TITLE: PDEs - F.O.L. M.o.C. - Theory: Back checking (non-hom.) solution to Transport Eq. leads to an issue!
QUESTION [0 upvotes]: I am having trouble differentiating integrals in $\mathbb{R}^n$.
This particular problem requires the Fundamental Theorem of Calculus and the Chain Rule in $\mathbb{R}^n$.
$$---------------------------------------$$
Allow me to set the context. We have a First Order Linear (Non-homogeneous) PDE:
$$F\big(x,\{\partial_i\}_{i\in\{1,...n\}}\big)u = \bigg(\sum\limits_{i=1}^n \lambda_i(x)\cdot\partial_i \bigg)u - \lambda_{n+1}(x) = 0$$
with boundary condition:
$$u\big|_{\partial U}(x)\equiv g(x).$$
$$---------------------------------------$$
The instance PDE is called the "(Non-hom.) Transport Equation". And looks like:
$$u_t+\boldsymbol b\cdot Du = f. \text{ }\text{ }\text{ }(\star)$$
$$---------------------------------------$$
Painstakingly understanding and applying the Method of Characteristics, we come across the solution:
$$u(\boldsymbol x,t) = g(\boldsymbol x- \boldsymbol bt) + \int_0^tf( \boldsymbol x+(s-t) \boldsymbol b,s)ds$$
$$---------------------------------------$$
And my problem is: For $u$ to be a "solution", it must satisfy $(\star)$.
But, how does one compute $\partial_t u$ and $\partial_i u$? (The term with $g$ I don't care about). Surely it helps to see also:
$$\boldsymbol x, \boldsymbol b\in\mathbb{R}^n\text{ and }s,t\in \mathbb{R}.$$

REPLY [0 votes]: I assume you denote with $Du $ taking the gradient $\nabla u$. But first, define \begin{align} \boldsymbol y(\boldsymbol x, t) &:= \boldsymbol x - \boldsymbol b t \\
\boldsymbol h (\boldsymbol x, t, s) &:= \boldsymbol x + (s-t) \boldsymbol b\end{align}
such that you can formulate the solution as
$$u(\boldsymbol x, t) = g\big (\boldsymbol y(\boldsymbol x, t) \big) + \int_0^t f\big (\boldsymbol h (\boldsymbol x, t, s), s\big) \mathrm d s. $$
Let's take the gradient now (and assume that $f\big (\boldsymbol h (\boldsymbol x, t), s\big)$ is continuosly differentiable in the first argument). Furthermore, I am going to use index notation to not have to worry about the order of multiplication etc.
\begin{align} \nabla u(\boldsymbol x, t) = \partial^x_i u(\boldsymbol x, t) &=
\partial^x_i \Bigg[ g\big (\boldsymbol y(\boldsymbol x, t) \big) + \int_0^t f\big (\boldsymbol h (\boldsymbol x, t, s), s\big) \mathrm d s \Bigg] \\
&\overset{\text{Chain rule}}{=} \partial_j^y g\big (\boldsymbol y(\boldsymbol x, t) \big) \frac{\partial y_j}{\partial x_i}+ \int_0^t \partial_j^h f\big (\boldsymbol h (\boldsymbol x, t), s\big) \frac{\partial h_j}{\partial x_i} \mathrm d s \\
&= \partial_j^y g\big (\boldsymbol y(\boldsymbol x, t) \big) \delta_{ij} + \int_0^t \partial_j^h f\big (\boldsymbol h (\boldsymbol x, t, s), s\big) \delta_{ij} \mathrm d s \\
&= \partial_i^y g\big (\boldsymbol y(\boldsymbol x, t) \big) + \int_0^t \partial_i^h f\big (\boldsymbol h (\boldsymbol x, t, s), s\big)  \mathrm d s
 \end{align}
Now the temporal derivative, for which we require the Leibniz integral rule, as mentioned in the comments by Sal.
\begin{align}
\partial_t u (\boldsymbol x, t) &\overset{\text{Chain rule}}{=} \partial_j^y g\big (\boldsymbol y(\boldsymbol x, t) \big) \frac{\partial y_j}{\partial t}+ \partial_t \int_0^t f\big (\boldsymbol h (\boldsymbol x, t, s), s\big) \mathrm d s \\
&= \partial_j^y g\big (\boldsymbol y(\boldsymbol x, t) \big) (-b_j ) + f\big (\boldsymbol h (\boldsymbol x, t, t), t\big) \frac{\mathrm d }{\mathrm d t} t - (\boldsymbol h (\boldsymbol x, t, 0), 0\big) \frac{\mathrm d }{\mathrm d t} 0 + \int_0^t \partial_t f\big (\boldsymbol h (\boldsymbol x, t, s), s\big) \mathrm d s \\
&\overset{\text{Chain rule}}{=} -b_i \partial_i^y g\big (\boldsymbol y(\boldsymbol x, t) \big) + f\big (\boldsymbol x + (t -t) \boldsymbol b, t\big) + \int_0^t \partial^h_i f\big (\boldsymbol h (\boldsymbol x, t, s), s\big) \frac{\partial h_i}{\partial t} \mathrm d s \\
&= -b_i \partial_i^y g\big (\boldsymbol y(\boldsymbol x, t) \big) + f\big (\boldsymbol x, t\big) + \int_0^t \partial^h_i f\big (\boldsymbol h (\boldsymbol x, t, s), s\big) (-b_i) \mathrm d s \\
&= -b_i \partial_i^y g\big (\boldsymbol y(\boldsymbol x, t) \big) + f\big (\boldsymbol x, t\big) -b_i \int_0^t \partial^h_i f\big (\boldsymbol h (\boldsymbol x, t, s), s\big) \mathrm d s
\end{align}
Rewrite $\boldsymbol b \cdot \nabla u(\boldsymbol x, t) = b_i \partial_i^x u(\boldsymbol x, t)$ and put things together; you see that the PDE is exactly satsified.
Regarding the "boundary condition", this looks to me more like a initial condition (what should be the boundary of $\boldsymbol x \in \mathbb R^n$?):
$$u(\boldsymbol x, t_0 = 0) \overset{!}{=} g(x) = g(\boldsymbol x - \boldsymbol b * 0) + \int_0^0 f\big (\boldsymbol h (\boldsymbol x, t, s), s\big) \mathrm d s $$


TITLE: determining if a curve passes through a loop
QUESTION [0 upvotes]: I have two curves described by parametric equations, and one is a closed loop. How do I analytically determine whether or not the other curve passes through the loop? That is, without graphing and visually inspection the graph. Also assuming that both paths are smooth, continuous, and have no singularities.

REPLY [1 votes]: If the two curves are defined by $$\gamma_1(t)=(x_1(t),y_1(t))\quad\text{and}\quad\gamma_2(t)=(x_2(t),y_2(t))$$
so you have to verify that there's $t_0,t_1,t_2$ s.t.
$$x_1(t_1)<x_2(t_0)<x_1(t_2)\quad\text{and}\quad y_1(t_1)<y_2(t_0)<y_1(t_2)\,\text{or}\,y_1(t_1)>y_2(t_0)>y_1(t_2)$$


TITLE: Did $\int dx^4 \partial ^\mu \phi(x) \partial_\mu \phi(x)= -\int dx^4 \phi(x) \partial^2 \phi(x)$ in the path integral formalism?
QUESTION [1 upvotes]: In the canonical quantization, one assumed the condition that the field $\phi(x)$ vanished at both space and time infinity.
However, in the path integral formalism, thought the field $\phi(x)$ was assumed to be vanished at $t\rightarrow \pm \infty $, it wasn't assumed to be so in the spatial direction, i.e. part of the strength of the path integral in the non-perturbative region.
Thus, consider the following equality which hold in the canonical quantization
$$\int dx^4 \partial ^\mu \phi(x) \partial_\mu \phi(x)= \int dx^4 \phi(x) \partial^2 \phi(x)$$
through the integral by path.
However, in the path integral formalism,
$$\int dx^3 \int dy \partial_y \phi(x) \partial_y\phi(x)=\int dx^3 \phi(x)\partial_y\phi(x)|_{y=-\infty}^{y=\infty} -\int dx^4 \phi(x)\partial_y\partial _y  \phi(x) $$
since both the $\phi(x)$ and $\partial_y \phi(x)$ did not have to vanish at the spatial boundary,
$$\int dx^3 \phi(x)\partial_y\phi(x)|_{y=-\infty}^{y=\infty}$$
didn't have to be zero. But one still saw the texts exchange those two different expressions, without specify which one was more "fundamental" or considered to be the definition.
Did $\int dx^4 \partial ^\mu \phi(x) \partial_\mu \phi(x)= -\int dx^4 \phi(x) \partial^2 \phi(x)$ in the path integral formalism?

REPLY [1 votes]: One needs to impose appropriate$^1$ boundary or fall-off conditions at both temporal and spatial infinity in field theory. For starters, to ensure that the action $S$ has a mathematically well-defined variational/functional derivative $\delta S/ \delta\phi$.
--
$^1$ To assume that the field $\phi$ vanishes might be too strong to describe relevant physical field configurations.


TITLE: Explain why Mandelbrot set escape radius is 2 to a dummy
QUESTION [5 upvotes]: I'm curious, in the Mandelbrot set, why is the escape radius $2$? I've seen few proofs of that on the internet, but i can't understand them enough.
Why is the bailout value of the Mandelbrot set 2?
Mandelbrot sets and radius of convergence
https://mrob.com/pub/muency/escaperadius.html
Some of the statements in them seem "out of the blue" for me.
For example, in the second in-site link I gave above: $ |c|≤2 \Rightarrow|z_n+1|≥|z_n|2−|c|>2|z_n|−2$
Where does $2|z_n|−2$ come from?

REPLY [3 votes]: From the third link I interpeted the proof like this :
What we want to proof is the criteria $|z_n| \le 2$.
Let $|z_n|>2$ and $|z_n|>|c|$. This is done to create a situation where the ratio from $|z_{n+1}|$ to $|z_n|$ is always greater 1 so the next term in the sequence always gets bigger than the one before (sequence is unbound) : $\frac{|z_{n+1}|}{|z_n|}>1$.
So we now have to prove $\frac{|z_{n+1}|}{|z_n|}>1$ :
In the linked article they first use the triangle inequality on the term (https://en.wikipedia.org/wiki/Triangle_inequality) :
$\frac{|z_{n+1}|}{|z_n|} = \frac{|z_n^2+c|}{|z_n|} \ge \frac{|z_n|^2-|c|}{|z_n|} = |z_n|-\frac{|c|}{|z_n|}$
From $|z_n|>|c|$ we get $\frac{|c|}{|z_n|}<1$ so :
$|z_n|-\frac{|c|}{|z_n|} > |z_n|-1$
From $|z_n|>2$ we get :
$|z_n|-1 > 1$ , so all in all we have proven $\frac{|z_{n+1}|}{|z_n|}>1$
This inequality is now true (and the sequence unbound) for all  $|c| \le 2 < |z_n|$.
But what if $|c|>2$ (second case)? Then we can show that the sequence always "escapes" at least after 2 iterations : $z_{n=2}=|c^2+c| \ge |c|^2-|c| > 2$.
If we put this all togehter we get the criteria for the sequence to be unbound if  $|z_n| > 2$ , so the criteria for the sequence to be bound (number in the mandelbrot set) is : $|z_n| \le 2$.
PS : I'm a student for myself and not an expert. As I mentioned above this is just my interpretation of the proof in the linked article. If some expert comes along this thread pls have a quick look over my explaination and if everything is correct.
PPS : Sorry for my English im from Germany xD


TITLE: A question on the formulation of core-shell quantum dot Boundary conditions in K.P method
QUESTION [0 upvotes]: When studying the energy levels of core-shell quantum dots with eight-band k.p method, an important part is the formulation of boundary conditions for wavefunctions. In the Supporting Information of the paper:
http://pubs.acs.org/doi/suppl/10.1021/acs.nanolett.6b00066/suppl_file/nl6b00066_si_001.pdf
the authors (Roman Vaxenburg et al.) summarized this as below

However, I have difficulty in understanding the concept of the vectorial valence-band envelope function $\Psi_v$ here. We know that wavefunction is a scalar, not a vector, while the above equations seem to indicate that $\Psi_v$ is a vector. So my question is:
How do we interpret this vectorial wavefunction?
In fact, this question might be trivial for people in the field. In case anyone is familiar with the calculation of quantum levels of core-shell quantum dot with eight-band k.p method, would you please help me out with this issue?  Thanks!

REPLY [0 votes]: The wave function is a scalar quantity. However, it is often convenient to expand it in other bases. k.p theory can be understood intuitively as follows. The energy bands in a crystal are described by Bloch functions. According to the Bloch theorem and in the single particle picture, you can write the wave function, $\psi_e(\mathbf{r})$, of an electron in the conduction band as
\begin{equation}
\psi_e(\mathbf{r}) = u_{e,\mathbf{k}}(\mathbf{r})\mathrm{e}^{i\mathbf{k}\cdot\mathbf{r}},
\end{equation}
where $\mathbf{k}$ is the wave vector of the electron. It can be shown that the Bloch functions belonging to all the energy bands in the solid (at a specific $\mathbf{k}$-point) form a complete and orthogonal basis set.
It is thus convenient to expand the (unknown) wavefunction of the quantum dot into the basis set created by the Bloch functions in bulk. And since all the action happens at the $\Gamma$ point in optics (i.e., $\mathbf{k}=0$), one can thus write
for, e.g., the hole wave function
\begin{equation}
\psi_{h,\mathrm{QD}}(\mathbf{r}) = \sum_{i=1}^\infty u_{i,\mathbf{0}}(\mathbf{r})\Psi_{i}(\mathbf{r}),
\end{equation}
where $\Psi_i(\mathbf{r})$ are the expansion coefficients, also called slowly varying envelopes. As you can see, the quantum-dot wave function, $\psi_{h,\mathrm{QD}}(\mathbf{r})$, inherits the properties of all the bands in the unperturbed crystal. In practice, this infinite sum can be truncated because $\Psi_{i}(\mathbf{r})$ are significant for a couple of $i$-s only. This is because the hole wave function, $\psi_{h,\mathrm{QD}}(\mathbf{r})$, is not perturbed significantly compared to bulk and thus inherits most of the properties from a few bands nearby in energy. In 8x8 k.p theory, you consider three valence bands (because the valence-band states inherit the symmetry of the $p_{x,y,z}$ orbitals and are thus three-fold degenerate) and one conduction band. So 4 bands in total but if you include spin then it's 8 (hence 8x8). You can thus write the slowly varying envelope of the hole wave function as
\begin{equation}
\vec{\Psi_{h,\mathrm{QD}}} = (\Psi_x, \Psi_y, \Psi_z, \Psi_e) = (\vec{\Psi_v},\Psi_e),
\end{equation}
In reality, you also include spin, see for instance https://journals.aps.org/prb/pdf/10.1103/PhysRevB.67.155312 . I hope this helps.


TITLE: What are all the even positive integers $n$ such that for $1 \le r \le n-1 $ , $2n \choose r$ is odd for exactly one $r$ ?
QUESTION [1 upvotes]: Motivated from this To find all odd integers $n>1$ such that $2n \choose r$ , where $1 \le r \le n$ , is odd only for $r=2$     ; what are all the even positive integers $n$ such that for $1 \le r \le n-1 $ , $2n \choose r$ is odd for exactly one $r$ ?

