,knwing that  $ 0 \le x,y < K$ . The total area of the square with sides K is $K^2$ , but I am strugging to find the probability of $x+y > K$

REPLY [0 votes]: You can think of this in terms of probability. Let $X, Y \overset{i.i.d.}{\sim}\mathcal{U}([0, K])$. Then $Z = X + Y$ is a symmetric triangular distribution on the interval $[0, 2K]$, with mean $\operatorname{E}(Z) = K$.
Because the distribution of $Z$ is symmetric, the events $Z > \operatorname{E}(Z)$ and $Z < \operatorname{E}(Z)$ are equiprobable and $P(Z > \operatorname{E}(Z)) = P(Z < \operatorname{E}(Z)) = 0.5$.


TITLE: Noetherian module and noetherian ring
QUESTION [5 upvotes]: If $R$ is a noetherian ring then also $R[x]$ is a noetherian ring, i.e. $R[x]$ is noetherian as $R[x]$-module. Is $R[x]$ also noetherian as $R$-module?

REPLY [2 votes]: Let $R$ be a commutative ring and $X$ an indeterminate. Then $R[X]$ is a noetherian $R$-module if and only if $R$ is the zero ring.
The other answers contain a proof (but not a statement) of this fact.
Edit. In the first version, I had written "$R[X]$ is a noetherian $R$-module if and only if $R$ is not the zero ring". Thanks to quid for having pointed out this mistake!


TITLE: Hyperbolic Geometry: Representation of the metric in a chart
QUESTION [2 upvotes]: Let $\mathbb{R}^{1,n}$ denote Lorentzian $n$-space, i.e., $\mathbb{R}^n$ equipped with the Lorentzian inner product $$\langle x,y \rangle = - x_0 y_0 + x_1 y_1 + \cdots + x_n y_n.$$
Define $\mathbb{H}^n : = \left \{ \xi \in \mathbb{R}^{1,n} : \langle \xi, \xi \rangle = -1, \ \xi_0 > 0 \right \}$ to be hyperbolic $n$-space. I have been battling with the following seemingly standard/trivial exercises in Jöst's Riemannian Geometry and Geometric Analysis and have had little success:
5) Show that $\langle \cdot, \cdot \rangle$ induces a Riemannian metric on the tangent spaces $T_p \mathbb{H}^n \subset T_p \mathbb{R}^{n+1}$ for $p \in \mathbb{H}^n$.
6) Let $s = (-1, 0, ..., 0) \in \mathbb{R}^{n+1}$, and define $$f(x) = s - \frac{2(x-s)}{\langle x-s, x-s \rangle}.$$ Show that $f : \mathbb{H}^n \longrightarrow \{ \xi \in \mathbb{R}^n : | \xi | < 1 \}$ is a diffeomorphism, and that in this chart, the metric assumes the form $$\frac{4}{(1 - | \xi |^2)^2} d\xi_i \otimes d\xi_i.$$
Thoughts/Progress:
The fact that $\langle \cdot, \cdot \rangle$ defines an inner product is clear, but I cannot convince myself that $\langle \cdot, \cdot \rangle$ is positive definite. For Q6., I am aware that a similar question has been posted on here, but the solution is not clear to me.
Note that I do not want answers to these problems, just guidance.
Thanks in advance.

REPLY [5 votes]: We will prove $5)$. As differential manifold, $\mathbb{H}^n$ is a regular submanifold of $\mathbb{R}^{n+1}$ so we can identify tangent vectors in $\mathbb{H}^n$ as velocities of differentiable curves in $\mathbb{H}^n$. Take a point $p=(p_0,\ldots,p_n)$ in $\mathbb{H}^n$ and a differentiable curve $\sigma : (-\varepsilon,\varepsilon)\to \mathbb{H}^n$ such that $\sigma(0)=p$. Write the coordinates of $\sigma$ as $\sigma_0,\ldots,\sigma_n$ and let $v_i=\frac{d\sigma_i}{dt}(0)$ for every $i$. So we want to prove that $\sum_{i=1}^nv_i^2>v_0^2$.
We have that $\langle \sigma(t),\sigma(t) \rangle =-1$ for every $t$. So in particular at $0$ we have that
$$
\sum_{i=1}^np_i^2 = p_0^2-1
$$
On the other hand, by the product rule for derivatives we have that
$$
\sum_{i=1}^nv_ip_i = v_0p_0
$$
So, by the Cauchy-Schwarz inequality and the first equation, it follows that
$$
v_0^2p_0^2=\bigg{(}\sum_{i=1}^nv_ip_i\bigg{)}^2\leq \bigg{(}\sum_{i=1}^nv_i^2\bigg{)} \bigg{(}\sum_{i=1}^np_i^2\bigg{)} = \bigg{(}\sum_{i=1}^nv_i^2\bigg{)} \bigg{(}p_0^2-1\bigg{)} < \bigg{(}\sum_{i=1}^nv_i^2\bigg{)} p_0^2
$$
We know that $p_0>0$ by definition of $\mathbb{H}^n$ so the proposition follows.
EDIT: Sketch of $6)$. Let $\mathbb{D}^n=\{\xi\in\mathbb{R}^n:|\xi|<1\}$. Note that $f:\mathbb{H}^n\to\mathbb{D}^n$ is defined by the formula
$$
f(x_0,\ldots,x_n)=\bigg{(}\frac{x_1}{1+x_0},\ldots,\frac{x_n}{1+x_0}\bigg{)}
$$
and it is differentiable. The inverse $f^{-1}:\mathbb{D}^n\to\mathbb{H}^n$, defined by the formula
$$
f^{-1}(\xi_1,\ldots,\xi_n)=\bigg{(}\frac{1+|\xi|^2}{1-|\xi|^2},\frac{2\xi_1}{1-|\xi|^2},\ldots,\frac{2\xi_n}{1-|\xi|^2}\bigg{)},
$$
is differentiable so $f$ is a diffeomorphism. Prove that the pullback by $f^{-1}$ of the metric defined in $\mathbb{H}^n$ is equal to the metric given on $\mathbb{D}^n$, that is
$$
\bigg{\langle} \frac{\partial f^{-1}}{\partial \xi_i}(\xi),\frac{\partial f^{-1}}{\partial\xi_j}(\xi) \bigg{\rangle} = \frac{4\delta_{ij}}{(1-|\xi|^2)^2}
$$
for every $i,j\in \{1,\ldots,n\}$ and every $\xi\in\mathbb{D}^n$.
Note that $6)$ implies $5)$ because the pullback by a diffeomorphism of a riemannian metric is a riemannian metric.


TITLE: How to check positive-definiteness of this function?
QUESTION [2 upvotes]: Consider a real random vector $\vec X =(X_1, X_2)$ with characteristic function $\phi(\vec t) \equiv \mathbb{E} \big[ e^{i \vec t \cdot \vec X} \big]$ (where $\vec{t}=(t_1, t_2) \in \mathbb{R}^2$) given by
$$
\phi(\vec t) =\exp \bigg\{ - \int_{S^1} \mu(d \vec{s}) \, | \vec{t} \cdot \vec{s}| \Big( 1 + \frac{2i}{\pi} \, \text{sign} ( \vec{t} \cdot \vec{s}) \log | \vec{t} \cdot \vec{s}| \Big) \bigg\} \, ,
$$
where $S^1$ is the 1-sphere $S^1 = \{ \vec{s} \in \mathbb{R}^2 \mathrel: \vec{s}^2 = 1 \}$ and $\mu(d \vec{s})$ is a measure on $S^1$ ($\mu \geq 0$). I know that $\phi$ is a characteristic function of some probability distribution*. Therefore, according to Bochner's theorem, $\phi$ is a positive-definite function, i.e. it satisfies
$$
0 \leq \sum_{k,l=1}^N a_k^* a_l \, \phi(\vec{t}_l - \vec{t}_k) \, ,  \qquad \forall a_1, ..., a_N \in \mathbb{C} \, , \, \, \forall \,  \vec{t}_1, ..., \vec{t}_N \in \mathbb{R}^2 \, , \, \text{ and } \, \forall N \in \mathbb{N} \, .
$$
My question is: how do I check this condition for the particular $\phi$ given above?

*In particular, $\phi$ is the characteristic function of a bivariate Cauchy distribution, as parametrized by Samorodnitsky and Taqqu - Stable non-Gaussian random processes, but this is not relevant to the question.

REPLY [3 votes]: $\newcommand{\vpi}{\varphi}\newcommand{\R}{\mathbb R}\newcommand{\s}{\vec s}\newcommand{\ttt}{\vec t}$For any real $c>0$, the function $\vpi_c$ given by the formula
\begin{equation*}
    \vpi_c(t):=\exp\{-c|t|(1+\tfrac{2i}\pi\,\text{sign}(t)\ln|t|)\}
\end{equation*}
for real $t$ is the characteristic function (c.f.) of the (univariate) stable distribution with parameters $\alpha=1$, $\beta=1$, $c$, and $\mu=\frac2\pi\,c\ln c$, so that
\begin{equation*}
    \vpi_c(t)=Ee^{itY_c}
\end{equation*}
for some real-valued random variable $Y_c$ and all real $t$.
So, for each real $c>0$, each $\s\in\R^2$, and all $\ttt\in\R^2$,
\begin{equation*}
    f_{c,\s}(\ttt):=\vpi_c(\s\cdot\ttt)=Ee^{i(Y_c\s)\cdot\ttt},
\end{equation*}
so that $f_{c,\s}$ is a c.f.; specifically, $f_{c,\s}$ is the c.f. of the random vector $Y_c\s$ in $\R^2$.
Approximating now the integral in the definition of the function $\phi$ in the OP by corresponding integral sums, we see that $\phi$ is the pointwise limit of functions of the form
\begin{equation*}
    \prod_{j=1}^n f_{c_j,\s_j} \tag{1}
\end{equation*}
for some natural $n$, some real positive $c_j$'s, and some $\s_j$'s in $\R^2$.
All functions of the form (1) are c.f.'s (of bivariate distributions), because the product of c.f.'s of distributions is a c.f. (namely, the c.f. of the convolution of the distributions).
So, obviously all functions of the form (1) are positive definite.
Also, obviously the pointwise limit of positive-definite functions is positive definite.
Thus, $\phi$ is indeed positive definite.


TITLE: Number of ways to fill 4 cells in a $4 \times 4$ array under constraints
QUESTION [3 upvotes]: Two undistinguishable letters $a$ and two undistinguishable letters $b$ are filled into $16$ cells of a square matrix. It is required that each cell contains at most $1$ letter and each row or column cannot contain the same letters. What is the number of ways which the matrix can be filled in?
My attempt:
I considered the case of first placing the two '$a$' s in $\frac{\binom{16}{1}\binom{9}{1}}{2} =72$ ways
and then the two '$b$'s in also $72$ ways.
Now we need to find the overlapping cases.
$(i)$
In the same cell we have got $72 +$ the case where one $b$ is in the same cell as that of $a$ and the other $b$ is in it's proper place. This is where I am getting confused.
Please help me out.

