TITLE: Deforming contour in complex plane integration and wave equation
QUESTION [2 upvotes]: I'm studying d'Alembert's equation with scalar field in a homogeneous medium in time and space. In particular the calculation of the Green function, that is the field irradiated by an impulsive source in space and time:
$$
\left(\nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right) g(\mathbf{R},\tau)= \delta(\mathbf{R}) \delta(\tau)
$$
Obviously, uniqueness also requires the initial conditions on the function and the first derivative.
For the resolution, the book does the space-time Fourier transform of the equation, obtaining:
$$
[-K^2 + k^2] \tilde{G}(\mathbf{K},\omega)=1
$$
where $K^2=\mathbf{K}\cdot \mathbf{K}$, $k=\omega/c$.
Then it does the antitransformation, getting
$$
g(\mathbf{R},\tau)= \frac{1}{(2\pi)^4} \int_{-\infty}^{+\infty} d\omega \int d\mathbf{K} \frac{i(\mathbf{K}\cdot\mathbf{R}- \omega \tau)} {-K^2+k^2}
$$
Now, the text notes that, contrary to appearances, the latter expression does not unambiguously describe the green function, since we have not used the initial conditions anywhere. This is due to the improper nature of the integral for the presence of the poles $k=\omega/c=\pm K$. To give meaning to the integral it is necessary to deform the integration path to avoid the poles.
Then, without demonstrating how to get there, it verifies that, choosing as integration path not the real axis but a straight line parallel to it with positive imaginary part, the integral, solved by exploiting the residual theorem and the jordan lemmas, is a causal green function, that is with null initial conditions.
Could you explain why we can deform contour integration and why it corresponds to change initial conditions?
I probably have complex analysis gaps that prevent me from understanding. What is the theory necessary to understand and from where to study it?
Thank you

REPLY [1 votes]: We must view $\omega$ as a complex variable and the integrand as a meromorphic function of a complex variable, with simple poles at $\omega_{\pm} = \pm cK$.
\begin{equation}
\int_{-\infty}^{\infty}\frac{c^2e^{it\omega}}{(\omega - cK)(\omega + cK)}d\omega
\end{equation}

There are two distinct parts to the integral along $\mathbb{R}$:

the principal value ($\mathsf{PV}$) of the integral is what we get when we integrate along the real line, except for the radius-$\epsilon$ semicircles, and then let $\epsilon$ shrink to 0;
the numbers we pick up from integrating along the radius-$\epsilon$ semicircles and letting $\epsilon$ shrink to 0.

\begin{equation}
\begin{split}
&~\oint_{\gamma}\frac{c^2e^{itz}}{(z - cK)(z + cK)}dz\\
=&~ \textsf{PV}\int_{-\infty}^{\infty}\frac{c^2e^{it\omega}}{(\omega - cK)(\omega + cK)}d\omega \pm i\pi\frac{e^{itcK}}{2cK} \pm i\pi\frac{e^{-itcK}}{-2cK} + 0,
\end{split}
\end{equation}
where $\gamma$ is the limit curve of curves that that lie along the real line from $-R$ to $R$ but make small semicircles above or below the singularities, and then go along a large arc from $R$ back to $-R$. Jordan's Lemma ensures that the limit of the integral along the large arc is 0. The sign of $t$ determines whether one chooses an arc in the upper half-plane or the lower half-plane.
Each sign attached to a latter term depends on whether the semicircle went into the upper half-plane (clockwise around singularity) or into the lower half-plane (counter-clockwise around the singularity).

Moving the line away from the real line gives another expression for the integral and works because integrating around a box containing both $\mathbb{R}$ (deformed to avoid singularities) and $\mathbb{R}\pm i\epsilon$ (in the opposite direction) is 0.
Let $\Gamma$ be a box with two indentations that goes

from $-R$ to $R$ along the real line,
but avoids the singularities via semicircles,
goes from $R$ to $R \pm i\epsilon$,
goes from $R \pm i\epsilon$ to $-R + \pm i\epsilon$,
and then from $-R \pm i\epsilon$ to $-R$.

The integrand has no singularities inside this box, so integrating around it yields 0 by Cauchy's Integral Theorem. As $R\to\infty$, the integrals along vertical edges drop to 0, so the $\mathsf{PV}$ of the integral along the real line (from $-\infty$ to $\infty$) plus the integral along $\mathbb{R} + i\epsilon$ (from $\infty + i\epsilon$ to $-\infty + i\epsilon$) is 0.
This means that the integral along $\mathbb{R} \pm i\epsilon$ is equal to the $\mathsf{PV}$ plus the terms picked up from the semicircles over or under the singularities at $\omega_{\pm} = \pm cK$:
\begin{equation}
\begin{split}
&~\int_{-\infty\pm i\epsilon}^{\infty \pm i\epsilon}\frac{c^2 e^{itz}}{(z - cK)(z+cK)}dz\\
=&~\textsf{PV}\int_{-\infty}^{\infty}\frac{c^2e^{it\omega}}{(\omega - cK)(\omega + cK)}d\omega \pm i\pi\frac{e^{itcK}}{2cK} \pm i\pi\frac{e^{-itcK}}{-2cK}
\end{split}
\end{equation}
This might also be written as
\begin{equation}
\begin{split}
&~\int_{-\infty}^{\infty}\frac{c^2 e^{it(\omega \pm i\epsilon)}}{(\omega \pm i\epsilon - cK)(\omega \pm i\epsilon + cK)}d\omega\\
=&~\textsf{PV}\int_{-\infty}^{\infty}\frac{c^2e^{it\omega}}{(\omega - cK)(\omega + cK)}d\omega \pm i\pi\frac{e^{itcK}}{2cK} \pm i\pi\frac{e^{-itcK}}{-2cK}
\end{split}
\end{equation}

From what I see in a textbook, some authors note that as $\epsilon\to 0$, $f(\omega \pm i\epsilon)\to f(\omega)$ for $f$ continuous at $\omega$. Then they cheat by dropping the $\pm i\epsilon$ in the numerator. In our case, we get the incorrect expression
\begin{equation}
\int_{-\infty}^{\infty}\frac{c^2 e^{it\omega}}{(\omega \pm i\epsilon - cK)(\omega \pm i\epsilon + cK)}d\omega
\end{equation}
because $\pm i\epsilon$ has been removed from the argument of the exponential. This notation gives the reader a warning that there are singularities encountered along $\mathbb{R}$, but the OP's experience shows that it must be used with care or not at all.

For more examples with diagrams, see Principal Value of an Integral in Mathematical Physics by Sadri Hassani.


TITLE: Integral $\int_{1}^{\infty}\frac{-4}{(2 \cos{x} - 2) x^3}\ \mathrm dx.$
QUESTION [0 upvotes]: I'm not sure how to proceed with the following integral:
$$I=\int_{1}^{\infty}\frac{-4}{(2 \cos{x} - 2) x^3}\ \mathrm dx.$$
Mathematica could not find a closed form solution for it and I really have no idea how to go about computing it.

REPLY [0 votes]: $$I=\int_{1}^{\infty}\frac{-4}{(2 \cos{x} - 2) x^3}\ \mathrm dx$$

The integral here is divergent.
In the denominator,  $\cos x -1 = 0$ has zeroes at $x=2n\pi, \,\,\,\, n=0,1,2,..$
So the integrand has singularities at $x=2n\pi, \,\,\,\, n=0,1,2,..$ in the given interval and quite obviously $I \to \infty$.


TITLE: Simultaneous Diophantine approximation: multiple solutions required
QUESTION [2 upvotes]: Suppose we have $n$ linearly independent (over $\mathbb{Q}$) irrational numbers $\{ \alpha_i | 1\leq i \leq n \}$. For the simultaneous Diophantine approximation problem
$$ |q \alpha_i - p_i |  < \epsilon , $$
where $q$ and the $p$'s are all integers, we have the LLL algorithm. The problem is, by this algorithm, for each precision $\epsilon$,  we get only one $q$. But we know that such $q$ have a density of $\epsilon^{n-1} $.
So, how can we get a series of $q$ with the prescribed precision $\epsilon$?

REPLY [3 votes]: One very practical method of finding the $q$'s you are looking for is presented here:
Kovács, Attila, and Norbert Tihanyi. "Efficient computing of n-dimensional simultaneous Diophantine approximation problems." Acta Universitatis Sapientiae, Informatica 5.1 (2013): 16-34.
In what has to be one of the coolest applications of simultaneous Diophantine approximation ever, the authors use their algorithm to find many (but not all) values of the Riemann-Siegel Z function where Z(t) >500.  See: http://riemann-siegel.com/index.php for more details.
Another choice is the method described in:
Bosma, Wieb, and Ionica Smeets. "Finding simultaneous Diophantine approximations with prescribed quality." The Open Book Series 1.1 (2013): 167-185.
From the abstract: "We give an algorithm that finds a sequence of approximations with Dirichlet coefficients bounded by a constant only depending on the dimension. The algorithm uses LLL lattice basis reduction."
Added (3/27):
You may find this paper helpful:
Lagarias, J. C. "Geodesic multidimensional continued fractions." Proceedings of the London Mathematical Society 3.3 (1994): 464-488.
Doug Hensley references the Lagarias algorithm in this paper (book chapter?):
http://www.math.tamu.edu/~doug.hensley/SimultaneousDiophantine.pdf
If you are not too concerned with the quality of the approximations this book presents many algorithms that work (w/o LLL) and may be useful:
Brentjes, Arne Johan. "Multi-dimensional continued fraction algorithms." MC Tracts 145 (1981): 1-183.
This book may still be available for purchase through ntis.gov and is well worth the cost.
A complete answer to your question depends on how the $q$'s you are looking for compare to the given $\epsilon$ and the desired computational complexity of the process for finding them.  In the extremes, I don't know if there is a completely satisfying answer.


TITLE: Question about uniform continuous proof.
QUESTION [1 upvotes]: If $f(x)$ define as: $f(x) = x\cos(\log(x))$, when $x  > 0$. $f(x) = 0$, when $x = 0$.
How we prove that $f$ is uniform continuous in $[0, +\infty[$?

REPLY [2 votes]: $ f $ is continuous at $ x=0$ since
$$(\forall x>0)\;|f(x)|\le |x|$$
$ f $ is differentiable at $ (0,+\infty) $ and
$$(\forall x>0)\;$$
$$f'(x)=\cos(\ln(x))-\sin(\ln(x))$$
thus
$$(\forall x>0)\;\;|f'(x)|\le 2$$
So, By MVT,
$$(\forall x,y\ge0)\;\;(\exists c>0) \;\;:$$
$$|f(x)-f(y)|=|x-y||f'(c)|$$
$$\le 2|x-y|$$
$ f $ is then UC at $ [0,+\infty)$.
You just can take $ \delta=\frac{\epsilon}{2}$.


TITLE: geodesic of infinite length without self-intersections
QUESTION [0 upvotes]: I'm curious, does a geodesic of infinite length without self-intersections exist in a Riemannian manifold (M, g) which has finite diameter? There shouldn't be one rigth?

REPLY [4 votes]: Take a line in $\mathbb{R}^2$ with irrational slope, then its image in the flat two torus $\mathbb{R}^2/\mathbb{Z}^2$ is a geodesic that never intersects itself and has infinite length.


TITLE: Uniform distribution in Ball with radius $\sqrt{n}$ is sub-gaussian
QUESTION [2 upvotes]: I have to show that a random vector $X$ who ist uniformly distributed on the Ball with Radius $\sqrt{n}$ is sub-gaussian with
$$\lVert X \rVert_{\psi_2}\leq C$$
I already know that the same result does hold for a random vector on the sphere with radius $\sqrt{n}$ (1).
I tried to show that $r Y$ is uniformly distributed on the Ball with Radius 1 for r uniformly distributed in $[0,1]$ and $Y$ uniformly distributed on the sphere with radius 1. Then I could use this to prove the claim in a similar way (1) is proved. However I am not sure if this is the right way?

REPLY [4 votes]: Let $X$ be uniformly distributed on the ball $B_{\sqrt n}$ of radius $\sqrt n$ in $\mathbb R^n$. Then
$$X=RY,$$
where $R:=|X|/\sqrt n$ and $Y:=\sqrt n\,X/|X|$ is uniformly distributed on the sphere $S_{\sqrt n}$ of radius $\sqrt n$ in $\mathbb R^n$.
Note that $0\le R\le1$ and hence $E\exp\{c(X\cdot t)^2\}=E\exp\{cR^2(Y\cdot t)^2\}\le E\exp\{c(Y\cdot t)^2\}$ for all real $c>0$ and $t\in\mathbb R^n$. So, $\|X\|_{\psi_2}\le\|Y\|_{\psi_2}$.
Also, you know that $\|Y\|_{\psi_2}\le C$. Thus,
$$\|X\|_{\psi_2}\le C,$$
as desired.


TITLE: characteristic function of a convolution of measures
QUESTION [2 upvotes]: Take the probability measures $\mu,\nu$ on $\mathbb{R}$ and denote $\varphi_{\mu}$ (the same for $\nu$) its characteristic function. Why holds
$$\varphi_{\mu *\nu}(t)=\varphi_{\mu}(t)\cdot\varphi_{\nu}(t)$$
where $\mu*\nu$ denotes the convolution of $\mu$ and $\nu$?

REPLY [0 votes]: Neither do we need to assume that $\mu$ and $\nu$ admit densities with respect to a common reference measure, nor do we need to restrict this result to $\mathbb R$.
Remember that if $(E,\mathcal E)$ is a measurable group and $$\tau:E^2\to E\;,\;\;\;(x,y)\mapsto xy$$ denotes the group operation, then $\mu\ast\nu$ is the pushforward $\tau(\mu\otimes\nu)$ of the product measure $\mu\otimes\nu$ under $\tau$.
Assuming that $E$ is a $\mathbb R$-Banach space (consider as a group with the group operation being the addition) and $\mathcal E=\mathcal B(E)$, we immediately obtain \begin{equation}\begin{split}\varphi_{\mu\ast\nu}(x')&=\int\tau(\mu\otimes\nu)({\rm d}x)e^{{\rm }i\langle x',\:x\rangle}\\&=\int\mu({\rm d}x)\int\nu({\rm d}y)e^{{\rm }i\langle x',\:\tau(x,\:y)\rangle}\\&=\int\mu({\rm d}x)e^{{\rm }i\langle x',\:x\rangle}\int\nu({\rm d}y)e^{{\rm }i\langle x',\:y\rangle}=\varphi_\mu(x')\varphi_\nu(x')\end{split}\tag1\end{equation} for all $x'\in E'$.


