TITLE: Why Pauli exclusion instead of electrons canceling out?
QUESTION [5 upvotes]: To quote Wikipedia,

The Pauli exclusion principle is the quantum mechanical principle that no two identical fermions (particles with half-integer spin) may occupy the same quantum state simultaneously. A more rigorous statement is that the total wave function for two identical fermions is anti-symmetric with respect to exchange of the particles.

As far as I can tell, this means that two identical electrons in identical states have wave functions that cancel each other out.
My question is, if this is the case, why do we have the Pauli exclusion principle? It seems that getting two electrons too close should annihilate both of them (or cancel them out, or however you want to say it), but instead we have the Pauli exclusion principle, where somehow the electrons remain separate. Why?

REPLY [2 votes]: Putting two fermions in the same state doesn't cause the two fermions to cancel out and disappear; it causes the entire universe to cancel out and disappear. Or, more prosaically but accurately, a state with two fermions in the same single-particle state simply isn't a valid state in the fermionic Hilbert space. The point is that the badness isn't just isolated to the two fermions themselves - it incurably corrupts the entire system. Fermions are weird, only semi-local beasts.
However, there are particles that behave in exactly the way you're describing: putting two of them in the same state causes them to cancel each other out, but leaves the rest of the system completely unharmed. They're misleading called "Majorana fermions" in the condensed-matter community, although this name is extremely unfortunate, because (a) they're not actually fermions, but rather "anyons", and (b) they're not the same kind of particle that high-energy physicists refer to as "Majorana fermions", which really are fermions.


TITLE: $x(e^x - e^{-x})-e^x = 0$ has two reals solutions but am I supposed to use the intermediate value theorem?
QUESTION [0 upvotes]: Recently I've stumbled across a problem asking how many real solutions there was to $x(e^x-e^{-x})-e^x = 0 $, the answer is at least two, and it's easy to verify when plotting the function. However, in exams conditions, I would not have the opportunity to plot and was wondering if there is an "easy" way to find it?
Personally, I did use the intermediate values theorem. But here's the problem. I found that there was a $x_0 \in \mathbb{R}$ where $f(x_0)$ was negative and two values such as $ a < x_0 < b$ where $f(a) > 0$ and $f(b) > 0$
I think it's okay but I would be glad if there was a better way, not relying on roughly knowing how much $e^2$, etc is equal to. In this case it was fairly easy to find the values but it's always stressing me out to think I could've not think of a specific value.

REPLY [1 votes]: Your problem is equivalent than showing that $$f(x):=x(1-e^{-2x})=1,$$
has several solutions. It's rather clear that  $f(0)=0$ and $f(x)\to +\infty$ whenever $x\to \pm \infty $. So, oblviously, $f(x)=1$ has at least two solutions. It's also quite clear that $f(x)=1$ has exactly two solutions since $f'(x)<0$ whenever $x<0$ and $f(x)>0$ whenever $x>0$.


TITLE: Identity in quantum operator tutorial
QUESTION [3 upvotes]: I'm reading this tutorial by Ben Simons entitled Operator methods in
quantum mechanics in connection with his course in advanced QM, and I'm a bit puzzled by an identity in page 25, a bit above relation (3.3):
With the momentum operator $\textbf{p}=-i\nabla$ and the vector $\textbf{a}$ we have
$e^{-i\textbf{a}\cdot\textbf{p}}=e^{\ \textbf{a}\cdot\nabla}=\displaystyle\sum_{n=0}^{\infty}\frac{1}{n!}a_{i_1}\cdots a_{i_n}\nabla_{i_1}\cdots\nabla_{i_n}\ ,$
where repeated indices are summed over. What confuses me here is that (in 3D) $\textbf{a}$ and $\nabla$ have 3 components while this expression seems to refer to component $i_n$ where $n\rightarrow\infty$, and I don't recognize the usual expansion of the exponential.
What am I missing here?

REPLY [1 votes]: In this case the index can vary on the number of spatial dimensions (three if you are in 3D). The $i_n$ notation refers to the fact that you are "repeating" the scalar product "$\mathbf{a}\cdot\nabla$" n times:
$$
(\mathbf{a}\cdot\nabla)^n = (a_i\nabla_i)^n = (a_{i_1}\nabla_{i_1})(a_{i_2}\nabla_{i_2})\dots (a_{i_n}\nabla_{i_n}) = a_{i_1}a_{i_2}\dots a_{i_n}\nabla_{i_1}\nabla_{i_2}\dots \nabla_{i_n}
$$

REPLY [1 votes]: Notice that
\begin{align}
  e^{\mathbf a\cdot\nabla} = \sum_{n=0}^\infty \frac{1}{n!} (\mathbf a\cdot\nabla)^n
\end{align}
and
\begin{align}
  (\mathbf a \cdot\nabla)^n
&= \underbrace{(\mathbf a\cdot\nabla)\cdots(\mathbf a\cdot\nabla)}_{\text{$n$ factors}}
= (a_{i_1}\nabla_{i_1})\cdots (a_{i_n}\nabla_{i_n})
= a_{i_1}\cdots a_{i_n}\nabla_{i_1}\cdots \nabla_{i_n}
\end{align}
where we have made sure to use a different dummy index for each $\mathbf a\cdot \nabla$, hence the extra index on each index.


TITLE: How to solve a "foraging" problem on a directed graph?
QUESTION [6 upvotes]: I'm a little squirrel living in a forest with, say, eight trees $A,B,C,\ldots,H$. I can jump from tree to tree, though not all jumps are possible. (The fact that I can jump from $A$ to $B$, for example, doesn't guarantee that I can jump from $B$ to $A$.) Basically, I live in a directed graph whose adjacency matrix contains a few (non-symmetrically distributed) zeros.
I have stored five sorts of nuts $a,b,c,d,e$ in my little forest, each tree containing some portion of these. For example, tree $A$ contains nuts $a,d,e$ (quantities don't matter).
Now my question is, does there exist an efficient algorithm to calculate a path (always starting in my home tree $A$) through the forest so that I will be able to pick up a given sequence of nuts along the way, say $d,c,e,a,a,b,c,d,c$? (I can only pick up one kind of nut per tree.)
Same question, but with weighted edges (that measure energy expenditure). In that case, how can I find the path with least total weight?
Final question, how long does an algorithm have to run as a function of the nut sequence length $N$?

REPLY [2 votes]: $\color{brown}{\textbf{Parametrization.}}$
Denote
$T=8$ - the quantity of trees in the forest,
$a = 4$ - the average quantity of the available next trees,
$K=5$ - the quantity of kinds of the nuts in the forest,
$k=3$ - the average quantity of kinds of the nuts in the trees,
$D=9$ - the length of the given chain of the kinds of nuts,
$p=\dfrac kK = \dfrac35$ - the probability that the required kind of nuts is in the arbitrary tree,
$b=pa=\dfrac{12}{5}$ - the expectation of the quantity of the next trees with the required kind of nuts,
$C$ - the complexity of the algorithm for calculating the path, or the total number of checks "tree - sort of nuts."
$\color{brown}{\textbf{Calculating a path.}}$
Total quantity of sequences equals to $kK^{D-1}\approx=11\,71875.$
If $b>1,$ then the quantity of the variants in the ordinary search algorithms increases exponentially,
$$C\approx a\sum\limits_{d=0}^{D-1} b^d = a\dfrac{b^D-1}{b-1} \approx 7545 < kK^{D-1}.\tag1$$
This means that the searching can not be broken quickly.
At the same time, required sequence "dceaabcdc" can be splitted to the subsequences in the form of ("dceaa","abc","cdc"), wherein the second and the third subsequences can start from the any tree.
Calculating paths for the subsequences require
$$C_s \approx a\dfrac {b^{5}-1}{b-1}+2Tb\dfrac {b^{3}-1}{b-1} \approx 576\tag2$$
operations, with the approximate quantities of subpaths
$$g_1 \approx \dfrac {b^{5}-1}{b-1} \approx 56,\quad g_2 \approx g_3 \approx Tp\dfrac {b^{3}-1}{b-1} \approx 44,\tag3$$
and that corresponds with the quantity of possible sequences
$$N\approx \dfrac{g_1g_2g_3}{K^2} \approx 4337.$$
Subpaths $g_i$ can be placed in three arrays,wherein the special first index of the array $g_1$ corresponds to the number of the last tree, and the special first index of the array $g_3$ corresponds to the number of the first tree. The array $g_2$ has not special indexes.
Since the first tree in the next subpath should coinside with the last tree in the previous subpath, proposed structure allows to get any subpath of $g_2$ and to add the "head" from $g_1$ and "tail" from $g_3$ by the first indexes.
This means, that the first task has the complexity $(2),$ wherein the last step of this algorithm allows to calculate "the tail" by the entrance, from the given tree. This possibly accelerates the algorithm.
On the other hand, this approach allows to solve weighted edges task, if to place weights of the subpath in the same arrays. As minimum, it provides searching of the all variants. On the other hand, the subpaths can be sorted by the weights, and that accelerates the algorithm.


TITLE: Doubt on limit of sum of sequences. Two procedures leads to different answers.
QUESTION [5 upvotes]: I have the following problem.
Determine the convergence or divergence of the sequence $(x_n)$ where $$x_n=\frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n}.$$
My first approach was:
Well, since $(x_n)$ is the sum of the sequences $\left(\frac{1}{n+1}\right),\ldots,\left(\frac{1}{2n}\right)$ and each of them are convergent, in fact they converges to $0$, then the limit of $(x_n)$ must be $0$.
My second approach was:
If we make the calculations, we have that $$x_{n+1}-x_n=\frac{1}{(2n+1)(2n+2)}>0$$ then $$x_{n+1}>x_n$$ i.e., $x_n$ is an increasing sequence.
Also, as in $x_n$ we sum $n$ elements which are less or equan than $\frac{1}{n+1}$,  then $$x_n=\frac{1}{n+1} + \frac{1}{n+2} + \ldots + \frac{1}{2n}<\frac{n}{n+1}<1$$ i.e., $(x_n)$ is a bounded sequence.
Now, since $(x_n)$ is increasing and bounded sequence, it follows that its limit is the supremum of $(x_n)$.
Annoying question.
But how $(x_n)$ can converge to $0$ (as shown in my first approach) and at the same time converges to its supremum when it is an increasing sequence? Can be 0 its supremum? It is not a contradiction?
Any help would be appreciated.

REPLY [1 votes]: If you use harmonic numbers, you have $$x_n=H_{2 n}-H_n$$  For large values of $k$, the Wikipedia page gives $$H_k=\gamma +\log (k)+\frac{1}{2 k}-\frac{1}{12
   k^2}+O\left(\frac{1}{k^3}\right)$$ which makes $$x_n=\log (2)-\frac{1}{4 n}+\frac{1}{16 n^2}+O\left(\frac{1}{n^3}\right)$$ which shows the limit and also how it is approached.
For illustration purposes, let us us $n=10$; the exact result is $\frac{155685007}{232792560}\approx 0.66877140$ while the above approximation would give $\log(2)-\frac{39}{1600}\approx 0.66877218$


TITLE: Tossing Dice repeatedly, probability that 2nd trial had more tosses than 1st one
QUESTION [1 upvotes]: John repeatedly tosses a die until a six occurs for the first time. Alice then repeats the experiment. What is the probability that Alice made more tosses
than John?

REPLY [1 votes]: Outline: Let's first find the probability that Alice and John have the same number of tosses. They can either each get their first $6$ on their first toss, or they can each get their first $6$ on their second toss, or each on their third toss, and so on.
The probability it occurs in  the first round is $\frac{1}{6^2}$. The probability it occurs in the second is $\frac{1}{6^2}\frac{5^2}{6^2}$. The probability it occurs in the third is $\frac{1}{6^2}\frac{5^3}{6^3}$. And so on. This is an infinite geometric progression. Calculate the sum, and call it $p$.
Let $a$ be the probability Alice made more tosses. Then $2a+p=1$.  For by symmetry either John made fewer tosses, or Alice made fewer tosses (probabilities equal by symmetry) or there was a tie.
Added: The infinite geometric sequence
$$\sum_0^\infty \frac{1}{6^2}\left(\frac{5^2}{6^2}\right)^n$$
has sum $\frac{1}{6^2}\frac{1}{1-\frac{25}{36}}$, which simplifies to $\frac{1}{11}$.
It follows that the probability Alice makes more tosses than John is $\frac{5}{11}$.


TITLE: What is the probability of drawing 6 unique cards from 6 different decks?
QUESTION [2 upvotes]: I am working through some problems while studying and would like to validate my thinking with this problem.
There are six distinct deck of cards and one card is drawn from each deck. What is the probability that each card drawn is different?
I approached it as 1/52* 1/51 *... *1/47 because each time a card is drawn there is one less that is available to choose from if we don't want to match. I figured this is similar to 52 choose 1 times 51 choose 1 etc
Is this the correct way to approach it?

REPLY [2 votes]: Your answer to this problem is not correct. In fact, the correct answer should be $$ \frac{52}{52} * \frac{51}{52} * \frac{50}{52} * \frac{49}{52} * \frac{48}{52} * \frac{47}{52} $$
This is how I got this answer - Consider event A to be the set of all outcomes in which all of the six cards are pairwise-distinct. Using combinatorics, this is equivalent to $$ 52*51*50*49*48*47 $$
The total number of ways in which 6 cards can be drawn from 6 decks is $$ 52^{6} $$
Thus by probability theory, the answer should be equal to the number of favourable outcomes divided by total outcomes. Hence $$ \mathbb{P}(A) = \frac{52}{52} * \frac{51}{52} * \frac{50}{52} * \frac{49}{52} * \frac{48}{52} * \frac{47}{52} $$
Hope it helps you!!!