REPLY [2 votes]: I think you've got a good idea. Let's say we start with $a$ and $b$ in a common cell, and place $a$ and $b$ freely in other cells that aren't in the same row/column, but also so that the second $a$ and $b$ end up in distinct cells. There are $16$ places to put the $a$ and $b$ together, then $9$ cells to place the second $a$, then $8$ cells to place the second $b$. In total,
$$16 \times 9 \times 8 = 1152.$$
We also need to exclude the cases where both pairs of $a$ and $b$ end up together, which is simply $72$, as you calculated previously. So, the total should be
$$72^2 - 1152 - 72 = 3960.$$


TITLE: Stationary aether with respect to Earth?
QUESTION [0 upvotes]: Michelson & Morley's experiment implies that either the aether does not exist OR that the Earth is stationary with respect to aether.
Now, one consequence of relativity is that we CAN physically regard the Earth as stationary, with all the universe moving with respect to it. So, we could say that in our Earth-stationary frame the aether is stationary as well.
This would mean that Michelson & Morley's experiment does not imply by itself the lorentz transfomation apparatus (since this experiment was performed on aether-stationary-earth).
It seems to me that some different experiment should be required in order to prove the Lorentz transformations, considering some reference frames IN MOTION with respect to Earth. Michelson & Morley's one is insufficient in this light.
Ok well, I'm pretty sure I have gone wrong conceptually in some statement, but I can't see where. Can you please correct me?
And moreover,  can you please state any experiment which effectively tested relativistic effects (time dilation and length contraction) in reference frames in motion with respect to Earth?

REPLY [1 votes]: It seems to me that some different experiment should be required in order to prove the Lorentz transformations

Yes, you are completely correct. The Michelson and Morley experiment is insufficient by itself to experimentally deduce the Lorentz transform.
Robertson (Rev. of Mod. Phys. 21, pg 378, 1949) showed that you could deduce the Lorentz transform from the combination of the Michelson and Morley experiment, the Kennedy and Thorndike experiment, and the Ives and Stilwell experiment. When you do that you empirically obtain the Lorentz transform to within 0.1% without assuming the postulates of relativity. Of course, with more recent and more precise experiments, the accuracy is far better than 0.1% now.
If you are not looking for a quantitative measurement of the Lorentz transform, then qualitatively you can say the the Michelson Morley experiment is incompatible with a traditional rigid aether, and the Sagnac experiment is incompatible with a dragged aether. Only the Lorentz aether remains compatible with experiment since it is the only aether that is designed to be experimentally identical to there being no aether.


TITLE: $f_n$ converges to $f$ in $L^3(X,\mathcal{M},\mu)$ , prove that $f_n^3$ converges to $f^3$ in $L^1(X,\mathcal{M},\mu) $
QUESTION [2 upvotes]: Suppose $(X,\mathcal{M},\mu)$ is a complete measure space. If $f_n$ converges to $f$ in $L^3(X,\mathcal{M},\mu)$ , prove that  $f_n^3$ converges to $f^3$ in $ L^1 (X,\mathcal{M},\mu) $.
Actually my issue is the following :
$$\int_X |f^3-f_n^3|d\mu=\int_X |f-f_n||f^2+ff_n+f_n^2|d\mu$$
If I know that $f^2+ff_n+f_n^2$ belongs to $L^{3/2}(X,\mathcal{M},\mu)$ I can use the Holders inequality and I am done (in fact I need to show that $ff_n \in L^{3/2}(X,\mathcal{M},\mu)$)
I'd appreciate your hints/answers

REPLY [2 votes]: in fact I need to show that $ff_n \in L^{3/2}(X,\mathcal{M},\mu)$

This is Cauchy-Schwarz:
$$
\left(\int\left|ff_n\right|^{3/2}\mathrm d\mu\right)^2\leqslant\int\left|f\right|^{3}\mathrm d\mu\cdot\int\left|f_n\right|^{3}\mathrm d\mu.
$$


TITLE: Linear independence of matrices $I, A, A^2$
QUESTION [1 upvotes]: I want to prove that $I,A,A^2\:$matrices $\in M_{2\times 2}$ are $\textit {linearly independent}$.
I consider the following matrices and their "corresponding" vectors:
$I=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\to (1,0,0,1)$
$A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}\to (a,b,c,d)$
$A^2=\begin{pmatrix} a^2+ b c & ab+db \\ ac+dc & d^2+bc\end{pmatrix}\to (a^2+bc, ab+db, ac+dc,d^2+bc)$
What I am trying to do, is to prove that these vectors are $\textit{linearly independent}$.
For that direction, I want to prove that (at least) one $3\times 3$ determinant is $\neq 0$.
It seems that all $3\times 3$ determinants $\left( e.g. \begin{vmatrix} 1 & 0 & 0\\ a & b & c\\ a^2+bc & ab+db & ac+dc\end{vmatrix} \right)$ are equal to zero! What am I really missing?

REPLY [1 votes]: First of all your matrix is a root of its characteristic polynomial . The bad news is that its square is linear dependant from its lower order terms. The good news is that the algebra generated by this matrix (if its characteristic polynomial is irreducible which it's mostly is) is the splitting field of its roots.


TITLE: Question: Well-Defined Operator between Hilbert Spaces
QUESTION [0 upvotes]: Let $\;\mathcal B(H,F)\;$ be the set of all bounded linear operators $\;S:H\rightarrow F\;$ where $H,F\;$ are some Hilbert spaces.
To begin with, we consider $\;T_n \in \mathcal B(H,F)\;\forall n\;$ to be a Cauchy sequence. Then $\forall x \in H\;\;\vert\vert T_nx-T_mx \vert\vert \le \vert \vert T_n-T_m \vert \vert\cdot \vert \vert x \vert \vert \;$ which implies $\;T_nx\;$ is a Cauchy sequence on $F\;$. Since $\;F$ is Banach space, one can find $\;y \in F\;$ such that $T_nx \rightarrow y$ as $n \to \infty\;$
I want to show there is $\;x \in H\;$ such that $T_x=y\;$. This is why I define $T:H \rightarrow F\;$ as $Tx=\lim_{n\rightarrow \infty} T_nx\;\forall x \in H\;$.
My question is: How can I show the above is well defined?
I took $x,z \in H\;$ such that $\;T_x=T_z\; \Rightarrow \lim_{n \to \infty} T_nx=\lim_{n \to \infty} T_nz\;$
How do I proceed? I believe the answer is obvious but I 've been stuck. I would appreciate any help. Thanks in advance!!

REPLY [1 votes]: What you're saying is you want to show $\lim_{n\rightarrow\infty} T_n x$ exists and is unique for each $x$.  You found the limit using completeness, and it is unique because of uniqueness of limits in $F$ (since it's a metric space and has all of those nice properties that topological spaces can have).


TITLE: A quick question on the subring generated by a finite set
QUESTION [2 upvotes]: Let $R$ be a commutative unital ring and let $r_1,\ldots,r_n \in R$.
Let $S$ be the unital subring of $R$ generated by $r_1,\ldots,r_n$.
Let $\varphi:\mathbb{Z}[X_1,\ldots,X_n]\to R$ be the unique homomorphism sending $1\mapsto 1_R$ and each $X_i\mapsto r_i$.
Am I right in thinking that $S=\text{im}(\varphi)$?
Many thanks!

REPLY [2 votes]: Yes. It is clear that $\text{im}(\phi) \subseteq S$. On the other hand, $\text{im}(\phi)$ is a subring of $R$ that contains all the $r_i$'s, and since $S$by definition is the smallest subring of $R$ containing the $r_i$'s, we must have $S \subseteq \text{im}(\phi)$. Hence, $S = \text{im}(\phi)$.


TITLE: How to calculate the following limit
QUESTION [3 upvotes]: $$ \lim_{x\to 0} \frac{(1+2x)^{1/x}-(1+x)^{2/x}}{x} $$ I tried calculating it with the e limit but I end up with and undefined limit.I found its $$ -e^2 $$ by putting in numeric values like 1/2 and 1/3 and I saw it gets closer and closer to that but how could I solve it in an algebraic way?Someone suggested I sould simply use the $$ f^g=e^{g\ln f} $$ and then give common factor $$ e^{(1/x)\ln(1+2x)} $$ any ideas?

REPLY [2 votes]: Let's try with standard limits $$\lim_{x \to 0}\frac{\log(1 + x)}{x} = \lim_{x\to 0}\frac{e^{x} - 1}{x} = 1\tag{1}$$ We can proceed as follows:
\begin{align}
L &= \lim_{x \to 0}\frac{(1 + 2x)^{1/x} - (1 + x)^{2/x}}{x}\notag\\
&= \lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + 2x)}{x}\right) - \exp\left(\dfrac{2\log(1 + x)}{x}\right)}{x}\notag\\
&= \lim_{x \to 0}\exp\left(\dfrac{2\log(1 + x)}{x}\right)\cdot\dfrac{\exp\left(\dfrac{\log(1 + 2x)}{x}-\dfrac{2\log(1 + x)}{x}\right) - 1}{x}\notag\\
&= e^{2}\lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + 2x)}{x}-\dfrac{2\log(1 + x)}{x}\right) - 1}{\dfrac{\log(1 + 2x)}{x}-\dfrac{2\log(1 + x)}{x}}\cdot\frac{\log(1 + 2x) - 2\log(1 + x)}{x^{2}}\notag\\
&= e^{2}\lim_{x \to 0}\frac{\log(1 + 2x) - 2\log(1 + x)}{x^{2}}\notag\\
&= e^{2}\lim_{x \to 0}\dfrac{\log\left(\dfrac{1 + 2x}{1 + 2x + x^{2}}\right)}{x^{2}}\notag\\
&= e^{2}\lim_{x \to 0}\dfrac{\log\left(1 - \dfrac{x^{2}}{1 + 2x + x^{2}}\right)}{- \dfrac{x^{2}}{1 + 2x + x^{2}}}\cdot\frac{-1}{1 + 2x + x^{2}}\notag\\
&= e^{2}\cdot 1\cdot (-1)\notag\\
&= -e^{2}\notag
\end{align}


TITLE: extension of regular logic
QUESTION [5 upvotes]: The internal language of some categories (such as abelian categories) support only a fraction of first order logic. But every category can be fully faithfully embedded into a category of presheaves, and any such presheaf category is a topos and supports full higher order logic. If I am able to proof that some sentence $\phi$ using HOL and if $\phi$ is a sentence which is in the regular first order fragment of HOL, does then also $\vdash \phi$ hold in first order regular logic?
Is there some other way to exploit the embedding of a regular category into its presheaf category on a logical level?