TITLE: Positive and Negative Real numbers
QUESTION [0 upvotes]: I stumbled upon this question on Yahoo answers here :

Prove that for every positive real number x, l/x is also a positive real number.?

Answering to the question is [now] closed, but I was thinking about the answer, when I sketched this argument below, but am not sure if am right, and whether this could be a part of the required proof (or whether this is a mere echo of obvious statements?)
if $p$ is positive real number then this is true:
$p \ge 0 \implies |p| - p = 0$
A similar argument for negative number $n$ would be:
$n \lt 0 \implies |n| - n = 2|n|$
With the above truths, I can then say:
If $x \ge 0$ then ${1 \over x} \ge 0$ since
$|{1 \over x} | - {1 \over x} = {0 \over x} = 0$
Is this right?

REPLY [0 votes]: What about this: x . 1/x = 1, so x and 1/x must have the same sign.


TITLE: Non-disjointness of hom-sets
QUESTION [4 upvotes]: In the category $\bf Set$ of sets, $\mathcal C$ is the class of all sets. Given
sets $A$, $B$ the set $\text{Hom}(A, B)$ consists of all functions from $A$ to
$B$.
This is a standard example of a category but I did not understand how they ensure that $\text{Hom}$ is disjoint, that is
    For any $(A,B)\neq(C,D)$ then $\text{Hom}(A,B)\cap\text{Hom}(C,D)=\emptyset,$
where $A,B.C,D\in\mathcal C.$ Let $1_A$ be an identity map so $1_A\in\text{Hom}(A,A)$. Since $A\subset B$, we have the inclusion map $i\colon A\to B$ with $i\in\text{Hom}(A,B).$ Thus $1_A$ and $i$ are mathematically the same, so $$\text{Hom}(A,A)\cap \text{Hom}(A,B)\neq \emptyset.$$ I know I am wrong since $\mathcal C$ is definitely a category. Can anyone help me understand what I did wrong?
Also, what is a good reference as a starting point to get a better understanding of the notation of category theory?

REPLY [4 votes]: Here is the bigger picture.
It is important in category theory that every morphism $f$ has a well-defined domain $\mathrm{dom}(f)$ as well as a well-defined codomain $\mathrm{cod}(f)$. When you define a category as a tuple consisting of two (large) sets $\mathrm{Ob}(\mathcal{C})$, $\mathrm{Mor}(\mathcal{C})$ and functions $\mathrm{dom},\mathrm{cod} : \mathrm{Mor}(\mathcal{C}) \to \mathrm{Ob}(\mathcal{C})$ as well as identity and composition (these are not important here, all what I say applies equally to directed multigraphs) this holds by definition.
However, another way of defining categories is by giving $\mathrm{Ob}(\mathcal{C})$ and a function which associates to every pair of objects $A,B \in \mathrm{Ob}(\mathcal{C})$ a set $\mathrm{Hom}(A,B)$. In order to get a category as defined above, we define $\mathrm{Mor}(\mathcal{C})$ as the disjoint union of these sets, i.e. as
$$\mathrm{Mor}(\mathcal{C}) := \bigcup_{(A,B) \in \mathrm{Ob}(\mathcal{C})^2} \mathrm{Hom}(A,B) \times \{(A,B)\}.$$
Conversely, given $\mathrm{Mor}(\mathcal{C})$, we define $\mathrm{Hom}(A,B) := \{f \in \mathrm{Mor}(\mathcal{C}) : \mathrm{dom}(f)=A,\, \mathrm{cod}(f)=B\}$.
These constructions are inverse to each other, but only in a certain sense: $\mathrm{Mor}(\mathcal{C})$ is only isomorphic to the disjoint union of its Hom-sets, and conversely each Hom-set is isomorphic to the set of morphisms with given domain and codomain. It is not true that they are equal, but of course this does not matter at all in mathematical practice. When we start with $\mathrm{Mor}(\mathcal{C})$, the Hom-sets will be disjoint, but this is not true for other other approach.
Generally speaking, if we have a family of sets $(X_i)_{i \in I}$, their disjoint union $X := \coprod_{i \in I} X_i$ does not have the $X_i$ as subsets in the traditional sense of ZFC, but there are only injections $X_i \to X$. The images of the $X_i$ are disjoint, but the $X_i$ themselves are not in general.
But we have to ask ourselves the following question: does it actually make sense to ask if two abstract sets $X,Y$ are disjoint or not? Of course this is a very meaningful question if $X,Y$ are both subsets of a given set, but when there is no relation between $X$ and $Y$, what does it even mean that an element of $X$ is equal to an element of $Y$? All this works fine with ZFC where everything is a set and we have intersections and unions of all pairs of sets, but it enables us also to ask nonsense questions such as "Is the set of transcendental numbers disjoint to the set of smooth plane curves?". As you, see this is a "type error". This is fixed by structural set theories such as ETCS and even more by type theoretic foundations such as HoTT.
In structural set theories, there is no way of even saying that two random sets are disjoint or not. (We can say that two morphisms $A \to X$, $B \to X$ are disjoint, namely $A \times_X B = 0$.) In other words, the whole problem asked here just does not exist. And that being said, it also does not exist in mathematical practice. The reason is that almost all meaningful statements in category theory not involve "loose" morphisms without any known domain or codomain, but rather morphisms with a given domain and codomain. Secondly, we should never compare two morphisms $f : A \to B$ and $g : A' \to B'$ unless it is known that $A = A'$ and $B = B'$. If we want to do that, we should use commutative squares instead.
I hope all this explains a bit more David's comment above, saying "Disjointness of homs is irrelevant".


TITLE: In how many ways can $2$ doors be selected from $3$ doors for entering and leaving a room?
QUESTION [1 upvotes]: In how many ways can $2$ doors be selected from $3$ doors for entering and leaving a room?
Options for answers are: $1, 3, 6, 9, 12$.
Since we have 3 doors and we should pick 2 doors for ENTERING and LEAVING then the answer is $C_3^2\times 2!=3!=6.$ That it my answer. However, the correct answer is $3$. But I suppose that $3$ is incorrect.
Can anyone explain it please?

REPLY [1 votes]: If you choose $2$ out of the  $3$ doors then you can enter through  $1$ and leave through the other.   So $\binom3{2}=3$.  They say nothing about the order.  So it's combinations, not permutations. ..


TITLE: Is this way of finding $\lim\limits_{x\to +\infty}(x-\ln(x^2+1))$ valid?
QUESTION [3 upvotes]: I needed to find: $$\lim\limits_{x\to +\infty}(x-\ln(x^2+1))$$
So here are the steps I took:
Step 1: Replace $x$ with $\ln(e^x)$:
$$\lim\limits_{x\to +\infty}\left(\ln(e^x)-\ln(x^2+1)\right)$$
$$\lim\limits_{x\to +\infty}\ln\left(\frac{e^x}{x^2+1}\right)$$
Step 2: Bring the limit inside of the natural log function since it is continuous on the required interval.
$$\ln\left(\lim_{x\to +\infty}\frac{e^x}{x^2+1}\right)$$
Step 3: Apply L'Hospital's rule twice and evaluate:
$$\ln\left(\lim_{x\to +\infty}e^x\right)$$
$$\ln(+\infty) = +\infty$$
My question is whether step 2 is valid here because $\lim\limits_{x\to \infty}\frac{e^x}{x^2 + 1}$ doesn't exist (its $+\infty$), and in order to move the limit operator inside the function the limit $\lim\limits_{x\to \infty}\frac{e^x}{x^2 + 1}$ must exist according to this theorem in a book about Calculus (ISBN 978-0-470-64769-1):
If it's not valid, what would be a valid way to find the limit?

REPLY [1 votes]: In step 2 all you need is the fact that $\ln (y) \to \infty$ as $y\to \infty$. Since $\frac {e^{x}} {x^{2}+1} \to \infty$ it follows that $\ln (\frac {e^{x}} {x^{2}+1}) \to \infty$. To prove that $\ln (y) \to \infty$ as $y\to \infty$ assume that this is false. Since $\ln \, x$is an increasing function, if it doesn't not tend to infinity, it would be bounded for $x>1$, say $\ln\, x <C$ for all $x>1$. You get a contradiction from this if you take $x=e^{C}$.


TITLE: Selecting From a Question Pool (Combination Question)
QUESTION [1 upvotes]: Alice is taking her probability exam tomorrow and she will be assigned  5 questions out of a pool
consisting of  13 questions. In particular, the pool includes  5 easy, 2 medium, and 6 difficult
questions.
If 5 questions are selected at random at the beginning of the exam, find the probability that Alice
will be assigned: both of the medium questions and at least 1 difficult question.

I've been trying hard to find a solution to this question but I couldn't. Here is what I did,
(nCr(2,2) * nCr(6,1))/nCr(13, 5) + (nCr(2,2) * nCr(6,2))/nCr(13, 5) + ... + (nCr(2,2) *
nCr(6,6))/nCr(13, 5) = 0.04895
But it turned out to be wrong, what am I doing wrong? Any help would be appreciated!

REPLY [2 votes]: Total number of unrestricted selections of $5$ questions $ = \displaystyle {13 \choose 5} = 1287$
Total number of possible selections where both medium questions are selected and at least one difficult question is selected
$\displaystyle = {11 \choose 3} - {5 \choose 3} = 155$
So the desired probability $ = \displaystyle \frac{155}{1287}$.
In your working, you also need to select remaining questions from easy. So it should be $\displaystyle \small \bigg[{6 \choose 1}{5 \choose 2} + {6 \choose 2} {5 \choose 1} + {6 \choose 3}\bigg] \bigg/ {13 \choose 5}$


text:
{\bf Problem.} Steve has an isosceles triangle with base 8 inches and height 10 inches. He wants to cut it into eight pieces that have equal areas, as shown below. To the nearest hundredth of an inch what is the number of inches in the greatest perimeter among the eight pieces? [asy]
size(150);
defaultpen(linewidth(0.7));
draw((0,0)--(8,0));
for(int i = 0; i < 9; ++i){
draw((4,10)--(i,0));
}
draw((0,-0.5)--(8,-0.5),Bars(5));
label("$8''$",(0,-0.5)--(8,-0.5),S);
[/asy]
{\bf Level.} Level 4
{\bf Type.} Geometry
{\bf Solution.} To make the $8$ triangles have the same area, the base must be divided into $8$ segments of length $1$ inch each.  Define points $A$, $B_0$, $B_1$, $B_2$, $B_3$, and $B_4$ as in the figure.  For $0\leq k\leq 3$, the perimeter $P(k)$ of triangle $A B_k B_{k+1}$ in inches is \[
P(k)=1+\sqrt{10^2+k^2}+\sqrt{10^2+(k+1)^2},
\]where each distance $A B_k$ is calculated by applying the Pythagorean theorem to right triangle $A B_0 B_k$.  Since $P(k)$ increases as $k$ increases, its largest value is $P(3)=1+\sqrt{100+3^2}+\sqrt{100+4^2}$, which to the nearest hundredth is $\boxed{22.21}$ inches.  [asy]
size(200);
defaultpen(linewidth(0.7)+fontsize(10));
draw((0,0)--(8,0));
for(int i = 0; i < 9; ++i){
draw((4,10)--(i,0));
if(i>=4)

label("$B_"+string(i-4)+"$",(i,0),S);
}
label("$A$",(4,10),N);
[/asy]

set_name: MATH


TITLE: What is the Gelfand-Naimark representation of functions that don't vanish at infinity?
QUESTION [4 upvotes]: The Gelfand-Naimark theorem says that every commutative C*-algebra is isometrically isomorphic to $C_0(X)$, the set of continuous functions $f:X\rightarrow\mathbb{C}$ that vanish at infinity, for some locally compact Hausdorff space $X$.
What happens when we apply this construction to a commutative C*-algebra that looks almost like $C_0(X)$, but we relax one of the requirements? For example, we can consider the C*-algebra $C_b(Y)$, the set of bounded continuous functions $f:Y\rightarrow\mathbb{C}$, with no requirement they vanish at infinity. Is there any relation between the the space $Y$ of the original C*-algebra and the space $X$ that we generate using Gelfand-Naimark?
Bonus followup: What if Y is not Hausdorff? Or not locally compact?

REPLY [1 votes]: Here is a slight elaboration of the previous answer. Lets us look at a system $\mathcal F$ of (bounded) continuous functions on $X$ that do not vanish at infinity and "include" these in our algebra, what space do we get with the Gelfand isomorphism?
To be more concrete, let $A$ be the $C^*$ sub-algebra of $C_b(X)$ generated by $C_0(X)$ and $\mathcal F$, what is $\sigma(A)$?
There are two such completions I can always write down. The simplest case is $\mathcal F = \{1\}$ (the constant function $1$) from which $A=C_0(X)\oplus\Bbb C\Bbb1$ follows, the second ist $A=C_b(X)$. The first case will give you the one-point compactification of $X$, ie $\sigma(C_0(X)\oplus\Bbb C\Bbb1)\cong X\cup\{\infty\}$ (either this is clear to you or its a very illustrative exercise). Meanwhile the second will return the Stone-Cech compactification, to see this note that any bounded function $f:X\to\Bbb C$ extends to a function $\beta X\to \Bbb C$ by the universal property of $\beta X$ (note that $\mathrm{im}(f)$ is contained in a compact set), furthermore the extension is unique by denseness of $X$ in $\beta X$. From this $C_b(X)\subset C(\beta X)$ follows and the other direction is clear.
Both of the spaces we got back were compact, which is not a coincidence. Note that in both cases we added the constant functions, which results in the unit of $C(X)$ being added to the algebra. I believe that it is more or less clear that the Gelfand isomorphism results in a compact space iff the original algebra was unital, ie it gives a correspondence
$$\text{unital commutative $C^*$ algebras} \leftrightarrow \text{compact locally compact Hausdorff spaces}$$
in particular we find via this correspondence another correspondence (for $X$ locally compact Hausdorff):
$$\text{unitisations of $C_0(X)$}\leftrightarrow\text{compactifications of $X$}.$$
So basically when you add functions at infinity you are compactifying $X$, provided you add enough functions so you can reconstruct $1$. Lets consider another example.
For $X=(0,1)$ you can choose to add the function $x\mapsto x$ and the function $x\mapsto 1-x$. If you do this you will get a unital algebra and indeed you will get all polynomials, so by Stone-Weierstrass your algebra will have all functions in $C([0,1])$. Since you also don't have more functions than that you see that adding these two functions results in taking the closure of $(0,1)$ in $\Bbb R$.
Indeed if $X$ is a bounded open subset of $\Bbb R^n$, the compactification you get by considering its closure will correspond to the unitsation you get by adding the $n$ coordinate functions $x_i$ to the algebra.
What if you don't add enough functions to recover identity? In our example $X=(0,1)$ if we just add $x\mapsto x$ the space we get is $(0,1]$. Indeed this is what always will happen: You will get an open subspace of a compactification of $X$.
The general lesson is thus: If you add functions at infinity you are basically extending $X$ by points at infinity.