REPLY [3 votes]: The binomial coefficient $\binom{m}{k}$ is odd if and only if $k$ and $m-k$ have no common terms in their binary expansions. That is the case if and only if the set of $1$-bits in the binary expansion of $k$ is a subset of the set of $1$-bits in the binary expansion of $m$.
To see that, let us more generally consider with what power a prime $p$ divides $n!$. Using Legendre's formula in the form
$$v_p(n!) = \frac{n - s_p(n)}{p-1},$$
where $v_p(r)$ is the exponent of $p$ in the prime factorisation of $r$ and $s_p(r)$ is the sum of the digits in the $p$-ary expansion of $r$, from
\begin{align}
v_p\Biggl[\binom{m}{k}\Biggr] &= v_p(m!) - v_p(k!) - v_p((m-k)!)\\
&= \frac{\bigl(m-s_p(m)\bigr) - \bigl(k-s_p(k)\bigr) - \bigl((m-k) - s_p(m-k)\bigr)}{p-1}\\
&= \frac{s_p(k) + s_p(m-k) - s_p(m)}{p-1}
\end{align}
it immediately follows that $\binom{m}{k}$ is not divisible by $p$ if and only if $s_p(m) = s_p(k) + s_p(m-k)$. If we write
$$k = \sum_{\lambda = 0}^\ell a_\lambda p^\lambda,\quad m-k = \sum_{\lambda = 0}^\ell b_\lambda p^\lambda\quad\text{and}\quad m = \sum_{\lambda = 0}^\ell c_\lambda p^\lambda,$$
where $\ell$ is "sufficiently large" (so large that $m < p^{\ell+1}$) and $a_\lambda,\,b_\lambda,\,c_\lambda \in \{0,1,\dotsc,p-1\}$ for $0 \leqslant \lambda \leqslant \ell$, then we see
$$s_p(k) = \sum_{\lambda = 0}^\ell a_\lambda,\quad s_p(m-k) = \sum_{\lambda = 0}^\ell b_\lambda\quad\text{and}\quad s_p(m) = \sum_{\lambda = 0}^\ell c_\lambda.$$
The addition $k + (m-k)$ has no carry in base $p$ if and only if $a_\lambda + b_\lambda < p$ for $0 \leqslant \lambda \leqslant \ell$, and then we see that $c_\lambda = a_\lambda + b_\lambda$ for all $\lambda$, hence $s_p(m) = s_p(k) + s_p(m-k)$. If on the other hand the addition has carries, that is, $a_\lambda + b_\lambda \geqslant p$ for some $\lambda$, say the first carry occurs at the digit with index $\delta$, then we have $c_\lambda = a_\lambda + b_\lambda$ for $0 \leqslant \lambda < \delta$, and $c_\delta = a_\delta + b_\delta - p$. If no further carries occur, then we have $c_{\delta+1} = a_{\delta+1} + b_{\delta+1} + 1$ and $c_\lambda = a_\lambda + b_\lambda$ for $\lambda > \delta+1$ and $s_p(m) = s_p(k) + s_p(m-k) - (p-1)$. If further carries occur, each carry reduces the digit sum by $p-1$, so we have
$$\frac{s_p(k) + s_p(m-k) - s_p(m)}{p-1} = \text{ number of carries in the base-$p$ addition}.$$
For base $2$, a carry occurs if and only if $k$ and $m-k$ have a common $1$-bit.
Thus there is exactly one odd binomial coefficient $\binom{m}{k}$ for $1 \leqslant k \leqslant \left\lfloor \frac{m}{2}\right\rfloor$ if and only if $m$ has exactly two $1$-bits in its binary expansion. If $m$ is a power of $2$ (has only one $1$-bit), then every nontrivial addition $k + (m-k) = m$ has carry in base $2$, and if $m$ has at least three $1$-bits, then at least two of the powers of $2$ in the binary expansion of $m$ are $\leqslant \left\lfloor\frac{m}{2}\right\rfloor$ and the corresponding binomial coefficients are both odd.
So the even $n$ such that $\binom{2n}{r}$ is odd for exactly one $1 \leqslant r \leqslant n-1$ are precisely the numbers $n = 2^a + 2^b$ with $1 \leqslant a < b$. Then the binomial coefficient $\binom{2n}{2^{a+1}}$ is odd, and all other are even.


TITLE: Presentation matrix for module in terms of other presentation matrices
QUESTION [1 upvotes]: Let $R$ be a commutative ring, and let $A$ and $B$ be right $R$-modules. Suppose that the following two sequences are exact:
$$
R^u\stackrel{Q}{\to} R^s\stackrel{\pi}{\to}A\to 0 \\ R^r\stackrel{f}{\to} A\to B\to 0
$$
and there exists a map $P:R^r\to R^s$ such that $\pi\circ P=f$. Also $P$ and $Q$ can be considered to be matrices of appropriate dimensions (and are multiplied on the right. That they are right-modules and matrix multiplication is on the right is not super important to the problem, but I'm too tired to convert it to left)
(Sorry, I don't know how to draw commutative diagrams here, but I've included a picture of the commutative diagram.)
Question: What is a presentation matrix for $B$?
Secondary Question: Is there a way to get a formula for $f$?
Known: $A=R^s/R^uQ$, that is $Q$ is a presentation matrix for $A$, and $B=A/f(R^r)=A/\pi(R^rP)$. Combining these two facts, we obtain $$B=\dfrac{R^s/R^uQ}{\pi(R^rP)}$$
My suspicion is that $B$ is presented by the block matrix
$$V=\begin{pmatrix}
P\\
Q
\end{pmatrix}
$$
and $R^r\oplus R^u\stackrel{V}{\to}R^s\to B\to 0$ is exact.
My reasoning is, $B=A/f(R^r)$ and $f=\pi\circ P$. $A$ has $s$ generators and $u$ relations with presentation matrix $Q$. So $B$ has at most $s$ generators. Given $x\in R^r$, $xP$ gives $f(x)$ in terms of the generators of $A$. So $R^rP$ gives $r$ relations for $B$. But these aren't enough, they can be further simplified with the relations $R^uQ$. So $B=R^s/(R^rP)/(R^uQ)=R^s/(R^{r+u}V)$. I don't know if this makes any sense. I'm feeling very stupid right now. Any help would be much appreciated.

REPLY [1 votes]: $\require{AMScd}\DeclareMathOperator{\cok}{cok}\newcommand{\psm}[2]{\left[\genfrac{}{}{0pt}{}{#1}{#2}\right]}$Your suspicion is correct, that's $V$ is the representation matrix of $B$.
This means that $\pi g:R^s\to B$ (with diagrammatic order composition) is the cokernel of $V$.
To prove this, consider the following commutative diagram with exact rows
\begin{CD}
R^u@>Q>>R^s@>\pi>>A@>>>\{0\}\\
@.@APAA@|\\
@.R^r@>>f>A@>>g>B@>>>\{0\}
\end{CD}
Then $\pi$ is the cokernel of $Q$ and $g$ is the cokernel of $f$.
Note that the cokernel of $V$ is the cointersection of cokernels of $P$ and $Q$, that's the diagonal of the right-handed square in the following commutative diagram (where both squares are pushout):
\begin{CD}
R^r@>P>>R^s@>\pi>>A\\
@VVV@VV\cok PV@VVgV\\
\{0\}@>>>\bullet@>>>B
\end{CD}
hence the assertion follows.

If you don't know this property of cokernels argue as follows.
Clearly, we have
$$\psm PQ\pi g=\psm{P\pi g}{Q\pi g}=\psm{fg}{Q\pi g}=0$$
for $fg=0$ because $g$ is the cokernel of $f$ (by exactness of bottom row) and $Q\pi=0$ by exactness of upper row.
On the other hand let $k:R^s\to Y$ such that
$$\psm PQk=0$$
that's $Pk=0$ and $Qk=0$.
From $Pk=0$ we get $k=(\cok P)u$ where $\cok P$ denote the cokernel of $P$, for some $u$ and $k=\pi v$ for some $v$, because $\pi$ is the cokernel of $Q$.
Since $(\cok P)u=k=\pi v$, by universal property of pushout, there exists one and only one morphism $h:B\to Y$ such that $v=gh$, hence $k=\pi gh$.


TITLE: Does a certain contractive mapping have a fixed point?
QUESTION [7 upvotes]: Let $f:X\rightarrow X$ be a contractive mapping of a complete metric space satisfying
$$d(f(x),f(y))\leq\alpha(d(x,y))d(x,y)$$
where $\alpha:\mathbf{R}^+\rightarrow [0,1)$, and $\alpha(t_n)\rightarrow 1$ implies $t_n\rightarrow 0$
Does $f$ have a fixed point?
Thank you.

REPLY [0 votes]: Fix $x\in X$ and let $x_n=f^n(x)$, $n=1,2, \ldots$ yet again we break the argument into two steps.
Step 1: $\lim\limits _{n\rightarrow\infty}d(x_n,x_{n+1})=0$
Since $f$ is contractive the sequence $\{d(x_n,x_{n+1})\}$ is monotone decreasing and bounded below, so $$\lim\limits _{n\rightarrow\infty}d(x_n,x_{n+1})=r\geq0$$
Assume that $r>0$. Then by the contractive condition:
$$\frac{d(x_{n+1},x_{n+2})}{d(x_n,x_{n+1})}\leq \alpha(d(x_n,x_{n+1}))\,\,\,,n=1,2,3, \ldots$$
Then $n\rightarrow\infty\implies 1\leq\lim\limits _{n\rightarrow\infty}\alpha(d(x_n,x_{n+1}))\implies r=0$, contradiction.
Step 2: $\{x_n\}$ is a Cauchy sequence:
Assume $\lim\limits_{n,m\rightarrow\infty}\sup d(x_n,x_m)>0$.
$$d(x_n,x_m)\leq d(x_n,x_{n+1})+d(x_{n+1},x_{m+1})+d(x_{m+1},x_m)$$
So by the contractive condition :
$$d(x_n,x_m)\leq {(1-\alpha d(x_n,x_{m}))}^{-1}[d(x_{n},x_{n+1})+d(x_{m+1},x_m)]$$
Under the assumption $\lim\limits_{n,m\rightarrow\infty}\sup d(x_n,x_m)>0$,step 1 implies $\lim\limits_{n,m\rightarrow\infty}\sup {(1-\alpha (d(x_n,x_m)))}^{-1}=+\infty$
From which $\lim\limits_{n,m\rightarrow\infty}\sup \alpha (d(x_n,x_m))=1$, that implies $\lim\limits_{n,m\rightarrow\infty}\sup d(x_n,x_m)=0$, again a contradiction.


TITLE: Why is $S = X^2 + Y^2$ distributed $Exponential(\frac{1}{2})$?
QUESTION [2 upvotes]: Let $X$ and $Y$ be independent, standard normally distributed random variables ($\sim Normal(0, 1)$). Why is $S = X^2 + Y^2$ distributed $Exponential(\frac{1}{2})$?
I understand that an exponential random variable describes the time until a next event given a rate at which events occur. However, I do not understand what this has to do with $S = X^2 + Y^2$.

REPLY [7 votes]: Let $W=X^2+Y^2$. We find the cumulative distribution function $F_W(w)$ of $w$.
This is the probability that $W\le w$.  Clearly $F_W(w)=0$ if $w\le 0$. Now we deal with $w\gt 0$.
So we want $\Pr(X^2+Y^2 \le w)$.  This is the integral over the disk with centre the origin, and radius $\sqrt{w}$, of the joint density function of $X$ and $Y$. So for positive $w$,
$$F_W(w)=\iint_D \frac{1}{2\pi}e^{-(x^2+y^2)/2}\,dx\,dy.$$
Change to polar coordinates. Then $dx\,dy=r\,dr\,d\theta$, and our integral is
$$\int_0^{2\pi}\frac{1}{2\pi}\left(\int_0^{\sqrt{w}}re^{-r^2/2}\,dr           \right)\,d\theta.$$
The integration is easy. For the inner integral, let $u=r^2$. The outer integral is even simpler. Quickly we get $\dfrac{1}{2}e^{-w/2}$, which is the cdf of the exponential with parameter $\dfrac{1}{2}$.


TITLE: Gauge orbit orthogonality in a gauged linear sigma model on $\mathbb{C}P^{N}$
QUESTION [1 upvotes]: I am back with another question from the book Mirror Symmetry, this time from Section $15.1.1$.
Consider the gauged linear sigma model for $N$ complex scalar fields and the Lagrangian:
$$
L=-\sum_{i=1}^{N}
|D_{\mu}\phi_{i}|^{2}
-
\frac{e^{2}}{2}\left(\sum_{i=1}^{N}|\phi_{i}|^{2}-r\right)
$$
where $r>0$, and $D_{\mu}\phi_{i}=(\partial_{\mu}+iv_{\mu})\phi_{i}$ for the $U(1)$ gauge field $v_{\mu}$.
This Lagrangian is invariant under $U(1)$ gauge transformations acting as:
$$
\phi_{i}\mapsto e^{i\gamma(x)}\phi_{i}\text{ , and }
v_{\mu}\mapsto v_{\mu}-\partial_{\mu}\gamma(x)
$$
The vacuum manifold consists of field configurations for which $\sum_{i=1}^{N}|\phi_{i}|^{2}=r$, and since these fields are also related by local phase rotations, this is nothing other than $\mathbb{C}P^{N-1}$.
Since there is no kinetic term for the gauge field, it can easily be integrated out using the equation of motion:
$$
\sum_{i=1}^{N}
\left(
\overline{D_{\mu}\phi^{i}}\phi_{i}
-
\overline{\phi_{i}}D_{\mu}\phi_{i}
\right)
=
0
$$
In the low energy ($e\to\infty$), limit, a fixed field configuration $\{\phi_{i}(x)\}$ determines a map from the worldsheet (where $x$ lives), to $\mathbb{C}P^{N-1}$. We can now pick a tangent vector $\xi^{\mu}$ on the worldsheet, and use this map to obtain a tangent vector on $\mathbb{C}P^{N-1}$. The claim is then that the resulting vector is orthogonal to the gauge group orbit. We now arrive at my question.
Firstly, it seems to me that the resulting tangent vector should be the push-forward of $\xi^{\mu}$ along $\phi$, namely $\xi^{\mu}\frac{\partial\phi^{i}}{\partial x^{\mu}}$. Instead, the book seems to state that it is either $\xi^{\mu}$ itself, or even $\xi^{\mu}D_{\mu}\phi^{i}$.
Secondly, the gauge group orbit is clearly $\delta\cdot(\phi_{i},\bar{\phi}_{i})=(i\phi_{i},-i\bar{\phi}_{i})$, so the claim is that $\langle\xi^{\mu}D_{\mu}\phi,i\phi\rangle=0$. This is apparently able to be derived from the $v_{\mu}$ equation of motion, but all this equation of motion says to me is that:
$$
\langle\xi^{\mu}D_{\mu}\phi^{i},i\phi\rangle
=
-
\overline{\langle\xi^{\mu}D_{\mu}\phi^{i},i\phi\rangle}
$$
i.e. that this expression is purely imaginary. (Note that I have also had to assume that $\xi^{\mu}$ is real to get to this point, I am not quite sure this is the case either).
Any pointers would be much appreciated.

