TITLE: Uniform convergence of $\frac{1}{x^3}\cos\left(\frac{x}{n}\right)$
QUESTION [1 upvotes]: Determine if $\frac{1}{x^3}\cos\left(\frac{x}{n}\right)$ uniformly converges on $E_1$ and $E_2$, where $E_1 = (0, 1)$ and $E_2 = (1, +\infty)$.
The pointwise convergence is trivial, since $\lim \limits_{n \to \infty} \frac{1}{x^3}\cos(\frac{x}{n}) = \frac{1}{x^3}$ for both $E_1$ and $E_2$.
I could also prove that it does not uniformly converge on $E_1$ by picking $x_n = \frac{1}{n^2}$ and plugging it into the original function.
However, I don't know how to prove uniform convergence on $E_2$, because I don't know how to evaluate $\sup f_n(x)$.
What are the methods to prove or disprove uniform convergence on $E_2$?

REPLY [1 votes]: As regards $E_2=(1, +\infty)$ , note that by letting $t=x/n$,
$$\sup_{x\in(1, +\infty)} \left|\frac{1}{x^3}-\frac{1}{x^3}\cos\left(\frac{x}{n}\right)\right|=\sup_{t\in(1/n, +\infty)} \frac{1-\cos(t)}{n^3t^3}\leq \frac{1}{2n^3}\sup_{t\in(1/n, +\infty)} \frac{1}{t}=\frac{n}{2n^3}=\frac{1}{2n^2}$$
where we used the fact that for $\frac{1-\cos(t)}{t^2}\leq \frac{1}{2}$ for any $t>0$ (it suffices to show that this function is bounded).


TITLE: Trace zero matrix
QUESTION [1 upvotes]: I have seen a problem where for a real matrix $A$ if $A^2=-I$ then trace $A$ is zero but I didn't understand why it is not true in general for a complex matrix $A.$

REPLY [2 votes]: because a complex matrix can be:
$$A=
\begin{bmatrix}
i&0\\
0&i
\end{bmatrix}
$$
So:
$$
A^2=\begin{bmatrix}
i&0\\
0&i
\end{bmatrix}
\begin{bmatrix}
i&0\\
0&i
\end{bmatrix}=
\begin{bmatrix}
-1&0\\
0&-1
\end{bmatrix}=-I
$$
and note that this is true also for an $n \times n$ diagonal matrix that has all diagonal elements $a_{hh}=i$


text:
\section{Asymptotic analysis: Part II}\label{Section6}  The purpose of the present section is to prove Proposition \ref{PropTermBound} . In Section \ref{Section6.1} we summarize various results that will be used in the proof, which is given in Section \ref{Section6.2}.

\subsection{Technical lemmas}\label{Section6.1} In this section we state several lemmas that go into the proof of Proposition \ref{PropTermBound}. The proofs of these lemmas can be found in Section \ref{Section8.2}.

The first result we require is as follows.
\begin{lemma}\label{LCDetBound} Suppose that $0 < r < R$ are real numbers and $\alpha \in [0, (r/R)^{1/2})$. Then for any $N \in \mathbb{N}$ and $z_i, w_i \in \mathbb{C}$ with $|w_i| = r$ and $|z_i| = R$ for $i =1, \dots, N$ we have
\begin{equation}\label{CDetBound}
\left|\det \left[ \frac{1}{z_i - w_j}\right]_{i,j=1}^N \right| \leq \frac{N^{N/2}}{(R\alpha - r\alpha^{-1})^N} \cdot \left(\frac{R \alpha + r\alpha^{-1} }{r + R} \right)^{N^2}.
\end{equation}
\end{lemma}

The second result we require is as follows.
\begin{lemma}\label{S5techies}
Let $t \in (0, 1)$ and suppose that $U \in \mathbb{R}$ satisfies $0 < U \leq [-\log t]/4$. Suppose further that $z \in \mathbb{C}$ is such that $Re(z) \in [U, [-\log t]/2]$. Then there exists a constant $C_t^1 > 0$, depending on $t$ alone, such that the following holds
\begin{equation}\label{S5yellow1}
 \sum_{k \in \mathbb Z} \left| \frac{1}{\sin(-\pi [z + 2\pi\iota k]/ [-\log t])}\right| \leq \frac{C_t^1}{U}.
\end{equation}
\end{lemma}

The third result we require is as follows.
\begin{lemma}\label{descentLemma2} Let $S_a(z)$ be as in (\ref{FunExp}). There exist universal constants $C_1, C_2> 0$ such that the following holds. Let $a \in (0,e^{-2\pi}]$ and $\delta \in [0, \pi]$ be given. Then for $y\in [-\pi, \pi]$ we have
\begin{equation}\label{SaBoundVert}
\pm Re \left[ S_a(\pm \delta + \iota y  )  \right] \leq - a \cdot \delta \cdot \left[ C_1 \cdot y^2 - C_2 \cdot \delta^2   \right] .
\end{equation}
\end{lemma}

The fourth result we require is as follows.
\begin{lemma}\label{descentLemma3} Let $R_a(z)$ be as in (\ref{FunExp}). Suppose that $a \in (0, e^{-2\pi}]$ and $z = x + \iota y$ with $x, y \in [-\pi, \pi]$. There is a universal constant $C_3 > 0$ such that
\begin{equation}\label{RBound2}
-C_3 \cdot a \cdot|z|^2 \leq Re[R_a(z)]  \leq  C_3 \cdot a \cdot  |z|^2.
\end{equation}
\end{lemma}

The fifth result we require is as follows.
\begin{lemma}\label{descentLemma4} There is a universal constant $C_4 > 0$ such that for all $\delta \in [-\pi, \pi]$ we have
\begin{equation}\label{HypSineBound}
\frac{e^{\delta/2} + e^{-\delta/2} }{ e^{\delta} + e^{-\delta} } \leq e^{-C_4 \delta^2}.
\end{equation}
\end{lemma}

The sixth result we require is as follows.
\begin{lemma}\label{descentLemma5} For any $x \in [0, 1/2]$ we have
\begin{equation}\label{RatFracIneq}
\frac{1 + x}{1 - x} \leq e^{3x}.
\end{equation}
\end{lemma}

The seventh result we require is as follows.
\begin{lemma}\label{LemmaCDet2} There exists a function $f: (0,\infty) \rightarrow (0,\infty)$ such that the following holds. Let $\delta \in (0, 1],$ $c > 0$ and $N \in \mathbb{N}$. Suppose that $x_i, y_i \in \mathbb{R}$ for $i = 1, \dots ,N$  are such that
\begin{equation}\label{squaresBound}
\sum_{i = 1}^N (x_i^2 +  y_i^2) \leq N c^2 \delta^2 .
\end{equation}
Then
\begin{equation}\label{CDetSquare}
\left| \det \left[ \frac{1}{e^{\delta} e^{\iota x_i} - e^{-\delta} e^{\iota y_j}} \right]_{i,j = 1}^N\right| \leq  \frac{e^{-f(c) N^2} N^{N/2}}{(e^{\delta/2} - e^{-\delta/2})^N}.
\end{equation}
\end{lemma}

The eighth result we require is as follows.
\begin{lemma}\label{BoundMixedS} There exists a universal constant $C_5 > 0$ such that the following holds. For any $t \in (0, e^{-2\pi}]$ we have that
\begin{equation}
\frac{(-te^2;t)_\infty}{(t e^2;t)_\infty} \leq \exp \left( tC_5 \right).
\end{equation}
\end{lemma}

The ninth result we require is as follows.
\begin{lemma}\label{LBoundQ} Let $z = x+ iy$ with $|x| \leq 2\pi$, $|y| \leq 2\pi$ and $t \in (0, e^{-4\pi}]$. For such a choice of $z$ and $t$ define the function
\begin{equation}\label{DefQ}
Q_t(z) = \log (te^z;t)_\infty - \log(t;t)_\infty + z \cdot \sum_{n = 1}^\infty \frac{t^n}{1 - t^n}.
\end{equation}
Then we have
\begin{equation}\label{QBound}
-2 t|z|^2 \leq Re[Q_t(z)]  \leq 2t |z|^2.
\end{equation}
\end{lemma}

\subsection{Proof of Proposition }\label{Section6.2} In this section we give the proof of Proposition \ref{PropTermBound}. We will follow the notation from Section \ref{Section4.1} and Section \ref{Section6.1} above. For clarity we split the proof into several steps. In the first step we specify $a_2,t_2$, fix the parameters $a, t, x_1, x_2, s_1, s_2$ and specify $M_2$ as in as in the statement of the proposition. In the second step we formulate a certain inequality for $|I_M(N_1, N_2,k)|$ as (\ref{BoundEachTerm}), where we recall that $I_M(N_1, N_2,k)$ were defined in equation (\ref{INK}). Assuming the inequality (\ref{BoundEachTerm}) we deduce the statement of the proposition. The next three steps establish (\ref{BoundEachTerm})  by considering the cases when $N$ is bigger, roughly of the same size as or much smaller than $\Delta = N_1 + N_2$. \\

{\bf \raggedleft Step 1.} In this step we formulate our choice of $a_2$, $t_2$ and $M_2$ as in the statement of the proposition. Let us denote for simplicity
\begin{equation}\label{defp}
p =\frac{2^{1/2} + 2^{-1/2}}{5/2},
\end{equation}
which is a fixed constant in $(0,1)$ that will be used later in the arguments. We set $c =  \sqrt{\frac{8(C_2+C_5)}{C_1}}$, where $C_1, C_2$ are as in Lemma \ref{descentLemma2} and $C_5$ is as in Lemma \ref{BoundMixedS}. We also pick $R > 1$ sufficiently large so that
\begin{equation}\label{defR}
p^{-1/8} \geq e^{18/R} \mbox{ and } \frac{2C_2 + 2C_5}{R} \leq \frac{1}{8} \cdot f\left( c\right),
\end{equation}
where the function $f$ is the one afforded to us by Lemma \ref{LemmaCDet2}.

We fix $a_2, t_2  \in (0,1)$ sufficiently small so that the following conditions hold:
\begin{equation}\label{Cond1}
t_2 \leq e^{-4\pi} \mbox{ and }a_2 \leq e^{-2\pi};
\end{equation}
\begin{equation}\label{Cond2}
 \left(\frac{(-4t_2;t_2)_\infty}{(4t_2;t_2)_\infty} \right) \leq p^{-1/8};
\end{equation}
\begin{equation}\label{Cond3}
t_2 \leq R^{-1}.
\end{equation}

The above fixes our definition of $a_2$ and $t_2$ and we fix $a \in (0, a_2]$ and $t \in (0, t_2]$ as well as $x_1, x_2  \in \mathbb{R}$ and $s_1, s_2 \in \mathbb{R}$ such that $s_1 > s_2$. For this choice of parameters we let $n_1, n_2, u_1, u_2$ be the $M$-dependent functions from Definition \ref{constants}. We proceed to specify $M_2$ as in the statement of the proposition.
We pick $M_2$ sufficiently large depending on $x_1, x_2, s_1, s_2, a ,t$ so that the following inequalities all hold for $M \geq M_2$:
\begin{equation}\label{M1}
\begin{split}
M \cdot f_1 +   M^{1/3}  \sigma_a  x_i + f_1'   [M - n_i] + (1/2) f''_1 s_i^2 M^{1/3} \geq 0 \mbox{ for $i = 1,2$};
\end{split}
\end{equation}
\begin{equation}\label{M3}
\begin{split}
2M  \geq n_i  \mbox{ for $i = 1,2$};
\end{split}
\end{equation}
\begin{equation}\label{M4}
\begin{split}
M^{-1/6} \leq (s_1-s_2)(\epsilon_1/2);
\end{split}
\end{equation}
\begin{equation}\label{M2}
\begin{split}
& C_1 (Ra)^{-1/3} M^{5/6} - 2 C_3 |M-n_i| \geq 0, \mbox{ for $i = 1,2$}
\end{split}
\end{equation}
where $\epsilon_1$ is as in Lemma \ref{LemmaTaylor}, $C_2$ and $C_3$ are the universal positive constants from Lemmas \ref{descentLemma2} and \ref{descentLemma3} respectively and $R$ is as in (\ref{defR}). We recall that $f_1, f_1', f_1'', \sigma_a$ were all defined in (\ref{eqnConst}) and that $M-n_i = O(M^{2/3})$, which is why the choice of $M_2$ is possible. This fixes our choice of $M_2$.

