TITLE: Classical Mechanics contradicts Conservation of energy?
QUESTION [6 upvotes]: Imagine a Stanford torus rotating with 1 rpm so that centripetal/reactive centrifugal acceleration provides about 1.0g of artificial gravitational acceleration inside the ring. The picture below shows a Stanford torus in a cutaway view.

The rotating torus has a specific rotational energy (let's say 100 J). A mass of 1 kg is located 1 m above the ground inside the torus.
Step 1:
The mass is dropped. The torus' moment of inertia increases, its angular velocity decreases. The rotational energy of the torus does not change.
Rotational energy of the torus after step 1: 100 J.
Step 2:
¾ of the torus' rotational energy is converted into electric energy. We assume no heat losses in the process. The electric energy is temporarily stored.
Rotational energy of the torus after step 2: 25 J.
Stored electric energy: 75 J.
Step 3:
The mass is lifted again. The torus' moment of inertia decreases, its angular velocity increases. The rotational energy of the torus does not change.
Rotational energy of the torus after step 3: 25 J.
Stored electric energy: 75 J.
Step 4:
The stored electric energy is converted into rotational energy of the torus. Again, we assume no heat losses in the process.
Rotational energy of the torus after step 4: 100 J.
Stored electric energy: 0 J.
The sum of rotational energy and electric energy is equal in all steps.
The critical actions are accelerating/decelerating the torus. When rotational energy is invested in or taken out of the torus, not only does the rotational energy of the torus change but also the potential energies of the masses inside the torus. In step 2 potential energy "disappears" and in step 4 potential energy is "created". Energy is gained in this example since the mass is lifted under less acceleration and is dropped when acceleration is greater. But the process can be done the other way around.
This would indicate a contradiction of classical mechanics and energy conservation. Do you see any mistakes or something that I missed?

REPLY [1 votes]: Step 1
  The mass is dropped ... The rotational energy of the torus does not change.

Going to stop you right there. Although total energy is indeed conserved, rotational energy does not need to be conserved. Remember that the potential energy of dropped objects is generally converted to heat when they go thud on the ground. Angular momentum is however conserved in this step.
In fact, it is simply impossible for rotational energy ($L^2/2I$) to stay the same when a rotating object adjusts its mass distribution without interacting with other objects (changing moment of inertia $I$ while keeping constant angular momentum $L$). When the mass drops, energy is released and if not captured that energy goes to heat.


TITLE: Pre-images under the multiplication map $(x,y) \mapsto xy$ of open intervals are open
QUESTION [0 upvotes]: Let $f: \mathbb{R^2} \to \mathbb{R}$ be $f(x,y) = xy$. Find the pre-image $f^{-1}((a,b))$ of an open interval $(a, b) \subseteq \mathbb{R}$, and show that this pre-image is open in $\mathbb{R^2}$.

I can't seem to conceptualize this, and I didn't know what the graph looked like during the exam either. If we have an interval $(a, b)$, then the pre-image will be a bunch of ordered pairs/coordinates on the $\mathbb{R^2}$ plane. So the pre-image of an interval like $(2, 3)$ would have some coordinates whose first and second components multiply to some number in the interval $(2, 3)$. How can I generalize this, and also show this generalization is an open set in $\mathbb{R^2}$?

REPLY [1 votes]: If you consider $(-1,1)\subset \mathbb{R}$ then you can easily verify that the preimage of $(-1,1)$ under $f$ is the region between the graph of the function $g(x)=\frac{1}{x}$.

REPLY [1 votes]: First, addressing your visualization question. You might first ask yourself: what is the set of points satisfying $f(x,y)=c$, where $c$ is some constant, say 1. Well, $f(x,y)=1$ is satisfied by the poitns on the arcs $y=1/x$. $f(x,y)=2$ is satisfied by the points on the arcs $y=2/x$. Draw these arcs on the same diagram. Now, what is the set of points satisfying $f(x,y)=c$ where $1<c<2$? It is the "open band of arcs" between what you currently have drawn on your diagram. Intuitively, it looks open (read: intuitively).
Now, we prove rigorously that $f$ is continuous, hence solving the problem. That multiplication is continuous as a map from $\mathbb{R}^2 \to \mathbb{R}$ can be found easily on the web. Here's a proof.
Proof: Fix the point $(x_0,y_0)$ and $0<\epsilon<1$. Take $\delta = \min(\frac{\epsilon} {|x_0|+|y_0|+1},1)$. Then for $(x,y)$ within $\delta$ distance from $(x_0,y_0)$, we have (by definition of norm on $\mathbb{R}^2$ as max of coordinate difference) that $|x-x_0|,|y-y_0|<\delta$ and therefore:
$$|xy-x_0y_0|\\
=|(x-x_0)(y-y_0)+x_0(y-y_0)+y_0(x-x_0)|\\
\le|x-x_0||y-y_0|+|x_0||y-y_0|+|y_0||x-x_0|\\
<\delta^2 + \delta |x_0| + \delta |y_0|\\
\le \delta(1+|x_0|+|y_0|)\\
\le\epsilon$$


TITLE: Radial solutions to Possion equation with Robin condition in annulus
QUESTION [2 upvotes]: Let $\Omega$ be an annulus with inner radius $R_1$ and outer radius $R_2$, and $\beta>0$. Let $u$ solve the equation
$$
\begin{cases}
-\Delta u=1\quad &\mbox{in $\Omega$}\\
\frac{\partial u}{\partial\nu}+\beta u=0\quad &\mbox{on } \partial \Omega
\end{cases}
$$
Is $u$ necessarily radial for every $\beta>0$? If so, is it possible to explicitly write down the solution?

REPLY [4 votes]: Solutions in $C^2(\Omega) \cap C^1(\bar{\Omega})$ are indeed unique (though, of course, this can get swapped for Sobolev spaces or others if you prefer, but the basic idea is similar).  If you have two solutions $u_1$ and $u_2$, let $u = u_1 - u_2$, multiply the PDE they satisfy by $u$ and integrate by parts.  You'll find that
$$
\int_\Omega |\nabla u|^2 + \int_{\partial \Omega} \beta |u|^2 =0
$$
and hence that $0 = u = u_1 - u_2$ in $\Omega$.
Once you have uniqueness in hand, you can just propose a radial ansatz and see if it works.  If it does, then that's the solution.  A radial solution $u(x) = v(|x|)$ must satisfy
$$
v''(r) + \frac{(n-1)}{r} v'(r) = -1
$$
and thus
$$
(r^{n-1} v'(r)) ' = - r^{n-1}.
$$
Integrate twice to get the general form with two free coefficients, and then plug into the BCs to determine the coefficients.  I'll leave this to you to work out.


TITLE: Age of the universe in years
QUESTION [0 upvotes]: It seems to be commonly accepted that the Big Bang occurred roughly 13.7 billion years ago.  My question is what is the meaning of the year in this context?
When I type year definition into google, it returns this:

The age of the earth likewise is accepted 4.54 billion years.  I'm assuming that by years old, we mean Earth years.
Now, if the earth only existed for 4.54 billion years

What is the meaning of the year before the Earth existed?
Why is the age of the universe even expressed in years as opposed to something like distance?

It appears that 13.7 billion (Earth) years is completely meaningless since a year depends on the existence of the Earth.

REPLY [1 votes]: There is just a caveat in expressing the age of the universe in terms of earth years. This is implicitly assuming that the clock being used to measure is at rest with respect to earth or at least has the same astronomical conditions as the earth. Furthermore, you would be implicitly assuming that this clock holds this same condition throughout the time evolution of the universe. What does that even mean? When we try to measure the total time it took for the universe to evolve, (I could be wrong here), its as if we assume we have a clock "outside of the universe" with conditions identical to that of earth, starts the clock at the Big Bang, then waited for the universe to come to its present condition then checked the elapsed time on this hypothetical clock. It is absolutely necessary to assume that the clock has to have the same condition as earth, otherwise, the time measurement would not be in "earth time". I not sure if this kind of hypothetical clock is possible even in principle. We know that clocks don't run at the same rate and is affected by both gravity (general relativity) and special relativity.
One possible way is to define a clock then allow it to evolve along with the universe. But then, how do you transform the time measurement of this clock with respect to earth years? This can get really complicated.