REPLY [0 votes]: I do not know if you still care about this question but I tried something and decided that maybe is worth sharing. To focus better on the geometric picture, I went for a simplification and decided to try to derive a simplified model, where instead of a world-sheet a world-line is used. I found the model fairly instructive, so maybe you would too and maybe it can give you some intuition into the world-sheet case. According to it, you cannot expect the tangent vector to be pushed forward to a vector orthogonal to the $\text{U}(1)-$orbits (in fact that non-orthogonality is where the gauge field $v$ arises). Furthermore, the covariant derivative of the motion however is orthogonal to the  $\text{U}(1)-$orbits.
Analysis of a world-line rather than a world-sheet. To simplify things for my intuition, I represent $\mathbb{C}^N$ together with the standard Hermitian dot product as $\mathbb{R}^{2N}$ together with the standard real dot product paired with the standard block-digonal symplectic matrix $J$. Then for any $\phi \, \in \, \mathbb{C}^N = \mathbb{R}^{2N}$, we have the following identity between complex and real notations $i \, \phi_{\mathbb{C}} \, = \, J \, \phi_{\mathbb{R}}$. Furthermore, there is a dot product plus symplectic product interpretation of the Hermitian dot product $$\bar{\phi}_{\mathbb{C}}^{\,T} \psi_{\mathbb{C}} \, =\, \phi_{\mathbb{R}}^T \psi_{\mathbb{R}} \, + \, \phi^T_{\mathbb{R}} J \, \psi_{\mathbb{R}}$$ From now on, we can think that the norm is simply the real norm (wich is equal to the Hermitian one anyway) $$| \,\phi \,|^2 \,=\, \phi^T \phi$$ We are interested in the dynamics of the lagrangian $$L \, = \, \frac{1}{2} \left|\,\frac{d\phi}{dt} + v J \, \phi \,\right|^2 \, + \, \frac{\lambda}{2}\big(\,|\phi|^2 - 1\,\big)$$ where $v$ is real valued scalar function.  The Euler-Lagrange equations of this Lagrangian are
\begin{align}
&\frac{d}{dt}\left(\frac{d\phi}{dt} + vJ\phi \right) \, =\, -\,vJ\left(\frac{d\phi}{dt} + vJ\phi \right) \, + \, \lambda \phi \,\,\,\,\,\text{ (the equation for $\phi$)}\\
&|\phi|^2 \, = \, \phi^T \phi \,=\,1 \,\,\,\,\,\text{ (the equations for $\lambda$)}\\
&\left(\frac{d\phi}{dt} + vJ\phi \right)^TJ\,\phi \, =\, 0  \,\,\,\,\,\text{ (the equations for $v$)}
\end{align}
Because $J\phi^T J\phi = |J\phi|^2 = |\phi|^2 = 1$, we can rewrite the last equation as $v = -\,\frac{d\phi}{dt}^T J\phi$.
Aslo,
I am going to use the notation $$\frac{\nabla}{dt} \, =\, \frac{d}{dt} \, +\, vJ$$ Then, the Euler-Lagrange equations can be written as
\begin{align}
&\frac{\nabla^2 \phi}{dt^2}  \, = \, \lambda \phi \,\,\,\,\,\text{ (the equation for $\phi$)}\\
&|\phi|^2 \, = \, \phi^T \phi \,=\,1 \,\,\,\,\,\text{ (the equations for $\lambda$)}\\
&v = -\,\frac{d\phi}{dt}^T J\phi  \,\,\,\,\,\text{ (the equations for $v$)}
\end{align}
Differentiate the second equation with respect to $t$
$$0 \, =\, \frac{d}{dt} \big(\,\phi^T\phi\,\big) \, =\, 2\, \phi^T\frac{d\phi}{dt}$$ which represent the fact that the trajectories' velocities are always tangent to the unit sphere. Due to the fact that $J$ is symplectic, $$\phi^TJ\,\phi \, = \, 0$$ and so we can rewrite the latter equation as
$$0 \, =\, \phi^T\left(\frac{d\phi}{dt} + vJ\phi \right) \, =\, \phi^T\frac{\nabla \phi}{dt}$$ Now, differentate covariantly the latter identity
$$0 \, =\, \frac{\nabla}{dt} \left( \phi^T\frac{\nabla \phi}{dt}\right) \, =\, \left|\frac{\nabla \phi}{dt}\right|^2 \, +\,  \phi^T\frac{\nabla^2 \phi}{dt^2}$$
Plugging the first Euler-Lagrange equation is this latter identity, we get an equation for $\lambda$
$$0 \, =\, \left|\frac{\nabla \phi}{dt}\right|^2 \, +\,  \phi^T\frac{\nabla^2 \phi}{dt^2} \, =\, \left|\frac{\nabla \phi}{dt}\right|^2 \, +\,  \phi^T \big(\lambda \phi\big)  \, =\, \left|\frac{\nabla \phi}{dt}\right|^2 \, +\,  \lambda  \big(\phi^T\phi\big)$$
and since $\phi^T\phi=1$
$$\lambda \, =\,  -\,\left|\frac{\nabla \phi}{dt}\right|^2$$
Now, the Euler-Lagrange equations take the form
\begin{align}
&\frac{\nabla^2 \phi}{dt^2}  \, = \, -\,\left|\frac{\nabla \phi}{dt}\right|^2 \phi \\
&\\
&\text{where }\\
&v \,=\, -\,\frac{d\phi}{dt}^T J\phi\\
&\frac{\nabla}{dt} \, =\, \frac{d}{dt} \, +\, vJ
\end{align}
with the unit-sphere constriant $|\phi|^2 \,=\,1$ authomatically satisfied as long as the intital filed (at $t=0$) is on the unit sphere, i.e. $|\phi_0|^2 = 1$. This is because with the choice of $\lambda$ we arrenged for $|\phi|^2$ to be a conserved quantity.
Now, let us look at the geometric consequences. From the action of $\text{U}(1)$ on the unit sphere as $e^{i\theta} \phi$ in the complex picture, we can see, by simply differentating with respect to $\theta$, that the tangent vector to the unit sphere that is also tangent to the circular orbit $e^{i\theta} \phi$ at the point $\phi$ is $i\phi$. In the real picture, the latter tangent vector is $J\phi$.  As you can see, $$v \,=\, -\,\frac{d\phi}{dt}^T J\phi$$  tells us that in general the trajectories of the motion are not perpendicular to the circular orbits of the $\text{U}(1)$ action. In fact, since $|\phi| = 1$, scalar function $v$ is negative of the magnitude of the projection of $\frac{d\phi}{dt}$ on the $\text{U}(1)$ orbits. On the other hand,
$$\frac{\nabla \phi}{dt}^T J \phi \, = \, \left(\frac{d\phi}{dt} + vJ\phi\right)^T J\phi \,=\, \frac{d\phi}{dt}^T J\phi + v \, \phi^TJ^T J\phi \, =\,  \frac{d\phi}{dt}^T J\phi + v \, = \, 0$$ so the covariant derivative $\frac{\nabla \phi}{dt}$ is orthogonal to the $\text{U}(1)$ orbits. In fact I can rewrite the covariant derivative of $\phi$ as follows
\begin{align}
\frac{\nabla \phi}{dt} \, =\, \frac{d\phi}{dt} \, +\, v\,  J\phi \, =& \, \frac{d\phi}{dt} \, -\,\left(\frac{d\phi}{dt}^T J\phi \right)\,  J\phi \, =\,  \frac{d\phi}{dt} \, -\,\left( \big(J\phi\big)^T  \frac{d\phi}{dt} \right)\,  J\phi \\
 =& \, \frac{d\phi}{dt} \, -\, \left( J\phi \right)\,(J\phi\big)^T   \frac{d\phi}{dt} \\
=& \, \Big(\, \text{Id} \, -\,  \big(J\phi\big)\,\big(J\phi\big)^T \,\Big)  \frac{d\phi}{dt}
\end{align}
where the linear operator $\text{Id} \, -\,  \big(J\phi\big)\,\big(J\phi\big)^T $ is the orthogonal projector onto the orthogonal complement of the tangent vector $J\phi$ to the $\text{U}(1)-$orbit $e^{i\theta}\phi$. So basically, the gauge field $v$ is chosen so that we differentiate the trajectory with respect to time and then projecti it orthogonally to the $\text{U}(1)-$orbit.


text:
\section{Power of Complex Modulus equals Complex Modulus of Power}
Tags: Complex Modulus, Complex Powers

\begin{theorem}
Let $z \in \C$ be a [[Definition:Complex Number|complex number]].
Let $\left\vert{z}\right\vert$ be the [[Definition:Complex Modulus|modulus]] of $z$.
Let $n \in \Z_{\ge 0}$ be a [[Definition:Positive Integer|positive integer]].
Then:
:$\left\vert{z^n}\right\vert = \left\vert{z}\right\vert^n$
\end{theorem}

\begin{proof}
The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
:$\left\vert{z^n}\right\vert = \left\vert{z}\right\vert^n$
$P \left({0}\right)$ is the case:
{{begin-eqn}}
{{eqn | l = \left\vert{z^0}\right\vert
      | r = \left\vert{1}\right\vert
      | c =
}}
{{eqn | r = 1
      | c =
}}
{{eqn | r = \left\vert{z}\right\vert^0
      | c =
}}
{{end-eqn}}
Thus $P \left({0}\right)$ is seen to hold.
=== Basis for the Induction ===
$P \left({1}\right)$ is the case:
{{begin-eqn}}
{{eqn | l = \left\vert{z^1}\right\vert
      | r = \left\vert{z}\right\vert
      | c =
}}
{{eqn | r = \left\vert{z}\right\vert^1
      | c =
}}
{{end-eqn}}
Thus $P \left({1}\right)$ is seen to hold.
This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
=== Basis for the Induction ===
$P \left({2}\right)$ is the case:
: $\left\vert{z^2}\right\vert = \left\vert{z}\right\vert^2$
which is demonstrated in [[Square of Complex Modulus equals Complex Modulus of Square]].
Thus $P \left({2}\right)$ is seen to hold.
This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
=== Induction Hypothesis ===
Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true.
So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
:$\left\vert{z^k}\right\vert = \left\vert{z}\right\vert^k$
from which it is to be shown that:
:$\left\vert{z^{k + 1} }\right\vert = \left\vert{z}\right\vert^{k + 1}$
=== Induction Step ===
This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
{{begin-eqn}}
{{eqn | l = \left\vert{z^{k + 1} }\right\vert
      | r = \left\vert{z^k \cdot z}\right\vert
      | c =
}}
{{eqn | r = \left\vert{z^k}\right\vert \cdot \left\vert{z}\right\vert
      | c = [[Complex Modulus of Product of Complex Numbers]]
}}
{{eqn | r = \left\vert{z}\right\vert^k \cdot \left\vert{z}\right\vert
      | c = [[Power of Complex Modulus equals Complex Modulus of Power#Induction Hypothesis|Induction Hypothesis]]
}}
{{eqn | r = \left\vert{z}\right\vert^{k + 1}
      | c =
}}
{{end-eqn}}
So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
Therefore:
:$\forall n \in \ Z_{\ge 0}: \left\vert{z^n}\right\vert = \left\vert{z}\right\vert^n$
{{qed}}
\end{proof}

config: wiki


text:
\begin{document}
\title[Mean dimension and an embedding theorem for real flows]
{Mean dimension and an embedding theorem\\for real flows}
\author[Y. Gutman]{Yonatan Gutman}
\address{Yonatan Gutman:
Institute of Mathematics, Polish Academy of Sciences, ul. \'Sniadeckich 8, 00-656 Warszawa, Poland}
\email{y.gutman@impan.pl}
\author[L. Jin]{Lei Jin $^*$}\let\thefootnote\relax\footnote{* Corresponding author.}
\addtocounter{footnote}{-1}\let\thefootnote\svthefootnote
\address{Lei Jin:
Center for Mathematical Modeling, University of Chile and UMI 2807 - CNRS}
\email{jinleim@mail.ustc.edu.cn}
\subjclass[2010]{37B05; 54H20.}\keywords{Mean dimension, real flow, equivariant embedding, band-limited function}\begin{abstract}
We develop mean dimension theory for $\mathbb{R}$-flows.
We obtain fundamental properties and examples and prove an embedding theorem:
Any real flow $(X,\mathbb{R})$ of mean dimension strictly less than $r$ admits
an extension $(Y,\mathbb{R})$ whose mean dimension is equal to that of $(X,\mathbb{R})$
and such that $(Y,\mathbb{R})$ can be embedded in the $\mathbb{R}$-shift on the compact function space
$\{f\in C(\mathbb{R},[-1,1])|\;\supp(\hat{f})\subset [-r,r]\}$, where $\hat{f}$ is the Fourier transform of $f$ considered as a tempered distribution. These canonical embedding spaces appeared previously as a tool in embedding results for $\mathbb{Z}$-actions.
\end{abstract}\maketitle\section{Introduction}\label{sec:intro}
Mean dimension was introduced by Gromov \cite{Grom} in 1999,
and was systematically studied by Lindenstrauss and Weiss \cite{LW}
as an invariant of topological dynamical systems (t.d.s).
In recent years it has extensively been investigated with relation to the so-called
embedding problem, mainly for $\mathbb{Z}^k$-actions ($k\in\mathbb{N}$).
For $\mathbb{Z}$-actions, the problem is which $\mathbb{Z}$-actions $(X,T)$ can be embedded in the shifts on the Hilbert cubes
$(([0,1]^N)^\mathbb{Z},\sigma)$, where $N$ is a natural number and the shift $\sigma$ acts on $([0,1]^N)^\mathbb{Z}$ by
$\sigma((x_n)_{n\in\mathbb{Z}})=(x_{n+1})_{n\in\mathbb{Z}}$ for $x_n\in [0,1]^N$.
Under the conditions that $X$ has finite Lebesgue covering dimension and the system
$(X,T)$ is aperiodic, Jaworski \cite{J} proved in 1974 that
$(X,T)$ can be embedded in the shift on $[0,1]^\mathbb{Z}$.
Using Fourier and complex analysis, Gutman and Tsukamoto showed that
if $(X,T)$ is minimal and has mean dimension strictly less than $N/2$
then it can be embedded in $(([0,1]^N)^\mathbb{Z},\sigma)$ (see a more general result in \cite{GQT}).
We note that the value $N/2$ is optimal since
a minimal system of mean dimension
$N/2$ which cannot be embedded in $(([0,1]^N)^\mathbb{Z},\sigma)$
was constructed in \cite[Theorem 1.3]{LT}. More references for the embedding problem are given in
\cite{A,K,L,G11,G,GT14,GLT,G16,G17,GQS}.

In this paper, we develop the mean dimension theory for $\mathbb{R}$-actions
and investigate the embedding problem in this context.
Throughout this paper, by a {\bf flow}
we mean a pair $(X,\mathbb{R})$, where $X$ is a compact metric space
and $\Gamma:\mathbb{R}\times X\to X,(r,x)\mapsto rx$ is a continuous map
such that $\Gamma(0,x)=x$ and $\Gamma(r_1,\Gamma(r_2,x))=\Gamma(r_1+r_2,x)$
for all $r_1,r_2\in\mathbb{R}$ and $x\in X$.
Let $(X,\mathbb{R})=(X,(\varphi_r)_{r\in\mathbb{R}})$ and
$(Y,\mathbb{R})=(Y,(\phi_r)_{r\in\mathbb{R}})$ be flows.
We say that $(Y,\mathbb{R})$ can be {\bf embedded} in $(X,\mathbb{R})$ if there is
an $\mathbb{R}$-equivariant homeomorphism of $Y$ onto a subspace of $X$; namely,
there is a homeomorphism $f:Y\to f(Y)\subset X$ such that
$f\circ\phi_r=\varphi_r\circ f$ for all $r\in\mathbb{R}$.