REPLY [2 votes]: The correct way of doing this is to embed regular categories into categories of sheaves.
We have the following fact:
Proposition.
Let $\mathcal{C}$ be a small regular category and let $\mathcal{E}$ be the category of all functors $X : \mathcal{C}^\textrm{op} \to \textbf{Set}$ such that, for every regular epimorphism $p : A \twoheadrightarrow B$ in $\mathcal{C}$, $X (p) : X (B) \to X (A)$ is the equaliser of the evident parallel pair $X (A) \rightrightarrows X (A \times_B A)$.
Then:

$\mathcal{E}$ is a Grothendieck topos.
The Yoneda embedding $\mathcal{C} \to [\mathcal{C}^\textrm{op}, \textbf{Set}]$ factors through $\mathcal{E}$, i.e. every representable presheaf on $\mathcal{C}$ is in $\mathcal{E}$.
The Yoneda embedding $\mathcal{C} \to \mathcal{E}$ preserves regular epimorphisms.

From the above fact we deduce that any sentence in the internal language of $\mathcal{C}$ as a regular category is true in $\mathcal{C}$ if and only if it is true in $\mathcal{E}$.
Here, it is crucial that we use $\mathcal{E}$ and not $[\mathcal{C}^\textrm{op}, \textbf{Set}]$: the Yoneda embedding $\mathcal{C} \to [\mathcal{C}^\textrm{op}, \textbf{Set}]$ does not preserve regular epimorphisms in general.
Since this holds for all (small) regular categories $\mathcal{C}$, we deduce that topos HOL is a conservative extension of regular logic.
Since you mentioned abelian categories, I would be remiss if I did not mention the Freyd–Mitchell embedding theorem, which is a much stronger result: it says every small abelian category $\mathcal{A}$ embeds as an exact full subcategory of the category of $R$-modules, for some (non-commutative) ring $R$ (depending on $\mathcal{A}$, of course).


TITLE: Local coordinates identities
QUESTION [1 upvotes]: Given $(M,g)$ Riemannian manifold, $f \in \mathcal{C}^{\infty}(M)$, I want to prove that the following holds (denoting by $D\,df $ the Hessian of the function $f$ and $\text{grad} f$ its gradient):
$$2D\,df (\text{grad} f,\text{grad} f) = \langle \text{grad}  \,|\text{grad} f|^2, \text{grad} f\rangle$$
I tried doing this:
Start from $$D\,df (\text{grad} f,\text{grad} f) = D_{\text{grad} f}D_{\text{grad} f}(f)$$ Then since we can write in coordinates $\text{grad} f = g^{ij}\frac{\partial f}{\partial x^i} \, \frac{\partial f}{\partial x^j}$ then we have:
$$D_{\text{grad} f} \, (g^{ij}\frac{\partial f}{\partial x^i} D_{\frac{\partial }{\partial x^j}}f)$$
Now, since $f$ is a function, the covaiant derivative of $f$ along $\frac{\partial }{\partial x^j} $ is $ \frac{\partial f}{\partial x^j}$ so we get:
$$D_{g^{ab}\frac{\partial f}{\partial x^a}\frac{\partial f}{\partial x^b}}g^{ij}\frac{\partial f}{\partial x^i}\frac{\partial f}{\partial x^j}$$
getting
$$g^{ab}\frac{\partial f}{\partial x^a}\left[\frac{\partial}{\partial x^b}\left(g^{ij}\frac{\partial f}{\partial x^i}\frac{\partial f}{\partial x^j}\right)\right]$$
In an analogous fashion, if I develop in coordinates this term $\langle \text{grad} f \,|\text{grad} f|^2, \text{grad} f\rangle$ I get:
$$\langle \text{grad} \left( g^{i \alpha}\frac{\partial f}{\partial x^i}\frac{\partial f}{\partial x^\alpha} \right), \text{grad} f\rangle$$ and tracing with the metric we end up with:
$$g^{aA}\frac{\partial f}{\partial x^A}\left[\frac{\partial}{\partial x^a}\left(g^{i\alpha}\frac{\partial f}{\partial x^i}\frac{\partial f}{\partial x^\alpha}\right)\right]$$
First I ask if I am doing the computations correctly, and then why I do not get a $2$, I think I misssed something. Thanks!

REPLY [2 votes]: The $\textit{Hessian}$ ($\nabla df$) of a differentiable function $f\,:\,M\to \mathbb{R}$ on a Riemannian manifold $M$ is
\begin{equation}
  \nabla df=\left(\frac{\partial^2f}{\partial x^k \partial x^a}-\Gamma^{m}_{ka}\frac{\partial f}{\partial x^m}\right)dx^k\otimes dx^a \tag{1}
\end{equation}
It is easy enough to plug in $\mathrm{grad} f$ as the arguments (and multiply by 2)
\begin{equation}
2\nabla df(\mathrm{grad}f,\mathrm{grad}f)=2\left(\frac{\partial^2f}{\partial x^k \partial x^a}-\Gamma^{m}_{ka}\frac{\partial f}{\partial x^m}\right)\frac{\partial f}{\partial x^b}g^{ab}\frac{\partial f}{\partial x^v}g^{kv}\tag{2}
\end{equation}
Now that we know what we are looking for we move to the right side of the equation
\begin{equation}
|\mathrm{grad}f|^2=g^{ab}\frac{\partial f}{\partial x^a}\frac{\partial f}{\partial x^b} \tag{3}
\end{equation}
\begin{equation}
<\mathrm{grad}|\mathrm{grad}f|^2,\mathrm{grad}f>=g^{kv}\frac{\partial }{\partial x^k}\left(g^{ab}\frac{\partial f}{\partial x^a}\frac{\partial f}{\partial x^b}\right)\frac{\partial f}{\partial x^v} \tag{4}
\end{equation}
Before we evaluate the tedious partial derivatives we remind the reader that an affine connection $\nabla$ is a Riemann connection if it preserves the metric $\nabla g=0$ or equivalently $\nabla_X g=0$ for all vector fields $X$ on $M$. We say that the metric is "compatible" with the connection. In component notation this means that the (partial) covariant derivative of the symmetric nonsingular (2,0) tensor $g^{ab}$ vanishes identically:
\begin{equation}
\nabla_k g^{ab}=\frac{\partial g^{ab}}{\partial x^k}+\Gamma^{a}_{mk}g^{mb}+\Gamma^{b}_{mk}g^{am}=0
\end{equation}
\begin{equation}
\frac{\partial g^{ab}}{\partial x^k}=-\Gamma^{a}_{mk}g^{mb}-\Gamma^{b}_{mk}g^{am}\tag{5}
\end{equation}
notice how $k$ is an index in the above expression, not a vector.
Now we are ready to expand the partial derivatives (and please remember the Leibniz (product) rule).
\begin{equation}
\frac{\partial }{\partial x^k}\left(g^{ab}\frac{\partial f}{\partial x^a}\frac{\partial f}{\partial x^b}\right)=\frac{\partial g^{ab}}{\partial x^k}\frac{\partial f}{\partial x^a}\frac{\partial f}{\partial x^b}+ g^{ab}\frac{\partial^2 f}{\partial x^k \partial x^a}\frac{\partial f}{\partial x^b}+ g^{ab} \frac{\partial f}{\partial x^a}\frac{\partial^2 f}{\partial x^k \partial x^b}\tag{6}
\end{equation}
Obviously the last two terms of $(6)$ are the same and after some dummy index renaming, using $(5)$ and the symmetric property of the metric we end up with
\begin{equation}
\frac{\partial }{\partial x^k}\left(g^{ab}\frac{\partial f}{\partial x^a}\frac{\partial f}{\partial x^b}\right)=2g^{ab}\frac{\partial^2 f}{\partial x^k \partial x^a}\frac{\partial f}{\partial x^b}-2 g^{ab} \frac{\partial f}{\partial x^m}\frac{\partial f}{\partial x^b}\Gamma^m_{ak} \tag{7}
\end{equation}
Putting this back in $(4)$ we have exactly $(2)$


text:
\begin{document}
\title{A proof of two conjectures on APN functions.}
\author{Elodie Leducq\footnote{IMJ, PhD thesis under direction of Jean-François Mestre}}
\date{}
\maketitle
\section{Introduction}

In \cite{Hell_new_families}, after a computer search, the authors give a complete list of power mappings almost perfectly nonlinear (APN) on $\F_{p^n}$, for $$p^n\in\{2^2,2^3,\ldots,2^{11},3^2,\ldots,3^7,5^2,\ldots,5^5,7^2,7^3,7^4,11^2,11^3\}.$$ Their article contains many theorems showing that some of these functions are members of an infinite family of APN functions. In \cite{MR2679075}, Zha and Wang give theorems that explain several new cases. In \cite{DOB_apn_odd_char}, Dobbertin, Mills, Müller, Pott et Willems try to find families containing values not yet explained in \cite{Hell_new_families}. They make conjectures that we prove here.
\\\\First we recall the definition of APN function:

\begin{defi}Let $q=p^n$, $p$ being a prime number and $n$ an integer. Let $f:\F_{q}\rightarrow\F_{q}$; for all $a$ and $b\in\F_q$, we denote by $N_f(a,b)$ the number of solutions in $\F_q$ of the equation $f(x+a)-f(x)=b$. We say that $f$ is APN if $$\Delta_f:=\max(N_f(a,b),a,b\in\F_{q},a\neq0)=2.$$\end{defi}

\begin{remarque}

\begin{itemize}\item On $\F_{p^n}$, if we denote $\Delta_{x^d}$ by $\Delta_d$, we have $\Delta_{dp^i}=\Delta_d$, for $0\leq i\leq n-1$.
\item We also have $\Delta_d=\max(N_{x^d}(1,b),b\in\F_{q})$.\end{itemize}\end{remarque}

In the following table, we give all cases not yet explained in \cite{Hell_new_families} :
\begin{center}\begin{tabular}{|c|c|c|c|}
\hline
 & $p^n$ & $d$ &$dp^i$\\
 \hline
 I & $3^5$ & 134 & (134,160,238,230,206) \\
 II & $3^5$ & 152 & (152,214,158,232,212)\\
 III & $3^7$ & 40 & (40,120,360,1080,1054,976,742)\\
 IV & $3^7$ & 224 & (224,672,2016,1676,656,1968,1532)\\
 V & $3^7$ & 274 & (274,822,280,840,334,1002,820)\\
 VI & $5^3$ & 14 & (14,70,102)\\
 VII & $5^5$ & 843 & (843,1091,2331,2283,2043)\\
 \hline
 \end{tabular}\end{center}

We put $m=\frac{n+1}{2}$.