TITLE: Finding the limit $\lim_{t \to \infty} \frac{t-t\sqrt{t}}{2t^{3/2}+3t-5} $
QUESTION [1 upvotes]: Can someone help me solve this limit?
$$\lim_{t \to \infty} \frac{t-t\sqrt{t}}{2t^{3/2}+3t-5} $$

REPLY [1 votes]: Thx @amWHy
$$\lim_{t \to \infty} \frac{t-t\sqrt{t}}{2t^{3/2}+3t-5}
\implies
\lim_{t \to \infty} \frac{t-t^{3/2}}{2t^{3/2}+3t-5}
\implies
\lim_{t \to \infty} \frac{\frac{t}{t^{3/2}}-\frac{t^{3/2}}{t^{3/2}}}{\frac{2t^{3/2}}{t^{3/2}}+\frac{3t}{t^{3/2}}-\frac{5}{t^{3/2}}}
\implies
\lim_{t \to \infty} \frac{0-1}{2+0-0}
\implies
\lim_{t \to \infty}-\frac{1}{2}$$


TITLE: How to take the derivative of this function?
QUESTION [0 upvotes]: After 2-3 hours thinking about taking derivative of this function:

And finding its answer on the net, don't get how they've taken its derivative using Euler equation. the drivative of this function is this:

This is the Euler equation:

Would you please explain me step by step how they take its drivative.
Added:
This is the whole of that problem and its solution which I'm trying to solve:

REPLY [0 votes]: Here is the formally correct way to think about the Lagrange equations. First, note that your Lagrangian depends on $x$ only through the dependent "variables" $y$ and $y'$ (I put variables in quotes because $y'$ isn't really what a mathematician would consider a variable). Your Lagrangian
$$f=f(y,y')$$
can be rewritten as a function of two independent variables
$$f=f(v,w):\mathbb{R}^2\to\mathbb{R}$$
$$(v,w)\mapsto v\sqrt{1+w^2}$$
Your problem is asking you to first differentiate $f$ with respect to the second variable,
$$\frac{\partial f}{\partial v}(u,v)$$
then obtain a function of $x$ by plugging in $y(x)$ and $y'(x)$ into the two slots
$$g(x)=\frac{\partial f}{\partial v}(y(x),y'(x))$$
and finally differentiate $g$ with respect to $x$
$$\frac{d g}{dx} = \frac{d}{dx}\Bigg[\frac{\partial f}{\partial v}(y(x),y'(x))\Bigg]$$
Note that I used "$d$" instead of "$\partial$" because $g$ is a function of one variable. Follow this procedure and you'll get the correct answer.
Physicists use incorrect notation sometimes. The notation used in the Lagrange equations is mathematically unsound and confusing for students, although once you get used to it it's quicker than everything I typed out. In your problem, $\frac{\partial ^2}{\partial x \partial z}$ is being used inappropriately. If you evaluated $\frac{\partial ^2}{\partial x \partial z}f$, by equality of mixed partials you could differentiate with respect to $x$ first, and since $f$ is independent of $x$, the whole thing would be zero. On the other hand, applying $\frac{\partial ^2}{\partial x \partial z}$ to $f(x,y(x),v(x))$ makes no sense either, since this is a function of one variable.


TITLE: $f(x ,y)$ differentiable all over the plain. $g(u, v) = f(u^2 - v^2, u^2v).$ if $\nabla f (-3, 2) = 2 \vec i + \vec j$ , calculate $ \nabla g(1,2)$
QUESTION [1 upvotes]: unfortunately, I had to miss the lecture that gradients leant and I don't know how to solve this question.

Let $f(x ,y)$ differentiable all over the plain.
Let $g(u, v) = f(u^2 - v^2, u^2v).$
if $\nabla f (-3, 2) = 2 \vec i + \vec j$ , calculate $ \nabla g(1,2)$
($\vec i$ is unit vector $[1, 0, 0]$ and $\vec j$ is unit vector $[0, 1, 0]$.

I read about gradients, but still don't know how to calculate $\nabla g(1,2).$
Can you please help me solve this question? I'd rather to get help such as hints and general explaination, than a final solution.
Thanks in advance!

REPLY [0 votes]: $g_u=2uf_u(u^2-v^2,u^2v)+2uvf_v(u^2-v^2,u^2v)$
$g_v=-2vf_u(u^2-v^2,u^2v)+u^2f_v(u^2-v^2,u^2v)$
$g_u(1,2)=2f_u(-3,2)+4f_v(-3,2)=4+4=8$
$g_v(1,2)=-4f_u(-3,2)+f_v(-3,2)=-7$
since $\nabla f(-3,2)=(2,1)$ , means that $f_u(-3,2)=2, f_v(-3,2)=1$
So $\nabla g(1,2)=(g_u(1,2),g_v(1,2))=(8,-7)$


TITLE: Why are the $\alpha_i$ trivial on $T \cap R(G)$?
QUESTION [1 upvotes]: The reference is Linear Algebraic Groups by A. Borel.  Fix a Borel subgroup $B_0$ of a connected linear algebraic group $G$.  Let $T$ be a maximal torus of $G$.  Then there exists an embedding $G \rightarrow \textrm{GL}(V)$ and a $0 \neq w \in V$ such that $B_0$ consists of the set of $x \in G$ for which $xw$ is a scalar multiple of $w$.  Then $G/B_0$ is isomorphic to the orbit $X$ of $[w]$
 in $\mathbb{P}(V)$.
We can choose $V$ so that $X$ is not contained in any hyperplane in $\mathbb{P}(V)$.  There exits a basis $v_1, ... , v_n$ of $V$ together with characters $\alpha_1, .. , \alpha_n$ of $T$ such that $\alpha_i(t)v_i = tv_i$ for all $i$ and all $t \in T$.

I'm confused on a point in part (3) of the proposition.  One can show that there is a unique index $i_{\lambda}$ such that $\langle \alpha_{i_{\lambda}}, \lambda \rangle > \langle \alpha_i, \lambda \rangle$ for all $i \neq i_{\lambda}$.  One then shows that $[v_{i_{\lambda}}]$ is an element of $X$ which is fixed by $T$, and corresponds to the desired Borel subgroup $B(\lambda)$.
Now, Borel takes the characters $\beta_i$ in (3) to be the differences $\alpha_{\lambda_i} - \alpha_i$ for $i \neq i_{\lambda}$.  I'm confused on the argument for why $\beta_i$ is trivial on $T \cap R(G)$.  Borel says the following: "Since $\alpha_i$ are characters of $T$, trivial on $T \cap R(G)$, this proves (3)."
What I don't understand is why the $\alpha_i$ are trivial on $T \cap R(G)$.
I mean, I think I can argue that the $\beta_i$ are trivial on $T \cap R(G)$.  It's just that Borel seems to be claiming something stronger.
My proof that $\beta_i$ is trivial on $T \cap R(G)$:
More generally, I can prove that the differences $\alpha_i - \alpha_j$ for all $i, j$ are trivial on $T \cap R(G)$.  If $i = j$, there's nothing to prove, so suppose $i \neq j$.  Then there exists a point $[v] = [c_1v_1 + \cdots + c_nv_n]$ in $X$ for which both $c_i$ and $c_j$ are nonzero: otherwise, $X$ is contained in the union of two hyperplanes $H_i = \{ [a_1v_1 + \cdots + a_nv_n] : a_i = 0 \}$ and $H_j$, which we assumed not to be the case.
Now, let $t \in T \cap R(G)$.  Now $t$ fixes a point of $X$ if and only if $t$ lies in the corresponding Borel subgroup of that point.  So $R(G)$ is the set of $g \in G$ which fix every point of $X$.  In other words, $tv = \lambda v$ for some $\lambda \neq 0$.  Then $$\lambda c_1v_1 + \cdots + \lambda c_nv_n = \alpha_1(t)c_1v_1 + \cdots + \alpha_n(t) c_n v_n $$ so $\lambda c_iv_i = \alpha_i(t)c_iv_i$, and $\lambda c_jv_j = \alpha_j(t) c_jv_j$.  This implies $\alpha_i(t) = \lambda = \alpha_j(t)$.
This implies $(\alpha_i - \alpha_j)(t) = 1$, for any $t \in T \cap R(G)$.  $\blacksquare$
Is Borel correct that $\alpha_i$ is trivial on $T \cap R(G)$?

REPLY [0 votes]: I may be missing something, but as far as I understand, Borel constructs the representation of $G$ in $GL(V)$ in such a way that it factors over $G/R(G)$. Thus elements of $R(G)$ act trivially on $V$, so the weights $\alpha_i$ must map $R(G)$ to $1$.


TITLE: Is the special orthogonal group really rationally homotopy commutative?
QUESTION [4 upvotes]: It is a classical result that the rational cohomology of $SO(n)$ is given by:
$$H^*(SO(2m); \mathbb{Q}) = \begin{cases}
S(\beta_1, \dots, \beta_{m-1}, \alpha_{2m-1}) & n = 2m \\
S(\beta_1, \dots, \beta_m) & n = 2m+1
\end{cases}$$
where the $\beta_i$ have degree $4i-1$, $\alpha_{2m-1}$ has degree $2m-1$, and $S(V)$ is the free graded commutative algebra on the generators given by the graded vector space $V$. Since all the generators have odd degree, this happens to be an exterior algebra here.
Since $SO(n)$ is a (Lie) group, there is an induced Hopf algebra structure on $H^*(SO(n);\mathbb{Q})$, and the generators $\beta_i$ (and $\alpha_{2m-1}$ if applicable) happen to be primitive. But then, a free (graded) symmetric algebra on primitive generators is (graded) cocommutative, so in this case $H^*(SO(n))$ is cocommutative.
Now, like all Lie groups, $SO(n)$ is formal, and all its rational homotopy theory is contained in its cohomology ring. This is where things start to get muddy for me: if I understand correctly, the coproduct on $H^*(SO(n))$ should be a model for the product on $SO(n)$. Since $H^*(SO(n))$ is cocommutative, then the product on $SO(n)$ is rationally homotopy cocommutative.
Is this correct? If not, where's my mistake? If yes, how to interpret that fact? The group $SO(n)$ doesn't really "look" commutative at all. Maybe I'm assigning a stronger meaning than I should to the commutativity up to rational homotopy?

