TITLE: What are the exact holomorphic Lagrangians in complex 2-space?
QUESTION [8 upvotes]: In an exact symplectic manifold, i.e. where the symplectic form can be written $\omega = d \lambda$, it's natural to look for exact Lagrangians, i.e. $L$ on which  $\lambda_L = df$.  One reason is that, any disk ending on such a Lagrangian has vanishing area:
$$ \int_D \omega = \int_{\partial D} \lambda = \int_{\partial \partial D} f = 0$$
On the other hand, the search for such manifold-without-boundary Lagrangians in cotangent bundles of manifolds-without-boundary is expected to be fruitless -- Arnol'd conjectured that all such are homotopic to the zero section, and Abouzaid has proven that they're at least homotopy equivalent to the zero section.
I am interested in understanding how hard such things are to come by in the cotangent bundle of an open manifold.  For cotangent bundles of one dimensional manifolds, the question is trivial -- a Lagrangian encloses area if it's a closed curve, and otherwise has no topology.
So the first case is $T^* \mathbb{C}$.  What I really want to know is what the holomorphic exact Lagrangian manifolds are; this is motivated in part by the article by Xin Jin which explains that such exact holomorphic Lagrangians give rise to perverse sheaves under the Nadler-Zaslow correspondence.
All this is the motivation for the following entirely elementary question:

Characterize all algebraic curves $C \subset \mathbb{C}^2$ on which the one-form $y dx$ is exact.

REPLY [5 votes]: I'm not sure what you mean by 'characterize'.  For example, here is a trivial characterization: Let $C$ be any Riemann surface, with meromorphic functions $a$ and $b$ with $da\not=0$.  Then let $(x,y) = (a, db/da)$.  The image of $C$ (minus a polar divisor) in $\mathbb{C}^2$ will be an algebraic curve on which $y\ dx = db$.  Thus, there are tons of these.


TITLE: Discrete Mathematics Logic Based
QUESTION [0 upvotes]: For our maths assignment it is heavily Logic based, with some of the questions being quite confusing, such as the one below.
My lecturer has asked the following:
In the 1730’s, the “Grande Loge” of Freemasons in Paris was a
highly secretive society following some rather bizarre rules. Each of the
freemasons in the lodge had shaved one other member. No freemason
in the lodge had ever shaved himself. Furthermore, no freemason was
ever shaved by more than one member of the lodge. There was one
freemason who had never been shaved by any other member of the
lodge. The number and identity of the freemasons in the lodge was
kept secret. One rumour circulating in Paris at that time was that there
were less than a hundred freemasons in the “Grande Loge.” Another
rumour put the number at over a hundred. Which one of the two
rumours is true? Justify your answer.
I have no idea what the answer is and would love a bit of help in solving it thanks.

REPLY [0 votes]: HINT: Let $F_0$ be the freemason who had never been shaved by any other member of the lodge. Given a freemason $F_n$ in the lodge, let $F_{n+1}$ be the member whom $F_n$ had shaved. If $m\ne n$, can $F_m=F_n$, or must $F_m$ and $F_n$ be different members of the lodge?


text:
\section{Freely available Python code}
\label{sec:pycode}

The pseudocode described in \cref{alg:pred_limiter,alg:corr_limiter_1,alg:corr_limiter_2,alg:kriv_limiter,alg:full_method}
has been implemented into source code using the Python (version 2.7) programming language   \cite{article:python}, with use of the NumPy  library for basic mathematical functions and numerical
linear algebra tools \cite{book:numpy}, the PyPy  implementation of Python that makes use of just-in-time compilers to drastically improve runtime \cite{code:pypy}, and Matplotlib  for
visualization \cite{article:matplotlib}. The resulting code is publicly available for download \cite{code:Pythonic_LxW_DG}.