To address your follow up: Continuous functions from a non-Hausdorff space into $\Bbb C$ cannot distinguish two non-separated points, ie if $x,y$ are such that every neighbourhood of $x$ intersects every neighbourhood of $y$ then $f(x)=f(y)$ for any continuous $f$. For that reason $C(X)=C(X/\sim)$ if $X$ is not Hausdorff and $\sim$ is the relation that glues all points that are not separated together. So basically continuous functions to $\Bbb C$ do not actually the non-Hausdorff parts of your space.
For spaces that are not locally compact I believe $C_0(X)$ will be entirely inadequate. Notice for example that $C_0(V)=0$ if $V$ is an infinite dimensional Banach space (for me the model of non locally compact spaces). I have no idea what the space associated to $C_b(V)$ looks like.


text:
\begin{document}
\maketitle

\vspace{-1.5 truecm}

\centerline{\large Michel Bauer\footnote[1]{ Institut de Physique
    Th\'eorique de Saclay, CEA-Saclay, 91191 Gif-sur-Yvette, France
    and Laboratoire de Physique Th\'eorique, Ecole Normale
    Sup\'erieure, 24 rue Lhomond, 75005 Paris, France.  {\small \tt
      <michel.bauer@cea.fr>}}, Denis Bernard\footnote[2]{Member of the
    CNRS; Laboratoire de Physique Th\'eorique, Ecole Normale
    Sup\'erieure, 24 rue Lhomond, 75005 Paris, France.  {\small \tt
      <denis.bernard@ens.fr>}}, Tom Kennedy \footnote[3]{Department of
    Mathematics, University of Arizona, Tucson, AZ 85721. {\small \tt
      <tgk@math.arizona.edu>}}}

\vspace{.3cm}

\begin{abstract}
We discuss properties of dipolar SLE$_{\kappa}$ under conditioning.
We show that $\kappa=2$, which describes continuum limits
of loop erased random walks, is characterized as being
the only value of $\kappa$ such that dipolar SLE conditioned
to stop on an interval coincides with dipolar SLE on
that interval. We illustrate this property by computing
a new bulk passage probability for SLE$_2$.
\end{abstract}

\newpage
\section{Introduction}

Schramm-Loewner evolution (SLE) comes in several flavors.
Chordal SLE describes random curves between two points on
the boundary of a simply connected domain. Radial SLE describes
random curves between an interior point and a boundary point.
Dipolar SLE describes random curves between a boundary point and
a subarc of the boundary. These various types of SLE are closely
related to each other and to the more general SLE$_{\kappa,\rho}$
processes \cite{Schramm-Wilson, Werner, Dubedat}

We consider dipolar SLE from a boundary point $z$ to a subarc $A$ of
the boundary and ask what happens if we condition it to end in
a subarc $A^\prime \subset A$, or to end at a point $w \in A$.
We will show that the result is an SLE$_{\kappa,\rho}$ process. In
the particular case of $\kappa=2$, when the dipolar SLE is conditioned
to end at a fixed point the distribution of the resulting curve is
just chordal SLE.

Let us observe however that even if a discrete model is known to (or
conjectured to) converge to chordal SLE$_{\kappa}$ in the continuum
limit, there is at the moment very little understanding how to modify
it to a discrete process candidate to converge to an
SLE$_{\kappa,\rho}$ in the continuum limit. Even the definition of
``dipolar self avoiding walk'' escapes us at the moment.

Certain lattice models of random curves have special properties that
allow us to identify the associated $\kappa$ in the continuum limit
assuming the continuum limit is an SLE. The restriction property of
the self avoiding walk implies that $\kappa=8/3$ as the only possible
candidate for the scaling limit.  The locality property of percolation
identifies $\kappa=6$ as the only possible scaling limit.

The loop-erased random walk (LERW) has another property that together
with our result for $\kappa=2$ implies that if the continuum limit
is SLE, then $\kappa$ must be $2$.
For concreteness, we consider the model in the infinite  strip.
Consider a random walk starting at the
origin and conditioned to exit the strip on the upper boundary.
Loop-erasing this model gives what we will refer to as the dipolar LERW.
If we take the original random walk which was conditioned to exit
through the upper boundary and further condition it to exit through
a point $w$ in the upper boundary, the result is the same
as conditioning a random walk starting at the origin to exit the
strip through $w$. So if we condition the dipolar LERW to end at $w$,
then the result is the same as the loop erasure of a random walk
starting at $0$ and conditioned to exit at $w$. The latter is just
the chordal LERW. In short, conditioning the dipolar LERW to end
at a particular point gives the chordal LERW. As we will see, SLE has
this property only for $\kappa=2$.

The LERW has been proved to converge to SLE$_2$ in the scaling limit
\cite{LSW,Zhan}. Since the LERW trivially has the property
that conditioning the dipolar version to end at a fixed point
give the chordal version, the proofs of convergence of the LERW to
SLE$_2$ have as a corollary the result that conditioning dipolar SLE$_2$
to end at a fixed point gives chordal SLE$_2$.
Nonetheless, it is desirable to have a direct proof that SLE$_2$
has this property.

Properties that SLE satisfies for only one value of the parameter
$\kappa$ are useful for identifying the scaling limit of
discrete or lattice models that satisfy the property.
Recently, claims that SLE samples describe lines in a variety of
2d systems have appeared : these claims concern zero vorticity
lines in turbulence \cite{Bernard et al}, nodal lines of chaotic or
random wave function \cite{Bogomolny et al} and interfaces in ground
states of spin glasses \cite{Hartmann et al,Bernard et other al}. The
first two cases seem to involve $SLE_6$ (i.e. percolation).  This is
far from fully justified, even if some heuristic theoretical arguments
make this plausible. The case of spin glasses is even more intricate.
The value of $\kappa$ computed numerically is not too far (but
apparently reliably different) from $2$. Moreover there are several
definitions of the notion of interface. For some of the definitions
conformal invariance has been ruled out so that only for one
definition is the relationship with SLE still a plausible guess.

Of course numerical studies are made in finite geometries (even with
finite systems) and the problem of finite size effects has to be taken
into account. In particular dipolar SLE, which is well adapted to a
strip geometry, seems to give much more accurate numerical evaluations
of $\kappa$ and of the quadratic variation of the driving process than
chordal SLE.

SLE has been studied by researchers coming from a variety of disciplines
ranging from probability to conformal field theory.
To make this article accessible to these various groups, we
derive our results in several ways. In addition, for several results
that have already appeared in the literature we give another derivation
from a different point of view.

\section{Avatars of dipolar and chordal SLE}

\subsection{Preliminaries}

In the half plane geometry $\mathbb{H}\equiv\{z,\; \text{Im}\; z
\in]0,+\infty[\}$, a growth process of a piece of curve starting from
a point on the lower boundary $\text{Im}\; z=0$ can be uniformized by
a map $g(z)$ having the point at infinity as a fixed point with the
normalization: $g(z) \simeq z+2c/z+o(1/z)$ for $z \rightarrow \infty$. It is
customary to call this the hydrodynamic normalization. The number $c$
is real, nonnegative, and increases during the growth. Then, with a
time parameterization $s$ of the growing trace such that $c_s\equiv
s$, the uniformizing map satisfies
\[
\frac{dg_s(z)}{ds}=\frac{2}{g_s(z)-V_s}, \qquad g_0(z)=z
\]
Chordal SLE from $0$ to $\infty$ is the stochastic growth
process obtained when $V_s=\sqrt{\kappa}B_s$ where $B_s$ is a standard
Brownian motion.

\vspace{.3cm}

In the strip geometry $\mathbb{S}\equiv\{z,\; \text{Im}\; z
\in]0,\pi[\}$ a growth process of a piece of curve starting from a
point on the lower boundary $\text{Im}\; z=0$ can be uniformized by a
map $g(z)$ fixing the two points at infinity. This leaves the
possibility to translate $g$ by a real constant. But $g(z) \simeq z
+c_{\pm}+o(1)$ for $z \rightarrow \pm \infty$, and $g$ can be fully
normalized by requiring that $c_++c_-=0$. We propose to call this the
strip symmetric normalization. Then, with a time parameterization $s$
such that $c_+=s$ of the growing trace, the uniformizing map satisfies

\[
\frac{dg_s(z)}{ds}=\frac{1}{\tanh (g_s(z)-V_s)/2}, \qquad g_0(z)=z
\]
Throughout the paper $\tanh x/2$ stands for $\tanh (x/2)$. Dipolar
SLE$_{\kappa}$ from $0$ to the upper boundary $\text{Im}\; z =\pi$ (as
defined in \cite{Bauer Bernard Houdayer}) is the stochastic growth
process obtained when $V_s=\sqrt{\kappa}B_s$ where $B_s$ is a standard
Brownian motion.

\subsection{Change of domain} \label{subsec:change of domain}

Chordal SLE is particularly simple in the half plane geometry with the
hydrodynamic normalization and dipolar SLE is particularly simple in
the strip geometry with the symmetric normalization. But both are
conformally invariant processes, and as such could be described in any
domain (with two marked points for the chordal case, or with one
marked boundary interval and one marked point in its boundary
complement in the dipolar case) and moreover with any choice of
normalization of the Loewner map.

We illustrate this issue with four examples. The first one is well
known. We treat the second one in detail, using a commutative diagram
technique introduced in the study of locality and restriction for SLE,
see e.g. \cite{Lawler et other al}, and sketch the treatment of the
fourth. The same method can be used in all four cases, leading to
straightforward but slightly painful computations. We shall see later
how ideas from statistical mechanics and conformal field theory
followed by a simple application of Girsanov's theorem allow to
simplify the computations.

\vspace{.2cm}

\textbf{Chordal SLE from $0$ to $a$ in the upper half plane} with the
hydrodynamic normalization is described by the system of stochastic
differential equations:
\[\frac{dg_s(z)}{ds}=\frac{2}{g_s(z)-V_s}, \quad
\frac{dA_s}{ds}=\frac{2}{A_s-V_s},\]
\[ dV_s=\sqrt{\kappa}dW_s+(\kappa-6)\frac{ds}{V_s-A_s},\]
with initial conditions $g_0(z)=z$, $V_0=O$ and $A_0=a$.

\vspace{.2cm}

\textbf{Chordal SLE from $0$ to $i\pi+a$ in the strip} with the
symmetric normalization is described by the system of stochastic
differential equations:
\[\frac{dg_s(z)}{ds}=\frac{1}{\tanh (g_s(z)-V_s)/2} , \quad
\frac{dA_s}{ds}=\tanh (A_s-V_s)/2,\]
\[ dV_s=\sqrt{\kappa}dW_s+(\kappa/2-3)ds\tanh (V_s-A_s)/2  \]
with initial conditions $g_0(z)=z$, $V_0=O$ and $A_0=a$.

\vspace{.2cm}

\textbf{Dipolar SLE from $0$ to $[a,b]$ in the upper half plane} with the
hydrodynamic normalization is described by the system of stochastic
differential equations:
\[\frac{dg_s(z)}{ds}=\frac{2}{g_s(z)-V_s} , \quad
\frac{dA_s}{ds}=\frac{2}{A_s-V_s} , \quad \frac{dB_s}{ds}=\frac{2}{B_s-V_s},\]
\[ dV_s=\sqrt{\kappa}dW_s+(\kappa/2-3) \left(\frac{ds}{V_s-A_s} +
\frac{ds}{V_s-B_s}\right),\]
with initial conditions $g_0(z)=z$, $V_0=O$, $A_0=a$ and $B_0=b$.

\vspace{.2cm}

\textbf{Dipolar SLE from $0$ to $[i\pi+a,i\pi+b]$  in the strip} with the
symmetric normalization is described by the system of stochastic
differential equations:
\begin{eqnarray*} \frac{dg_s(z)}{ds} & = & \frac{1}{\tanh
    (g_s(z)-V_s)/2}, \\
\frac{dA_s}{ds}& = & \tanh (A_s-V_s)/2, \quad \frac{dB_s}{ds}\; = \;\tanh
(B_s-V_s)/2,\\
dV_s & = & \sqrt{\kappa}dW_s+(\kappa/2-3)ds\frac{\tanh (V_s-A_s)/2 + \tanh
(V_s-B_s)/2}{2}\end{eqnarray*}
with initial conditions $g_0(z)=z$, $V_0=O$, $A_0=a$ and
$B_0=b$.

\vspace{.4cm}

In all the above systems, $W_s$ is a normalized Brownian
motion. Before we explain the second case, let us make a few remarks.\\
-- The drifts all vanish at $\kappa=6$, which is a
manifestation of the locality of percolation.\\
-- For $\kappa \leq 4$ the above equations give a complete description.
For $\kappa >4$, the first two equations give only a partial
description : the first is a description of chordal SLE from $0$ to
$a$ in the upper half plane only up to the first time the process
separates $a$ from $\infty$ by hitting the real line in the interval
$[a,\infty]$ that does not contain $0$; the second is a description of
chordal SLE from $0$ to $i\pi+a$ in the strip only up to the first
time the process hits the line $\text{Im}\; z =\pi$. In both cases,
the trouble comes solely from the choice of normalization.