REPLY [4 votes]: In fact every connected Lie group is rationally homotopy commutative, in the sense that every connected Lie group is rationally homotopy equivalent (as a rational H-space) to a product of odd spheres, which are rationally homotopy commutative. This is an extremely weak form of commutativity, though.
Note that the stable orthogonal and unitary groups are already commutative in the strong sense that they form infinite loop spaces, even though they don't look particularly commutative either. So, loosely speaking, rational homotopy theory has a hard time telling apart the orthogonal and stable orthogonal groups, which means the most important differences are in torsion. (This is easiest to see for the homotopy groups.)


TITLE: Expected value of pairs remaining after draw
QUESTION [1 upvotes]: Problem: Someone removed $m$ shoes from a pile with $n$ distinct pairs of shoes ($2n$ shoes total). Assuming that every shoe has an equal chance of $1/(2n)$ to be removed, what is the mean number of complete pairs of shoes to remain in the pile.
Solutions:
Defining $K$, a random variable denoting the number of shoes left in the pile, I argued that the probability of $k_i$ pairs of shoes remaining in the pile would be given by:
\begin{align}
p_{k_i} & = \cfrac{{2n - 2k_i \choose m}}{2n \choose m} & \text{therefore,} \\
E(K) & = \sum\limits_{k=0}^n k\cfrac{{2n - 2k \choose m}}{2n \choose m}
\end{align}
However, I was given another solution, that defined the random variable $X$, such that, $x_i = 1$, when the $i$th pair remains in the pile. Then,
\begin{align}
p_{x_i = 1} & = \cfrac{{2n - 2 \choose m}}{2n \choose m} & \text{and,} \\
E(X) & = \sum\limits_{i=1}^n x_i\cfrac{{2n - 2 \choose m}}{2n \choose m} \\
& = n\cfrac{{2n - 2 \choose m}}{2n \choose m}
\end{align}
Are either solution correct? And most importantly, why?
Edit With some real numbers, it appears the second answer is correct, but I still do not understand the reasoning (or the error in reasoning for the first proposed answer).
ans1 <- function(n, m) {
  sum(sapply(1:n, function(k) k*choose(2*n - 2*k, m)/choose(2*n, m)))
}
ans2 <- function(n, m) {
  n*choose(2*n - 2, m)/choose(2*n, m)
}
list('ans1' = ans1(10, 4), 'ans2' = ans2(10, 4))
# $ans1
# [1] 2.73808
#
# $ans2
# [1] 6.315789

Clearly, if we remove 4 shoes, the maximum number of pairs destroyed is 4, so an expected value less than 6 is impossible.

REPLY [1 votes]: The latter solution appears largely correct. To be clear, if $X_{i}=1$ if the $i$th pair of shoes remains after the selection, and is 0 otherwise, then $X_{i}\sim$Bernoulli($p$), where $p=\binom{2n-2}{m}/\binom{2n}{m}$. Then $X=\sum_{i=1}^{n}X_{i},$ so $E(X)=\sum_{i=1}^{n}E(X_{i})=np.$ Note that despite the similarity in computation, $X$ is not Binomial, however, since the $X_{i}$ are not independent.
One problem with the first solution is that you're saying that the probability of $k_{i}$ pairs of shoes remaining is $\binom{2n-2k_{i}}{m}/\binom{2n}{m},$ so the numerator is essentially counting the number of ways to select $m$ shoes from the $2n-2k_{i}$ shoes left over, given that $k_{i}$ pairs are left intact. But this does not account for the fact that there are multiple ways to select the $k_{i}$ pairs from the $n$ possible. To see an example, suppose we have shoes labeled $1,2,3,4,$ with $1,2$ being a pair, and $3,4$ being a pair. Then if $k_{i}=1,$ $m=1$, then $\binom{4-2}{1}=2$, as you have counted. But we may remove any of the four shoes, and exactly one pair will remain, so the correct number should be $4=\binom{2}{1}\binom{4-2}{1}$.


text:
{\bf Problem.} Let $x$ be a positive real number. Find the maximum possible value of $$\frac{x^2+2-\sqrt{x^4+4}}{x}.$$
{\bf Level.} Level 5
{\bf Type.} Intermediate Algebra
{\bf Solution.} Rationalizing the numerator, we get

\begin{align*}
\frac{x^2+2-\sqrt{x^4+4}}{x}\cdot\frac{x^2+2+\sqrt{x^4+4}}{x^2+2+\sqrt{x^4+4}}&=\frac{(x^2+2)^2-(x^4+4)}{x(x^2+2+\sqrt{x^4+4})}\\
&=\frac{4x^2}{x(x^2+2+\sqrt{x^4+4})}\\
&=\frac{4}{\frac{1}{x}(x^2+2+\sqrt{x^4+4})}\\
&=\frac{4}{x+\frac{2}{x}+\sqrt{x^2+\frac{4}{x^2}}}.
\end{align*}Since we wish to maximize this quantity, we wish to minimize the denominator. By AM-GM, $x+\frac{2}{x}\geq 2\sqrt{2}$ and $x^2+\frac{4}{x^2}\geq 4$, so that the denominator is at least $2\sqrt{2}+2$. Therefore, $$\frac{x^2+2-\sqrt{x^4+4}}{x}\leq \frac{4}{2\sqrt{2}+2}=\boxed{2\sqrt{2}-2},$$with equality when $x=\sqrt{2}$.

set_name: MATH


TITLE: Definite integration of a trigonometric function
QUESTION [2 upvotes]: How to integrate $$\int_0^{\pi/2}\!\dfrac{2a \sin^2 x}{a^2
\sin^2 x +b^2 \cos^2 x}\,dx $$

my first step is $$\frac{2}{a} \int_0^{\pi/2}\!\dfrac{a^2 \sin^2 x}{a^2 +(b^2 - a^2) \cos^2 x}\, dx $$
I would kind of want to do some sort of $u=\cos x$ substitution, to get at $\arctan u $ but no idea what to do with the sine in the numerator.

REPLY [1 votes]: Often, this type of integral is workable using the substitution
$$
z=\tan(x/2)\quad\text{and}\quad\mathrm{d}x=\frac{\mathrm{2\,d}z}{1+z^2}\\
\sin(x)=\frac{2z}{1+z^2}\quad\text{and}\quad\cos(x)=\frac{1-z^2}{1+z^2}
$$
A couple of identities that simplify this integral are
$$
\sin^2(x)=\frac{1-\cos(2x)}{2}\quad\text{and}\quad\cos^2(x)=\frac{1+\cos(2x)}{2}
$$
Then, assuming $a,b\gt0$,
$$
\begin{align}
\int_0^{\pi/2}\frac{2a\sin^2(x)}{a^2\sin^2(x)+b^2\cos^2(x)}\,\mathrm{d}x
&=\int_0^{\pi/2}\frac{2a(1-\cos(2x))}{a^2(1-\cos(2x))+b^2(1+\cos(2x))}\,\mathrm{d}x\\
&=\int_0^\pi\frac{a(1-\cos(x))}{(b^2+a^2)+(b^2-a^2)\cos(x)}\,\mathrm{d}x\\
&=\int_0^\infty\frac{az^2}{b^2+a^2z^2}\frac{2\mathrm{d}z}{1+z^2}\\
&=\int_{-\infty}^\infty\frac{az^2}{b^2+a^2z^2}\frac{\mathrm{d}z}{1+z^2}\\
&=2\pi i\left(\frac1{2i}\frac{a}{a^2-b^2}-\frac1{2i}\frac{b}{a^2-b^2}\right)\\
&=\frac\pi{a+b}
\end{align}
$$
where we use the contour running along the real axis from $-\infty$ to $\infty$, circling back in the upper half-plane. The residue at $z=i$ is $\dfrac1{2i}\dfrac{a}{a^2-b^2}$ and at $z=i\dfrac{b}{a}$ is $-\dfrac1{2i}\dfrac{b}{a^2-b^2}$.
Removing the assumption on the signs of $a$ and $b$, we get
$$
\int_0^{\pi/2}\frac{2a\sin^2(x)}{a^2\sin^2(x)+b^2\cos^2(x)}\,\mathrm{d}x
=\mathrm{sgn}(a)\frac\pi{|a|+|b|}
$$


TITLE: Quotient vs Ratio vs Fraction vs Rational
QUESTION [0 upvotes]: I can't seem to differentiate between a Quotient and a Ratio and a Fraction and a Rational. From what I know a rational is a number like $2/3$ or $5.4/7$  whereas quotient, ratio and fraction all are just three names for one same thing $1+x/1-x^3$ , and that is just same as rational number except that a variable is used instead of a number. Am I correct ?

