TITLE: inverse limit of $k[x]/(x^n)$
QUESTION [2 upvotes]: I know the inverse limit of $k[x]/(x^n)$ is $k[[x]]$ but i can't show the homomorphism between these is onto.
For example
$\alpha:k[[X]]\to\varprojlim k[x]/(x^n)$ by F(x)=family of$ F(x)+(x^n)$. for example $(1+(x),0,5+7x^2+(x^3),3+11x+132x^2+(x^4),0,...)$
is image of things in k[[x]]?

REPLY [2 votes]: The element
$$(1+(x),0,5+7x^2+(x^3),3+11x+132x^2+(x^4),0,...)$$
that you're looking at is not in $\varprojlim k[x]/(x^n)$ because it's components do not respect the transition maps.  For example $5 + 7x^2 + (x^3)$ goes to $5 + (x^2)$ under the map $k[x]/(x^3) \to k[x]/(x^2)$, it doesn't go to $0$ as you have in your family.
Once you write down an element that does respect all the transition maps you'll see that if you look at the coefficient of some $x^m$ in the polynomial $f \in k[x]/(x^n)$ and $m < n$ then that coefficient can't change in any of the further elements of your tuple, so it uniquely identifies the coefficient of $x^m$ of a polynomial in $k[[x]]$.


TITLE: Why is this set closed?[metric-spaces]
QUESTION [0 upvotes]: I am reading a note, where part of it is this:

Why is S' closed?
I have tried to argument like this, but I am not able to finish the argument:
Let $\{x_n\}$ be a convergent sequence in S', then it is also a convergent sequence in S, since S is closed the point it converges to must be in S. But how do we know that this point is in S'?
Another way I thought is that the dfiniton of S' might look like a closed ball, however it might not be that sincewe only choose values of S it might be more complicated.

REPLY [0 votes]: Your idea works. Let $(x_n)$ be a sequence in $S'$, such that $x_n \to x \in \mathbb R^n$. Then $x \in S$, as $x$ is closed, moreover, $\|\cdot - z\| \colon S \to \mathbb R$ is continuous, even Lipschitz, as
$$ \bigl| \|a-s\| - \|b - s\|\bigr| \le \|a-b\| $$
by the triangle inequality. So $\|x_n - z\| \to \|x-z\|$ and hence $\|x-z\| \le d_S + 1$. So $x \in S'$.


TITLE: recurrence relation with min-max
QUESTION [2 upvotes]: While solving a puzzle I got stuck with this recurrence
\begin{align*}  &A(n) = \min_x \left \{ \max \left [ x,  1+A(n-x)
\right ] \right \} \\ &\text{where }A(1) = 1,A(0) = 0 \text{ and } 1 \leq x
\leq n , \text{ x is natural }\\ \end{align*}
Please give some idea how to proceed ?
Thanks !

REPLY [1 votes]: Hint
Here are a few initial steps:

Show by induction that $A(n) \le n$
Show that when $n>1$, $\displaystyle\min_{1\le x\le n} = \min_{1\le x < n}$
Use this to prove by induction that $A(n)\le A(n+1)$
Prove that $A(n)\le A(n+1)\le 1 + A(n)$, so that $A(n+1) = A(n) \text{ or } 1+A(n)$


TITLE: Are momentum and energy together conserved?
QUESTION [0 upvotes]: Are momentum and energy together conserved?

REPLY [1 votes]: Your question says that momentum and energy are together conserved.
The question does not specify if you are asking about a closed system's (like a galaxy cluster) total energy level or about micro interactions (particle) energy level as per QM.

First, let's take the micro interactions as per QM:

This is only true for an elastic interaction, like elastic scattering. It is the typical case when a photon hits an atom and the photons energy is conserved, only its direction is changed. Rayleigh scattering is like that.
For an inelastic interaction, like inelastic scattering, Compton scattering, only momentum can be conserved, but energy not. When a photon hits an atom, and the photon gets inelastically scattered, its energy will change, but its momentum could be conserved.
Please see here:
http://teacher.nsrl.rochester.edu/phy_labs/Conservation_Laws/Conservation_Laws.html

Now let's take a closed system's total energy level:

Though our universe is not closed, so its energy level is not conserved, we can select areas of space, like a galaxy cluster, which we can say is a local area as per GR, and so it's total energy level will be conserved.


TITLE: Uniqueness up to isometric isomorphism of predual of $(\sum_{\lambda\in\Lambda} H_\lambda)_{l_\infty}$ where $H_\lambda$ are Hilbert spaces
QUESTION [8 upvotes]: This fact is an easy consequence of results of the paper by Leon Brown and Takashi Ito, but it looks like an overkill. Does anyone know a simpler proof?

REPLY [6 votes]: In this particular case, you can proceed similarly as in Example 2.1 here. For more general results please consult a fantastic survey on unique preduals by G. Godefroy:

G. Godefroy. Existence and uniqueness of isometric preduals: a survey. In: BorLuh Lin, editor, Banach Space Theory. Proc. of the Iowa Workshop on Banach Space Theory 1987, 131–193. Contemp. Math. 85, 1989.


TITLE: Book suggestions for Combinatorics for IMO
QUESTION [2 upvotes]: I'm currently preparing Combinatorics for IMO (International Mathematical Olympiad). I searched through the web (including some threads in MSE), for the best book recommendations. After a lot of "research", I decided to go with $$ Problem-Solving \  Methods \  in  \ Combinatorics \ by  \ Pablo \ Soberon. $$
Some other books I went through were either very difficult for me, or way too easy.
I started with the book, then I find it really difficult to understand the examples illustrated in it. It's not that I am a beginner in combinatorics, but I don't know why I was finding it very difficult.
I am just not able to get a proper book since a couple of days.
And the only way for me to study is through a good book.
Can anyone suggest me the best books for combinatorics, highly relevant for Mathematical Olympiads? I will be very helpful.
I hope I have cleared things properly.
I know that a lot of people have asked the same question here already, but many of them were for Research related, and for just problem solving.
Thanks.
EDIT: AFTER A LOT OF CONSIDERATIONS, I HAVE DECIDED TO GO WITH THE BOOK $$ A \ Path \ to \ Combinatorics \ for \ Undergraduates \ $$ by $ Titu \ Andreescu. $
I just wanted to ask if this book is Olympiad oriented, or Olympiad relevant, as it's title doesn't seem to suggest that?

REPLY [2 votes]: Introduction to Combinatorics - Gerry Leversha and Dominic Rowland

Topics in Combinatorics - Gerry Leversha and Dominic Rowland

They are aimed at national maths contestants.


TITLE: When $\overline{H} \cap \overline{K} = \overline{H \cap K}$?
QUESTION [1 upvotes]: Let $G$ be a group, $N \trianglelefteq G$ and $H, K \leq G$.
When we have $\overline{H} \cap \overline{K} = \overline{H \cap K}$
(where $\overline{X}$ means the image of $X$ with natural map in $G / N$)?
I guess we should have $N \leq H \cap K$. Is there any weaker condition?

REPLY [3 votes]: $\newcommand{\Span}[1]{\langle #1 \rangle}$Prove instead that this is not true. Take for instance
$$
G = S_{3},\quad
N = \Span{(123)},\quad
H = \Span{(12)},\quad
K = \Span{(13)}.
$$
Of course there could be something special about group homomorphisms, but in general direct images of maps preserve unions, but not intersections. Whereas inverse images preserve both.

This was the answer to the previous question, which required to prove that the identity holds in general.
For the current question, all I am able to say is the following.
The identity is equivalent to $(H \cap K) N = H N \cap K N$. Clearly the left-hand side is contained in the right-hand side. By Dedekind's identity, we have
$$
H N \cap K N = (H N \cap K) N = (H \cap K N) N.
$$
So for the identity to hold we need $H N \cap K \le (H \cap K) N$ (or equivalently $H \cap K N \le  (H \cap K) N$). As noted in a comment, certainly $N \le H$ or $N \le K$ guarantee this.


TITLE: Is the Legendre transformation just simple substitution?
QUESTION [2 upvotes]: I am confused by the Legendre transform. I did the math and carried it out for a function $f(x,y)$, but I don't understand what good it does.
What is the purpose? Didn't I merely just redefine my variables in terms of other variables?
It seems to me just a fancy way of rewriting the function so instead of $x$ and $y$ I now have $u$ and $w$ which are called the conjugate variables.
Can someone explain why the Legendre transform is useful?

REPLY [0 votes]: "Didn't I merely just redefine my variables in terms of other variables?"
Well, yes. That right there is the point of the Legendre transform. It can be thought of in the same way as integrating by parts (in fact, that's really what you're doing). The only thing that integrating by parts does to an equation is swap around the terms, but often that simplifies the integral. In the same way, the Legendre transform can simplify some equations.


TITLE: Is there an easier way to integrate $\int \frac{d\, x}{\sin^2{x}}$?
QUESTION [0 upvotes]: So, hey.
I've made it like this: $ \tan{\frac{x}{2}} = t, x=2\arctan{t}, d\, x =\frac{2t\,d\,t}{1+t^2}$, therefore $\int\frac{d\, x}{\sin^2{x}} = \int\frac{1+t^2}{2t^2}d\,t$, which gives us $\frac{t^2-1}{2t}+ C$...
And then I had to make some truly unevident transformations in order to conclude that it was  just $(-\cot{x} +C)$
Tangent half-angle substitution seems a little...excessive here. Any tips to make it in a simpler way?
And, no, I have to prove that $\int \frac{d\, x}{\sin^2{x}}$ is $-\cot{x} +C$, and not just tell "Well, okay, $(-\cot{x})'=\frac{1}{\sin^2{x}}$, that's why $\int \frac{d\, x}{\sin^2{x}}= -\cot{x} +C$
Thanks in advance

REPLY [3 votes]: If you want it so badly, then:
$$\int\frac{dx}{\sin^2x}=\int\frac{\sec^2xdx}{\tan^2x}=\int\frac{d(\tan x)}{\tan^2x}=-\frac1{\tan x}+c=-\cot x+c$$

REPLY [0 votes]: $$\int dx\, \frac{1}{\sin^2 x} = \int dx\, \frac{\cos^2 x + \sin^2 x}{\sin^2 x} = \int dx\, -\frac{ \cos x \sin' x - \sin x \cos'x}{\sin^2 x}=\int \frac{d}{dx}\left( -\frac{\cos x}{\sin x} \right) = -\frac{\cos x}{\sin x}.
$$
The third equality is from the quotient rule.


TITLE: How to divide polynomials?
QUESTION [2 upvotes]: The title says it all : How to divide polynomials?
I don't understand the method below taught in my school.
Would any of you mind explaining it or even better, suggest an alternative way to solve this?