\vspace{.5cm}

With this proviso in mind, we go on to write down chordal SLE in
the strip with the symmetric normalization. The starting point will be
$0$ and the end point will be $i\pi+a$ where $a$ is a real number.  An
invaluable tool to make computations straightforward is a commutative
diagram, just as in the study of locality and restriction for SLE
\cite{Lawler et other al}.

\begin{figure}
\begin{center}
\includegraphics[height=7cm]{commut.eps}
\end{center}
\caption{The basic commutative diagram}
\end{figure}

Uniformize chordal SLE in the upper-half plane from $0$ to $\infty$
with trace $\gamma_{[0,t]}$ up to time $t$ by
$h_t:\mathbb{H}\backslash \gamma_{[0,t]} \rightarrow \mathbb{H}$ in the
hydrodynamical normalization:
\[ \frac{dh_t(w)}{dt}=\frac{2}{h_t(w)-\xi_t}, \quad
\xi_t=\sqrt{\kappa}B_t.\] Map the upper half plane to the strip by any
conformal map $F:\mathbb{H}\rightarrow \mathbb{S}$ such that $F(0)=0$
and $F(\infty)=i\pi+a$ to get SLE in $\mathbb{S}$ from $0$ to
$i\pi+a$. The formula for $F$ is
\[ \tanh (F(w)-a)/2 +\tanh a/2 =lw.\]
where $l \in \mathbb{R}^{+*}$ is arbitrary (we have given only two
conditions to normalize $F$).
Let $g_t$ be the uniformizing map $g_t:\mathbb{S}\backslash
F(\gamma_{[0,t]}) \rightarrow \mathbb{S}$ with the symmetric normalization
as above except that the time parameterization is not at our disposal
\[ \frac{dg_t(z)}{dt}=\frac{a_t}{\tanh (g_t(z)-\tilde{\xi}_t)/2}.\]
Finally let $F_t:\mathbb{H}\rightarrow
\mathbb{S}$ be the map ``closing the square'',
\[ g_t \circ F = F_t \circ h_t.\]

\vspace{.1cm}

Take the time derivative
of $g_t \circ F = F_t \circ h_t$ and afterwards substitute $w$ for
$h_t$ to get
\[\frac{a_t}{\tanh (F_t(w)-\tilde{\xi}_t)/2}=
\frac{dF_t(w)}{dt}+F'_t(w)\frac{2}{w-\xi_t}.\]
Now $F_t(w)$ is non singular at $w=\xi_t$ so that the pole on the
right hand-side is cancelled by a pole on the left hand-side:
\[\frac{a_t}{\tanh (F_t(w)-\tilde{\xi}_t)/2}=F'_t(\xi_t)\frac{2}{w-\xi_t}
+ O(1) \text{ when } w\rightarrow \xi_t,\]
leading to $\tilde{\xi}_t=F_t(\xi_t)$ and
$a_t=F'_t(\xi_t)^2$. Continuing the expansion one step further yields
\[-a_tF''_t(\xi_t)/F'_t(\xi_t)^2=\frac{dF_t}{dt}(\xi_t)+2F''_t(\xi_t)
\text{ i.e. } \frac{dF_t}{dt}(\xi_t)=-3F''_t(\xi_t).\]
Now use Ito's formula (in a slightly extended context as usual in this
computation) to get
\[d\tilde{\xi}_t=d(F_t(\xi_t))=dF_t(\xi_t)+F'_t(\xi_t)d\xi_t+
\frac{\kappa}{2} F''_t(\xi_t)dt,\]
i.e. $d\tilde{\xi}_t=F'_t(\xi_t)d\xi_t+(\kappa /2-3)F''_t(\xi_t)dt$.
Define $s(t)\equiv\int_0^t F'_u(\xi_u)^2du$ and
$\sqrt{\kappa}W_{s(t)}\equiv \int_0^t F'_u(\xi_u)d\xi_u$ so that
$W_{s}$ is a standard Brownian motion with time parameter $s$. Set
$V_{s(t)} \equiv F_t(\xi_t)$ so
\[dV_s=\sqrt{\kappa}dW_s+(\kappa
/2-3)\frac{F''_{t(s)}(\xi_{t(s)})}{F'_{t(s)}(\xi_{t(s)})^2}ds.\]

This equation is in fact valid for all four cases, but of course $F_t$
closes a different commutative diagram in each case. Computations are
simplified a little bit when one realizes that if $F(w)=z$ has inverse
$\Phi(z)=w$ then
\[ \frac{F''(w)}{F'(w)^2}=-(\log \Phi'(z))'.\] In this identity, the $'$
indicates derivative with respect to $w$ on the left hand side but with
respect to $z$ on the right hand side.  So if we write $\Phi_s(z)=w$
for the inverse of $F_t(w)=z$ (where $s$ and $t$ are related by the
time change) we get the important intermediate formula :
\begin{equation} \label{eq:drift1}
dV_s=\sqrt{\kappa}dW_s+(3-\kappa /2)(\log \Phi_s'(z))'_{|z=V_s} ds.
\end{equation}

\vspace{.1cm}

Let $A_s \equiv g_{t(s)}(i\pi+a)-i\pi$ be the trajectory of $a$
under the Loewner evolution. From $g_t \circ F = F_t \circ h_t$,
$h_t(\infty)=\infty$ and $F(\infty)=i\pi+a$ we learn that
$g_t(i\pi+a)=F_t(\infty)=i\pi+A_s$. Combined with
$F_t(\xi_t)=V_{s(t)}$ we find
\[\tanh (F_{t(s)}(w)-A_s)/2 +\tanh (A_s-V_s)/2 =l_s(w-\xi_{t(s)}),\]
or
\[\tanh (z-A_s)/2 +\tanh (A_s-V_s)/2 =l_s(\Phi_s(z)-\xi_{t(s)}),\]
where $l_s$ is an unspecified but irrelevant parameter.

The computation of the drift is now easy (the reader should compare
with a direct use of the formula involving $F_t$) :
\[((\log \Phi_s'(z))'_{|z=V_s}=-\tanh (V_s-A_s)/2.\]

\vspace{.1cm}

This finishes to establish the formula announced above for chordal SLE in
the strip from $0$ to $i\pi+a$ in the symmetric normalization.

\vspace{.3cm}

We end this section with some remarks on how to write down dipolar SLE
from $0$ to $[i\pi+a,i\pi+b]$ in the strip with the symmetric
normalization. The steps above could be reproduced, leading to
equation (\ref{eq:drift1}), but with a function $\Phi_s$ mapping the
strip to the strip, $V_s$ to $\xi_{t(s)}$, $i\pi+A_s$ to $-\infty$ and
$i\pi+B_s$ to $+\infty$. This fixes $\Phi_s$ uniquely. A convenient
representation is
\[ \coth (\Phi_s(z)-\xi_{t(s)})/2=\nu + \mu \coth (z-V_s)/2\]
with
$-1=\nu + \mu \tanh (A_s-V_s)/2$, $1= \nu + \mu \tanh (B_s-V_s)/2.$

To compute the drift we have to be able to take two derivatives with
respect to $z$ and then put $z=V_s$ so it is enough to use the second
order expansion:
\[\frac{\Phi_s(z)-\xi_{t(s)}}{2}=\frac{z-V_s}{2\mu}-
\nu\left(\frac{z-V_s}{2\mu}\right)^2+\cdots\]
for $z \rightarrow V_s$. We infer that
\[(\log \Phi_s'(z))'_{|z=V_s}=-\nu/\mu = \frac{\tanh (V_s-B_s)/2 +
  \tanh (V_s-B_s)/2 }{2},\]
leading to the announced formula.

\section{Conditioning}

We aim at a description of dipolar SLE in a domain (call it $D$) from
a boundary point (call it $0$) to a boundary interval (call it $I$)
conditioned to hit either a subinterval $J$ of $I$ or, via a limiting
procedure, a point in $I$.

\subsection{Na\"{\i}ve considerations}
\label{subsec:naive}

Girsanov's theorem is the general tool to tackle these problems, but
in this section we want to illustrate what goes on using only
elementary manipulations, with the hope that the explicit argument
shows in the clearest way what is involved, compensating at least
partly the lack of elegance of the approach.

\vspace{.2cm}

We could in principle make computations in any domain, with any
normalization of the uniformizing maps. For illustration, we choose
$D$ to be the strip $\mathbb{S}$, $I$ to be upper boundary
$\text{Im}\; z =\pi$, and $J$ to be the interval $]i\pi+a,i\pi+b[$
where $a < b$ are real numbers. Before conditioning, the process is
described by the equations
\[
\frac{dg_s(z)}{ds}=\frac{1}{\tanh (g_s(z)-V_s)/2}, \qquad g_0(z)=z
\]
where $V_s=\sqrt{\kappa}W_s$ and $W_s$ is a standard
Brownian motion. The inverse image $g_s^{-1}(\mathbb{S})$ is
$\mathbb{S}\backslash \gamma(]0,s])$. In particular, the inverse
image of $g_s^{-1}(V_s)$ --defined as $\lim_{\varepsilon \mapsto
  0^+}g_s^{-1}(V_s+i\varepsilon)$-- is $\gamma(s)$.

\vspace{.2cm}

\textbf{The martingale} We condition on the event that the curve ends
in an interval $J$. We will denote this event by $J$ as well, no
confusion should arise. The first thing we have to know is the
probability of $J$. This is a routine computation if one uses
martingales, but this is also the key to make contact with Girsanov's
theorem later. Write $p(J)$ for the hitting probability of dipolar SLE
and observe that $p(J)=f(a)-f(b)$ where $f(c)$ is the probability to
hit on the right of $c$. So it is enough to deal with the case when
$J=]i\pi+a,i\pi+\infty[$. Observe that $A_s\equiv g_s(i\pi+a)-i\pi$
satisfies
\[
\frac{dA_s}{ds}=\tanh (A_s-V_s)/2 , \quad A_0=a. \]

Our claim is the following: for $t \geq 0$,
$\Expect{1_J|\mathcal{F}_t}=f(A_t-V_t)$. Recall that an
$\mathcal{F}_t$-measurable event is an event whose realization can be
decided from the knowledge of the process $W_{\bullet}$ for times up
to $t$.  So the left hand side is the probability to hit in $J$
knowing the process $W_{\bullet}$ up to time $t$. Intuitively, this
knowledge is equivalent to the knowledge of $g_t$. Dipolar SLE is
conformally invariant, and in particular, for $s \geq t$, the
statistics of $g_t(\gamma(]t,s])-V_t$, which is supposed to hit on the
right of $A_t-V_t$, is the same as the statistics of a
$\gamma(]0,s-t])$. The hitting probability of this process is
$f(A_t-V_t)$ as announced.

The equation $\Expect{1_J|\mathcal{F}_t}=f(A_t-V_t)$ means that
$f(A_t-V_t)$ is a --so called ``closed''-- martingale. In particular
the Ito derivative of the stochastic process $f(A_t-V_t)$ has no drift. A
routine use of Ito's formula yields
\[\frac{\kappa}{2}f''(a)+ f'(a) \tanh a/2 =0,\]
whose only solution with an acceptable probabilistic interpretation is
proportional to $\int_a^{+\infty}dx \cosh ^{-4/\kappa} x/2$, leading to
\[p(J)=\frac{\int_a^b \, dx \cosh ^{-4/\kappa} (x/2)}
{\int_{-\infty}^{+\infty}\, dx \cosh ^{-4/\kappa} (x/2)} \]
for the hitting probability of the interval $J=]i\pi+a,i\pi+b[$, a
formula established in \cite{Bauer Bernard Houdayer}.

\vspace{.2cm}

\textbf{Markov property} Let $p$ denote the probability distribution for
dipolar SLE (more precisely for the driving process $V_s$, which is
proportional to a Brownian motion), and let $\tilde{p}$ denote the
probability distribution for dipolar SLE conditioned to hit the upper
boundary in $J$ (again more precisely for the driving process
$V_s$, which after conditioning is not simply a Brownian motion
anymore). Define $J_s \equiv g_s(J)=]i\pi+A_s,i\pi+B_s[$.

Observe that the knowledge of $V_t,A_t,B_t$ for all $t$'s is a
(redundant) description of the growth process. It is redundant because
the knowledge of $V_t$ for all $t$'s would suffice. For $A < B$ and $t
\geq 0$, let $E_{t,V,A,B}$ denote the set of
driving functions $V_{\bullet}$ such that $V_t\in [V,V+dV]$ $A_t\in
[A,A+dA]$ and $B_t \in [B,B+dB]$. This is an
$\mathcal{F}_t$-measurable set. Let $E$ be any
$\mathcal{F}_t$-measurable subset of $E_{t,V,A,B}$, and for $t \leq s$
compute $\tilde{p}(E \text{ and } E_{s,V',A',B'})$. By definition,
this is
\[\frac{p(E \text{ and } E_{s,V',A',B'} \text{ and } \gamma \text{ hits in }
]a,b[)}{p(\gamma \text{ hits in }]a,b[)}.\]
But under $p$, the process $(V_{\bullet},A_{\bullet},B_{\bullet})$ is
Markovian, so that
\[p(E \text{ and } E_{s,V',A',B'} \text{ and }
\gamma \text{ hits in } ]a,b[)=p(E)p(E')p(J')\] where $E'$ is the
event that starting from $V,A,B$ at time $t$,
$V_s\in[V',V'+dV'],A_s\in[A',A'+dA'],B_s\in[B',B'+dB']$ and $J'$ is
the event that dipolar SLE started at $V_s$ on the lower boundary hits
the upper boundary in the interval $J'=]A',B'[$, or by translation
invariance, that  dipolar SLE started at $0$  on the lower boundary hits
the upper boundary in the interval $J'=]A'-V',B'-V'[$. We have already
computed $p(J')$. Hence
\[\tilde{p}(E \text{ and } E_{s,V',A',B'})=p(E)p(E')\frac{p(\gamma
  \text{ hits in }]A'-V',B'-V'[)}{p(\gamma \text{ hits in }]a,b[)}.\]
A simpler argument yields
\[\tilde{p}(E)=p(E)\frac{p(\gamma\text{ hits in
  }]A-V,B-V[)}{p(\gamma \text{ hits in }]a,b[)},\]
and by comparison,
\[ \tilde{p}(E \text{ and } E_{s,V',A',B'})=
\tilde{p}(E)p(E')\frac{p(\gamma \text{ hits in
  }]A'-V',B'-V'[)}{p(\gamma \text{ hits in }]A-V,B-V[)}\]
for any $\mathcal{F}_t$-measurable event $E \subset E_{t,V,A,B}$.