TITLE: Conjectured bound on Bessel function of the first kind
QUESTION [1 upvotes]: Looking at the asymptotic expansion of the Bessel function of the first kind $J_0(x)$, the following bound seems an obvious conjecture
$$
\left | J_0(x) \right | \le \min \left (1, \sqrt{\frac{2}{\pi x}} \right)
$$
I would also be interested in any bound of the form $J_0(x) \le A/ \sqrt{x +B}$ for some constants $A>0,\, B\ge0$. The bound looks tight when plotted on short segments and seems an obvious result. However I was a bit surprised when coming across this article.
Notes that the bounds are meant to hold for all real $x$, though $J_0(x)$ is even so one can restrict oneself to $x\ge0$.

REPLY [1 votes]: This bound is known, and one citation given for it is G.N. Watson's book A Treatise On The Theory Of Bessel Functions, pp 205-8. Even accounting for the book's quite unfriendly writing style, it's not clear to me that the bound actually occurs therein.... Here is a complete proof.
Modulo a computation on the interval $[0,2]$, it suffices to show that $\frac{\pi x}2 J_0(x)^2 \le 1$ for $x\ge2$. At the top of page 206, equation (1), we have
$$
J_0(x) = \sqrt{\frac2{\pi x}} \big( \cos(x-\tfrac\pi4) P(x,0) - \sin(x-\tfrac\pi4) Q(x,0) \big),
$$
where $P$ and $Q$ are certain functions. Then on page 208, we take $p=2$ in equation (1) to get
$$
1 - \frac{1^23^2}{2!(8x)^2} \le P(x,0) \le 1 - \frac{1^23^2}{2!(8x)^2} + \frac{1^23^25^27^2}{4!(8x)^4},
$$
and we take $p=0$ in equation (2) to get
$$
-\frac1{8x} \le Q(x,0) \le 0.
$$
Therefore by Cauchy's inequality,
\begin{align*}
\frac{\pi x}2J_0(x)^2 &\le \big( \cos(x-\tfrac\pi4)^2 + \sin(x-\tfrac\pi4)^2 \big) \big( P(x,0)^2 + Q(x,0)^2 \big) \\
&\le 1 \cdot \bigg( \bigg( 1 - \frac{1^23^2}{2!(8x)^2} + \frac{1^23^25^27^2}{4!(8x)^4} \bigg)^2 + \frac1{(8x)^2} \bigg) \\
&\le \bigg( \bigg( 1 - \frac{1^23^2}{2!(8x)^2} + \frac{1^23^25^27^2}{4!(8x)^216^2} \bigg)^2 + \frac1{(8x)^2} \bigg) \\
&= 1-\frac{4517}{65536 x^2}+\frac{30702681}{17179869184 x^4},
\end{align*}
where the final inequality used $x\ge2$. It's now easy to verify that this last bound is less than $1$ for $x\ge2$.
As for the interval $[0,2]$: for $x$ in this interval, the series
$$
J_0(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m!^2} \bigg(\frac x2 \bigg)^2
$$
satisfies the hypotheses of the alternating series test, and thus the series's truncations alternately form upper and lower bounds for $J_0(x)$ on this interval. The first two truncations give $1 \ge J_0(x) \ge 1-x^2/4 \ge 0$. The next truncation gives
$$
J_0(x) \le 1-\frac{x^2}4 + \frac{x^4}{64} < \sqrt{\frac2{\pi x}},
$$
where we need to verify the last inequality to finish the proof. Taking square roots of both sides and subtracting, it suffices to show that
$$
\sqrt[4]{\frac2{\pi x}} - \bigg( 1-\frac{x^2}8 \bigg) > 0
$$
on $[0,2]$. Standard calculus shows that the minimum of this function occurs at $x=(2/\pi)^{1/9}$ and has value $9/(2^{25/9} \pi ^{2/9})-1$; verifying that this minimum value is positive boils down to confirming that $\pi < 3^9/2^{25/2}$, which in turn (since $\pi<22/7$) follows from the inequality $3^{18}7^2 > 2^{27} 11^2$.


TITLE: Probability problem that involves number theory
QUESTION [5 upvotes]: Let $k \in Z^+$. Assume integers 1, 2, 3, . . . , 3k+ 1 are written down randomly. Calculate the probability that at no time during this process, the sum of the integers is a positive integer divisible by 3?
Attempt: I am trying to approach this by finding the complement of what's being asked which is the number times the sum of the integers is divisible by 3. The sample space I think is $\prod_{i = 0}^{3k+1}(3(i)+1)!$ since that's I think the number of trees we can generate by doing this process.
I think my sample space is off. The right way is to
 figure out how many sequences can we have at some time i where $1 \leq i \leq 3k+1$ during the process. This is:
$(3k+1) +(3K+1)(3k) + (3k+1)(3k)(3k-1)+ ... + (3k+1)!$
I also have the feeling that this is done by using states. There are just three state where the sum can be at any time and these are: 0mod3, 1mod3 and 2mod3. We have to find all the possible ways we can reach the state 0mod3 somehow.

REPLY [2 votes]: For the problem where repitition is not allowed, i.e. we are looking at permutations of the numbers $1,2,3,\dots,3k+1$
Looking at the numbers modulo3 for now, every arrangement will be of the form $$1.1.2.1.2.1.2.1.\dots2.$$
where those numbers which are multiples of three will be placed somewhere where the dots are (not the first number of the sequence)
The reason should be clear: if the first non-multiple of three is $1$mod3, the next non-multiple of three cannot be a $2$, else we will have a sub-sum that is a multiple of three.  Therefore, the second must be a $1$ as well.  Similarly, the next must be a $2$mod3 to avoid a sub-sum being a multiple of three again, and so on.  Alternatively, if the first non-multiple of three is $2$mod3, we will have it begin $221212121\dots$, but this would require one or two more numbers which are $2mod3$ than numbers which are $1$mod3, an impossibility.  Finally, it cannot start with a multiple of three for obvious reasons.
There are $\binom{3k}{k-1}$ ways to insert the $k$ copies of $0$ into the sequence above by using stars and bars as there are $k$ 0's to place into $2k+1$ available spots.
Now, replacing the $1$'s with an arrangement of all of the numbers which are $1$mod3, replacing all of the $2$'s with an arrangement of the numbers which are $2$mod3, and similarly for the $0$'s, we have a sequence of the numbers $1,2,3,\dots,3k+1$ satisfying your conditions.
There are $(k+1)!,k!,$ and $k!$ ways to do this respectively.
There are then $(k+1)!k!k!\binom{3k}{k-1}$ ways you can arrange the numbers satisfying your desired conditions.  This is out of the $(3k+1)!$ equally likely arrangements, making the probability:
$$\frac{(k+1)!k!k!\binom{3k}{k-1}}{(3k+1)!}$$


TITLE: Understanding Integration by Substitution 'Scale factor'
QUESTION [1 upvotes]: I was hoping to gain some intuition as to the process of integration by substitution and how exactly it works when evaluating a definite integral in one dimension.
So when evaluating a definite integral in one dimension, $\int_a^b f(x) dx$, you divide the $x$-axis between $a$ and $b$ into $n$ equal sections of length $\delta x$ and then you have that $\int_a^b f(x) dx = lim_{n \rightarrow \infty} \sum_{i=1}^{n}f(x_i)\delta x$
So then when you use integration by substitution to evaluate the equivalent integral, you use a change of variables, $u(x)$ which means you are now integrating along a new $u$-axis between limits $u(b)$ and $u(a)$. You compute the integral in the exact same way to obtain $\int_{u(a)}^{u(b)} f(x(u)) du$. However you then have to multiply by the 'scale factor' $\frac{dx}{du}$, and I'm having difficulty understanding this. I guess that the problem is that since the difference between $u(b)$ and $u(a)$ on the $u$-axis is in general different from the difference between $b$ and $a$ on the $x$-axis that this means that $du$ and $dx$ are not equal in size and thus multiplying by the 'scale factor' basically has the effect of making every section of length $du$ back into the 'proper length' $dx$. Is this the case?

REPLY [1 votes]: Yes, intuitively you're right.
So if $u = g(x)$.
Then $\frac{du}{dx} = g'(x)$
Then we can intuitively think in terms of differentials
$du = g'(x)dx$
So $dx = \frac{du}{g'(x)}$
So you can substitute this $dx$ back into the integrand to get
$\int_a^b f(x) dx$
$\int_{u(a)}^{u(b)} f(x(u)) dx$
$=\int_{u(a)}^{u(b)} f(x(u)) \frac{du}{g'(x)}$
If you want to be completely rigorous though, we're using the chain rule in reverse... the "integration by substitution" theorem is proven here:
https://en.wikipedia.org/wiki/Integration_by_substitution
In reality the cross multiplying by $dx$ is just a mnemonic... a book-keeping tool that works well to accomplish what the theorem is saying. I guess there is a theory of differentials that could make this multiplying by dx rigorous.


TITLE: Why is the efficiency of human cells less than the efficiency of an Otto engine?
QUESTION [41 upvotes]: I always used to think (I don’t know why!) that the efficiency of human (and animal and plant) cells should be equal to or near the efficiency of a Carnot engine or at least should be the highest efficiency among all practical engines. But I wondered when I saw the answers given to this question. They are talking about an efficiency of $18-26\%$. But you can see here that the efficiency of an Otto engine is between $56\%$ and $61\%$. Is there any explanation for this? Which cycle do human cells work with? Can we compare living cells with heat engines at all?

REPLY [2 votes]: Let's calculate engine efficiency by using the ratio of work and gasoline energy (assuming it can be completely converted to CO2 and H2O). And let's calculate cell (muscular) efficiency by the ratio of work (from gym equipment) and fat energy (assuming it can be completed converted to CO2 and H2O). These two efficiency are comparable.
A gasoline engine efficiency is about 30%. If the oil processing is included, it can be even lower. Cell efficiency 18-26% is not too low. The question might be where does the rest 82%-74% energy go? My thinking it can be used to accelerating circulation system (increase heart beating frequency, transporting material like oxygen etc) and it is used up as heat is removed from the body. A demographic plot should be a study topic for bio engineering scientists.


text:
\section*{Supplementary material}

The following sections discuss in more detail the theoretical guarantees of our approach.  \S\ref{appendix:SecExpectationApproximation} presents the optimality conditions underlying our shooting formulation and it is shown how these optimality equations  can be approximated via a collection of particles. \S\ref{appendix:Barron} proposes different regularizations, whose choice is key for practical and theoretical results. We show that, under some conditions, the Rademacher complexity of the set of flows can be bounded and apply our results in \S\ref{SecAppendixConsequencesUpDown} to the {\UpDown} model. \S\ref{appendix:freeparams} discusses the number of free parameters of our shooting approach in relation to the number of free parameters for direct optimization. \S\ref{appendix:automatic_shooting} explains how the shooting equations can be automatically derived via automatic differentiation.  \S\ref{appendix:updown_universal} shows the universality of our {\UpDown} model. \S\ref{appendix:experimental_settings} provides details on our experimental setup. Lastly, \S\ref{appendix:figures_results} shows some additional experimental results.

\section{Expectation approximation of optimality equations}
\label{appendix:SecExpectationApproximation}

We first discuss a general variational setup of supervised learning including regularization.

\subsection{Variational setup}
Suppose the data consists of input $X \in \R^d$.
Let $f(\cdot,\theta(t))$ be a vector field on $\mathbb \R^d$, e.g. the single hidden layer of \eqref{Eq:SingleHiddenLayer} or a linear (in parameter) layer. Consider the flow $\varphi := \varphi(T,\cdot)$ generated by $f$ according to
\begin{equation}\label{EqDiffeomorphicFlows}
\begin{cases}
\frac{\ud}{\ud t} \varphi(t,\x) = f(\varphi(t,\x),\theta(t))\,,\\
\varphi(0,\x) = \x\,.
\end{cases}
\end{equation}

We consider the general task of minimizing,
\begin{equation}
\on{Reg}(\varphi) + \gamma \mathbb{E}[\ell(\varphi(X))]\,,
\label{eq:variational_task}
\end{equation}
where $\gamma$ is a positive regularization parameter.

\par
We now consider the particular case of a \ResNet\ model where each layer is given by an \UpDown\ model \eqref{updownODE} or even a single hidden layer \eqref{Eq:SingleHiddenLayer}.

\par
Without loss of generality, set the terminal time to $T=1$.
Letting $\rho_0$ denote the probability density of $X$, minimizing \eqref{eq:variational_task} is equivalent to minimizing
\[
\inf_{\varphi}\left[ \on{Reg}(\varphi) + \gamma \int_{\R^d} \ell(\varphi(\x)) \rho_0(\x) \ud \x\ \right]\enspace.
\]
This can be rewritten as
\[
\inf_{\varphi} \left[\on{Reg}(\varphi) + \gamma  \int_{\R^d} \ell(\x') \rho_1(\x') \ud \x'\right]\enspace,
\]
where $\rho_1(\x):=\rho(1,\x)$ is the flow of the continuity equation
\[
\partial_t \rho(t,\x) + \on{div}(\rho(t,\x) f(\x,\theta)) = 0\,,\rho(0,\x)=\rho_0(\x)\,,
\]
where $\on{div}$ is the divergence operator on vector fields. Note that $\rho_1$ can be regarded as the density representing the data at time $1$.
In the following, we deal with a general regularization term $\on{Reg}(\varphi) = \int_0^1 R(\theta(t),\rho(t)) \ud t$, where the $R$ term can depend on the density of data at time $t$. A particular though important case is when the regularization $R$ does not depend on $\rho$,
$\int_0^1 R(\theta(t)) \ud t$.