The numerical time-step as described in \cref{alg:full_method}, at least in principle, can be made arbitrarily high-order. For practical reasons, the implementation in \cite{code:Pythonic_LxW_DG}
is currently limited to orders of accuracy from $\morder=1$ to $\morder=5$. One limitation with
very high-order schemes is that the maximum Courant-Friedrichs-Lewy number
for which the scheme is still stable,
\begin{equation}
\text{CFL} = |\lambda|_{\text{max}} \, \frac{\Delta t}{\Delta x},
\end{equation}
{\it decreases} with increasing $\morder$.
In the above expression, $|\lambda|_{\text{max}}$ is a bound on the maximum spectral radius of the flux Jacobian over the entire mesh and over the current time-step, $[t^n,t^n+\Delta t]$. The maximum allowable CFL number decreases roughly
as the inverse of $\morder$ (e.g., see \cite{article:GuRo2017} and references therein).
We catalog the the CFL numbers used in the implementation of
\cite{code:Pythonic_LxW_DG} in \cref{table:cfl_numbers}.

The Pythonic Lax-Wendroff DG code \cite{code:Pythonic_LxW_DG} we developed
is structured so that the top-level directory contains all of
the application-specific sub-directories
(Burgers, shallow water, and Euler), as well as the {\tt lib} sub-directory that contains
the main LxW-DG and plotting functions. In each application-specific sub-directory there
are several specific numerical example sub-directories, as well as another
{\tt lib} directory that contains information about the fluxes and limiters for that particular
equation. Each numerical example sub-directory contains the following
three files that are required to run the main code:
\begin{enumerate}
\item \texttt{parameters.py}: \, set parameters for all values needed in the simulation;
\item \texttt{run\_example.py}: \, set initial conditions for simulation and execute main routine;
\item \texttt{plot\_example.py}: \, using Matplotlib, create plots of desired variables.
\end{enumerate}
Inside each numerical example sub-directory, the Pythonic Lax-Wendroff DG code can be run
by executing the following Makefile commands:
\begin{enumerate}
\item \texttt{make run}: \, executes code with parameters set in \texttt{parameters.py} and
initial conditions set in \texttt{run\_example.py};
\item \texttt{make plot}: \, executes main plotting routine with options set in \texttt{plot\_example.py}.
\end{enumerate}

\begin{table}[!t]
\begin{center}
\begin{Large}
\begin{tabular}{|c||c|c|c|c|c|}
\hline
{\normalsize $\morder$} & {\normalsize 1} & {\normalsize 2} & {\normalsize 3} & {\normalsize 4} & {\normalsize 5} \\
\hline\hline
{\normalsize $\text{CFL} = |\lambda|_{\text{max}} \frac{\Delta t}{\Delta x}$} &
{\normalsize 0.90} & {\normalsize 0.30} & {\normalsize 0.14} & {\normalsize 0.10} & {\normalsize 0.06} \\
\hline
\end{tabular}
\caption{List of CFL numbers as used in actual simulations
 for locally-implicit Lax-Wendroff DG schemes of
various orders as implemented in \cite{code:Pythonic_LxW_DG}.
These CFL numbers are slightly reduced from the values on the linear stability boundary. Note that these CFL numbers decrease roughly as the inverse of the method order $\morder$.\label{table:cfl_numbers}}
\end{Large}
\end{center}
\end{table}

config: arxiv


text:
{\bf Problem.} Two circles have the same center $C.$ (Circles which have the same center are called concentric.) The larger circle has radius $10$ and the smaller circle has radius $6.$ Determine the area of the ring between these two circles. [asy]
import graph;
filldraw(circle((0,0),10), lightgray, black+linewidth(1));
filldraw(circle((0,0),6), white, black+linewidth(1));
dot((0,0));
label("$C$",(0,0),NE);
[/asy]
{\bf Level.} Level 4
{\bf Type.} Prealgebra
{\bf Solution.} The area of a circle with radius $r$ is $\pi r^2.$

So the area of the larger circle is $\pi(10^2)=100\pi$ and the area of the smaller circle is $\pi(6^2)=36\pi.$

The area of the ring between the two circles is the difference of these two areas.