For future use, we let $C_\alpha$ be sufficiently large depending on $x_1, x_2, s_1, s_2, a$ so that
\begin{equation}\label{Calpha}
2\pi \cdot [ (|x_1| + |x_2| )  \cdot \sigma_a + |f_1''| (s_1^2 + s_2^2)]\leq C_\alpha.
\end{equation}

{\bf \raggedleft Step 2.} Recall from (\ref{DefIN}) that
\begin{equation}\label{S6INKD}
I_M(N_1,N_2) = \sum_{k = 0}^{\min(N_1,N_2)} I_M(N_1, N_2, k),
\end{equation}
where $I_M(N_1,N_2,k)$ are as in (\ref{INK}) -- we recall this formula here for the reader's convenience in a form that will be easier to work with below.
\begin{equation}\label{S6INK}
\begin{split}
&I_M(N_1, N_2, k) = \frac{(-1)^{N_1 + N_2}}{k! (N_1- k)! (N_2- k)!}  \int_{\gamma_{1}^{N_1-k}} \int_{\gamma^{N_1-k}_3}  \int_{\gamma_2^{N_2-k}}\int_{\gamma^{N_2-k}_{4}} \int_{\gamma^{k}_1}\int_{\gamma^{k}_3} \int_{\gamma^k_4} \\
& A_1^M  A_2^M A_3^M A_4^M\prod_{i = 1}^{k}\frac{d\hat{w}^2_i}{2\pi \iota}\prod_{i = 1}^{k}\frac{d{w}^2_i}{2\pi \iota}\prod_{i = 1}^{k}\frac{dz^2_i}{2\pi \iota}\prod_{i = 1}^{N_2 - k}\frac{d\hat{w}^1_i}{2\pi \iota} \prod_{i = 1}^{N_2 - k} \frac{d\hat{z}_i}{2\pi \iota} \prod_{i = 1}^{N_1 - k}\frac{dw^1_i}{2\pi \iota}\prod_{i = 1}^{N_1- k}\frac{dz^1_i}{2\pi \iota}.
\end{split}
\end{equation}
In the above formula we have that $A_1^M = \det B$ with $B$ being the matrix from (\ref{BlockMatrix}). Also we have
\begin{equation}\label{C1DefA2}
\begin{split}
&A^M_2 = \prod_{i = 1}^{N_1-k} \prod_{j = 1}^{N_2-k}\frac{(t\hat{z}^1_j/ z^1_i; t)_\infty (t\hat{w}^1_j / w^1_i; t)_\infty}{(t\hat{w}^1_j/ z^1_i;t)_\infty (t\hat{z}^1_j/ w^1_i;t)_\infty} \cdot  \prod_{i = 1}^{k} \prod_{j = 1}^{N_2-k} \frac{(t\hat{z}^1_j / z^2_i;t)_\infty (t\hat{w}^1_j/ w^2_i;t)_\infty}{(t\hat{w}^1_j/ z^2_i;t)_\infty (t\hat{z}^1_j/ w^2_i; t)_\infty} \cdot \\
&   \prod_{i = 1}^{N_1-k}\prod_{j = 1}^{k} \frac{(tw^2_j/ z^1_i;t)_\infty (t\hat{w}^2_j/ w^1_i;t)_\infty}{(t\hat{w}^2_j/ z^1_i;t)_\infty (tw^2_j/ w^1_i; t)_\infty} \cdot \prod_{i = 1}^{k} \prod_{j = 1}^{k} \frac{(tw^2_j/ z^2_i)_\infty (t\hat{w}^2_j/ w^2_i;t)_\infty}{(t\hat{w}^2_j/ z^2_i;t)_\infty (tw^2_j/ w^2_i;t)_\infty};
\end{split}
\end{equation}
\begin{equation}\label{C1DefA3}
\begin{split}
A^M_3 = &\prod_{i =1}^{N_2-k}  [-u_2]^{[\log(\hat{w}_i^1) - \log(\hat{z}_i^1)][\log t]^{-1} }\left( \frac{1 + a\hat{z}^1_i}{1 + a \hat{w}_i^1}\right)^M \cdot \left(\frac{1 + a/\hat{w}^1_i}{1 + a/\hat{z}_i^1} \right)^{n_2} \cdot \\
& \prod_{i =1}^{N_1-k} [-u_1]^{[\log(w_i^1) - \log(z_i^1)][\log t]^{-1} }  \left( \frac{1 + az_i^1}{1 + a w_i^1}\right)^M  \left(\frac{1 + a/w_i^1}{1 + a/z_i^1} \right)^{n_1} \cdot \\
& \prod_{i =1}^{k}[-u_2]^{[\log(\hat{w}_i^2) - \log(w_i^2)][\log t]^{-1} } [-u_1]^{[\log(w_i^2) - \log(z_i^2)][\log t]^{-1}}  \cdot  \\
& \left( \frac{1 + az_i^2}{1 + a \hat{w}_i^2}\right)^M  \left(\frac{1 + a/\hat{w}^2_i}{1 + a/ z_i^2} \right)^{n_2} \left(\frac{1 + a/z_i^2}{1 + a/w_i^2} \right)^{n_2- n_1};   \\
\end{split}
\end{equation}
\begin{equation}\label{C1DefA4}
\begin{split}
&A^M_4 =  \prod_{i =1}^{N_2-k} \frac{\hat{S}(\hat{w}_i^1,\hat{z}^1_i; u_2, t)}{ \log t \cdot \hat{w}_i^1 } \cdot \prod_{i =1}^{k} \frac{\hat{S}(\hat{w}_i^2,w^2_i; u_2, t)}{ \log t \cdot \hat{w}_i^2}  \frac{\hat{S}(w_i^2,z_i^2; u_1, t)}{ \log t \cdot w_i^2}   \cdot \prod_{i =1}^{N_1-k} \frac{\hat{S}(w^1_i,z_i^1; u_1, t)}{\log t \cdot w_i^1 } , \\
& \mbox{ where } \hat{S}(w,z;u,t) =  \sum_{m \in \mathbb{Z}} \frac{\pi \cdot [ - u ]^{-  2m \pi  \iota [\log t ]^{-1}}}{\sin(-\pi [[\log w - \log z] [\log t]^{-1} -  2m \pi  \iota [\log t ]^{-1}])}.
\end{split}
\end{equation}

We claim that there is a constant $\tilde{C}_2> 0$ that depends on $a,t, x_1, x_2, s_1,s_2$  such that for each $M, N_1, N_2 \in \mathbb{N}$ with $M \geq M_2$ as in Step 1 and $k \in \{1 , \dots, \min(N_1, N_2)\}$ we have
\begin{equation}\label{BoundEachTerm}
|I_M(N_1, N_2, k)| \leq \tilde{C}_2^{\Delta} \cdot  \Delta^{-\Delta/4}, \mbox{ with $\Delta = N_1 + N_2$. }
\end{equation}
From (\ref{S6INKD}) and (\ref{BoundEachTerm}) we conclude that $|I_M(N_1, N_2) | \leq \Delta \cdot \tilde{C}_2^{\Delta} \cdot  \Delta^{-\Delta/4}$, which clearly implies the statement of the proposition.

In the remainder of the proof we prove (\ref{BoundEachTerm}) by considering the three cases: (1) $\Delta \geq RaM$, (2) $\Delta  \in [M^{1/2}, RaM]$ and (3) $\Delta  \in [1, M^{1/2}],$ where $R$ is the universal constant from (\ref{defR}).\\

{\bf \raggedleft Step 3.}  In this and all the steps below we will denote by $C$ a generic constant that depends on $a$ and $t$ alone, whose value may change from line to line. The purpose of this step is to prove (\ref{BoundEachTerm}) if $\Delta \geq RaM$.\\

Let $C_+, C_-$ and $C_m$ be the positively oriented circles, centered at the origin, of radius $1/2$, $2$ and $1$ respectively.
Observe that by Lemma \ref{S3Analyticity} and Cauchy's theorem we may deform the $z_i^1$, $\hat{z}_i^1$ and $z_i^2$ contours in  (\ref{S6INK}) to $C_+$, the $w_i^1$, $\hat{w}_i^1$ and $\hat{w}_i^2$ contours to $C_-$ and the $w_i^2$ contours to $C_m$ without affecting the value of the integral. Here we used $a \in (0, a_2]$ with $a_2 \leq e^{-2\pi}$ -- see (\ref{Cond1}) -- which ensures we do not cross any poles in the process of the deformation. We now proceed to find appropriate upper bounds for each of the four terms $A^M_1, A^M_2, A^M_3$ and $A^M_4$ along these contours.\\

Let us first analyze $A^M_4$. Notice that if $u \in (-\infty, 0)$ we have that
$$\left| [ - u ]^{-  2m \pi  \iota [\log t ]^{-1}}\right| = 1$$
for all $m \in \mathbb{Z}$. Consequently, if $1/2= |w|$ and $|z| = 2$ we have from Lemma \ref{S5techies} applied to $U = 2 \log 2$
$$|\hat{S}(w,z;u_i,t)|  \leq \pi \cdot \sum_{k \in \mathbb Z} \left| \frac{1}{\sin(-\pi[\log z - \log w + 2\pi\iota k]/[-\log t])}\right| \leq  \frac{\pi C_t^1}{2 \log 2} \mbox{ for $i =1 ,2$},$$
where $C_t^1$ is as in Lemma \ref{S5techies} and depends on $t$ alone. In applying the above lemma we used that $t \leq t_2 \leq e^{-4\pi}$ -- see (\ref{Cond1}) -- this ensures that $U = 2 \log 2 \leq [-\log t]/4$.
The above inequality (using our definition of $C_+$ and $C_-$) implies that
$$\left| \prod_{i =1}^{N_2-k} \frac{\hat{S}(\hat{w}_i^1,\hat{z}^1_i; u_2, t)}{ \log t \cdot \hat{w}_i^1} \cdot \prod_{i =1}^{N_1-k} \frac{\hat{S}(w^1_i,z_i^1; u_1, t)}{\log t \cdot w^1_i} \right| \leq C^{N_1 +N_2 - 2k},$$
where we recall that $C$ stood for a generic constant that depends on $a$ and $t$ alone, whose value changes from line to line. We will not mention this further. We may again apply Lemma \ref{S5techies} for $U = \log 2$ to conclude that if $1/2 = |w| < |z| = 1$ or $1 = |w| < |z| = 2$ we have
$$|\hat{S}(w,z;u_i,t)|  \leq   \frac{\pi C_t^1}{\log 2} \mbox{ for $i =1 ,2$}.$$
The above inequality (using our definition of $C_{+}, C_-$ and $C_m$) implies that
$$\left|  \prod_{i =1}^{k} \frac{\hat{S}(\hat{w}_i^2,w^2_i; u_2, t)}{ \log t \cdot \hat{w}_i^2 }  \frac{\hat{S}(w_i^2,z_i^2; u_1, t)}{ \log t \cdot w_i^2}  \right| \leq C^{2k}.$$
Combining the above two inequalities we conclude that
\begin{equation}\label{C1BoundH4}
\begin{split}
&|A^M_4|  \leq C^{\Delta}.
\end{split}
\end{equation}

We next analyze $A^M_2$. Recall from (\ref{S3PochBound}) that if $|\zeta| \leq \alpha < 1$ we have that
\begin{equation*}
(\alpha;t)_\infty = \prod_{n = 0}^\infty( 1 - \alpha t^n) \leq \prod_{n = 0}^\infty | 1 - \zeta t^n| = | (\zeta;t)_\infty | \leq \prod_{n = 0}^\infty( 1 + \alpha t^n)  = (-\alpha; t)_\infty.
\end{equation*}
Using the latter inequalities and our definition of $C_-, C_+, C_m$ we conclude that
\begin{equation}\label{C1BoundH2}
\begin{split}
&|A^M_2|  \leq \left( \frac{(-4t; t)_\infty}{(4t; t)_\infty}\right)^{2N_1N_2} \leq \left( \frac{(-4t; t)_\infty}{(4t; t)_\infty}\right)^{\Delta^2/2}.
\end{split}
\end{equation}
In deriving the above inequality we used that $t \leq t_2 \leq e^{-4\pi}$ -- see (\ref{Cond1}) -- which ensures that $4t <1$. \\