In \cite{DOB_apn_odd_char}, Dobbertin, Mills, Müller, Pott and Willems make conjectures which explain cases I, II, III, V and VII of the table :

\begin{conjecture}\label{2.1}For $n\geq5$ an odd integer, the function $x\mapsto x^d$ is APN over $\F_{3^n}$ for $$d=\left\{\begin{array}{ll} \frac{3^m-1}{2} & \textrm{if $n\equiv3 \mod4$}\\ \frac{3^m-1}{2}+\frac{3^n-1}{2}&\textrm{if $n\equiv1\mod4$}\end{array}\right..$$\end{conjecture}

\begin{conjecture}\label{2.2}For $n\geq5$ an odd integer, the function $x\mapsto x^d$ is APN over $\F_{3^n}$ for $$d=\left\{\begin{array}{ll} \frac{3^{n+1}-1}{8} & \textrm{si $n\equiv3 \mod4$}\\ \frac{3^{n+1}-1}{8}+\frac{3^n-1}{2}&\textrm{si $n\equiv1\mod4$}\end{array}\right..$$\end{conjecture}

\begin{conjecture}\label{5}Let $n$ be an odd integer. The function $x\mapsto x^d$ is APN over $\F_{5^n}$ for $$d=\frac{5^n-1}{4}+\frac{5^m-1}{2}.$$ \end{conjecture}

Now we recall two theorems proved in \cite{MR2679075} (theorem 4.1 and 4.4) that explain cases IV and VII of the table :

\begin{theoreme}(Zha, Wang) On $\F_{3^n}$, the function $f:x\mapsto x^d$ satisfies $\Delta_d\leq2$ for $d$ such that $(3^k+1)d-2=u(3^n-1)$ where $u$ is odd and $\gcd(n,k)=1$. Furthermore, $f$ is APN if $2k<n$.\end{theoreme}

\begin{theoreme}\label{4.4}(Zha, Wang) The function $x\mapsto x^d$ is APN over $\F_{5^n}$ for $d$ such that $(5^k+1)d-2=u(5^n-1)$ where $\gcd(n,k)=1$, $u$ is odd and $k$ is even.\end{theoreme}

Using theorem \ref{4.4} Zha and Wang prove conjecture \ref{5}.

The following theorems are proved in \cite{DOB_apn_odd_char} (Theorem 2.1 and 2.2) :
\begin{theoreme}\label{3}(Dobbertin, Mills, Müller, Pott, Willems) Let $n$ be an odd integer. In $\F_{3^n}$, the function $x\mapsto x^d$ satisfies $\Delta_d\leq2$ for $$d=\left\{\begin{array}{ll} \frac{3^m-1}{2} & \textrm{if $n\equiv3 \mod4$}\\ \frac{3^m-1}{2}+\frac{3^n-1}{2}&\textrm{if $n\equiv1\mod4$}\end{array}\right..$$\end{theoreme}

\begin{theoreme}\label{4}(Dobbertin, Mills, Müller, Pott, Willems) Let $n$ be an odd integer. In $\F_{3^n}$, the function $x\mapsto x^d$ satisfies $\Delta_d\leq2$ for $$d=\left\{\begin{array}{ll} \frac{3^{n+1}-1}{8} & \textrm{si $n\equiv3 \mod4$}\\ \frac{3^{n+1}-1}{8}+\frac{3^n-1}{2}&\textrm{si $n\equiv1\mod4$}\end{array}\right.$$\end{theoreme}

~\\In the next section, we prove conjectures \ref{2.1} and \ref{2.2}. In section 3, we prove the following theorem, which gives a new infinite family of APN functions :

\begin{theoreme}\label{n5}Let $l\geq 2$ and $n$ an integer such that $n\equiv -1\mod2^l$ then the function $x\mapsto x^d$ is APN over $\F_{5^n}$ for $$d=\frac{1}{2}\frac{5^{n+1}-1}{5^{\frac{n+1}{2^l}}+1}+\frac{5^n-1}{4}.$$ \end{theoreme}
Finally, in the last section, we make some remarks about Zha and Wang theorems. In particular we give a new proof of conjecture \ref{5} in the case where $n\equiv3\mod4$.

\section{Proof of conjectures \ref{2.1} and \ref{2.2}}

In this part $p=3$. Using theorems \ref{3} and \ref{4} , we only have to show that $\Delta_d\neq1$ which means that $(x+1)^d-x^d$ is not a permutation polynomial. We do that with Dickson and Hermite's criterion which gives a necessary and sufficient condition for a polynomial with coefficients in $\F_q$ to be a permutation polynomial over $\F_q$ :
\begin{theoreme}(Hermite an Dickson's criterion, see \cite{lidl_finite_fields} p. 349) The following propositions are equivalent :
\begin{enumerate}\item $f(x)\in\F_q[x]$ is a permutation polynomial over $\F_q$.
\item $f(x)$ has exactly one root in $\F_q$ and
\\ $\forall t$, $1\leq t\leq q-2$, $t\not\equiv 0\mod p$, the reduction $(f(x))^t\mod x^q-x$ has degree less than $q-1$\end{enumerate}\end{theoreme}

We also need Lucas' theorem :
\begin{theoreme}(Lucas see \cite{Lucas}, p.230) Let $p$ be a prime number, $n$ and $r$ integers. We consider p-adic decomposition of $n$ and $r$ :
\\$n=n_0+n_1p+\ldots+n_kp^k$ with $0\leq n_i\leq p-1$,
\\$r=r_0+r_1p+\ldots+r_kp^k$ with $0\leq r_i\leq p-1$.
\\Then $$\binom{n}{r}\equiv\prod_{i=0}^k\binom{n_i}{r_i}\mod p.$$\end{theoreme}

Now we can prove the conjectures \ref{2.1} and \ref{2.2}.

Since in each case $\gcd(d,3^n-1)=2$, we have to find $t$ in Hermite and Dickson's criterion such that the degree of $((x+1)^d-x^d)^t\mod x^q-x$ is $q-1$.

\begin{align*}\left((x+1)^d-x^d\right)^t&=\sum_{k=0}^t\binom{t}{k}(x+1)^{dk}(-1)^{t-k}x^{d(t-k)} \\&=\sum_{k=0}^t\binom{t}{k}(-1)^{t-k}\left(\sum_{j=0}^{dk}\binom{dk}{j}x^{dk-j}\right)x^{d(t-k)}\\&=\sum_{k=0}^t\binom{t}{k}(-1)^{t-k}\sum_{j=0}^{dk}\binom{dk}{j}x^{dt-j} \\&=\sum_{j=0}^{dt}x^{dt-j}\sum_{k=\lceil\frac{j}{d}\rceil}^t(-1)^{t-k}\binom{t}{k}\binom{dk}{j}\end{align*}
The degree of $x^k \mod x^q-x$ is $q-1$ if and only if $k\equiv0\mod q-1$ and $k\neq0$. For $v\in\mathbb{R}$ we denote by $\lceil v\rceil$ the ceilling of $x$ and by $\lfloor v\rfloor$ its floor. Then the coefficient of $x^{q-1}$ in $\left((x+1)^d-x^d\right)^t \mod x^q-x$ is $$C=\sum_{i=1}^{\lfloor\frac{dt}{q-1}\rfloor}\sum_{k=\lceil\frac{dt-i(q-1)}{d}\rceil}^t(-1)^{t-k}\binom{t}{k}\binom{dk}{dt-i(q-1)}.$$

\paragraph*{Case where $n\equiv 1\mod 4$ in conjecture \ref{2.1}}
~~\\\\We have $d=\frac{3^m-1}{2}+\frac{3^n-1}{2}$. We choose $t=2$ then, since $n\ne1$, $\lfloor\frac{dt}{q-1}\rfloor=1$  and $\lceil\frac{dt-(q-1)}{d}\rceil=1$. So $$C=\binom{3^n+3^m-2}{3^m-1}+\binom{\frac{3^n-1}{2}+\frac{3^m-1}{2}}{3^m-1}$$
On one side, $3^n+3^m-2=3^n+\displaystyle\sum_{k=1}^{m-1}2\times3^k+1$ and $3^m-1=\displaystyle\sum_{k=0}^{m-1}2\times3^k$ so by Lucas' theorem, $\binom{3^n+3^m-2}{3^m-1}\equiv 0\mod 3$. \\On the other side, $\frac{3^n-1}{2}+\frac{3^m-1}{2}=\displaystyle\sum_{k=0}^{n-1}3^k+\sum_{k=0}^{m-1}3^k=\sum_{k=m}^{n-1}3^k+\sum_{k=0}^{m-1}2\times3^k$. So $\binom{\frac{3^n-1}{2}+\frac{3^m-1}{2}}{3^m-1}\equiv1\mod 3$.