Therefore, the area of the ring is $100\pi - 36\pi = \boxed{64\pi}.$

set_name: MATH


TITLE: What can we say about the series: $-1+1-1+1-1+1-1+\cdots=\sum_{n=1}^{\infty}(-1)^n$?
QUESTION [0 upvotes]: What can we say about this series? $$-1+1-1+1-1+1-1+\cdots=\sum_{n=1}^{\infty}(-1)^n$$
Intuitively, the sum of it seems to converge to zero, as each term cancels the one before it, although none of the convergence tests seem to work on the series. Also I have no idea as for how to calculate the limit     $\lim\limits_{n \to \infty}a_{n} = \lim\limits_{n \to \infty}(-1)^n$, which would have enabled me to at least some degree to determine what is going to happen with this sum.

REPLY [0 votes]: Just take a look at the definition of a limit. We say that:
$$\lim_{n\rightarrow\infty}a_n=a$$
IF for any $\varepsilon>0$, there is an $N$ such that for all $n>N$, we have:
$$|a_n-a|<\varepsilon$$
This is not the case for $0$, $1$ or $-1$.


TITLE: Use the Karush-Kuhn-Tucker conditions to find the optimal solution
QUESTION [0 upvotes]: Use the Karush-Kuhn-Tucker conditions to find the optimal solution to the following non-linear programming problem.
$$
\begin{gathered}
\min z=\left(x_{1}-1\right)^{2}+\left(x_{2}-2\right)^{2} \\
\text { s.t. }-x_{1}+x_{2}=1 \\
x_{1}+x_{2} \leq 2 \\
x_{1}, x_{2} \geq 0
\end{gathered}
$$

I followed this site to solve this problem,

Note: at the optimum, it is known that the inequality constraint is satisfied but not
binding. Take advantage of this information.

Question: I couldn't understand the note mentioned that site.
Dropping the inequality constraint, we get,
$$
\begin{align}
\pmatrix{2x_1-2\\2x_2-4}-\lambda_1\pmatrix{-1\\2}&=\pmatrix{0\\0}\\
\pmatrix{2x_1\\2x_2}+\pmatrix{\lambda_1\\-2\lambda_1}&=\pmatrix{2\\4}
\end{align}
$$
Use the equality constraint $-x_1+x_2+0.\lambda_1=1$,
$$
A=\pmatrix{-1&1&0\\2&0&1\\0&2&-2}\quad b=\pmatrix{1\\2\\4}
$$
$$
\pmatrix{x_1\\x_2\\\lambda_1}=A^{-1}b=\pmatrix{1\\2\\0}
$$
Hence, the $z_{\min}=(1-1)^2+(2-2)^2=0$.

But the solution provided in my textbook was $z_{\min}=\frac12,x_1=\frac12,x_2=\frac32$. Did their procedure is wrong? I used that because it seems short. But now it wasn't giving the exact solution.

REPLY [1 votes]: The information given to you is wrong. The inequality is binding as we can see that the circle touches the halfspace.

To make use of your working, you can assume that you were considering the case where it is not binding and you found a contradiction. Hence we know that it is binding, now you can solve the problem by solving $-x_1+x_2=1, x_1+x_2=2$.


TITLE: square roots algebra
QUESTION [0 upvotes]: I'm trying to follow along with this: https://dandavison.github.io/fibonacci-eigenbasis.html
And I'm getting stuck on what's presumably very basic algebra. The step that's giving me trouble is going from:
$$
\frac{-1}{2\sqrt5}
\begin{bmatrix}
-5-\sqrt5 & 0 \\
0 & 5-\sqrt5 \\
\end{bmatrix}
$$
to:
$$
\frac{1}{2}
\begin{bmatrix}
1+\sqrt5 & 0 \\
0 & 1-\sqrt5 \\
\end{bmatrix}
$$
the closest i can get is:
$$
\frac{1}{2}
\begin{bmatrix}
5(1+\sqrt5) & 0 \\
0 & 5(1-\sqrt5) \\
\end{bmatrix}
$$
and I don't understand how the extra factor of 5 comes out. What am I missing here?