\subsection{Optimality equations and Hamiltonian ensemble approximation}
We detail the optimality equations when data points are represented by a probability measure.
As mentioned above, the regularity of the map is enforced via a penalty on the weights at each timepoint and is  the integral $\int_0^1 R(\theta(t)) \ud t$ or even more generally $\int_0^1 R(\theta(t),\rho(t)) \ud t$.
Using Lagrange multipliers, this constraint can be enforced and minimizers of the energy should be saddlepoints of the energy
\begin{multline*}
\mathcal{L}(\rho,\theta,p) := \gamma \int_{\R^d} \ell(\x) \rho_1(\x) \ud \x+ \int_{0}^1 R(\theta(t),\rho(t))\ud t \\ + \int_0^1 \int_{\R^d}  p(t,\x) (\partial_t \rho(t,\x) + \on{div}(\rho(t,\x) f(\x,\theta(t))))  \ud \x \ud t\,,
\end{multline*}
where $p(t,\x)$ is a time and space dependent function.
The optimality equations are then
\begin{equation}\label{EqOptimalityEquations_density}
\begin{cases}
\partial_t \rho(t,\x) + \on{div}(\rho(t,\x) f(t,\x,\theta(t))) = 0\,,\\
\partial_t p(t,\x) + \nabla p(t,\x) \cdot f(t,\x,\theta(t)) = \frac{\delta R}{\delta \rho}(\theta(t),\rho(t))\,,\\

 \partial_{\theta}R
 (\theta(t),\rho(t)) - \int_{\R^d}

 \partial_{\theta}f(\x,\theta(t))^\top(\nabla p(t,\x) \rho(t,\x)) = 0\,,
\end{cases}
\end{equation}

where $\nabla p$ is the gradient w.r.t. $\x$ of $p(t,\x)$ and $\delta$ denotes differentiation w.r.t. the indicated parameter. The notation $\frac{\delta R}{\delta \rho}$ means the Fréchet derivative of the penalty w.r.t. the density $\rho$. Note that in our current work, $R$ is independent of $\rho$. However, this more general setup encompasses optimal transport models, see Section \ref{SecNonlinearInParameters}.

In practice, one does not have access to the full distribution and the variational setup needs to be approximated. As proposed in the main text, we approximate it using a collection of particles that follow the optimality equations which are Hamiltonian evolution equations for this collection of particles. The collection of particles $\{(\q_j,\p_j)\}$ are defined by their state and costate. We estimate $\rho$ using the empirical measure $\frac{1}{K}  \sum_{j = 1}^K \delta_{\q_j(t)}(\cdot)$. Writing the optimality equation for this particular empirical measure leads to the equation \eqref{EqShootingForward}.

When the number of particles tends to infinity, we can hope to recover the optimal trajectory.
However, we do not explore this question formally here. We simply remark that this question is directly connected to expressiveness and generalization properties of the constructed neural network and is also probably data dependent.

\section{Choice of regularization}
\label{appendix:Barron}
The simplest regularization on the flow $\varphi$ is given by
\begin{equation}
	\on{Reg}(\varphi) = \int_0^1 R(\theta(t)) \ud t\,,
\end{equation}
where $R$ does not depend on $\rho(t,\cdot)$.

The first possibility is a quadratic penalty for the single-hidden-layer vector field of \eqref{Eq:SingleHiddenLayer}, where $R(\theta(t)) = \frac 12 \| \theta(t) \|^2_2$ is the Frobenius norm of the parameter $\theta$.

Since the space of vector fields is a \emph{finite} dimensional linear space, it can be endowed with a scalar product, which turns this space into a Reproducing Kernel Hilbert Space (RKHS). Therefore, the linear in parameter - quadratic penalty setting of \S\ref{SecChoiceOfReg} is a particular case of vector fields encoded by $f(\cdot,\theta(t))\in H$, with $H$ a RKHS embedded in $W^{1,\infty}$ vector fields.

This setting

leverages strong analytical and geometrical foundations \cite{laurentbook,CompletenessDiffeomorphismGroup}:

1) When the activation function is smooth, the resulting vector field is smooth\footnote{I.e smoothness asks for Lipschitz regularity vector field, which ensures existence and uniqueness of the flow.}, and consequently the associated flow map $\varphi$ is guaranteed to be a one-to-one smooth map (i.e., a diffeomorphism). For instance, with the \UpDown\ model, it is a homeomorphism in $(\x,\v)$.
Moreover, the quadratic penalty induces a right-invariant distance on the set of flows generated by \eqref{EqDiffeomorphicFlows} and the distance to identity of the resulting flow can be bounded by  $\on{Reg}(\varphi)$ (see~\cite{laurentbook,CompletenessDiffeomorphismGroup} for more details in a Sobolev setting). 2) When the activation function is of {\ReLU} type, the resulting map is still a $W^{1,\infty}$ one-to-one map (i.e., a homeomorphism) and has Lipschitz regularity.

\par
Another type of regularization for the single-hidden-layer vector field of \eqref{Eq:SingleHiddenLayer} we discuss is based on the Barron norm \cite{e2019barron}:
\begin{equation*}
\| \theta \|_{\mathcal{B}}^2 :=  \frac 1{d'}  \sum_{j = 1}^{d'}\| \theta_1^j \|_2^2 (\| [\theta_2]_j \|_1 + \| b_2^j \|_1)^2\enspace,
\end{equation*}
where $\theta_1^j$ denotes the $j^{\text{th}}$ column of $\theta_1$ and $[\theta_2]_j$ denotes the $j^{\text{th}}$ row of $\theta_2$.
As discussed in the main text, the reason we might consider a Barron norm penalty for the single-hidden-layer vector field in \eqref{Eq:SingleHiddenLayer} rather than the quadratic penalty is because of its theoretical results. Indeed, the Rademacher complexity is bounded

for the combination of a single-hidden-layer vector field with a Barron norm penalty, but not when combined with a quadratic penalty.

\subsection{Linear in parameters - quadratic energy}
\label{appendix:quadratic_energy}
Now let us examine in detail models that are \emph{linear} in parameters and have \emph{quadratic} energy on parameters: this case is the simplest to be studied, and computationally not as demanding as the nonlinear case. As mentioned above, the set of possible vector fields $f(\cdot,\theta(t))$ is a finite dimensional linear space, which is a reproducing kernel Hilbert space when endowed with an $L^2$ norm. Since all Hilbert norms in finite dimensions are equivalent, this choice of regularization is universal in this class of quadratic penalties.
\begin{enumerate}
	\item The vector field is $f(\cdot,\theta(t)) = \theta \cdot \mathbf{\sigma}$, where $\mathbf{\sigma}$ is a vector of maps.
	In this case, the optimality equation reads
	\begin{equation*}

	\partial_{\theta}f(\x,\theta(t))^\top
	(\nabla p(t,\x) \rho(t,\x)) = \int_{\R^d} \mathbf{\sigma}(x)^\top(\nabla p(t,\x)  \rho(t,\x))\ud \x \,.
	\end{equation*}
	\item If the penalty $R$ only depends on $\theta$ and is quadratic: $R(\theta(t)) = \frac{1}{2}\int_0^1 \| \theta(t) \|^2 \ud t$, then one has $\frac{\delta R}{\delta \theta}(\theta(t),\rho(t)) = \theta(t)$.
\end{enumerate}
Thus, under these two conditions, the parameters are \emph{explicit} in terms of $p$, $\rho$ and $\sigma$:
\begin{equation}\label{EqOptimalityEquationLinearInParameter}
\theta(t) =  \int_{\R^d} \sigma(\x)^\top(\nabla p(t,\x) \rho(t,\x))\ud \x\,.
\end{equation}

Two observations are warranted. First, if, instead of quadratic regularization on the parameters, we were to choose a RKHS norm (in the infinite dimensional case) as penalty, it would result in the introduction of the kernel applied to the R.H.S. of \eqref{EqOptimalityEquationLinearInParameter}.
Second, from \eqref{EqOptimalityEquationLinearInParameter}, one could be tempted to derive an evolution equation for $\theta$. This equation is known as the EPDiff equation \cite{laurentbook} and is unfortunately not a closed equation on the set of parameters $\theta(t)$ themselves. Therefore, our approach is a possible way to approximate it.

An important property of this simple setting is that the norm of the vector field is preserved by the forward model defined by the collection of Hamiltonian particles and it also holds in the continuous setting. As stated in Section \ref{SecChoiceOfReg}, the Hamiltonian is given by $R(\theta(t)) = \frac{1}{2} \Tr\left(A(t)^\top M_{A} A(t)\right) + \frac{1}{2} b(t)^\top M_{b} b(t)$ where $A,b$ are the optimal parameters given by
\begin{equation}
\begin{cases}
A(t) \! &= {M_{A}}^{-1}(-\sum_{j=1}^K \p_j(t) \sigma(\q_j(t))^\top)\\
b(t) \! &= {M_{b}}^{-1}(-\sum_{j=1}^K \p_j(t))\,.
\end{cases}
\end{equation}
The Hamiltonian $R(\theta(t))$ being constant gives a constant norm vector field.

\subsection{Nonlinear in parameters - energy which depends on the distribution}\label{SecNonlinearInParameters}
For exposition purposes, we present two cases of interest which we have not well explored numerically.

\textbf{Example of the Barron norm.}
Obviously, the single-hidden-layer vector field in \eqref{Eq:SingleHiddenLayer} is not linear in parameters. We have already discussed that it is proper in this case to endow the space with norms such as the Barron norm \cite{e2019barron}.
For simplicity, consider the single-hidden-layer vector field in \eqref{Eq:SingleHiddenLayer} without $b_1$, i.e., $f(\x(t),\theta(t)) = \theta_1 \sigma(\theta_2(\x) + b_2)$. A simple upper bound for the Barron norm\footnote{The actual Barron norm is defined as the infimum of the r.h.s. in \eqref{EqBarronNormNotOptimized} on all the possible representations of the function $f(\cdot,\theta)$ as a single-hidden-layer.} is
\begin{equation}\label{EqBarronNormNotOptimized}
\| f(\cdot,\theta)  \|_{\mathcal{B}}^2 :=  \frac 1{d'}  \sum_{j = 1}^{d'}\| \theta_1^j \|_2^2 (\| [\theta_2]_j \|_1 + \|[b_2]_j\|_{1})^2\enspace.
\end{equation}

Again, $\theta_1^j$ denotes the $j^{\text{th}}$ column of $\theta_1$ and $[\theta_2]_j$ denotes the $j^{\text{th}}$ row of $\theta_2$.

Let us consider the case of $R(\theta(t)) = \frac 1 2 \| f(\cdot,\theta)  \|_{\mathcal{B}}^2$.
In this case, one has the following optimality equations to solve
\begin{align*}
& \theta_1^j (\| [\theta_2]_j \|_1 + \| |[b_2]_j \|_1)^2 =  \int_{\R^d} \sigma([\theta_2]_j\x + [b_2]_j)^\top(\nabla p(t,\x) \rho(t,\x))\ud \x\,,\\
& \| \theta_1^j \|^2_2  (\| [\theta_2]_j \|_1 + \| [b_2]_j\|_1) \partial \| [\theta_2]_j^k \|_1 = \int_{\R^d}
[\ud \sigma([\theta_2]_j\x + [b_2]_j)(\x_k)]^\top(\nabla p(t,x) \rho(t,\x))\ud \x \,,\\
& \| \theta_1^j \|^2_2  (\| [\theta_2]_j \|_1 + \| [b_2]_j \|_1) \partial \|  [b_2]_k^j \|_1 = \int_{\R^d}
[\ud \sigma([\theta_2]_j\x + [b_2]_j)(\x_k)]^\top(\nabla p(t,\x) \rho(t,\x))\ud \x \,.
\end{align*}

These equations involve the subdifferential of the $L^1$ norm, and optimization of this type of functions, which involves sparsity, is a well-explored field~\cite{MAL-015}.
We leave experiments with this norm for future work. Note that in this case the norm of the vector field is not equal to the Hamiltonian and it is not a constant of the flow.
\subsection{$L^2$ regularization, optimal transport}
Last, we briefly mention a model that is part of our framework which has the advantage of not specifying the penalty on the space of parameters encoding the vector field. In case there is no obvious norm to be used on the space of vector fields, it is possible to use an $L^2$ type of penalty on the vector fields themselves \mnrev{instead of on the parameters.}

Indeed, one way to be rather independent of the choice of the parameterization of the map consists in introducing a cost that represents the $L^2$ norm of the map. However, $L^2$ depends on the choice of a measure and this measure can be  chosen as the density of the data, $\rho(t,\x)$. More precisely, one can use
\begin{equation}\label{EqOTLike}
R(f,\rho(t))  = \frac 12 \int_{\R^d} \| f(\x,\theta)\|^2 \rho(t,\x) \ud \x\,.
\end{equation}
In such a case, this formulation resembles finding an optimal transport (OT) map between $\rho_0$ and $\rho_1$. Specifically,
optimal transport is an optimization problem  which can be solved via a fluid dynamic formulation \cite{benamou2000computational} introducing the kinetic penalty above. However, the two models (OT and the one defined by the regularization \eqref{EqOTLike}) differ since the optimization set for optimal transport is the set of $L^2$ vector fields with respect to measure $\rho$ and the above formulation is a parameterized approximation of this set.

This parameterized approximation needs to retain generalization properties of the optimized map. Note however, that in the limit where the number of neurons goes to infinity, optimal transport will be well-approximated since the optimization is performed on a dense subset of all vector fields. Obviously, fixing the choice to a single-hidden-layer design implies a choice for $d'$ in $\theta_1(t) \in L(\R^{d'},\R^d)$ and $\theta_2(t) \in L(\R^d,\R^{d'})$ of ~\eqref{Eq:SingleHiddenLayer}, which thus gives a regularization of the computed approximation of the optimal transport map.