We continue with the analysis of $A^M_3$. Using the definition of $u_1$ and $u_2$ from (\ref{ScaleU}) and the definition of the contours $C_-, C_+$ and $C_m$ we have for each $i = 1, \dots, N_2 -k$
$$\left|[-u_2]^{[\log(\hat{w}_i^1) - \log(\hat{z}_i^1)][\log t]^{-1} }\hspace{-0.5mm} \right| \hspace{-0.5mm}=\hspace{-0.5mm} \exp \left( \hspace{-0.5mm} -2 \log 2  \hspace{-0.5mm} \left( \hspace{-0.5mm} M  f_1 +   M^{1/3}  \sigma_a  x_2 + f_1'   [M - n_2] \hspace{-0.5mm} +\hspace{-0.5mm}  (1/2) f''_1 s_2^2 M^{1/3}\hspace{-0.5mm} \right) \hspace{-1mm} \right) \hspace{-0.5mm} \leq 1;$$
for each $i = 1, \dots, N_1 - k$
$$ \left| [-u_1]^{[\log(w_i^1) - \log(z_i^1)][\log t]^{-1} }\hspace{-0.5mm} \right| \hspace{-0.5mm} =\hspace{-0.5mm} \exp \hspace{-0.5mm}\left(\hspace{-0.5mm}  - 2\log 2  \left(  M f_1 +   M^{1/3}  \sigma_a  x_1\hspace{-0.5mm} +\hspace{-0.5mm}  f_1'   [M - n_1] \hspace{-0.5mm} +\hspace{-0.5mm}  (1/2) f''_1 s_1^2 M^{1/3}\hspace{-0.5mm} \right)\hspace{-1mm} \right) \hspace{-0.5mm} \leq 1; $$
for each $i = 1, \dots , k$
$$\left| [-u_2]^{[\log(\hat{w}_i^2) - \log(w_i^2)][\log t]^{-1} } \hspace{-0.5mm}\right| \hspace{-0.5mm}= \hspace{-0.5mm}\exp\hspace{-0.5mm} \left( \hspace{-0.5mm} - \log 2  \left( \hspace{-0.5mm}  M  f_1 +   M^{1/3}  \sigma_a  x_2 + f_1'   [M - n_2] \hspace{-0.5mm} +\hspace{-0.5mm}  (1/2) f''_1 s_2^2 M^{1/3} \hspace{-0.5mm} \right) \hspace{-1mm} \right) \hspace{-0.5mm} \leq 1;$$
$$\left| [-u_1]^{[\log(w_i^2) - \log(z_i^2)][\log t]^{-1}}\hspace{-0.5mm} \right| \hspace{-0.5mm} = \hspace{-0.5mm} \exp \hspace{-0.5mm}\left( \hspace{-0.5mm} - \log 2   \left( M  f_1 +   M^{1/3}  \sigma_a  x_1 + f_1'   [M - n_1] + (1/2) f''_1 s_1^2 M^{1/3}\hspace{-0.5mm} \right) \hspace{-1mm} \right) \hspace{-0.5mm} \leq 1, $$
where in all of the above inequalities we used that $M \geq M_2$ and $M_2$ was chosen sufficiently large so that the exponents above are negative -- see (\ref{M1}). Furthermore, by the triangle inequality
$$|1 + z| \leq 1 + |z| \mbox{ and } |(1- z)^{-1}| \leq (1- |z|)^{-1}$$
whenever $|z| < 1.$ Utilizing the latter and the inequality $n_i \leq 2 M$, which follows from (\ref{M3}), we see that for each $i = 1, \dots, N_2 - k$
$$\left| \left( \frac{1 + a\hat{z}^1_i}{1 + a \hat{w}_i^1}\right)^M \cdot \left(\frac{1 + a/\hat{w}^1_i}{1 + a/\hat{z}_i^1} \right)^{n_2}\right| \leq \left( \frac{1 + 2a}{1- 2a}\right)^{3M}.$$
Similarly, we have for each $i = 1, \dots, N_1 -k$ that
$$ \left| \left( \frac{1 + az_i^1}{1 + a w_i^1}\right)^M  \left(\frac{1 + a/w_i^1}{1 + a/z_i^1} \right)^{n_1} \right| \leq \left( \frac{1 + 2a}{1- 2a}\right)^{3M},$$
and for each $i = 1, \dots, k$ that
$$ \left|\left( \frac{1 + az_i^2}{1 + a \hat{w}_i^2}\right)^M  \left(\frac{1 + a/\hat{w}^2_i}{1 + a/ z_i^2} \right)^{n_2} \left(\frac{1 + a/z_i^2}{1 + a/w_i^2} \right)^{n_2- n_1} \right| \leq \left( \frac{1 + 2a}{1-2 a}\right)^{3M}.$$
Note that $2a \leq 1/2$ by our choice of $a \leq a_2\leq e^{-2\pi}$, see (\ref{Cond1}), and so by Lemma \ref{descentLemma5} we conclude
$$\frac{1 + 2a}{1- 2a} \leq \exp \left( 6 a \right).$$
Summarizing all of the above inequalities we conclude that for $M \geq M_2$ we have
\begin{equation}\label{C1BoundH3}
\begin{split}
&|A^M_3| \leq \exp \left(9 M \Delta a  \right).
\end{split}
\end{equation}

We finally analyze $A^M_1$. Observe that by our definition of $C_-, C_+, C_m$ and Lemma \ref{LCDetBound} applied to $R =2$, $r = 1/2$ and $\alpha = 2^{-1/2}$ we have
\begin{equation}\label{C1BoundH1}
\begin{split}
&|A^M_1|  \leq \frac{(N_1 +N_2 - k)^{(N_1 + N_2 -k)/2}}{(2^{1/2} - 2^{-1/2})^{N_1+ N_2- k}} \cdot \left(\frac{2^{1/2} + 2^{-1/2}}{5/2}\right)^{(N_1 + N_2 - k )^2} \leq C^{\Delta} \Delta^{\Delta/2}p^{\Delta^2/4} ,
\end{split}
\end{equation}
where in the last inequality we used that $\Delta \geq N_1 +N_2 - k \geq 2^{-1}\Delta$ as $k \leq \min (N_1, N_2)$. We also recall that $p$ was defined in (\ref{defp}).\\

Combining (\ref{C1BoundH4}), (\ref{C1BoundH2}), (\ref{C1BoundH3}) and (\ref{C1BoundH1}) we conclude that
\begin{equation*}
|A^M_1 A^M_2 A^M_3 A^M_4| \leq C^{\Delta} \cdot \Delta^{\Delta/2} \cdot \left(p^{1/2} \cdot \frac{(-4t;t)_\infty}{(4t;t)_\infty}\right)^{\Delta^2/2} \cdot \exp (9M \Delta a).
\end{equation*}
Finally, using that $\Delta \geq RaM$ we conclude that
\begin{equation}\label{C1BoundAll}
|A^M_1 A^M_2 A^M_3A^M_4| \leq  C^{\Delta} \cdot \Delta^{\Delta/2} \cdot \left(p^{1/2} \cdot e^{18/R} \cdot  \frac{(-4t;t)_\infty}{(4t;t)_\infty}\right)^{\Delta^2/2} \leq C^{\Delta} \cdot \Delta^{\Delta/2} \cdot p^{\Delta^2/8},
\end{equation}
where in the last inequality we used (\ref{Cond2}) and (\ref{defR}). It is now clear that (\ref{S6INK}) and (\ref{C1BoundAll}) together imply (\ref{BoundEachTerm}) since $p \in (0,1)$. This concludes Step 3.\\

{\bf \raggedleft Step 4.} The purpose of this step is to prove (\ref{BoundEachTerm}) if $\Delta \in [M^{1/2}, RaM]$. We put $\delta = \left( \Delta / RaM \right)^{1/3}$ and note that $\delta \in [M^{-1/6}(Ra)^{-1/3}, 1]$. We start with the same argument as in Step 3 except that the contours $C_+$ and $C_-$ that we deform to have radii $e^{\delta}$ and $e^{-\delta}$ respectively. Notice that our assumption that $a \leq a_2 \leq e^{-2\pi}$, see (\ref{Cond1}), implies that in the process of deforming to these contours we do not cross any poles and so the value of the integral does not change by Cauchy's theorem. As in Step 3 we proceed to derive estimates for $A^M_1, A^M_2, A^M_3, A^M_4$.

We first analyze $A^M_4$. Observe that $\delta \leq 1 \leq \frac{-\log t}{8}$ by our assumption that $t \leq t_2 \leq e^{-4\pi}$ from Step 1, see (\ref{Cond1}). Similarly to our work in Step 3, we may apply Lemma \ref{S5techies} with $U = 2\delta$ to get that for all $i = 1, \dots, N_2 - k$ we have
$$\left| \frac{\hat{S}(\hat{w}_i^1,\hat{z}^1_i; u_2, t)}{ \log t\cdot \hat{w}_i^1 } \right| \leq \frac{\pi \cdot e \cdot C_t^1}{2\delta [-\log t]}.$$
Analogously for all $i = 1, \dots, N_1 - k$ we have
$$\left|\frac{\hat{S}(w^1_i,z_i^1; u_1, t)}{\log t \cdot w^1_i } \right| \leq \frac{\pi \cdot e \cdot C_t^1}{2\delta[-\log t] }.$$
Also we may apply Lemma \ref{S5techies} with $U = \delta$ to get that for all $i = 1, \dots,  k$ we have
$$\left| \frac{\hat{S}(\hat{w}_i^2,w^2_i; u_2, t)}{ \log t \cdot \hat{w}_i^2}  \frac{\hat{S}(w_i^2,z_i^2; u_1, t)}{ \log t \cdot w_i^2} \right| \leq  \left( \frac{\pi \cdot e \cdot C_t^1}{\delta  [-\log t] } \right)^2.$$
Combining all of the above inequalities we arrive at
\begin{equation}\label{C2BoundH4}
\begin{split}
&|A^N_4|  \leq \delta^{-\Delta} C^{\Delta} \leq C^{\Delta} \exp \left( \frac{ \Delta \log \Delta }{3} \right),
\end{split}
\end{equation}
where in the last inequality we used the definition of $\delta$ and the fact that $M \leq \Delta^2$. \\

Next we analyze $A^M_2$. Notice that by (\ref{S3PochBound}) and (\ref{C1DefA2}) we have that
\begin{equation}\label{C2BoundH2}
\begin{split}
&|A^M_2| \leq \left( \frac{(-te^{2\delta};t)_\infty}{(t e^{2\delta}; t)_\infty } \right)^{2N_1N_2} \leq \left( \frac{(-te^{2};t)_\infty}{(t e^{2}; t)_\infty } \right)^{2N_1N_2} \leq \exp \left( tC_5 \Delta^2/2 \right),
\end{split}
\end{equation}
where in the first inequality we used that $\delta \in (0,1]$ and that $t \leq t_2 \leq e^{-4\pi}$, see (\ref{Cond1}). In the last inequality we used Lemma \ref{BoundMixedS} and the constant $C_5$ is as in the statement of that lemma. \\

We next turn our attention to $A^M_3$ and it will be convenient to consider the following change of variables. Let us put
$$\log (\hat{z}^1_i) = \hat{Z}^1_i = \delta + \iota \hat{X}^1_i, \hspace{2mm}  \log(\hat{w}^1_i) = \hat{W}^1_i =- \delta + \iota \hat{Y}^1_i \mbox{ with $\hat{X}^1_i,\hat{Y}^1_i  \in [-\pi, \pi]$ for $i = 1,\dots, N_2 - k$};$$
$$\log (z^1_i) = Z^1_i = \delta + \iota X^1_i, \hspace{2mm}  \log(w^1_i) = W^1_i =- \delta + \iota Y^1_i \mbox{ with $X^1_i,Y^1_i  \in [-\pi, \pi]$ for $i = 1,\dots, N_1 - k$};$$
$$\log(\hat{w}^2_i) = \hat{W}^2_i =-\delta + \iota \hat{Y}^2_i, \hspace{2mm} \log(z^2_i) = Z^2_i = \delta + \iota X^2_i, \hspace{2mm} \log(w^2_i) = W^2_i = \iota Y^2_i $$
$$ \mbox{ with $\hat{Y}^2_i,Y^2_i , X^2_i \in [-\pi, \pi]$ for $i = 1,\dots, k$ }.$$
 Recall from (\ref{DefH3V2}) and (\ref{DefH3V22}) that in these variables we have $A^M_3 =  \prod_{i = 1}^7 A^M_{3,i}$ where
\begin{equation}\label{S6DefH3V2}
\begin{split}
& A^M_{3,1}=  \prod_{i =1}^{N_2-k}e^{ F^M_{2}(\hat{Z}_i^1)}, \hspace{2mm}  A^M_{3,2} = \prod_{i =1}^{N_2-k}e^{- F^M_2(\hat{W}_i^1)},\hspace{2mm} A^M_{3,3} =\prod_{i =1}^{N_1-k} e^{F^M_1({Z}_i^1)}, \\
& A^M_{3,4} = \prod_{i =1}^{N_1-k} e^{-F_1^M(W_i^1)}, A^M_{3,5}= \prod_{i =1}^{k} e^{F_1^M(Z_i^2)},  A^M_{3,6}= \prod_{i =1}^{k} e^{-F_2^M(\hat{W}_i^2)}, A^M_{3,7} = \prod_{i =1}^{k} e^{F_3^M(W_i^2)},
\end{split}
\end{equation}
where $F_1^M, F_2^M, F_3^M$ are given by
\begin{equation}\label{S6DefH3V22}
\begin{split}
&F_1^M(Z) = \exp\left({M S_a(Z) -  M^{1/3}\sigma_a x_1 Z + (M-n_1)R_a(Z) - (Z/2) f''_1 s_1^2 M^{1/3}}\right), \\
& F_2^M(Z) = \exp \left({MS_a(Z) - M^{1/3} \sigma_a x_2 Z} + (M-n_2)R_a(Z) - (Z/2) f''_1 s_2^2 M^{1/3} \right)  \\
& F_3^M(W) = \exp \left(  M^{1/3} \sigma_a (x_1 - x_2)W  + (n_1-n_2)R_a(W) + (W/2)M^{1/3} f''_1 [s_1^2 - s_2^2]  \right).
\end{split}
\end{equation}