This paper is organized as follows:
In Section \ref{sec:mdim}, we present basic notions and properties
of mean dimension theory for flows. In Section \ref{sec:construction} we construct minimal real flows with arbitrary mean dimension. In Section \ref{sec:conj}, we propose an embedding conjecture for flows and discuss its relation to the Lindenstrauss-Tsukamoto embedding conjecture for $\mathbb{Z}$-systems.
In Section \ref{sec:embedding}, we state the main embedding theorem
and prove it using a key proposition.
In Section \ref{sec:prod embed}, we prove the key proposition.

\bigskip

\noindent\textbf{Acknowledgements.}
Y. Gutman was partially supported by the NCN (National Science Center, Poland) Grant 2016/22/E/ST1/00448. Y. Gutman and L. Jin were partially supported by the NCN (National Science Center, Poland) Grant 2013/08/A/ST1/00275. L. Jin was supported by Basal funding PFB 170001 and Fondecyt Grant No. 3190127. This work owes greatly to previous work by Y. Gutman and M. Tsukamoto. We are grateful to the anonymous reviewer for a careful reading and many useful suggestions.

\bigskip

\section{Mean dimension for real flows}\label{sec:mdim}
We first introduce the definition of mean dimension for $\mathbb{R}$-actions.
Let $(X,d)$ be a compact metric space.
Let $\epsilon>0$ and $Y$ a topological space. A continuous map $f:X\to Y$ is called a \textbf{$(d,\epsilon)$-embedding} if for any $x_1,x_2\in X$
with $f(x_1)=f(x_2)$ we have $d(x_1,x_2)<\epsilon$. Define $$\mathrm{Widim}_\epsilon(X,d)=\min_{K\in\mathcal{K}}\dim(K),$$
where $\dim(K)$ is the Lebesgue covering dimension of the space $K$
and $\mathcal{K}$ denotes the collection of compact metrizable spaces $K$ satisfying that
there is a $(d,\epsilon)$-embedding $f:X\to K$.
Note that $\mathcal{K}$ is always nonempty since we can take
$K=X$ which is a compact metric space and $f=id$ which is the identity map from $X$ to itself.

Let $(X,\mathbb{R})$ be a flow.
For $x,y\in X$ and a subset $A$ of $\mathbb{R}$ let $$d_A(x,y)=\sup_{r\in A}d(rx,ry).$$
For $R>0$ denote by $d_R$ the metric $d_{[0,R]}$ on $X$.
Clearly, the metric $d_R$ is compatible with the topology on $X$.
\begin{Proposition}
For any $\epsilon>0$, we have

(1) $\mathrm{Widim}_\epsilon(X,d)\le\dim(X)$;

(2) if $0<\epsilon_1<\epsilon_2$ then $\mathrm{Widim}_{\epsilon_1}(X,d)\ge\mathrm{Widim}_{\epsilon_2}(X,d)$;

(3) if $0\le R_1<R_2$ then $\mathrm{Widim}_\epsilon(X,d_{R_1})\le\mathrm{Widim}_\epsilon(X,d_{R_2})$;

(4) $\mathrm{Widim}_\epsilon(X,d_{[r_1,r_2]})=\mathrm{Widim}_\epsilon(X,d_{[r_0+r_1,r_0+r_2]})$ for any $r_0,r_1,r_2\in\mathbb{R}$;

(5) $\mathrm{Widim}_\epsilon(X,d_{N+M})\le\mathrm{Widim}_\epsilon(X,d_N)+\mathrm{Widim}_\epsilon(X,d_M)$ for any $N,M\ge0$.
\end{Proposition}\begin{proof}
Since $(X,d)$ is a compact metric space that belongs to $\mathcal{K}$,
we have (1). Points (2) and (3) follow from the definition.
Let $\epsilon>0$.
If $K$ is a compact metrizable space and $f:X\to K$ is a continuous map
such that for any $x_1,x_2\in X$ with $f(x_1)=f(x_2)$ we have $d_{[r_1,r_2]}(x_1,x_2)<\epsilon$,
then $f\circ r_0:X\to K$ is a continuous map
such that for any $x_1,x_2\in X$ with $f\circ r_0(x_1)=f\circ r_0(x_2)$
we have
$d_{[r_1,r_2]}(r_0x_1,r_0x_2)<\epsilon$ which implies that
$d_{[r_0+r_1,r_0+r_2]}(x_1,x_2)<\epsilon$.
This shows (4).

To see (5),
let $\epsilon>0$,
$K$ (resp. $L$) be a compact metrizable space and $f:X\to K$ (resp. $g:X\to L$) be
a continuous map such that for any $x_1,x_2\in X$ with $f(x_1)=f(x_2)$ (resp. $g(x_1)=g(x_2)$)
we have $d_N(x_1,x_2)<\epsilon$ (resp. $d_M(x_1,x_2)<\epsilon$).
Define $F:X\to K\times L$ by $F(x)=(f(x),g(Nx))$ for every $x\in X$.
Clearly, $K\times L$ is a compact metrizable space and the map $F$ is continuous.
For $x,y\in X$,
if $F(x)=F(y)$ then $f(x)=f(y)$ and $g(Nx)=g(Ny)$, thus we have $d_N(x,y)<\epsilon$
and $d_M(Nx,Ny)<\epsilon$, and hence $d_{N+M}(x,y)<\epsilon$.
It follows that
$\mathrm{Widim}_\epsilon(X,d_{N+M})\le\dim(K\times L)\le\dim(K)+\dim(L)$.
Thus,
$\mathrm{Widim}_\epsilon(X,d_{N+M})\le\mathrm{Widim}_\epsilon(X,d_N)+\mathrm{Widim}_\epsilon(X,d_M)$.
\end{proof}We define the {\bf mean dimension} of a flow $(X,\mathbb{R})$ by:
$$\mathrm{mdim}(X,\mathbb{R})=\lim_{\epsilon\to0}\lim_{N\to\infty}\frac{\mathrm{Widim}_\epsilon(X,d_N)}{N}.$$

The limit exists by the Ornstein-Weiss lemma \cite[Theorem 6.1]{LW} as subadditivity holds.

Next we recall the definition of mean dimension for $\mathbb{Z}$-actions in \cite[Definition 2.6]{LW}.
Let $(X,T)$ be a $\mathbb{Z}$-action.
For $x,y\in X$ and $N\in\mathbb{N}$, denote
$$d^\mathbb{Z}_N(x,y)=\max_{n\in\mathbb{Z}\cap[0,N-1]}d(T^n(x),T^n(y)).$$
Define the mean dimension of $(X,T)$ by:
$$\mathrm{mdim}(X,\mathbb{Z})=\mathrm{mdim}(X,T)=\lim_{\epsilon\to0}
\lim_{N\to\infty(N\in\mathbb{N})}\frac{\mathrm{Widim}_\epsilon(X,d^\mathbb{Z}_N)}{N}.$$
\begin{Proposition}
Let $(X,\mathbb{R})$ be a flow.
If $X$ is finite dimensional
then $\mathrm{mdim}(X,\mathbb{R})=0$.
\end{Proposition}\begin{proof}
We have $\mathrm{Widim}_\epsilon(X,d_N)\le\dim(X)<+\infty$.
The result follows.\end{proof}
Although the definition of mean dimension for $\mathbb{R}$-actions
depends on the metric $d$,
the next proposition shows that the mean dimension of a flow has the same value for
all metrics compatible with the topology.
Therefore mean dimension is an invariant of $\mathbb{R}$-actions.

\begin{Proposition}
Let $(X,\mathbb{R})$ be a flow.
Suppose that $d$ and $d'$ are compatible metrics on $X$.
Then $\mathrm{mdim}(X,\mathbb{R};d)=\mathrm{mdim}(X,\mathbb{R};d')$.
\end{Proposition}

\begin{proof}
Since $d$ are $d'$ are equivalent, the identity map $id:(X,d')\to(X,d)$ is uniformly continuous.
Thus, for every $\epsilon>0$ there is $\delta>0$ with $\delta<\epsilon$ such that
for any $x,y\in X$ with $d'(x,y)<\delta$ we have $d(x,y)<\epsilon$
which implies that
$\mathrm{Widim}_\epsilon(X,d_N)\le\mathrm{Widim}_\delta(X,d'_N)$
for every $N\in\mathbb{N}$.
Noting that $\epsilon\to0$ yields $\delta\to0$ we obtain that
$\mathrm{mdim}(X,\mathbb{R};d)\le\mathrm{mdim}(X,\mathbb{R};d')$.
In the same way we also obtain
$\mathrm{mdim}(X,\mathbb{R};d')\le\mathrm{mdim}(X,\mathbb{R};d)$.
\end{proof}

\begin{Proposition}[{\cite[Def. 2.6]{LW}}]\label{ddz}
Let $(X,\mathbb{Z})$ be a t.d.s. If $d$ and $d'$ are compatible metrics on $X$
then we have $\mathrm{mdim}(X,\mathbb{Z};d)=\mathrm{mdim}(X,\mathbb{Z};d')$.
\end{Proposition}

Note that a flow $(X,(\varphi_r)_{r\in\mathbb{R}})$
naturally induces a ``sub-$\mathbb{Z}$-action'' $(X,\varphi_1)$.
\begin{Proposition}\label{rz}
Let $(X,(\varphi_r)_{r\in\mathbb{R}})$ be a flow.
Then $\mathrm{mdim}(X,(\varphi_r)_{r\in\mathbb{R}})=\mathrm{mdim}(X,\varphi_1)$.
\end{Proposition}\begin{proof}
Recall that for any compatible metric $D$ on $X$ and $R>0$, we denote
$D_R=D_{[0,R]}$. For a flow $(X,d;\mathbb{R})$ and $N\in\mathbb{N}$,
we have
$$(d_1)^\mathbb{Z}_N=(d^\mathbb{Z}_N)_1=d_N.$$
Thus,
$$\mathrm{mdim}(X,\mathbb{R};d)=\mathrm{mdim}(X,\mathbb{Z};d_1).$$
Since $d_1$ and $d$ are compatible metrics on $X$,
by Proposition \ref{ddz} we have
$$\mathrm{mdim}(X,\mathbb{Z};d_1)=\mathrm{mdim}(X,\mathbb{Z};d).$$
Combining the two equalities we have as desired $$\mathrm{mdim}(X,(\varphi_r)_{r\in\mathbb{R}})=\mathrm{mdim}(X,\varphi_1).$$
\end{proof}Thus if the space is not metrizable then we may take
$\mathrm{mdim}(X,\varphi_1)$ as the definition of mean dimension.
\begin{Proposition}
Let $(X,(\varphi_r)_{r\in\mathbb{R}})$ be a flow.
If the topological entropy of $(X,(\varphi_r)_{r\in\mathbb{R}})$ is finite
then the mean dimension of $(X,(\varphi_r)_{r\in\mathbb{R}})$ is zero.
\end{Proposition}\begin{proof}
By \cite[Proposition 8.3.6]{HK} we have
$h_{top}(X,\varphi_1)=h_{top}(X,(\varphi_r)_{r\in\mathbb{R}})$ which is finite.
By \cite[Theorem 4.2]{LW} we have $\mathrm{mdim}(X,\varphi_1)=0$.
By Proposition \ref{rz}, $\mathrm{mdim}(X,(\varphi_r)_{r\in\mathbb{R}})=0$.
\end{proof}The following proposition directly follows from the definition.
\begin{Proposition}
For any flow $(X,(\varphi_r)_{r\in\mathbb{R}})$ and $c\in\mathbb{R}$,
$$\mathrm{mdim}(X,(\varphi_{cr})_{r\in\mathbb{R}})=|c|\cdot\mathrm{mdim}(X,(\varphi_r)_{r\in\mathbb{R}}).$$
\end{Proposition}
\section{Construction of minimal real flows with arbitrary mean dimension}\label{sec:construction}
By defintion $\mathrm{mdim}(X,\mathbb{R})$ belongs to $[0,+\infty]$.
In this section we will show that for every $r\in[0,+\infty]$,
there is a minimal flow $(X,\mathbb{R})$ with $\mathrm{mdim}(X,\mathbb{R})=r$.

Recall that there are natural constructions for passing from a $\mathbb{Z}$-action to a flow,
and vice versa \cite[Section 1.11]{BS}.
Let $(X,T)$ be a $\mathbb{Z}$-action and $f:X\to(0,\infty)$ be a continuous function (in particular bounded away from $0$).
Consider the quotient space (equipped with the quotient topology) $$S_fX=\{(x,t)\in X\times\mathbb{R}^+:0\le t\le f(x)\}/\sim,$$
where $\sim$ is the equivalence relation $(x,f(x))\sim(Tx,0)$.
The {\bf suspension} over $(X,T)$ generated by the {\bf roof function} $f$ is
the flow $(S_fX,(\psi_t)_{t\in\mathbb{R}})$ given by
$$\psi_t(x,s)=(T^nx,s') \text{ for } t\in\mathbb{R} \text{ and } (x,s)\in S_fX,$$
where $n$ and $s'$ satisfy
$$\sum_{i=0}^{n-1}f(T^ix)+s'=t+s, \,\; 0\le s'\le f(T^nx).$$
In other words, flow along $\{x\}\times\mathbb{R}^+$ to $(x,f(x))$ then continue from
$(Tx,0)$ (which is the same as $(x,f(x))$) along $\{Tx\}\times\mathbb{R}^+$ and so on.
When $f\equiv 1$, then $S_fX$ is called the \textbf{mapping torus} over $X$.

Let $d$ be a compatible metric on $X$. Bowen and Walters introduced a compatible metric $\tilde{d}$ on $S_fX$ \cite[Section 4]{BW} known today as the \textbf{Bowen-Walters metric}\footnote{Note that in \cite{BW} it is assumed  that $\mathrm{diam}(X)<1$ but this is unnecessary.}. Let us recall the construction. First assume $f\equiv1$.
We will introduce $\tilde{d}_{S_1X}$ on the space $S_1X$. First, for $x,y\in X$ and $0\le t\le1$ define
the length of the horizontal segment $((x,t),(y,t))$ by:
$$d_h((x,t),(y,t))=(1-t)d(x,y)+td(Tx,Ty).$$
Clearly, we have $d_h((x,0),(y,0))=d(x,y)$ and $d_h((x,1),(y,1))=d(Tx,Ty)$.
Secondly, for $(x,t),(y,s)\in S_1X$ which are on the same orbit define
the length of the vertical segment $((x,t),(y,t))$ by:
$$d_v((x,t),(y,s))=\inf\{|r|:\psi_r(x,t)=(y,s)\}.$$
Finally, for any $(x,t),(y,s)\in S_1X$ define the distance
$\tilde{d}_{S_1X}((x,t),(y,s))$ to be the infimum of the lengths of paths
between $(x,t)$ and $(y,s)$ consisting of a finite number of horizontal and vertical segments. Bowen and Walters showed this construction gives rise to a compatible metric on $S_1X$.
Now assume a continuous function $f:X\to(0,\infty)$ is given. There is a natural homeomorphism $i_f:S_1X\to S_fX$ given by $(x,t)\mapsto (x,tf(x))$. Define $\tilde{d}_{S_fX}=(i_f)_{*}(\tilde{d}_{S_1X})$.