REPLY [0 votes]: Here is another method that you asked for.  Its main advantage is that you can easily do this in a text editor, which helps immensely in accurately tracking your work.  It is completely done with algebraic manipulation.
So, with $A$ as the highest term in numerator and $B$ as the sum of all remaining terms.   And $C$ as the highest term in denominator, and $D$ is the remaining terms.  The goal is to eliminate $A$ and get an answer in terms of $m + \frac{r}{C + D}$.  In other words, $m$ is a multiple of the denominator, and $r$ is the remainder.  So you start with two things being divided.
Here we justify an algebraic procedure.  Imagine some example:
$$
\color{red}{A = 4x^2}
$$
$$
\color{green}{B = 3x + 5}
$$
$$
\color{blue}{C = 2x^2}
$$
$$
\color{purple}{D = 9x - 7}
$$
So the polynomial being divided is like this, at a high level.
So ignore their values and see what happens when we perform one step:
$$
\frac{A + B}{C + D} =
$$
And note that you can add and subtract a ratio of the highest terms $\frac{A}{C}$ without changing the answer.
$$
\frac{A}{C} - \frac{A}{C} + \frac{A + B}{C + D} =
$$
And you can multiply and divide by the denominator $C + D$ without changing the answer as well.  Specifically, you can do this to the subtracted term.  It just happens that $A$, which represents the highest term will always cancel out when you simplify this.  This is actually the only thing you need to remember.  You can derive everything else with this.
$$
\frac{A}{C} - \frac{A}{C}*\frac{C+D}{C+D} + \frac{A + B}{C + D} =
$$
If you start there, you can usually jump right to that step in your head.
So now, we can figure out what it simplifies to, just to see what happens with $A$:
$$
\frac{A}{C} + \frac{-AC -AD}{CC + CD} + \frac{AC + BC}{CC + DC} =
$$
$$
\frac{A}{C} + \frac{BC - AD}{C(C + D)}
$$
So, let's apply it to the example values:
$$
\frac{\color{red}{4x^2} + \color{green}{(3x + 5)}}{\color{blue}{2x^2}+ \color{purple}{(9x - 7)}} =
$$
$$
\frac{\color{red}{4x^2}}{\color{blue}{2x^2}} + \frac{(\color{green}{3x + 5})(\color{blue}{2x^2}) - (\color{red}{4x^2})(\color{purple}{9x-7})}{(\color{blue}{2x^2})(\color{blue}{2x^2}+ \color{purple}{(9x - 7)})} =
$$
$$
\frac{4x^2}{2x^2} + \frac{6x^3 + 10x^2 - 36x^3 + 28x^2}{(2x^2)(2x^2+ (9x - 7))} =
$$
$$
\frac{4x^2}{2x^2} + \frac{3x + 5 - 18x + 14}{2x^2+ (9x - 7)} =
$$
$$
2 + \frac{-15x + 19}{2x^2+ (9x - 7)}
$$
So, the current answer is $2$ plus remainder $(-15x+19)$.  If $\frac{-15x}{2x^2} $ has a positive $x$ exponent, then we apply this method to the remainder to get more terms.  Let's check the answer:
$$
2(2x^2+ (9x - 7)) + (-15x + 19) =
$$
$$
(4x^2+ 18x - 14) + (-15x + 19) =
$$
$$
\color{red}{4x^2} + \color{green}{3x +5}
$$
Which is our numerator $\color{red}{A} + \color{green}{B}$.
Let's apply this to a different polynomial that requires more than one reduction:
$$
\frac{\color{red}{5x^2} + \color{green}{2x + 3}}{\color{blue}{x} + \color{purple}{1}} =
$$
$$
\frac{5x^2}{x} + \frac{(2x + 3)x - 5x^2}{x(x + 1)} =
$$
$$
5x + \frac{2x + 3 - 5x}{x + 1} =
$$
$$
5x + \frac{\color{red}{-3x} + \color{green}{3}}{\color{blue}{x} + \color{purple}{1}} =
$$
$$
5x  - 3 + \frac{3x + 3x}{x(x + 1)} =
$$
$$
5x  - 3 + \frac{6}{x + 1} =
$$
So, let's check that the answer is $5x - 3$ with remainder $6$.
$$
(5x - 3)(x + 1) + 6 = 5x^2 + 5x - 3x - 3 + 6 = 5x^2 + 2x + 3
$$


TITLE: System Availability of a server cluster in terms of MTTF, MTTR and RMT
QUESTION [2 upvotes]: Consider the System Availability (A) of a server cluster in terms of
  three parameters: namely the mean time to failure (MTTF), the mean
  time to repair (MTTR), and a regular maintenance time (RMT). The MTTF
  reflects the average uptime between two adjacent natural failures. The
  MTTR is the downtime due to natural failure. The RMT refers to
  scheduled down time for hardware/software maintenance or updates. (a)
  Given a cloud system with a demanded availability A = 98%. If the MTTF
  is known to be two years (or 365 × 24 × 2 = 17,520 hours) and the MTTR
  is known 24 hours, what is the value of RMT in hours per month that
  you can schedule for this cloud system?

So if $$A = \frac{MTBF}{MTBF - MTTR} $$
$$A = \frac{17520}{17520+24}$$
$$A = \frac{17520}{17544} = .998632$$
Then 24 hours times 30 days: $$24 x 30 = 720$$
$$1-.998632 = .001368$$
$$.001368 x 720 = 0.98496$$
$$0.98496x24 = 23.63$$
So is the answer 23.63 hours? That doesn't seen right.

(b)  Consider a cloud cluster of three servers. The cluster is
  considered available (or acceptable with a satisfactory performance
  level), if at least k servers are operating normally where k ≤ 3.
  Assume that each server has an availability rate of p (or a failure
  rate of 1 − p). Derive a formula to calculate the total cluster
  availability A (i.e., the probability that the cluster is available
  satisfactorily). Note that A is a function of k and p.

So for all three servers to be available, the availability of $A = p*p*p$; the probablity of one and the only one server is available is $A = 3p(1-p)(1-p)$, so the probability of one server being up and the other two being down. Any one of three servers can be the one that is up, so I'm supposed to add the three cases together?
The cited formula looks like this which I'm struggling to understand.

REPLY [1 votes]: Intuitively, you have spent almost none of your inavailability budget on random failures because $0.998$ is so much closer to $1$ than $0.98$.  You are allowed to spend $0.02$ of the time down, which would be $0.02 \cdot 720=14.4$ hours per month.  You should reduce the $0.02$ to account for the random failures.  Your $0.001368\cdot 720$ gives the average number of hours per month that the system is down due to random failure.  There is no sense multiplying that by $24$.
For 2, the chance of exactly one server being up is $3p(1-p)^2$ as you say.  That is exactly the $i=1$ term of the sum.  If you understand where $3p(1-p)^2$ came from, you should be able to identify the terms in the sum.  The $i=2$ term is ${3\choose 2}p^2(1-p)^1$


TITLE: Removed Archer example from wikipedia.
QUESTION [4 upvotes]: I have question about example illustration Convergence of random variables in probability but not almost surely.

Suppose a person takes a bow and starts shooting arrows at a target. Let $X_n$ be his score in $n$-th shot. Initially he will be very likely to score zeros, but as the time goes and his archery skill increases, he will become more and more likely to hit the bullseye and score $10$ points. After years of practice the probability that he hit anything but $10$ will be getting increasingly smaller and smaller and will converge to $0$. Thus, the sequence $X_n$ converges in probability to $X = 10$.
Note that $X_n$ does not converge almost surely however. No matter how professional the archer becomes, there will always be a small probability of making an error. Thus the sequence $(X_n)$ will never turn stationary: there will always be non-perfect scores in it, even if they are becoming increasingly less frequent.

https://en.wikipedia.org/w/index.php?title=Convergence_of_random_variables&diff=879355996&oldid=879219919
For me this example is true. But not for 69.181.249.190. Why? You have any idea?

69.181.249.190 talk 36,722 bytes -957 →Convergence in probability: removed false archer example

REPLY [4 votes]: The archer example incorrectly suggests that if there is always a small-but-positive probability of missing the target, then the archer misses infinitely often.
Counter-example: Consider $\{X_n\}_{n=1}^{\infty}$ with:
$$X_n=\left\{\begin{array}{}
10  & \mbox{ with prob $1-1/2^n$}\\
0& \mbox{ else (miss target)}
\end{array}\right.$$
We don’t care about independence but you can assume the variables are mutually independent if you want. We observe
$$\sum_{n=1}^{\infty}P[X_n\neq 10]<\infty$$
So by the Borel-Cantelli lemma we conclude that (with prob 1) we miss only finitely often. So $X_n\rightarrow10$ both in probability and with prob 1.

On the other hand, if the variables are mutually  independent with
$$P[X_n=10]=1-1/n, P[X_n=0]=1/n$$
then we can conclude (either by Borel-Cantelli or by direct calculation) that we miss infinitely often with prob 1. This is a classic example where $X_n\rightarrow10$ in probability but not with prob 1.
Overall, it depends on the rate at which the miss probabilities converge to zero.


TITLE: Showing ((A→B)→A)→A. and A,B ⊢ ¬(A→¬B) using Deduction Theorm, etc.
QUESTION [2 upvotes]: Using the Deduction Theorem and the principles of Ex Falso, Reductio ad Absurdum, and Indirect Proof, I am to show that:

(a) $$((A→B)→A)→A$$

and

(b)   $$A,B ⊢ ¬(A→¬B)$$

I know how to forumlate the principles mentioned and I also intuitively see that (a) and (b) are valid. My problem, however, is that I actually can't figure out how to use the principle mention to show this ...
I would love it if someone could offers some assistance in showing how we apply the principle in order to show (a) and (b).

REPLY [2 votes]: For (a) [Peirce's law] see this post : it is enough to omit steps 1) and 13).
For (b) :

$A,B \vdash ¬(A → ¬B)$

we have the following proof :
1) $A$ --- premise
2) $B$ --- premise
3) $A → ¬B$ --- assumed [a]
4) $¬B$ --- from 1) and 3) by $\to$-elimination (modus ponens)
5) $\bot$ --- from 2) and 4) by $\lnot$-elimination

6) $\lnot (A → ¬B)$ --- from 3) and 5) by $\lnot$-introduction, discharging [a].

Thus, steps 3) to 6) are nothing more than an Indirect Proof : assume  the negation of the sought conclusion and derive a contradiction.


TITLE: Rings of formal power series is finitely generated as module.
QUESTION [0 upvotes]: The question rises when I was reading Kemper's "A course in Commutative Algebra."
Let $K$ be a field and consider the ring of formal power series $K[[x]]$. In an exercise I proved $K[[x]]$ is not Jacobson, hence not finitely generated as a K-algebra. But since $K$ is Noetherian and $K[[x]]$ is a K-module, $K[[x]]$ is Noetherian is equivalent to $K[[x]]$ is finitely generated as a K-module. There is a proof of $K[[x]]$ being Noetherian. I can't see why $K[[x]]$ is a finitely generated K-module but not a finitely generated K-algebra. Is there a mistake in my understanding? Any help is appreciated.

REPLY [2 votes]: $K[[x]]$ is a Noetherian ring, not a Noetherian $K$-module.  In other words, every ideal in the ring $K[[x]]$ is finitely generated, rather than every $K$-submodule of $K[[x]]$ being finitely generated.


TITLE: Determine the length and angle of a hypotenuse starting from a rectangle at a different angle
QUESTION [1 upvotes]: Lumber support for Plywood
I'm trying to orient a piece of lumber over a piece of plywood as shown. The dimensions of the plywood and the width of the lumber are known. The lumber needs to be oriented and cut as a parallelogram. I have to determine the length of the lumber (l), the distance from the end to trim off (t), and the angle of the cut in degrees (d). I think I can figure out the trim off length but I'm not sure.
Referring to the diagram:
a, o, h - is known
w - is known
l, t, d - is unknown
I need a formula to determine:

the length (l) of the (second) rectangle created from filling out the parallelogram
the length (t) of a triangle created from filling out the parallelogram
the parallelogram's lesser angle in degrees (d)

My math isn't too good and I don't understand Greek characters. I would GREATLY appreciate a reply using something like Excel functions.

REPLY [0 votes]: Here's what I've got for you:
https://drive.google.com/file/d/1kxf3H83sP6BI3BP7CBGj5u3EMs4RGY8z/view?usp=sharing
$$ l^2 + w^2 = h^2 \\
l = \sqrt{h^2 - w^2}$$
Knowing $l$, you can use similar triangles XYZ, NMZ to show,
$$ t = h' \text{ cos}(d) \\
t = \frac{w}{\text{ tan}(d)} \\
a' = l \text{ cos}(d) \\
\frac{t}{a'} = \frac{w}{o} \\
\text{therefore, } h' = \frac{wl}{o}$$
Knowing $h'$ we can solve for $d$, and consequently $t$.
$$w = h' \text{ sin}(d) \\
d = \text{ arcsin}(\frac{w}{h'}) \\
t = h' \text{ cos}(d)$$
I may have made some mistake here having done this so quickly, but this can be easily implemented into excel for calculations by varying $a,o,$ and $w$.


TITLE: What is the shortest and most preferred way of writing the result?
QUESTION [0 upvotes]: Let's say I have to calculate 6/4 (i know, it's simple). Would the preferred answer be 3/2 or 1.5?
Same with square roots: is sqrt(2) a better answer than 2^(1/2)?