REPLY [1 votes]: $$\frac AB$$ is a ratio, whatever $A$ and $B$ denote. Fraction is a quasi synonym, that also conveys the graphical representation with a division bar.
A quotient is the result of a division, especially when there can be a remainder (as in integer and polynomial divisions).
A rational (number)  is any number that can be expressed as the ratio of two integers.

Addendum:
A rational fraction is the ratio of two polynomials.


TITLE: Indicator equation $f(n) = n$ for $n \in \Bbb{N}$ such that $n \pm 1$ is a pair of twin primes.
QUESTION [0 upvotes]: Consider the formula:
$$
f : \Bbb{N} \to \Bbb{C} \\
f(n) := 2 + \sum_{a = 1}^{n-2} \exp\left({\dfrac{2\pi i}{\gcd(a, n^2-1)}}\right)
$$
If $n \pm 1$ is a pair of twin primes, then $f(n) = n$.  This is because the power of $e$ will be an integer $m$ and thus $e^{i2\pi m} = e^{0} = 1$ gives us in the sum a total number of $n-2$ ones.
Question.

Conversely, if $f(n) = n$ does it follow that $n \pm 1$ is a pair of twin primes?

Another map $g : \Bbb{N} \to \Bbb{N}$ is the related:
$$
g(n) := 2 + \sum_{a = 1}^{n-2} \gcd(a, n^2 -1)
$$
and it happens to be the case that $g(n) = n$ is satisfied if and only if $n\pm 1$ is a pair of twin primes, or in other words:
$$
n \pm 1 \text{ are twin primes} \\
\iff \gcd(a, n^2 -1) = 1 \ \ \forall a = 1..(n-2)
$$
Evidence for conjecture:
from sympy import isprime
from cmath import exp, pi
from math import gcd

N = 100000

def f(n):
    s = 2
    for a in range(1, n-2 + 1):
        s += exp((2*pi*1j)/gcd(a, n*n - 1))
    return s

for n in range(1, N):
    p = n - 1
    q = n + 1

    if isprime(p) and isprime(q):
         x = f(n)
         if (abs(x.imag) <= 0.00000001 and int(x.real) != n):
             print("f(", n, ")=", x)
             print("False")
         else:
             print("f(", n, ")=", n)

which prints:
f( 4 )= 4
f( 6 )= 6
f( 12 )= 12
f( 18 )= 18
f( 30 )= 30
f( 42 )= 42
f( 60 )= 60
f( 72 )= 72
....
f( 34512 )= 34512
f( 34590 )= 34590
f( 34650 )= 34650
f( 34758 )= 34758
f( 34842 )= 34842
f( 34848 )= 34848
f( 34962 )= 34962
f( 35052 )= 35052
....

REPLY [1 votes]: Note that, if $n-1$ is not $1$ or prime, there is some prime factor $p|n-1$ so that $1\leq p\leq n-2$, and if $n+1$ is not $1$, $4$, or prime, there is some prime factor $p|n+1$ so that $1\leq p\leq n-2$, so we have that
$$\gcd(a,n^2-1)=1\text{ for all }1\leq a\leq n-2$$
if and only if $n-1$ and $n+1$ are both primes, or $n\in\{2,3\}$. This should immediately give your second result (or, at least, a true version of it). Your first conjecture is also true for the same reason, since a sum of $n-2$ roots of unity is $n-2$ if and only if they are all $1$.


TITLE: Definition of a map of (pre)spectra in HoTT
QUESTION [0 upvotes]: I'm trying to pick up some homotopy theory by reading various HoTT sources (I find it much easier than reading classical homotopy theory books). In Floris Van Doorn's PhD thesis he defines (pre)spectra as shown below:

It seems that a map of (pre)spectra is not required to be degreewise pointed. Is this a typo (he missed the star on the highlighted arrow?), or is this intentional?

REPLY [2 votes]: This is surely a typo.  Pretty much everything about spectra takes place in the pointed category and every map in sight is pointed.  I'm not familiar with this particular exposition in the context of HoTT, but I've never seen any similar definition in any other context that would use unpointed maps.  (And if the $f$ maps were not pointed, it is not clear what $\Omega f_{n+1}$ would even mean in the specification of $p$.)


TITLE: Smoothness of critical elliptic problem
QUESTION [4 upvotes]: I am convinced I have seen results along the lines of: if $ u \ge 0$ is an $H_0^1(\Omega)$ solution of
$$-\Delta u = u^{q-1}$$ in $\Omega$ with $ u=0$ on $ \partial \Omega$ (here $\Omega$ is a smooth bounded domain in $R^N$ and $ q=2^*$,  then $u$ is smooth.  Any idea how one proves the regularity result?   Since $ q=2^*$ it appears the standard iteration method fails.   Maybe one uses some $ \epsilon$ regularity approaches (which I don't know).  Or maybe I am wrong and the result is false. thanks

REPLY [5 votes]: The result is true. As you suggest, the starting regularity assumption $u \in H^1_0$ is critical in the sense that, if you additionally knew $u \in L^p$ with $p>2^*$, then you could bootstrap to smoothness (at least, when $2^*$ is an integer, as in dimension $N=3$, for which $2^*=6)$. In the case $p=2^*$, the key observation is that $H^1_0$ solutions to $-\Delta u = Vu$ do satisfy higher integrability in the critical case $V \in L^{N/2}$. That is, such solutions belong to all $L^p$, $p < +\infty$. The idea is to split the potential $V$ into a bounded part and a part which is small in $L^{N/2}$ and run Moser's iteration. The argument is given in the book Elliptic Partial Differential Equations by Qing Han and Fanghua Lin, see Theorem 4.4, p. 76.


TITLE: Taylor expansion for $\frac{1}{x^2 + 2x + 2}$
QUESTION [3 upvotes]: I'm trying to find the Taylor expansion of the function:
$$ f(x) = \frac{1}{x^2 + 2x + 2} $$
about the point $ x = 0 $. I have worked out the terms up to the fourth derivative, which was very tedious. I found:
$$ f(x) = \frac{1}{2} - \frac{1}{2} x + \frac{1}{4} x^2 + 0 x^3 - \frac{1}{8} x^4 + O \left( x^5 \right) $$
I notice powers of two in the denominator, but I'm not sure of the pattern (and to calculate the next term to confirm would entail another tedious product rule).
Any ideas? Thanks!

REPLY [1 votes]: There is a much simpler way involving no contrived derivatives, binomial expansions, etc., that unfortunately does not easily generalize to any other quadratic reciprocal.
If we notice that $x^2+2x+2$ occurs in the factorization of a certain binomial:
$$ x^8 - 16 = \left( x^4 - 4 \right) \left( x^4 + 4 \right) = \left( x^4 - 4 \right) \left( x^2 - 2x + 2 \right) \left( x^2 + 2x + 2 \right) = \left( x^6 - 2x^5 + 2x^4 - 4x^2 + 8x - 8 \right) \left( x^2 + 2x + 2 \right) $$
Rearranging:
$$ \frac{1}{x^2 + 2x + 2} = \frac{x^6 - 2x^5 + 2x^4 - 4x^2 + 8x - 8}{x^8 - 16} $$
We can bring the right-hand side to the form of a simple geometric series:
$$ \frac{1}{x^2 + 2x + 2} = \frac{\frac{8}{16} - \frac{8}{16} x + \frac{4}{16} x^2 - \frac{2}{16} x^4 + \frac{2}{16} x^5 - \frac{1}{16} x^6}{\frac{16}{16} - \frac{x^8}{16}} = \frac{\frac{1}{2} - \frac{1}{2} x + \frac{1}{4} x^2 - \frac{1}{8} x^4 + \frac{1}{8} x^5 - \frac{1}{16} x^6}{1 - \frac{x^8}{16}} = \left( \frac{1}{2} - \frac{1}{2} x + \frac{1}{4} x^2 - \frac{1}{8} x^4 + \frac{1}{8} x^5 - \frac{1}{16} x^6 \right) \left( 1 + \frac{x^8}{16} + \frac{x^{16}}{256} + \cdots \right) $$
Clearly, there is no overlap between the terms when multiplying out, so if we write only the coefficients, we see that they group together in sets of eight:
$$ \frac{1}{x^2 + 2x + 2} = \left[ \frac{1}{2}, -\frac{1}{2}, \frac{1}{4}, 0, -\frac{1}{8}, \frac{1}{8}, -\frac{1}{16}, 0, \frac{1}{16}, -\frac{1}{32}, \frac{1}{32}, 0, -\frac{1}{64}, \frac{1}{64}, -\frac{1}{128}, 0, \cdots \right] $$
in an elegant pattern!


TITLE: What is the probability of selecting at least two aces given deck of 52 cards?
QUESTION [0 upvotes]: What is the probability of selecting at least two aces given a deck of 52 cards. We are selecting 4 cards randomly and the question is what is the probability of selecting at least 2 aces?
( referring to the same exercise) What is the probability of choosing maximum of 3 clubs?