Recall from \cite[Definition 4.1]{LW} that for a $\mathbb{Z}$-action $(X,T)$,
the {\bf metric mean dimension} $\mathrm{mdim}_M(X,d)$ of $X$ with respect to
a metric $d$ compatible with the topology on $X$ is defined as follows.
Let $\epsilon>0$ and $n\in\mathbb{N}$.
A subset $S$ of $X$ is called $(\epsilon,d,n)$-spanning if for every
$x\in X$ there is $y\in S$ such that $d_n^\mathbb{Z}(x,y)\leq\epsilon$.
Set $$A(X,\epsilon,d,n)=\min\{\#S: S\subset X \text{ is } (\epsilon,d,n)\text{-spanning}\}$$
and define
$$\mathrm{mdim}_M(X,T,d)=\liminf_{\epsilon\to0}\frac{1}{|\log\epsilon|}
\limsup_{n\to\infty}\frac{1}{n}\log A(X,\epsilon,d,n).$$

Similarly one may define metric mean dimension for flows but we will not pursue this direction.
\begin{Theorem}[Lindenstrauss-Weiss {\cite[Theorem 4.2]{LW}}]\label{lw42}
For any $\mathbb{Z}$-action $(X,T)$ and any metric $d$ compatible with the topology
on $X$, $$\mathrm{mdim}(X,T)\le\mathrm{mdim}_M(X,T,d).$$
\end{Theorem}\begin{Theorem}[Lindenstrauss {\cite[Theorem 4.3]{L}}]\label{lin43}
If a $\mathbb{Z}$-action $(X,T)$ is an extension of an aperiodic minimal system
then there is a compatible metric $d$ on $X$ such that
$\mathrm{mdim}(X,T)=\mathrm{mdim}_M(X,T,d)$.
\end{Theorem}For related results we refer to \cite[Appendix A]{G17}.
\begin{Proposition}\label{suspmdim}
Let $(Y,(\varphi_r)_{r\in\mathbb{R}})$ be the mapping torus over $(X,T)$ (the suspension generated by the roof function $1$).
Assume that there is a compatible metric $d$ on $X$ with $\mathrm{mdim}_M(X,T,d)=\mathrm{mdim}(X,T)$. Then $$\mathrm{mdim}(X,T)=\mathrm{mdim}(Y,(\varphi_r)_{r\in\mathbb{R}})=\mathrm{mdim}_M(Y,T,\tilde{d}).$$
\end{Proposition}\begin{proof}
By Proposition \ref{rz} we have
$\mathrm{mdim}(Y,(\varphi_r)_{r\in\mathbb{R}})=\mathrm{mdim}(Y,\varphi_1)$.
Since $(X,T)$ is a subsystem of $(Y,T)=(Y,\varphi_1)$,
we have
$\mathrm{mdim}(X,T)\le\mathrm{mdim}(Y,\varphi_1)$.
Note that for every $r\in[0,1)$, $\varphi_r(X)$ is a $\varphi_1$-invariant closed subset of $Y$,
and $(\varphi_r(X),\varphi_1)$ can be regarded as a copy of $(X,T)$.
Let $\epsilon>0$ and $n\in\mathbb{N}$. If $d_{n+1}^\mathbb{Z}(x,y)\leq \frac\epsilon2$ and $|t-t'|\leq \frac\epsilon2$ for $0\le t,t'<1$ then $\tilde{d}_n^{\,\mathbb{Z}}((x,t),(y,t'))\leq \epsilon$. Thus it is easy to see $A(Y,\epsilon,\tilde{d},n)\le ([1/\epsilon]+1)\cdot A(X,\epsilon/2,d,n+1)$.
In particular $$\limsup_{n\to\infty}\frac{1}{n}\log A(Y,\epsilon,\tilde{d},n)\leq \limsup_{n\to\infty}\frac{1}{n}\log A(X,\epsilon/2,d,n)$$ and we obtain that $\mathrm{mdim}_M(Y,\tilde{d})\le\mathrm{mdim}_M(X,d)$.
By Theorem \ref{lw42} we know that $\mathrm{mdim}(Y,\varphi_1)\le\mathrm{mdim}_M(Y,\tilde{d})$.
Summarizing, we have
$$
\mathrm{mdim}(X,T)\le\mathrm{mdim}(Y,\varphi_1)\le\mathrm{mdim}_M(Y,\varphi_1,\tilde{d})$$$$
\le\mathrm{mdim}_M(X,T,d)=\mathrm{mdim}(X,T).
$$
This ends the proof.\end{proof}
We note that for general roof functions Proposition \ref{suspmdim} does not hold.
Indeed Masaki Tsukamoto has informed us that he has constructed
an example of a minimal topological dynamical system $(X,T)$ with compatible metric $d$ and $f\not\equiv 1:X\to(0,\infty)$ such that
$\mathrm{mdim}(X,T)=\mathrm{mdim}_M(X,d)=0$ but $\mathrm{mdim}_M(S_f X,\varphi_1,\tilde{d})>0$ (\cite{TsuPer}).
\begin{Problem}\label{prob:mdim1suspension=mdim}
Is Proposition \ref{suspmdim} always true
without assuming that there is a compatible metric $d$ on $X$
with $\mathrm{mdim}_M(X,d)=\mathrm{mdim}(X,T)$?
\end{Problem}\begin{Problem}
Is it possible to find a topological dynamical system $(X,T)$ with compatible metric $d$ and $f:X\to(0,\infty)$ such that
$\mathrm{mdim}(X,T)=0$ and $\mathrm{mdim}(S_f X,(\varphi_r)_{r\in\mathbb{R}})\neq 0.$
\end{Problem}In Proposition \ref{suspmdim},
if $(X,T)$ is minimal then $(Y,(\varphi_r)_{r\in\mathbb{R}})$ is minimal.
In particular, by Theorem \ref{lin43} we have the following:
\begin{Proposition}\label{mnm}
Suppose that $(X,T)$ is minimal and $(Y,\mathbb{R})$ is be the mapping torus over $(X,T)$ (the suspension generated by the roof function $1$). Then $(Y,\mathbb{R})$ is also minimal
and $\mathrm{mdim}(X,T)=\mathrm{mdim}(Y,\mathbb{R})$.
\end{Proposition}\begin{Proposition}
For every $c\in[0,+\infty]$ there is a minimal flow
$(X,(\varphi_r)_{r\in\mathbb{R}})$ such that
$\mathrm{mdim}(X,(\varphi_r)_{r\in\mathbb{R}})=c$.
\end{Proposition}\begin{proof}
By the $\mathbb{Z}$-version result due to Lindenstrauss and Weiss \cite[Proposition 3.5]{LW} there is
a minimal $\mathbb{Z}$-action $(Y,\mathbb{Z})$ such that $\mathrm{mdim}(Y,\mathbb{Z})=c$.
By Proposition \ref{mnm} we obtain a minimal flow $(X,\mathbb{R})$ with $\mathrm{mdim}(X,\mathbb{R})=c$.
\end{proof}

\section{An embedding conjecture}\label{sec:conj}
We now state the main embedding theorem of this paper.
We recall some necessary notions and results in Fourier analysis. A $C^{\infty}$ function $f:\mathbb{R}\to\mathbb{C}$, is said to be rapidly decreasing if there are constants $M_{n,m}>0$ such that $|f^{(m)}(x)| < M_{n,m} |x|^{-n}$ as $x\rightarrow \infty$,
for all $n,m\in \mathbb{N}$. The space of such function is called the \textit{Schwartz space} and is denoted by $\mathcal{S}$. For $f\in \mathcal{S}$
the definitions of the Fourier transform and its inverse are given by:
$$
\mathcal{F}(f)(\xi)=\int_{-\infty}^\infty e^{-2\pi it\xi}f(t)dt, \,\,\,
\overline{\mathcal{F}}(f)(t)=\int_{-\infty}^\infty e^{2\pi it\xi}f(\xi)d\xi.
$$
One has $\mathcal{F}(\mathcal{S})=\mathcal{S}$, $\overline{\mathcal{F}}(\mathcal{S})=\mathcal{S}$ and for all $f\in\mathcal{S}$, $\overline{\mathcal{F}}(\mathcal{F}(f))=\mathcal{F}(\overline{\mathcal{F}}(f))=f$.
The operators
$\mathcal{F}$ and $\overline{\mathcal{F}}$ can be extended to tempered distributions in a standard way
(for details see \cite[Chapter 7]{Schwartz} and \cite[Chapters 3 \& 4]{Strichartz}).
The tempered distributions include in particular bounded continuous functions.

Let $a<b$ be real numbers.
We define $V[a,b]$ as the space of bounded continuous functions $f:\mathbb{R}\to\mathbb{C}$
satisfying $\mathrm{supp}\mathcal{F}(f)\subset[a,b]$. We denote $B_1(V[a,b])=\{f\in V[a,b]:\,||f||_{\infty}\le1\}$ and $B_1(V^{\mathbb{R}}[-a,a])=\{f\in B_1(V[-a,a]):\,f(\mathbb{R})\subset\mathbb{R}\}$.
One may show that $B_1(V[a,b])$ is a compact metric space
with respect to the distance:
$$\boldsymbol{d}(f_1,f_2)=\sum_{n=1}^\infty\frac{||f_1-f_2||_{L^\infty([-n,n])}}{2^n}.$$
This metric coincides with the standard topology of tempered distributions
(for details see \cite[Chapter 7, Section 4]{Schwartz}).
Let $\mathbb{R}=(\tau_r)_{r\in\mathbb{R}}$ act on $B_1(V[a,b])$ by the shift:
for every $r\in\mathbb{R}$ and $f\in B_1(V[a,b])$,
$(\tau_rf)(t)=f(t+r)$ for all $t\in\mathbb{R}$.
Thus we obtain a flow $(B_1(V[a,b]),\mathbb{R})$.

In \cite[Conjecture 1.2]{LT},
Lindenstrauss and Tsukamoto posed
the following conjecture:
\begin{Conjecture}\label{conj:Z}
Let $(X,T)$ be a $\mathbb{Z}$ dynamical system and $D$ an integer. For $r\in \mathbb{N}$, define $P_r(X,T)=\{x\in X:
\, rx=x\}$.
Suppose that for every $r\in\mathbb N$ it holds that
$\dim P_r(X,T)<\frac {rD}{2}$ and $\mathrm{mdim}(X,T)<\frac D2$.
Then $(X,T)$ can be embedded in the system $(([0,1]^D)^\mathbb{Z},\sigma)$.
\end{Conjecture}
By \cite[Proposition 3.3]{LW}, $\mathrm{mdim}(([0,1]^D)^\mathbb{Z},\sigma)= D$. It is not hard to see that for $r\in \mathbb{N}$, $$\dim P_r(([0,1]^D)^\mathbb{Z},\sigma)= rD.$$ Thus the above conjecture may be rephrased as if
$$\dim P_r(X,T)<\frac {\dim P_r(([0,1]^D)^\mathbb{Z},\sigma)}{2}$$ for all $r\in \mathbb{N}$ and $$\mathrm{mdim}(X,T)<\frac {\mathrm{mdim}(([0,1]^D)^\mathbb{Z},\sigma)}{2}$$ then $(X,T)\hookrightarrow(([0,1]^D)^\mathbb{Z},\sigma)$.
We expect that a similar phenomenon holds for flows where the role of $(([0,1]^D)^\mathbb{Z},\sigma)$ is played by $(B_1(V^{\mathbb{R}}[-a,a]),\mathbb{R})$. By \cite[Footnote 4]{GQT}, $\mathrm{mdim}(B_1(V^{\mathbb{R}}[-a,a]),\mathbb{R})=2a$.
For $r\in\mathbb{R}_{> 0}$ denote $$P_r(X,\mathbb{R})=\{x\in X:\,rx=x\}.$$
We now calculate $\dim P_r(B_1(V^{\mathbb{R}}[-a,a]),\mathbb{R})$.
\begin{Proposition}
Let $r>0$ then  $\dim P_r(B_1(V^\mathbb{R}[-a,a]))=2\lfloor ar\rfloor+1$.
\end{Proposition}
\begin{proof}
  Let $f\in B_1(V^\mathbb{R}[-a,a])$ with $f(x)=f(x+r)$ for all $x\in\mathbb{R}$.
In particular we have a periodic $f\in C^\infty(\mathbb{R},\mathbb{R})$, being a restriction of a holomorphic function,
and hence the Fourier series representation of $f$, $f(x)=\sum_{k=-\infty}^\infty c_ke^\frac{2\pi ikx}{r}$, converges uniformly to $f$
and $c_{-k}=\overline{c_k}$ for all $k$. Since $\mathcal{F}(f)=c_0\mathcal{F}(1)+\sum_{k=1}^\infty
c_k\mathcal{F}(e^\frac{2\pi ikt}{r})+\overline{c_k}\mathcal{F}(e^\frac{-2\pi ikt}{r})$
is supported in $[-a,a]$, we have $c_k=0$ for $|k|>ar$. Let $N=\lfloor ar \rfloor$.
Choose $x_0<x_1<x_2<\dots<x_N$ so that $e^\frac{2\pi i\cdot x_i}{r}\neq e^\frac{2\pi i\cdot x_j}{r}$ for $i\neq j$. The Vandermonde matrix formula indicates that $\det\left(e^\frac{2\pi i\cdot k x_l}{r}\right)_{l,k=0}^{N}\neq 0$.
This implies that  the functions $e^\frac{2\pi ikx}{r}$, $0\le k\le N$ are linearly independent. Thus, we conclude that $\dim P_r(B_1(V^\mathbb{R}[-a,a]))=2\lfloor ar\rfloor+1$.
\end{proof}
We now conjecture:

\begin{Conjecture}\label{conj:R}
Let $(X,\mathbb R)$ be a flow and $a>0$ a real number. Suppose that $\mathrm{mdim}(X,\mathbb{R})<a$ and for every $r\in\mathbb R$, $\dim P_r(X,\mathbb{R})<\lfloor ar\rfloor+\frac{1}{2}$. Then $(X,\mathbb R)$ can be embedded in
the flow $(B_1(V^{\mathbb{R}}[-a,a]),\mathbb{R})$.
\end{Conjecture}
\begin{Problem}
Does Conjecture \ref{conj:R} imply Conjecture \ref{conj:Z}? Does Conjecture \ref{conj:Z} imply Conjecture \ref{conj:R}?
\end{Problem}
We give a very partial answer:

\begin{Proposition}
Assume Conjecture \ref{conj:R} holds. Let $(X,T)$ be a t.d.s such that:
\begin{enumerate}[i.]
  \item $\exists D\in \mathbb N$, $\mathrm{mdim}(X,T)<\frac D2$,
  \item $\exists b\in \mathbb R$, $b<\frac D2$ and $\forall r> \frac{3}{D-2b}, \dim P_r(X,T)<br$,
  \item $\forall r\leq \frac{1}{D-2b}$, $P_r(X,T)=\emptyset$.
      \item
      $\mathrm{mdim}(S_1 X,\mathbb{R})=\mathrm{mdim}(X,T)$
\end{enumerate}
 Then $(X,T)$ can be embedded in the system $(([0,1]^D)^\mathbb{Z},\sigma)$.
\end{Proposition}
\begin{proof}
 Note that the periodic orbits of the suspension $(S_1 X,\mathbb{R})$ have positive integer lengthes and orbits of length $r\in \mathbb{N}$ in $S_1 X$ corresponds to the $r$-periodic points of $(X,T)$ so that $P_r(X,T)=\emptyset$ implies $P_r(S_1 X,\mathbb{R})=\emptyset$ and $P_r(X,T)\neq\emptyset$ implies:
$$\dim P_r(S_1 X,\mathbb{R})=\dim P_r(X,T)+1.$$
Consider the following sequence of embeddings:
$$(X,T)\stackrel{(1)}{\hookrightarrow}  (S_1 X, \psi_1)\stackrel{(2)}{\hookrightarrow} (B_1(V^{\mathbb{R}}[-c,c]),\sigma)\stackrel{(3)}{\hookrightarrow} (([-1,1]^{D})^{\mathbb{Z}},\sigma).$$
Embedding (1) is the trivial embedding from  $(X,T)$  into $(S_1 X, \psi_1)$ where $\psi_1$ is the time-$1$ map. Embedding (3) is a consequence of \cite[Lemma 2.4]{GQT} as long as $c<\frac D2$. We now justify Embedding (2). This $\mathbb{Z}$-embedding is induced from an $\mathbb{R}$-embedding $(S_1 X,\mathbb{R})\hookrightarrow (B_1(V^{\mathbb{R}}[-c,c]),\mathbb{R})$ whose existence follows from Conjecture \ref{conj:R} which we assume to hold. We need to verify the conditions appearing in Conjecture \ref{conj:R}. Let $c$ be a real number such that $\mathrm{mdim}(X,T)<c<\frac D2$. Thus $\mathrm{mdim}(S_1 X,\mathbb{R})=\mathrm{mdim}(X,T)<c$. Let $r$ be an integer such that $r> \frac{3}{D-2b}$, then $\dim P_r(X,\mathbb{R})<br+1$, whereas $\frac 12\dim P_r(B_1(V^{\mathbb{R}}[-c,c]),\shift)=\lfloor rc\rfloor+\frac 12=cr-t_r +\frac 12$, where $0\leq t_r<1$. Note $cr-t_r +\frac 12\geq br+1$ if $(c-b)r\geq\frac 32>t_r+\frac 12$, i.e if $r\geq \frac {3}{2(c-b)}$. Thus it is enough to check it for the minimal integer $r_0$ such that $r_0> \frac{3}{D-2b}=\frac{3}{2(\frac D2-b)}$. We thus choose $b<c<\frac D2$ such that $r_0\geq \frac{3}{2(c-2b)}> \frac{3}{2(\frac D2-2b)}$ and this ends the proof.
\end{proof}