This proves that under $\tilde{p}$, the process $V_t,A_t,B_t$ is
Markovian and, from the definition of $E'$, that the transition
probability density under $\tilde{p}$ to go from $(V,A,B)$ at time $t$
to $(V',A',B')$ at time $s$ is simply the transition probability
density under $p$ to go from $(V,A,B)$ at time $t$ to $(V',A',B')$ at
time $s$ weighted by the known ratio $\frac{p(\gamma \text{ hits in
  }]A'-V',B'-V'[)}{p(\gamma \text{ hits in }]A-V,B-V[)}$. Under $p$,
$V_t$ alone is a Markov process, so there is some price to pay to save
the Markov property.

\vspace{.2cm}

\textbf{Computation of the drift} Note that the transition probability
density under $p$ to go from $(V,A,B)$ at time $t$ to $(V',A',B')$ at
time $s$ is translation invariant and time homogeneous, so that we can
assume without loss of generality that $V=0$ and $t=0$. Now we assume
that $s=\varepsilon$ and $V'=\Delta$ are small and we compute the
probability distribution of $\Delta$. Under $p$, the density of
$\Delta$ is $\frac{1}{\sqrt{2\pi\kappa\varepsilon}}
e^{-\Delta^2/(2\kappa\varepsilon)}$, and the argument in the
exponential is of order $1$ only when $\Delta \sim
\sqrt{\varepsilon}$. The conditioning ratio $\frac{p(\gamma \text{
    hits in }]A'-V',B'-V'[)}{p(\gamma \text{ hits in }]A-V,B-V[)}$ can
be expanded using this scaling, and only terms up to order $\Delta$
need to be kept. As $A'-A$ and $B'-B$ are of order $\varepsilon$ we
find
\[
\frac{p(\gamma \text{ hits in }]A'-V',B'-V'[)}{p(\gamma \text{ hits in
  }]A-V,B-V[)}=1-\Delta F(A,B)+0(\varepsilon,\Delta^2),
\]
where
\[ F(A,B)=\frac{\cosh ^{-4/\kappa} B/2 - \cosh ^{-4/\kappa} A/2}
{\int_A^B \, dx \cosh ^{-4/\kappa} x/2} \]

So the transition probability density under $\tilde{p}$ of
$V_{\varepsilon}$ starting from $V_0=0$ is
\[\frac{1}{\sqrt{2\pi\kappa\varepsilon}}
e^{-(\Delta+\kappa\varepsilon F(A,B))^2/(2\kappa\varepsilon)}
(1+0(\varepsilon,\Delta^2)).\]
The meaning of this is that for infinitesimal $\varepsilon$,
$V_{\varepsilon}$ is a Gaussian random variable of mean
$-\kappa\varepsilon F(A,B)$ and standard deviation
$\sqrt{\kappa\varepsilon}$. Using translation invariance and time
homogeneity, this is exactly the meaning of the
stochastic differential equation
\[dV_s=\sqrt{\kappa}dW_s-\kappa F(A_s-V_s,B_s-V_s)ds.\]

When $B\rightarrow A$, $F(A,B)$ has a finite limit,
$-\frac{2}{\kappa}\tanh A/2$.

\vspace{.2cm}

We summarize the central results of this section:

\textbf{Dipolar SLE from $0$ to the upper boundary in the strip,
  conditioned to hit in $[i\pi+a,i\pi+b]$} with the symmetric
normalization is described by the system of stochastic differential
equations:
\begin{eqnarray*}
\frac{dg_s(z)}{ds}& = &\frac{1}{\tanh (g_s(z)-V_s)/2}, \\
\frac{dA_s}{ds} & = &\tanh (A_s-V_s)/2 , \quad \frac{dB_s}{ds}\; =
\; \tanh
(B_s-V_s)/2,\\
dV_s & = & \sqrt{\kappa}dW_s-\kappa\left(\frac{\cosh ^{-4/\kappa} (B_s-V_s)/2 -
    \cosh ^{-4/\kappa} (A_s-V_s)/2} {\int_{A_s-V_s}^{B_s-V_s} \, dx
    \cosh ^{-4/\kappa}
    (x/2)}\right) \end{eqnarray*}
with initial conditions $g_0(z)=z$, $V_0=O$, $A_0=a$ and $B_0=b$.

\vspace{.2cm}

\textbf{Dipolar SLE from $0$ to the upper boundary in the strip
  conditioned to hit at $i\pi+a$} with the symmetric normalization is
described by the system of stochastic differential equations:
\[\frac{dg_s(z)}{ds}=\frac{1}{\tanh (g_s(z)-V_s)/2} , \quad
\frac{dA_s}{ds}=\tanh (A_s-V_s)/2,\]
\[ dV_s=\sqrt{\kappa}dW_s-2\tanh (V_s-A_s)/2  \]
with initial conditions $g_0(z)=z$, $V_0=O$ and $A_0=a$.

\vspace{.2cm}

As a corollary, dipolar SLE conditioned to hit in a subinterval is
dipolar SLE if and only if $\kappa=2$, and dipolar SLE conditioned to
hit at a given point is chordal SLE if and only if $\kappa=2$.

This is obtained by direct comparison with the formul\ae\ in
subsection \ref{subsec:change of domain}.

\subsection{An elementary application at $\kappa=2$}

The above result is a nice characterization of the value $\kappa=2$.
We show in this section that it allows us to compute in an elementary way
the probability that a dipolar SLE$_2$ trace passes to the right of a
given bulk point. The same strategy can be used to compute other
physical observables of interest, see \cite{Hagendorf}.

\vspace{.2cm}

The probability that a dipolar SLE$_{\kappa}$ trace (or hull) ends to
the right of a given boundary point of the hitting interval is
obtained by routine martingale techniques, see e.g. \cite{Bauer
  Bernard Houdayer}.

For $\kappa < 8$, the probability that the chordal SLE$_{\kappa}$
trace passes to the right of a given bulk point $z$ (for $4 < \kappa <
8$ we means that when $z$ is swallowed it goes to the negative real
axis) is again obtained via a routine martingale technique see
\cite{Schramm}.

For $\kappa \geq 4$, the probability that a dipolar SLE$_{\kappa}$
hull contains the bulk point $z$ and the probability that it passes to
the right of $z$ can be computed (somewhat miraculously) again by
routine martingale techniques by making the ansatz that they are
harmonic functions of $z$ and then checking that appropriate boundary
conditions can be imposed, see again \cite{Bauer Bernard Houdayer}.
But the harmonic ansatz fails for $\kappa < 4$.

\vspace{.2cm}

For $\kappa=2$, we obtain the probability that a dipolar SLE$_2$ trace
passes to the right of a given bulk point by the following steps.

We start from Schramm's result at $\kappa=2$ for the chordal case in
the upper half plane: if $\vartheta\in [0,\pi]$ is the argument of
$z\in \mathbb{H}$, the probability that chordal SLE$_2$ passes to the
right of $z$ is
\[\frac{\vartheta}{\pi}-\frac{\sin 2\vartheta}{2\pi}.\]

We consider a conformal map from $\mathbb{H}$ to the strip
$\mathbb{S}$ sending $0$ to $0$ and $\infty$ to the point $a+i\pi$ on
the hitting interval, to get the probability that chordal SLE$_2$ in
the strip from $0$ to $a+i\pi$ passes to the right of $w\in
\mathbb{S}$. Cumbersome but elementary algebra shows that this
probability is obtained by substitution of $\vartheta (a,w)$ for
$\vartheta$ in the above formula, where
\[ \vartheta (a,w) \equiv \arctan \frac{\sin v}{\sinh u-(\cosh u -
  \cos v) \tanh a/2},\] the function $\arctan$ takes values in
$[0,\pi]$ and $w=u+iv$ is the decomposition into real and imaginary
parts.

The trick now is that because  dipolar SLE$_2$ conditioned to hit
at $a$ is nothing but chordal SLE$_2$ aiming at $a$, the probability
$p_2(w)$ that dipolar SLE$_2$ passes to the right of $w$ can be
represented as the integral over $a$ of the dipolar hitting point
density at $a$ times the probability that chordal SLE$_2$ aiming at
$a$ passes to the right of $w$. Explicitly:
\[p_2(w)=\int_{-\infty}^{+\infty}\frac{da}{4\cosh^2 a/2}
\left(\frac{\vartheta(a,w)}{\pi}-\frac{\sin
    2\vartheta(a,w)}{2\pi}\right).
\]
Choosing $\tanh a/2$ as the new integration variable leads to an
elementary integral because the differential of $x \arctan 1/x$ is
exactly $dx (\arctan 1/x-\frac{1}{2} \sin 2 \arctan 1/x)$. Finally:
\[p_2(w)=\frac{(e^u -\cos v) \arctan \frac{\sin v}{e^u -\cos v} -
  (\cos v -e^{-u}) \arctan \frac{\sin v}{\cos v -e^{-u}}}{2\pi (\cosh
  u - \cos v) }.\]
with $w=u+iv$ the decomposition into real and imaginary parts.

\vspace{.2cm}

For other values of $\kappa$, we would have to work with dipolar SLE
conditioned to hit a point, and compute probability that the trace
passes to the right of $w$ for this process. This is likely to be even
more complicated than $p_{\kappa}(w)$, the probability we are aiming
at: it involves four boundary points (the starting point, the hitting
interval and the conditioned endpoint) whereas $p_{\kappa}(w)$
involves only three. The miracle of $\kappa=2$ is that after
conditioning the hitting interval plays no role anymore. Another
natural explanation for this comes from conformal field theory.

\subsection{Another computation using martingales}

\label{sec:sw}

Reference \cite{Schramm-Wilson} contains several computations
analogous to the above ones (some are even assigned as
exercises). In this section give another derivation of the result from
subsection \ref{subsec:naive}.

For definiteness, we consider dipolar SLE in the half plane from $0$
to $[-1,1]^c$.  (By $[-1,1]^c$ we mean the complement of $[-1,1]$ with
respect to the real axis.) The hitting density is proportional to
\begin{eqnarray*}
  p(x)=x^{-4/\kappa} (x^2-1)^{-1 + 2/\kappa}
  \end{eqnarray*}

We use the following martingale. For real forcing points
$x_1,\cdots,x_n$ with weights $\rho_1,\cdots,\rho_n$, it is
\begin{eqnarray*} M_t(x_1,\cdots,x_n;\rho_1,\cdots,\rho_n) =
  \prod_{j=1}^n \, L(x_j,\rho_j) \prod_{1 \le i < j \le n}
  Q(x_i,x_j;\rho_i,\rho_j)  \end{eqnarray*}
where
\begin{eqnarray*} L(x_j,\rho_j)= |g^\prime_t(x_j)|^{\alpha_j}
  |X^{x_j}_t|^{\rho_j/\kappa} \end{eqnarray*}
\begin{eqnarray*}
  Q(x_i,x_j;\rho_i,\rho_j) = |X^{x_i}_t - X^{x_j}_t|^{\rho_i \rho_j
    /(2 \kappa)} \end{eqnarray*}
\begin{eqnarray*} \alpha_j=
  (8-2\kappa+2 \rho_j) \rho_j/(8 \kappa)
\end{eqnarray*}
Here $X^x_t$ denotes $g_t(x) - W_t$.  Let $\hat{M}_t$ be this
martingale normalized so its expectation is $1$.  Note that $M_0$ is a
constant, so $\hat{M}_t = M_t/M_0$.  If we take chordal SLE in the
half plane from $0$ to $\infty$ and weight it by $\hat{M}_t$, then we
get the SLE$_{\kappa,\rho}$ process with forcing points at the $x_i$
and weights $\rho_i$. This martingale appears in
\cite{Schramm-Wilson,Kytola}\footnote{Schramm and Wilson comment that the
  martingale was also discoverd independently by M. Biskup.}.