REPLY [0 votes]: Answer from @Wuestenfux in comments:
No really best way here. As a hint, if the outcome is an algebraic number, write it down as such and not as a decimal number


TITLE: Expected value of function of exponential random variable
QUESTION [0 upvotes]: Let $X \sim Exp(\lambda)$ (exponential distribution) and $Y=X^a$. For what  values of $a\in R$ is $E[Y]< \infty$?
I computed $E[Y] = \lambda \int x^a e^{-\lambda x}dx$. How to proceed? Does the answer depend on $\lambda$?

REPLY [0 votes]: Hint:
(1) $a\leq-1$
$\int_0^{1}  x^{a}e^{-x} dx$ diverges.
Since For $0<x\leq 1$
$$e^{-x}\geq e^{-1} \hspace{.5cm} and  \hspace{.5cm} x^{a} \geq x^{-1}$$
so
$$\int_0^{1}  x^{a}e^{-x} dx \geq \int_0^{1} e^{-1} x^a dx \geq \int_0^{1} e^{-1} \frac{1}{x} dx  $$  and
$\int_0^{1} \frac{1}{x} dx $ diverges.
(2) For $a>-1$
$$\int_0^{+\infty}  x^{a}e^{-x} dx=\int_0^{+\infty}  x^{(a+1) -1}e^{-x} dx=\Gamma(a+1)$$
see Incomplete_gamma_function.
Gamma distribution is also useful.
For $k>0$ (equal with a>-1) and $\lambda>0$
$$\int_0^{+\infty} \frac{\lambda^k }{\Gamma(k)} x^{k-1}e^{-\lambda x} dx=1$$
so for any $k>0$
$$\int_0^{+\infty} x^{k-1}e^{-\lambda x} dx= \frac{\Gamma(k)}{\lambda^k}$$


TITLE: Noetherian integral domain and fractional ideal
QUESTION [2 upvotes]: Let $a$ be a non-zero ideal of an integral doman $A$,whose field of fractions is $K$. Let $a'$ denote the set of $x \in K$, the field of fractions of an integral domain $A$, such that $xa \subset A$. In Local Fields: Pierre-Serre, it says that if $A$ is Noetherian, then $a'$ is a fractional ideal. Isn't $a'$ a fractional ideal regardless of whether $A$ is Noetherian or not?
My attempt: Let $y$ be a non-zero element of $a$, then by definition of $a'$, $ya' \subset A \implies a'$ is a fractional ideal (by the Wikipedia definition of fractional ideal).

REPLY [0 votes]: The issue is that Serre is taking a different definition of "fractional ideal" than Wikipedia; if you look at page 11, you'll see that to Serre, a fractional ideal is an $A$-submodule of $K$ which is finitely generated over $A$. Serre's definition implies the definition you see on Wikipedia, and while the converse is not true in general, the two definitions are equivalent when $A$ is Noetherian.


TITLE: Binomial coefficient modularity conjecture
QUESTION [2 upvotes]: I am not sure if I have seen this before:
$$\dbinom{n}{p}\equiv\big\lfloor{\dfrac{n}{p}}\big\rfloor\mod p$$
where $p$ is prime and $\lfloor x\rfloor$ is the floor function.
So, for example, $\dbinom{34}{7}\equiv\lfloor\dfrac{34}{7}\rfloor=4\mod 7$.

REPLY [2 votes]: Your statement is a consequence of Lucas' theorem for binomial coefficients, which says that if $k$ is a positive integer and $0\leq a_i, b_i\leq p-1$ are integers for $i=0,1,\ldots,k$, then
$$
{b_0+b_1 p+\ldots+b_k p^k\choose a_0+a_1p+\ldots+a_k p^k}\equiv {b_0\choose a_0}\cdots {b_k\choose a_k}\mod p.
$$
If we write $n$ in base $p$ as $n=b_0+b_1p+\ldots +b_kp^k$, then
$$
{n\choose p}\equiv {b_0\choose 0}{b_1\choose 1}{b_2\choose 0}\cdots {b_k\choose 0}= b_1\equiv \left\lfloor\frac{n}{p}\right\rfloor\mod p.
$$


TITLE: The logarithmic inequality
QUESTION [2 upvotes]: $$ \log_{3x+7}{(9+12x+4x^2)}+ \log_{2x+3}{(6x^2+23x+21)} \ge 4$$
The logarithmic inequality is defined for: $ x \in (-3/2, -1) \cup(-1, \infty)$.
First, I supposed that my solutions are in the interval $(-3/2, -1)$. Following this interval for which the logarithmic function $t$ are decreasing and multiplying both sides of the inequality by $\log_{2x+3}{(3x+7)}$, I obtained:
$$ 2+ [1+\log_{2x+3}{(3x+7)}-\log_{2x+3}{6}]\cdot\log_{2x+3}{(3x+7)} \le4\log_{2x+3}{(3x+7)}\\ \log_{2x+3}{(3x+7)}=t<0\\t^2-(3+\log_{2x+3}{6})t+2 \le0$$
And here I got stuck. This inequailty is intended to be solved witouth a calculator, is it possible to continuie doing the inequality withouth it? And what I did wrong?

REPLY [2 votes]: Hint:
$$9+12x+4x^2=(2x+3)^2$$
$$ 6x^2+23x+21=(2x+3)(3x+7)$$
$$4\le\log_{3x+7}(2x+3)^2+\log_{2x+3}(2x+3)(3x+7)=2\log_{3x+7}(2x+3)+1+\log_{2x+3}(3x+7)$$
As $\log(y)$ is real for $y>0,$
we need $2x+3, 3x+7>0\implies x>$max$(-3/2,-7/3)=-3/2$
Now if $\log_{3x+7}(2x+3)=a,$
$$4\le2a+1+\dfrac1a$$
$$\iff0\le2a+\dfrac1a-3=\dfrac{2a^2-3a+1}a=\dfrac{(2a-1)(a-1)}a$$
If $\dfrac{(2a-1)(a-1)}a=0, a=1$ or $\dfrac12$
Else we need $a(2a-1)(a-1)>0$


text:
\begin{document}
\maketitle

\begin{abstract}\noindent
We define Lie algebroids over infinite jet spaces
and establish their equivalent representation through
homological evolutionary vector fields.

\noindent\textbf{Keywords:} {Lie algebroid, BRST\/-\/differential, Poisson structure, integrable systems, string theory.}

\end{abstract}

\paragraph*
{Introduction.}
The construction of Lie algebroids over smooth manifolds is important in differential geometry (particularly, in Poisson geometry) and appears in various models of mathematical physics (e.g., in Poisson sigma\/-\/models).
We extend the classical definition of Lie algebroids over smooth manifolds \cite{Vaintrob} to the construction of variational Lie algebroids over infinite jet spaces. We define these structures in a standard way via vector bundles and also
through homological vector fields $Q^2=0$ on infinite jet super\/-\/bundles, then proving the equivalence. Our generalization of the classical construction manifestly respects the geometry which appears under mappings between smooth manifolds. For this reason, the variational picture, which we develop here, more fully grasps the geometry of strings in space\/-\/time \cite{AKZS}.

First, we very briefly recall
the definition of classical Lie algebroids;

we refer to \cite{Vaintrob}
for more details (see also \cite{Voronov} or
the surveys \cite{
YKSDB
} and references therein).
The standard examples of Lie algebroids over usual manifolds are the tangent bundle or the Poisson algebroid structure of the cotangent bundle to a Poisson manifold.

At the same time,
Lie algebras are toy examples of Lie algebroids over a point.
Likewise,
we view smooth manifolds $M^m$ as (very poor) `infinite jet spaces' for mappings of the point into them, so that the manifolds become `fibres' in the `bundles' $\pi\colon M^m\to\{\text{pt}\}$.
This very informal (but equally productive\,!) understanding is our starting point.

We shall show what becomes of the Lie algebroids over $M^m$
under the extension of the base point to an $n$-\/dimensional
manifold $\varSigma^n$, which yields a true jet
bundle $J^\infty(\varSigma^n\to M^m)$.

Our results are structured in this paper as follows.
It is readily seen that a literal transfer of the classical
definition, see p. \pageref{DefLieAlgd},
over smooth manifolds

to the infinite jet bundles is impossible because the Leibniz
rule \eqref{LeibnizInAlg} is lost \textit{ab initio} (to the same extent as it is lost for the commutators of evolutionary vector fields or for the variational Poisson bracket \cite{ClassSym}). For this reason,
in section \ref{SecDef} we take, as
the new definition, an appropriate consequence of the classical one: namely, the commutation closure for the images of the anchors.

(Remarkably, this

is often \textsl{postulated} [for convenience, rather than derived]
as a part of the definition of a Lie algebroid, e.g.,
see \cite{Vaintrob,Voronov} vs \cite{Herz,YKSMagri}.)
In these terms, our main result of \cite{SymToda} states that
the non\/-\/periodic 2D Toda equations associated with semi\/-\/simple complex Lie algebras \cite{Leznov}, being the representatives for the vast class of hyperbolic Euler\/--\/Lagrange systems of Liouville type \cite{SokolovUMN}, are variational Lie algebroids.

\begin{rem}
We duly recall
that Lie algebroids over the spaces of finite jets of sections for the \textsl{tangent} bundle $\pi\colon T\varSigma\to \varSigma^n$ were defined in \cite{Kumpera}. However, in this paper we let the vector bundle $\pi$ be arbitrary.
Instead of the tangent bundle (as in \cite{Kumpera}), we use model examples of a principally different nature. Namely, our illustrations come from the geometry of differential equations: e.g., we address Hamiltonian (non-)\/evolutionary systems \cite{Topical} or the 2D Toda chains \cite{Leznov},

the solutions of which are $r$ functions in two variables for every semi\/-\/simple complex Lie algebra of rank $r$. In other words, we analyse the general case when the base and fibre dimensions in the bundle $\pi$ are not related.
\end{rem}

In section \ref{SecVarQ}, we represent the structure of a variational Lie algebroid $\gA$, defined in the above sense, by the homological evolutionary vector field $Q$ on the infinite jet bundle to $\Pi\gA$ viewed as the vector bundle with parity reversed fibres. This is again nontrivial due to the use of the Leibniz rule (now missing) in the equivalence proof for the classical definition.
We also impose the natural nondegeneracy assumption on the variational anchors so that the odd field $Q$ is well defined.

To
illustrate our assertion, we let the variational anchors be
Hamiltonian differential operators and obtain the fields $Q$ which are
themselves Hamiltonian with respect to the corresponding variational Poisson bi\/-\/vectors and the canonical symplectic structure of \cite{KuperCotangent}.

In the sequel, the ground field is $\BBR$ and all mappings are supposed to be $C^\infty$-\/smooth. Likewise, we suppose that all total
differential operators are local (i.e., polynomial in total derivatives).
We assume

that all spaces are purely even,
i.e., that the manifolds are not superized, unless we state the opposite
explicitly when using the parity reversion $\Pi$.
We employ the standard techniques from the geometry of differential equations \cite{ClassSym,Olver}; for convenience, we summarize the necessary notions in the beginning of section \ref{SecDef}.

Because all our reasonings are local,
we can conveniently restrict to local charts

and, instead of the jet bundles for mappings $\varSigma^n\to M^m$ of manifolds, we consider the jet spaces for $m$\/-\/dimensional vector bundles $\pi\colon E^{m+n}\to\varSigma^n$.

By default, we define all structures
on the empty infinite jet spaces, identifying

autonomous evolution equations with evolutionary vector fields, which
is standard.
The delicate interplay between the definitions of all structures at hand on the empty jet spaces and on differential equations would require a much longer

description; this will be the object of another paper.
In the meantime,
we refer to the review~\cite{Topical}.

We follow the notation of \cite{SymToda,TMPhGallipoli,d3Bous},
borrowing it in part from \cite{YKSDB,
ClassSym,
Olver
}.

This paper is a continuation of the papers \cite{SymToda,d3Bous},

to which we refer for motivating examples of variational Lie algebroids.