\section{An embedding theorem}\label{sec:embedding}
For every $n\in\mathbb{N}$ denote by $S_n$ the circle of circumference $n!$ (identified with $[0,n!]$).
Let $\mathbb{R}$ act on
$\prod_{n\in\mathbb{N}}S_n$ as follows:
$(x_i)_i\mapsto(x_i+r\;(\text{mod }i!))_i$, $r\in\mathbb{R}$.
Define the {\bf solenoid} (\cite[V.8.15]{NS})
$$S=\{(x_n)_n\in\prod_{n\in\mathbb{N}}S_n:x_n=x_{n+1}\;(\text{mod }n!)\}.$$
It is easy to see that $(S,\mathbb{R})$ is a (minimal) flow.

The following definitions are standard: A continuous surjective map $\psi:(X,\mathbb{Z})\rightarrow (Y,\mathbb{Z})$ is called an \textbf{extension} (of t.d.s) if for all $n\in \mathbb{Z}$ and $x\in X$ it holds $\psi(n.x)=n.\psi(x)$. A continuous surjective map  $\psi:(X,\mathbb{R})\rightarrow (Y,\mathbb{R})$ is called an \textbf{extension} (of flows) if for all $r\in \mathbb{R}$ and $x\in X$ it holds $\psi(r.x)=r.\psi(x)$.

The following embedding result, which is the main result of this paper, provides a partial positive answer to Conjecture \ref{conj:R}. This result may be understood as an analog for flows of \cite[Corollary 1.8]{GT14} which states that Conjecture \ref{conj:Z} is true for any $\mathbb{Z}$-system which is an extension of an aperiodic subshift, i.e. an aperiodic subsystem of a symbolic shift $\left(\{1,2,\dots,l\}^{\mathbb{Z}},\sigma\right)$ for some $l\in \mathbb{N}$.

\begin{Theorem}\label{thm0}
Let $a<b$ be two real numbers.
If $(X,\mathbb{R})$ is an extension of
$(S,\mathbb{R})$ and $\mathrm{mdim}(X,\mathbb{R})<b-a$,
then $(X,\mathbb{R})$ can be embedded in
$(B_1(V[a,b]),\mathbb{R})$.
\end{Theorem}

\begin{cor}
Conjecture \ref{conj:R} holds for $(X,\mathbb{R})$ which is an extension of
$(S,\mathbb{R})$.
\end{cor}
\begin{proof}
Suppose $\mathrm{mdim}(X,\mathbb{R})<a$ for some $a>0$. As $(X,\mathbb{R})$  is an extension of an aperiodic system, it is aperiodic and in particular  for every $r\in\mathbb R$, $\dim P_r(X,\mathbb{R})=0$. We have to show that  $(X,\mathbb R)$ may be embedded in the flow $(B_1(V^{\mathbb{R}}[-a,a]),\mathbb{R})$. Indeed by Theorem \ref{thm0} $(X,\mathbb{R})$ may be embedded in
$(B_1(V[0,a]),\mathbb{R})$. It is now enough to notice that one has the following embedding:
$$B_1(V[0,a])\to B_1(V^{\mathbb{R}}[-a,a]),\,\;\,
\varphi\mapsto\frac{1}{2}(\varphi+\overline{\varphi}).$$
\end{proof}
Since for any flow $(X,\mathbb{R})$,
the product flow $(X\times S,\mathbb R\times\mathbb R)$
is an extension of the flow $(S,\mathbb{R})$,
the following result is a direct corollary of Theorem \ref{thm0}.
\begin{Theorem}
For every flow $(X,\mathbb{R})$ with
$\mathrm{mdim}(X,\mathbb{R})<b-a$
(where $a<b$ are real numbers)
there is an extension $(Y,\mathbb{R})$
with $\mathrm{mdim}(X,\mathbb{R})=\mathrm{mdim}(Y,\mathbb{R})$
that can be embedded in $(B_1(V[a,b]),\mathbb{R})$.
\end{Theorem}
In our proof of Theorem \ref{thm0},
the key step is to embed $(X,\mathbb R)$
in a product flow (Theorem \ref{thm71}):\begin{Theorem}\label{thm71}
Suppose that $a<b$, $\mathrm{mdim}(X,\mathbb{R})<b-a$ and
$\Phi:(X,\mathbb{R})\to (S,\mathbb{R})$ is an extension.
Then for a dense $G_\delta$ subset of $f\in C_\mathbb{R}(X,B_1(V[a,b]))$ the map
$$(f,\Phi):X\to B_1(V[a,b])\times S,\,\;\,x\mapsto(f(x),\Phi(x))$$
is an embedding.\end{Theorem}
\begin{remark}
It is possible to prove a similar theorem where $(S,\mathbb{R})$ is replaced by a solenoid defined by circles of circumference $r_n\rightarrow_{n\rightarrow \infty} \infty$  but we will not pursue this direction.
\end{remark}

The proof is given in the next section. We start by an auxiliary result:
\begin{Proposition}\label{prop:sol}
There is an embedding of $(S,\mathbb{R})$ in $(B_1(V[0,c]),\mathbb{R})$
for any $c>0$.
\end{Proposition}
 \begin{proof}
Define a continuous and $\mathbb{R}$-equivariant map $$\phi:(S,\mathbb{R})\to(B_1(V[0,c]),\mathbb{R})$$ by:
$$S\ni x=(x_n)_n\mapsto f_x(t)=
\sum_{n\ge m(c)}\frac{1}{2^n}\cdot
e^{2\pi i(t+x_n)/n!}=\sum_{n\ge m(c)}(\frac{1}{2^n}\cdot e^{\frac{2\pi i}{n!}x_n})\cdot
e^{\frac{2\pi i}{n!}t}$$
where $m(c)\in\mathbb{N}$ it taken to be sufficiently large so that the (RHS) belongs to $B_1(V[0,c])$.

Assume $f_x(t)=f_y(t)$ for some $x=(x_n)_n,y=(y_n)_n\in S$. We claim $x=y$. This implies that the map is an embedding. Indeed it is enough to show that for all $n$, $\frac{1}{2^n}\cdot e^{\frac{2\pi i}{n!}x_n}=\frac{1}{2^n}\cdot e^{\frac{2\pi i}{n!}y_n}$. This is a consequence of the following more general lemma:

\begin{Lemma}
Let $a_n$ be an absolutely summable series ($\sum |a_n|<\infty$). Let $\lambda_n$ be a pairwise distinct sequence of real numbers bounded in absolute value by $M>0$ ($|\lambda_n|\leq M$). Then $f(z)=\sum a_n e^{i\lambda_n z}$, $z\in \mathbb{C}$, defines an entire function such that $f\equiv 0$ iff $a_n=0$ for all $n$.
\end{Lemma}
\begin{proof} (Compare with the proof of \cite[Theorem I.3.1]{Mandelbrojt72})
We claim $$\lim_{T\rightarrow \infty}\frac{1}{T} \int_0^T f(t)e^{-i\lambda_m t}dt=a_m$$ for all $m$. Thus $f\equiv 0$ implies $a_m=0$ for all $m$. Indeed $$\frac{1}{T} \int_0^T f(t)e^{-i\lambda_m t}dt=\frac{1}{T} \int_0^T \sum_{n\neq m} a_n e^{i(\lambda_n-\lambda_m) t}dt+\frac{1}{T} \int_0^T a_n dt.$$ For $n\neq m$ as $\lambda_n-\lambda_m\neq 0$, we have $$\lim_{T\rightarrow \infty}\frac{1}{T}\int_0^T e^{i(\lambda_n-\lambda_m) t}dt=0.$$ As absolute summability implies one may reorder the limiting operations one has $$\lim_{T\rightarrow \infty}\frac{1}{T} \int_0^T \sum_{n\neq m} a_n e^{i(\lambda_n-\lambda_m) t}dt= \sum_{n\neq m} \lim_{T\rightarrow \infty}\frac{1}{T} \int_0^T a_n e^{i(\lambda_n-\lambda_m) t}dt=0,$$ This  completes the proof.
\end{proof}
\begin{comment}
Alternative proof:

 It is enough to show that if $(b_n)_{n\in\mathbb{N}}\in \ell^1(\mathbb{N})\subset \ell^2(\mathbb{N})$, then:

$$\forall k,m\in\mathbb{N}\,\, \int_{0}^{m!}\sum_{n=1}^{\infty} b_n e^{2\pi it/n!}\cdot e^{-2\pi it/k!}dt=0\Rightarrow \forall n\in\mathbb{N}\,\, b_n=0.$$

Indeed fix $m>k$. We estimate $\int_{0}^{m!} e^{2\pi it/n!}\cdot e^{-2\pi it/k!}dt$ for $n<k$, $n=k$, $k<n\leq m$ and $n>m$. For $n<k$  and $k<n\leq m$ we have under the change of variable $s=\frac{t}{m!}$:
$$\int_{0}^{m!} e^{2\pi it/n!}\cdot e^{-2\pi it/k!}dt=\int_{0}^{1} e^{2\pi is\big((n+1)(n+2)\cdots m -(k+1)(k+2)\cdots m\big)} m!ds=0.$$
For $n=k$ we have:
$$\int_{0}^{m!} e^{2\pi it/k!}\cdot e^{-2\pi it/k!}dt=m!.$$
For $n>m$  we have:
$$\left|\int_{0}^{m!} e^{2\pi it/n!}\cdot e^{-2\pi it/k!}dt\right|\leq m!.$$
Thus:
$$\int_{0}^{m!}\sum_{n=1}^{\infty} b_n e^{2\pi it/n!}\cdot e^{-2\pi it/k!}dt=0\Rightarrow |b_k|\leq \sum_{m+1}^{\infty} |b_n|.$$
Letting $m\to\infty$, we conclude $b_k=0$ for all $k$ as desired.
\end{comment}
\end{proof}
Now we show Theorem \ref{thm0} assuming Theorem \ref{thm71}.
\begin{proof}[Proof of Theorem \ref{thm0}
assuming Theorem \ref{thm71}]
We take $a<c_1<c_2<b$ with $\mathrm{mdim}(X,\mathbb{R})<c_1-a$.
By Theorem \ref{thm71}, $(X,\mathbb{R})$ can be embedded in
$(B_1(V[a,c_1])\times S,\mathbb{R}\times\mathbb{R})$,
which, by Proposition \ref{prop:sol}, can be embedded in
$(B_1(V[a,c_1])\times B_1(V[c_2,b]),\mathbb{R}\times\mathbb{R})$,
and finally embedded in $(B_1(V[a,b]),\mathbb{R})$ by the following embedding:
$$B_1(V[a,c_1])\times B_1(V[c_2,b])\to B_1(V[a,b]),\,\;\,
(\varphi_1,\varphi_2)\mapsto\frac{1}{2}(\varphi_1+\varphi_2).$$
This ends the proof.\end{proof}\section{Embedding in a product}\label{sec:prod embed}
Let $C_\mathbb{R}(X,B_1(V[a,b]))$ be
the space of $\mathbb{R}$-equivariant continuous
maps $f:X\to B_1(V[a,b])$.
This space is nonempty because it contains the constant $0$.
The metric on $C_\mathbb{R}(X,B_1(V[a,b]))$ is chosen to be the uniform distance
$\sup_{x\in X}\boldsymbol{d}(f(x),g(x))$.
This space is completely metrizable and hence is a Baire space
(see \cite[Theorem 48.2]{M}).