\paragraph*{Case where $n\equiv 3\mod 4$ in conjecture \ref{2.1}}
~~\\\\We have $d=\frac{3^m-1}{2}$. Let $s=2\times(3^{m-1}+1)$, we choose $t=2s+4$. For $n>3$, $0<t<q-1$ and $t\not\equiv0\mod3$. Furthermore, $\lfloor\frac{dt}{q-1}\rfloor=2$, $\lceil\frac{dt-(q-1)}{d}\rceil=s+6$ and $\lceil\frac{dt-2(q-1)}{d}\rceil=7$.
So $$C=\sum_{k=s+6}^t(-1)^k\binom{t}{k}\binom{dk}{dt-(q-1)}+\sum_{k=7}^t(-1)^k\binom{t}{k}\binom{dk}{dt-2(q-1)}.$$
We have $t=4\times3^{m-1}+8=3^{m}+3^{m-1}+2\times3+2$.
\\So by Lucas' theorem, $\binom{t}{k}\not\equiv0\mod3$ if and only if $k=a3^{m}+b3^{m-1}+c3+d$ where $a$, $b\in\{0,1\}$ and $c$, $d\in\{0,1,2\}$.
\\Then $$dk= a3^n+(a+b)\sum_{j=m+1}^{n-1}3^j+(a+b+c)3^{m}+(b+c+d)3^{m-1}+(c+d)\sum_{j=1}^{m-2}3^j+d$$
Furthermore, $dt-2(q-1)=3^{m+1}+\displaystyle\sum_{j=1}^{m-2}2\times3^j+1$ and \\$dt-(q-1)=3^n+3^{m+1}+\displaystyle\sum_{j=1}^{m-2}2\times3^j$.
\\So $\binom{dk}{dt-(q-1)}\not\equiv0\mod3$ if $a=1$ and $c+d=2$.
\\Assume that $b=1$, then $b+c+d=3$ and $5\geq a+b+c+1\geq3$. So the coefficient of $3^{m+1}$ in 3-adic decomposition of $dk$ is 0 which means that $\binom{dk}{dt-2(q-1)}\equiv0\mod3$.
Hence the only $k$ remaining in the first sum are $3^{m}+2\times3$, $3^{m}+3+1$ and $3^{m}+2$.
\\Now we consider the second sum : $\binom{dk}{dt-2(q-1)}\not\equiv0\mod3$ if   $d\in\{1,2\}$ and $c+d=2$, namely if $c=1$ and $d=1$ or if $c=0$ and $d=2$.
Since $k\geq7$, $a$ and $b$ can't be both 0.
Assume that $a=b=1$ then $b+c+d=3$ and $4\geq a+c+b+1 \geq3$. So the coefficient of $3^{m+1}$ in 3-adic decomposition of $dk$ is 0, which means that $\binom{dk}{dt-2(q-1)}\equiv0\mod3$. Hence the only $k$ remaining in this sum are $3^{m}+3+1$, $3^{m}+2$, $3^{m-1}+3+1$ and $3^{m-1}+2$.
\\Finally :
\begin{align*}C&=-\binom{t}{3^{m}+2\times3}\binom{d(3^{m}+2\times3)}{dt-(q-1)}  -\binom{t}{3^{m}+3+1}\binom{d(3^{m}+3+1)}{dt-(q-1)}
\\&\hspace{0.5cm}-\binom{t}{3^{m}+2}\binom{d(3^{m}+2)}{dt-(q-1)}
-\binom{t}{3^{m}+3+1}\binom{d(3^{m}+3+1)}{dt-2(q-1)}
\\&\hspace{0.5cm}-\binom{t}{3^{m}+2}\binom{d(3^{m}+2)}{dt-2(q-1)}
-\binom{t}{3^{m-1}+3+1}\binom{d(3^{m-1}+3+1)}{dt-2(q-1)} \\&\hspace{3cm}-\binom{t}{3^{m-1}+2}\binom{d(3^{m-1}+2)}{dt-2(q-1)}\end{align*}
Now, we have $t=3^{m}+3^{m-1}+2\times3+2$, so by Lucas' theorem, $\binom{t}{3^{m}+2\times3}\equiv1\mod3$. Moreover $d(3^m+2\times3)=3^n+\displaystyle\sum_{j=m+2}^{n-1}3^j+2\times3^{m+1}+2\times3^{m-1}+2\sum_{j=1}^{m-2}3^j$ and $dt-(q-1)=3^n+3^{m+1}+\displaystyle\sum_{j=1}^{m-2}2\times3^j$, so $\binom{d(3^{m}+2\times3)}{dt-(q-1)}\equiv2\mod3$.
\\We do the same for all binomials and we get $C\equiv1\mod3.$
\paragraph*{Case where $n\equiv 1\mod 4$ in conjecture \ref{2.2}}
~~\\\\We have $d= \frac{3^{n+1}-1}{8}+\frac{3^n-1}{2}$. We choose $t=2$ then $\lfloor\frac{dt}{q-1}\rfloor=1$  and $\lceil\frac{dt-(q-1)}{d}\rceil=1$. So $$C=\binom{\frac{3^{n+1}-1}{4}+3^n-1}{\frac{3^{n+1}-1}{4}}+\binom{\frac{3^n-1}{2}+\frac{3^{n+1}-1}{8}}{\frac{3^{n+1}-1}{4}}$$
On one side, we have $\frac{3^{n+1}-1}{4}+3^n-1=\displaystyle\sum_{k=1}^{\frac{n-1}{2}}2\times3^{2k}+3^n+1$
and $\frac{3^{n+1}-1}{4}=2\displaystyle\sum_{k=0}^{\frac{n-1}{2}}3^{2k}$.
So by Lucas' theorem, $\binom{\frac{3^{n+1}-1}{4}+3^n-1}{\frac{3^{n+1}-1}{4}}\equiv0\mod3$
\\On the other side, $\frac{3^n-1}{2}+\frac{3^{n+1}-1}{8}=\displaystyle\sum_{k=0}^{\frac{n-1}{2}}2\times 3^{2k}+\displaystyle\sum_{k=0}^{\frac{n-1}{2}-1} 3^{2k+1}$.
\\So, $\binom{\frac{3^n-1}{2}+\frac{3^{n+1}-1}{8}}{\frac{3^{n+1}-1}{4}}\equiv1\mod3$.
\paragraph*{Case where $n\equiv 3\mod 4$ in conjecture \ref{2.2}}
~~\\\\We have $d=\frac{3^{n+1}-1}{8}$. We choose $t=26$, for $n\geq 5$, $1\leq t\leq q-2$ and $t\not\equiv 0\mod 3$. Furthermore $$\frac{dt}{q-1}=\frac{13}{4}(2\frac{3^n}{3^n-1}+1)$$
For $n\geq5$, $\frac{243}{242}\geq\frac{3^n}{3^n-1}\geq1$.
So $\lfloor\frac{dt}{q-1}\rfloor=9$.
\\We have $t=2+2\times3+2\times9$ so by Lucas' theorem, $\binom{t}{k}\not\equiv0\mod3$ if and only if $k=a+3b+9c$, $a$, $b$, $c\in\{0,1,2\}$.
\\\\In addition, $d=\displaystyle\sum_{k=0}^{\frac{n-1}{2}}3^{2k}$, so $dk=a+\displaystyle\sum_{k=1}^{\frac{n-1}{2}}(a+c)3^{2k}+\sum_{k=0}^{\frac{n-1}{2}}b3^{2k+1}+c3^{n+1}$.
\\We have $dt-(q-1)=\displaystyle\sum_{k=1}^{\frac{n-1}{2}} 2\times3^{2k}+2\times3^n+2\times3^{n+1}$.
So, if we write $j=\alpha+3\beta$, $\alpha$, $\beta\in\{0,1,2\}$, $$dt-(j+1)(q-1)=\alpha+\beta3+\displaystyle\sum_{k=1}^{\frac{n-1}{2}} 2\times3^{2k}+(2-\alpha)3^n+(2-\beta)3^{n+1}.$$
So $\binom{dk}{dt-(j+1)(q-1)}\not\equiv0\mod3$ if and only if $a\geq\alpha$, $b\geq\beta$, $b\geq2-\alpha$, $c\geq2-\beta$ and $a+c=2$.
\\Finally, \begin{align*}C&\equiv\binom{26}{24}-\binom{26}{16}+\binom{26}{24}-\binom{26}{13}+\binom{26}{16}+\binom{26}{8}+\binom{26}{16}\\&\hspace{3,5cm}+\binom{26}{24}-\binom{26}{16}+\binom{26}{8}+\binom{26}{8}\mod3\\&\equiv1\mod3\end{align*}
In all cases we have proved that $C\not\equiv0\mod3$; so by Hermite and Dickson's criterion $\Delta_d\neq1$.

\section{Proof of theorem \ref{n5}}

We give first some preliminary results :
\begin{lemme}\label{inverse}(see \cite{DOB_apn_odd_char} p. 97) If $\pgcd(d,q-1)=1$ then $\Delta_d=\Delta_{d^{-1}}$ where $d^{-1}$ is the inverse of $d$ modulo $q-1$.\end{lemme}

\begin{proposition}\label{HE}(see \cite{Hell_new_families}, corollary 1 p.484) Let $n$ and $k$ be integers such that $\gcd(2n,k)=1$. Then $x\mapsto x^d$ is APN over $\F_{5^n}$ for $d=\frac{5^k+1}{2}$.\end{proposition}

\begin{corollaire}\label{6}Let $n$ be a integer such that $n\equiv2^l-1\mod 2^{l+1}$ then $x\mapsto x^d$ is APN over $\F_{5^n}$ for $d=\frac{5^{\frac{n+1}{2^l}}+1}{2}$.
 \end{corollaire}

\begin{preuve} If $n\equiv2^l-1\mod2^{l+1}$ then $\frac{n+1}{2^l}\equiv1\mod2$. So $\pgcd(2n,\frac{n+1}{2^l})=1$ and by proposition \ref{HE} we get the result.
\end{preuve}

Now we are able to prove theorem \ref{n5}.
\\\\First we consider the case where $n\equiv2^l-1\mod2^{l+1}$. We have
\begin{align*}\frac{5^{\frac{n+1}{2^l}}+1}{2}d&=\frac{5^{n+1}-1}{4}+\frac{5^{\frac{n+1}{2^l}}+1}{2}\frac{5^n-1}{4}\\&= 5^n+\frac{5^n-1}{4}\frac{5^{\frac{n+1}{2^l}}+3}{2}\end{align*}
but if $n\equiv2^l-1\mod2^{l+1}$, $\frac{n+1}{2^l}\equiv1\mod 2$ and $5^{\frac{n+1}{2^l}}+3\equiv 0\mod8$.
So $$\frac{5^{\frac{n+1}{2^l}}+1}{2}d\equiv1\mod 5^n-1$$
Hence $d$ is invertible modulo $5^n-1$ and $d^{-1}=\frac{5^{\frac{n+1}{2^l}}+1}{2}$. By corollary \ref{6}, $x\mapsto x^{d^{-1}}$ is APN over $\F_{5^n}$, so by lemma \ref{inverse}, we get the result for $n\equiv2^l-1\mod2^{l+1}$.
\\\\If $n\equiv-1\mod 2^{l+1}$ then $(2^l+1)n\equiv2^l-1\mod2^{l+1}$. We put $$e=\frac{1}{2}\frac{5^{(2^l+1)n+1}-1}{5^{\frac{(2^l+1)n+1}{2^l}}+1}+\frac{5^{(2^l+1)n}-1}{4}.$$ By the first case $x\mapsto x^e$ is APN over $\F_{5^{(2^l+1)n}}$. Furthermore,
\begin{align*}e&=\frac{1}{2}\sum_{k=0}^{2^l-1}(-1)^{k+1}(5^{\frac{(2^l+1)n+1}{2^l}})^k+\frac{5^{(2^l+1)n}-1}{4}\\&=\frac{1}{2}\sum_{k=0}^{2^l-1}(-1)^{k+1}5^{nk}(5^{\frac{n+1}{2^l}})^k+5^n\frac{5^{2^ln}-1}{4}+\frac{5^n-1}{4}\\&=5^n(5^n-1)\frac{1}{4}\sum_{j=0}^{2^l-1}5^{nj}+\frac{5^n-1}{4}+(5^n-1)\frac{1}{2}\sum_{k=1}^{2^l-1}(-1)^{k+1}\sum_{j=0}^{k-1}5^{nj}(5^{\frac{n+1}{2^l}})^k\\&\hspace{2cm}+\frac{1}{2}\sum_{k=0}^{2^l-1}(-1)^{k+1}(5^{\frac{n+1}{2^l}})^k\end{align*}
Since $\displaystyle\sum_{k=0}^{2^l-1}5^{nj}\equiv 0\mod 4$ and $\displaystyle\sum_{k=1}^{2^l-1}(-1)^{k+1}\sum_{j=0}^{k-1}5^{nj}(5^{\frac{n+1}{2^l}})^k\equiv0\mod 2$ for $l\geq2$, we have :
$$e\equiv\frac{1}{2}\frac{5^{n+1}-1}{5^{\frac{n+1}{2^l}}+1}+\frac{5^n-1}{4}\mod5^n-1$$
Since $x\mapsto x^e$ is APN over $\F_{5^{(2^l+1)n}}$, for all $b\in\F_{5^n}$ the equation $(x+1)^e-x^e=b$ has at most two solutions in $\F_{5^{(2^l+1)n}}\supset\F_{5^n}$. So, since $e\equiv d\mod 5^n-1$, for all $b\in\F_{5^n}$, $(x+1)^d-x^d=b$ has at most two solutions in $\F_{5^n}$ and $\Delta_d\leq 2$ for $n\equiv -1\mod 2^{l+1}$.
\\Furthermore since $n$ is odd, $\frac{5^n-1}{4}\equiv1\mod2$. In addition $\displaystyle\sum_{k=0}^{2^l-1}(-1)^{k+1}(5^{\frac{n+1}{2^l}})^k\equiv 0\mod4$, so $d\equiv1\mod2$. Then both 0 and -1 are solutions of $(x+1)^d-x^d=0$ and we get the result for $n\equiv-1\mod2^{l+1}$.