By Lemmas \ref{descentLemma2} and \ref{descentLemma3} we have that
\begin{equation*}
\begin{split}
&|A^M_{3,1}| \leq \exp \left(C_2 M (N_2 -k)a \delta^3  - [C_1 a M  \delta - C_3 a |M-n_2|]  \sum_{i = 1}^{N_2 - k} (\hat{X}^1_i)^2  + C_\alpha [N_2- k] M^{1/3} \right)  \\
& |A^M_{3,2}|  \leq \exp \left(C_2 M (N_2 -k)a \delta^3  - [C_1 a M  \delta - C_3 a |M- n_2|]  \sum_{i = 1}^{N_2 - k} (\hat{Y}^1_i)^2  + C_\alpha [N_2 - k]  M^{1/3} \right), \\
&| A^M_{3,3}| \leq  \exp \left(C_2 M (N_1 -k)a \delta^3  - [C_1 a M  \delta -C_3 a |M-n_1|]  \sum_{i = 1}^{N_1 - k} (X^1_i)^2  + C_\alpha [N_1 - k]  M^{1/3} \right),  \\
&|A^M_{3,4}| \leq\exp \left(C_2 M (N_1 -k)a \delta^3  - [C_1 a M \delta -C_3a |M-n_1|]     \sum_{i = 1}^{N_1 - k} (Y^1_i)^2  + C_\alpha [ N_1 - k ] M^{1/3} \right),  \\
&| A^M_{3,5}| \leq\exp \left(C_2 M k a \delta^3  - [C_1 a M\delta  -C_3a |M-n_1|]    \sum_{i = 1}^{k} (X^2_i)^2  + C_\alpha k  M^{1/3} \right) ,\\
&|A^M_{3,6}| \leq  \exp \left(C_2 M k a \delta^3  - [C_1 a N  \delta - C_3 a |M-n_2| ]  \sum_{i = 1}^{k} (\hat{Y}^2_i)^2  + C_\alpha k   M^{1/3} \right).
\end{split}
\end{equation*}
Furthermore by (\ref{K3}) we know $Re[(R_a(\iota Y^2_i) ] \leq Re[R_a(0)]= 0$ for $ i  =1 ,\dots, k$. This implies that
$$ |A^M_{3,7}| \leq \exp \left(  C_\alpha M^{1/3} k  \right).$$
Combining all of the above estimates for $A^M_{3,i}$ for $i =1, \dots , 7$  we conclude that
\begin{equation*}
\begin{split}
|A^M_3| \leq &\exp \left( 2C_2 M\Delta  a \delta^3  + 2C_\alpha \Delta M^{1/3}-  [C_1 a M  \delta - C_3 a |M-n_2|]  \sum_{i = 1}^{N_2 - k} [(\hat{X}_i^1)^2 + (\hat{Y}_i^1)^2]  \right) \cdot \\
&\exp\left(- [C_1 a M \delta  - C_3 |M-n_1|] \sum_{i = 1}^{N_1 - k} [(X_i^1)^2 + (Y_i^1)^2] )  \right) \cdot\\
& \exp\left(- [C_1 aM \delta - C_3a|M-n_1|] \sum_{i = 1}^{k} (X_i^2)^2 - [C_2 a M  \delta - C_3 a |M-n_2| ]  \sum_{i = 1}^{k} (\hat{Y}^2_i)^2    \right).
\end{split}
\end{equation*}
The above upper bound for $|A^M_3|$ can be simplified once we use the fact that $M \geq M_2$ and our choice of $M_2$ in Step 1 as well as the definition of $\delta$. Namely, recall from the beginning of the step that $\delta \in [M^{-1/6} (Ra)^{-1/3}, 1]$ and so we have
$$C_1 a M  \delta - C_3 a  | M-n_i| \geq (C_1/2)a M  \delta + [(C_1/2)a (Ra)^{-1/3} M^{5/6} - C_3 a |M-n_i| ] \geq (C_1/2)a M  \delta,$$
where in the last inequality we used (\ref{M2}). Utilizing the last inequality, together with $M \leq \Delta^2$ and the fact that $\delta^3 M = (Ra)^{-1} \Delta$ we see that
\begin{equation}\label{C2BoundH3}
\begin{split}
&|A^M_3| \leq \exp \left( 2C_2 R^{-1}\Delta^2    + 2C_\alpha \Delta^{5/3} \right) \cdot \exp \left( -  (C_1/2) a M \delta Q \right), \mbox{ where } \\
&Q = \sum_{i = 1}^{N_2 - k} [(\hat{X}_i^1)^2 + (\hat{Y}_i^1)^2] + \sum_{i = 1}^{N_1 - k} [(X_i^1)^2 + (Y_i^1)^2]   +  \sum_{i = 1}^{k} [ (X_i^2)^2 + (\hat{Y}_i^2)^2] .
\end{split}
\end{equation}

We next consider bounding $|A^M_1|$ and consequently $|A_1^M A_2^M A_3^M A_4^M|$ in the following two cases:
\begin{equation}\label{cases}
\begin{split}
(1) \hspace{2mm}  Q \geq  \frac{4(C_2 + C_5) }{C_1}\cdot \delta^2 \Delta \mbox{ and } (2) \hspace{2mm} Q \leq  \frac{4(C_2 + C_5) }{C_1}\cdot \delta^2 \Delta,
\end{split}
\end{equation}
where $C_1, C_2$ are as in Lemma \ref{descentLemma2} and $C_5$ is as in Lemma \ref{BoundMixedS}.  \\

The first case in (\ref{cases}) is somewhat trivial and we simply use Hadamard's inequality to get
\begin{equation}\label{C2C1BoundH1}
\begin{split}
&|A^M_1|  \leq \frac{(N_1 +N_2 - k)^{(N_1 + N_2 -k)/2}}{(e^{\delta} - e^{-\delta})^{N_1+ N_2- k}} \leq \delta^{-\Delta}\Delta^{\Delta/2} \leq C^{\Delta}\exp \left( \frac{5 \Delta \log \Delta}{6} \right) ,
\end{split}
\end{equation}
where in the first inequality we used that $e^x - e^{-x} \geq 2x$ for $x > 0$ and the fact that $\Delta \geq N_1 +N_2 - k \geq 2^{-1}\Delta$ as $k \leq \min (N_1, N_2)$. In the second inequality we used the definition of $\delta$ and the fact that $M \leq \Delta^2$. Combining the inequalities (\ref{C2BoundH4}), (\ref{C2BoundH2}), (\ref{C2BoundH3}) and (\ref{C2C1BoundH1}) we conclude that
\begin{equation}\label{C2C1BoundAll}
\begin{split}
&|A^M_1A^M_2A^M_3A^M_4| \leq C^{\Delta} \exp \left( (tC_5/2 - 2C_5 R^{-1}) \Delta^2  +   \frac{7 \Delta \log \Delta}{6}  + 2C_{\alpha} \Delta^{5/3}  \right),
\end{split}
\end{equation}
where we used that $(C_1/2) a M\delta Q \geq 2(C_2+C_5) R^{-1} \Delta^2 $ in the first case in (\ref{cases}). \\

The second case in (\ref{cases}) is more complicated and to bound $|A_1^M|$ we apply Lemma \ref{LemmaCDet2} with $N = N_1 + N_2 - k$ and $c = \sqrt{\frac{8(C_2 + C_5)}{C_1}}.$ Indeed, from (2) in (\ref{cases}) and the fact that $N_1 + N_2 - k \geq \Delta/2$ as $k \leq \min(N_1, N_2)$ we see that
$$Q \leq \delta^2 \cdot c^2 \cdot [N_2 +N_1 - k],$$
which in view of Lemma \ref{LemmaCDet2} implies
\begin{equation}\label{C2C2BoundH1}
\begin{split}
&|A^M_1|  \leq \frac{e^{-f(c) (N_1 +N_2 - k)^2} (N_1 +N_2 - k)^{(N_1 + N_2 -k)/2}}{(e^{\delta} - e^{-\delta})^{N_1+ N_2- k}}\leq C^{\Delta}e^{-f(c) \Delta^2/4}\exp \left( \frac{5 \Delta \log \Delta}{6} \right) ,
\end{split}
\end{equation}
where $f(c)$ is as in Lemma \ref{LemmaCDet2} and the second inequality is derived as in (\ref{C2C1BoundH1}). Combining the inequalities (\ref{C2BoundH4}), (\ref{C2BoundH2}), (\ref{C2BoundH3}) and (\ref{C2C2BoundH1}) we conclude that
\begin{equation}\label{C2C2BoundAll}
\begin{split}
&|A^M_1A^M_2A^M_3A^M_4| \leq C^{\Delta} \exp \left( (tC_5/2 + 2C_2 R^{-1}- f(c)/4) \Delta^2  +   \frac{7 \Delta \log \Delta}{6}  + 2C_{\alpha} \Delta^{5/3}\right),
\end{split}
\end{equation}
where we used that $Q \geq 0$.

 Recall from (\ref{defR}) and (\ref{Cond3}) that
$$tC_5/2 + 2C_2 R^{-1}- f(c)/4 \leq R^{-1} (C_5/2 + 2C_2) \leq -f(c)/8 \mbox{ and }tC_5/2- 2C_5 R^{-1} \leq - (3/2) R^{-1} C_5.  $$.
 So if we combine (\ref{C2C1BoundAll}) and (\ref{C2C2BoundAll}) we conclude that for $\epsilon = \min (f(c)/8, (3/2) R^{-1} C_5)$ we have
\begin{equation}\label{C2BoundAll}
\begin{split}
&|H^N_1H^N_2H^N_3H^N_4| \leq C^{\Delta} \exp \left( -\epsilon \Delta^2  +   \frac{7 \Delta \log \Delta}{6}  + 2C_{\alpha} \Delta^{5/3}\right),
\end{split}
\end{equation}
which together with (\ref{S6INKD}) clearly implies (\ref{BoundEachTerm}). This concludes Step 4.\\

{\bf \raggedleft Step 5.} The purpose of this step is to prove (\ref{BoundEachTerm}) if $\Delta \in [1, N^{1/2}]$. In this step we use the formula (\ref{INK3}) for $I_M(N_1, N_2, k)$, which we recall here for the reader's convenience.
\begin{equation}\label{S5INK3}
\begin{split}
&I_M(N_1, N_2, k) = \frac{(-1)^{N_1 + N_2}}{k! (N_1- k)! (N_2- k)!}  \int_{\gamma_z^{N_1-k}}  \int_{\gamma_z^{N_2-k}} \int_{\gamma_z^{k}} \int_{\gamma_w^{N_1-k}}\int_{\gamma_w^{N_2-k}} \int_{\gamma_w^k}\int_{\gamma_m^{k}}      \\
&H^M_1  H^M_2 H^M_3  H^M_4  \prod_{i = 1}^{k} \frac{{\bf 1}\{ |w_i^2| \leq \pi L\} d{w}^2_i}{2\pi \iota  L} \prod_{i = 1}^{k} \frac{{\bf 1}\{ |w_i^2| \leq \pi \rho_A  L \} d\hat{w}^2_i }{2\pi \iota  L} \prod_{i = 1}^{N_2 - k} \frac{{\bf 1}\{ |\hat{w}_i^1| \leq \pi \rho_A  L\} d\hat{w}^1_i}{2\pi \iota L} \\
& \prod_{i = 1}^{N_1 - k}  \frac{{\bf 1}\{ |w_i^1| \leq \pi \rho_A  L \} dw^1_i}{2\pi \iota  L}  \prod_{i = 1}^{k}   \frac{{\bf 1}\{ |z_i^2| \leq \pi \rho_A  L \} e^{{z}^2_i/ L}dz^2_i}{2\pi \iota  L}  \\
&\prod_{i = 1}^{N_2 - k}\frac{{\bf 1}\{ |\hat{z}_i^1| \leq \pi \rho_A  L \} e^{\hat{z}^1_i/L}d\hat{z}_i}{2\pi \iota L} \prod_{i = 1}^{N_1- k}\frac{{\bf 1}\{ |z_i^1| \leq \pi \rho_A   L \}  e^{{z}^1_i/L}dz^1_i}{2\pi \iota  L} ,
\end{split}
\end{equation}
We recall that $H_1^M$, $H_2^M$, $H_3^M$ and $H_4^M$ are as in (\ref{TildeBlockMatrix}), (\ref{DefH2V2}), (\ref{DefH3V2}) and (\ref{DefH4V2}) respectively. In addition, $\gamma_w, \gamma_z, \gamma_m$ and $A$ are as in \ref{Defcontours}, $L = M^{1/3}$ and $\rho_A = \sqrt{1 + A^2}$. As in Step 3 in Section \ref{Section5.2} we let $E$ denote  the set of points $(\vec{z}^1, \vec{\hat{z}}^1, \vec{z}^2, \vec{w}^1, \vec{\hat{w}}^1, \vec{\hat{w}}^2, \vec{w}^2) \in \mathbb{C}^{2N_1 + 2N_2 - k}$ that satisfy the inequalities in the indicator functions in (\ref{S5INK3}). Then we know from (\ref{BoundH1H4}) that
 \begin{equation}\label{S5BoundH1H4}
\begin{split}
&|{\bf 1}_{E} H^M_1|  \leq   C^{N_1+N_2} (N_1 +N_2 - k)^{(N_1 + N_2 - k)/2} L^{N_1 +N_2 -k}, \mbox{ and } \\
&|{\bf 1}_{E} H^M_4| \leq C^{N_1 + N_2} L^{N_1 +N_2},
\end{split}
\end{equation}
where as earlier $C$ is a constant that depends on $a,t$ alone. Furthermore, from (\ref{DefH3V2})  and (\ref{BoundH3})
\begin{equation}\label{S5BoundH3}
\begin{split}
&|{\bf 1}_{E} H_3^M| \leq  \exp \left( - \epsilon_1 \sum_{i = 1}^{N_2 - k} \left[ |\hat{z}_i^1|^3 + |\hat{w}_i^1|^3   \right] -\epsilon_1\sum_{i = 1}^{N_1 - k} \left[ |{z}_i^1|^3 + |{w}_i^1|^3 \right] \right) \times  \\
& \exp \left(  - \epsilon_1 \sum_{i = 1}^{k} \left[ |{z}_i^2|^3 + |\hat{w}_i^2|^3   \right] -  \epsilon_1 (s_1 - s_2) \sum_{i = 1}^{k}  |{w}_i^2|^2   \right) \times \exp \left( \tilde{c}(N_1 +N_2 + \tilde{Q}) \right), \mbox{ where }\\
&\tilde{Q} = \sum_{i = 1}^{N_2 - k} \left[   |\hat{z}_i^1|^2 +|\hat{w}_i^1|^2   \right]  + \sum_{i = 1}^{N_1 - k} \left[ |{z}_i^1|^2 +|{w}_i^1|^2   \right]  + \sum_{i = 1}^{k}  \left[  |{z}_i^2|^2 + |\hat{w}_i^2|^2  + |{w}_i^2|  \right],
\end{split}
\end{equation}
where $\tilde{c} > 0$ is a constant that depends on $a,t,x_1, x_2, s_1, s_2$ and $\epsilon_1$ is as in Lemma \ref{LemmaTaylor}.