\textbf{Computational burden.}
In either case of the Barron norm or the optimal transport type of penalty, the implicit equation corresponding to the third equation in \eqref{EqOptimalityEquations_density} has to be solved at each layer of the discretization.

We experimented with a simple strategy of unrolling the related minimization scheme. An efficient approach to solve such implicit equations will be necessary for practical implementations.

\subsection{Rademacher complexity of bounded energy flows. }\label{SecRademacher}
In this section, given a set of vector fields with bounded Rademacher complexity, we show that the resulting flows also have bounded Rademacher complexity.

The flow of a vector field $f(\cdot,\theta(t))$ is a vector valued map denoted by $\varphi$.
Let us first treat the case of the Rademacher complexity of a component of the flow map $\varphi^k$.

\begin{theorem}\label{ThRademacher}
	Let $\mathcal{F}$ be a space of vector fields defined on a compact space $C \subset \R^d$. Assume that the Rademacher complexity on $n$ points in $C$ of each component of the vector fields $f^k(t,\cdot)$ for $k = 1,\ldots,d$ is controlled by $M(n,t)$ which depends on $n$, then the Rademacher complexity of each component of the flows at time $1$ is bounded by $\int_0^1 M(n,t) \ud t$.
\end{theorem}

\begin{proof}

Recall that Rademacher complexity, see \cite{wainwright2019high}, of a class of functions $\mathcal{F}$ is defined as, for $\mathbf{Z} = (\z_1,\ldots,\z_n) \in C$,
\begin{equation*}
\on{Rad}_{\mathbf{Z}}(\mathcal{F})\eqdef \mathbb{E}\left[\sup_{g \in \mathcal{F}} \sum_{i = 1}^n \varepsilon_i g(\z_i)\right]\,,
\end{equation*}
where the $\{\varepsilon_i\}_{i = 1}^n$ are i.i.d. Rademacher random variables. Our hypothesis ensures
$\on{Rad}_{\mathbf{Z}}(\mathcal{F}) \leq M(n)$. Apply the definition of the flow to get
\begin{equation*}
\varphi(1,\x) = \x + \int_0^1 f(\varphi(t,\x),\theta(t)) \ud t\,.
\end{equation*}
Therefore,
\begin{align*}
\mathbb{E}\left[\sup_{\varphi \in \mathcal{F}} \sum_{i = 1}^n \varepsilon_i \varphi^k(\z_i)\right] &\leq \on{Rad}_{\mathbf{Z}}(\{\on{Id}\}) + \int_0^1 \mathbb{E}\left[\sup_{f(\cdot,\theta(t))}\sum_{i = 1}^n \varepsilon_i f^k(\varphi(t,\z_i),\theta(t))\right] \ud t\,,\\
&\leq  0 + \int_0^1 M(n,t) \ud t \,.
\end{align*}
In the previous formula, we used the fact that the Rademacher complexity of a set comprised of a single map is zero.
\end{proof}

\begin{corollary}
Let $H$ be a RKHS of vector fields whose kernel $\mathsf{k}$ is bounded on the diagonal $\| \mathsf{k}(\x,\x)^k \|_\infty < \infty$ , then, the set of flows denoted by $\mathcal{F}$ at time $1$ of time-dependent vector fields in $\mathcal B(0,R)$, the ball of radius $R$ centered at the origin satisfies $\on{Rad}_{\mathbf{Z}}(\mathcal{F}) \leq \frac{2R \sqrt{\| \mathsf{k}(\x,\x)^k \|_\infty}}{\sqrt{n}}$, where $\on{Rad}_{\mathbf{Z}}(\mathcal{F})$ is the Rademacher complexity for $n$ points.
\end{corollary}

\begin{proof}
The Rademacher complexity of the ball of radius $R$ in the RKHS $H$ \cite[Lemma~22]{bartlett} is upper bounded: $\on{Rad}_{\mathbf{Z}}(\mathcal{B}(0,R)) \leq \frac{2R \sqrt{\| \mathsf{k}(\x,\x)^k \|_\infty}}{\sqrt{n}}$. We then directly apply Theorem \ref{ThRademacher}.
\end{proof}

A similar result also holds for vector fields generated by the single-hidden-layer vector field in \eqref{Eq:SingleHiddenLayer}, see \cite{e2019barron}.

Last, we note that the result and its proof also hold if one uses the following Rademacher complexity for vector valued functions  \cite{RademacherVectorValued},
\begin{equation*}
\on{Rad}_{\mathbf{Z}}(\mathcal{F})\eqdef \mathbb{E}\left[\sup_{g \in \mathcal{F}} \sum_{i = 1}^n\sum_{j =1}^d \varepsilon_j \varepsilon_i g_j(\z_i)\right]\,,
\end{equation*}
for $g = (g_j)_{j = 1,\ldots,d} \in \mathcal{F}$.

\subsection{Consequences for the  \texttt{UpDown} model}\label{SecAppendixConsequencesUpDown}

We put together the previous results on the \texttt{UpDown} model. First, the space of vector fields endowed with the quadratic penalty on the parameters forms a RKHS.
The variational formulation implies that the norm of the velocity field generated by a given collection of Hamiltonian particles $\{(\q_j(0),\p_j(0))\}$ is preserved. In addition, this norm can be explicitly computed since the parameters at time $0$ can be computed in terms of $\{(\q_j(0),\p_j(0))\}$. Last, the generated space of maps has a Rademacher complexity which is linearly bounded by this norm. In order to be fully explicit on the constant for the Rademacher complexity, we need to compute $\sup_{x \in C} \| \mathsf{k}(x,x) \|$ where $\mathsf{k}$ is the kernel associated with the RKHS. Without making this quantity explicit here, we simply mention that the bound degrades (i.e. increases) with increasing inflation factor $\alpha$, as it can be expected.

\section{Analysis of the number of free parameters}
\label{appendix:freeparams}

It is instructive to understand the number of parameters for a shooting approach in comparison to the typical approach of optimizing a neural network (where the parameter-dependency at optimality is only considered implicitly at convergence of the numerical solution rather than explicitly during the shooting). We focus on the cases of affine and convolutional layers for illustration.

Consider a DNN with a depth of $L$ layers, where each hidden layer has $P$ parameters. The number of free parameters is then $LP$, compared to $2KS$ where $K$ is the number of active particles, each of them of size $2S$\footnote{For example, $S$ for our {\UpDown} model simply corresponds to the dimension of its state space: $S=(\alpha+1)d$, where $\alpha$ is the inflation factor and $d$ the data dimension. Note that in our experiments with the {\UpDown} model we also learned an affine map from the initial conditions $\x(0)$ to the initial conditions $\v(0)$. Such a map has $\alpha d(d+1)$ parameters. These parameters are included in the table of Fig.~\ref{fig:spiral_short_long_range} and in Tables~\ref{tab:number_of_parameters_simple_functions}/\ref{tab:number_of_parameters_spiral} summarizing the number of mode parameters. However, we will not consider parameters in our discussion here, as they would equally apply to both a shooting and a direct optimization approach and could also be avoided by simply initializing $\v(0)$ to zero. A similar initialization to zero approach is, for example, commonly taken in {\ResNets} when increasing the number of feature channels~\cite{he2016deep}.}. Hence, solutions with less than $LP/(2S)$ particles provide benefits in the number of free parameters. Therefore, as the number of particles is reduced, we may parameterize the DNN with a smaller number of parameters. \emph{Most remarkably, the number of free parameters is always $2KS$ regardless of the number of parameters of a particular layer as the layer parameters are obtained via the shooting equations based on the particle states.} This is a consequence of regularizing the parameters in our loss which couples them across time at optimality. We make this clearer in what follows.

\textbf{Affine layers.} Recall that in our simple example of \S\ref{SecChoiceOfReg} the parameters $\theta(t)=[A(t),b(t)]$ of our affine\footnote{In this section, we mean affine with respect to $(\sigma(\q_j(t)))_{j \in 1,\ldots,K}$.} model are given as
\begin{equation}
  A(t) \! = {M_{A}}^{-1}\left(-\sum_{j=1}^K \p_j(t) \sigma(\q_j(t))^\top\right),\quad
  b(t) \! = {M_{b}}^{-1}\left(-\sum_{j=1}^K \p_j(t)\right)\enspace.
  \label{eq:parameters_for_affine_model}
\end{equation}
Here, $A(t)$ and $b(t)$ have $d^2$ and $d$ parameters, respectively; these parameters are indirectly given by the set of particles $\{(\q_j(t),\p_j(t))\}$ at any given time. Hence, for this model $S=d$ and $P=d(d+1)$. If we assume we have $K$ particles and compare to a discrete layer implementation of this model then the particle-based approach will have less free parameters if
\begin{equation*}
  2Kd < L d(d+1)\enspace.
\end{equation*}
Importantly, the state-space dimension, $d$, only enters the number of free parameters linearly for the particle approach ($2Kd$), while there is a quadratic dependence for direct optimization ($Ld(d+1)$). This is a direct consequence of the optimality condition which couples the parameters $\theta(t)$ across time. One can see this phenomenon in action in~\eqref{eq:parameters_for_affine_model}, where the matrix $A$ is expressed as the sum of matrices $\p_j(t) \sigma(\q_j(t))^\top$ with rank $\leq 1$. Concretely, a particle-based shooting approach uses less parameters if the number of particles $K<L(d+1)/2$. Another interesting observation based on this example is that even if we would have only considered a linear model (i.e., without the bias term, $b(t)$) the number of parameters for the particles would have still remained at $2KS$. This is again a consequence of optimality and of our parameter regularization. Note that this also means that even though our {\UpDown} model
\begin{equation*}
\dot{\x}(t) = \theta_1(t) \sigma(\v(t)) + b_1(t),\quad
\dot{\v}(t) = \theta_2(t) \x(t) + b_2(t) + \theta_3(t) \sigma(\v(t))\,,
\end{equation*}
has significantly more parameters $\theta(t)=[\theta_1(t),b_1(t),\theta_2(t),b_2(t),\theta_3(t)]$ when directly optimized, this has no direct impact on the number of free parameters of its particle-based parameterization. Only the state-space dimension matters. Concretely, if we were to instead consider a model of the form
\begin{equation*}
\dot{\x}(t) = \theta_1(t) \sigma(\v(t)),\quad
\dot{\v}(t) = \theta_2(t) \x(t)\,,
\end{equation*}
the particle-based parameterization would stay \emph{unchanged!} Only the way how one infers $\theta(t)$ from the particles changes.

\textbf{Convolutional layers.} Shooting approaches for convolutional models can also be derived. We did not experiment with such models in this work. However, we show here that the number of free parameters may also be decreased with a particle-based approach. This will be interesting to explore in future work. Specifically, for convolutional layers a particle-based parameterization could be particularly effective as one typically has quadratic complexity in the number of filters between convolutional layers (i.e., if a layer with $N$ feature channels is followed by a layer with $M$ feature channels, this will induce the estimation of $N\times M$ convolutional filters and hence will drastically influence the number of parameters for large $N$ or $M$). In contrast, a particle-based shooting approach does not increase the number of parameters as it ties them together via the optimality conditions expressed by the shooting equations. As a rough estimate for a standard convolutional \ResNet\, for $L = 50$, $P = 100^2 \times 16$, $LP \approx 8.10^6$. Thus, if particles have size $40$, we end up with at most $10^5$ active particles.

\textbf{General remarks.} Nevertheless, all model parameters (e.g., $[A(t),b(t)]$ or all convolutional filters for a convolutional layer) are still instantiated during computation. It is important to note that regardless of the chosen number of particles, a shooting neural network solution is a possible optimal solution (for a given data set) at any given time, not only at convergence. One optimizes over the family of possible neural network models with the goal of finding the element within this family that best matches the observations.

\section{Automatic shooting}
\label{appendix:automatic_shooting}

The general shooting equations were presented in \eqref{EqOptimalityEquations}. We then proceeded to explicitly derive the shooting equations for a continuous DNN with linear-in-parameter layers and \UpDown\ layers in \S\ref{SecChoiceOfReg} and \S\ref{sec:updownNODE}, respectively. While this was instructive, it is somewhat cumbersome, in particular, for more complex models or when moving to convolutional networks. Fortunately, in practice these shooting equations do not need to be derived by hand. Indeed, they are completely specified by the Hamiltonian
$$
H(\p,\x,\theta) = \p^\top (\dot{\x} - f(t,\x,\theta)) + R(\theta)\enspace,
$$
in the sense that the shooting equations in \eqref{EqOptimalityEquations} are computed via differentiation of $H$. Specifically, the shooting equations in \eqref{EqOptimalityEquations} are equivalently given by
\begin{align*}
	\begin{cases}
		\dot{\x} = \frac{\partial H(\p,\x,\theta)}{\partial p}\,,  \\
		\dot{\p} = -\frac{\partial H(\p,\x,\theta)}{\partial x}\,,\\
		\theta \in \argmin_{\theta} H(\p,\x,\theta)\,.
	\end{cases}
\end{align*}
As discussed above, the last equation can be replaced by solving
$$
\partial_\theta R(\theta) - \sum_{i = 1}^N\partial_\theta f(t,\x_i,\theta)^T(\p_i) = 0\enspace.
$$
Automatic differentiation can be used to automatically obtain the shooting equations.
As fitting a shooting model requires differentiating the shooting equations, we in effect end up with differentiating twice. This can be done seamlessly using  modern deep learning libraries, such as \texttt{PyTorch}.