We denote by $d$ the metric on $X$.
To prove Theorem \ref{thm71},
it suffices to show that the set
$$\bigcap_{n=1}^\infty\big\{f\in C_\mathbb{R}(X,B_1(V[a,b])):(f,\Phi)\text{ is a }
\frac1n\text{-embedding with respect to }d\big\}$$
is a dense $G_\delta$ subset of $C_\mathbb{R}(X,B_1(V[a,b]))$.
It is obviously a $G_\delta$ subset of $C_\mathbb{R}(X,B_1(V[a,b]))$. Therefore it remains to prove the following:
\begin{Proposition}\label{ppppp}
For any $\delta>0$ and $f\in C_\mathbb{R}(X,B_1(V[a,b]))$,
there is $g\in C_\mathbb{R}(X,B_1(V[a,b]))$ such that:

(1) for all $x\in X$ and $t\in\mathbb{R}$, $|f(x)(t)-g(x)(t)|<\delta$;

(2) $(g,\Phi):X\to B_1(V[a,b])\times S$ is a $\delta$-embedding with respect to $d$.
\end{Proposition}

\medskip

To show Proposition \ref{ppppp},
we prove several auxiliary results.
We start by quoting
\cite[Lemma 2.1]{GT14}:
\begin{Lemma}\label{11111}
Let $(X,d')$ be a compact metric space, and let $F:X\to[-1,1]^M$
be a continuous map.
Suppose that positive numbers $\delta'$ and $\epsilon$ satisfy
the following condition:
\begin{equation}\label{eq:lemma42}
d'(x,y)<\epsilon\Longrightarrow
||F(x)-F(y)||_{\infty}<\delta',
\end{equation}
then if $\mathrm{Widim}_\epsilon(X,d')<M/2$
then there is an $\epsilon$-embedding
$G:X\to[-1,1]^M$ satisfying:
$$\sup_{x\in X}||F(x)-G(x)||_{\infty}<\delta'.$$
\end{Lemma}
We say that a holomorphic  function $g$ in $S\subset\mathbb{C}$ is of \textbf{exponential type} if for all $z\in S$, $|g(z)|\leq C e^{T|z|}$ for some $C,T>0$. The following classical theorem is proven in \cite[Section 3.1.7]{DM}.
\begin{Theorem}[Phragm\'{e}n--Lindel\"{o}f principle]\label{thm:pl}
Let $g$ be a function of exponential type that is holomorphic in the sector
$$S=\left\{z\in \mathbb{C}\,{\big |}\,\alpha <\arg z<\beta \right\}$$ of angle $\beta -\alpha <\pi$, and continuous on its boundary. If $|g(z)|\leq 1$ for $z\in \partial S$ then $|g(z)|\leq 1$ for $z\in S$.
\end{Theorem}

According to the classical Paley-Wiener theorem (\cite[Theorem 19.3]{Rud87}), if $f\in L^2(\mathbb{R})$ extends to an entire function $F$ such that there exist $A,C>0$ such that for all $z=x+iy\in\mathbb{C}$,
$ |f(x+yi)| \leq C e^{2\pi A|y|}$, then $\mathcal{F}(f)\in L^2(\mathbb{R})$ is supported in $[-A,A]$. We will need a generalized version:
\begin{Theorem}\label{thm:gpw}
Let $f\in L^{\infty}(\mathbb{R})$ be a function which extends to an entire function $F:\mathbb{C}\rightarrow\mathbb{C}$ ($F_{|\mathbb{R}}=f$) such that there exist $A,C>0$ and $M\in\mathbb{N}$ such that for all $z=x+iy\in\mathbb{C}$
$$|F(z)|\leq C(1+|z|)^{M}\cdot e^{2\pi A|y|}.$$
Then $f\in V[-A,A]$.
\end{Theorem}
\begin{proof}
See\footnote{While reading the proof in the reference one should note that in \cite{Strichartz} the Fourier transform is defined as $\mathcal{F}(f)(\xi)=\int_{-\infty}^\infty e^{it\xi}f(t)dt$.} \cite[Theorem 7.2.3]{Strichartz}.
\end{proof}
Let $\rho>0$ and $N\in\mathbb N$ so that
$\rho N!\in\mathbb N$. Define:
\begin{comment}
$$L(\rho,N)=\{0,\frac1\rho,\frac2\rho,\dots,N!-\frac1\rho\},$$
\end{comment}
$$L(\rho)=\{\frac k\rho\}_{k\in\mathbb{Z}},\,\, L^*(\rho)=L(\rho)\setminus\{0\}.$$

 In the next lemma we write $x\lesssim y$ for two real numbers $x$ and $y$ if there exists a constant $C>0$ which depends only on $\rho$ and $N$ such that $x\le Cy$.
\begin{Lemma}\label{lem:prod_function}
Let $$f(z)=\lim_{A\to\infty}\prod_{\lambda\in L(\rho),0<|\lambda|<A}\left(1-\frac{z}{\lambda}\right).$$ Then $f$
defines a holomorphic function in $\mathbb{C}$ satisfying
$$f(0)=1,\quad\quad f(\lambda)=0,\quad\forall\;\lambda\in L^*(\rho).$$
Moreover, for all $z\in\mathbb{C}$ we have
$$|f(z)|\lesssim(1+|z|)^{5\rho N!}\cdot e^{\pi\rho|y|},$$
where $y$ is the imaginary part of $z$.
\end{Lemma}
\begin{proof}
We first show the convergence of $f(z)$. Notice
$$f(z)=\lim_{A\to\infty}\prod_{\lambda\in L(\rho),0<\lambda<A}\left(1-\frac{z^2}{\lambda^2}\right)$$
As $\sum_{\lambda\in L(\rho),0<\lambda}\frac{1}{\lambda^2}$ converges, the limit above converges locally uniformly (see \cite[\S 29, Theorems 6 \& 7]{kno51}. Thus, $f(z)$ is a holomorphic function which satisfies
$$f(0)=1,\quad\quad f(\lambda)=0,\quad\forall\;\lambda\in L^*(\rho).$$
Next we shall estimate the growth of $f$ on the real line. Suppose $x>0$ and let $k$ be the integer with $kN!\le x<(k+1)N!$. We may assume $k>0$, as the case $k=0$ is easier and can be dealt with in a similar way. For $n\in\mathbb{Z}$, set
$$L_n=L(\rho)\cap[nN!,(n+1)N!).$$
For $\lambda\in L_n$ with $n\le-2$ or $n\ge k+1$ we have
$$|1-x/\lambda|\le1-x/(n+1)N!$$
and hence
$$\prod_{\lambda\in L_n}\left|1-\frac{x}{\lambda}\right|\le\left|1-\frac{x}{(n+1)N!}\right|^{\rho N!}.$$
For $\lambda\in L_n$ with $1\le n<k$ we have
$$|1-x/\lambda|\le x/(nN!)-1$$
and hence
$$\prod_{\lambda\in L_n}\left|1-\frac{x}{\lambda}\right|\le\left|1-\frac{x}{nN!}\right|^{\rho N!}.$$
The factors for $n=-1,0,k$ need to be treated separately. Recall Euler's sine product formula (\cite{C15}):
$$\frac{\sin z}{z}=\lim_{A\to\infty}\prod_{0<|n|<A}\left(1-\frac{z}{n\pi}\right)$$
Using this it is easy to see that $|f(x)|$ is bounded by
\begin{align*}
&\prod_{0\ne\lambda\in L_{-1}\cup L_0\cup L_k}\left|1-\frac{x}{\lambda}\right|\cdot\lim_{A\to\infty}\prod_{|n|<A,n\ne0,k,k+1}\left|1-\frac{x}{nN!}\right|^{\rho N!}\\
=&\prod_{0\ne\lambda\in L_{-1}\cup L_0\cup L_k}\left|1-\frac{x}{\lambda}\right|\cdot\left|\frac{\sin\frac{\pi x}{N!}}{\frac{\pi x}{N!}\left(1-\frac{x}{kN!}\right)\left(1-\frac{x}{(k+1)N!}\right)}\right|^{\rho N!}.
\end{align*}
The first factor is easy to estimate:
$$\prod_{0\neq \lambda\in L_{-1}\cup L_0\cup L_k}\left|1-\frac{x}{\lambda}\right|\lesssim(1+x)^{3\rho N!}.$$
Set $t=x/N!$,
$$\frac{\sin\frac{\pi x}{N!}}{\frac{\pi x}{N!}\left(1-\frac{x}{kN!}\right)\left(1-\frac{x}{(k+1)N!}\right)}=\frac{k(k+1)\sin\pi t}{\pi t(k-t)(k+1-t)}.$$
By the mean value theorem,
$$\left|\frac{\sin\pi t}{t}\right|\le\pi,\quad\left|\frac{\sin\pi t}{k-t}\right|\le\pi,\quad\left|\frac{\sin\pi t}{k+1-t}\right|\le\pi.$$
Thus,
$$\left|\frac{k(k+1)\sin\pi t}{\pi t(k-t)(k+1-t)}\right|\lesssim k(k+1)\lesssim(1+x)^2.$$
Therefore
$$|f(x)|\lesssim(1+x)^{5\rho N!}.$$
The case $x<0$ is similar so we get
$$|f(x)|\lesssim(1+|x|)^{5\rho N!}.$$
We now turn to estimating $|f(yi)|$ for $y\in \mathbb{R}\setminus\{0\}$. For $r>0$ we set
$$n(r)=\#(L^*(\rho)\cap(-r,r)).$$
We have
$$n(r)< 2\rho r,$$
Note that for $0<r\leq \frac{1}{\rho}$, one has $n(r)=0$. Since $$|f(yi)|^2=\prod_{\lambda\in L^*(\rho)}(1+y^2/\lambda^2),$$
As $n(r)$ is monotonic increasing, we may use the Riemann–Stieltjes integral to write:
$$\log|f(yi)|=\frac{1}{2}\sum_{\lambda\in L^*(\rho)}\log\left(1+\frac{y^2}{\lambda^2}\right)=\frac{1}{2}\int_{\frac{1}{\rho}}^\infty\log\left(1+\frac{y^2}{r^2}\right)dn(r).$$
Using integration by parts for the Riemann–Stieltjes integral (\cite[Theorem 12.14]{G94}), we see that for all $R\geq \frac{1}{\rho}$ it holds:
$$\frac{1}{2}\int_{\frac{1}{\rho}}^{R}\log\left(1+\frac{y^2}{r^2}\right)dn(r)=\frac{1}{2}\left( \log\left(1+\frac{y^2}{r^2}\right)n(r)\bigg|_{\frac{1}{\rho}}^{R}-\int_{\frac{1}{\rho}}^{R}n(r)d\,\log\left(1+\frac{y^2}{r^2}\right)\right).$$
Taking $R\rightarrow \infty$, we conclude:
$$\log|f(yi)|=y^2\int_{\frac{1}{\rho}}^\infty \frac{n(r)}{r(r^2+y^2)}dr$$
Since $n(r)\le 2\rho r$, we deduce
$$\log|f(yi)|\le 2\rho y^2\int_{\frac{1}{\rho}}^\infty\frac{dr}{r^2+y^2}.$$
It is a standard exercise to show:
$$\int_{0}^\infty\frac{dr}{r^2+y^2}=\frac{1}{|y|}\int_{0}^\infty\frac{dr}{1+r^2}=\frac{\pi}{2|y|}.$$
It follows that
$$|f(yi)|\leq e^{\pi\rho |y|}.$$
Finally we show that $|f(z)|$ grows at most exponentially. Let $z=x+yi$. We may assume $x,y>0$, as all the other cases are similar. Let $k$ be the integer with $kN!\le x<(k+1)N!$. Set
$$L'=L(\rho)\setminus(L_{k-1}\cup L_k\cup L_{k+1}).$$
We estimate
$$\prod_{0\ne\lambda\in L_{k-1}\cup L_k\cup L_{k+1}}\left|1-\frac{z}{\lambda}\right|\lesssim(1+|z|)^{3\rho N!}.$$$$\lim_{A\to\infty}\prod_{0\ne\lambda\in L',|\lambda|<A}\left|1-\frac{z}{\lambda}\right|^2=\lim_{A\to\infty}\prod_{0\ne\lambda\in L',|\lambda|<A}\left\{\left(1-\frac{x}{\lambda}\right)^2+\frac{y^2}{\lambda^2}\right\}$$$$=\left\{\lim_{A\to\infty}\prod_{0\ne\lambda\in L',|\lambda|<A}\left(1-\frac{x}{\lambda}\right)^2\right\}\cdot\prod_{0\ne\lambda\in L'}\left\{1+\frac{y^2}{(\lambda-x)^2}\right\}.$$
As in the proof of $|f(x)|\lesssim(1+|x|)^{5\rho N!}$ we estimate
$$\lim_{A\to\infty}\prod_{0\ne\lambda\in L',|\lambda|<A}\left(1-\frac{x}{\lambda}\right)^2\lesssim(1+x)^{12\rho N!}.$$
As in $|f(yi)|\leq e^{\pi\rho|y|}$,
$$\prod_{0\ne\lambda\in L'}\left\{1+\frac{y^2}{(\lambda-x)^2}\right\}\leq e^{2\pi\rho|y|}.$$
Thus, we deduce that $|f(z)|$ grows at most exponentially.

We have thus shown that $f(z)$ has exponential type and satisfies $|f(x)|\lesssim(1+|x|)^{5\rho N!}$ and $|f(yi)|\leq e^{\pi\rho|y|}$. By the Phragm\'en--Lindel\"of principle of Theorem \ref{thm:pl} (e.g. in the first quadrant $x,y\ge0$) applied  to $(1+z)^{-5\rho N!}e^{\pi\rho iz}f(z)$), the claim follows.
\end{proof}

Next we construct an interpolation function based on \cite[pp. 351--365]{Beurling}:
\begin{Proposition}\label{prop:interpolation}
Let $a<b$. Let $\rho>0$ with $\rho\in\mathbb Q$ and $\rho<b-a$. There exists
$\varphi\in V[a,b]$ rapidly decreasing so that $\varphi(0)=1$ and for all $\lambda\in L^*(\rho)$,
$\varphi(\lambda)=0$.
\end{Proposition}\begin{proof}
Fix $\tau>0$ so that $\rho+\tau<b-a$. Let $\psi(\xi)\in \mathcal{S}$ be a nonnegative smooth function in $\mathbb{R}$ satisfying
\[  \supp(\psi)\subset \left[-\frac{\tau}{2}, \frac{\tau}{2}\right], \quad \int_{-\infty}^\infty \psi(\xi)d\xi = 1.\]
Define the function $h:\mathbb{C}\rightarrow \mathbb{C}$ by $$h(z)=\int_{-\frac{\tau}{2}}^{\frac{\tau}{2}} \psi(\xi)e^{2\pi i z \xi}d\xi.$$
It is easy to see that $h$ is  an entire function which satisfies:
\begin{equation}\label{eq:h}
h_{|\mathbb{R}}=\overline{\mathcal{F}}(\psi)\in \mathcal{S}, \quad h(0)=1,\quad
\left|h(x+yi)\right|\le e^{\pi\tau|y|},
\quad\forall x,y\in\mathbb{R}.
\end{equation}
Let $$g(z)=\lim_{A\to\infty}  \prod_{\lambda\in L(\rho), 0<|\lambda|<A}\left(1-\frac{t}{\lambda}\right).$$
By Lemma \ref{lem:prod_function}, $g(z)$ is an entire function. Thus we may define the following entire functions:
$$\tilde{\varphi}(z)=h(z)g(z),\,\, \varphi(z)=e^{\pi iz(a+b)}\tilde{\varphi}(z).$$
It is easy to see that $\varphi(0)=1$ and for all $\lambda\in L^*(\rho)$, $\varphi(\lambda)=0$. By Lemma \ref{lem:prod_function}, $g_{|\mathbb{R}}$ has polynomial growth. Therefore as $\overline{\mathcal{F}}(\psi)$ is rapidly decreasing, so are $\varphi_{|\mathbb{R}}$ and $\tilde{\varphi}_{|\mathbb{R}}$. By Lemma \ref{lem:prod_function} and (\ref{eq:h}) (recall the convention $z=x+iy$):
$$|\tilde{\varphi}(z)|\lesssim(1+|z|)^{5\rho N!}\cdot e^{\pi(\rho+\tau)|y|}$$
As in addition $\tilde{\varphi}_{|\mathbb{R}}$ is bounded (as it is rapidly decreasing), it follows from Theorem \ref{thm:gpw} that
 $\tilde{\varphi}\in V[\frac{-\rho-\tau}{2},\frac{\rho+\tau}{2}]\subset V[\frac{a-b}{2},\frac{b-a}{2}]$. This immediately implies $\varphi\in V[a,b]$ which finishes the proof.
\end{proof}