Some of the results, which we present here in
detail, were briefly reported in the preprint \cite{Twente}.

\paragraph*
{Classical Lie algebroids: a review.}\label{SecAlgd}
\noindent
Let $M^m$ be a smooth real $m$-\/dimensional manifold
($1\leq m\leq+\infty$)
and $TM\to M^m$ be its tangent bundle.

\begin{define}[\cite{Vaintrob}]\label{DefLieAlgd}
A \textsl{Lie algebroid} over a
manifold $M^m$ is a vector
bundle $\xi\colon\Omega^{d+m}\to M^m$
the space of sections $\Gamma\Omega$ of which
is equipped with a Lie algebra
structure $[\,,\,]_A$ together with a morphism of the
bundles $A\colon\Omega\to TM$,
called the \textsl{anchor}, such that the Leibniz rule

\begin{equation}\label{LeibnizInAlg}
[f\cdot\cX,\cY]_A=f\cdot[\cX,\cY]_A
  -\bigl(A(\cY)f\bigr)\cdot\cX
\end{equation}
holds for any
$\cX,\cY\in\Gamma\Omega$ and any $f\in C^\infty(M^m)$.
\end{define}

Essentially,
the anchor in a Lie algebroid is a specific
fiberwise\/-\/linear mapping from a given vector bundle
over a smooth manifold $M^m$ to its tangent bundle.

\begin{lemma}[\cite{Herz}
]\label{PropMorphismFollows}

The anchor $A$ maps the bracket $[\,,\,]_A$ for sections
of the vector bundle $\xi$

to the Lie bracket $[\,,\,]$ for sections of the tangent bundle
to the manifold $M^m$.
\end{lemma}

This property is a consequence
of the Leibniz rule \eqref{LeibnizInAlg} and the Jacobi identity
for the Lie algebra structure $[\,,\,]_A$ in $\Gamma\Omega$.

\begin{lemma}[\cite{Vaintrob}]\label{RemHVF}

Equivalently, a Lie algebroid structure on $\Omega$ is
a homological vector field $Q$ on $\Pi\Omega$
(take the fibres of $\Omega$, reverse their parities, and thus obtain
the new super\/-\/bundle $\Pi\Omega$ over $M^m$).
These homological vector fields,

which are differentials on
$C^\infty(\Pi\Omega)=\Gamma(\bigwedge^\bullet\Omega^*)$, equal
\begin{equation}\label{HomVF}
Q=A_i^\alpha(u)b^i\frac{\dd}{\dd u^\alpha}
 -\frac{1}{2}b^i\,c^k_{ij}(u)\,b^j\frac{\dd}{\dd b^k},
\qquad [Q,Q]=0\ \Leftrightarrow\ 2Q^2=0,
\end{equation}
where
\begin{itemize}
\item
$(u^\alpha)$ is a system of local coordinates near a point $u\in M^m$,
\item
$(p^i)$ are local coordinates along the $d$-\/dimensional fibres of $\Omega$
and $(b^i)$ are the respective coordinates on $\Pi\Omega$, and
\item
$[e_i,e_j]_A=c^k_{ij}(u)e_k$ give
the structural constants for a $d$-\/element
local basis $(e_i)$ of sections
in $\Gamma\Omega$ over the point~$u$ and
$A(e_i)=A_i^\alpha(u)\cdot\dd/\dd u^\alpha$ is the image of $e_i$
under the anchor $A$.
\end{itemize}
\end{lemma}

\begin{proof}[Sketch of the proof]
The anti\/-\/commutator $[Q,Q]=2Q^2$ of the odd vector field $Q$ with itself is a vector field. Its coefficient of $\dd/\dd u^\alpha$ vanishes because $A$ is the Lie algebra homomorphism by
Lemma \ref{PropMorphismFollows}.
The equality to zero of the coefficient of $\dd/\dd b^q$ is achieved in three steps. First, we notice that the application of the second term in \eqref{HomVF} to itself by the graded Leibniz rule for vector fields yields the overall numeric factor $\tfrac{1}{4}$, but it doubles due to the skew\/-\/symmetry of the structural constants $c^k_{ij}$. Second, we recognize the right\/-\/hand side of the
Leibniz rule \eqref{LeibnizInAlg}, with $c^k_{ij}(u)$ for $f\in C^\infty(M^m)$, in the term
\begin{equation}\label{BeforeTriple}
\frac{1}{2}\sum\nolimits_{ijn}b^ib^jb^n\left(
-A_n^\alpha\,\frac{\dd}{\dd u^\alpha}\bigl(c^q_{ij}(u)\bigr)
 +c^\ell_{ij}c^q_{\ell n}
\right)\cdot\frac{\dd}{\dd b^q}.
\end{equation}
Third, we note that the \textsl{cyclic} permutation $i\to j\to n\to i$ of the odd variables $b^i$,\ $b^j$, $b^n$ does not change the sign of \eqref{BeforeTriple}. We thus triple it by taking the sum over the permutations (and duly divide by three) and, as the coefficient of $\dd/\dd b^q$, obtain the Jacobi identity for $[\,,\,]_A$,
\begin{equation}\label{JacobiUsual}
\tfrac{1}{2}\cdot\tfrac{1}{3}\cdot\sum\nolimits_{\circlearrowright}
\bigl[[e_i,e_j]_A, e_n\bigr]_A=0.
\end{equation}
The zero in its right\/-\/hand side calculates the required coefficient.
Thus, $Q^2=0$.
\end{proof}

\section{Variational Lie algebroids}\label{SecDef}
\noindent
For consistency, we first summarize the notation.
Let $\varSigma^n$ be an $n$-\/dimensional orientable smooth real manifold,
and let $\pi\colon E^{m+n}\xrightarrow[M^m]{}\varSigma^n$ be a vector
bundle over it
with $m$-\/dimensional fibres $M^m\ni u=(u^1,\ldots,u^m)$.
By $J^\infty(\pi)$ we denote the infinite jet space over
$\pi$, and
we set $\pi_\infty\colon J^\infty(\pi)\to \varSigma^n$. We denote by $u_\sigma$,
$|\sigma|\geq0$, its fibre coordinates. Then $[u]$ stands for
the differential dependence on $u$ and its derivatives up to some finite order,
and we put $\cF(\pi)\mathrel{{:}{=}}C^\infty(J^\infty(\pi))$,
understanding it in the standard way
(as the inductive limit of filtered algebras, see \cite{ClassSym}).

Let $\xi\colon N\to \varSigma^n$
be another vector
bundle over the same base $\varSigma^n$ and take its pull\/-\/back $\pi_\infty^*(\xi)\colon N
\mathbin{{\times}_{\varSigma^n}}J^\infty(\pi)\to J^\infty(\pi)$ along $\pi_\infty$.
By definition, the $\cF(\pi)$-\/module of sections
$\Gamma\bigl(\pi_\infty^*(\xi)\bigr)=\Gamma(\xi)\mathbin{{\otimes}_{C^\infty(\varSigma^n)}}\cF(\pi)$ is called \textsl{horizontal}, see \cite{Lstar}.

\begin{example}\label{CanonicalInduced}

Three horizontal $\cF(\pi)$-\/modules are canonically associated with
every jet space $J^\infty(\pi)$, see, e.g., \cite{Lstar}.

First, let $\xi\mathrel{{:}{=}}\pi$,
whence we obtain the $\cF(\pi)$-\/module $\Gamma(\pi_\infty^*(\pi))
=\Gamma(\pi)\mathbin{{\otimes}_{C^\infty(\varSigma^n)}}\cF(\pi)$.

The shorthand notation
is $\varkappa(\pi)\equiv\Gamma(\pi_\infty^*(\pi))$.
Its sections $\vph\in\varkappa(\pi)$
are in a one\/-\/to\/-\/one correspondence with
the $\pi$-\/vertical evolutionary derivations
$\cEv_\vph=\sum_\sigma \frac{\Id^{|\sigma|}}{\Id x^\sigma}
   (\vph)\cdot\dd/\dd u_\sigma$
on $J^\infty(\pi)$, here
$\tfrac{\Id}{\Id x}=\tfrac{\dd}{\dd x}+u_x\,\tfrac{\dd}{\dd u}+\cdots$
is the total derivative.

For all $\psi$ such that $\cEv_\vph(\psi)$ makes sense,
the linearizations $\ell_\psi^{(u)}$ are defined by
$\ell_\psi^{(u)}(\vph)=\cEv_\vph(\psi)$, where $\vph\in\varkappa(\pi)$.

Second, let $\xi$ be the $q$th exterior power of the cotangent bundle $T^*\varSigma$, here $q\leq n$. Taking the pull\/-\/back $\pi_\infty^*(\xi)$, we obtain the horizontal $\cF(\pi)$-\/module $\bar{\Lambda}^q(\pi)$ of its sections, which are the \textsl{horizontal $q$-\/forms} on the jet space $J^\infty(\pi)$. In local coordinates, they are written as $h\bigl(x,[u]\bigr)\cdot\Id x^{i_1}\wedge\dots\wedge\Id x^{i_q}$, where $h\in\cF(\pi)$ and $1\leq i_1<\dots<i_q\leq n$.

Thirdly, take the
module $\bar{\Lambda}^n(\pi)$
of highest $\pi$-\/horizontal forms on $J^\infty(\pi)$.
Then we denote by $\widehat{\varkappa}(\pi)=
\Hom_{\cF(\pi)}\bigl(\varkappa(\pi),\bar{\Lambda}^n(\pi)\bigr)$ the horizontal $\cF(\pi)$-\/module dual to $\varkappa(\pi)$.

\end{example}

\begin{rem}\label{RemComposite}
The structure of fibres in the bundle $\xi$ can be composite

and their dimension infinite. With the following canonical example
we formalize
the 2D Toda geometry \cite{SymToda}.

Namely, let $\zeta\colon I^{r+n}\xrightarrow[W^r]{}\varSigma^n$ be a
vector
bundle with $r$-\/dimensional fibres in which $w=(w^1$,\ $\ldots$,\ $w^r)$ are local coordinates. Consider the infinite jet bundle $\xi_\infty\colon J^\infty(\xi)\to \varSigma^n$ and, by definition, set either $\xi=\zeta_\infty\circ\zeta_\infty^*(\zeta)\colon\varkappa(\xi)\to \varSigma^n$ or
$\xi=\zeta_\infty\circ\zeta_\infty^*(\widehat{\zeta})\colon\widehat{\varkappa}(\xi)\to \varSigma^n$.

To avoid an inflation
of formulas, we shall always briefly denote
by $\Gamma\Omega\bigl(\xi_\pi\bigr)$
the horizontal $\cF(\pi)$-\/module at hand.
\end{rem}

By definition, mappings between
$\cF(\pi)$-\/modules are called \textsl{total differential operators} if they are sums of compositions of $\cF(\pi)$-\/linear maps and liftings of vector fields from the base $\varSigma^n$ onto $J^\infty(\pi)$ by the Cartan connection $\dd/\dd x\mapsto\Id/\Id x$.
The main objects of our study are total differential operators
(i.e., matrix, linear differential operators in total derivatives)
that take values
in the Lie algebra $\gothg(\pi)=\bigl(\varkappa(\pi),[\,,\,]\bigr)$.

\begin{define}\label{DefFrob}
Let the above assumptions and notation hold.
Consider a total differential operator
$A\colon\Gamma\Omega\bigl(\xi_\pi\bigr)\to\varkappa(\pi)$
the
image of which is closed under the commutation
in $\gothg(\pi)$:
\begin{equation}\label{EqDefFrob}
[\text{im}\,A, \text{im}\,A]\subseteq\text{im}\,A
\
\Longleftrightarrow
\
\bigl[A(\bp),A(\bq)\bigr]=A\bigl([\bp,\bq]_A\bigr),\quad
\bp,\bq\in\Gamma\Omega\bigl(\xi_\pi\bigr).
\end{equation}
The operator $A$ transfers the Lie algebra structure
$[\,,\,]\bigr|_{\mathrm{im}\,A}$ in a Lie subalgebra of $\gothg(\pi)$
to the
bracket $[\,,\,]_A$
in the quotient $\Gamma\Omega\bigl(\xi_\pi\bigr)/\ker A$.