Dipolar SLE in the upper half plane from $0$ to $[-1,1]^c$ is the same
as SLE$_{\kappa,\rho}$ with two force points at $-1$ and $+1$ with both
weights equal to $(\kappa-6)/2$.  This is the same as taking chordal
SLE in the upper half plane from $0$ to $\infty$ and weighting it by
the martingale $M_t(-1,1;(\kappa-6)/2,(\kappa-6)/2)$.  We want to show
that if we condition this to end at some $a \in [-1,1]$, then we get
an SLE$_{\kappa,\rho}$ process with three force points.  One is at $a$
with weight $-4$ and the other two are at $\pm 1$, both with weight
$\frac{\kappa-2}{2}$.  We will show this by showing that
\begin{eqnarray*}
  \int_{[-1,1]^c} \,
  \hat{M}_t(a,-1,1;-4,\frac{\kappa-2}{2},\frac{\kappa-2}{2}) \,
  p(a) \, da
  = \hat{M}_t(-1,1;\frac{\kappa-6}{2},\frac{\kappa-6}{2})
\end{eqnarray*}
This equation is trivially true at $t=0$.  We note that $
M_0(a,-1,1;-4,\frac{\kappa-2}{2},\frac{\kappa-2}{2}) = c p(a)$.  for
some constant $c$ which only depends on $\kappa$, and
$M_0(-1,1;\frac{\kappa-6}{2},\frac{\kappa-6}{2})$ only depends on
$\kappa$. So with some rearranging, the above is the same as
\begin{eqnarray}
  && \int_{[-1,1]^c} \, L(a,-4) \, Q(a,-1;-4,\frac{\kappa-2}{2}) \,
  Q(a,1;-4,\frac{\kappa-2}{2}) \, \, da
  \nonumber \\
  &=& C \, \frac{L(-1,\frac{\kappa-6}{2}) \, L(1,\frac{\kappa-6}{2}) \,
    Q(-1,1;\frac{\kappa-6}{2},\frac{\kappa-6}{2})
    }{
    L(-1,\frac{\kappa-2}{2}) \, L(1,\frac{\kappa-2}{2}) \,
    Q(-1,1;\frac{\kappa-2}{2},\frac{\kappa-2}{2}) }
\label{inteq}
\end{eqnarray}

For $\rho=(\kappa-2)/2$ and $\rho=(\kappa-6)/2$ we get the same value
of $\alpha$. Thus
\begin{eqnarray*}
  \frac{L(p,\frac{\kappa-6}{2})}{L(p,\frac{\kappa-2}{2})}
  =  |X^p_t|^{-2/\kappa}
\end{eqnarray*}
for $p= \pm 1$.  We find that
\begin{eqnarray*} \frac{
    Q(-1,1;\frac{\kappa-6}{2},\frac{\kappa-6}{2})}{
    Q(-1,1;\frac{\kappa-2}{2},\frac{\kappa-2}{2})} = |X_t^{-1} -
  X_t^1|^{-1+4/\kappa}
\end{eqnarray*}
We also have
\begin{eqnarray*}
  L(a,-4)  = |g^\prime_t(a)| \, |X^a_t|^{-4/\kappa}
\end{eqnarray*}
and for $p= \pm 1$,
\begin{eqnarray*}
  Q(a,p;-4,\frac{\kappa-2}{2}) = |X^a_t-X^p_t|^{(2-\kappa)/\kappa}
\end{eqnarray*}

Thus eq.(\ref{inteq}) becomes
\begin{eqnarray*}
  && \int_{[-1,1]^c} \, |g^\prime_t(a)| \, |X^a_t|^{-4/\kappa} \,
  |X^a_t-X^{-1}_t|^{(2-\kappa)/\kappa} \,
  |X^a_t-X^1_t|^{(2-\kappa)/\kappa} \, da
  \\
  &=& c \, |X^{-1}_t|^{-2/\kappa} \, |X^1_t|^{-2/\kappa} \,
  |X_t^{-1} - X_t^1|^{-1+4/\kappa}
\end{eqnarray*}
Note that $|g^\prime_t(a)|=\frac{dX^a_t}{da}$.  $X^a_t$ is a
differentiable function of $a$ and we can do a change of variables to
change the above to
\begin{eqnarray*}
  \int_{[P,Q]^c} \,
  X^{-4/\kappa} \,
  \left[ (X-P) (Q-X)\right]^{(2-\kappa)/\kappa}
  \, d X
  = c \, P^{-2/\kappa} \, Q^{-2/\kappa} \, (Q-P)^{-1+4/\kappa}
\end{eqnarray*}
where $X$ is shorthand for $X_t^a$, $P$ is for $X^{-1}_t$ and $Q$ for
$X^1_t$. This identity follows from the substitution
\begin{eqnarray*}
  X=PQ \frac{z+1}{Pz+Q}
\end{eqnarray*}
followed by some nice cancellations.

\section{ Girsanov's theorem}

Girsanov's theorem gives a unified view of the computations of the
two preceding sections. Let us first briefly recall the theorem and
then go on with illustrations.

Take a probability space with a filtration $(\Omega,{\mathcal
  F},{\mathcal F}_t,p)$ and let $M_t,{\mathcal F}_t$ be a nonnegative
martingale on it such that $M_0=1$.

If $X$ is ${\mathcal F}_s$-measurable and $t\geq s$ then basic rules
of conditional expectations yield $\Expect{XM_t}=\Expect{XM_s}$ so
that one can make a consistent definition $\tExpect{X}\equiv
\Expect{XM_t}$ whenever $X$ is ${\mathcal F}_t$ measurable. Then
$\tExpect{\cdots}$ is easily seen to be a positive linear functional
with $\tExpect{1}=1$. Hence the definition $\tilde{p}_t(A)\equiv
\tExpect{{\mathbf 1}_A}$ for $A \in {\mathcal F}_t$ makes
$(\Omega,{\mathcal F}_t,\tilde{p}_t)$ a probability space.

\vspace{.2cm}

Now take for $(\Omega,{\mathcal F},p)$ a probability space carrying a
(continuous) Brownian motion $B_t$ with its filtration ${\mathcal
  F}_t$. If $M_t$ is a continuous martingale, then It\^o's
representation formula says $M_t$ satisfies the stochastic integral
equation $M_t=1+\int_0^t M_sX_sdB_s$ for some process $X_s$ (which
satisfies a number of technical conditions, in particular $X_s$ is
${\mathcal F}_s$-measurable for each $s$). In fact one can write
$M_t=\exp \big[\int_0^t X_s\, dB_s-\frac{1}{2}\int_0^t X_s^2\,
ds\big]$ as shown by an application of It\^o's formula.

Girsanov's theorem states that the process $V_t\equiv B_t-\int_0^t
X_sds$ is a Brownian motion on $[0,T]$ for $(\Omega,{\mathcal
  F}_T,\tilde{p}_T)$ for each $T >0$ (see for instance \cite{Oksendal}
for a readable mathematical introduction, \cite{Karatzas} for a more
exhaustive mathematical reference or \cite{Bauer Bernard PR} for a
heuristic proof).

\vspace{.2cm}

A simple special case is $M_t\equiv e^{xB_t-tx^2/2}$, which is
a martingale on the Brownian motion space satisfying the conditions
above for a constant  $X_s=x$. Girsanov's theorem yields that under
$\tilde{p}_T$, $V_t=B_t-xt$ is a Brownian motion on $[0,T]$, so that under
$\tilde{p}_T$ the original process $B_t$ is now a Brownian motion with a
constant drift. This makes clear that the restriction to finite $T$
in Girsanov's theorem is crucial: if we could take $T \rightarrow
\infty$, we would have
$B_t/t\rightarrow x$ for large $t$  with probability $1$ for
$\tilde{p}$ while this event has
probability $0$ for $p$.

\subsection{Comparison between different kinds of SLEs}

Our computations in subsection \ref{subsec:change of domain} all dealt
with the following situation : for a certain variant of SLE on a
certain domain and with a certain normalization of the Loewner
uniformizing map the driving function satisfied a known stochastic
differential equation (in fact it was a Brownian motion), and we aimed
at the stochastic differential equation for another variant.

\vspace{.2cm}

Physical arguments allow to get this from Girsanov's theorem.  SLE
curves can be seen in general as interfaces in conformally invariant
field theories which are continuum limits of statistical mechanics
systems where an interface can already be identified in the discrete
setting due to boundary conditions.  Statistical mechanics relies on the
computation of partition functions, and in the case at hand, the
partition function when an initial segment of the interface is fixed
is (up to normalization) the probability that the interface starts
with this initial segment.  Changing the boundary conditions changes
the partition function i.e. the distribution of the interface. The
ratio of two such partition functions for the same initial segment of
interface then appears as a discrete version of a Radon-Nykodim
derivative, and indeed a slight extension of the arguments in
\cite{Bauer Bernard Houdayer} yields the fact the ratio of partition
functions is a martingale when the initial segment gets larger
\footnote{This extension however points to the importance that the
  variant of interface involved in the numerator is absolutely
  continuous with respect to the interface involved in the
  denominator.}.

A discrete partition function in an arbitrary geometry is hard to
compute, to say the least. However the conformally invariant continuum
limit (when it exists) is much more accessible.  Its relation to the
discrete system involves removing certain divergences, and the limit
in arbitrary geometry (pieces of the boundary are SLEs with some
change of boundary conditions at the tip) can be quite singular. But
ratios of partition functions behave smoothly exactly when one variant
of interface involved is absolutely continuous with respect to the
other, and the limiting ratio is expected to be the Radon-Nykodim
derivative for the corresponding probability measures in the
continuum.

\vspace{.2cm}

Let us illustrate this on an example and then say a few words on the
general framework.

The partition function $Z^{\text{dip}}_{\mathbb{D}}(x,a,b)$ for
dipolar SLE in domain $\mathbb{D}$ from a boundary point $x$ to a
boundary interval $[a,b]$ (not containing $x$) is given by a three
point correlation function
$\statav{\phi_{h_{0,1/2}}(a)\phi_{h_{1,2}}(x)\phi_{h_{0,1/2}}(b)}_{\mathbb{D}}$
where $\phi_h$ is a customary notation for a primary conformal
(boundary) field of weight $h$, and $h_{r,s}\equiv \frac{(\kappa r
  -4s)^2-(\kappa -4)^2} {16\kappa}$. So
$h_{0,1/2}=\frac{(\kappa-2)(6-\kappa)}{16\kappa}$ and
$h_{1,2}=\frac{(6-\kappa)}{2\kappa}$.

Such three point functions are completely fixed up to
normalization by conformal invariance because all quadruples
$(\mathbb{D},x,a,b)$ are conformally equivalent. If $\mathbb{D}$ is
the upper half plane $\mathbb{H}$ (then $a,b,x$ are real numbers),
\[\statav{\phi_{h}(a)\phi_{h'}(x)\phi_{h''}(b)}_{\mathbb{H}}\propto
|b-a|^{h'-h-h''}|x-a|^{h''-h-h'}
|x-b|^{h-h''-h'}.\]
More generally, if the boundary of $\mathbb{D}$ is smooth at $a,b,x$ and
$x$ and $g$ maps $\mathbb{D}$ to $\mathbb{H}$,
\[\statav{\phi_{h}(a)\phi_{h'}(x)\phi_{h''}(b)}_{\mathbb{D}}=
\statav{\phi_{h}(g(a))\phi_{h'}(g(x))\phi_{h''}(g(b))}_{\mathbb{H}}
|g'(a)|^{h}|g'(x)|^{h'}|g'(b)|^{h''},\]
emphasizing the fact that partition functions are sections of tensor
density bundles.

In the case at hand,
\[Z^{\text{dip}}_{\mathbb{H}}(x,a,b)\propto
|b-a|^{(\kappa-6)^2/(8\kappa)}|x-a|^{(\kappa-6)/(2\kappa)}
|x-b|^{(\kappa-6)/(2\kappa)}\]
and
\[Z^{\text{dip}}_{\mathbb{D}}(x,a,b)=Z^{\text{dip}}_{\mathbb{H}}
(g(x),g(a),g(b)) |g'(a)|^{\frac{(\kappa-2)(6-\kappa)}{16\kappa}}
|g'(x)|^{\frac{(6-\kappa)}{2\kappa}}
|g'(b)|^{\frac{(\kappa-2)(6-\kappa)}{16\kappa}}.\]

Chordal SLE in $\mathbb{D}$ from $x$ to $y$ has partition function
$Z^{\text{chord}}_{\mathbb{D}}(x,y)=\statav{\phi_{h_{1,2}}(x)
  \phi_{h_{1,2}}(y)}_{\mathbb{D}}$ which is even simpler. Taking
$\mathbb{D}$ to be the upper half plane $\mathbb{H}$,
\[Z^{\text{chord}}_{\mathbb{H}}(x,y)=
|x-y|^{-2h_{1,2}}=|x-y|^{(\kappa-6)/(\kappa)},\] and in general
\[Z^{\text{chord}}_{\mathbb{D}}(x,y)=
Z^{\text{chord}}_{\mathbb{H}}(g(x),g(y))
|g'(x)|^{\frac{(6-\kappa)}{2\kappa}}
|g'(y)|^{\frac{(6-\kappa)}{2\kappa}}.\] Note that if $y \rightarrow
\infty$, one has to change local coordinate to express the section,
and $Z^{\text{chord}}_{\mathbb{H}}(x,\infty)=1$.

\vspace{.2cm}

In the identity
\[
\frac{Z^{\text{chord}}_{\mathbb{D}}(x,y)|g'(y)|^{\frac{(\kappa-6)}{2\kappa}}}
{Z^{\text{chord}}_{\mathbb{H}}(g(x),g(y))}=
\frac{Z^{\text{dip}}_{\mathbb{D}}(x,a,b)
  |g'(a)|^{\frac{(\kappa-2)(\kappa-6)}{16\kappa}}
  |g'(b)|^{\frac{(\kappa-2)(\kappa-6)}{16\kappa}}}{Z^{\text{dip}}_{\mathbb{H}}
  (g(x),g(a),g(b))},\] the derivative of $g$ at $x$ disappears, and
the formula makes sense even if $\mathbb{D}$ is not smooth at $x$.