\section{Universality of the \UpDown\ model}
\label{appendix:updown_universal}
In this section, we set out to demonstrate that the \UpDown\ model is universal in the sense that its associated flow can come $\varepsilon$-close to the flow of any well behaving time-dependent vector field.

Recall the single-hidden-layer vector field in \eqref{Eq:SingleHiddenLayer} with time-varying parameters $\theta(t)=(\theta_1(t),\theta_2(t),b_1(t), b_2(t))$.
While shooting with the single hidden layer vector field is theoretically appealing as it is universal \cite{Cybenko1989}, it would result in implicit shooting equations.

We first show that the \UpDown\ model introduced in \ref{sec:updownNODE} can give the same flow as the single hidden layer (Lemma \ref{lem:updown_shl}) and then leverage this relationship to show that the \UpDown\ model inherits the universality of the single hidden layer (Proposition \ref{prop:updown_universal}).

 \begin{lemma}\label{ThLemma1}
 Consider the single-hidden-layer vector field in \eqref{Eq:SingleHiddenLayer} with $\theta_2(t)$ and $b_2(t)$ being piecewise $C^1$  and $\theta_1(t),b_1(t)$ continuous. Then, there exists a parameterization of the {\UpDown} model that gives the same flow at a fixed time, $T=1$.
 \end{lemma}
 \begin{proof}
 We rewrite the differential equation
 \begin{equation*}
 \dot{\q}(t) = \theta_1(t) \sigma (\theta_2(t) \q + b_2(t)) + b_1(t)\,,
 \end{equation*}
 by introducing the additional state variable $\v(t)  = \theta_2(t) \q(t)  + b_2(t)$ which we differentiate w.r.t. time.
 We obtain $\dot{\v}(t)  = \dot{\theta}_2(t) \q(t)  + \dot{b}_2(t) + \theta_2(t) \dot{\q}(t)  \,.$ Replacing $\dot{\q}(t)$ by its formula, we get
 \begin{equation*}
 \dot{\v}(t)  = \dot{\theta}_2(t) \q(t)  + \dot{b}_2(t) + \theta_2(t) \theta_1(t)( \sigma (\v(t)) + b_1(t))\,.
 \end{equation*}
 The system can be rewritten as

 \begin{equation}
 \begin{cases}\label{EqSystemUpdown}
 \dot{\q}(t)  =  \theta_1(t) \sigma (\v(t)) + b_1(t)\,,\\
 \dot{\v}(t)  = \theta_3(t) \q(t)  + \theta_4(t)\sigma (\v(t)) + b_3(t)\,.
 \end{cases}
 \end{equation}
Therefore, with the initial condition $\v(0) = \theta_2(0)\q(0)$ and $\q(0) = \q_0$, the two systems of ordinary differential equations are equivalent.
 \end{proof}

Note that the key point in Lemma \ref{ThLemma1} is the loss of regularity in the evolution of $\theta_2$ since we differentiated once in time. For that reason, we now show that adding more dimensions using the inflation factor $\inflation$ alleviate this issue.
It is likely possible that one could prove a universality result using only $\inflation = 1$ but we shall leave this question for future work\footnote{Note that the case $\inflation =1$ is similar in its formulation to a second-order model on $\q$.}. However, experimentally, the inflation factor has a crucial effect on the performance of the optimization, as discussed in \S\ref{section:experiments}. Lemma \ref{ThLemma1} helps us establish the next result.

\begin{lemma}\label{lem:updown_shl}
	Consider the single-hidden-layer vector field in \eqref{Eq:SingleHiddenLayer} with $\theta(t)$ being piecewise continuous. Then, there exists a parameterization of the \UpDown\ model that gives the same flow.
\end{lemma}

\begin{proof}
Without loss of generality, we only treat the case of one discontinuity in time of the parameterization; We thus assume that $\theta(t)$ is continuous on $[0,t_1[$ and $[t_1,1]$. We consider $\q,\v_1,\v_2 \in \R^d$ such that $\q,\v_1$ are defined as in Lemma~\ref{ThLemma1}.
We now define, up to time $t_1$, $\v_2(t) = \theta(t_1) \v_1(t) + \theta_2(t_1)$ which implies (differentiating w.r.t. time) that $\v_2$ follows an evolution equation similar to $\v_1$ and thus can be encoded in the general form of~\eqref{EqSystemUpdown}.
Now, $\q(t),\v_2(t)$ are defined on $[t_1,1]$ by the evolution \eqref{EqSystemUpdown} in order to coincide with the flow of single-hidden-layer vector field on $[t_1,1]$, $ \dot{\q}(t)  =  \theta_1(t) \sigma (\v_2(t)) + b_1(t)$ and $\dot{\v}_2(t)  = \theta_3(t) \q(t)  + \theta_4(t)\sigma (\v_2(t)) + b_3(t)$ for well chosen parameters as in  Lemma~\ref{ThLemma1}. Since the value of $\v_1(t)$ is not used in the evolution equation of $\q(t) $, we can simply extend it by $\v_1(t) = \v_1(t_1)$ which is a valid evolution equation for~\eqref{EqSystemUpdown}.
\par
In the general case, we decompose the time interval $[0,1]$ into $k$ intervals $[t_i,t_{i + 1}[$ on which $\theta(t)$ is continuous and the proposed method can be directly extended using an inflation factor $\inflation = k$, introducing $\v_k \in \R^d$.
\end{proof}
Note that the result of this lemma gives an equality between the two flows defined on the \emph{whole} space $\R^d$.
The next result is an approximation result which holds on a compact domain $C \subset \R^d$. For a function $f: \R^d \to \R$,  we denote $\| f \|_{C,\infty} = \sup_{x \in C} |f(x)|$.

\begin{proposition}\label{prop:updown_universal}
	The \UpDown\ model is universal in the class of time-dependent vector fields. Let $C \subset \R^d$ be a compact domain. For every time-dependent vector field (such that it is time continuous and is Lipschitz in space) $w: [0,1] \times \R^d \mapsto \R^d$ and its associated flow $\varphi(t,\x)$ there exist time dependent parameters of the \UpDown\ model such that

	\begin{equation*}
	\begin{cases}
	\dot{\q}(t)  = \theta_1(t)  \sigma(\v(t)) + b_1(t) \,,\\
	\dot{\v}(t)  = \theta_2(t) (\q(t) ) + b_2 + \theta_3(t)  \sigma(\v(t) )\,,
	\end{cases}
	\end{equation*}

	is $\varepsilon$-close to the solution $\varphi(1,\x)$, e.g. $\| \varphi(1,\x) - \q(1,\x) \|_{C,\infty} \leq \varepsilon$.
\end{proposition}

\begin{proof}
	The proof is standard and we include it here for self-containedness. It is the consequence of \cite{Cybenko1989} and Lemma \ref{lem:updown_shl}. Let $B(0,r)$ a ball of radius $r$ in $\R^d$ which contains $\varphi(t,\x)$ for all time $t \in [0,1]$. The flow associated with a given time-dependent vector field $v(t,\cdot)$ can be approximated by a vector field which is piecewise constant in time; i.e. let $\varepsilon >0$ be a positive real, (by continuity in time of $v(t,\cdot)$) there exists a decomposition of $[0,1]$ into $k$ intervals $[t_i,t_{i + 1}]$ and Lipschitz vector fields  $v_i(\x) = f(\x,\theta_i)$ where $f$ is the single hidden layer in \eqref{Eq:SingleHiddenLayer} such that
$\| v_i(\x) - v(t,\x) \|_{B(0,r),\infty} \leq \varepsilon$ for $t \in [t_i,t_{i + 1}]$. Denote by $w(t,\cdot)$ the time-dependent vector field defined by $w(t,\cdot) = v_i(\cdot)$  for all $t \in [t_i,t_{i + 1}]$. Thus, denoting the flow of $v(t,\cdot)$ by $\varphi^v$ and the flow of $w(t,\cdot)$ by $\varphi^w$, we get
\begin{align*}
&\|\varphi^v(1,\x) - \varphi^{w}(1,\x) \| \leq \int_0^1 \| v(t,\varphi^v(t,\x)) - v(t,\varphi^{w}(t,\x))\|  + \| v(t,\varphi^{w}(t,\x)) - w(t,\varphi^{w}(t,\x))\| \ud t \,\\
&\|\varphi^v(1,\x) - \varphi^{w}(1,\x) \|_{C,\infty}  \leq \int_0^1 \on{Lip}(v) \|\varphi^v(t,\x) - \varphi^{w}(t,\x) \|_{C,\infty} +  \| v(t,\cdot) - w(\cdot)\|_{B(0,r),\infty} \ud t\\
& \phantom{\|\varphi^v(1,\x) - \varphi^{w}(1,\x) \|_{K,\infty} } \leq  \int_0^1 \on{Lip}(v) \|\varphi^v(t,\x) - \varphi^{w}(t,\x) \|_{C,\infty} \ud t + \varepsilon\,,
\end{align*}
where $\on{Lip}(v)$ denotes a bound on the Lipschitz constant of $v(t,\x)$ w.r.t. $\x \in B(0,r)$ for all $t\in [0,1]$.
Then, the Gr\"onwall lemma \cite{laurentbook} gives
\begin{equation}
\|\varphi^v(1,\x) - \varphi^{w}(1,\x) \|_{C,\infty} \leq \varepsilon e^{\on{Lip}(v)}\,.
\end{equation}
By Lemma~\ref{lem:updown_shl}, $\varphi^{w}(1,\x)$ can be approximated by the flow of the \UpDown\ and the result is obtained via the triangle inequality.
\end{proof}

In this section, we focused on a universality result in the space of time-dependent vector fields. Interestingly, due to the additional dimensions, it is likely that the model is universal in the space of functions as well. This conjecture is supported by the quadratic 1D function regression example which shows that the \UpDown\ model is able to capture some maps which are not homeomorphic.
We leave this question for future work.

\section{Experimental settings}
\label{appendix:experimental_settings}

This section describes our experimental settings. We use our {\UpDown} model for all experiments and simply use a weighted Frobenius norm penalty for all parameters. Specifically, we weigh this penalty for all parameters with $1$ except for, $\theta_3$ which we penalize by 10. In our experiments, we have observed better convergence properties for higher penalties on $\theta_3$. This might be due to the special role that $\theta_3$ plays in the model as it subsumes a quadratic term in the original derivation of the {\UpDown} model (see \S\ref{sec:updownNODE}). In all experiments, we also optimize over the affine map from $x(0)$ to $v(0)$ for the data evolution.

\textbf{Simple function regression.} We use 500 epochs for all experiments. For all particle-based experiments we freeze the positions of the particles for the first 50 epochs. We use a {\ReLU} activation function and the MSE loss. We weigh the MSE loss by 100 and the parameter norm loss by 1. We use 500 training samples, 1,000 testing samples and 1,000 validation samples and a batch size of 50. Note that for these simple examples there is, in practice, no real difference between the training, testing, and validation data, as the number of samples is large and the domain is $[-1.5,1.5]$. We initialize the particle positions uniformly at random in $[-1.5,1.5]$ and draw the momenta from a Gaussian distribution with zero mean and standard deviation $0.1$. All time-integrations are done via a fourth-order Runge-Kutta integrator with time-step 0.1. For optimization, we use {\Adam} with a learning rate of 0.01 and the \texttt{ReduceLROnPlateau} learning rate scheduler of {\PyTorch} with a learning rate reduction factor of 0.5.

\textbf{Spiral.} The spiral data is generated between time $t=0$ and $t=10$ with 200 uniformly spaced timepoints. Training is only on small time snippets with an approximate length of $0.25$ time-units. Evaluation is on these short time snippets as well as on the entire trajectory by pasting together solutions for these short time snippets, i.e., an individual short solution starts where the previous one ends. Settings for the spiral are the same as for the simple function regression with the following exceptions. We use 1,500 epochs and the step-size for the fourth-order Runge-Kutta integrator is 0.05. The MSE loss is still weighted by 100, but the parameter norm loss only by 0.01. We randomly draw 100 new training samples during each epoch and use 100 evaluation samples and 1,000 short range samples and 1 long-range testing sample. All samples are randomly drawn from the trajectory. However, as the trajectory is traversed at highly nonuniform speed the samples are drawn from a uniform distribution across the trace of the spiral. As for the simple function regression experiment, there is little practical difference between the training, validation, and testing data as the problem is so simple. However, this is not of concern in these experiments as the prime objective is to study the fitting behavior of the different models. We use a batch size of 100.

\textbf{Rotating MNIST.} We use the data provided by the authors of \cite{Yildiz19a} and follow the same autoencoder architecture, except that our encoder maps into a 20-dimensional
representation space. The
number of particles is set to 100 and the inflation factor $\alpha$ is set to 10. For optimization, we use
{\Adam} with a learning rate of 0.001 and the \texttt{CosineAnnealingLR} learning rate scheduler of {\PyTorch}. We train for 500 epochs with a batch size of 25 and the parameter norm loss set to 0.1.

\textbf{Bouncing balls.} As in the rotating MNIST experiment, we rely on the data provided by the authors of \cite{Yildiz19a},
follow their autoencoder architecture and set the dimensionality
of the representation space of the encoder to 50. The first three images of each sequence are provided to the encoder by concatenating the images along the channel dimension. The inflation factor $\alpha$ is set to 20 and we use 100 particles.
We optimize over 100 epoch using {\Adam} with the \texttt{CosineAnnealingLR} learning rate scheduler of {\PyTorch}, the initial learning rate is set to 0.001 and the parameter norm loss is set to 0.0001.

\section{Additional results}
\label{appendix:figures_results}

\input{cubic_quadratic_and_spiral}

config: arxiv


text:
\section{Discussion and Future Works}
In this paper, we present a novel algorithm for the identification of individuals that have high probability to spread the infection during the SARS-CoV-2 outbreak.