\section{Some remarks about Zha and Wang theorems}

First we notice that theorem \ref{4.4} gives another proof of the case $n\equiv-1\mod2^{l+1}$ of theorem \ref{n5}.
\\\\Now, we give another proof of conjecture \ref{5}. In the case where $n\equiv1\mod 4$ it is the same that in theorem 4.5 of Zha and Wang. But the proof of the case where $n\equiv 3\mod 4$ doesn't use theorem 4.4.
\\\\First we recall the idea of the proof in case where $n\equiv1\mod4$ :
$$\frac{5^{m}+1}{2}d\equiv 1\mod (5^n-1).$$
So by lemma \ref{inverse} and corollary \ref{6} we get the result.
\\\\ If $n\equiv 3\mod 4$ then $3n\equiv 1\mod 4$. So
\begin{align*}\frac{5^{3n}-1}{4}+\frac{5^{\frac{3n+1}{2}}-1}{2}-\frac{5^n-1}{4}-\frac{5^{m}-1}{2}&=(5^n-1)\frac{2.5^m+5^n(5^n+1)}{4} \\&\equiv 0\mod (5^n-1)\end{align*}
So $e=\frac{5^{3n}-1}{4}+\frac{5^{\frac{3n+1}{2}}-1}{2}\equiv\frac{5^n-1}{4}+\frac{5^{m}-1}{2}\mod 5^n-1$ and, by the case where $n\equiv1\mod4$, $x\mapsto x^e$ is APN over $\F_{5^{3n}}$ and $\Delta_d\leq2$ for $n\equiv3\mod4$. Furthermore $d$ is odd, so $(0+1)^d-0^d=1=(-1+1)^d-(-1)^d$ and $\Delta_d=2$.
\\\\We finish by a remark on theorem 4.1 of \cite{MR2679075} :
\\For $l\in\mathbb{N}$ and $n\equiv-1\mod 2^l$, if we take $k=\frac{n+1}{2^l}$ and $u=3$ in theorem \ref{4.4}, we get that $x\mapsto x^d$ is APN for  $$d=\frac{3^{n+1}-1}{3^{\frac{n+1}{2^l}}+1}.$$
For $l=2$, this gives an explicit family to explain case IV of the table. Actually this family contains $656=224\times 5^4$.

\nocite{Li_perm_pol_deg_6_7}

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TITLE: How can only one line be drawn through 4 collinear points?
QUESTION [1 upvotes]: There are 10 points in a plane no three of which are in the same straight line except 4 points, which are collinear. Find the probability that the number of straight lines obtained from the pairs of these points.
Given solution: total lines between 10 non collinear points = $^{10}C_2$ = 45
total lines between 4 non collinear points = $^{4}C_2$ = 6
Since the 4 points are non collinear and only one line can be drawn through all four:
Total lines possible = $45 - 6 + 1 = 40$
My doubt: if I were to mark the 4 collinear points as a, b , c ,d then can’t I draw more than one line such as ab, bc, cd, bd etc? So how come can only one line be drawn?
Also the question asks for the probability which would be possible outcomes divided by total outcomes so how come is this the correct solution? Isn’t this just the possible outcomes?

REPLY [0 votes]: Part 1:As how you said,ab,ac,ad,bc,bd,cd are the same line because the points are collinear.
Part 2:Yes you are right.The question doesn't imply randomness.It asks "Find the probability that the number of straight lines obtained from the pairs of these points" and it is either wrong or incomplete


TITLE: Contour integrals in complex coordinates in 2D CFT
QUESTION [9 upvotes]: To my understanding, in a 2D CFT with complex coordinates, the coordinates $z$ and $\bar{z}$ are to be treated as independent, and only at the end of the calculation should one take $\bar{z}=z^*$. But I'm not sure what this means for a contour integral. What exactly is the meaning of integrating a function of both coordinates over just one? For example, in
$$\oint \phi(z,\bar{z}) dz,$$
do I integrate only over $z$ so that the result of the integral is a function of $\bar{z}$, or do I get a number (i.e. the result is independent of $\bar{z}$)? And in the latter case, how is the integral calculated?
Also, how is this integral different from the integral
$$\oint_\gamma dz \oint_\delta d\bar{z}\phi(z,\bar{z}),$$
where $\gamma$ is the contour in $z$ and $\delta$ is the contour in $\bar{z}$? And in particular, does $\delta$ need to be the conjugate of $\gamma$, i.e. $\delta=\gamma%*$? (For an example of an integral of this kind, see Di Francesco Eq. 6.7)

REPLY [4 votes]: This is a pretty interesting question, since there are two different ways of looking at this. Let me try to give an idea of both viewpoints:
(for whoever can't be bothered to read the whole thing, just read the ``Note'' in both cases as a representative example)

The mathematician's point of view (which is also the viewpoint that I'm familiar with). Then $z$ and $\bar z$ are not in any way independent, as one would have expected on first sight. We just work with the complex plane, originally defined in the two real coordinates $x$ and $y$, and any point can equivalently be described by $z := x+iy$. The complex conjugate is then simply a function of $z$, defined by $\Re (z) - i \Im(z) = x-iy$ and denoted $\bar z$. Moreover, one defines the derivative $\frac{\partial}{\partial z} := \frac{\partial}{\partial x} - i\frac{\partial}{\partial y}$ and its complex conjugate $\frac{\partial}{\partial \bar z} := \frac{\partial}{\partial x} + i\frac{\partial}{\partial y}$. These definitions are not chosen arbitrarily, indeed one can derive that e.g. $\frac{\partial}{\partial z} z = 1$ and $\frac{\partial}{\partial \bar z} z = 0$. If we then have a function on the complex plane, then in principle we could just write $f(z)$ (since it is only a function of $z$), but sometimes it is helpful to write $f(z,\bar z)$ to denote the fact that we don't presume that $\frac{\partial}{\partial \bar z} f = 0$. So in this case writing $f$ as a function of both $z$ and $\bar z$ is not literal, and it's not implying we treat them as independent variables. It is just to denote that it is a very general function. If our function $f$ is special such that $\frac{\partial}{\partial \bar z} f = 0$, we call $f$ holomorphic and denote it as $f(z)$. If we want to integrate over the complex plane, we can do this by the most obvious way $\int f dxdy$. Note that this is a two-dimensional integral. Equivalently, since $z = x+iy$ we have that $dz = dx + idy$ and $d\bar z = dx - idy$, such that $dz d\bar z = -2i dxdy$ (where we use $(dx)^2 = 0$), we can also write $\frac{i}{2} \; \int f \; dz d\bar z$. We might also want to look at contour integrals, in which case we integrate over a 1-dimensional subset of the complex plane. Let $\gamma(t)$ be such a curve. Then writing $\int_\gamma f(z,\bar z) dz$ is defined to mean $\int f(\gamma(t),\bar \gamma(t)) \; \gamma'(t) dt$. Moreover, if we consider contour integrals of holomorphic functions, then there is the whole branch of complex analysis with some beautiful results like the Cauchy integral formula $f(z_0) = \int_\gamma \frac{f(z)}{z-z_0} dz$ where $\gamma$ is any contour encircling $z_0$. [Note: so in this way of looking at things, the expression $\oint \phi(z,\bar z) dz$ would just be a number, and not a function of $\bar z$, since by definition it's $\int \phi(\gamma(t),\bar \gamma(t)) \gamma'(t) dt$ where $\gamma(t)$ is the curve integrated over.]
And apparently the CFT point of view. There it seems $z$ and $\bar z$ are treated ``independently''. The main question is then: how exactly, and how does it compare to the previous viewpoint? After all if you flip open a math book you will most likely see the treatment following (1), so it's good to know how one should compare things. The main point seems to be the following: while $z$ and $\bar z$ are strictly speaking dependent on each other (as discussed in viewpoint 1), they are not algebraically dependent! That is, you cannot write $\bar z$ as some polynomial in $z$. This has an important consequence: if one decomposes a function $f(z,\bar z)$ algebraically in terms of $z$ and $\bar z$ (e.g. through a power series), then this is not ambiguous. E.g. once I write $f(z) = z + \bar z ^2 + 2 z \bar z$, I know there is no other way of writing it in terms of $z$ and $\bar z$. (This would be entirely different if $\bar z$ could be expressed as $g(z)$ where $g$ is algebraic.). This means that it is entirely consistent of me to consider $f(z,\bar z)$ as a function of two independent variables, as long as I only consider algebraic operations. This way of looking at things is really nice: whereas in viewpoint (1) a general function $f(z,\bar z)$ does not have nice methods for it (unlike holomorphic functions), in viewpoint (2) we can consider $f(z,\bar z)$ as being decomposed into holomorphic functions of our two variables. Example: the function $f(z,\bar z) = z + 2 z \bar z$ uniquely defines the function $g(u,v) = u + 2uv$ which is holomorphic in the two complex coordinates $u$ and $v$, and $f(z,\bar z) = g(z, \bar z)$. The point is that we can now use our holomorphic tricks (Cauchy etc) on this function $g$. That is now what is implicitly happening in expressions such as (6.7) of Di Francesco. I.e. there we really act as if $z$ and $\bar z$ are two complex numbers that have nothing to do with each other, such that we can decompose general functions $\phi(z,\bar z)$ into a power series where the coefficients are determined by the Cauchy integral formula, just as we are used to for holomorphic functions of one variable. [Note: so in this way of looking at things, the expression $\oint \phi(z,\bar z) dz$ is indeed still a function of $\bar z$, with the expression being defined by first expanding $\phi(z,\bar z)$ into a power series of $z$ and $\bar z$, then temporarily replacing $z \to u$ and $\bar z \to v$ and then employing two-variable complex analysis (with a contour integral for the first variable) and then plugging back in $u \to z$ and $v\to \bar z$. Of course once one gets the point, this is not done explicitly and is merely implied in the expressions.]