REPLY [4 votes]: The author distributed the $-\frac{1}{\sqrt{5}}$ into each component of the matrix. This gives us
$$-\frac{1}{\sqrt{5}}(-5-\sqrt{5}) = \frac{5}{\sqrt{5}}+\frac{\sqrt{5}}{\sqrt5} = \sqrt5+1$$
and
$$-\frac{1}{\sqrt{5}}(5-\sqrt{5}) = -\frac{5}{\sqrt{5}}+\frac{\sqrt{5}}{\sqrt5} = -\sqrt5+1$$


TITLE: In coding the math, how would you calculate how many times the size of one cities population is to another? Like Quebec. pop to Copenhagan pop?
QUESTION [1 upvotes]: I have a Quebec population : $8,164,361$
And Copenhagen population : $775,033$

How would I be able to determine the how many times the size of one is to another?
For example I know for Quebec the inclusive inverse would read:
Quebec is $1.0$ times the size of Quebec
and then for two other city populations it should read like:
Beijing is $31.56334$ times the size of Washington D.C.
I imagine this is very simple math. Would I just divide one by the other if so which one goes in the denominator and numerator? Or is there something more to this than meets the eye.
Thank you

REPLY [1 votes]: The simpler and standard way is to calculate the following ratio
$$\frac{8164361}{775033}\approx 10.5$$
and say that Quebec, as population, is 10 times larger than Copenaghen.
As an alternative we can take for example
$$\frac{8164361-775033}{8164361}\approx 0.905=90.5\%$$
and say that Copenaghen has $90.5\%$ poplation less than Quebec or
$$\frac{8164361-775033}{775033}\approx 9.53=953\%$$
and say that Quebec has 953% more pupulation than Copenaghen.

REPLY [0 votes]: City $A$'s population is $\dfrac{p_A}{p_B}$ times the size of that of city $B$, where $p_A$ and $p_B$ denote the population of cities $A$ and $B$ respectively.


TITLE: Contravariant associate to volume form
QUESTION [3 upvotes]: The volume form $\epsilon$ on a pseudo-Riemannian manifold is a top-degree form given in coordinates by $\epsilon = \sqrt{|g|}dx^1 \wedge \cdots \wedge dx^n$, where $|g|$ is the absolute value of the determinant of the matrix representation of the metric $g$ in coordinates $(x^1, \ldots, x^n)$.
Is there a concise form for the $n$-contravariant tensor $\epsilon^{a_1\cdots a_n}$ obtained by raising all the indices on $\epsilon$? I have seen that it should be related to $1/\sqrt{|g|}$ but cannot find a proof.

REPLY [1 votes]: With upper indices, the expression in coordinates is
$$
  \epsilon^{a_1\cdots a_n}
  = \frac{(-1)^s}{\sqrt{\lvert g \rvert}} \mathrm{sgn}(k \mapsto a_k)
$$
where $(-1)^s$ is $g/\lvert g \rvert$ (e.g. $1$ if Riemannian, $-1$ if Lorentzian) and $\mathrm{sgn}(k \mapsto a_k)$ is the sign of the permutation.
One way to think of things is that the volume form $\epsilon$ (a.k.a. Levi-Civita tensor) is normalized so that
$$
  \frac{1}{n!} g^{a_1 b_1} \cdots g^{a_n b_n} \epsilon_{a_1 \cdots a_n} \epsilon_{b_1 \cdots b_n} = (-1)^s.
$$
We can solve this equation to find that $\epsilon_{1\cdots n} = \sqrt{\lvert g \rvert}$ and $\epsilon_{a_1 \cdots a_n} = \sqrt{\lvert g \rvert} \mathrm{sgn}(k \mapsto a_k)$; the determinant $g$ makes its appearance because we're summing over permutations. And since the left-hand side of the equation can also be though of as
$$
  \frac{1}{n!} \epsilon^{a_1 \cdots a_n} \epsilon_{a_1 \cdots a_n},
$$
we immediately also find $\epsilon^{a_1 \cdots a_n}$.
(Wald, General Relativity, Appendix B essentially takes the view just outlined, but stops short of explicitly stating the formula $\epsilon^{a_1 \cdots a_n}$. Carroll, Spacetime and Geometry, Section 2.8 states the formula, although with not as much proof. Note that the lecture notes which formed the basis of Carroll's book are openly available on his website.)