We show how to use the contact tracing apps and the data stored therein to provide with a suggestion on the individuals to test for the positivity to the virus.

This is crucial task for the containment of the contagion, due to the limited number of available tests the public health organizations can perform each day.

Our algorithm suggests the set of individuals that should be tested to limit the spread by an innovative AI-based algorithm which communicates in a decentralized fashion only by means of token generated by the different devices, thus preserving the private information of the app users.

As immediate future work, we will evaluate our algorithm in realistic scenarios comparing it with testing policies currently employed by governments.

However, simulating realistic scenarios is a hard task since the actual values of some parameters of our epidemic model are still unknown to the scientific community, such as the probability of being asymptomatic, the contagion probability given a contact.

For this reason, in addition to resort to new works in the scientific literature, we will design methods to estimate them by combining aggregated data about this epidemics.

Furthermore, another interesting improvement will be the development of a framework in which these parameters can be estimated online gathering information trough the tests.


TITLE: Prove that $W_1 + W_2$ is the smallest subspace of $V$ containing both $W_1$and $W_2$.
QUESTION [2 upvotes]: So I've been looking at the proof for this, and there's one thing I don't understand:

Let $U$ be some subspace of $V$, we need to show if $W_1 \subseteq U$,
  and if $W_2 \subseteq U$. Then $W_1 + W_2 \subseteq U$.

Thing is, I don't see WHY we need to show $W_1 + W_2 \subseteq U$. That is, I don't understand how showing that proves $W_1 + W_2$ is the smallest subspace of $V$ containing $W_1$and $W_2$?

REPLY [3 votes]: Consider any subspace D of V containing $W_1$ and $W_2$. Then elements of the subspaces $W_1$ and $W_2$ are contained in D and therefore any linear combination of elements of $W_1$ and $W_2$ is contained in D since D is closed under addition and scalar multiplication. Therefore any v in $L(W_1 \cup W_2)$ is contained in D which implies that $W_1+W_2=L(W_1 \cup W_2)$ is contained in $D$.
This means that if any subspace of $U$ contains $W_1$ and $W_2$ then it must contain $W_1+W_2$ which implies that $W_1+W_2$ is the smallest subspace containing $W_1$ and $W_2$.


text:
\begin{document}

 \title{Defining Curvature as a Measure via Gauss-Bonnet on Certain Singular Surfaces}

\author[  ]{Robert S Strichartz}

\affil[  ]{Mathematics Department, Cornell University, Ithaca, NY 14853, str@math.cornell.edu}

\maketitle

\begin{abstract}
 We show how to define curvature as a measure using the Gauss-Bonnet Theorem on a family of singular surfaces obtained by gluing together smooth surfaces along boundary curves. We find an explicit formula for the curvature measure as a sum of three types of measures: absolutely continuous measures, measures supported on singular curves, and discrete measures supported on singular points. We discuss the spectral asymptotics of the Laplacian on these surfaces.

 \end{abstract}

\section{Introduction}

 Curvature on a surface $M$ with Riemannian metric $g$ is traditionally defined as a pointwise function $K_g(x)$ on $M$, or the interior of $M$ if $M$ has a boundary $\partial M$. However, the classical Gauss-Bonnet Theorem says
 \begin{equation}
 \int_N K_g dA + \int_{\partial N} \kappa ds = 2 \pi \chi (N)
 \end{equation}
 for any smooth subsurface $N$ of $M$, where $dA$ is the Riemannian area element, $ds$ is the arclength element on $\partial N$, $\kappa$ is the one-dimensional curvature function on $\partial N$, and $\chi(N)$ is the Euler characteristic. We see here that the absolutely continuous signed measure
 \begin{equation}
 \mu_{K} (N) = \int_N K_g dA
 \end{equation}
 is the key ingredient. We can always obtain the pointwise curvature $K_g(x)$ from the measure curvature $\mu_{K}$, so we really don't lose any information by identifying curvature as a measure. We call this the measure-centered viewpoint. Notice that the Gauss-Bonnet Theorem also adopts this viewpoint for the curvature of the boundary $\partial N$.

 This is philosophy, not mathematics. The point of this paper is that this philosophy leads to some interesting mathematics. Specifically, we would like to extend the definition of curvature to more general spaces than smooth Riemannian surfaces so that an analog of Gauss-Bonnet continuous to hold. We can already point to a simple example that is well-known. Let $M$ be a convex polyhedron, with vertices $V$. On $M \backslash V$, we have a flat Riemannian metric, so we only see curvature at the vertices. It is natural to think of the curvature of $M$ as a measure supported on $V$, a sum of delta masses multiplied by the angle defect $2 \pi - \sum \alpha_j,$ where $\alpha_j$ are the angles of the faces that come together at $v \in V$. (In the case of a polyhedron that is  not convex, the same formula is valid and yields a signed measure.) And the analog of Gauss-Bonnet is simply
 \begin{equation}
 \chi(M) = \sum_{v \in V}( 2 \pi - \sum \alpha_j).
 \end{equation}

 In this paper we concentrate on surfaces with $1$-dimensional singularities obtained by gluing together smooth surfaces with smooth boundaries along the boundaries. In the simplest example we take $M$ to be $M_1 \sqcup M_2$ factored by an identification map $\varphi: \partial M_1 \to \partial M_2$ that is one-to-one, onto, and smooth.

 \begin{theorem}
 It is possible to define a curvature measure on $M$ using a version of the Gauss-Bonnet Theorem such that
 \begin{equation} \label{eq:1.4}
 \mu_{ K }(M) = \mu_{K}(M_1) + \mu_{ K }(M_2) + \nu_1 + \nu_2
 \end{equation}
 where $\nu_1$ and $\nu_2$ are supported on the identified boundary, with
 \begin{equation}
 \nu_j = \kappa ds \text{ on } \partial M_j \text{ for } j=1,2.
 \end{equation}
 \end{theorem}

 We give the proof of the Theorem in section 2, along with some straightforward generalizations. In section 3, we discuss a broader class of examples in which the curvature measure is allowed to have both one-dimensional and discrete parts. (A work in progress \cite{DPS} describes an example of a convex surface whose curvature measure is a fractal measure.) In section 4 we discuss the spectral asymptotics of a natural Laplacian on these surfaces.

 According to the standard philosophy of mathematics, this paper seems to be putting the cart before the horse. After all, shouldn't we try to describe a category of objects that might be called $\textit{singular surfaces}$, and then give a definition of curvature measures on all singular surfaces, proving a version of Gauss-Bonnet? I agree that this would be far better than what I am presenting here. But since I don't know how to do this, it will have to wait for the future. And don't forget Hamlet's famous comment to Horatio!

 There are, of course, many other theories of objects that might be called singular surfaces; for example, Alexandrov spaces \cite{Sh}. Some, but not all of our surfaces are Alexandrov spaces, and the kind of curvature associated with Alexandrov spaces is qualitative, not quantitative.

 An entirely different approach to understanding the curvature measure on the identified boundary is to examine the asymptotics of the area of $D_r(x)$ as $r \to 0$ for points $x$ on the boundary, where $D_r(x)$ is the disc of radius $r$ centered at $x$ in $M$. Clearly $D_r(x) = D^1_r(x_1) \cup D^2_r(x_2)$ where $D^j_r(x_j)$ is the disc of radius $r$ centered at $x_j$ in $M_j$, and $x_j$ are the boundary points in $M_j$ identified at $x$. But we know
 \begin{equation}
 \text{Area } D^j_r(x_j) = \frac{ \pi }{2} r^2 + \frac{1}{3} \kappa(x_j) r^3 + O(r^4) .
 \end{equation}
 If we agree that the curvature measure along the identified boundary should be $\rho (x)ds$ where $\rho (x)$ satisfies
 \begin{equation}
  \text{Area }D_r(x) =  \pi r^2 + \frac{1}{3} \kappa(x_j) r^3 + O(r^4)
\end{equation}
then we obtain (\ref{eq:1.4}). See p. 35 of \cite{H} for an example of this approach.

 \section{ Proof of Theorem}
\begin{proof}
 We assume that $M_j$ are $C^2$ surfaces with $C^2$ boundaries, with Riemannian metrics $g_j$ that are $C^2$, and the gluing identification map $\varphi: \partial M_1 \to \partial M_2$ is also $C^2$.

To simplify the exposition we first assume that the surfaces $(M_j, g_j)$ are flat. Let $x$ and $y$ denote points on the identified boundaries, so $x_j$ and $y_j$ are the corresponding points on $\partial M_j$, with $x_2 = \varphi(x_1)$ and $y_2 = \varphi (y_1)$. We assume that $x$ and $y$ are sufficiently close that the construction in Figure 2.1 is possible.

\begin{figure}[h!]
\begin{tikzpicture}

\draw (0,7) .. controls (5,6) and (2,1) .. (5,0.5);
\draw  (0.5,4.6) -- (2.15,6);
\draw  (1,2.3) -- (3.45,2.4);
\draw (0.5,4.6) --(1,2.3);

\draw  (2.15,5.5) -- (2.4,5.7);
\draw (2.15,5.5) -- (1.89,5.75);

\draw  (2.95,2.76) -- (3.4,2.8);
\draw  (2.95,2.76) -- (3,2.4);

\draw (0.55,4.3) arc (-80:35:3.5mm);

\draw (1.3,2.3) arc (0:95:3.5mm);

\draw (0.3,7.3) node
{ $\partial M_1$ };

\draw (-0.5,5) node
{ $ M_1$ };
\draw (3.5,4.3) node
{ $ A_1$ };
\draw (2.2,3.8) node
{ $ N_{1}$ };

\draw (2.6,6.2) node
{ $ x_1$ };
\draw (3.9,2.35) node
{ $ y_1$ };

\draw (1.2,5.7) node
{ $ L_{11}$ };
\draw (2.2,2) node
{ $ L_{12}$ };
\draw (0.4,3.5) node
{ $ L_{1}$ };

\draw (1.2,4.5) node
{ $ a_{11}$ };
\draw (1.6,2.7) node
{ $ a_{12}$ };

\draw (9,7) .. controls (8,6) and (10,5)   .. (10,4);
\draw (10,4) .. controls (10,3) and (9,2)   .. (12,1);

\draw  (8.9,5.8) -- (10.7,7.0);
\draw  (9.85,2.6) -- (11.6,2.8);
\draw (10.7,7.0) -- (11.6,2.8);

\draw  (9.1,5.5) -- (9.4, 5.68);
\draw (9.42, 5.68) --(9.22, 5.98);

\draw  (9.85,2.88) --(10.1, 2.93);
\draw (10.1, 2.93) -- (10.15, 2.63);

\draw (10.3,6.75) arc (-155:-58:3.5mm);
\draw (11.55,3.15) arc (90:185:3.5mm);

\draw (12.5,5.5) node
{ $ M_{2}$ };
\draw (9,7.4) node
{ $\partial M_2$ };
\draw (10.5,5) node
{ $ N_{2}$ };

\draw (9.6,6.8) node
{ $ L_{21}$ };
\draw (11.5,5) node
{ $ L_{2}$ };
\draw (11,2.3) node
{ $ L_{22}$ };

\draw (8.6,5.7)  node
{ $ x_{2}$ };
\draw (9.5,2.7)  node
{ $ y_{2}$ };

\draw (9.6,4.1)  node
{ $ A_{2}$ };

\draw (10.45,6.3) node
{ $a_{21}$};
\draw (11.1,3.3) node
{ $a_{22}$};

\end{tikzpicture}

\caption{Figure 2.1}
\label{Figure 2.1}
\end{figure}

Here $L_{11}$ and $L_{12}$ are line segments in $M_1$ perpendicular to $\partial M_1$ at $x_1$ and $y_1$, and similarly $L_{21}$ and $L_{22}$ are line segments perpendicular to $\partial M_2$ at $x_2$ and $y_2$. Then we may interpret the unions $L_{11} \cup L_{21}$ and $L_{12} \cup L_{22}$ as geodesic line segments in $M$. It is not significant how long we take the perpendicular line segments, as long as the line segment $L_1$ connecting the endpoints of the perpendiculars $L_{11}$ and $L_{12}$ stay within $M_1$, and similarly for $L_2$ in $M_2$. We denote the arc joining $x$ and $y$ in the identified boundary by $A$, and let $A_1$ and $A_2$ denote the corresponding arcs in $\partial M_1$ and $\partial M_2$.

Let $Q$ denote the quadrilateral in $M$ whose sides are the straight line segments $L_1$, $L_{12} \cup L_{22}$, $L_2$ and $L_{11} \cup L_{22}$, enclosing the region $N= N_1 \cup N_2$. For the Gauss-Bonnet Theorem to hold for $N$ in $M$, we need to have

\begin{equation}
\mu_{K}(N) = a_{11} + a_{12} + a_{21} +a_{22} - 2 \pi . \label{eq:2.1}
\end{equation}

Note that the individual angles depend on the choices of the lengths of the perpendicular segments, but elementary geometry shows that $a_{11} + a_{12}$ and $a_{21}+a_{22}$ are independent of these choices. Thus, we may take  \eqref{eq:2.1} as the definition of $\mu_{K} (N)$. Because $M_1$ and $M_2$ are flat, the support of the measure $\mu_{K}$ must be the identified boundary. But now we can apply Gauss-Bonnet to the regions $N_j$ in $M_j$ to obtain

\begin{equation} \label{eq:2.2}
\begin{split}
a_{11} + a_{12} - \pi & = \nu_1(A) \\
a_{21}+ a_{22} - \pi & = \nu_2(A)
\end{split}
\end{equation}

 Adding these and comparing with \eqref{eq:2.1} we obtain \eqref{eq:1.4} in the flat case.