REPLY [1 votes]: Assuming that the selection is without bias and without replacement, given $N$ total objects, $K$ of which are of one type and $N-K$ of which are of the remaining type, the probability of getting exactly $k$ objects of the one type when choosing $n$ objects will be:
$$\dfrac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}$$
This is the hypergeometric distribution.  On the top, we counted the number of possible outcomes where we have $k$ objects selected of the first type and the remaining $n-k$ objects needed to bring the total number of selected objects up to $n$ of the second type and divided by the total number of ways to select $n$ objects.
Now, for your case, you simply use $N=52,K=4,n=4$ and calculate for each of $k=2,3,4$.  For the clubs problem, you instead have $K=13$ and instead have $k$ ranging over $0,1,2,3$.


text:
\begin{document}
\newcommand{\pt}{\partial}
\newcommand{\lp}{\langle}
\newcommand{\rp}{\rangle}
\newcommand{\ra}{\rightarrow}
\newcommand{\rat}{\mapsto}
\newcommand{\lra}{\longrightarrow}
\newcommand{\ul}{\underline}
 \renewcommand{\o}[2]{\frac{#1}{#2}}
\newcommand{\hf}{\o{1}{2}}

\newcommand{\qed}{\hspace*{\fill}\rule{3mm}{3mm}\quad}

 \newcommand{\Pf}{\noindent {\em Proof.} }
\newcommand{\Ex}{\noindent {\em Example.} }
\newcommand{\Rk}{\noindent {\em Remark.} }
\newcommand{\Def}{\noindent {\em Definition.} }
\newcommand{\heta}{\hat{\eta}}
\newcommand{\reta}{\overline{\eta}}
\newcommand{\ch}{\mbox{\rm ch}}
\newcommand{\sgn}{\mbox{\rm sgn}}
\newcommand{\tr}{\mbox{\rm tr}}
\newcommand{\Tr}{\mbox{\rm Tr}}
\newcommand{\DET}{\mbox{\rm DET}}
\newcommand{\hol}{\mbox{\rm hol}}
\newcommand{\End}{\mbox{\rm End}}
\newcommand{\spec}{\mbox{\rm spec}}
\renewcommand{\sf}{\mbox{\rm sf}}
\newcommand{\ind}{\mbox{\rm ind}}
\newcommand{\rk}{\mbox{\rm rk}}
\newcommand{\re}{\mbox{\rm Re}}
\newcommand{\Res}{\mbox{\rm Res}}
\newcommand{\HH}{{\rm H}}
\newcommand{\im}{\mbox{\rm Im}}
\newcommand{\inte}{\mbox{\rm int}}
\newcommand{\zn}{\stackrel{\circ}{\nabla}}
\newcommand{\bn}{\bar{\nabla}}
\newcommand{\zg}{\stackrel{\circ}{g}}

\newcommand{\Dirac}{\mathcal{D}}
\newcommand{\R}{\mathcal{R}}

\renewcommand{\theequation}{\arabic{section}.\arabic{equation}}
\newcommand{\sect}[1]{\section{#1} \setcounter{equation}{0}}

\newtheorem{theo}{Theorem}[section]

\newtheorem{lem}[theo]{Lemma}
\newtheorem{prop}[theo]{Proposition}
\newtheorem{coro}[theo]{Corollary}

\title{\sc A Note on Positive Energy Theorem for Spaces with Asymptotic SUSY Compactification}
\author{Xianzhe Dai}
\maketitle

\begin{abstract} We extend the positive mass theorem in \cite{d} to the Lorentzian
setting. This includes the original higher dimensional Positive
Energy Theorem whose spinor proof is given in \cite{Wi} and
\cite{PT} for dimension $4$ and in \cite{Z} for dimension $5$.
\end{abstract}

\section{Introduction and statement of the result}
In this note, we formulate and prove the Lorentzian version of the
positive mass theorem in \cite{d}. There we prove a positive mass
theorem for spaces which asymptotically approach the product of a flat Euclidean
space with a compact manifold which admits a nonzero parallel spinor (such as a
Calabi-Yau manifold or any special honolomy manifold except the
quaternionic K\"ahler). This is motivated by string theory,
especially the recent work \cite{HHM}. The application of the positve mass theorem of \cite{d} to the study of stability of
Ricci flat manifolds is discussed in \cite{dww}.

In general relativity, a spacetime is modeled by a Lorentzian
$4$-manifold $(N, \ g)$ together with an energy-momentum tensor
$T$ satisfying Einstein equation \be \label{ee} R_{\alpha\beta} -
\hf g_{\alpha \beta} R = 8\pi T_{\alpha \beta}. \ee The positive
energy theorem \cite{SY1}, \cite{Wi} says that an isolated
gravitational system with nonnegative local matter density must
have nonnegative total energy, measured at spatial infinity. More
precisely, one considers a complete oriented spacelike
hypersurface $M$ of $N$ satisfying the following two conditions:

a). {\em $M$ is asymptotically flat}, that is, there is a compact set
$K$ in $M$ such that $M-K$ is the disjoint union of a finite
number of subsets $M_1, \dots , M_k$ and each $M_l$ is
diffeomorphic to $({\Bbb R}^3 -B_R(0))$. Moreover, under this
diffeomorphism, the metric of $M_l$ is of the form \be \label{afm}
g_{ij}=\delta_{ij} + O(r^{-\tau}), \ \ \  \partial_k
g_{ij}=O(r^{-\tau -1}), \ \ \
\partial_k \partial_l g_{ij}=O(r^{-\tau -2}). \ee Furthermore, the
second fundamental form $h_{ij}$ of $M$ in $N$ satisfies \be
\label{affsff} h_{ij}= O(r^{-\tau -1}), \ \ \ \  \partial_k
h_{ij}=O(r^{-\tau -2}). \ee Here $\tau>0$ is the asymptotic order
and $r$ is the Euclidean distance to a base point.

b). {\em $M$ has nonnegative local mass density}: for each point $p \in
M$ and for each timelike vector $e_0$ at $p$, $T(e_0, \ e_0) \geq
0$ and $T(e_0, \ \cdot)$ is a nonspacelike co-vector. This implies
the dominant energy condition
\be \label{dec} T^{00} \geq |
T^{\alpha \beta} |, \ \ \ T^{00} \geq (-T_{0i}T^{0i})^{\hf}. \ee

The total energy (the ADM mass) and the total (linear) momentum of
$M$ can then be defined as follows \cite{ADM}, \cite{PT} (for
simplicity we suppress the dependence here on $l$ (the end $M_l$))
\ba \label{admm} E = \lim_{R \ra \infty} \frac{1}{4 \omega_n}
\int_{S_R} (\partial_i g_{ij} - \pt_j g_{ii} ) * dx_j , \nonumber \\
 P_k =\lim_{R \ra \infty} \frac{1}{4 \omega_n} \int_{S_R} 2(h_{jk} -
\delta_{jk} h_{ii} ) * dx_j \ea
Here $\omega_n$ denotes the volume
of the $n-1$ sphere and $S_R$ the Euclidean sphere with radius $R$
centered at the base point.

\begin{theo}[Schoen-Yau, Witten] With the assumptions as above and assuming that $M$ is spin, one has
\[ E - |P| \geq 0\]
on each end $M_l$. Moreover, if $E=0$ for some end $M_l$, then $M$
has only one end and $N$ is flat along $M$.
\end{theo}

Now, according to string theory \cite{CHSW}, our universe is really ten
dimensional, modelled on  $\mathbb R^{3,1} \times X$ where $X$ is a
Calabi-Yau 3-fold. This is the so called Calabi-Yau
compactification, which motivates the spaces we now consider.

Thus, we consider a Lorentzian manifold $N$ (with signature $(-,
+, \cdots, +)$) of $\dim N=n+1$, with a energy-momentum tensor
satisfying the Einstein equation. Then let $M$ be a complete
oriented spacelike hypersurface in $N$. Furthermore the Riemannian
manifold $(M^n, g)$ with $g$ induced from the Lorentzian metric
decomposes $M=M_0 \cup M_{\infty}$, where $M_0$ is compact as
before but now $M_{\infty} \simeq ({\Bbb R}^k - B_R(0)) \times X$
for some radius $R>0$ and $X$ a compact simply connected spin
manifold which admits a nonzero parallel spinor. Moreover the
metric on $M_{\infty}$ satisfies \be \label{afmsu} g=\zg + u, \ \
\ \zg=g_{{\mathbb R}^k} + g_X, \ \ \ u= O(r^{-\tau}), \ \ \ \zn\!
u=O(r^{-\tau -1}), \ \ \ \zn \zn \! u=O(r^{-\tau -2}), \ee and the
second fundamental form $h$ of $M$ in $N$ satisfies \be
\label{afsfsu} h=O(r^{-\tau-1}), \ \ \ \zn\! h=O(r^{-\tau -2}).
\ee Here $\zn$ is the Levi-Civita connection of $\zg$ (extended to
act on all tensor fields), $\tau
>0$ is the asymptotical order.