TITLE: What is a reasonable bound for integrating the standard normal distribution with a single tail?
QUESTION [2 upvotes]: This is a question I'm trying to figure out how to solve because my calculator does not allow me to evaluate improper integrals. I am trying to find the z that produces a small enough p to be negligible. In general I think I will need about 5 digits of accuracy maximum but I would also be interested in the general solution for some natural n. The problem boils down to solving this inequality:
Find the least value of $a$ for which:
$$\int_a^\infty \frac{\exp(-x^2/2)}{\sqrt{2\pi}} < 10^{-n}$$
Any insights on how I might go about this? This function doesn't even have an elementary antiderivative so this looks like a pretty tricky problem.

REPLY [2 votes]: Admitting that you know the error function, you are looking for $a$ such that
$$\frac{1}{2}-\frac{1}{2} \text{erf}\left(\frac{a}{\sqrt{2}}\right) < 10^{-n}$$that is to say
$$\text{erf}\left(\frac{a}{\sqrt{2}}\right)> 1-2\times 10^{-n}$$
When $a$ is large
$$\text{erf}\left(\frac{a}{\sqrt{2}}\right)=1-e^{-\frac{a^2}{2}} \left(\frac{\sqrt{\frac{2}{\pi
   }}}{a}+O\left(\frac{1}{a^3}\right)\right)$$ and then we search for $a$ such that
$$e^{-\frac{a^2}{2}} <  \sqrt{2\pi}\, 10^{-n}\, a$$ The solution of the equation is given in terms of Lambert function
$$a=\sqrt{W\left(\frac{10^{2n}}{2\pi}\right)}$$
Some results
$$\left(
\begin{array}{ccc}
 n & \text{estimate} & \text{solution} \\
1 & 1.43165 &  1.28155 \\
 2 & 2.37533 &  2.32635 \\
 3 & 3.11528 &  3.09023 \\
 4 & 3.73464 &  3.71902 \\
 5 & 4.27575 &  4.26489 \\
 6 & 4.76151 &  4.75342 \\
 7 & 5.20565 &  5.19934 \\
 8 & 5.61710 &  5.61200 \\
 9 & 6.00204 &  5.99781 \\
 10 & 6.36492 &  6.36134 \\
 11 & 6.70910 &  6.70602 \\
 12 & 7.03717 &  7.03449 \\
 13 & 7.35116 &  7.34882 \\
 14 & 7.65274 &  7.65056 \\
 15 & 7.94324 &  7.94324
\end{array}
\right)$$
Playing with the inverse error function, we could also have, as an approximation,
$$a=\sqrt{\log \left(\frac{2}{\pi  t^2}\right)-\log \left(\log \left(\frac{2}{\pi
   t^2}\right)\right)}\quad \text{where} \qquad t=2\times 10^{-n}$$ which looks to be slightly better.
$$\left(
\begin{array}{ccc}
 n & \text{estimate} & \text{solution} \\
 1 & 1.32266 &  1.28155 \\
 2 & 2.31834 &  2.32635 \\
 3 & 3.08133 &  3.09023 \\
 4 & 3.71139 &  3.71902 \\
 5 & 4.25847 &  4.26489 \\
 6 & 4.74795 &  4.75342 \\
 7 & 5.19461 &  5.19934 \\
 8 & 5.60785 &  5.61200 \\
 9 & 5.99413 &  5.99781 \\
 10 & 6.35805 &  6.36134 \\
 11 & 6.70305 &  6.70602 \\
 12 & 7.03178 &  7.03449 \\
 13 & 7.34632 &  7.34882 \\
 14 & 7.64835 &  7.65056 \\
 15 & 7.93923 &  7.94324
\end{array}
\right)$$
Now, if you need better, I think that the only solution would be given by a couple of iterations of Newton method.


TITLE: Whitney stratification and affine grassmanian
QUESTION [7 upvotes]: Let $G$ a simply connected group over $\mathbb{C}$ and $Gr:=G(\mathbb{C}((t)))/G(\mathbb{C}[[t]])$ the affine grassmannian. By Cartan decomposition we have a partition of stratas indexed by $\lambda\in X_{*}(T)^{+}$.
Let $Gr_{\lambda}$ such a strata and $\overline{Gr}_{\lambda}$ the closure in $Gr$.
Then, $\overline{Gr}_{\lambda}=\coprod\limits_{\mu\leq\lambda}Gr_{\mu}$, is it a Whithney stratification?

REPLY [5 votes]: The point is following: $\overline{Gr^\lambda}$ is a finite dimensional variety  acted upon by the pro-algebraic group $G(\mathbb{C}[[t]])$.
This action factors  through the action of some finite dimensional algebraic group called $G(\mathcal{J}^l)$. The strata you describe are just the orbits of this group action and there are finitely many of them. (See explanation below)
It is well known, that if a complex algebraic group $G$ acts on a complex variety $X$ algebraically with only finitely  many orbits, then the $G$-orbits form a Whitney stratification of $X$, see Algebraic Stratifications of $G$-varieties for details.
Remark: Let $\mathcal{J}^l:= \mathbb{C}[[t]]/t^l \mathbb{C}[[t]]$ for $l\in \mathbb{N}$. Then $G(\mathcal{J}^l)$ is a finite dimensional algebraic group. Let $G(\mathcal{O}^l)$ be the kernel of the canonical homomorphism $G(\mathbb{C}[[t]])\to G(\mathcal{J}^l)$.
If you choose an embedding $G\hookrightarrow GL_n(\mathbb{C})$ you can think of $G(\mathcal{O}^l)$ as the subgroup of $G(\mathbb{C}[[t]])$ where all entries on the diagonal are in the form of $1+t^l\cdot f$ with $f\in \mathbb{C}[[t]]$ and all other entries are in the form of $t^l f$. Anyway it is straight forward to show that for any $\lambda$ there exists an $l\gg 0$ such that $G(\mathcal{O}^l)$ operates trivially on $\overline{Gr^\lambda}$, in other words the $G(\mathbb{C}[[t]])$ action factors through
 $G(\mathcal{J}^l)$


TITLE: Combinatorial Proof of Combinatorial Identity involving $(-1)^k \binom {n-1}{k}$
QUESTION [7 upvotes]: Given the following identity:
$$\sum_{i=0}^k (-1)^i \binom ni = (-1)^k \binom {n-1}{k}.$$
This is provable by induction. However, I wonder if there is a way to prove this in a combinatorial fashion (something like subsets, groups, choosing $n$ balls from a box with $k$ balls, et cetera.)
Thank you.

REPLY [7 votes]: The left side of your equation is a weighted count of the subsets of $\{1,\dots,n\}$ of all sizes $i$ from $0$ to $k$, the weights being $1$ or $-1$ according as $i$ is even or odd.  We can pair up most of these subsets as follows: If a set $X$ contains $n$, pair it with $X-\{n\}$; if it doesn't contain $n$, pair it with $X\cup\{n\}$.  Note that this is truly a pairing; if we paired $X$ with $Y$, then we also paired $Y$ with $X$.  Any two paired sets contribute with opposite signs to your sum, so their contributions cancel out, and the sum reduces to the unpaired sets.  But these are just the sets that have $k$ elements and don't contain $n$.  Any smaller set can be paired, as described, by adjoining or removing $n$; and a set of size $k$ that contains $n$ can be paired as described, by removing $n$.  Only when a set doesn't contain $n$ (so we want to adjoin $n$) and has size $k$ (so that adjoining an element would make its size impermissibly large) is there a failure of pairing.  The unpaired sets, the sets of size $k$ that don't contain $n$, all contribute with sign $(-1)^k$, and there are $\binom {n-1}k$ of them. Hence the right side of your equation.


TITLE: All prime ideals of $\mathbf{R}[x,y]$ and $\mathbf{C}[x, y]$.
QUESTION [5 upvotes]: What are the all the prime ideals of $\mathbf{R}[x,y]$ and $\mathbf{C}[x, y]$? (Also, how do you prove that you've found all of them?) I'm trying to understand what the $\mathbf{R}$-algebraic vs $\mathbf{C}$-algebraic subsets of $\mathbf{C}^2$ are, defined as the zeros (in $\mathbf{C}^2$) of an ideal in $\mathbf{R}[x, y]$ and $\mathbf{C}[x,y]$, respectively.
Edit: I'm not that familiar with dimension theory, so I'd prefer to see an argument that does not rely on it.

REPLY [2 votes]: Here's an answer for $\mathbf C[x,y]$:  Hilbert's Nullstellensatz, asserts the maximal ideals  have the form $(x-\alpha,y-\beta)$ for some $\;\alpha, \beta\in \mathbf C$.
On the other hand, $\mathbf C[x,y]$ is a U.F.D. of (Krull) dimension $2$. So the prime ideals of height $1$ are principal, generated by irreducible polynomials.
Add to this list the only ideal of height $0$, and you've finished.


TITLE: Order of a group and cyclic group theory connection
QUESTION [1 upvotes]: Just stuck on a problem. If group $G$ of order $6$ contains an element of order $6$, then prove that $G$ is a cyclic group of order $6$. Any hint will be appreciated.