 In the case of curved surfaces we again use the construction in Figure $2.1$, where the segments $L_j$ and $L_{jk}$ are geodesics. Then in place of \eqref{eq:2.2} we have
 \begin{equation} \label{eq:2.3}
\begin{split}
a_{11} + a_{12} - \pi & = \nu_1(A)  + \mu_{K}(N_1)\\
a_{21}+ a_{22} - \pi & = \nu_2(A) + \mu_{K}(N_2)
\end{split}
\end{equation}
  and this again leads to \eqref{eq:1.4}.

\end{proof}

We note that, although the definition \eqref{eq:2.1} gives the Gauss-Bonnet Theorem for one type of domain in $M$, if we combine it with the Gauss-Bonnet Theorem on domains in $M_1$ and $M_2$ we obtain the result on all domains in $M$.

It is straightforward to extend the theorem to the setting where $M$ is obtained from a finite collection of surfaces $ \{ M_j \}_{j \in J}$ glued together along components of the boundaries $\{ \partial M_j \}$. In fact we could allow gluing of more than two boundaries together. The resulting object would no longer be a surface, of course.

The construction leads to measures on the identified boundaries that are absolutely continuous with respect to arclength measure on the boundaries of $M_j$ (because of the assumptions on the identification map the arclength measures on $\partial M_1$ and $\partial M_2$ are mutually absolutely continuous). We can also identify certain subarcs on the boundary where the measure is positive, if both $\partial M_1$ and $\partial M_2$ are positive on the subarc, and similarly for negativity.

  \section{Examples with point singularities}

  In this section we discuss briefly some examples of surfaces with both line and point singularities, again obtained by gluing. We will simplify the discussion to look at examples with just a single point singularity at the intersection of several line singularities, but of course everything is local so with some more gluing we could handle a finite number of point singularities.

  The set-up is shown in Figure 3.1.

\begin{figure}[h!]
\centering
\begin{tikzpicture}
\draw (5,7) .. controls (4,6) and (7,5.8)   .. (6,4);
\draw (5,7.5) node { $L_{41}$};

\draw (6,4) .. controls (5,3) and (4,1)   .. (6,1);
\draw  (6,0.5) node { $L_{23}$};

\draw (3,4) .. controls (3.2,5) and (4.1, 5) .. (5, 4);
\draw (5,4) .. controls (5.4,3.6) and (5.7, 4.7) .. (6, 4);
\draw  (2.5,4) node { $L_{34}$};

\draw (6,4)--(6.4, 4);
\draw  (6.4,4) .. controls (6.8,4) and (7.5, 3) .. (8.5, 4);
\draw  (9, 4) node { $L_{12}$};

\draw (6.4,4) arc (0:76: 3.5mm);
\draw  (6.6,4.3) node { $\theta_1$};
\draw (7.6,5.3) node { $M_1$};

\draw (5.4,4) arc (180:60: 5mm);
\draw (5.6,4.7)  node{ $\theta_4$};
\draw (4.7,5.4)  node{ $M_4$};

\draw (5.45,4.03) arc (180:270: 3.5mm);
\draw (5.35,3.65) node { $\theta_3$};
\draw (4.35,3.1) node { $M_3$};

\draw (6.3,4) arc (0:-122: 3.5mm);
\draw (6.3,3.5) node { $\theta_2$};
\draw (6.8,2.5) node { $M_2$};
\end{tikzpicture}

\caption{Figure 3.1}
\label{Figure 3.1}
\end{figure}

  The curves $L_{j (j+1)}$ are identified pieces of the boundaries of $M_j$ and $M_{j+1}$. The angles $\theta_j$ are between the boundary pieces in $M_j$ of $L_{(j-1)j}$ and $L_{j(j+1)}$. If we delete a small neighborhood of the intersection point, then the method in section 2 shows that the curvature measure has the absolutely continuous part supported in the interior of $M_j$ and the 1-dimensional parts supported on the identified curves $L_{j(j+1)}$ so the only issue is what happens in the neighborhood of the intersection point.

  To be more specific, suppose we have smooth manifolds $M_1, M_2, \ldots M_n$ with boundaries that are smooth except for one corner singularity $y_j$ with angle $\theta_j$ (in Figure $3.1$ we have $n=4$). Let $\gamma_{j1}$ and $\gamma_{j2}$ denote the portions of $\partial M_j$ on either side of $y_j$. Suppose we are given $C^2$ identification maps $\varphi_j : \gamma_{(j-1)2} \to \gamma_{j1}$ with $\varphi_j(y_{j-1}) = y_j$. Then we let $M$ be the union of the $M_j$ modulo the identifications. Note that all the corner points $y_j$ are identified in $M$ in the single intersection point we call $y$. The curves $L_{(j-1)j}$ in Figure 3.1 are the identified boundary pieces $\gamma_{(j-1)2}$ and $\gamma_{j1}$ (of course we use cyclic notation in the variable $j$).

  In Figure 3.2 we zoom in on one of the surfaces $M_j$ in Figure 3.1 and perform a construction analogous to that in Figure 2.1. We choose points $x_{j1}$ on $\gamma_{j1}$ and $x_{j2}$ on $\gamma_{j2}$, and we require that $x_{(j-1)2}$ is identified with $x_{j1}$.

  \begin{figure}[h!]
\centering
\begin{tikzpicture}
\draw (0.5,2) .. controls (2,4.2) and (4,4.8)   .. (6,5);
\draw (1.5,3.75) node { $ \gamma_{j1}$};
\draw (0.5,2) .. controls (2,1) and (4,0.5)   .. (5.6,0.5);
\draw (1.5,1.1) node { $ \gamma_{j2}$};

\draw (0.15,2) node { $y_i$ };
\draw (5,3) node {$M_j$};
\draw (2.2,2.8) node {$N_j$};

\draw ( 3,0.9) -- (3.3, 2);
\draw ( 4,3.1) -- (3.6, 4.5);
\draw (3.3, 2) --( 4,3.1);
\draw (3.5, 2.3) arc (50:260: 3mm);
\draw (2.8, 2.2) node { $b_j$ };
\draw ( 3.95,3.35) arc (100:230: 3mm);
\draw ( 3.5,3.2) node { $a_j$ };

\draw ( 2.68,0.95)-- (2.78, 1.25);
\draw (2.78, 1.25) --(3.08, 1.16);
\draw (3, 0.6) node { $x_{j2} $};

\draw (3.7, 4.2) -- (3.4, 4.1);
\draw (3.3, 4.4) -- (3.4, 4.1);
\draw (3.75, 4.75) node { $x_{j1} $};

\draw (0.85,1.8) arc (-40:60: 3.5mm);
\draw (1.2,2) node { $\theta_j$ };

\end{tikzpicture}

\caption{Figure 3.2}
\label{Figure 3.2}
\end{figure}

We also require that the distances of the points $x_{j1}$ and $x_{j2}$ to $y_j$ are bounded above by $\epsilon$, where $\epsilon$ is a parameter that will eventually approach zero. We take geodesic segments perpendicular to $\gamma_{j1}$ at $x_{j1}$ (and similarly for $\gamma_{j2}$ and $x_{j2}$), and connect them up by another geodesic segment at angles $a_j$ and $b_j$. When then obtain a pentagon in $M_j$ with three geodesic sides and the curved sides that are segments of $\gamma_{j1}$ and $\gamma_{j2}$, and let $N_j$ be the subset of $M_j$ bounded by this pentagon.

Let $N$ be the union of the $N_j$ with the identification of the $\gamma_{(j-1)2}$ and $\gamma_{j1}$ sides. This will be a neighborhood of $y$ in $M$. We note that perpendicular geodesic segments at $x_{j1}$ in $M_j$ and $x_{(j-1)2}$ in $M_{j-1}$ combine to form a single geodesic segment in $M$. Thus, $N$ is a $2n$-gon in $M$ bounded by geodesic segments with interior angles $\{ a_j \}$ and $\{ b_j \}$. If we are to have Gauss-Bonnet hold on $M$ we need to define

\begin{equation}
\mu_{K}(N) = \sum_{j=1}^n(a_j +b_j) -2(n-1) \pi
\end{equation}

On the other hand, from Gauss-Bonnet on $N_j$ in $M_j$ we have

\begin{equation}
\mu_{K}(N_j) + \int_{y_j}^{x_{j1}} \kappa (\gamma_{j1}) ds +  \int_{y_j}^{x_{j2}} \kappa (\gamma_{j2}) ds +a_j+b_j+ \theta_j -2 \pi
\end{equation}

We may compute $\mu_{\kappa}( \{ y \} )$ by taking the limit as $\epsilon \to 0$ of $\mu_{K}(N) - \sum_{j} \mu_{\kappa}(N_j)$. Clearly the limits of the curvature integrals over the shrinking line segments will vanish, so

\begin{equation}
\begin{split}
\mu_{K}( \{ y \} & = \lim_{ \epsilon \to 0} \sum_{j=1}^n (a_j + b_j) - 2(n-1) \pi - \sum_{j=1}^n (a_j +b_j+ \theta_j - 2 \pi) \\
& = ( \sum_{j=1}^n \theta_j) -2 \pi.
\end{split}
\end{equation}

\begin{corollary}
In this setting we have
\begin{equation}
\mu_{K}(M) = \sum_{j=1}^n \mu_{K} (M_j) + \sum_{j=1}^n ( \nu_{j1} + \nu_{j2}) + (( \sum_{j=1}^n \theta_j) -2 \pi) \delta_y
\end{equation}
where $\nu_{j1}$ and $\nu_{j2}$ are the curvature measures along $\gamma_{j1}$ and $\gamma_{j2}$ as in the Theorem.

\end{corollary}

 \section{Spectrum of the Laplacian}
 It is possible to define a Laplacian on $M$, via the weak formulation $\Delta u =f$ means $u \in dom \mathcal{E} $, $f \in L^2(M)$ and
 \begin{equation}
 \mathcal{E} (u,v) = - \int_{M} f v dA \text{ for all } v \in dom \mathcal{E}
 \end{equation}

 Here the area measure $dA$ is just the sum of the area measures on $M_j$ and the energy form is the sum of energy forms on $M_j$
 \begin{equation}
 \mathcal{E} (u,v) = \sum_j \int_{M_j} \nabla u \cdot \nabla v dA
 \end{equation}
 However, the domain $dom( \mathcal{E})$ of the energy form requires a more careful explanation. On a surface, the finiteness of the energy $\mathcal{E}(u,u)$ does not imply that $u$ is continuous, but the discontinuities cannot be too pervasive. Thus the finiteness of $\int_{M_j} | \nabla u|^2 dA$ implies that there is a well defined trace of $u$ on $\partial M_j$ that belongs to the Sobolev space $H^{1/2}( \partial M_j)$. So $dom( \mathcal{E} )$ is defined to be the functions with $\mathcal{E} (u,u)$ finite and whose traces are equal on the identified boundaries. By considering test functions $v$ supported in the interior of $M_j$, it follows that $\Delta u =f$ means that the pointwise formula holds in the interior of $M_j$. It is the continuity condition on the identified boundaries that requires the above careful formulation.

 Assume $M$ is compact. Then the Laplacian is a negative definite self-adjoint operator with compact resolvent. The spectrum consists of a discrete sequence
 \begin{equation}
 \lambda_1 \le  \lambda_2 \le   \lambda_3 \le \ldots \to \infty
 \end{equation}
 of eigenvalues of $- \Delta$ with eigenfunctions
 \begin{equation}
 - \Delta u_j = \lambda_j u_j
 \end{equation}
 with $\{ u_j \}$ giving an orthonormal basis of $L^2(M)$ (note that $\lambda_1 =0$ and $u_1$ is constant). The famous paper of Mark Kac \cite{K} discusses the relationships between the spectrum and the geometry of $M$ in the case of smooth manifolds. It is thus a fundamental problem to examine similar relationships for the surfaces described in this paper.

 We consider here the special case of the ``double" of a smooth manifold $M_1$ with boundary $\partial M_1$. That is, we take $M_2$ isometric to $M_1$ and take essentially the identity map to glue the boundaries. Note that we have a natural symmetry $\sigma$ of $M$ that interchanges the isometric points of $M_1$ and $M_2$, and $\sigma$ is the identity on the identified boundaries.

 \begin{lemma}
 The spectrum of $M$ is simply the union of the Dirichlet and Neumann spectra of $M_1$. A Dirichlet eigenfunction on $M_1$ yields an eigenfunction on $M$ by odd reflection under $\sigma$. Similarly, a Neumann eigenfunction on $M_1$ yields an eigenfunction on $M$ by even reflection under $\sigma$. And there are no other eigenfunction on $M$.
 \end{lemma}

 \begin{proof}
 Since the symmetry $\sigma$ commutes with the Laplacian, we can break up the eigenspaces on $M$ into odd and even functions under the symmetry $\sigma$. The odd functions vanish on $\partial M_1$ so they correspond to Dirichlet eigenfunctions on $M_1$. Similarly, the even functions have vanishing normal derivative on $\partial M_1$, so they correspond to Neumann eigenfunctions on $M_1$.
 \end{proof}

 Consider the eigenvalue counting function $N_{M}(t) = \# \{ \lambda_j \le t \}$ on $M$, and similarly the Dirichlet and Neumann eigenvalue counting functions $N_M(t)$ and $N_N (t)$ on $M_1$. The lemma says $N_M(t) = N_D(t) +N_N(t)$.