Suppose now that $\mathbb{D}$ is the domain obtained after the
interface has grown for some time (and $x$ is at the tip), capacity is
used to measure time, and the uniformizing map is taken in the
hydrodynamical normalization i.e. $\mathbb{D}=\mathbb{H}_t\equiv
\mathbb{H}\backslash \gamma_{]0,t]}$ is uniformized by $g_t$
satisfying $dg_t(z)=2dt/(g_t(z)-V_t)$ where $V_t\equiv
g_t(\gamma_t)$. Let $A_t\equiv g_t(a)$, $B_t\equiv g_t(b)$. Take
$y=\infty$ and observe that $g_t(\infty)=\infty$ and $g'_t(\infty)=1$.
Then the above ratio becomes
\[ Z^{\text{chord}}_{\mathbb{H}_t}(\gamma_t,\infty)=
Z^{\text{dip}}_{\mathbb{H}_t}(\gamma_t,a,b)
\frac{|g_t'(a)|^{\frac{(\kappa-2)(\kappa-6)}{16\kappa}}
  |g'_t(b)|^{\frac{(\kappa-2)(\kappa-6)}{16\kappa}}}
{Z^{\text{dip}}_{\mathbb{H}} (V_t,A_t,B_t)}\]

\vspace{.2cm}

Two strategies are now available. The first one looks at the ratio
\[
\frac{Z^{\text{dip}}_{\mathbb{H}_t}(\gamma_t,a,b)}
{Z^{\text{chord}}_{\mathbb{H}_t}(\gamma_t,\infty)}\equiv
\frac{Z^{\text{dip}}_{\mathbb{H}}
  (V_t,A_t,B_t)}{|g_t'(a)|^{\frac{(\kappa-2)(\kappa-6)}{16\kappa}}
  |g'_t(b)|^{\frac{(\kappa-2)(\kappa-6)}{16\kappa}}}.\] A general
argument from conformal field theory guaranties that this ration is a
martingale of chordal SLE, call it $M_t$ (of course this can also be
checked by a straightforward computation). The naive continuum limit
argument given above suggests that $M_t$ is exactly the Radon-Nykodim
derivative of dipolar SLE from $x=V_0$ to $[a,b]$ with respect to
chordal SLE.  For chordal SLE, $\kappa^{-1/2}V_t$ is a Brownian
motion, and the other building blocks in the formula for $M_t$ are of
bounded variation (i.e. do not contribute to $dV_t$ terms in $dM_t$),
so that
\[dM_t=M_t\partial_x \log Z^{\text{dip}}_{\mathbb{H}}
(x=V_t,A_t,B_t)dV_t. \] Then Girsanov's theorem yields the stochastic
differential equation of dipolar SLE:
\[dV_t=\sqrt{\kappa}dW_t+\kappa \partial_x \log
Z^{\text{dip}}_{\mathbb{H}} (x=V_t,A_t,B_t) dt, \] which coincides
with the already obtained formula.  From this viewpoint, the
appearance of a logarithmic derivative in the drift is totally
natural, and comparison with the initial formula implies an amusing
relationship between the partition function
$Z^{\text{dip}}_{\mathbb{H}} (x,a,b)$ and any conformal map $\Phi(z)$
mapping the upper-half plane to the strip $a$ to $-\infty$ and $b$ to
$+\infty$ (two such maps differ by an additive constant):
$Z^{\text{dip}}_{\mathbb{H}} (x,a,b)\Phi'(x)^{h_{1,2}}$ is
$x$-independent.

\vspace{.2cm}

The second one is to look at the inverse ratio
\[
\frac{Z^{\text{chord}}_{\mathbb{H}_t}(\gamma_t,\infty)}
{Z^{\text{dip}}_{\mathbb{H}_t}(\gamma_t,a,b)}.\]
This time CFT ensures that this is a martingale of dipolar SLE.
Assuming that dipolar is absolutely continuous with respect to chordal
i.e. that $\kappa^{-1/2}V_t$ is a Brownian motion plus a drift, this
drift is fixed by the martingale property. Of course the result is in
agreement with the other approaches. The computation is more painful
because one needs to look at the second order term in Ito's
formula. However, there is one benefit: only the martingale property
is used, and not the fact that the ratio is a Radon-Nykodim
derivative. Indeed one could replace the ratio
$Z^{\text{chord}}_{\mathbb{H}_t}(\gamma_t,\infty)/
Z^{\text{dip}}_{\mathbb{H}_t}(\gamma_t,a,b)$
by any correlation function of dipolar SLE in $\mathbb{H}_t$ with
hitting interval $[a,b]$, which is again a tautological martingale, to
compute the drift.

This second strategy was used in \cite{Bauer Bernard Kytola} to get
the form of the stochastic differential system describing multiple
SLE's, and in \cite{Kytola} to give the conformal field theory
approach to SLE($\kappa,\rho$'s) and generalizations thereof. Indeed,
the martingale used in section \ref{sec:sw} is the simplest partition
function\footnote{In the coulomb gas formalism of conformal field
  theory, it involves no screening charges, hence appears as a simple
  product.} describing the effect of marked points (in physics
language, these could be impurities) on SLE interfaces and was
identified as such in \cite{Kytola}. From the CFT viewpoint, this
partition function leads to an SLE$_{\kappa,\rho}$ tautological
martingale by a general argument, and no explicit computation is
required.

\subsection{Conditioning}

Take a probability space with a filtration $(\Omega,{\mathcal
  F},{\mathcal F}_t,p)$ and an event $J \subset \Omega$ such that
$p(J)=\Expect{\mathbf{1}_J}  \neq 0$ and let $\tilde{p},
\tExpect{\cdots}$ be the probability and expectation on
$(\Omega,\mathcal{F})$ obtained by conditioning $p$ on $J$.

If $Y$ is a random variable, $\tilde{\mathbb{E}}(Y)=
\frac{\Expect{Y\mathbf{1}_J}} {\Expect{\mathbf{1}_J}}.$ In
particular, if $Y$ is $\mathcal{F}_s$ measurable
$\Expect{Y\mathbf{1}_J}=\Expect{\Expect{Y\mathbf{1}_J|
\mathcal{F}_s}}=\Expect{Y\Expect{\mathbf{1}_J|\mathcal{F}_s}}$.
Then $M_s\equiv \frac{\Expect{\mathbf{1}_J|\mathcal{F}_s}}
{\Expect{\mathbf{1}_J}}$ is a closed bounded positive martingale
with $M_0=1$ and $\tExpect{Y}=\Expect{YM_s}.$ From the previous
subsection, it is natural so view the numerator as a partition
function $Z_t$.

If $(\Omega,{\mathcal F},p)$ is a probability space carrying a
(continuous) Brownian motion $B_t$ with its filtration ${\mathcal
  F}_t$ and  $M_s \equiv \frac{\Expect{\mathbf{1}_J|\mathcal{F}_s}}
{\Expect{\mathbf{1}_J}}$ is a continuous martingale, Girsanov's
theorem applies.

In our illustrative example, $J$ is ``the dipolar SLE process from $0$
to the upper boundary in the strip hits in the interval
$[i\pi+a,i\pi+b]$''. We have computed the corresponding martingale:
$M_t=Z(V_t,A_t,B_t)/Z(0,a,b)$ where
\[
Z(V,A,B)\equiv \int_{A-V}^{B-V} \, dx \cosh ^{-4/\kappa} x/2.
\]
The martingale $M_t$ is obviously continuous and Girsanov's theorem
avoids the previous elementary but clumsy discussion to yield the
drift in one stroke, and then the Markov property follows immediately. \\
Again, the processes $A_t$ and $B_t$ are differentiable with respect
to $t$ so that in the equation $M_t=1+\int_0^t M_sX_sdV_s$ only the
variation of $Z(V_s,A_s,B_s)$ with respect to $V_s$ contributes to $X_s$
and we find $X_s =\partial_V \log Z(V_s,A_s,B_s)$: the drift is again
the variation of the appropriate free energy with respect to the
starting point of the curve.

\section{Conclusions}

In this paper we have shown that if $\kappa=2$, conditioning dipolar
SLE to hit a subinterval yields dipolar SLE$_2$ towards that
subinterval. As a limiting case, conditioning dipolar SLE$_2$
to hit at a point yields chordal SLE$_2$. We have also shown that
these properties characterize the value $\kappa=2$ by computing
explicitly the drift term that arises when dipolar SLE$_\kappa$ for
general $\kappa$ is conditioned to hit a subinterval. This drift term
corresponds to some SLE$_{\kappa,\rho}$. We have shown that this drift
can be computed by standard martingale techniques or by conformal
field theory inspired methods.

This can be seen as a characterization of $\kappa=2$ analogous
to the characterization of $\kappa=8/3$ by the restriction property or
$\kappa=6$ by locality.

The relationship with lattice models leads to puzzling facts however.
First of all, there is no obvious interpretation of the dipolar
geometry for all lattice models. Loop erased random walks, the Ising
model for spin clusters and percolation are among the few favorable
cases. For loop erased random walks, the fact that conditioning of
dipolar to hit a subinterval is dipolar to the subinterval is true
almost by definition. For the Ising model and percolation, the
boundary condition on the hexagonal lattice are $+$ and $-$ on each
side of the starting point and \textit{free} on the boundary interval.
It is easy to understand then that conditioning to hit a subinterval
is a non trivial operation. But the case of self avoiding walks is
irritating.  The obvious candidate for the dipolar geometry would be
to consider all the simple walks joining a boundary point to a
boundary interval and weight each step by the critical fugacity
$\mu_c$. This prescription has the property that conditioning to end
at a given point yields the chordal case. So we conclude that this
naive definition cannot converge to dipolar SLE$_{8/3}$, most likely
because it corresponds to non-conformally invariant boundary
conditions.

\vspace{1cm}

\emph{Aknowledgements:} This research began during a visit to the Kavli
Institute for Theoretical Physics in September, 2006. The research of
TK was supported in part by the National Science Foundation under
grants DMS-0501168 and DMS-0758649.  The research of MB and DB was
supported in part by ANR-06-BLAN-0058.

config: arxiv


TITLE: What is topology of $X^{\mathbb{Z}}$ where $X$ is a topological space.
QUESTION [1 upvotes]: Let $(X, \tau)$ be a topological space and $I$ be an infinite set. I want to define a topology on $X^I$ by open sets in $X$. Is it true that a non-empty set $\mathcal{U}\subseteq X^I$ is open in $X^I$ if there is finite set $A$ such that $\mathcal{U}=\prod_{\alpha\in A} U_\alpha$, where $U_\alpha=X$ if $\alpha\notin A$ and for $\alpha\in A$, $U_\alpha\in \tau$.`
Please help me to know is it true?

REPLY [1 votes]: There are all kinds of topologies on $X^{I}$. The collection of sets you have defined is not a topology. You have to all possible unions of those sets to get a  topology. The resulting topology is called the product topology on $X^{I}$. [For  another topology you can look at Wikipedia for the 'box topology']


TITLE: Prove that $\sum_{1}^{\infty} a_{n} \lt \infty \rightarrow \sum_{1}^{\infty} \arcsin(a_{n}) \lt \infty$
QUESTION [2 upvotes]: Let be $\{a_{n}\}$ a sequence of real number / $0\lt a_{n} \le 1 \; \forall n \in \Bbb N $. Prove that $\sum_{1}^{\infty} a_{n} \lt \infty \rightarrow \sum_{1}^{\infty} \arcsin(a_{n}) \lt \infty$
So, what I did:
$$\text{If} \; \sum_{1}^{\infty} a_{n} \lt \infty \; \rightarrow \lim_{x\to \infty} a_{n} = 0$$
So, by the Limit comparison test, which states that if $\lim_{x\to \infty} \frac{a_{n}}{b_{n}} \gt 0 \; \text{then} \sum_{1}^{\infty} a_{n} \lt \infty \iff \sum_{1}^{\infty} b_{n} \lt \infty$
Then, $\lim_{x\to \infty} \frac{a_{n}}{b_{n}} = \lim_{t\to 0} \frac{\arcsin(t)}{t} = \frac{d}{dt} \arcsin(t)|_{t=0}=1 \gt 0$.
So that would be it. I don´t know if it's done correctly. I don't know why the hypothesis that  $0\lt a_{n} \le 1$. This is actually a two-part exercise but I'll put the second part in another Question. Thanks in advance.

REPLY [1 votes]: Defining the arcsine function as $\arcsin(x)=\int_0^x\frac1{\sqrt{1-t^2}}\,dt$, we have the following estimate for $a_n\in (0,1)$.
$$\begin{align}
|\arcsin(a_n)|&=\int_0^{a_n}\frac1{\sqrt{1-x^2}}\,dx\\\\
&\le \frac{a_n}{\sqrt{1-a_n^2}}
\end{align}$$
Inasmuch as $a_n\to 0$ as $n\to \infty$, there exists a number $N$ such that for all $n>N$,  $\sqrt{1-a_n^2}>\frac12$.
Hence we assert that
$$\left|\sum_{n=N+1}^\infty \arcsin(a_n)\right|\le 2\sum_{n=N+1}^\infty a_n<\infty$$
and the series of interest converges.


TITLE: Second reading on set theory? Any recommendations?
QUESTION [10 upvotes]: I have in past six-ish months studied through the Herbert Enderton's Elements of set theory book.
Up to the point the book is great,I loved most parts of it and learned almost everything up to the section on cardinal arithmetic.
Next I plan on reading Jechs Introduction to set theory and/or Set theory for working mathematician.
What I have wondered about is,what do I do next?Where can I master cardinals and see examples of relatively simple applications of axiom of choice(most examples are inclusive of advanced mathematics and the point of axioms use is obscured)?
Topics I am mostly interested in are:
      -Applications of axiom of choice
      -Cardinals and ordinals
      -Material on countable and uncountable infinities
      -Perhaps a easy introduction to descriptive set theory
      -Morse-kelly and Bernays-Von Neumann set theory
      -Transfinite induction

So what do you guys recommend?Share your wisdom

REPLY [12 votes]: Intermediate level set theory texts: [1], [2], [4], [8], [10], [12].
More advanced than these, but possibly within your present reach are [6] and [11].
Descriptive set theory can be found in some of these (e.g. [6] and [12]), and more fully in [9] and [14].
A lot of useful material on cardinal and ordinal numbers is in [5] and [13].
[3] is a good elementary reference for topics that are often omitted in standard elementary texts (e.g. arithmetic operations on linear orderings) and for historical/bibliographical information.
Finally, Handbook of Set Theory [see also here] might have some useful articles, but at this point I have not had a chance to look at it very carefully.
[1] Krzysztof [Chris] Ciesielski, Set Theory for the Working Mathematician, London Mathematical Society Student Texts #39, Cambridge University Press, 1997, xi + 236 pages. table of contents
[2] Frank Robert Drake and Dasharath Singh, Intermediate Set Theory, John Wiley and Sons, 1996, x + 234 pages. table of contents book review
[3] Abraham Adolf [Adolph] Halevi Fraenkel, Abstract Set Theory, 3rd revised edition, Studies in Logic and the Foundations of Mathematics, 1966, viii + 297 pages. review of 1953 edition
[4] András Hajnal and Peter Hamburger, Set Theory, London Mathematical Society Student Texts #48, Cambridge University Press, 1999, viii + 316 pages.
[5] Michael Holz, Karsten Steffens, and Edmund [Edi] Weitz, Introduction to Cardinal Arithmetic, Birkhäuser Advanced Texts, Birkhäuser Verlag, 1999, viii + 304 pages. book review
[6] Thomas J. Jech, Set Theory, Springer Monographs in Mathematics, Springer-Verlag, 2003, xiv + 769 pages. 2006 edition 2011 edition
[7] Winfried Just and Martin Weese, Discovering Modern Set Theory. I. The Basics, Graduate Studies in Mathematics #8, American Mathematical Society, 1996, xviii + 210 pages. list of topics
[8] Winfried Just and Martin Weese, Discovering Modern Set Theory. II. Set-Theoretic Tools for Every Mathematician, Graduate Studies in Mathematics #18, American Mathematical Society, 1997, xiv + 224 pages. list of topics
[9] Alexander Sotirios Kechris, Classical Descriptive Set Theory, Graduate Texts in Mathematics #156, Springer-Verlag, 1995, xviii + 402 pages. corrections and updates (4 October 2013)
[10] Péter Komjáth and Vilmos Totik, Problems and Theorems in Classical Set Theory, Problem Books in Mathematics, Springer, 2006, xii + 514 pages. table of contents book review
[11] Herbert Kenneth Kunen, Set Theory. An Introduction to Independence Proofs, Studies in Logic and the Foundations of Mathematics #102, North-Holland, 1980, xvi + 313 pages. 2011 edition review of 1983 edition
[12] Azriel Levy, Basic Set Theory, Perspectives in Mathematical Logic, Springer-Verlag, 1979, xiv + 391 pages. Reprinted by Dover Publications in 2002. review
[13] Waclaw Franciszek Sierpinski, Cardinal and Ordinal Numbers, 2nd edition revised, Monografie Matematyczne #34, PWN--Polish Scientific Publishers, 1965, 491 pages. review of 1958 edition
[14] Sashi Mohan Srivastava, A Course on Borel Sets, Graduate Texts in Mathematics #180, Springer-Verlag, 1998, xvi + 261 pages.