The total energy and total momentum for such a space can then be
defined by
\ba \label{admkkm}
E= \lim_{R \ra \infty} \frac{1}{4 \omega_k vol(X)} \int_{S_R \times X} (\partial_i g_{ij} - \pt_j g_{aa} ) * dx_j dvol(X),
\nonumber \\
P_k = \lim_{R \ra \infty} \frac{1}{4 \omega_k vol(X)} \int_{S_R
\times X} 2(h_{jk} - \delta_{jk} h_{ii} ) * dx_j dvol(X). \ea Here
the $*$ operator is the one on the Euclidean factor, the index $i,
j$ run over the Euclidean factor while the index $a$ runs over the
full index of the manifold.

Then we have

\begin{theo}
Assuming that $M$ is spin, one has
\[ E - |P| \geq 0\]
on each end $M_l$. Moreover, if $E=0$ for some end $M_l$, then $M$
has only one end. In this case, when $k=n$, $N$ is flat along $M$.
\end{theo}

In particular, this result includes the original higher
dimensional Positive Energy Theorem whose spinor proof is given in
\cite{Wi} and \cite{PT} for dimension $4$ and in \cite{Z} for
dimension $5$.

{\em Acknowledgement:} This work is motivated and inspired by the
work of Gary Horowitz and his collaborators \cite{HHM}. The author
is indebted to Gary for sharing his ideas and for interesting
discussions. The author would also like to thank Xiao Zhang and Siye Wu for
useful discussion.

\section{The hypersurface Dirac operator}

We will adapt Witten's spinor method \cite{Wi}, as given in \cite{PT}, to our situation.
The crucial ingredient here is the hypersurface Dirac operator on
$M$, acting on the (restriction of the) spinor bundle of $N$. Let
$S$ be the spinor bundle of $N$ and still denote by the same
notation its restriction on (or rather, pullback to) $M$. Denote by $\nabla$
the connection on $S$ induced by the Lorentzian metric on $N$. The Lorentzian
metric on $N$ also induces a Riemannian metric on $M$, whose Levi-Civita
connection gives rise to another connection, $\bar{\nabla}$ on $S$.
The two, of course, differ by a term involving the second fundamental form.

There are two choices of metrics on $S$, which is another subtlety
here. Since part of the treatment in \cite{PT} is special to
dimension $4$, we will give a somewhat detailed account here.

Let $SO(n,1)$ denote the identity component of the groups of
orientation preserving isometries of the Minkowski space $\mathbb
R^{n,1}$. A choice of a unit timelike covector $e^0$ gives rise to
injective homomorphisms $\alpha$, $\hat{\alpha}$,  and a
commutative diagram \be
\label{injh} \begin{array}{cccc} \alpha: \ & SO(n) & \ra & SO(n,1) \\
& \uparrow & & \uparrow \\
\hat{\alpha}: \ & Spin(n) & \ra & Spin(n,1). \end{array} \ee

We now fix a choice of unit timelike normal covector $e^0$ of $M$
in $N$. Let $F(N)$ denote the $SO(n,1)$ frame bundle of $N$ and
$F(M)$ the $SO(n)$ frame bundle of $M$. Then
$i^*F(N)=F(M)\times_{\alpha} SO(n,1)$, where $i: \ M
\hookrightarrow N$ is the inclusion. If $N$ is spin, then we have
a principal $Spin(n,1)$ bundle $P_{Spin(n,1)}$ on $N$, whose
restriction on $M$ is then $i^*P_{Spin(n,1)}=P_{Spin(n)}
\times_{\hat{\alpha}} Spin(n,1)$, where $P_{Spin(n)}$ is the
principal $Spin(n)$ bundle of $M$. Thus, even if $N$ is not spin,
$i^*P_{Spin(n,1)}$ is still well-defined as long as $M$ is spin.

Similarly, when $N$ is spin, the spinor bundle $S$ on $N$ is the
associated bundle $P_{Spin(n,1)} \times_{\rho_{n,1}} \Delta$,
where $\Delta=\mathbb C^{2^{[\o{n+1}{2}]}}$ is the complex vector
space of spinors and \be \rho_{n,1}: \ Spin(n,1) \ra GL(\Delta)
\ee is the spin representation. Its restriction to $M$ is given by
$i^*P_{Spin(n,1)} \times_{\rho_{n,1}} \Delta=
P_{Spin(n)} \times_{\rho_{n}} \Delta$ with \be \rho_{n}: \
Spin(n) \stackrel{\hat{\alpha}}{\hookrightarrow} Spin(n,1)
\stackrel{\rho_{n,1}}{\longrightarrow} GL(\Delta) \ee Again, the restriction
is still well defined as long as $M$ is spin.

Let $e^0, e^i$ ($i=1, \cdots, n$ will be the range for the index $i$ in this section) be an orthonormal basis of the
Minkowski space $\mathbb R^{n,1}$ of dimension $n+1$ such that $|e^0|=-1$.

\begin{lem} There is a positive definite hermitian inner product $\lp \ , \ \rp$  on
$\Delta$ which is $Spin(n)$-invariant.  Moreover, $(s, s')= \lp
e^0 \cdot s, s' \rp$ defines a  hermitian inner product which is
also $Spin(n)$-invariant but not positive definite. In fact
\[ (v\cdot s, \ s')=(s, \ v\cdot s') \]
for all $v\in \mathbb R^{n,1}$.
\end{lem}

\Pf Detailed study via $\Gamma$ matrices \cite[p10-11]{CBDM} shows that
there is a positive definite hermitian inner product $\lp \ , \
\rp$ on $\Delta$ with respect to which $e^i$ is skew-hermitian
while $e^0$ is hermitian. It follows then that $\lp \ , \ \rp$ is
$Spin(n)$-invariant. We now show that $(s, s')=\lp e^0 \cdot s, s'
\rp $ defines a $Spin(n)$-invariant hermitian inner product. Since
$e^0$ is hermitian with respect to $\lp \ , \ \rp$, $(\ , \ )$ is
clearly hermitian. To show that $(\ , \ )$ is $Spin(n)$-invariant,
we take a unit vector $v$ in the Minkowski space: $v=a_0 e^0 + a_i
e^i$,  $a_0, a_i \in \mathbb R$ and $-a_0^2 + \sum_{i=1}^{n} a_i^2
=1$. Then
\ban (vs, vs') & =  & \lp e^0 vs, vs' \rp \\
& =  & a_0^2 \lp e^0 e^0s, e^0s' \rp + a_i a_0 \lp e^0 e^is, e^0s' \rp + a_0 a_i \lp e^0 e^0s, e^is' \rp +  a_ia_j \lp e^0
e^is, e^js' \rp\\
 & = & a_0^2 \lp s, e^0 s' \rp - a_ia_j \lp e^je^0 e^is, s' \rp \\
& = & a_0^2 \lp e^0 s, s' \rp + a_ia_j \lp e^0 e^je^is, s' \rp \\
& = & a_0^2 \lp e^0 s, s' \rp - a_i^2 \lp e^0 s, s' \rp \\
& = & -(s, s') \ean Consequently, $(\ , \ )$ is
$Spin(n)$-invariant.  The above computation also implies that
$v\cdot$ acts as hermitian operator on $\Delta$ with respect to
$(\ , \ )$. \qed

Thus the spinor bundle $S$ restricted to $M$ inherits an hermitian
metric $(\ , \ )$ and a positive definite metric $\lp \ , \ \rp$.
They are related by the equation \be (s, s')=\lp e^0 \cdot s, s'
\rp . \ee

Now the hypersurface Dirac operator is
defined by the composition

\be \label{hd} \Dirac: \ \Gamma(M, S)
\stackrel{\nabla}{\longrightarrow} \Gamma(M, T^*M \otimes S)
\stackrel{c}{\longrightarrow} \Gamma(M, S), \ee where $c$ denotes
the Clifford multiplication. In terms of a local orthonormal basis
$e_1, e_2, \cdots, e_n$ of $TM$,
\[ \Dirac \psi = e^i \cdot \nabla_{e_i} \psi,\]
where $e^i$ denotes the dual basis.