Then the triad
\begin{equation}\label{EqVarLieAlgd}
\bigl(\Gamma\Omega\bigl(\xi_\pi\bigr),[\,,\,]_A\bigr)
\xrightarrow{\ A\ {}}\bigr(\varkappa(\pi),[\,,\,]\bigr)
\end{equation}
is the \textsl{variational Lie algebroid} $\gA$ over the infinite jet space $J^\infty(\pi)$,
and the Lie algebra homomorphism $A$ is the \textsl{variational anchor}.
\end{define}

Essentially,
the variational anchor in a variational Lie algebroid is a specific linear mapping
from a given horizontal module of sections of an induced bundle over
the infinite jet space $J^\infty(\pi)$
to the prescribed horizontal module of generating sections for evolutionary derivations on $J^\infty(\pi)$.

\begin{example}\label{ExKdV}
Consider the Hamiltonian operator
$A_2=-\tfrac{1}{2}\,\tfrac{\Id^3}{\Id x^3}+
w\,\tfrac{\Id}{\Id x}+\tfrac{\Id}{\Id x}\circ w$,
which yields the second Poisson structure for the
Korteweg\/--\/de Vries equation $w_t=-\tfrac{1}{2}w_{xxx}+3ww_x$.
The image of $A_2$ is closed under commutation:

the bracket $[\,,\,]_{A_2}$

is \cite{SokolovUMN}
\begin{equation}\label{BrA2}
[p,q]_{A_2}=\cEv_{A_2(p)}(q)-\cEv_{A_2(q)}(p)
  +
{\tfrac{\Id}{\Id x}(p)\cdot q -p\cdot\tfrac{\Id}{\Id x}(q)}.
\end{equation}
\end{example}

\begin{example}\label{Exd3Bous}
In \cite{d3Bous} we demonstrated that the dispersionless $3$-\/component
Boussinesq system of hydrodynamic type admits a two\/-\/parametric family of

finite deformations $[\,,\,]_{\boldsymbol{\epsilon}}$ for the
standard bracket $[\,,\,]$ of its symmetries $\sym\cE$.
For this, we used
two
recursion differential operators
$R_i\colon\sym\cE\to\sym\cE$, $i=1,2$,
the images of which are closed under commutation
and which are \textsl{compatible}
in this sense, spanning the two\/-\/dimensional space of the
operators $R_{\boldsymbol{\epsilon}}$ with involutive images.
We obtain the new brackets $[\,,\,]_{\boldsymbol{\epsilon}}$
via \eqref{EqDefFrob} (c.f. \cite{
Contractions} and references therein):

$\bigl[R_{\boldsymbol{\epsilon}}(p),R_{\boldsymbol{\epsilon}}(q)\bigr]=
R_{\boldsymbol{\epsilon}}\bigl([p,q]_{\boldsymbol{\epsilon}}\bigr)$ for $p,q\in\sym\cE$ and $\boldsymbol{\epsilon}\in\BBR^2\setminus\{0\}$.
\end{example}

\begin{rem}
We standardly regard autonomous evolution equations as
the $\pi$-\/vertical evolutionary vector fields on
the infinite jet bundles $J^\infty(\pi)$ for
vector bundles $\pi$ over $\varSigma^n\ni x$.
We thus reduce the construction at hand to the
``absolute'' case of empty jet spaces.

Generally, there is no Frobenius theorem for the involutive distributions
spanned by the images of the variational anchors.
Nevertheless, the Euler\/--\/Lagrange systems $\cE$
of Liouville type provide an exception \cite{SymToda,SokolovUMN} because of their outstanding symmetry structure:
the variational anchors
yield the involutive distributions
of evolutionary vector fields that are tangent to the \textsl{integral
manifolds}, namely, to the differential equations $\cE$ themselves.
In these terms,
for a \textsl{given} differential equation $\cE\subset J^\infty(\pi)$, we accordingly restrict the definition of variational Lie algebroid
to the infinite\/-\/dimensional manifold $\cE$ such that the variational
anchors yield its
symmetries, and, if necessary, we shrink the fibres of $\xi$.

\end{rem}

\begin{example}

The generators of the Lie algebra of point symmetries for the
$(2+1)$-\/dimensional
`heavenly' Toda equation $u_{xy}=\exp(-u_{zz})$, see \cite{BoyerFinley},
are
$\vph^x=\widehat{\square}^x\bigl(p(x)\bigr)$ or
$\vph^y=\widehat{\square}^y\bigl(\bar{p}(y)\bigr)$, where
$p$ and $\bar{p}$ are arbitrary smooth functions and

$\widehat{\square}^x=u_x+\tfrac{1}{2}z^2\,\tfrac{\Id}{\Id x}$,

$\widehat{\square}^y=u_y+\tfrac{1}{2}z^2\,\tfrac{\Id}{\Id y}$.

The image of each operator is
closed under commutation such that
$
[p,q]_{\widehat{\square}^x}=\cEv_{\widehat{\square}^x(p)}(q)
  -\cEv_{\widehat{\square}^x(q)}(p)
+p\cdot\tfrac{\Id}{\Id x}(q) - \tfrac{\Id}{\Id x}(p)\cdot q
$
for any $p(x)$ and $q(x)$, and similarly for $\widehat{\square}^y$.

\end{example}

\begin{example}\label{ExVarCoTan}
Two distinguished constructions of the variational Lie algebroids emerge from the variational (co)\/tangent bundles over $J^\infty(\pi)$, see Example \ref{CanonicalInduced} and Remark \ref{RemComposite}.
These two cases involve an important intermediate component,
which we regarded as the Miura substitution in \cite{SymToda}.

Namely, consider the induced fibre bundle $\pi_\infty^*(\zeta)$ and fix its section $w$.
Making no confusion, we continue denoting by the same letter $w$
the fibre coordinates in $\zeta$ and the fixed
section $w[u]\in\Gamma\bigl(\pi_\infty^*(\zeta)\bigr)$,
which is a nonlinear differential operator in $u$.

Obviously, the substitution $w=w[u]$ converts
the horizontal $\cF(\zeta)$-\/modules
to the submodules of horizontal $\cF(\pi)$-\/modules.\footnote{
For instance,
we have $\zeta=\pi$ for the recursion
operators $\varkappa(\pi)\to\varkappa(\pi)$,
see Example \ref{Exd3Bous}.
It is standard to identify the domains of Hamiltonian operators with

$\widehat{\varkappa}(\pi)$, see Example \ref{ExKdV}.
Here we set $w=\id\colon\Gamma(\pi)\to\Gamma(\zeta)$ in both cases;
this is why the auxiliary bundle $\zeta$ becomes ``invisible''
in the standard exposition (e.g.\ in~\cite{Lstar,JKGolovko2008}).}

By this argument, we obtain the modules
\[

\varkappa(\zeta){\bigr|}_{w[u]\colon J^\infty(\pi)\to\Gamma(\zeta)}\quad
\text{and}\quad

\widehat{\varkappa}(\zeta){\bigr|}_{w[u]\colon J^\infty(\pi)\to\Gamma(\zeta)},
\]
where the latter is the module of sections of the pull\/-\/back by $\pi_\infty^*$ for the $\bar{\Lambda}^n(\zeta)$-\/dual to the induced bundle $\zeta_\infty^*(\zeta)$.
We emphasize that by this approach we preserve the correct

transformation rules for the sections
in $\varkappa(\zeta)$ or $\widehat{\varkappa}(\zeta)$
under (unrelated\,!) reparametrizations of the fibre coordinates $w$ and $u$ in the bundles $\zeta$ and $\pi$, respectively.

We say that the linear operators $A\colon\varkappa(\zeta){\bigr|}_{w[u]}\to
\varkappa(\pi)$ and $A\colon\widehat{\varkappa}(\zeta){\bigr|}_{w[u]}\to\varkappa(\pi)$ subject to \eqref{EqDefFrob} are the variational anchors of \textsl{first} and \textsl{second kind}, respectively.

Under
differential reparametrizations
$\tilde{u}=\tilde{u}[u]\colon
J^\infty(\pi)\to\Gamma(\pi)$ and $\tilde{w}=\tilde{w}[w]\colon
J^\infty(\zeta)\to\Gamma(\zeta)$,

the operators $A$ of first kind are transformed according to the formula

$A
\mapsto\tilde{A}=\ell_{\tilde{u}}^{(u)}\circ A\circ\ell_w^{(\tilde{w})}
  \Bigr|_{\substack{w=w[u]\\u=u[\tilde{u}]}}$.

Respectively, the operators of second kind obey the rule
$A
\mapsto\tilde{A}=\ell_{\tilde{u}}^{(u)}\circ A\circ
 \bigl(\ell_{\tilde{w}}^{(w)}\bigr)^\dagger
  \Bigr|_{\substack{w=w[u]\\u=u[\tilde{u}]}}$,

where the symbol $\dagger$ denotes the adjoint operator.

The recursion operators with involutive images, which we addressed
in \cite{d3Bous}, are examples of the anchors of first kind.

All Hamiltonian operators and, more generally, Noether operators
with involutive images are variational anchors of second
kind \cite{d3Bous}.
The operators $\square$,\ $\overline{\square}$
that yield symmetries of
the Euler\/--\/Lagrange Liouville\/-\/type systems
are also anchors of second kind \cite{SymToda,TMPhGallipoli};

they are `non\/-\/skew\/-\/adjoint generalizations

of Hamiltonian operators' \cite{SokolovUMN} in exactly this sense.
\end{example}

In the remaining part of this section we
analyse the standard structure of the induced bracket $[\,,\,]_A$ on the quotient of $\Gamma\Omega\bigl(\xi_\pi\bigr)$ by $\ker A$.
By the Leibniz rule, two sets of summands appear in the bracket of
evolutionary vector fields $A(\bp),A(\bq)$ that belong to the image of the variational anchor $A$:
\begin{equation}\label{LeibnizForOp}
\bigl[A(\bp),A(\bq)\bigr]=A\bigl(\cEv_{A(\bp)}(\bq)-\cEv_{A(\bq)}(\bp)\bigr)
  +\bigl(\cEv_{A(\bp)}(A)(\bq)-\cEv_{A(\bq)}(A)(\bp)\bigr).
\end{equation}
In the first summand we have used the permutability of
evolutionary derivations and total derivatives.
The second summand hits the image of $A$ by construction.
Consequently, the
Lie algebra structure $[\,,\,]_A$ on the domain of $A$
equals
\begin{equation}\label{EqOplusBKoszul}
[\bp,\bq]_A=\cEv_{A(\bp)}(\bq)-\cEv_{A(\bq)}(\bp)+ \{\!\{\bp,\bq\}\!\}_A.
\end{equation}
Thus, the bracket $[\,,\,]_A$, which is defined up to $\ker A$,
contains the two standard summands and
the bi\/-\/differential skew\/-\/symmetric part
$\{\!\{\,,\,\}\!\}_A\in\CDiff\bigl(\Gamma\Omega(\xi_\pi)\times\Gamma\Omega(\xi_\pi)
\to\Gamma\Omega(\xi_\pi)\bigr)$.