Finally, we bound $|{\bf 1}_E H^M_2|$. By Lemma \ref{LBoundQ} we have that
\begin{equation}\label{C3BoundH2}
\begin{split}
&|{\bf 1}_EH^M_2|  \leq \prod_{i = 1}^{N_1 -k}e^{16t N_2 |{z}^1_i|^2 /L^2} \prod_{i = 1}^{k}e^{16t N_2 |{z}^2_i|^2/L^2}\prod_{i = 1}^{N_1 -k}e^{16t N_2 |{w}^1_i|^2/L^2} \prod_{i = 1}^{N_2 -k}e^{16t N_1 |\hat{z}^1_i|^2/L^2} \\
& \prod_{i = 1}^{N_2 -k}e^{16t N_1 |\hat{w}^1_i|^2/L^2} \cdot \prod_{i = 1}^{k}e^{16t N_2  |{w}^2_i|^2/L^2}\prod_{i = 1}^{k}e^{16t N_2|\hat{w}^2_i|^2/L^2} \leq \\
& \exp\left(16 t \Delta  \tilde{Q}/L^2 + 16 t\Delta \sum_{i = 1}^{k}  |{w}_i^2|^2/L^2 \right)
\end{split}
\end{equation}

Let $H^M$ denote the integrand in (\ref{S5INK3}). Observe that by (\ref{S5BoundH1H4}), (\ref{S5BoundH3}) and (\ref{C3BoundH2}) we have that
\begin{equation}\label{C3HBoundAll}
\begin{split}
&|H^M|\leq  (N_1 +N_2 - k)^{(N_1 + N_2 - k)/2} \exp \left( - \epsilon_1 \sum_{i = 1}^{N_2 - k} \left[ |\hat{z}_i^1|^3 + |\hat{w}_i^1|^3   \right] -\epsilon_1\sum_{i = 1}^{N_1 - k} \left[ |{z}_i^1|^3 + |{w}_i^1|^3 \right] \right)  \\
& \exp \left(  - \epsilon_1 \sum_{i = 1}^{k} \left[ |{z}_i^2|^3 + |\hat{w}_i^2|^3   \right] -  (\epsilon_1/2) (s_1 - s_2) \sum_{i = 1}^{k}  |{w}_i^2|^2  +\tilde{C}(N_1 +N_2 + \tilde{Q})  \right) ,
\end{split}
\end{equation}
where we used that $\Delta \leq M^{1/2}$, $\tilde{C} = \tilde{c} + 1$, $16t \leq 1$ (this follows from $t \leq t_1 \leq e^{-4\pi}$ -- see (\ref{Cond1})), and $M^{-1/6}  \leq (\epsilon_1/2)(s_1 -s_2)$ (this follows from (\ref{M4})).

 We now note that the upper bound in (\ref{C3HBoundAll}) does not depend on $M$ and is a product of functions in each of the $2N_1 + 2N_2 - k$ variables and each of those functions is integrable on its corresponding contour. For all but the $w^2_i$ variables this is true by the negative cube in the exponential with a positive coefficient $\epsilon_1$, which dominates the quadratic terms in $\tilde{Q}$. Also for the $w^2_i$ variables this is true by the negative square in the exponential with a positive coefficient $(\epsilon_1/2) (s_1 - s_2)$, which dominates the linear terms in $\tilde{Q}$. From (\ref{C3HBoundAll}) and (\ref{S5INK3}) we conclude that
$$ |I_N(N_1, N_2, k)| \leq  \frac{\tilde{C}_1^{\Delta}(N_1 +N_2 - k)^{(N_1 + N_2 - k)/2} }{k! (N_1- k)! (N_2- k)!},$$
for some large enough $\tilde{C}_1$ that depends on $a,t,x_1, x_2, s_1, s_2$.

Finally, we note that
$$3^{N_1 + N_2 - k} = \sum_{i = 0}^{N_1 +N_2 -k } \sum_{j = 0}^{N_1 + N_2 -k - i} \frac{(N_1 +N_2 - k)!}{i! j! (N_1 +N_2 - k - i -j)!}.$$
Setting $i = k$ and $j = N_1 - k$ above we see that
$$\frac{3^{N_1 + N_2 - k} }{(N_1 +N_2 - k)!} \geq  \frac{1}{k! (N_1- k)! (N_2- k)!}.$$
Using the latter we conclude that
 $$ |I_N(N_1, N_2, k)| \leq  \frac{(3\tilde{C}_1)^{\Delta}(N_1 +N_2 - k)^{(N_1 + N_2 - k)/2}}{(N_1 +N_2 - k )!} \leq  $$
$$(3 \cdot e \cdot \tilde{C}_1)^{\Delta} \cdot  \exp\left( - (1/2) (N_1 +N_2 - k ) \log(N_1 +N_2 - k ) \right),$$
where we used $n! \geq  n^{n} e^{-n}$ -see \cite[Equation (1)]{Robbins}. The last inequality now clearly implies (\ref{BoundEachTerm}) once we utilizie that $N_1 +N_2 - k \geq 2^{-1}\Delta$ as $k \leq \min (N_1, N_2)$. This completes Step 5 and hence the proof of the proposition.

config: arxiv


TITLE: Extended Calendar Cube Question
QUESTION [4 upvotes]: The calendar cube puzzle is famous: using two six-sided cubes, label them such that any day of any month can be represented by positioning the cubes accordingly.
The solution involves allowing the "6" and "9" to be represented by the same character.
My question is an extension: is it possible, using 4 six-sided cubes, to represent the two digit month and two digit day of any day of the year? With the proviso that 6 and 9 not be represented by the same character.
I don't know if this is possible or not, and the best I've been able to do is:
cube1 = {1, 9, 5, 2, 3, 4};
cube2 = {1, 5, 0, 7, 8, 9};
cube3 = {1, 6, 0, 2, 3, 4};
cube4 = {1, 6, 0, 2, 7, 8};

This allows for any dates besides the following:
06/06; 06/07; 06/08;
07/06; 07/07; 07/08;
08/06; 08/07; 08/08.

Is it possible to do better?

REPLY [2 votes]: The previous solution (ommiting one of two digits $9$) is not the best one, it is the third best, ex aequo with ommitting any digit from set $\{4,5,...,8\}$
There is a family of the best solutions, for instance:
$$C_1 := \{0,2,3,4,5,6\}$$
$$C_2 := \{0,1,3,4,5,6\}$$
$$C_3 := \{0,1,2,7,8,9\}$$
$$C_4 := \{0,1,2,7,8,9\}$$
The last 4 digits of each cube could be shuffled between the cubes, of course such that no cube has repeated digits (especially $2$'s).
Proof:
We need at least:

two copies of digits $\{4,5,...,9\}$ because they are needed as units' digits for days and months and they cannot be tens' digits;
two copies of digit $3$ because there is no month or day number $33$ but there are $\text{Mar }03,13,23,30,31$);
three copies of digit $2$ because there is no month number $22$ but there are $\text{Feb }22$ and $\text{Dec }22$;
four copies of digit $1$ because there is $\text{Nov }11$;
four copies of digit $0$ because it needs to be combined with two sets of two cubes needed for each of nine digits $n\neq0$ (because of 14 dates: nine $0n/0n$ and five: $\text{Jan }10$, $\text{Feb }20$, $\text{Mar }30$, $\text{Oct }01$ and $\text{Oct }10$).

Therefore, we need at least $2\cdot7+3+4\cdot2=25$ sides of cubes which is just one side more than 4 cubes. That is why we cannot represent all dates using 4 cubes if we assume $6\neq9$.
But we could decide which of above conditions we could possibly omit. The least restrictive is condition for digit $1$ because it is necessary for only one day, namely $\text{Nov }11$. So, if we omit one of four $1$'s, we could arrange the rest of digits such to create any date except from $\text{Nov }11$.
Fortunately, in some European countries $\text{Nov }11$ is a holiday commemorating the end of WW I, so we do not necessarily need a calendar that day. We could also mark it specially as $\text{Nov }00$ or "the new origin" $00/00$.
The second least restrictive is the condition for digit $2$, necessary for two days.
The condition for each of digits $\{4,5,...,9\}$ is the third in the least-restrictive ranking because each pair of digit $n$ is necessary for three days $0n,1n,2n$ in month $0n$. However, we could make a rule that month $00$ means "take the day's units' digit as month number".


TITLE: Continuity of Dirichlet-like function
QUESTION [1 upvotes]: Let
$f(x) = \begin{cases}
x^2 & x \in \mathbb Q\\
     x+2 & x \notin \mathbb Q.
\end{cases}$
I want to find all points $x$ where $f(x)$ is continuous.
I'm not even sure what this question even asking
to find all the continuous points?
If x is not a rational number it should be irrational.
So should I say its continuous every $x+2$ and not continuous $x^2$

REPLY [2 votes]: My intuition is to set the two pieces of the functions together, $x^2 = x + 2$ implies $x = 2, x = -1$. If we exclude $(2,4)$ and $(-1,1)$, it should be clear that the function is not continuous anywhere. Every interval you choose will include a point greater than $\epsilon$ away for sufficiently small $\epsilon$.
For example, consider $x = 5$ and $\epsilon = 1$. The function is discontinuous because any interval around 5 contains an irrational number which will have a function value of approximately seven. 25 is not within one of seven. If you consider $x = 2.01$ you have to choose a smaller value of $\epsilon$, in this case $\epsilon = .01$. Similarly to the first case, every interval centered around 2.01 will contain an irrational number which is approximately 4.01 while $f(2.01) = 4.0401$. We can see that $4.0401 > 4.01 + .01$.
For our special points, I will just consider one of them. If our function tends to $x = 2$ and we pick an arbitrary $\epsilon$, any interval $(2-\frac{\epsilon}{8},2+\frac{\epsilon}{8})$ will contain no function value greater than $4+\frac{\epsilon}{2} + \frac{\epsilon^2}{64}$ (for the rational values) or no function value greater than $4 + \frac{\epsilon}{8}$ (irrational values). As long as $\frac{\epsilon^2}{64} < \frac{\epsilon}{2}$, our maximum function value in the interval will be less than $4+\epsilon$. I only proved this for one direction and for one of our points so you will have to complete the proof. Naturally, we can pick our $\delta = \mathrm{min}(\frac{\epsilon}{8},32)$ or $\delta = \mathrm{min}(\frac{\epsilon}{8},1)$ if we feel so inclined.