The two most important properties of hypersurface Dirac operator
are the self-adjointness with respect to the metric $\lp \ , \
\rp$ and the Bochner-Lichnerowicz-Weitzenbock formula \cite{Wi},
\cite{PT}.

\begin{lem} \label{fsa} Define a $n-1$ form on $M$ by $\omega=\lp \phi, e^i \cdot \psi \rp \inte(e_i) \, dvol$, where $dvol$
is the volume form of the Riemannian metric $g$. We have
\[ [ \lp \phi, \Dirac \psi \rp - \lp \Dirac \phi,  \psi \rp ]
dvol=d\omega. \] Thus $\Dirac$ is formally self adjoint with
respect to the $L^2$ metric defined by $\lp \ , \ \rp$ (and
$dvol$).
\end{lem}

\Pf Since $\omega$ is independent of the choice of the orthonormal basis, we do our computation locally using a preferred
basis. For any given point $p\in M$, choose a local orthonormal frame
$e_i$ of $TM$ near $p$ such that $\bar{\nabla}e_i=0$ at $p$.
Extend $e_0, e_i$ to a neighborhood of $p$ in $N$ by parallel
translating along $e_0$ direction. Then, at $p$,
$\nabla_{e_i}e^j=- h_{ij} e^0$ and $\nabla_{e_i}e^0=-h_{ij} e^j$.
Therefore (again at $p$), \ban d\omega & = & \nabla_{e_i} \lp
\phi, e^i \cdot \psi \rp \, dvol \\
& = & [((\nabla_{e_i}e^0)\cdot \phi, e^i \cdot \psi) + (e^0\cdot \nabla_{e_i} \phi, e^i \cdot \psi)
+ (e^0 \cdot \phi, (\nabla_{e_i}e^i) \cdot \psi) + (e^0 \cdot \phi, e^i \cdot \nabla_{e_i} \psi) ] dvol \\
& = & [ - h_{ij} (e^j \cdot \phi, e^i \cdot \psi) + (e^i\cdot
e^0\cdot \nabla_{e_i} \phi,  \psi ) - h_{ii} (e^0
\cdot \phi, e^0 \cdot \psi) + \lp \phi, \Dirac \psi \rp ] dvol \\
& = & [- h_{ij} (e^i \cdot e^j \cdot \phi,  \psi) - \lp \Dirac
\phi, \psi \rp - h_{ii} (e^0
\cdot \phi, e^0 \cdot \psi)   + \lp \phi, \Dirac \psi \rp ] dvol \\
& = & [- \lp \Dirac \phi, \psi \rp + \lp \phi, \Dirac \psi \rp ]
dvol \ean \qed

Now the Bochner-Lichnerowicz-Weitzenbock formula.

\begin{lem} One has
\ba \label{wlf} \Dirac^2=\nabla^* \nabla + \R,\hspace{1in} \\
\R=\o{1}{4}(R + 2R_{00} + 2R_{0i}e^0\cdot e^i \cdot ) \in End(S).
\nonumber \ea Here the adjoint $\nabla^*$ is with respect to the
metric $\lp \ , \ \rp$.
\end{lem}

\Pf We again do the computation in the frame as in the proof of
Lemma \ref{fsa}. Then \ban \Dirac^2 & = & e^i \cdot e^j \cdot
\nabla_{e_i} \nabla_{e_j} + e^i \cdot \nabla_{e_i}e^j \cdot
\nabla_{e_j} \\
& = & - \nabla_{e_i} \nabla_{e_i} + \o{1}{4} (R + 2R_{00} +
2R_{0i}e^0\cdot e^i \cdot ) - h_{ij} e^i \cdot e^0\cdot
\nabla_{e_j}. \ean Now \ban d [\lp \phi, \psi \rp \inte(e_i) \,
dvol ] & = & e_i \lp \phi, \psi \rp \, dvol \\
& = & (\nabla_{e_i}e^0 \cdot \phi, \psi) + \lp \nabla_{e_i} \phi,
\psi \rp + \lp  \phi, \nabla_{e_i} \psi \rp \\
& = & - h_{ij} (e^j \cdot \phi, \psi)
 + \lp \nabla_{e_i} \phi,
\psi \rp + \lp  \phi, \nabla_{e_i} \psi \rp \\
& = & - h_{ij} \lp e^0 \cdot e^j \cdot  \phi, \psi \rp + \lp
\nabla_{e_i} \phi, \psi \rp + \lp  \phi, \nabla_{e_i} \psi \rp
\ean This shows that $\nabla^*_{e_i}= - \nabla_{e_i} -h_{ij}e^j
\cdot e^0 \cdot$. The desired formula follows. \qed

\section{Proof of the Theorem}

By the Einstein equation,
\[ \R=4\pi (T_{00} + T_{0i}e^0\cdot e^i \cdot).\]
It follows then from the dominant energy condition (\ref{dec}) that

\be \label{decc} \R \geq 0. \ee

Now, for $\phi \in \Gamma(M, S)$ and a compact domain $\Omega
\subset M$ with smooth boundary, the Bochner-Lichnerowicz-Weitzenbock formula yields

\ba \label{iwf}
 \int_{\Omega} [  |\nabla \phi|^2 + \lp \phi, \R\phi \rp - |\Dirac \phi|^2
]\, dvol(g) & = & \int_{\pt \Omega} \sum \lp (\nabla_{e_a} +
e_a \cdot \Dirac)\phi, \ \phi \rp \, \inte(e_a) \, dvol(g) \ \ \ \ \ \\
& = & \int_{\pt \Omega} \sum \lp (\nabla_{\nu} + \nu \cdot \Dirac
)\phi, \ \phi \rp \, dvol(g|_{\pt \Omega}), \ \ \ \ \ \ea where
$e_a$ is an orthonormal basis of $g$ and $\nu$ is the unit outer
normal of $\pt \Omega$. Also, here $\inte(e_a)$ is the interior
multiplication by $e_a$.

Now let the manifold $M= M_0 \cup M_{\infty}$ with $M_0$ compact
and $M_{\infty} \simeq ({\Bbb R}^k - B_R(0)) \times X$, and $(X,
g_X)$ a compact Riemannian manifold with nonzero parallel spinors.
Moreover, the metric $g$ on $M$ satisfies (\ref{afmsu}). Let
$e^0_a$ be the orthonormal basis of $\zg$ which consists of
$\o{\pt}{\pt x_i}$ followed by an orthonormal basis $f_{\alpha}$
of $g_X$. Orthonormalizing $e^0_a$ with respect to $g$ gives rise
an orthonormal basis $e_a$ of $g$. Moreover, \be \label{onbasy}
e_a=e^0_a - \hf u_{ab} e^0_b + O(r^{-2\tau}). \ee This gives rise
to a gauge transformation
\[ A: \ SO(\zg) \ni e^0_a \ra e_a \in SO(g) \]
which identifies the corresponding spin groups and spinor bundles.

We now pick a unit norm parallel spinor $\psi_0$ of $({\Bbb R}^k,
g_{{\mathbb R}^k})$ and a unit norm parallel spinor $\psi_1$ of
$(X, g_X)$. Then $\phi_0=A(\psi_0 \otimes \psi_1)$ defines a
spinor of $M_{\infty}$.  We extend $\phi_0$ smoothly inside. Then
$\nabla^0 \phi_0 =0$ outside the compact set.

\begin{lem} \label{comt}
If a spinor $\phi$ is asymptotic to $\phi_0$: $\phi=\phi_0 + O(r^{-\tau})$, then we have
\[ \lim_{R \ra \infty} \Re \int_{S_R \times X} \sum \lp (\nabla_{e_a} + e_a \cdot D)
\phi, \ \phi \rp \, \inte(e_a) \, dvol(g) = \omega_k vol(X) \lp
\phi_0, \ E \phi_0 + P_k dx^0 \cdot dx^k \cdot \phi_0 \rp,
\]
where $\Re$ means taking the real part.
\end{lem}

\Pf Recall that $\bn$ denote the connection on $S$ induced from the
Levi-Civita connection on $M$. We have

\be \label{2con} \nabla_{e_a} \psi = \bn_{e_a} \psi -\hf h_{ab}e^0
\cdot e^b \cdot \psi. \ee By the Clifford relation,
\[ \lp (\nabla_{e_a} + e_a \cdot D)
\phi, \ \phi \rp=-\hf \lp [e^a\cdot, e^b\cdot]\nabla_{e_b} \phi, \
\phi \rp. \] Hence \ban \int_{S_R \times X} \sum \lp (\nabla_{e_a}
+ e_a \cdot D) \phi, \ \phi \rp \, \inte(e_a) \, dvol(g)  = \hspace{3in} \\
 -\hf \int_{S_R \times X}\lp [e^a\cdot, e^b\cdot]\bn_{e_b} \phi, \
\phi \rp\, \inte(e_a) \, dvol(g) +  \o{1}{4}\int_{S_R \times X}\lp
[e^a\cdot, e^b\cdot]h_{bc} e^0 \cdot e^c \cdot \phi, \ \phi \rp\,
\inte(e_a) \, dvol(g). \ean Using (\ref{onbasy}) and the asymptotic conditions (\ref{afsfsu}), the second term in the right
hand side can be easily seen to give us
\[ \lim_{R \ra \infty} \o{1}{4} \int_{S_R \times X} \lp 2(h_{ac}-\delta_{ac}h_{bb}) e^0 \cdot e^c \cdot \phi, \ \phi \rp\,
\inte(e_a)
\, dvol(g) = \omega_k vol(X) \lp \phi_0, \  P_k dx^0 \cdot dx^k
\cdot \phi_0 \rp. \]

The first term is computed in \cite{d} to limit to
\[ \omega_k vol(X) \lp \phi_0, \ E \phi_0 \rp.\]
\qed

The following lemma is standard \cite{PT}, \cite{Wi}.

\begin{lem} If
\[ \lp \phi_0, \ E \phi_0 + P_k dx^0 \cdot dx^k \cdot \phi_0 \rp
\geq 0 \] for all constant spinors $\phi_0$, then
\[ E - |P| \geq 0.\]
\end{lem}

As usual, the trick to get the positivity now is to find a
harmonic spinor $\phi$ asymptotic to $\phi_0$. Then the left hand
side of (\ref{iwf}) will be nonnegative since $\R \geq 0$. Passing
to the right hand side will give us the desired result.