 Now the Weyl-Ivrii asymptotics \cite{I} on $M_1$ say that

 \begin{equation}
 \begin{split}
 N_D(t) = & \frac{ \text{Area}(M_1)}{4 \pi} t - \frac{ \text{Length}( \partial M_1)}{4 \pi} t^{1/2} + o(t^{1/2}) \\
  N_N(t) = & \frac{ \text{Area}(M_1)}{4 \pi} t + \frac{ \text{Length}( \partial M_1)}{4 \pi} t^{1/2} + o(t^{1/2}),
  \end{split}
  \end{equation}
so
\begin{equation}
N_M(t) = \frac{ \text{Area}(M)}{4 \pi} t+ o(t^{1/2})
\end{equation}

According to the conjectures in \cite{St} and \cite{MS} we should do better in the case that $M_1$ has constant curvature if we average the errors. Let
\begin{equation}
\tilde{N}(t) =  \frac{ \text{Area}(M)}{4 \pi} t + \mu_{K}(M),
\end{equation}
  and
 \begin{equation}
 A(t) = \frac{1}{t} \int_0^t (N(s) - \tilde{N}(s)) ds
 \end{equation}

 \begin{conjecture}
 We have the average error estimate
 \begin{equation}
 A(t) = O (t^{-1/4}).
 \end{equation}
 \end{conjecture}

 There are some simple examples of $M_1$ for which the conjecture is known to be true ( rectangle, equilateral triangle, isosceles right triangle, see \cite{JS} \cite{St}) because we can compute $N_D(t)$ and $N_M(t)$ exactly. On the other hand, if $M$ is a flat disc, the conjecture is open, even though the eigenvalues are given explicitly in terms of zeros of Bessel functions and derivatives of Bessel functions.

 \bibliographystyle{annotate}


TITLE: Is there a topology which is coarser than the product topology?
QUESTION [0 upvotes]: Is there a topology which is coarser than the product topology on an infinite Cartesian product of topological spaces?

REPLY [6 votes]: Sure, the coarsest topology, i.e., the trivial topology.
Note that the product topology is the coarsest topology for which the projection maps are continuous. This is very important, and the reason why we care a lot more for this topology than the trivial one!


TITLE: Connected Subset of Finite Topological Space
QUESTION [3 upvotes]: What I want to know is this:
Am I correct in thinking that if we have a finite topological space (so no reals or anything here), that any connected subset contained in this space has only one element? For example, {x} is connected but {x,y} is disconnected.

REPLY [1 votes]: No; as a simple case, if any space $X$ has the indiscrete topology (only the empty set and $X$ are open), then any subset of $X$ is connected.
In more detail: Suppose $V \subset X$.  Then the relative topology on $V$ is also the indiscrete topology.  Thus $V$ has no open subsets other than the empty set and $V$, and is therefore connected.


TITLE: Time evolution of an electron in an homogeneous magnetic field: get rid of $S_y$ in the exponential
QUESTION [1 upvotes]: Context
We have a particle of spin $1/2$ and of magnetic moment $\vec{M} = \gamma\vec{S}$.
At time $t=0$, the state of the system is
$$
|\psi(t=0) \rangle = |+\rangle_z
$$
We let the system evolve under the influence of a magnetic field $B_0$ parallel to $Oy$.
We are asked to find the state of the system at time $t$.
Attempt
The energy of interaction between the $B$ field and the spin is
$$\hat{H} = E_m \equiv -\vec{\mu} \cdot \vec{B} = - \gamma \vec{S} \cdot B_0 \hat{y} = - \gamma B_0 y S_y = \omega_0 S_y$$
where $\gamma$ denotes the gyromagnetic ratio and $S_y$ is the spin operator in the $y$ direction.
We can write the time-dependance of $\psi$ by applying the time-evolution operator on $|\psi(t=0)\rangle$,
$$
\begin{align}
  |\psi(t)\rangle
    &= e^{-iHt/\hbar}|\psi(t=0)\rangle \\
    &= e^{-i \omega_0 S_y t/\hbar}|+\rangle_z
\end{align}
$$
Now I would like to express $|+\rangle_z$ in the $\{|+\rangle_y, |-\rangle_y\}$ basis.
We can express $S_y$ in term of the Pauli matrix $\sigma_y$,
$$
S_y = \frac{\hbar}{2} \underbrace{\begin{pmatrix}0 & -i \\ i & 0\end{pmatrix}}_{\sigma_y}
$$
Finding the eigenvalues of $S_y$,
$$
det(S_y - \lambda I) = 0 \implies \lambda_{1,2} = \pm \frac{\hbar}{2}
$$
Finding the eigenvectors,
$$
(S_y - \lambda_{1,2}I)(\chi) = \vec{0}
$$

$\lambda_1 = \hbar/2$
$$
\frac{\hbar}{2}
\begin{pmatrix}-1 & -i \\ i & -1\end{pmatrix} \begin{pmatrix}\chi_{_+} \\ \chi_{_-}\end{pmatrix}
=
\begin{pmatrix}0 \\ 0\end{pmatrix}
\implies
\begin{cases}
-\chi_{_+} - i\chi_{_-} = 0 \\ i\chi_{_+} - \chi_{_{-}} = 0
\end{cases}
$$

We find the normalised eigenvector
$$
\chi_{_+} = \frac{1}{\sqrt{2}} \begin{pmatrix}1 \\ i\end{pmatrix}
$$
We can write
$$
|+\rangle_y = \frac{1}{\sqrt{2}}\left(|+\rangle_z + i|-\rangle_z \right)
$$

$\lambda_1 = -\hbar/2$
$$
\frac{\hbar}{2}
\begin{pmatrix}1 & -i \\ i & 1\end{pmatrix} \begin{pmatrix}\chi_{_+} \\ \chi_{_-}\end{pmatrix}
=
\begin{pmatrix}0 \\ 0\end{pmatrix}
\implies
\begin{cases}
\chi_{_+} - i\chi_{_-} = 0 \\ i\chi_{_+} + \chi_{_{-}} = 0
\end{cases}
$$

We find the normalised eigenvector
$$
\chi_{_-} = \frac{1}{\sqrt{2}} \begin{pmatrix}1 \\ -i\end{pmatrix}
$$
We can write
\begin{equation}
|-\rangle_y = \frac{1}{\sqrt{2}}\left(|+\rangle_z - i|-\rangle_z \right)
\end{equation}
Inverting the $|+\rangle_y$ and $|-\rangle_y$ relations we get
$$
\begin{cases}
|+\rangle_z = \frac{1}{\sqrt{2}}\left(|+\rangle_y + i|-\rangle_y \right) \\
|-\rangle_z = \frac{1}{\sqrt{2}}\left(|+\rangle_y - i|-\rangle_y \right)
\end{cases}
$$
Therefore,
\begin{align}
|\psi(t)\rangle
    &= e^{-i \omega_0 S_y t/\hbar}|+\rangle_z \\
    &= e^{-i \omega_0 S_y t/\hbar} \frac{1}{\sqrt{2}}\left(|+\rangle_y + i|-\rangle_y \right)
\end{align}
Now my question is how can i get rid of $S_y$ knowing that $S_y =\frac{\hbar}{2}\begin{pmatrix}0 & -i \\ i & 0\end{pmatrix}$, I mean in class we saw that
$$
|\psi(t)\rangle = e^{-i \frac{\omega_0}{2} t} \frac{1}{\sqrt{2}}\left(|+\rangle_y + i|-\rangle_y \right)
$$
But I don't see how do we get that.

REPLY [1 votes]: In the basis of eigenstates of $S_y$, the matrix representation of $S_y$ is diagonal and of the form
\begin{align}
S_y\to \frac{\hbar}{2}\left(\begin{array}{cc} 1 & 0 \\ 0 & -1\end{array}\right)
\end{align}
so that the exponential of $S_y$ is a diagonal matrix
\begin{align}
e^{-i \omega_0 t S_y/\hbar} \frac{1}{\sqrt{2}}\left(\begin{array}{c} 1 \\ i \end{array}\right)= \frac{1}{\sqrt{2}}\left(\begin{array}{c} e^{i\omega_0 t/2} \\ i e^{-i\omega_0 t/2}\end{array}\right) = \frac{1}{\sqrt{2}}e^{i\omega_0 t/2}
\vert +\rangle_y +\frac{i}{\sqrt{2}}e^{-i\omega_0 t/2}\vert -\rangle_y\, .
\end{align}
This is all done in the basis of eigenstates of $S_y$.  There remains to convert to the eigenstates of $S_z$.


TITLE: What is the best way to show an isomorphism for small groups?
QUESTION [0 upvotes]: Suppose I want to show two groups are isomorphic to each other and the two groups have a small size, $4$ for example.
Would it be okay to simply show were each of the four elements get mapped to in the other group and then show the composition tables for each group and show that they are consistent with the defined map, or is that considered bad form?
It seems for small elements this is more efficient (and more obvious to me at least) what is going on rather than explicitly proving the required properties for a isomorphism. Of course when it is not feasible to show the composition tables and the map explicitly it will be necessary to show the isomorphism differently.
For example:
Say we have a group $G$ that is isomorphic to $V_4$ the Klein-$4$ group.
Would it be necessary to show that $\phi (ab)=\phi(a) \phi(b)$ for each $a,b$ or could you just draw the composition tables on both sides and show which element in $V_4$ the ($4$) elements of $G$ are mapped to.
Thanks!

REPLY [2 votes]: That is completely fine for small groups.  One benefit is that the composition tables then give the full isomorphism, rather than being explained indirectly.


TITLE: Reference for notions of angular mean
QUESTION [0 upvotes]: Intuitively I'd like to have a "well behaved" mean value M of a series of lines' angular coefficients $m_1$, $m_2$, .., $m_n$ such as for example
$$M=1/n (\sum_{k=1}^n m_k)$$

The problem is I'm not sure it is truly well behaved. Converting tangents to arctangents, I'd like to use a function:
$M: (-\pi/2,\pi/2)^n \to (-\pi/2,\pi/2)$  (limits not included) such that for all $x, y, z \in$ Domain:

$M(x, x)=x$ for every number of equal arguments. In particular $M(x)=x$
$M(x, y, z)=M(x,z,y)=M(y,x,z)=M(y,z,x)$, ..., etc for every permutation and number of arguments.
$ (x>y) \rightarrow (x>M(x,y)>y)$
$M$ isn't in general associative. That is $\lnot M(x,y,z)=M(x,M(y,z))$

Discounting the transformations cartesian cohordinates, radiants, polar, etc what problems might the previous formula for M have? Are there alternatives?
The notion of circular mean doesn't seem helpful as it includes the whole $(0,2\pi)$ and averages over $sin$/$cos$ values which doesn't seem equivalent to what I'm aiming for.
Any reference?
Should I ask MO?
Thank you.

REPLY [1 votes]: Given the slopes $m_k$ of lines passing through origin, not parallel to the $y$ axis,
$$y = m_k x$$
there are two obvious definitions for their mean or average.
The first one is the mean slope as shown by OP,
$$M = \frac{1}{N} \sum_{k=1}^{N} m_k \tag{1}\label{label1}$$
and the second one is the mean angle $\theta$ between each line and the $x$ axis that OP also mentioned (but did not show explicitly),
$$M = \tan{\theta}, \quad \theta = \frac{1}{N} \sum_{k=1}^{N} \arctan(m_k) \tag{2}\label{label2}$$
i.e.
$$M = \tan\left(\frac{1}{N}\sum_{k=1}^{N} \arctan(m_k)\right)$$
What is the difference between the two, then?
Consider a set of two lines with $m_1 = 0$ and $m_2 = 4$.
The first one yields $M = 2$, and the second one $M = 0.78$.
The angles of the two original lines are $0^\circ$ and about $76^\circ$.  The angle of the first mean is about $63^\circ$ and the angle of the second mean is about $38^\circ$.
So, the first one yields the mean slope, whereas the second one the mean angle.  Whichever is more proper for your use case, depends on what the lines and their slopes represent.
In simpler terms, if you consider the lines as the point set where they intersect $x = 1$, mean slope is equivalent to picking the point at $x = 1$ with the mean $y$ coordinate.  For the mean angle, you pick points on say the positive $x$ half of the unit circle, and measure their angles $\varphi$ in radians, along the circle from the intersection with the $x$ axis, with the signed distances being positive on the positive $y$ side, negative on the other side.
Now, let's assume $f$ is a monotonically increasing continuous function over all reals, i.e. $f(x) \lt f(x + \delta)$ for $x \in \mathbb{R}$ and $0 \lt \delta \in \mathbb{R}$.  Then, $f$ is invertible: $y = f(x) \iff f^{-1}(y) = x$.  With this, we can define an $f$-mean for the lines passing through origin:
$$M_f = f^{-1}\left(\frac{1}{N}\sum_{k=1}^{N} f(m_k)\right) \tag{3}\label{label3}$$
In the case of $f(x) = x$, this yields $\eqref{label1}$; in the case of $f(x) = \arctan(x)$, $f^{-1}(x) = \tan(x)$, this yields $\eqref{label1}$.
If we consider $\eqref{label1}$ to have a measurement line at $x = 1$, and $\eqref{label2}$ to have a measurement curve a circular arc (half-circle on the positive $x$ axis), then $f$ represents all other possible measurement curves.


TITLE: How to show that a local finite dimensional algebra is basic from definition?
QUESTION [1 upvotes]: A basis algebra is an algebra $A$ such that $e_i A \not\simeq e_j A$ for any $i \neq j$, where $e_1, \ldots, e_n$ is a complete set of primitive orthogonal idempotents. A local algebra is an algebra with only one maximal right ideal. How to show that a local finite dimensional algebra is basic from definition? Thank you very much.

REPLY [1 votes]: If $A$ is local, then $A$ has only two idempotents $0,1$ and $A=1A$. So by definition, $A$ is basic.