For any $\bp,\bq,\br
\in\Gamma\Omega(\xi_\pi)$,
the Jacobi identity for \eqref{EqOplusBKoszul} is
\begin{align}
0&=\sum\nolimits_\circlearrowright \bigl[ [\bp,\bq]_A,\br\bigr]_A =
 \sum\nolimits_\circlearrowright
  \bigl[ \cEv_{A(\bp)}(\bq)-\cEv_{A(\bq)}(\bp)
  +\ib{\bp}{\bq}{A}, \br\bigr]_A \notag\\

{}&=\sum\nolimits_\circlearrowright\Bigl\{
\cEv_{A\bigl(\cEv_{A(\bp)}(\bq)-\cEv_{A(\bq)}(\bp)\bigr)}(\br)
-\cEv_{A(\br)}\bigl(\cEv_{A(\bp)}(\bq)-\cEv_{A(\bq)}(\bp)\bigr)
  \notag \\
{}&\qquad{}
+\ib{\cEv_{A(\bp)}(\bq)-\cEv_{A(\bq)}(\bp)}{\br}{A}\notag\\
{}&\qquad{}
 +\cEv_{A\bigl(\ib{\bp}{\bq}{A}\bigr)}(\br)
 -\underline{\cEv_{A(\br)}\bigl(\ib{\bp}{\bq}{A}\bigr)}
 +\ib{\ib{\bp}{\bq}{A}}{\br}{A}\Bigr\}. \label{JacobiKoszul}
\end{align}
The underlined summand contains derivations of the coefficients of
$\ib{\,}{\,}{A}$, which belong to $\cF(\pi)$.
Even if the action of evolutionary fields
$\cEv_\vph^{(u)}=\vph\tfrac{\dd}{\dd u}+\ldots$ on
the the arguments of $A$

is set to zero (which makes sense, see below),
these summands may not vanish.

Note that the Jacobi identity for $[\,,\,]_A$ then amounts to

\begin{equation}\tag{\ref{JacobiKoszul}${}'$}\label{JacobiKoszulShort}
\sum\nolimits_{\circlearrowright}\Bigl(
-\cEv^{(\bu)}_{A(\br)}\bigl(\ib{\bp}{\bq}{A}\bigr)
+\ib{\ib{\bp}{\bq}{A}}{\br}{A}\Bigr)=0.
\end{equation}

Formulas \eqref{EqDefFrob} and \eqref{EqOplusBKoszul},
formula \eqref{JacobiKoszul}, and \eqref{JacobiKoszulShort} will play the same role in the proof of Theorem \ref{ThEvHomVF} in the next section as Lemma \ref{PropMorphismFollows} and, respectively, formulas \eqref{JacobiUsual} and \eqref{LeibnizInAlg} [or \eqref{BeforeTriple}] did for the equivalence of the two classical definitions.

\section{Homological evolutionary vector fields}\label{SecVarQ}\noindent
In this section we represent variational Lie algebroids using the homological evolutionary vector fields $Q$ on the infinite jet super\/-\/bundles that we naturally associate with Definition \ref{DefFrob}.
To achieve this goal, we make two preliminary steps. Namely, we first identify the odd \textsl{neighbour}\footnote{By Yu. I. Manin's definition, the \textsl{neighbours} of a Lie algebra $\gothg$ are $\gothg^*$,\ $\Pi\gothg$, and $\Pi\gothg^*$, where $\Pi$ is the parity reversion functor,
c.\,f. \cite{Voronov}.

}
$\overline{J^\infty}(\Pi\xi_\pi)$ of the infinite horizontal jet bundle $\overline{J^\infty}(\xi_\pi)$ over $J^\infty(\pi)$. Second, we impose the conditions upon the variational anchors $A$ such that the resulting homological field $Q$ becomes well defined
on $\overline{J^\infty}(\Pi\xi_\pi)$;
here we use the method which we already applied in \cite{SymToda}.

We recall that, by construction, the bundle $\xi_\pi\colon\Omega(\xi_\pi)\to J^\infty(\pi)$ is induced by $\pi_\infty$ from the vector
bundle $\xi$ over the base $\varSigma^n$.

Let us consider the horizontal infinite jet bundle $\overline{J^\infty}(\xi_\pi)$ for the vector bundle $\pi_\infty^*(\xi)$ over $J^\infty(\pi)$,
see \cite{Topical,Lstar,JKGolovko2008} for more details and examples.

Finally let us reverse the parity of the fibres in the bundle $\overline{J^\infty}(\xi_\pi)\to J^\infty(\pi)$,
we thus denote by $\Pi\colon\bp\mapsto\boldb$
the parity reversion; right below we recall what the variables
$\bp$ and $\boldb$~are.
This yields the horizontal infinite jet \textsl{super}\/-\/bundle, which we denote by $\overline{J^\infty}(\Pi\xi_\pi)$, over $J^\infty(\pi)$.

We now summarize the notation:\label{PList}
\begin{itemize}
\item $x$ is the $n$-\/tuple of local coordinate(s) on the base manifold $\varSigma^n$;
\item $u_{\boldsymbol{\sigma}}$, where $|\boldsymbol{\sigma}|\geq0$, are the jet coordinates in a fibre of the bundle $\pi_\infty\colon J^\infty(\pi)\to \varSigma^n$;

\item $\bp$ is the fibre coordinate in
the induced
bundle $\pi_\infty^*(\xi)$ over $J^\infty(\pi)$,

and

$\bp$,\ $\bp_x$,\ $\ldots$,\ $\bp_{\boldsymbol{\tau}}$
with any multi\/-\/index $\boldsymbol{\tau}$
are the fibre coordinates in the horizontal infinite jet bundle
$\overline{J^\infty}(\xi_\pi)$ over it;
\item $\boldb$,\ $\boldb_x$,\ $\ldots$,\ $\boldb_{\boldsymbol{\tau}}$,
where $|\boldsymbol{\tau}|\geq0$,
are the odd fibre coordinates in the horizontal infinite jet super\/-\/bundle $\overline{J^\infty}(\Pi\xi_\pi)\to J^\infty(\pi)$;

\item the variational anchor $A$, being a
linear differential operator in total derivatives,
is a fiberwise linear function on
either $\overline{J^\infty}(\xi_\pi)$ or $\overline{J^\infty}(\Pi\xi_\pi)$ and takes values in $\varkappa(\pi)$, which is the space of generating
sections for evolutionary vector fields on $J^\infty(\pi)$;
\item $[\,,\,]_A$ is the bracket \eqref{EqOplusBKoszul} on $\Gamma\Omega(\xi_\pi)$.

It is defined up to the kernel of $A$;
we now take \textsl{any} representative of
the bi\/-\/differential operation $\ib{\,}{\,}{A}$ from
the equivalence class and tautologically extend it to the
bi\/-\/linear function on the horizontal jet space.
\end{itemize}

Over each point $\theta^\infty\in J^\infty(\pi)$, the odd fibre of the horizontal infinite jet super\/-\/bundle $\overline{J^\infty}(\Pi\xi_\pi)$
contains the linear subspace $\ker A$ that is determined by the (infinite prolongation of the) linear equation $A{\bigr|}_{\theta^\infty}(\boldb)=0$.
It is important that the linear space of its solutions does not depend on the point $[\boldb]$ in the fibre, because the differential operator $A{\bigr|}_{\theta^\infty}$ depends only on the base point $\theta^\infty\in J^\infty(\pi)$ and the equation is linear. To eliminate the functional freedom in the kernel of $A$,
we require that the operator $A$ be \textsl{nondegenerate} in the sense
of \cite{SymToda}: $A\circ\nabla=0$ implies $\nabla=0$.

\begin{theor}\label{ThEvHomVF}
Let $A$ be a nondegenerate variational anchor and all above assumptions and notation hold.

Whatever be the choice of the bracket $\ib{\,}{\,}{A}\mod\ker A$,
the following odd evolutionary vector field on the horizontal infinite jet super\/-\/space $\overline{J^\infty}(\Pi\xi_\pi)$ is homological\textup{:}
\begin{equation}\label{VarHomVF}
Q=\cEv^{(\bu)}_{A(\boldb)}
 -\tfrac{1}{2}\cEv^{(\boldb)}_{\ib{\bp}{\bq}{A} {\bigr|}_{
\substack{
  \bp\mathrel{{:}{=}}\boldb\\
  \bq\mathrel{{:}{=}}\boldb
}
}
},
\qquad [Q,Q]=2Q^2=0.

\end{equation}
Moreover, either there is a unique canonical representative
in the equivalence class $Q\mod\cEv^{(\boldb)}_{\bn}$
of such fields in the fibre
of $\overline{J^\infty}(\Pi\xi_\pi)\to J^\infty(\pi)$
over each point $\theta^\infty\in J^\infty(\pi)$,
here $\bn\in\ker A$ so that $\cEv^{(\boldb)}_{\bn}$ is
a symmetry of the kernel,
or, when the anchor $A$ is close to degenerate in the sense below, the uniqueness is achieved by fixing $\ib{\,}{\,}{A}$ on \textup{(}at most\textup{)} the boundary of a star\/-\/shaped domain centered at the origin of the fibre over $\theta^\infty$ in some finite\/-\/order horizontal jet space $\overline{J^{|\boldsymbol{\tau}|}}(\Pi\xi_\pi)$, $|\boldsymbol{\tau}|<\infty$.

\end{theor}

\begin{example}
The second Hamiltonian operator $A_2$ for the KdV equation (see above)

determines the homological vector field

$Q=\cEv_{A_2(b)}^{(u)}+\cEv^{(b)}_{b b_x}$.

This is the simplest example that involves a nontrivial differential polynomial of degree two in the anti\/-\/commuting variable(s) $\boldb$ and its derivatives.

\end{example}

\begin{proof}[Proof of Theorem \protect{\textup{\ref{ThEvHomVF}}}]
The anticommutator $[Q,Q]=2Q^2$ of the odd evolutionary vector field is an evolutionary vector field itself. Hence it suffices to find only the coefficients of $\dd/\dd\bu$ and $\dd/\dd\boldb$ in it; by construction, all other terms are the respective total derivatives of these two.

We first consider the odd velocity of $\bu$; it equals
\begin{align*}
\cEv^{(\bu)}_{A(\boldb)}(A)(\boldb)&-\tfrac{1}{2}A\bigl(
\ib{\bp}{\bq}{A}\bigr) {\bigr|}_{
\substack{
  \bp\mathrel{{:}{=}}\boldb\\
  \bq\mathrel{{:}{=}}\boldb
}
}.
\\
\intertext{Since $\boldb$ is odd, we double the minuend as follows:}
{}={}&\frac{1}{2}\Bigl(
\cEv^{(\bu)}_{A(\bp)}(A)(\bq)
-\cEv^{(\bu)}_{A(\bq)}(A)(\bp)
-A\bigl(\ib{\bp}{\bq}{A}\bigr)
\Bigr) {\Bigr|}_{
\substack{
  \bp\mathrel{{:}{=}}\boldb\\
  \bq\mathrel{{:}{=}}\boldb
}
}.
\end{align*}
By the definition of $\ib{\,}{\,}{A}$ in (\ref{LeibnizForOp}--\ref{EqOplusBKoszul}), the expression in parentheses equals zero.

Second, the coefficient of $\dd/\dd\boldb$ is equal to
\begin{align*}
-\tfrac{1}{2}&\cEv^{(\bu)}_{A(\boldb)}\bigl(\ib{\bp}{\bq}{A}
\bigr) {\bigr|}_{
\substack{
  \bp\mathrel{{:}{=}}\boldb\\
  \bq\mathrel{{:}{=}}\boldb
}
}
+\tfrac{1}{4}\cEv^{(\boldb)}_{\ib{\bp}{\bq}{A} {\bigr|}_{
\substack{
  \bp\mathrel{{:}{=}}\boldb\\
  \bq\mathrel{{:}{=}}\boldb
}
}
}
\bigl(\ib{\br}{\bs}{A}\bigr) {\bigr|}_{
\substack{
  \br\mathrel{{:}{=}}\boldb\\
  \bs\mathrel{{:}{=}}\boldb
}
}.\\
\intertext{The bracket $\ib{\,}{\,}{A}$ is skew\/-\/symmetric (in the usual sense);
therefore, the substitution of an odd $\boldb$ for both arguments doubles it.
Besides, the odd derivation $\cEv^{(\boldb)}_{\ib{\bp}{\bq}{A}}$,

with $\boldb$ for $\bp$ and $\bq$,
acts on its argument by the graded Leibniz rule. By construction, it replaces every instance of a derivative of $\boldb$ with the respective derivative of $\ib{\boldb}{\boldb}{A}$. This yields}
{}={}&\frac{1}{2}\Bigl( -\cEv^{(\bu)}_{A(\boldb)}\bigl(\ib{\bp}{\bq}{A}\bigr)
+\ib{\ib{\bp}{\bq}{A}}{\br}{A} \Bigr) {\Bigr|}_{
\substack{
  \bp\mathrel{{:}{=}}\boldb\\
  \bq\mathrel{{:}{=}}\boldb\\
  \br\mathrel{{:}{=}}\boldb
}
}.
\end{align*}
We now triple this expression by taking the sum over the cyclic permutations
$\bp\to\bq\to\br\to\bp$, which does not contribute with any sign. The result
is one\/-\/sixth of the Jacobi identity \eqref{JacobiKoszulShort}. This proves our first claim.