Now we are ready to prove Proposition \ref{ppppp}.\begin{proof}[Proof of Proposition \ref{ppppp}]
We take $\delta>0$ and $f\in C_\mathbb{R}(X,B_1(V[a,b]))$.
Without loss of generality,
we assume that $|f(x)(t)|\le1-\delta$ for all $x\in X$ and $t\in\mathbb{R}$
(by replacing $f$ with $(1-\delta)f$ if necessary).
Fix $\rho\in\mathbb Q$ with
$$\mathrm{mdim}(X,\mathbb R)<\rho<b-a.$$ Let $\varphi$ be the function constructed in Proposition \ref{prop:interpolation}. As $\varphi$ is a rapidly decreasing
function, we may find $K>0$ such that:
\begin{equation}\label{eq75}
|\varphi(t)|\le\frac{K}{1+|t|^2}.
\end{equation}Let $\delta'>0$ be such that:
\begin{equation}\label{eq71}
\delta'\cdot\sum_{\lambda\in L(\rho)}\frac{K}{1+|t-\lambda|^2}<\delta
\text{ for all }t\in\mathbb{R}.
\end{equation}Fix $\epsilon\in(0,\delta)$. Let $N\in\mathbb N$ be such that
$\rho N!\in\mathbb N$, $\mathrm{Widim}_\epsilon(X,d_{N!})<\rho N!$, and such that
\begin{equation}\label{eq72}
d_{N!}(x,y)<\epsilon\text{ implies }
|f(x)(t)-f(y)(t)|<\frac{\delta'}{2}
\text{ for all }t\in[0,N!].
\end{equation}Define:
$$F:X\to[0,1]^{2\rho N!}=([0,1]^{2})^{\rho N!},\;\;\;
F(x)=(\text{Re}f(x)|_{L(\rho,N)},\text{Im}f(x)|_{L(\rho,N)}).$$
$$F^{\mathbb{C}}:X\to\mathbb{C}^{\rho N!},\;\;\; F^{\mathbb{C}}(x)=f(x)|_{L(\rho,N)}.$$
 Let $M=2\rho N!$, $d'=d_{N!}$.  Equation \eqref{eq72} implies that Equation \eqref{eq:lemma42} holds, so Lemma \ref{11111} implies,
there is an $(d_{N!},\epsilon)$-embedding $G:X\to[-1,1]^{2\rho N!}$ such that $\sup_{x\in X}||F(x)-G(x)||_{\infty}<\frac{\delta'}{2}$.
Similarly to $F^{\mathbb{C}}(x)(k)$, we introduce the notation $G^{\mathbb{C}}(x)(k)$, $k=0,\ldots,\rho N!-1$ in the natural way. Notice it holds:
\begin{equation}\label{deltaprime}
\sup_{x\in X}||F^{\mathbb{C}}(x)-G^{\mathbb{C}}(x)||_{\infty}<\delta'.\end{equation}
Take $x\in X$. Denote $\Phi(x)=(\Phi(x)_n)_{n\in\mathbb N}$,
where $\Phi(x)_n\in S_{n!}$.
For every $n\in\mathbb Z$
let $$\Lambda(x,n)=nN!-\Phi(x)_N+L(\rho,N),$$
$$\Lambda(x)=\bigcup_{n\in\mathbb{Z}}\Lambda(x,n)\subset\mathbb R.$$
\begin{center}\includegraphics[width=10.5cm]{draw}\\Figure 6.1. The set $\Lambda(x,n)$.\end{center}
Next we construct a perturbation $g$ of $f$:
$$g(x)(t)=f(x)(t)+h(x)(t),$$
where $h(x)(t)$ is defined by
$$
\sum_{n\in\mathbb Z}
\sum_{k=0}^{\rho N!-1}
\big(G^{\mathbb{C}}(T^{nN!-\Phi(x)_N}x)(k)-F^{\mathbb{C}}(T^{nN!-\Phi(x)_N}x)(k)\big)
\varphi(t-(\frac k\rho+nN!-\Phi(x)_N))
.$$
As $\varphi$ is rapidly decreasing the sum defining $g(x)$ for fixed $x$ converges in the compact open topology to a function in  $V[a,b]$. Moreover the mapping $x\mapsto g(x)$ is continuous. In order to see that $g(x)$ is $\mathbb{R}$-equivariant, it suffices to deal with $h(x)$ (because $f$ is already $\mathbb{R}$-equivariant). To see that $h(x)$ is $\mathbb{R}$-equivariant we first note that for $0\leq r<N!-\Phi(x)_N$ we have $\Phi(T^rx)_N=\Phi(x)_N+r$ and hence from the definition of $h$ it follows that $h(T^rx)(t)=h(x)(t+r)$. Similarly, if $N!-\Phi(x)_N\leq r<N!$ then $\Phi(T^rx)_N=r-(N!-\Phi(x)_N)$ and hence $(T^{nN!-\Phi(x)_N}T^rx)(k)=(T^{(n+1)N!-\Phi(x)_N-r}T^rx)(k)$. Using such information in each summand in the sum over $k$'s appearing in the definition of $h(T^rx)(t)$, and then substituting $n+1$ by $n$ when summing over $n\in\mathbb{Z}$, we get as desired $h(T^rx)(t)=h(x)(t+r)$ for $r$'s in this range. If $r=sN!$ where $s\in\mathbb{Z}$ then $\Phi(T^rx)_N=r-(sN!-\Phi(x)_N)$ and hence $(T^{nN!-\Phi(T^rx)_N}T^rx)(k)=(T^{(n+s)N!-\Phi(x)_N}x)(k)$. Using this information in each summand in the sum over $k$'s appearing in the definition of $h(T^rx)(t)$, and substituting $n+s$ by $n$ when summing over $n\in\mathbb{Z}$, we obtain as desired $h(T^rx)(t)=h(x)(t+r)$ for $r$'s in this range. Finally if $r=sN!+r'$ where $s\in\mathbb{Z}$ and $0<r'<N!$ we use the additivity properties of the terms involved in order to combine the two cases and get the desired result.
Note that by Equations \eqref{eq75} and \eqref{eq71}  for all $x\in X$ and $t\in \mathbb{R}$:
$$\sum_{n\in\mathbb Z}
\sum_{k=0}^{\rho N!-1}
\varphi(t-(\frac k\rho+nN!-\Phi(x)_N))<\frac{\delta}{\delta'}.$$
By Equation \eqref{deltaprime} for all $x\in X$, $k=0,\ldots,\rho N!-1$:
$$|G^{\mathbb{C}}(T^{nN!-\Phi(x)_N}x)(k)-F^{\mathbb{C}}(T^{nN!-\Phi(x)_N}x)(k)|<\delta'.$$
Combining the two last inequalities we have
$|g(x)(t)-f(x)(t)|<\delta$
for all  $x\in X$ and $t\in\mathbb{R}$.
Since $|f(x)(t)|\le1-\delta$,
we have $g(x)\in B_1(V[a,b])$.
Thus,
$g\in C_\mathbb{R}(X,B_1(V[a,b]))$.
It remains to check that the map
$$(g,\Phi):X\to B_1(V[a,b])\times S,\;\;\;x\mapsto(g(x),\Phi(x))$$
is a $\delta$-embedding with respect to $d$.
We take $x,x'\in X$ with $(g(x),\Phi(x))=(g(x'),\Phi(x'))$. We calculate for $k=0,\ldots,\rho N!-1$:
$$g(x)(-\Phi(x)_N+\frac k\rho)=f(x)(-\Phi(x)_N+\frac k\rho)+
\big(G^{\mathbb{C}}(T^{-\Phi(x)_N}x)(k)-F^{\mathbb{C}}(T^{-\Phi(x)_N}x)(k)\big)
.$$
As $F^{\mathbb{C}}(T^{-\Phi(x)_N}x)(k) =f(T^{-\Phi(x)_N}x)(\frac k\rho)=f(x)(-\Phi(x)_N+\frac k\rho)$, we conclude for $k=0,\ldots,\rho N!~-~1$ that $g(x)(-\Phi(x)_N+\frac k\rho)=G^{\mathbb{C}}(T^{-\Phi(x)_N}x)(k)$. Similarly $g(x')(-\Phi(x')_N+\frac k\rho)=G^{\mathbb{C}}(T^{-\Phi(x')_N}x')(k)$. Thus:
$$g(x)(-\Phi(x)_N+\frac k\rho)=g(x')(-\Phi(x)_N+\frac k\rho)=g(x')(-\Phi(x')_N+\frac k\rho)$$
implies
$$G^{\mathbb{C}}(T^{-\Phi(x)_N}x)(k)=G^{\mathbb{C}}(T^{-\Phi(x')_N}x')(k)=G^{\mathbb{C}}(T^{-\Phi(x)_N}x')(k).$$
Since
$G^{\mathbb{C}}:X\to[0,1]^{\rho N!}$
is an $(d_{N!},\epsilon)$-embedding,
we have
$$d_{N!}(T^{-\Phi(x)_N}x,T^{-\Phi(x)_N}x')<\epsilon<\delta$$ which implies $d(x,x')<\epsilon<\delta$.
This ends the proof.
\end{proof}

\medskip

config: arxiv


TITLE: Riehl 2.2 vii: Path functor and Yoneda Lemma
QUESTION [2 upvotes]: The path functor is defined here:

Now, it doesn't mention it, but it seems clear that Path functor would map continuous mapping $f: X \rightarrow Y$ to post composition by $f$.
Now, the problem asks

Natural automorphisms of the path functor $Path: Top \rightarrow Set$ would be given by
$$\alpha: Path \rightarrow Path$$ such that for topological space $X$, If $g$ is a path in $X$, and $f: X \rightarrow Y$ is a continuous map,
$$f\alpha_X(g) = \alpha_Y(fg)$$
The book says that this is called re-parametrization, but I'm not sure how this coincides with the intuitive definition. I'm not even sure how the naturality condition would say that $g$ and $\alpha_X(g)$ even have the same endpoints.
And to tackle the main question, would it be the right approach to consider the forgestful functor from Top to Set?
Thanks!

REPLY [2 votes]: By Yoneda, each natural transformation $\alpha:\textbf{Path}\to\textbf{Path}$
is induced by a map $\phi:[0,1]\to[0,1]$ in $\textbf{Top}$, that is a continuous map.
Then $\alpha$ is a natural isomorphism iff $\phi$ is a homeomorphism.
So take $\phi:[0,1]\to[0,1]$ to be a homeomorphism. The corresponding $\alpha$
takes a path $p:[0,1]\to X$ to the path $p_1=p\circ\phi$, that is
$$p_1(t)=p(\phi(t)).$$


TITLE: Do Standard Model Yukawa couplings depend on the gauge choice?
QUESTION [4 upvotes]: In the standard model and the Unitary gauge, we write the Higgs field as
$ \phi = \frac{1}{\sqrt{2}} \begin{pmatrix} 0 \\ v + H \end{pmatrix}$
and the Yukawa couplings (leaving out the neutrino terms) are
$
  \mathscr{L}_{\mathrm{Yuk}} =
  - \Gamma^u_{mn} \bar{q}_{\mathrm{L} m} \epsilon \phi u_{\mathrm{R}n}
  - \Gamma^d_{mn} \bar{q}_{\mathrm{L} m} \phi d_{\mathrm{R}n}
  - \Gamma^e_{mn} \bar{\ell}_{\mathrm{L} m} \phi e_{\mathrm{R}n} + \text{h.c.}
$
where $\epsilon$ is the antisymmetric tensor and the $\Gamma$ matrices are related to the mass matrices. I assume these terms are independent of the gauge choice (maybe I'm wrong to think this). Continuing on, we can write the Yukawa couplings in the Unitary gauge
$
  \mathscr{L}_{\mathrm{Yuk}}^{\mathrm{U}} =
  - \Gamma^u_{mn} \bar{u}_{\mathrm{L} m} (H + v) u_{\mathrm{R}n}
  - \Gamma^d_{mn} \bar{d}_{\mathrm{L} m} (H + v) d_{\mathrm{R}n}
  - \Gamma^e_{mn} \bar{e}_{\mathrm{L} m} (H + v) e_{\mathrm{R}n} + \text{h.c.}
$
So far so good.
But I'm confused by how the choice in gauge seems to dictate the Yukawa couplings. I've also seen the unitary gauge written (specifically in Srednicki's Quantum Field Theory) as
$ \phi' = \frac{1}{\sqrt{2}} \begin{pmatrix} v + H \\ 0 \end{pmatrix}$
which seems equally valid since it's also an expansion around the vacuum expectation value of $\phi$. But this would do very weird things to the Yukawa couplings:
$
  \mathscr{L}_{\mathrm{Yuk}}^{\text{?}} =
  - \Gamma^u_{mn} \bar{d}_{\mathrm{L} m} (H + v) u_{\mathrm{R}n}
  - \Gamma^d_{mn} \bar{u}_{\mathrm{L} m} (H + v) d_{\mathrm{R}n}
  - \Gamma^e_{mn} \bar{e}_{\mathrm{L} m} (0) e_{\mathrm{R}n} + \text{h.c.}
$
i.e. the charged leptons become massless and the quark mass terms don't make a lot of sense. Of course this would all be fine if I'd removed the $\epsilon$ in the first term of $\mathscr{L}_{\mathrm{Yuk}}$ and added it to the second two terms, but it seems weird to have this depend on the choice of gauge.
So what am I doing wrong here? Does the choice of gauge ($\phi$ or $\phi'$) depend on the author's choice of how the Yukawa couplings are defined? The way I see the unitary gauge presented in most cases doesn't make any reference to the eventual choice in Yukawa couplings.

REPLY [8 votes]: Very weird things are happening indeed.  Unless you take a step back and think...
By choosing the upper component to carry the vacuum expectation value, you will find that the electric charge $Q=T^3+Y$ appears to no longer be conserved by the vacuum (try it out by making the appropriate transformation).  Instead, the conserved charge is now $Q_\text{new} = -T^3 + Y$.
This means you have to reinterpret the meaning of electric charge as $Q_\text{new}$.  So what then are the electric charge assignments of the fermions?
The upper components of quarks have a new electric charge of $-1/3$ and the lower components have a new charge of $+2/3$.  Clearly, the down quark field must be occupying the upper component, and the up quark field must be occupying the lower component in this new scheme.
Let's look at the leptons: the lower components are what you had previously called the "charged leptons"... not anymore in this new scheme -- the lower components have $Q_\text{new}=0$, and the upper components have charge "$Q_\text{new}=-1$"... they got flipped around too.
Having correctly relabeled all the fields, you should find by inspecting the Yukawa couplings, that the fermions have their correct masses.

So what happened?  By putting the vacuum expectation value in the upper component, you have essentially done an isospin gauge rotation by $\pi/2$.  But when you make an isospin gauge rotation, you must remember to make the rotation for all fields that transform under isospin, and this includes the fermions too.  Since the $T^3$ in $Q=T^3+Y$ also transforms, the electric charge got transformed too.  That is why all the relabeling occurred, and the masses of fermions are unchanged.


TITLE: Associated non-minimal prime ideal
QUESTION [0 upvotes]: I am trying to find an example of a noetherian local ring with an associated prime of height greater or equal 1.
That is,

I want a noetherian local ring $R$ together with an associated prime $p$ such that $\text{ht}(p)\geq 1$.

I have tried thinking about examples such as $R = \left(k[x,y]/(y^2,xy)\right)_{(x,y)}$. Here $(y) = ann(x)$ is an associated prime but it has height 0, because $R$ is not a domain. Does anyone know of an example similar to this?
Also, the local hypothesis is somewhat superfluous as one can just find an example in the non-local case, then localise.

REPLY [0 votes]: Ok, I didn't think hard enough about my example. In $R = k[x,y]/(y^2,xy)$, $ann(y)$ contains $x$ and $y$ hence $(x,y)\subset ann(y)$ thus $ann(y) = (x,y)$ are the latter is maximal. But then $(y)\subset (x,y)$ and $(x,y)$ and $(y)$ are prime in $R$. Thus $(x,y)$ furnishes an example that I want. (note that we can localise $(k[x,y]/(y^2,xy))_{(x,y)}$ to get a local example).