TITLE: How many ways are there to use powers of $2$, each at most $3$ times, and sum up to $100$?
QUESTION [2 upvotes]: For $i=0,1,2,\ldots$, there're $3$ weights with mass $2^i$ gram(s). How many ways are there to weigh $100$ grams with them?

To solve with normal methods for counting doesn't seem possible. Although any way to select weights can be generated by starting from $100=2^6+2^5+2^2$ and replacing $2^{k+1}$ with $2^k+2^k$ numerous times, whether a replacement like that can take place depends on other replacements already made. For example, if there're already $2$ weights with mass $2^3$ used, you cannot replace $2^4$ with $2^3+2^3$ any more.
On the other hand, generating functions may help. I constructed the following function:$$\begin{aligned}f(x)=&\prod_{t=0}^6\left(1+x^{2^i}+x^{2\times2^i}+x^{3\times2^i}\right)\\=&(1+x+x^2+x^3)(1+x^2+x^4+x^6)\cdots(1+x^{64}+x^{128}+x^{192}).
\end{aligned}$$and we need to find the coeffitient of term $x^{100}$. How to compute this?
$\tiny\text{Why isn't there a tag for enumeration problems?}$

REPLY [0 votes]: This answer was inspired by my first answer that used the generating function, but this answer doesn't use generating functions.
Suppose $n_02^0+n_12^1+n_22^2+\cdots+n_k2^k=M$ (where $M$ can be $100$) with each $n_i\in\{0,1,2,3\}$. Then split up the even and odd powers, and for appropriate $i$ and $j$:
$$
\begin{align}
M
&=\left(n_02^0+n_22^2+n_42^4+\cdots+n_{2i}2^{2i}\right)+\left(n_12^1+n_32^3+n_52^5+\cdots+n_{2j+1}2^{2j+1}\right)\\
&=\left(n_0+n_24+n_44^2+\cdots+n_{2i}4^i\right)+2\left(n_1+n_34+n_54^2+\cdots+n_{2j}4^j\right)
\end{align}$$
So $M=(n_{2i}\cdots n_2n_0)+2(n_{2j}\cdots n_3 n_1)$ where the integers here are written in base $4$. So we've written $M$ as a sum of an even number with some other number (using base $4$). And this is reversible: write $M=a+2b$ and you can recover $M$ as a unique sum of the weights of size $2^i$.
So the answer to the OP is the same as how many non-negative even integers ($2b$) are less than or equal to $M$. If $M$ is even, this is $\frac{M+2}{2}$. If $M$ is odd, $\frac{M+1}{2}$. In the case of $M=100$, it's $51$.


TITLE: If $\alpha$ and $\beta$ are irrational, and $r\neq 0$ is rational, could $\alpha+\beta$ be rational? $\sqrt{\alpha}$? $\sqrt{r}$? $\alpha\beta$? etc
QUESTION [0 upvotes]: Little background, $\alpha$, $\beta$ are irrational, $r$ is the non-zero rational number. We are asked to tick the boxes where the result may end up being rational. The following has been declined as the correct answer for me.

I am fairly certain that my marked boxes are correct, which means I am missing one of the unticked options. I think it may be $\sqrt{\alpha + \sqrt{\beta}}$, though I cannot come up with an example...

REPLY [1 votes]: Yes, for the addition you can choose two numbers of opposite signs, for $\sqrt{r}$ you can choose $r = k ^ 2$, for multiplication you can choose $\sqrt{2} \cdot \sqrt{2}$ and for the last one , the example with $\pi$ and $-\sqrt{\pi}$ works.


TITLE: How to picture the Quantum Harmonic Oscillator with particle creation interpretation?
QUESTION [1 upvotes]: For the simple quantum harmonic oscillator we can solve Schrodinger's equation and derive the analytic form of the eigenstates of e.g. a non relativistic electron in a harmonic potential. We may then go on to define ladder operators which enable us to move between eigenstates of the potential. One interpretation of these ladder operators is that they 'create' or 'destroy' a photon of energy hw. One can go on to define the Number operator which enables us to determine how many times the oscillator has been excited or equivalently how many 'photons' there are in the system.
My question is: is it correct to say the wavefunction we derived for the different energy levels still just describes the single non-relativistic electron and doesn't describe the newly created/destroyed photons.
If this is correct then where does the interpretation of 'creation' of particles even come from because it seems like we are not describing the produced photons with any wavefunction we are merely postulating that they have been created? What is a good intuitive explanation for the interpretation in terms of photon creation?

REPLY [0 votes]: To me, the boson interpretation comes only clear when one considers a many particle situation.
When we deal with one single harmonic oscillator, one has one creation $a^{\dagger}$ and one annihilation operator $a$. Each time you act with $a^{\dagger}$ you are essential raising the total energy of the oscillator by one quanta of energy. One way to see this is that $H= \epsilon (n+ \frac{1}{2})$, i.e. your total energy only depends on how many times you acted with $a^{\dagger}$ as $H$ essentially counts how many times this happens and multiplies it by some energy gap $\epsilon$. The interpretation is that you have one single particle which can be excited upwards or downwards with the ladder ops. Note I didn't mention bosons yet as it wouldn't make sense as explained below.
Now consider a chain of independent harmonic oscillators. We are dealing with the same story as before as the oscillators are completely unaware of each other. Now we label each oscillator with an index $i$ to keep track of it. We also label their associated creation $a^{\dagger}_i$ and annihilation operators $a_i$. If you play a bit with the algebra, you can show that they satisfy the canonical commutation relationships,
$$
[a_i, a^{\dagger}_j] =\delta_{ij} \quad [a_i,a_j]=[a_i^{\dagger},a_j^{\dagger}]=0
$$
which is a defining property for bosonic particles (whether or not they are fundamental is a different story, they can, and are, quasiparticles).
The interpretation now is clear, at each site you can create/destroy a boson with its corresponding creation and annihilation operators. The energy of the system is simply the sum of the energies at each site, or the number of excitations/bosons we have at each site,
$$
H= \sum_i \epsilon (n_i+ \frac{1}{2}).
$$

To finish off, yes, the wave function that solves the Schrödinger equation for a particle in a harmonic potential will always be a valid wave function for such situation. Now, you can alternatively interpret it as bosonic quasiparticles exactly as explained above.


TITLE: An ultrafilter is a set of subsets containing exactly one element of each finite partition: reference request
QUESTION [21 upvotes]: There are probably dozens of ways of defining "ultrafilter".  The definition I've seen most often involves first defining "filter", then declaring an ultrafilter to be a maximal filter.
But there's another, shorter way to state the definition:

Let $X$ be a set.  An ultrafilter on $X$ is a set $\mathcal{U}$ of subsets such that for all partitions
  $$
X = X_1 \amalg \cdots \amalg X_n
$$
  of $X$ into a finite number $n \geq 0$ of subsets, there is exactly one $i$ for which $X_i \in \mathcal{U}$.

I'd be amazed if this wasn't in the literature somewhere, but I haven't been able to track it down.  Can anyone help?
Actually, there's an even more economical definition: instead of allowing $n$ to be any natural number, you take it to be 3.  Thus, the condition is that whenever $X = A \amalg B \amalg C$, exactly one of $A$, $B$ and $C$ is in $\mathcal{U}$.  (The same thing works with 4, or 5, etc., though not with 2.)  I'm mostly interested in the version with arbitrary $n$, which seems more natural, but if you've seen the $n = 3$ version in the literature then I'd like to hear about that, too.
Edit To be clear, when I use the word "partition" I don't mean to imply that the sets $X_i$ are nonempty.  I just mean a family of pairwise disjoint sets $X_i$ whose union is $X$.  They can be empty.

REPLY [12 votes]: You can find that characterization (even with n = 3), as well as the generalization to $\kappa$-complete ultrafilters, in: Fred Galvin and Alfred Horn, Operations preserving all equivalence relations, Proc. Amer. Math. Soc. 24 (1970), 521-523.


TITLE: Basis of adjoint operator
QUESTION [2 upvotes]: I have an exam coming up and I have a whole bunch of questions to study from, and I am realizing that I really don't have a good grasp on basis matrices for linear transformations. One of the questions is:

For a linear map $f$ in $Hom_F( V,W)$ the (operator) adjoint $f^*:W^*\rightarrow V^*$ is defined by $f^*(g)=g\circ f$. If $f$ has matrix $A$ in some basis, what is the matrix of $f^*$ in the dual basis?

Similarly, here is another question that Im struggling a bit with that would help if I could get some intuition on...

A linear map between finite dimensional vector spaces $f:V\rightarrow W$ corresponds to a matrix when bases for $V$ and $W$ are chosen. HOw does this matrix change if we change bases in $V$ and $W$.

REPLY [1 votes]: Let $(w_j)_{j=0}^{m-1},(g_j)_{j=0}^{m-1}$ be a basis-dual basis pair of $W$ and $W^*$. Let $(v_k)_{k=0}^{n-1},(f_k)_{k=0}^{n-1}$ be a basis-dual basis pair of $V$ and $V^*$. Suppose that $\Phi \in \operatorname{Hom}(V,W)$. Let $\mathbf M$ be the matrix of $\Phi$ w.r.t. $(w_j)_{j=0}^{m-1},(v_k)_{k=0}^{n-1}$ and $\mathbf N$ be the matrix of $\Phi^*$. By definition,
$$ \Phi^* g_j = \sum_{r=0}^{n-1} \mathit N_{j,r} f_r
$$
Hence
$$ (\Phi^* g_j)(v_k) = \sum_{r=0}^{n-1} \mathit N_{j,r} f_r(v_k) = \mathit N_{j,k}
$$
On the other hand,
$$ (\Phi^* g_j)(v_k) = g_j(\Phi v_k) = g_j \left(\sum_{r=0}^{m-1} \mathit{M}_{k,r} w_r\right) = \mathit{M}_{k,j}
$$
Hence $\mathbf M = \mathbf N^\intercal$.
Your second question can be solved with change of basis matrix. That is, the identity operator w.r.t. two different bases.


TITLE: Prove $2^n-1$ is divisible by $(2^m-1)^2$ iff...
QUESTION [1 upvotes]: Prove that for any $m,n\in\mathbb{Z}^+$, the number $2^n-1$ is divisible by $(2^m-1)^2$ iff the number $n$ is divisible by $m(2^m-1)$.
My work so far:
It's an if and only if statement, so the common approach is to split the proof into two "if then" proofs ($A\rightarrow B \text{ and } B\rightarrow A$).
Starting with $A\rightarrow B$, for a direct proof, I would assume $A$ and prove $B$. I can use the definition of divisibility:
Assume $n=m(2^m-1)k$ for some integer $k$
Then I need to prove that $2^n-1=(2^m-1)^2h$ for some integer $h$
So I'm thinking I need to start with the expression $2^n-1$ and transform it to $(2^m-1)^2h$, or vice versa, and I need to use the assumption $n=m(2^m-1)k$ somehow.
I notice that the LS of the equality contains the variable $n$ while the RS contains $m$, and the assumption relates $n$ to $m$, so that could be a way forward. For example, simple substitution gives $2^{m(2^m-1)k}-1$, but I'm not sure what I could do from here. Perhaps there's a divisibility theorem that could help?
Other things I notice: $2^n-1$ and $(2^m-1)^2$ are always odd. Also, the divisors in both cases are multiples of $2^m-1$.
Any help/hints appreciated.

REPLY [3 votes]: Hints.
$$\gcd(2^n-1,2^m-1) = 2^{\gcd(m,n)}-1,\tag{1} $$
hence in order that $(2^m-1)\mid (2^n-1)$, we must have $m\mid n$, and this is a sufficient condition, too, since:
$$ y^a-1 = (y-1)\cdot(y^{a-1}+\ldots+y+1).\tag{2} $$
Suppose now that $n=mh$. We have:
$$ 2^n-1 = (2^m-1)\cdot\left(2^{m(h-1)}+\ldots+2^{m}+1\right)\tag{3} $$
and:
$$ 2^{mu}=\left((2^m-1)+1\right)^{u}\equiv 1\pmod{(2^m-1)}\tag{4} $$
hence in order that $(2^m-1)^2\mid (2^n-1)$ we must have $(2^m-1)\mid h$, and it is also a sufficient condition. Putting all together:
$$ (2^m-1)^2\mid (2^n-1)\quad\Longleftrightarrow\quad m(2^m-1)\mid n. \tag{5} $$