\begin{lem} \label{hspinor}
There exists a harmonic spinor $\phi$ on $(M, \ g)$ which is
asmptotic to the parallel spinor $\phi_0$ at infinity:
\[ \Dirac \phi =0, \ \ \  \phi=\phi_0 + O(r^{-\tau}).  \]
\end{lem}

\Pf The proof is essentially the same as in \cite{d}. We use the
Fredholm property of $\Dirac$ on a weighted Sobolev space
and $\R \geq 0$ to show that it is an isomorphism. The harmonic
spinor $\phi$ can then be obtained by setting $\phi=\phi_0 + \xi$
and solving $\xi \in O(r^{-\tau})$ from the equation $\Dirac \xi =
-\Dirac \phi_0$. \qed

The rest of the Theorem follows as in \cite{PT}.

config: arxiv


TITLE: Proving a sequence converges.
QUESTION [1 upvotes]: Let $a_{1} >0$ and let $a_{n+1} = \frac{1}{2} ( a_{n} + \frac{1}{a_{n}})$ for all $n \geq 1$. Show that $a_{n}$ converges and find it's limit.
The Attempt: I am going to use the Monotone Convergence Theorem to be able to show the sequence converges. First I am going to show the sequence is monotonically decreasing by induction. However, I am having a hard time figuring out if the sequence is monotonically decreasing.
Please give me hints to solve the problem. Please do not solve the problem completely.
Thank you very much.

REPLY [0 votes]: First, if $x>0$, then $x+\frac1x-2=\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2\geq 0$, hence $\frac{1}{2}\left(x+\frac1x\right)\geq1$. And the equality holds only when $x=1$. This means that if $a_0>0$ and $a_0\neq 1$, then $a_1>1$, and then $a_n>1$ for all $n\geq1$.
Therefore, for $n>0$, you can define $u_n=\mathrm{argcoth}\; a_n$, i.e. $a_n=\coth u_n$, with $u_n>0$, and
$$a_{n+1}=\frac12(\coth u_n+\tanh u_n)=\coth (2u_n)$$
That is, $a_n=\coth (2^{n-1}u_1)=\coth (2^{n-1}\mathrm{argcoth}\; a_1)$.
Then, since $2^{n-1}u_1 \to \infty$ as $n\to\infty$, you get $a_n\to1$.


TITLE: Let $g$ be a function such that $\underline{\int_{y\in B}}f(x,y)\leq g(x)\leq \overline{\int_{y\in B}}f(x,y)$ for all $x\in A$. Show that if $f$
QUESTION [2 upvotes]: (a) Exercise 1 of that section is: Let $f,g: Q\to \mathbb{R}$ be bounded function such that $f(x)\leq g(x)$ for $x\in Q$. Show that $\underline{\int_{Q}}f\leq \underline{\int_{Q}}g$ and $\overline{\int_{Q}}f\leq \overline{\int_{Q}}f$,
I think of using the exercise to conclude that $\underline{\int_{x\in A}}\underline{\int_{y\in B}}f(x,y)\leq \underline{\int_{x\in A}}g(x)$ and $\overline{\int_{x\in A}}g(x)\leq \overline{\int_{x\in A}}\overline{\int_{y\in B}}f(x,y)$ for all $x\in A$, But fubini's theorem tells me that $\int_{Q}f=\int_{x\in A}\underline{\int_{y\in B}}f(x,y)=\int_{x\in A}\overline{\int_{y\in B}}f(x,y)=\underline{\int_{x\in A}}\underline{\int_{y\in B}}f(x,y)\overline{\int_{x\in A}}\overline{\int_{y\in B}}f(x,y)$, with which $\overline{\int_{x\in A}}g(x)=\int_{Q}f=\underline{\int_{x\in A}}g(x)$, with this I could not conclude that $g$ is integrable over $A$?
I need help for (b) and (c), could someone help me please? Thank you.

REPLY [2 votes]: Part (a) clarification
Given your first inequality for $g$ and the result of the exercise, we have
$$\tag{*}\underline{\int}_{x \in A}\,\underline{\int}_{y\in B}f \leqslant \underline{\int}_{x\in A}g \leqslant \overline{\int}_{x\in A}g  \leqslant \overline{\int}_{x \in A}\,\overline{\int}_{y\in B}f$$
If $f$ is integrable over $Q$ then by Fubini's theorem we have
$$\int_Q f = \int_{x \in A}\,\underline{\int}_{y\in B}f= \underline{\int}_{x \in A}\,\underline{\int}_{y\in B}f, \\ \int_Q f = \int_{x \in A}\,\overline{\int}_{y\in B}f= \overline{\int}_{x \in A}\,\overline{\int}_{y\in B}f$$
Using this result along with (*) we have
$$\int_Q f\leqslant \underline{\int}_{x\in A}g \leqslant \overline{\int}_{x\in A}g  \leqslant \int_Qf,$$
which proves both that $g$ is integrable over $A$ (since the lower and upper integrals are equal) and that
$$\int_Qf = \int_Ag$$
Part (b)
Consider  the example
$$
  f(x,y)=\begin{cases}
    0, & x \notin \mathbb{Q}\\
    0, & x \in \mathbb{Q}, \, y \notin \mathbb{Q}\\
    1/q, & x \in \mathbb{Q}, \, y = p/q \text{ in lowest terms}  \end{cases}$$
We have $\int_0^1 f(x,y) \, dy = 0$, since if $x$ is irrational $f(x,y) = 0$, and if $x$ is rational then $y \mapsto f(x,y)$ is the Thomae function whose integral is zero. Hence the iterated integral $\int_0^1 \left(\int_0^1 f(x,y) \, dy \right) \, dx$ exists.
However, if $y$ is rational, then $x \mapsto f(x,y)$ is a Dirichlet function and the Riemann integral $\int_0^1 f(x,y) \, dx$ does not exist.
For (c) see the answer given by David C. Ulrich here


TITLE: Converse of Thales' theorem
QUESTION [2 upvotes]: I'm trying to prove the following theorem from Venema's Foundations of Geometry (2nd edition).

Theorem 8.3.3. Let $\triangle ABC$ be a triangle and let $M$ be the midpoint of $\overline{AB}$. If $\angle ACB $ is a right angle, then $AM = MC$.

(This is the converse of Thales' theorem)
I know that there are different proofs of this, but I'm trying to do it according to the hint of the book, which is the following:

Hint: Chose a point $C'$ on $\overrightarrow{MC}$ so that $MC' = AM$. By Theorem 8.3.1 (Thales' theorem), $\angle AC'B$ is a right angle. Use the Exterior Angle Theorem in a proof by contradiction to show that $C = C'$.

Following this hint, I get the following sketch of a proof:

Create point $C'$ as described to get the triangle $\triangle ABC'$. By Thales' theorem the angle $\gamma' = \angle AC'B$ is a right angle.
Suppose $C \neq C'$. There are two possibilities: either $C$ lays between $M$ and $C'$ ($M * C * C'$), or $C'$ lays between $M$ and $C$ ($M * C' * C$). In the following, I'm only going to look at the first case, since I guess that the second case can be handled similarly. Thus, we are in the situation shown in the figure below.

Let $D$ be the point where the ray $\overrightarrow{BC}$ intersects $AC'$. Now this gives rise to a couple of triangles: $\triangle C'DB$ and $\triangle ADC$. See figure below.

We have an exterior angle $\delta$ to $\triangle C'DB$, which by the Exterior Angle Theorem must be greater than the interior angle $\gamma' = 90^\circ$. However, $\angle ADC$ is a vertical pair to the angle $\delta$, so $\angle ADC$ is also be greater than $90^\circ$. But this is a contradiction since $\angle ACD$ is $90^\circ$.

My questions
I am not confident in this proof, because there are many unfounded assumptions I have made that I'm not able to rigorously justify. For instance:

In order to create the point $D$, I (implicitly) used the Crossbar Theorem. However, in order to use the Crossbar Theorem, I need to know that $C$ lays in the interior of $\angle ABC'$. But how do I prove that?
How can I be certain that $C$ lays between $B$ and $D$ (needed in order to form the triangle $\triangle DCC'$)? It's conceivable that $D$ could be between $B$ and $C$. If so, then what? If not, how to rule it out? (I am able to rule out the possibility that $C = D$).
In the proof of the case $M * C * C'$, the Wikpedia page simply states that the angle $\angle AC'B$ must be acute. But how do you prove that? (Beware that this page uses different labels for the vertices than what I did here).

Note
Again, I know that there are other ways of proving this theorem. But I am not interested in those.

REPLY [1 votes]: I think you could avoid introducing point $D$ altogether. By the Exterior Angle Theorem you have
$$
\angle BCM\ >\ \angle BC'M
\quad\hbox{and}\quad
\angle ACM\ >\ \angle AC'M.
$$
We then get
$$
\angle BCM+\angle ACM\,>\,\angle BC'M+\angle AC'M,
\quad\hbox{that is}\quad
\angle ACB\,>\,\angle AC'B.
$$
But that is a contradiction.