\smallskip
We have established that $Q^2=0$ for any choice of the bracket
$\ib{\,}{\,}{A}+\bn([\boldb])$, where $\bn\in\ker A$ is arbitrary.
On one hand, the linear subspace $\ker A$ in the horizontal jet fibre over a point $\theta^\infty\in J^\infty(\pi)$ does not depend on a point $[\boldb]$ of that fibre. On the other hand, the representative $\bn\in\ker A$ can be nonzero and, in principle, depend on a point of the fibre. We claim that $\bn$ can be either trivialized or normalized in a well defined way uniformly on the entire fibre.

First, by the appropriate shift along the kernel, let us normalize the bracket
$\ib{\,}{\,}{A}$ by the condition $\ib{[0]}{[0]}{A}\equiv0$
at the origin $[\boldb]=0$ of the fibre over $\theta^\infty\in J^\infty(\pi)$
in the vector bundle $\overline{J^\infty}(\Pi\xi_\pi)\to J^\infty(\pi)$.
Now we notice that the remaining addend $\bn$, which, at all fibre points, belongs to
$\ker A$ that is common for them, is quadratic homogeneous in the horizontal jet coordinates $[\boldb]$:
we have $\bn=\bn\bigl([\boldb]\cdot[\boldb]\bigr)$.
The nondegeneracy assumption for the operator $A$ excludes the possibility of a free functional dependence on $[\boldb]\cdot[\boldb]$ for sections which belong to the domain of a nonzero operator $\nabla$ such that $A\circ\nabla\equiv0$.

Obviously, the differential order of the bracket $\ib{\,}{\,}{A}$
is finite and does not exceed the sum of differential orders of the operator $A$ and of its coefficients, which may depend on $[\bu]$. This implies that
$\bn=\bn\bigl((\boldb,\ldots,\boldb_{\boldsymbol{\tau}})\cdot
(\boldb,\ldots,\boldb_{\boldsymbol{\tau}})\bigr)$
with $|\boldsymbol{\tau}|<\infty$.
For the odd $\boldb$,\ $\ldots$,\ $\boldb_{\boldsymbol{\tau}}$, take
all pairwise products $\psi={}^{\textrm{t}}\bigl(
b^i\cdot b^j$,\ $\ldots$,\ $b_{\boldsymbol{\tau}}^i\cdot b_{\boldsymbol{\tau}}^j
\bigr)$ that are not identically zero,
and construct the $m$-\/tuple $\bn=\nabla(\psi)$ using the undetermined coefficients: let $\nabla^\alpha_\beta\in\BBR$ and $\bn^\alpha=\nabla^\alpha_\beta\cdot\psi^\beta$. The hypothesis $\bn\neq0$ formally does not contradict our initial assumption of $A$ being nondegenerate,

because the section $\psi$ thus obtained is not arbitrary (there may be fewer free parameters than the number of components in it due to the presence of polynomial relations between the components).
If there are no such nontrivial $\nabla$, the proof is over.

However, suppose there is a solution $\nabla\neq0$.
Then, under the rescaling by $\lambda\in\BBR$ in the fibre, the deviation
$\bn(\lambda)\in\ker A$ from the normalization $\bn\equiv0$ at the origin
grows quadratically in $\lambda$.
Therefore, to finally fix the bracket $\ib{\,}{\,}{A}$,
it is sufficient to assign the values to $\bn$ at the points
on the boundary of a central\/-\/symmetric star\/-\/shaped domain, centered around the origin in the finite\/-\/dimensional fibre over $\theta^\infty$
of the finite
jet space $\overline{J^{|\boldsymbol{\tau}|}}(\Pi\xi_\pi)$.

By this argument, we normalize $\ib{\,}{\,}{A}$ on the entire fibre
of $\overline{J^\infty}(\Pi\xi_\pi)$.

This defines the canonical representative $Q$ in the class \eqref{VarHomVF}
on this fibre.
The proof is complete.
\end{proof}

\begin{rem}
In the beginning of the proof, we used the postulated property of the variational anchor $A$ to be the Lie algebra homomorphism. The next part of the proof holds due to the skew\/-\/symmetry of $\ib{\,}{\,}{A}$, the universal Leibniz rule for derivations, and the Jacobi identity for $[\,,\,]_A$.
In a sense, it is possible to regard the summands in \eqref{JacobiKoszulShort} as the right\/-\/hand side of a \textsl{new} `Leibniz rule' for the bracket \eqref{EqOplusBKoszul} in the variational Lie algebroid.
\end{rem}

\begin{rem}
The homological evolutionary vector field \eqref{VarHomVF}
is quadratic in the odd variable $\boldb$ or its derivatives,
and contains no free term. Suppose, however, that this degree is arbitrary positive (possibly, infinite) and all other assumptions still hold.
This yields the Schlessinger\/--\/Stasheff
homotopy deformation \cite{Stasheff}
of the Lie algebra structure in $\bigl(\Gamma\Omega(\xi_\pi),[\,,\,]_A\bigr)$. Namely, the Jacobi identity $[Q,Q]=0$ involves now not only the binary bracket $[\,,\,]_A$, but also ternary, quadruple, and other $N$-\/ary
skew\/-\/symmetric brackets with the numbers of arguments $N\geq3$,
see \cite{AKZS,Voronov,Stasheff,ForKac} and references therein.

The non\/-\/existence of the variational $N$-\/vectors at $N\geq3$ for all evolutionary systems with invertible symbols\footnote{It is readily seen that
this assumption depends on the choice of local coordinates.
We conjecture that the same assertion holds for higher\/-\/order evolutionary systems with the nondegenerate linearizations of their right\/-\/hand sides.}
of differential order greater than one
was argued in \cite{JKVerb3Vect}, see references therein. This now states the rigidity, in the homotopy sense, of the variational Poisson algebroid structures determined by Hamiltonian operators for such equations.
However,
the variational Poisson algebroids for equations that can not be
cast to the above form (e.g., for gauge systems) may admit homotopy deformations.
\end{rem}

\paragraph*
{Variational master equation: an illustration.}
Let us study in more detail the properties of the homological vector
field \eqref{VarHomVF} if $A\colon\widehat{\varkappa}(\pi)\to\varkappa(\pi)$ is a Hamiltonian differential operator and thus its arguments $\bp$
are the \textsl{variational covectors}.\footnote{By definition, the variational covectors are homomorphisms from the space $\varkappa(\pi)=\Gamma\bigl(\pi_\infty^*(\pi)\bigr)$ of generating sections $\vph$ for evolutionary derivations $\cEv_\vph$ to the space of $n$-th horizontal differential forms on $J^\infty(\pi)$ fibred by $\pi_\infty$ over $\varSigma^n$. This implicitly requires to choose the volume form $\Id\boldsymbol{x}$ on the base and thus fix the coupling $\langle\,,\,\rangle$ between the variational covectors and evolutionary vector fields. In these terms, the homological vector field $Q$ for a Hamiltonian operator $A$ is a BV\/-\/field and not just a BRST\/-\/field, see \cite{AKZS}.}
Under this assumption,
the induced velocity $\dot{\bp}$ is known (see formula \eqref{EqDogma} below) whenever the evolution $\dot{u}=A(\bp)$ is given.

The evolutionary vector field \eqref{VarHomVF} on the horizontal jet super\/-\/bundle $\overline{J^\infty}(\Pi\widehat{\pi}_\pi)$ was recently obtained in \cite{JKGolovko2008} for Hamiltonian operators $A$ from exactly this initial standpoint:
the coefficient of $\dd/\dd\boldb$ was \textsl{postulated} once and forever by formula \eqref{EqDogma} and was not regarded as the bracket $\ib{\boldb}{\boldb}{A}$, whereas the quadratic equation $Q^2=0$ upon the skew\/-\/adjoint operators $A$ then yields variational Poisson structures.

In the variational Poisson case, the homological vector field \eqref{VarHomVF} can indeed be calculated explicitly

for every Hamiltonian operator $A$. Moreover, this field $Q$ is itself Hamiltonian with respect to the variational Poisson bi\/-\/vector $\bH=\tfrac{1}{2}A(\boldb)\cdot\boldb$ and the canonical symplectic structure \cite{KuperCotangent,DeDonderWeyl}
on the horizontal infinite jet super\/-\/space $\overline{J^\infty}(\Pi\widehat{\pi}_\pi)$.
We illustrate this standard algebraic fact by using techniques
from \cite{Topical}.

The bracket $\ib{\,}{\,}{A}$ for Hamiltonian operators $A$ emerges

from the Jacobi identity
for the Lie algebra $\bigl(\bar{H}^n(\pi),\{\,,\,\}_A\bigr)$
of the Hamiltonian functionals endowed by $A$ with the Poisson bracket.

Following the notation of the book~\cite{Opava}, we put

$
\ell^{(\bu)}_{A,\psi}(\vph)\mathrel{{:}{=}}\bigl(\cEv_{\vph}(A)\bigr)(\psi)
$
for any $\vph\in\varkappa(\pi)$,
$\psi\in\widehat{\varkappa}(\pi)$,

and a total differential operator $A\in\CDiff(\widehat{\varkappa}(\pi),
\varkappa(\pi))$.
We note that $\ell^{(\bu)}_{A,\psi}$ is an operator in total derivatives w.r.t.\ its
argument $\vph$ and w.r.t.\ $\psi$ (but not w.r.t.\ the coefficients of $A$),
and hence the adjoint $\bigl(\ell^{(\bu)}_{A,\psi}\bigr)^\dagger$ is well defined.

\begin{lemma}[\cite{Opava}, see a proof in \cite{d3Bous}]\label{AuxLemma}

A
skew\/-\/adjoint ope\-ra\-tor
$A$
is Hamiltonian if and only if the relation

$\bigl(\cEv_{A(\bp)}(A)\bigr)(\bq)-
  \bigl(\cEv_{A(\bq)}(A)\bigr)(\bp)=

A\bigl(\bigl(\ell_{A,\bp}^{(\bu)}\bigr)^\dagger(\bq)\bigr)$

holds for all $\bp,\bq\in\widehat{\varkappa}(\pi)$.

Consequently,
\begin{equation}\label{EqDogma}
\ib{\bp}{\bq}{A}=\left(\ell^{(\bu)}_{A,\bp}\right)^\dagger(\bq),\qquad \bp,\bq\in\widehat{\varkappa}(\pi),
\end{equation}
for a Hamiltonian operator $A$;

on the other hand,
this yields a
convenient criterion under which total differential operators are Hamiltonian.
\end{lemma}

We emphasize that the notation $\ell^{(\bu)}_{A,\bp}$ is \textsl{not}
the same as the linearization $\ell^{(u)}_{A(\bp)}$.
However, if $\bp$ is the fibre coordinate in the horizontal jet bundle

over $J^\infty(\pi)$,

then the notation becomes synonymic (e.g., see \cite{JKGolovko2008}).