TITLE: Is it true $P(\cap_{n=1}^\infty \cup_{m=n}^\infty A_m)=\lim_{n \to \infty}P(\cup_{m=n}^\infty A_m)$?
QUESTION [2 upvotes]: I just need some confirmation on the truth of this since I didn't see it in a textbook so far. I think it is true by continuity of probability. Am I right?

REPLY [3 votes]: Let $B_n:=\bigcup_{m\ge n}A_m$. Then $B_{n+1}\subseteq B_n$ and by the continuity from above,
$$
\mathsf{P}\left(\bigcap_{n\ge 1} B_n\right)=\lim_{n\to\infty} \mathsf{P}(B_n).
$$

REPLY [2 votes]: For each $n \in \mathbb N$, let $B_n=\bigcup_{m=n}^\infty A_m $. Clearly for each $n \in \mathbb N$ we have $B_n \supseteq B_{n+1}$, and so  $\{B_n\}_{n=1}^\infty$ is a decreasing sequence of sets. Moreover, $P(B_1)<\infty$ as a probability measure is finite. Therefore, $$P\left(\bigcap_{n=1}^\infty B_n\right)=\lim_{n\to \infty} P(B_n)$$ by the continuity from above property of a measure which is $(3')$ here.


TITLE: Question about Weyl character formula
QUESTION [7 upvotes]: In the book of Humphreys, page 139, Weyl character formula is $$\left(\sum_{w\in W} \operatorname{sn}(w)\epsilon_{w\delta}\right) * \operatorname{ch}_{\lambda} = \sum_{w\in W} \operatorname{sn}(w) \epsilon_{w(\lambda + \delta)}\tag{1} $$ where $*$ is convolution but not usual multiplication of functions. In other books, Weyl formula is given by $$ \operatorname{ch}_{\lambda} = \frac{\sum_{w\in W} \operatorname{sn}(w) \epsilon_{w(\lambda + \delta)}}{\sum_{w\in W} \operatorname{sn}(w)\epsilon_{w\delta}}.\tag{2}$$ Are these two formulas (1) and (2) equivalent? In some other books, Weyl formula is given by $$ \operatorname{ch}_{\lambda} = \frac{\sum_{w\in W} \operatorname{sn}(w) \epsilon_{w(\lambda + \delta)-\delta}}{\sum_{w\in W} \operatorname{sn}(w)\epsilon_{w\delta - \delta}}.\tag{3}$$ Is this because in (2) $w \in W$ act as dot pruduct and (3) $w\in W$ act as usual Weyl group action? If we are given a highest weight $\lambda$ explicitly, say, $\lambda = 3\omega_1 + 2\omega_2$, where $\omega_1, \omega_2$ are fundamental weights of $\mathfrak{g} = \mathfrak{sl}_3$ (type $A_2$), how can we compute explicitly the character of the irreducible $\mathfrak{g}$-module $V_{\lambda}$ with highest weight $\lambda$ using formula (2)? I am confused with the form of the formula since the character should be of the form $\operatorname{ch}(V_{\lambda}) = \sum_\mu\dim(V_\mu)\epsilon_\mu$ but the Weyl formula is not of this form. Thank you very much.