REPLY [1 votes]: A cyclic group $G$ is one such that there exists a $g \in G$ where every element of $G$ can be expressed as $g^n$ for some $n \in \mathbb{Z}$.  Essentially, given that $|g| = 6$, you just need to prove that $g^n$ is unique for all $n \in \{1, 2, ..., 6\}$.
To do this, assume that $g^n = g^m$ for $n \neq m$ and $n, m \in \{1, 2, ..., 6\}$ and arrive at a contradiction.


text:
\section{Numerical Results}
\label{sec:NumericalResults}
We demonstrate the quality of our bounds on the width of the transition region $\#\{k : \eps < \lambda_k < 1-\eps\}$ with some numerical computations. First, we  fix $W = \tfrac{1}{4}$ (a large value of $W$), and for each integer $2^4 \le N \le 2^{16}$ we use the method described in \cite{Gruenbacher94} to compute $\lambda_k$ for a range $k_{\text{min}} \le k \le k_{\text{max}}$ such that $\lambda_{k_{\text{min}}} > 1-10^{-13}$ and $\lambda_{k_{\text{max}}} < 10^{-13}$. From this, we can determine $\#\{k : \eps < \lambda_k < 1-\eps\}$ for $\eps = 10^{-3}, 10^{-8}, 10^{-13}$. We plot $\#\{k : \eps < \lambda_k < 1-\eps\}$ as well as the upper bound on $\#\{k : \eps < \lambda_k < 1-\eps\}$ from Theorem~\ref{thm:BoundWithoutW} in Figure~\ref{fig:GapsizeVsN}. We note that over this range of parameters, the difference between the bound in Theorem~\ref{thm:BoundWithoutW} and the true width of the transition region $\#\{k : \eps < \lambda_k < 1-\eps\}$ is between $1$ and $14$.

\begin{center}
\begin{figure}[ht]
\includegraphics[scale=0.5]{Theorem1Bound.pdf}
\caption{Plots of the width of the transition region $\#\{k : \eps < \lambda_k < 1-\eps\}$ vs. $N$ where $W = \tfrac{1}{4}$ and $\eps = 10^{-3}$(blue), $\eps = 10^{-8}$(green), and $10^{-13}$(red) are fixed. The dashed lines indicate the upper bound from Theorem~\ref{thm:BoundWithoutW}.}
\label{fig:GapsizeVsN}
\end{figure}
\end{center}

Next, we  fix $N = 2^{16}$ and for $10001$ logarithmically spaced values of $W$ between $2^{-14}$ and $2^{-2}$, we use the method described in \cite{Gruenbacher94} to compute $\lambda_k$ for a range $k_{\text{min}} \le k \le k_{\text{max}}$ such that $\lambda_{k_{\text{min}}} > 1-10^{-13}$ and $\lambda_{k_{\text{max}}} < 10^{-13}$. From this, we can determine $\#\{k : \eps < \lambda_k < 1-\eps\}$ for $\eps = 10^{-3}, 10^{-8}, 10^{-13}$. We plot $\#\{k : \eps < \lambda_k < 1-\eps\}$ as well as the upper bound on $\#\{k : \eps < \lambda_k < 1-\eps\}$ from Theorem~\ref{thm:BoundWithW} in Figure~\ref{fig:GapsizeVsW}. We note that over this range of parameters, the difference between the bound in Theorem~\ref{thm:BoundWithoutW} and the true width of the transition region $\#\{k : \eps < \lambda_k < 1-\eps\}$ is between $\approx 9.8$ and $\approx 30.7$.

\begin{center}
\begin{figure}
\includegraphics[scale=0.5]{Theorem2Bound.pdf}
\caption{Plots of the width of the transition region $\#\{k : \eps < \lambda_k < 1-\eps\}$ vs. $W$ where $N = 2^{16}$ and $\eps = 10^{-3}$(blue), $\eps = 10^{-8}$(green), and $10^{-13}$(red) are fixed. The dashed lines indicate the upper bound from \cref{thm:BoundWithW}.}
\label{fig:GapsizeVsW}
\end{figure}
\end{center}

In Figure~\ref{fig:GapsizeVsN}, we see that the plots of both $2\ceil{\tfrac{1}{\pi^2}\log(4N)\log(\tfrac{4}{\eps(1-\eps)})}$ (the bound in Theorem~\ref{thm:BoundWithoutW}) and the actual width of the transition region increase roughly linearly with $\log N$ and at roughly the same rate. However, the difference between the bound in Theorem~\ref{thm:BoundWithoutW} and the actual width of the transition region is noticeably larger for smaller values of $\eps$ than for larger values of $\eps$. This provides numerical evidence that for a large bandwidth $W$, the bound's dependence on $N$ is close to correct, but the dependence on $\eps$ has some room for improvement.

In Figure~\ref{fig:GapsizeVsW}, we see that the plots of both $\tfrac{2}{\pi^2}\log(100NW+25)\log\left(\tfrac{5}{\eps(1-\eps)}\right)+7$ and the actual width of the transition region increase roughly linearly with $\log(NW)$ and at roughly the same rate. However, the difference between the bound in Theorem~\ref{thm:BoundWithW} and the actual width of the transition region is quite noticeable. This provides numerical evidence that the leading constant of $\tfrac{2}{\pi^2}$ is indeed correct, but that the other constants leave significant room for improvement.

Finally, we note that for the range of parameters in both plots, the non-asymptotic bounds on the width of the transition region given by (\ref{eq:ZhuWakin}), (\ref{eq:Boulsane}), and (\ref{eq:FST}) (in Section~\ref{sec:PrevDPSSBounds}) would all be well above the range of the plots in Figures~\ref{fig:GapsizeVsN} and \ref{fig:GapsizeVsW}. The bounds in (\ref{eq:ZhuWakin}) and (\ref{eq:Boulsane}) are proportional to $\tfrac{1}{\eps(1-\eps)}$. Thus, they are only useful when $\eps$ is not too small. Also, the bound in (\ref{eq:FST}) is rather large since the leading constant $\tfrac{8}{\pi^2}$ being $4$ times larger than that in Theorems~\ref{thm:BoundWithoutW} and \ref{thm:BoundWithW}, and the trailing constant $12$ dominates $\tfrac{8}{\pi^2}\log(8N)$ when $N$ isn't too large. In particular, for $\eps = 10^{-3}$ and any $N \in \N$, if we impose the mild constraint that $NW \ge \tfrac{1}{2}$, then the bound in (\ref{eq:ZhuWakin}) is at least $\tfrac{4/\pi^2}{\eps(1-\eps)} \approx 405$, the bound in (\ref{eq:Boulsane}) is at least $\tfrac{0.45-1/6}{\eps(1-\eps)} \approx 283$, and the bound in (\ref{eq:FST}) is at least $(\tfrac{8}{\pi^2}\log(8)+12)\log(\tfrac{15}{\eps}) \approx 131$.


TITLE: Suppose that $f (x, y) = xe^{−x(y+1)}$, where $0 ≤ x < ∞$, $0 ≤ y < ∞$. Find marginal densities
QUESTION [0 upvotes]: This question comes from rice 3.14

Suppose that $$f (x, y) = xe^{−x(y+1)}$$ where  $0 ≤ x < ∞$, $0 ≤ y <
> ∞$
  a. Find the marginal densities of X and Y . Are X and Y independent?
  b. Find the conditional densities of X and Y

to find the marginal densities i have integrated out $x$ and $y$ such that:
\begin{align}
f_X(x) & = \int_0^\infty xe^{−x(y+1)} dx \\
& = x \int_0^\infty  e^{−x(y+1)} dx\\
& = x \Big[ \frac{e^{−x(y+2)}}{y+2} \Big]_0^\infty\\
& = -\frac{x}{y+2} \space  \text{for} 0\leq x < \infty
\end{align}
\begin{align}
f_Y(y) & = \int_0^\infty xe^{−x(y+1)} dy \\
\end{align}
but i get stuck here. How to solve the integral? is the first marginal distribution correct?

REPLY [1 votes]: $$f_Y (y)=\int_0^{\infty} xe^{-x(y+1)}dx$$ $$=-\frac x {y+1}e^{-x(y+1)}|_0^{\infty} +\frac 1 {y+1} \int_0^{\infty}e^{-x(y+1)}dx$$ $$=\frac 1{(y+1)^{2}}.$$ $f_X(x)$ is easier: $f_X (x)=\int_0^{\infty} xe^{-x(y+1)}dy=xe^{-x} \int_0^{\infty} e^{-xy} dy=e^{-x}$


TITLE: Is there a continous function which does not have a derivative in any of its points?
QUESTION [1 upvotes]: Possible Duplicate:
Are Continuous Functions Always Differentiable?

Is there a continous function (continous in every one of its points) which is not differentiable in any of its points?

REPLY [1 votes]: Yes, for example the Weierstraß function.
One can actually show that the set $A:= \{f \in C[0,1]; f$ has no right-derivative in any point in $[0,1)\}$ is dense in $C[0,1]$ and uncountable.


TITLE: Matrix bracket notation
QUESTION [2 upvotes]: I am reading a section in a book that talks about normal matrices, and I see the following:
A normal matrix is a matrix that commutes with its adjoint,

Eh? What do the brackets and the comma inside mean?

REPLY [8 votes]: The notation means that $AA^{T}-A^{T}A$=0.


TITLE: motivation of surgery
QUESTION [8 upvotes]: an $n$-surgery on m dim manifold M is to cut out $S^n\times D^{m-n}$and replace it by $D^{n+1}\times S^{m-n-1}$.
I want to know how this is invented?
I do know that the effect of passing a critical point of index $n$ in $m$-manifold is equivalent to attach an $n+1$-handle $D^{n+1}\times D^{m-n-1}$.Now the boundary of $D^n\times D^{m-n}$ is $S^n\times D^{m-n}\cup D^{n+1}\times S^{m-n-1}$,i think there must be some close relation between the special form of $n$-surgery and handle.can someone help make this clear?

REPLY [2 votes]: This question has already been answered, but there's a tiny piece of intuition which I'd like to make explicit:
If you're thinking about a manifold in the PL world, surgery might look a bit arbitrary- why cut out and glue in those pieces and not others? Surgery's natural setting is the smooth world, where you're equipping a manifold with a Morse function $f\colon\, M\to \mathbb{R}$, and using information about critical points of $f$ to encode $M$.
It's actually a bit more involved than you might think it might be, but when you pass a critical point of $f$ you add a handle to $M$, and the boundary changes by surgery. For details, see answers to this question.
So really, surgery isn't an a-priori construction which somebody pulled from a hat- it is rather an operation which stems naturally and inevitably from Morse theory.


TITLE: What are some good reasonably rigorous texts on the mathematics of infinity?
QUESTION [3 upvotes]: The Infinite Book is too light and not focused enough on the mathematics of infinity, and Everything and More: A Brief History of Infinity has too much focus on the history of infinity instead of the mathematics of it. Is it best to just read a text on set theory and learn about infinity from that?

REPLY [2 votes]: I assume that a book on set theory could kill your motivation to learn it, mostly because a lot of the books do not come with historical tidbits and reasons to do such activities. On these books, they assume that you're mature enough to know why you're doing that or that you're having lectures about their contents.
For a first book, I'd really recommend you to read Stillwell's: Road to Infinity. It teaches well, contains motivation and tells you the history of set theory. The book is also full of references, in which one of them is Thomas Jech's: Set Theory, a classical in the field of set theory (perhaps you can read it after reading Stillwell's book).
Another book I really liked is Moschovakis': Notes on Set Theory, although it's a bit harder than the first one. And the last one I have to recommend is Halbeisen's: Combinatorial Set Theory, it's a very unique and interesting book. It's a bit different of a set theory book because it employs the generalization of some results in combinatorics to set theory. All these books are packed with motivation and discussion about the need of these concepts (which in my opinnion, are very important to learn the subject).