Moreover, it is readily seen that\footnote{\label{VarS}
By the convention
$\delta S=\frac{\delta S}{\delta\eta}\delta\eta$, the variations $\delta\eta$ of (possibly, odd) variables $\eta$ are pushed through to the right.
This is rather unfortunate because it anti\/-\/correlates with Dirac's convention of bra-{} covectors standing on left and {}-ket vectors on right in any formula (e.g.\ in every ``bra-c-ket'')
and brings additional signs of a hardly explicable origin into the formulas.}
\[
\tfrac{1}{2}\left(\ell^{(\bu)}_{A,\boldb}\right)^\dagger(\boldb)=
 \tfrac{\delta}{\delta\bu}\bigl(\tfrac{1}{2}A(\boldb)\cdot\boldb\bigr),
\qquad
A(\boldb)=\tfrac{\delta}{\delta\boldb}\bigl(\tfrac{1}{2}A(\boldb)\cdot\boldb\bigr);
\]
the second identity holds because the Hamiltonian operator $A$ is skew\/-\/adjoint. We finally recall from \cite{KuperCotangent,DeDonderWeyl}
that the odd neighbour $\overline{J^\infty}(\Pi\widehat{\pi}_\pi)\to J^\infty(\pi)$
of the variational cotangent bundle is endowed with the canonical symplectic structure
$\boldsymbol{\omega}=\left(\begin{smallmatrix}\phantom{+}0 & 1\\ -1 & 0\end{smallmatrix}\right)$. Let us remember that the structure $\boldsymbol{\omega}$ is present even if the target space for local
sections of the bundle $\pi$ or, more generally, the target manifold for the mappings $\bu\colon\varSigma^n\to M^m$ is not symplectic. This is in contrast with \cite{AKZS} where a symplectic structure was induced on
the space of mappings ${\EuScript{E}}=\{\varSigma^n\to M^m\}$ from
the two\/-\/form on $M^m$, i.e., by using the requirement that the even\/-\/dimensional target manifold be symplectic.

Let us summarize the result. Now on the variational level,
it goes in parallel with the properties of the homological vector fields $Q$
within the finite\/-\/dimensional picture
of \cite{AKZS}.

\begin{state}
For every Hamiltonian differential operator $A$, the homological evolutionary vector field \eqref{VarHomVF} is itself Hamiltonian with respect to the variational Poisson bi\/-\/vector $\bH=\tfrac{1}{2}A(\boldb)\cdot\boldb$ and the canonical symplectic structure $\boldsymbol{\omega}$:
\begin{equation}\label{VarHomVFPoisson}
Q=\cEv^{(\bu)}_{\tfrac{\delta}{\delta\boldb}(\bH)}
+\cEv^{(\boldb)}_{-\tfrac{\delta}{\delta\bu}(\bH)}.
\end{equation}
The Lagrangian $\bL=\dot{\bu}\bp-\bH$, obtained from the Hamiltonian $\bH$ by the Legendre transform, equals $\bL=\tfrac{1}{2}A(\boldb)\cdot\boldb$
and satisfies the variational master equation $\lshad\bL,\bL\rshad=0$,
where $\lshad\,,\,\rshad$ is the variational Schouten bracket.
\end{state}

\begin{example}
Consider the Hamiltonian operator
$A_2=-\tfrac{1}{2}\,\tfrac{\Id^3}{\Id x^3}+
w\,\tfrac{\Id}{\Id x}+\tfrac{\Id}{\Id x}\circ w$
for the KdV equation; the bracket \eqref{EqDogma} is
$\ib{p}{q}{A_2}=p_xq-pq_x$, see \eqref{BrA2}.
The variational Poisson bi\/-\/vector is
$\bH=-\tfrac{1}{4}b_{xxx}b+wb_xb$, whence (see footnote \ref{VarS})
\[

{\delta\bH}/{\delta b}=-\tfrac{1}{2}b_{xxx}+2wb_x+w_xb
\quad \text{and}\quad

{\delta\bH}/{\delta w}=b_xb.
\]
The homological evolutionary vector field \eqref{VarHomVF} equals
$Q=\cEv^{(w)}_{A_2(b)}-\tfrac{1}{2}\cEv^{(b)}_{2b_xb}$,
which is precisely \eqref{VarHomVFPoisson}.
\end{example}

Let us finally remark that
the explicitly given\footnote{The Lagrangian $\bL$
and the Hamiltonian $\bH$ coincide,
whence the variational Poisson formalism can be viewed as a
very remote part of the theory of uniform rectilinear motion.}
Hamiltonian $\bH$ of $Q$ has nothing to do with a Hamiltonian $\cH$ of the actual motion $\dot{\bu}=A(\delta\cH/\delta\bu)$
of the image of $\varSigma^n$ in $M^m$ because
$\cH$ may not even exist\,!

\paragraph*
{Conclusion.}\noindent
We observe that the physically motivated geometry of \cite{AKZS} is manifestly \textsl{variational}, that is, the model of strings $\varSigma^n\to M^m$
requires us to use jet bundles that encode the information about the mappings together with all their derivatives. It is possible to introduce the constructions such as the
Poisson structures or homological
vector fields of zero differential order along $\varSigma^n$,
but, clearly, they neglect the full setup.
Physically speaking, the equality of all derivatives to zero means that either $n=0$ and $\varSigma=\{\text{pt}\}$ (hence there are no derivatives at all) or the image of $\varSigma^n$ under a constant mapping is a point in $M^m$ so that the point particle has $n$ hidden
dimensions.
In this paper, we introduced the definitions and established their properties along the lines of \cite{Vaintrob}, and thus we removed the limitation of \cite{AKZS}
for the particles to be only points. Furthermore, we notice that yet one more assumption of \cite{AKZS}, on the presence of a symplectic structure on the target manifold $M^m$, is excessive for the homological evolutionary BV-\/field to be Hamiltonian

with respect to
the canonical symplectic structure on the space of mappings
(although, of course, variational methods remain applicable to the Poisson sigma\/-\/models).

We hope that this will help in the study of the dimensional
reduction\/-\/free models for strings in Minkowski space\/-\/time $M^{3,1}$.

\paragraph*
{Acknowledgements.}

The authors thank
B.\,A.\,Dub\-ro\-vin, I.\,S.\,Kra\-sil'\-sh\-chik,
M.\,A.\,Ne\-ste\-ren\-ko, P. J. Olver, V. N. Roub\-tsov,
and V.\,V.\,So\-ko\-lov
for the discussions and remarks.
One of the authors (A.~K.) is grateful to the organizing committee of the International Workshop ``Nonlinear Physics: Theory and Experiment VI'' for the warm hospitality.
A.\,K.\ thanks the
$\smash{\mbox{IH\'ES}}$, SISSA, and CRM ($\text{\smash{Montr\'eal}}$)
for financial support and
hospitality.

This work was supported in part by the European Union
(FP6 Marie Curie RTN \emph{ENIGMA} Contract
No.\,MRTN-CT-2004-5652), the European Science Foundation Program
{MISGAM}, and the NWO (Grant Nos.\ B61--609 and VENI 639.031.623).

config: arxiv


TITLE: What is the Katz-Sarnak philosophy?
QUESTION [30 upvotes]: It has been recently mentioned by a speaker (his talk is completely not relevant to random matrix theory/RMT though) that modern statistics, especially random matrices theory, will help solving some number theoretic problems.
I was quite intrigued and ask for more explanation but the speaker himself said he did not understand the philosophy well enough but point me to a few keywords to search for, one among which is "Katz-Sarnak philosophy".
After a few searches, I figured out that all sources seem to point to [KS]. The major results in [KS] seems like saying that the class of general linear (compact) groups have the same n-level correlations. While some researchers [Kowalski][Miller] do mention that they applied KS philosophy, but what they have done seems very different...So for number theorists and experts on elliptic curves:

(1) When you refer to "Katz-Sarnak philosophy", what kind of
  thinking/technique do you actually mean?
(2) Is there a formalism/explanation of this philosophy in language of
  RMT? I asked this because it might be helpful to understand it in this
  perspective (at least to a probabilist).

Any inputs are highly appreciated.
Reference
[KS]Katz, Nicholas M., and Peter Sarnak, eds. Random matrices, Frobenius eigenvalues, and monodromy. Vol. 45. American Mathematical Soc., 1999.Google books
[Kowalski]http://blogs.ethz.ch/kowalski/2008/07/30/finding-life-beyond-the-central-limit-theorem/
[Miller]Miller, Steven J. "One-and two-level densities for rational families of elliptic curves: evidence for the underlying group symmetries." Compositio Mathematica 140.4 (2004): 952-992.

REPLY [18 votes]: I'm going to give an answer that discusses some things that the other answers don't go into as much detail on. In particular let me try to explain why the results you mention on classical groups having the same $n$-level correlations are in fact relevant to the number theory situation. It will build off KConrad's answer in some ways.
The Katz-Sarnak book proves theorems about zeroes of $L$-functions in function field setting. These are defined by exact analogues of the definitions of $L$-functions, over another field. Of course, as KConrad points out, these aren't the same as $L$-functions over a number field, and many of the families considered in the Katz-Sarnak book don't match the families of interest over number fields. But some match, and this has been rectified in later work.
One key advantage of the function field setting is that the situation is conceptually clearer. The $L$-function is a polynomial, and by the Grothendieck-Lefschetz trace formula it is the characteristic polynomial of some naturally defined matrix (the action of a Frobenius element on a Tate module or etale cohomology group). So we have no difficulty interpreting the zeroes as eigenvalues of a matrix. Moreover, this matrix lies in some matrix group, the monodromy group of the family.
These matrices are over an $\ell$-adic field and so can't naturally be viewed as unitary matrices. But Deligne (with Weil first in some cases) showed that if we base change to the complex numbers, they are conjugate to unitary matrices. So in fact the zeroes of the $L$-function are eigenvalues of elements of the maximal compact subgroup of some algebraic matrix group. So in this setting, the question of why we should expect the zeroes to behave like eigenvalues of random unitary matrices is not mysterious at all - it's the simplest possible behavior they could have, given that they are eigenvalues of unitary matrices.
In fact Deligne did more and showed that these Frobenius conjugacy classes are equidistributed in the maximal compact subgroup of the monodromy group. In other words, the statistics of any continuous function of the set of zeroes at all matches the statistics for random matrices in some compact group. Such a powerful result doesn't come without a cost, and it has two. First, we may not know what the group is. Second, this statistical equivalence only shows up in the limit as we take the finite field size to infinity. In this limit, the number of zeroes is fixed, so we can take the same compact group each time. However, in the number field settings, people always consider limits where some parameter controlling the number of zeroes grows.
The first cost was dealt with by Katz-Sarnak by computing the group in some important cases. They found that, while a priori it could be any group at all, in fact it was usually a classical group (which is something Katz had already noticed in some of his earlier investigations). So they focused attention on the maximal compact subgroups of classical groups.
The second was and is more difficult to deal with. However, since the transfer from the function field case to the number field case is conjectural anyways, one merely has to guess how to transfer statistics from finitely many zeroes to infinitely many zeroes, and then perform the calculations to do the transfer. They guessed that taking the limit of distribution of the smallest $k$ eigenvalues in a family of matrices in larger and larger groups would match the distribution of the smallest $k$ zeroes in $L$-functions over number fields. The first few chapters of the Katz-Sarnak book are then devoted to calculating these limits, which is the part that looks like honest random matrix theory.
So Katz and Sarnak combined algebraic geometry - the results of Deligne and the monodromy group calculations - with random matrix theory - these calculations in the limit as the size of the classical group grows - to obtain precise predictions for the statistics of zeroes of $L$-functions over number fields. These predictions have then been expanded and refined in later work.
I guess this is all a bit removed from your motivating question, which is how that random matrix theory can help in number theory. Unfortunately I don't know how much cutting edge research in random matrix theory is really needed, because we know of no rigorous relationship between $L$-functions over number fields and any kind of random matrix, so thus far random matrix theory has mostly been used to give predictions, and the predictions we already have are hard enough to try to prove that any property of random matrices found by more powerful techniques might be even further out of reach over number fields. However, on a positive note, I would suggest you check out Terry Tao's latest blog post, which is on a question involving random matrices, still unsolved, motivated by number theory (in fact, possibly motivated by a joke I made in a talk?).


TITLE: Fermat's Little theorem (Num Theory)
QUESTION [0 upvotes]: How can I compute $31^{1209}\equiv \mod (101)$ using Fermat's Little Theorem?

REPLY [3 votes]: Hint: $31^{1209} = 31^{1200}\cdot 31^{9}=(31^{100})^{12}\cdot 31^9$
What does Fermat's Little theorem say about $31^{100}\pmod{101}$?