text:
\begin{document}

\preprint{IFT-UAM/CSIC-20-82}
\title{\huge Anomalous diffusion and Noether’s second theorem\color{black}}
\author{Matteo Baggioli}
\email{matteo.baggioli@uam.es}
\affiliation{Instituto de Fisica Teorica UAM/CSIC, c/ Nicolas Cabrera 13-15, Cantoblanco, 28049 Madrid, Spain}
\author{Gabriele La Nave}
\email{lanave@illinois.edu }
\affiliation{Department of Mathematics, University of Illinois, Urbana, Il. 61801}
\author{Philip W. Phillips}
\email{dimer@illinois.edu }
\affiliation{Department of Physics and Institute for Condensed Matter Theory, University of Illinois
1110 W. Green Street, Urbana, IL 61801, U.S.A.}

\begin{abstract}
Despite the fact that conserved currents have dimensions that are determined solely by dimensional analysis (and hence no anomalous dimensions), Nature abounds in examples of anomalous diffusion~\cite{cells,anomdiffexpt} in which $L\propto t^\gamma$, with $\gamma\ne 1/2$, and heat transport in which the thermal conductivity diverges as $L^\alpha$~\cite{mcuentherm,therm2,therm3,therm4,therm5}.  Aside from breaking of Lorentz invariance, the true common link in such problems is an anomalous dimension for the underlying conserved current, thereby violating the basic tenet of field theory.  We show here that the phenomenological~\cite{klafter,fracdiff2016,nigmatullin} non-local equations of motion that are used to describe such anomalies all follow from Lorentz-violating gauge transformations arising from N\"other's second theorem.  The generalizations lead to a family of  diffusion and heat transport equations that systematize how non-local gauge transformations generate more general forms of Fick's and Fourier's laws for diffusive and heat transport, respectively. In particular, the associated Goldstone modes of the form $\omega\propto k^\alpha$, $\alpha\in \mathbb R$ are direct consequences of fractional equations of motion.
\end{abstract}

\maketitle

\section{Introduction}

Although Lorentz symmetry is a fundamental organizing principle of the universe, it is oftentimes violated.  For example, any collection of particles, complicated enough for a center of mass to be well defined, has a preferred reference frame and hence violates the boost-invariance of the Lorentz group.  The lack of boost-invariance rarely occurs as a solo act. The presence of a lattice and non-relativistic dispersions, graphene the exception, in condensed matter systems offer additional routes for the breakdown of Lorentz invariance. It comes as no surprise that low-energy phases of matter can be indeed classified \cite{Nicolis:2015sra} by looking at the different spontaneous symmetry breaking patterns of the Lorentz group. While continuous symmetry breaking such as rotational and translational give rise to Goldstone bosons of the form, $\omega\propto k^n$, where $n\in \mathbb Z$, the massless excitation arising from just the breaking of boost-invariance remains elusive.  Of course, such a problem would be rendered moot if phases of matter exist in which just \cite{Nicolis} boost-invariance were broken.  Nature, unfortunately is not so kind.

Our focus here is on two processes that break Lorentz invariance, diffusion and heat transport.  The equations for both are derivable from the continuity equation,
\begin{equation}
    \partial_\mu J^\mu=\partial_t\,n\,+\,\partial_i J^i=0\label{continuity}
\end{equation}
which arises from an integration by parts of the action
\beq
S=\int d^dx\, \left(F^2+J^\mu A_\mu\right)
\eeq
subject to the $U(1)$ gauge-invariant condition, $A_\mu\rightarrow A_\mu-\partial_\mu\Lambda$ with $F=dA$.  For simple diffusion, the inconsistency with Lorentz invariance occurs once Fick's law for the current,
\beq
J^i\,=\,-\,D\,\partial^i n \label{fick}
\eeq
is imposed.  Substituting \eqref{fick} into the continuity equation \eqref{continuity} results in the well known diffusion equation
\beq
\partial_t n - \,D\,\Delta\,n=0.
\eeq
The propagator for this equation has solutions even for space-like separations.   Such growth is inconsistent with special relativity and hence must be associated with a massless mode, in this case $\omega=-iDk^2$.  Arguments of this sort have motivated an extensive literature~\cite{goldstein,dudley,goldstein2} on squaring standard diffusion with special relativity.  A promising solution~\cite{hanggi} appears to be random walks based on non-Markovian processes, as relativistic Markov dynamics in space is a misnomer.  The inconsistency of diffusion with special relativity is also known in the context of linearized relativistic hydrodynamics where it manifests itself as the violation of causality -- superluminal propagation. A famous resolution to this problem was given by Israel and Stewart \cite{ISRAEL1979341} and it can be reformulated as an upper bound for the diffusion constant $D<v^2 \tau$ \cite{PhysRevLett.119.141601}, where $v$ is the lightcone speed and $\tau$ the equilibration time at which diffusion obtains.

What the previous derivation lays plain is that the form of the current imposed by Fick's law fixes the diffusion equation.  However, Fick's choice is not unique and, as a result, neither is the diffusion equation. In fact, there is an extensive literature, on replacing either the time derivative~\cite{cafffractime,nigmatullin} or the spatial Laplacian~\cite{klafter,zaslavsky,fracdiffcrit,fracfickslaw,caffdiff,fracdiff2016} in the diffusion equation with a pseudo-differential operator such as the fractional Laplacian.  What results from such analyses are non-local diffusive transport or L\'evy processes that are capable of describing the numerous experimental realizations of  anomalous diffusion~\cite{anomdiffexpt,cells,expt2,ramos} in which $L\propto t^\delta$,  $\delta\ne 1/2$.  \textit{Subdiffusion} corresponds to $\delta<1/2$, whereas \textit{superdiffusion} arises when $\delta>1/2$.

Anomalies also occur for heat transport which is derivable immediately from the continuity equation by substituting Fourier's law for the gradient of the temperature profile
\beq
J_H=-\kappa \,\frac{\partial T(x,t)}{\partial x},
\label{fourier}
\eeq
where $\kappa$ is the thermal conductivity. From dimensional analysis $J_H\propto L^{-1}$ and $\kappa$ scales as $L^0$.   However, since the pioneering paper of Fermi, Pasta and Ulam~\cite{fpu} showed that even in an anisotropic harmonic solid $\kappa$ diverges, numerous physical systems~\cite{dhar,povstenko1,Lepridivtc,heatcond2} as simple as single molecules~\cite{therm5} and nanotubes~\cite{mcuentherm,zettl,therm2,therm3,therm4} have been shown to violate Fourier's law as they exhibit a thermal conductivity diverging as $\kappa\propto L^\alpha$. Such a divergence of $\kappa$ implies that the heat current scales as $J_H\propto L^{\alpha-1}$. Only $\alpha=0$ corresponds to standard diffusive transport.  As in anomalous diffusion, fractional equations of motion~\cite{tf1,tf2,tf3} are also invoked to explain a heat current that scales with an anomalous power, $J_H\propto L^{\alpha-1}$. What is surprising with the heat current is that the scaling $\kappa^\alpha$ applies~~\cite{garrido} even to the archetypeal 1d diatomic hard-point gas implying that a fundamental non-locality underlies the heat transport even in this simple case.  From the Weidemann-Franz law, any anomaly in the heat transport is passed onto the electrical transport.  Equivalently, heat transport in charged fluids is governed by $U(1)$ invariance and the derivation presented in the first section applies.  The standard folklore~\cite{gross} is that any current tied to a $U(1)$ gauge field cannot acquire an anomalous dimension under any amount of renormalization.  Hence, squaring any of the non-local schemes~\cite{dhar,tf1,tf2,tf3} that have been devised to treat anomalous heat transport, that is $\kappa\propto L^{\alpha}$~\cite{garrido,zettl,mcuentherm,zettl,therm2,therm3,therm4} or anomalous diffusion~\cite{tf1,tf2,tf3,klafter,zaslavsky,fracdiffcrit,fracfickslaw,caffdiff} must account for how they get around the standard argument that the dimension of the conserved current is fixed solely by dimensional analysis.

It is precisely this problem that we address in this paper.  In previous work, we have shown~\cite{csforms,gl1} that N\"other's second theorem (NST) allows for non-traditional scaling of the current only if the action of the gauge group is changed fundamentally. However, all of our extensions of N\"other's Second theorem presented previously~\cite{csforms,gl1} preserve Lorentz invariance and hence are not directly applicable to anomalous heat or diffusive transport.  What we do here is generalize these arguments to the non-Lorentzian case. We focus on the minimal constraints that must be placed on the action of the gauge group to derive anomalous diffusive transport equations. We show that the phenomenological diffusion equations~\cite{tf1,tf2,tf3,klafter,zaslavsky,fracdiffcrit,fracfickslaw,caffdiff} have a rigorous basis in N\"other's second theorem.   This theorem stems from the simple observation that the continuity equation $\partial_\mu J^\mu=0$ is unique up to any operator, $\hat Y$ that commutes with the exterior derivative resulting in
\beq
    \partial_\mu \hat{Y}J^\mu=\partial_\mu \tilde{J}^\mu=0
\eeq
being an equally valid equation of motion (EOM).  To be consistent, the operator $\hat Y$ must also be present in the gauge-invariant condition.  Stated succinctly, the second theorem finds that the full family of generators of $U(1)$ invariance determines the dimension of the current not just the linear derivative term used earlier.  In general, the second theorem applies anytime there are either a collection of infinitesimal symmetries or one symmetry parameterized by an arbitrary number of functions. More recent examples that fall into this category are \textit{fractons} \cite{Gromov:2020yoc}, exotic excitations with confined mobility \cite{doi:10.1142/S0217751X20300033}.  Such anomalous behaviour was  connected to the conservation of higher-order \textit{charges} such as the dipole moment and to new tensorial symmetries \cite{PhysRevB.95.115139,Gromov:2018nbv,Pretko:2018jbi,Seiberg:2020bhn} and follow necessarily from NST.

We have exploited the degeneracy implied by N\"other's second theorem to show~\cite{csforms,gl1,varna} that the most general formulation of electricity and magnetism is one with fractional equations of motion with full Lorentz symmetry.  However, to describe diffusion, we need to restrict ourselves solely to a subset of operators that ultimately breaks Lorentz invariance explicitly. From our formulation, we see that the associated Goldstone is now $\omega\propto k^\alpha$ with $\alpha\in \mathbb R$.  Goldstone modes with fractional powers have been observed previously as ripplon modes in a 2D Wigner crystal~\cite{grimes,halperin}, which have the same dispersion as plasmons in a 2D electron gas, and domain-wall fluctuations at superfluid-superfluid interfaces~\cite{mazets} as they obey $k^{1/2}$ and $k^{3/2}$ dispersions, respectively.  In the context of the latter, a non-local action~\cite{haruki}  containing the fractional Laplacian acting on the displacement field for the interface was proposed to generate the corresponding gapless boson dispersion. The appearance of the fractional Laplacian was justified based on an integration of gapless degrees of freedom~\cite{haruki}.   We show here that such fractional dispersions result quite generally from N\"other's Second theorem.\\
The results presented in this work are very general and applicable to any system exhibiting a macroscopic anomalous diffusive dynamics, from soft-matter systems and biological structures (intracellular transport \cite{santamaria2009protein}, inhomogeneous media \cite{durang2015overdamped}, DNA polymer chains \cite{lomholt2005optimal}, monolayers of motile cells \cite{palmieri2015multiple}) to complex fluids, hard matter and cosmological evolution (protostellar birth \cite{vaytet2018protostellar}, viscoelastic systems \cite{mason1995optical}, electron-hole plasma in semiconductor quantum wells \cite{borges2006optical}, heterogeneous structures \cite{filipovitch2016infiltration}). Our framework, entirely based on symmetry principles, encompasses all these situations in a unified picture.\color{black}