REPLY [7 votes]: Hmm. IMHO convolution is the natural product on a group ring. May be you are used to viewing them as functions as opposed to elements of the group ring $\mathbf{Z}[\Lambda]$, where $\Lambda$ is the additive group of weights (= a free abelian group generated by the fundamental dominant weights)? A function $f:\Lambda\rightarrow \mathbf{Z}$ with a finite support is turned into an element of the group ring naturally as $\sum_{\lambda\in\Lambda}f(\lambda)e_{\lambda}$. The quantities $e_\lambda$ are multiplied by the natural formula $e_{\lambda}*e_{\mu}=e_{\lambda+\mu}$, and the convolution is just the linear extension of this (according to the distributive law).
Weyl character formula is actually very useful for the rank two cases, where you can just stare at the diagram. As I don't have a useful tool for drawing a sequence of 2-D diagrams of weight lattices, let's do an example of the case $A_1$, where you should know the formal character anyway (so we can use that to test the formula). In that case we can identify the weight $m\lambda_1$ with the integer $m$. In the group ring we then have $e_{m\lambda_1}=e_{\lambda_1}^m=z^m$, where I use the abbreviation $z=e_{\lambda_1}$. With this identification $\delta=\lambda_1=1, e_\delta=z$. The only non-trivial element of the Weyl group in this case is the negation of weights, and that has length =1 and hence a negative sign. Therefore
$$
\sum_{w\in W}sn(w)e_{w\delta}=e_\delta - e_{-\delta}= z -z^{-1}
$$
and
$$
\sum_{w\in W}sn(w)e_{w(\lambda+\delta)}=e_{\lambda+\delta} - e_{-\lambda-\delta}= z^{m+1} -z^{-m-1},
$$
where again $\lambda=m\lambda_1, m\ge0$. This time we know from $sl_2$-theory that
$$
ch_\lambda=\sum_{\mu}\dim(V_\mu)e_\mu=z^m+z^{m-2}+z^{m-4}+\cdots+z^{4-m}+z^{2-m}+z^{-m},
$$
because for the irreducible module $V(m)$ of highest weight $m$ the dimensions of the weight space $V(m)_k$ is equal to 1, if $k$ is in the arithmetic progression $m,m-2,m-4,\ldots,-m+2,-m$ and 0 otherwise. So in this case the Weyl character formula says that
$$
z^m+z^{m-2}+z^{m-4}+\cdots+z^{4-m}+z^{2-m}+z^{-m}=\frac{z^{m+1} -z^{-m-1}}{z -z^{-1}}.
$$
This identity in the group ring $\cong \mathbf{Z}[z,z^{-1}]$ is easy to verify and/or derive. Either you can write the l.h.s. as a geometric sum, or you can multiply both sides with the denominator $z-1/z$, and the usual avalanche of termwise cancellations gives you the result. Note that the group ring is an integral domain, so your equations (1) and (2) are equivalent.
In the case of $A_2$ we can proceed similarly. For example, if we write $z_i=e_{\lambda_i}$ and $s_i=s_{\alpha_i}$, then $\delta=\lambda_1+\lambda_2$,
$s_1(\delta)=\delta-\alpha_1=-\lambda_1+2\lambda_2=\alpha_2$,
$s_2(\delta)=\delta-\alpha_2=2\lambda_1-\lambda_2$, $s_1s_2(\delta)=-2\lambda_1+\lambda_2$, $s_2s_1(\delta)=\lambda_1-2\lambda_2$ and finally $s_1s_2s_1(\delta)=-\lambda_1-\lambda_2$. Therefore (taking the parity of the length of the group element into account)
$$
\sum_{w\in W}sn(w)e_{w\delta}=z_1z_2-z_1^{-1}z_2^2-z_1^2z_2^{-1}+z_1^{-2}z_2+z_1z_2^{-2}-z_1^{-1}z_2^{-1},
$$
where I wrote the terms (i.e. the elements of $W$) in the same order as above.
To do your example, you need to also compute the similar expansion for $\sum_{w\in W}sn(w)e_{w(\lambda+\delta)}$. Then the Weyl character formula tells you that $ch_{\lambda}$ is the element of the group ring $\mathbf{Z}[z_1,z_2,z_1^{-1},z_2^{-1}]$ that is the (unique) solution of the equation (1) or (of equation (2)).
This answer has just become unmanageably long, so I cannot continue. I recommend that you first work out the cases $\lambda=\lambda_1$, $\lambda=\lambda_2$ and $\lambda=\delta$ where you know the answer already: the first rep in this list is the natural 3-dimensional rep of $sl_3$. The second is its dual, and the last is the 8-dimensional adjoint representation as its highest weight is also the highest root. If your have problems with those, ask another question showing where you got stuck, and somebody here will take a look. Ping me, if necessary.
Edit: I just recalled that I already wrote an example on how to apply Weyl's formula in another answer.