\section{Background}

The power of NST in gauge theories can be understood from a simple degeneracy argument.  Consider the Maxwell action,
\beq\label{Maction}
S&=&\frac12\int \frac{d^dk}{2\pi^d}\, A_\mu(k)\,\left[k^2 \eta^{\mu\nu}-k^\mu k^\nu\right]\,A_\nu(k)\nonumber\\
&=&\frac12\int d^dk\, A_\mu(k)\,M^{\mu\nu}\,A_\nu(k).
\eeq
All gauge transformations  appear as zero-eigenvalues of $M$.  For example,
\beq
M^{\mu\nu}k_\nu=0,
\eeq
which yields the standard gauge-invariant condition in electromagnetism because  $ik_\nu$ is just the Fourier transform of $\partial_\nu $.
The ambiguity that leads to NST comes from noticing that if $k_\nu$ is an eigenvector, then so is $f(k) k_\nu$, where $f(k)$ is a scalar which depends on the momentum itself. Whence, there are a whole family of eigenvectors,
\beq\label{mgen}
M_{\mu\nu}\,f(k)\, k^\nu=0,
\eeq
that satisfy the zero eigenvalue condition, each characterizing a perfectly valid electromagnetism (EM).   It is for this reason that N\"other~\cite{nother} devoted the second half of her paper to the consequences retaining all possible integer derivatives,
\beq
\label{Nothergen}
A_\mu\rightarrow A_\mu +\partial_\mu\Lambda +\partial_\mu\partial_\nu G^\nu+\cdots,
\eeq
 in the gauge-invariant condition for $A_\mu$ has on the conservation laws for the current. All higher-order integer derivatives generate new constraints on the current and correspond to new conserved conserved quantities, dipoles for example \cite{PhysRevB.95.115139,Gromov:2018nbv,Pretko:2018jbi,Seiberg:2020bhn}.
To determine $f$, we note the following. We now take $\bf p$ to the the full 4-momentum.  1) $f$ must be rotationally and Lorentz-invariant.  2) $f$ cannot change the fact that $\Lambda$ is dimensionless; equivalently it cannot change the fact that $A$ is a 1-form.  3) $f$ must commute with the total exterior derivative; that is, $[f,p_\mu]=0$ just as $[d,{\hat Y}]=0$.  Hence, finding $f$ is equivalent to fixing $\hat Y$.  A form of $f$ that satisfies all of these constraints is $f\equiv f(p^2)$.   In momentum space, $k^2$ is simply the Fourier transform of the Laplacian, $-\Delta$.  As a result, the general form of $f(p^2)$ in real space is just the Box operator
raised to an arbitrary power, and the generalization in Eq. (\ref{mgen}) implies that there are a multitude of possible electromagnetisms  (in vacuum) that are
 invariant under the transformation,
 \beq
 A_\mu\rightarrow A_\mu + f(p^2)\,i\,p_\mu\,\Lambda,
 \eeq
 or in real space,
  \beq
  A_\mu\rightarrow A_\mu + (-\Box)^\gamma \partial_\mu\Lambda,
  \eeq
  resulting in $[A_\mu]=1+2[f]=\gamma$.  The definition of the fractional Laplacian and fractional Box operator we adopt here is determined by the general form of a differential operator $L$
  \beq\label{reisz2}
 (L_x)^\gamma f(x)= \frac{1}{\Gamma(\gamma)}\int_0^{+\infty} e^{-t\,L} f(t)\, \frac{dt}{t^{1-\gamma}}
 \eeq
 where $e^{-tL}$ is the "heat" flow of the operator $L$, i.e. $\beta :=e^{-tL}f$ is the unique solution to $\partial _t \beta + L\beta =0$ with Dirichlet boundary condition $\beta\mid_{t=0}= f$.
In the case of the Laplacian acting on functions, one can prove this coincides with the standard Riesz definition
\beq\label{reisz}
(-\Delta_x)^\gamma f(x)=C_{n,\gamma}\int_{\mathbb R^n}\frac{f(x)-f(\xi)}{\mid {x-\xi}\mid^{n+2\gamma}}\;d\xi
\eeq
for some constant $C_{n, \gamma}$.  Note, rather than just depending on the information of $f(x)$ at a point, the fractional Laplacian requires information everywhere in ${\mathbb R}^n$.  For the Laplacian acting on differential forms, there is a similar formula which holds componentwise, as it was shown by the authors in \cite{csforms}.
The standard Maxwell theory is just a special case in which $\gamma=1$.  In general, the theories that result for $\gamma\ne 1$ allow for the current to have an arbitrary dimension, not necessarily $d-1$, without spoiling its conservation nor violating Gross' argument \cite{gross}.  Identifying $\hat Y$ with the fractional Box operator yields the conservation law
\beq \label{fractchargecons}
\partial^\mu (-\Box)^{(\gamma-1)/2}J_\mu=0.
\eeq
\noindent
 As expected, this ambiguity shows up at the level of the Ward identities.  The current-current correlator for the photon
\beq
C^{ij}(k)  \propto (k^2)^{\gamma}\, \bigg( \eta^{ij} - \frac{k^ik^j}{k^2} \bigg)
\eeq
does not just satisfy  $k_\mu C^{\mu\nu}=0$ but also $k^{\gamma-1} k_\mu C^{\mu\nu}=0$. This translates into either $\partial_\mu C^{\mu\nu}=0$, the standard Ward identity, or
 \beq
 \partial_\mu (-\Box)^{\frac{\gamma-1}{2}}C^{\mu\nu}=0
 \eeq
which illustrates that the current conservation equation only specifies the current up to any operator that commutes with the total differential.    Conservation laws such as the one in Eq. \eqref{fractchargecons} are in some sense more fundamental, as one can infer the standard ones from them, but more importantly they can occur earlier~\cite{gl1,csforms} in the hierarchy of conservation laws that stem from N\"other's first theorem.\\

\section{Non-Lorentz Invariant Theory}

We start by discussing the form of the effective Lagrangian of a non-boost invariant theory (which may or may not preserve Galilean transformations). Such an effective Lagrangian can be written as
\beq
L= \frac{1}{2}\int \, d^d x \left( {\mathcal K}F_{ij}\, {\mathcal K}F^{i j}+ c\, F_{0i}^2\right) + L_m,
\eeq
where ${\mathcal K}F_{ij}= \int \,  K(x,y) \left( F_{ij}(x) -F_{ij}(y)\right)d^dy$ with $K(x,y)$ a function of the ~2 vector variable $x$ and $y$ (and we assume, naturally, that $K(x,y)=f(|x-y|)$ in case the Galilean symmetry is not broken and there is no loss of generality in assuming that $c$ is constant. For simplicity of calculations in recovering our eventual equations of motion, one should take $f(u)=u^{-n+1-\gamma},$ so that ${\mathcal K} F=(-\Delta)^{\frac{\gamma-1}{2}}$ to be consistent with Eq. (\ref{reisz}) and $d=n+1$. The matter-field Lagrangian $L_m$ also has to be of the form
\begin{widetext}
\begin{equation}
\resizebox{\hsize}{!}{$L_m= \int \, \left( \int \left(c_1(s,t) (\partial_t \phi)^2(x,s)\right) ds \right) d^dxdt+\int \left( \int \left(\,-\,c_2(x,y)\,(\partial_i \partial_j \phi)^2(y)-c_3(x,y) (\nabla^2 \phi)^2(y)+\dots \right)d^dy\right) d^dxdt+F_1(\nabla A-\phi)+F_2(A_0-\phi_t)$},
\end{equation}
\end{widetext}
and again without loss of generality, we can assume $\int \left(c_1(s,t) (\partial_t \phi)^2(y)\right) ds= (\partial_t \phi)^2$ and that in the case that Galilean symmetries are preserved, $c_k(x,y)=f_k(|x-y|^2).$
Further, the condition that boost symmetry be broken, is equivalent to requiring that the {\it pseudo-differential} operators $\int \left(c_1(s,t) (\partial_t \phi)^2(x,s)\right) ds$ and $ \int \left(\,-\,c_2(x,y)\,(\partial_i \partial_j \phi)^2(y)-c_3(x,y) (\nabla^2 \phi)^2(y)+\dots \right)d^dy$ be of different {\it order}.

Although most of our discussions apply to the general setting, nonetheless, for the sake of clarity, we will focus our attention on the case that the pseudo-differential operators are fractional Laplacians.
To apply this framework to diffusion, we break the Lorentz-invariant condition on $f(p^2)$.
In order to define the corresponding Lagrangian, we discuss a more general setting at first. We thus concern ourselves with the context in which we are given a mixed tensor $\Omega$ which for clarity of exposition we think of as being represented by a pair of tensors $A$ and $B$. In fact, since we are concerned with EM (or Yang-Mills theory more generally) we think of our mixed tensor $\Omega$ as being the sum of $2$ differential forms $A$ and $B$. We assume that the underlying manifold separates as $M=M_1\times M_2$. In this application, such a decomposition arises from the the switch from the Einsteinian to the Galilean perspective as a result of symmetry breaking.  Correspondingly, $M= \mathbb R\times M_2$ in this case (in particular this occurs for globally hyperbolic spacetimes).\begin{footnote}{ We don't really need the manifold to split as a product in the discussion that follows, but we just need the existence of a map $\pi: M\to M_2$ and a {\it horizontal} distribution of $TM$ relative to $\pi$.}\end{footnote}
This results in a decomposition on $r$-forms $\Omega ^r (M)=\bigoplus _{p+q=r}\, \Omega ^p  (M_1) \otimes \Omega ^q (M_2).$ Our notation here corresponds to regarding $ \pi_i^* \Omega ^s  (M_i) $ (for $i=1,2$) with $\Omega ^s  (M_i) $; that is an element of $\Omega ^s  (M_i) $ is understood as a linear combination $\alpha _{\ell _1 , \cdots \ell _s} dx^{\ell _1} \wedge \cdots \wedge dx^{\ell_s}$, where the $ \alpha _{\ell _1 , \cdots \ell _s} $'s are functions on the entirety of $M$ and $dx^{\ell _1},\cdots, dx^{\ell_s}$ arise from coordinates on $M_i$ only.
Letting $A$ and $B$ be represented in  coordinates as $A=A_I dx^I$ and $B=B_Jdx^J$, we use a multi-index notation, where $I=(i_1,\cdots , i_p)$ and $J=(j_1,\cdots , j_q)$, so $A= A_{i_1,\cdots, i_p} dx^{i_1}\wedge \cdots dx^{i_p}$, etc. The work horse of the non-Lorentzian picture will be the operator $D_\gamma$,
\begin{equation}
D_\gamma \Omega= d_1 \Omega +d_2\Delta _2 ^{\frac{\gamma -1}{2}} \Omega=  d_1 A+d_2\Delta _2 ^{\frac{\gamma -1}{2}} A+  d_1 B +d_2\Delta _2 ^{\frac{\gamma -1}{2}} B,
\end{equation}
where $d_i$ is the differential coming from $M_i$ and $\Delta _2= d_2d_2^{*_2}+ d_2^{*_2}d_2$ is the Hodge Laplacian on forms, where $*_2$ is the Hodge star operator on $\Omega ^r_2$  . This is clearly linear with respect to the decomposition of $\Omega$ when $\Omega= A+B$.  Thus we analyse it in terms of its coordinates for a given form. Let us indicate coordinates $x^i$ according to a separation arising from the split $M=M_1\times M_2$. So we write the coordinates $\{x^i\}_{i=1}^n$ for those coming from $M_1$ and $\{x^i\}_{i=n+1}^d$ for the ones coming from $M_2$. Whence, in terms of coordinates, for $A$ (and analogously for $B$), we have
\begin{equation} \begin{aligned} &D_\gamma A = \sum _{i+1}^n \frac{\partial } {\partial x^{i}} A_I\, d x^i\wedge dx^I+\sum _{i=n+1} ^d \frac{\partial } {\partial x^{\mu_i}}\left( \Delta _2^{\frac{\gamma-1}{2}}A_I\right) d x^{\mu_i}\wedge dx^I.\end{aligned}\end{equation}
In the special case (that we care about here) of breaking boost symmetry,
\begin{equation} \begin{aligned} &D_\gamma A = \frac{\partial } {\partial x^{0}}\, A_I \,d x^{0} \wedge dx^I+\sum _{i=1} ^d \frac{\partial } {\partial x^i}\left( \Delta _x^{\frac{\gamma-1}{2}}A_I\right) d x^i\wedge dx^I,\end{aligned}\end{equation}
where $\Delta _x^{\frac{\gamma-1}{2}}A_I$ is the fractional Laplacian on the spatial coordinates. In other words,
\begin{equation}
D_\gamma A = d_1 A+ d_{2,\gamma}\,A
\end{equation}
where $d_{2,\gamma}= d_2 \Delta _2^{\frac{\gamma-1}{2}} A$ is the fractional differential introduced in \cite{csforms} and can be computed using Eq (\ref{reisz}).

If we denote by $\star$ the usual Hodge Laplacian associated with a product metric $ds^2= ds_1^2 + ds_2^2$, it is a straightforward matter to show that
\beq
D_\gamma^*:= \star D_\gamma \star
\eeq
is an operator that takes a $p$-form to a $(p-1)$-form and it is the (formal) adjoint of $D_\gamma$ with respect to the standard scalar product on forms. We can show
\begin{theorem}
For any $p$-form $A = A_I \,dx^I$ we have
\begin{equation}
(D_\gamma ^* D_\gamma + D_\gamma D_\gamma ^* ) \,A= (\Delta _1 A_I )\,d^I + (\Delta _2 ^\gamma A_I)\, dx^I
\end{equation}
and in Lorentzian signature
\begin{equation}
(D_\gamma ^* D_\gamma + D_\gamma D_\gamma ^* ) \,A= (\Delta _1 A_I )\,d^I - (\Delta _2 ^\gamma A_I)\, dx^I
\end{equation}
In particular if we start with the mixed form $A=A_0\, dt+ A_{ij}\,dx^i\wedge dx^j$
\begin{equation}
\resizebox{\hsize}{!}{$(D_\gamma ^* D_\gamma + D_\gamma D_\gamma ^* ) A= \left( \frac{\partial ^2A_{0}}{\partial t^2}-\Delta_x ^\gamma A_ 0 \right)  dt  +\left( \frac{\partial ^2A_{ij}}{\partial t^2}  -\Delta _x^\gamma A_{ij}\right)dx^i \wedge dx^j\nonumber$}
\end{equation}
\end{theorem}

We now define a non-local $U(1)$-Gauge theory for a multi-form $\Omega$
\beq\label{non-lorentzEM}
D_\gamma \left( \star D_\gamma \Omega\right)=\star J,
\eeq
where $J$ is a mixed-form. We will specialize to the case that $\Omega = A_0 dt+ A_{ij} dx^i dx^j$ and then $J$ will be the form dual to the current $(d+1)$-vector $(\rho,J ^{ij})$. Equation \eqref{non-lorentzEM} is readily seen, after determining the Gauge fixing condition $D_\gamma ^* \Omega=0$, to be equivalent to
\beq
(D_\gamma ^* D_\gamma + D_\gamma D_\gamma ^* )\, \Omega= \star J.
\eeq
Writing as usual $ A_{ij}$ for the vector corresponding to the spatial components of $A$, Eq. \eqref{non-lorentzEM} is thus seen to be equivalent to
\begin{align} &\left( \frac{\partial ^2}{\partial t^2}-\Delta ^\gamma\right)  A_0 = \rho \\&  \left( \frac{\partial ^2}{\partial t^2}  -\Delta ^\gamma \right)\vec {A}= \vec {J}
\end{align}
the non-Lorentzian equations of motion for the conserved currents.

 The action of the Gauge group that yields such currents is easy to express in the case that $A$ is a Lie-valued $1$-form. We briefly described this without proofs since it follows from a simple generalization of what the authors did in \cite{csforms}. We describe it in the $U(1)$-case only. First, we note that the $U(1)$-action is determined by a diagonal action on a multi-form $\Omega=A_0+A_1$ (here $A_0$ and $A_1$ are both $1$-forms, so the decomposition is only meant to evoke the decomposition $M=M_1\times M_2$), meaning that given an element $e^{i\lambda}\in U(1)$, it will act on the individual components of $\Omega$ separately. We establish the following fact about our $U(1)$-Gauge action, denoting by $\Box_{(s_1,s_2)}:= \Delta _1^{s_1}-\Delta _2^{s_2}$ and in fact, as we argued, the most significant case is when $s_1=1$ and $s_2$ is arbitrary (as we can always reduce ourselves to this up to fields redefinitions). In what follows, we then write $\Box_s$ for $\Box_{1,s}.$
\begin{theorem}
We define an action of $U(1)$ on {\it local} sections of the principal bundle corresponding to $A$ as
\beq
e^{i\Lambda} \odot \phi= \Box_ \frac{1-\gamma}{2}\left( e^{i \Box_ \frac{\gamma-1}{2} \Lambda} \Box_ \frac{\gamma-1}{2}\phi\right),
\eeq
and we define the non-Lorentz invariant covariant derivative as
\beq
\nabla_{\gamma, A} \phi = (d+i\Box^ \frac{\gamma-1}{2}A) \Box_  \frac{\gamma-1}{2}\phi.
\eeq
The corresponding curvature will be
\beq
F_{\gamma, A} = (d+i\Box_ \frac{\gamma-1}{2}A) D_{\gamma, A} \phi.
\eeq
Then the equivariance condition
\beq
D_{\gamma, A} \left( e^{i\Lambda } \odot \phi\right)=  e^{i\Box_\frac{\gamma-1}{2}  \Lambda} D_{\alpha,\beta, A+d_\gamma \Lambda}\phi
\eeq
holds.
\end{theorem}
The proof here is a straightforward generalization of the equivariance condition presented in Sec. 4 of \cite{varna}. For the more general case of a multiform $\Omega$ which is not comprised of $1$-forms, it is easy to write the infinitesimal action of the Gauge group. In case $\Omega=A_0\,dt+ A_{ij} dx^i\wedge dx^j$ the infinitesimal action is given by
\begin{equation}
    A_0\to A_0 + \frac{\partial \Lambda }{\partial t}\qquad A_{ij} \to A_{ij} - \partial _i \partial _j \Lambda
\end{equation}

Writing the global (not infinitesimal) action in general is more complicated. One difficulty we mention is the fact that if $\Omega$ is even just a $p$-form with $p\neq 1$, there is no construction of a $G$-principal bundle of which $\Omega$ is the connection (or no covariant derivative obviously associated with it).  We leave this for later considerations.

We now specialize to the context where $\Omega$ is in fact a mixed tensor $\Omega= A_0 dt +  A_{ij}dx^i \wedge dx^j$.   We then use the transformation
\beq
 \partial_t \rightarrow \partial_t-A_t\,\quad \partial_i \partial_j \rightarrow \partial_i \partial_j -(- \Delta) ^{\frac{\gamma -1}{2}} A_{ij}
\eeq
and the coupling
\begin{equation}
    \sim\int  \left( (- \Delta) ^{\frac{\gamma -1}{2}} A_{ij}\right) J^{ij}= \int A_{ij}  \,(- \Delta) ^{\frac{\gamma -1}{2}} J^{ij}
\end{equation}
having integrated by parts so that $A_{ij}$ couples with $ (- \Delta) ^{\frac{\gamma -1}{2}} J^{ij}$.

Recall that the correlation functions of $\rho$ and $J$ are generated by
\beq
Z[A_t, A_{ij}]= \int exp \left[\,i \int d^{d+1} x \left( A_t \rho +(- \Delta) ^{\frac{\gamma -1}{2}} A_{ij} J^{ij}\right)\, \right]
\eeq
and the local symmetry implies that under a Gauge transformation $Z[A_t , A_{ij}]=  Z[A_t+ \partial _t \Lambda, A_{ij} -\partial_i \partial_j (- \Delta) ^{\frac{\gamma -1}{2}} \Lambda]$ .

Therefore, by Noether's first theorem
\begin{equation}
    \partial_t \rho +\partial_i \partial_j  \left( (- \Delta) ^{\frac{\gamma -1}{2}} \right) J^{ij}=0,
\end{equation}
meaning the ''current'' is
\begin{equation}
    \hat{J}^i= \partial_j \left( (- \Delta) ^{\frac{\gamma -1}{2}} J^{ij}\right).
\end{equation}
Given that no equation for the current of the form
\beq    J_{ij}= f(\rho) \,\delta_{ij}
\eeq
is valid as it is inconsistent\cite{gl1,csforms} with $U(1)$ symmetry,  the next  term is
\beq
    J_{ij}= \mathfrak{D}\, \partial_i \partial_j \,\rho \label{n1}
\eeq
or equivalently
 \beq
  (- \Delta) ^{\frac{\gamma -1}{2}}  J_{ij}= \mathfrak{D} \,\partial_i \partial_j  (- \Delta) ^{\frac{\gamma -1}{2}}\rho.
\eeq
Eq.\eqref{n1} is a direct generalization of the standard Fick's law.  From this we infer
\beq
    \partial_t \,\rho + \mathfrak{D}\,(-\Delta) ^\frac{\gamma +3}{2} \rho=0
\eeq
and therefore we find the associated diffusive  mode,
\begin{equation}
    \omega\,=\,-\,i\,\mathfrak{D}\,k^{\gamma +3}\,,
\end{equation}
which for $\gamma=-1$ reduces to the standard diffusion equation corresponding to the conservation of the total charge $\mathcal{Q}$ associated to the standard $U(1)$ invariance.  For the fracton case, we set $\gamma=1$ and we obtain the result in Ref.\cite{Gromov:2020yoc}. We see then that N\"other's Second Theorem entails a wide variety of modified diffusive behaviour indicative of anomalous diffusion \cite{klafter,fracdiff2016,nigmatullin}.\\

\section{Conserved Charges and Goldstone Effective Action}

Given the most general Lorentz violating Gauge principle, we can now investigate which are the corresponding conserved charges and the effective action for the Goldstone massless modes. Let us start with the example of  subdiffusive dynamics
\begin{equation}
    \langle x^2 \rangle \,\sim\,\mathcal{B}\,t^{1/2},
\end{equation}
which corresponds to the anomalous diffusive mode $\omega=-i \mathfrak{D}k^4$. This dynamics stems from the higher-order constitutive relation
\begin{equation}
   \hat{J}^i= \partial_j J^{ij}\,,\quad  J_{ij}= \mathcal{B} \,\partial_i \partial_j\,\rho
\end{equation}
where $\hat{J}^i$ is the spatial current and $\rho$ the charge density. It is simple to see that this modified constitutive relation follows from the conservation of the electric dipole,
\begin{equation}
    p\,=\,\int d^dx\, \,x\,\rho(x,t) \label{dip}.
\end{equation}
Moreover, it can be derived by gauging the following scalar effective action
\begin{equation}
    L\,=\,c_1 (\partial_t \phi)^2\,-\,c_2\,(\partial_i \partial_j \phi)^2-c_3 (\nabla^2 \phi)^2+\dots \label{act1}
\end{equation}
where terms $\sim (\partial_i \phi)^2$ are absent because they do not conserve the dipole \eqref{dip}. We gauge the action \eqref{act1} by introducing the covariant derivatives
\begin{equation}
    \partial_t \rightarrow \partial_t-A_t\,\quad \partial_i \partial_j \rightarrow \partial_i \partial_j - A_{ij}
\end{equation}
The tensorial current we just discussed comes directly from the minimal coupling with the gauge field, $\sim A_{ij}J^{ij}$. Notice that the conservation of the dipole implies that the dynamics of the emergent quasiparticles is confined along a subspace, in this case a line.

Borrowing from this simple example, we want to find now the general conserved charges and effective actions arising from the non-Lorentz invariant gauge symmetry defined in the previous section. Let us consider a field theory in which the generalized momentum of order $(\gamma+1)/2$ of a certain ''charge'' distribution $\rho(t,x)$ is a conserved quantity,
\begin{equation}
\frac{\partial}{\partial t}\,\mathfrak{m}_{\left(\frac{\gamma+1}{2}\right)}\,=\,\frac{\partial }{\partial t}\,\int\, \sum _i  a_i\, x_i^{\frac{\gamma+1}{2}}\,\rho(t,x)\,dx\,=\,0\,,\label{cont}
\end{equation}
where we remark that if $f(x_i)$ is a function of the sole variable $x_i$, then $(-\Delta )^\gamma f(x_i)= (-\Delta _{x_i})^\gamma f(x_i)$ where $ (-\Delta _{x_i})^\gamma$ is the (fractional) Laplacian in the variable $x_i$.  If $(\gamma+1)/2$ is an integer, then, $\mathfrak{m}_{\left(\frac{\gamma+1}{2}\right)}$ are the standard momenta, e.g. $\gamma=-1$ corresponds to total charge $Q$, $\gamma=1$ dipole $p$, etc. The continuity equation, giving rise to our anomalous diffusive mode, is simply a mathematical re-statement of the principle of conservation in Eq.\eqref{cont}. More precisely, the conservation of $\mathfrak{m}_{\left(\frac{\gamma+1}{2}\right)}$ can be re-written in differential form as the anomalous continuity equation,
\begin{equation}
\partial_t \,\rho + \mathfrak{D}\,(-\Delta) ^\frac{\gamma +3}{2} \rho=0
\end{equation}
In the same way, the effective action for the low-energy degrees of freedom takes the following form,
\begin{equation}
    L\,=\,D_\gamma \left( \star D_\gamma \phi\right)+\dots.
\end{equation}
and it represents a formal generalization of the example shown in Eq.\eqref{act1}.
\section{Conclusions}

It was the purpose of this paper to put under a single umbrella the myriad of diffusive-like equations that have proliferated recently \cite{anomlast,anomlast2} and in the past \cite{tf1,tf2,tf3,klafter,zaslavsky,fracdiffcrit,fracfickslaw,caffdiff} to describe anomalous diffusion.  What we have shown is that NST makes possible a myriad of diffusive-like equations of motion that only reduce to the traditional diffusive limit in special cases.  The source of this multiplicity is the high degeneracy of the eigenvalues of Eq. (\ref{mgen}) which generate a hierarchy of conserved charges.  At play here is the fact that Fick's law is not at all general.  Our work applies equally to Fourier's law for heat transport for which exceptions are well catalogued \cite{fpu}.  The key to this generalization is the non-Lorentzian formulation of the gauge symmetries and hence this work completes the fractional formulation of gauge theories we started previously \cite{csforms,varna,gl1}.

As a result of the work of Caffarelli and Silvestre \cite{cs} who have shown that second-order elliptic differential equations in the upper half-plane in ${\mathbb R}_+^{n+1}$ reduce to one with the fractional Laplacian, $(-\Delta)^\gamma$ at  ${\mathbb R}^n$, where a Dirichlet boundary condition is imposed, our work here has demonstrated that anomalous diffusion can be thought of as a problem in which collective excitations arise in a
lower-dimensional subspace. While it has been noticed that models requiring higher-rank $U(1)$ gauge theories \cite{PhysRevB.95.115139,PhysRevB.96.035119,PhysRevB.97.235102,Li:2019tje} lead to restricted particle motion in lower dimensional subspaces, the connection to fractional diffusive equations has not been made.  The work presented here suggests that for modes with dispersions (ripplons at superfluid interfaces\cite{mazets,haruki} of the form \begin{equation}
    \omega\,=\,-\,i\,D\,k^{\gamma+3},
\end{equation}
a deep connection exists between the parameter $\gamma$ and the dimensionality of the space for the confined motion in which the effective action is inherently non-local.
\vspace{0.25cm}
\section*{Acknowledgments}
MB thanks W.V.Liu, A.Lucas, A.Gromov, C.Morais-Smith, S.Moroz and D.N.Xuan  for enjoyable discussions and fruitful comments.
M.B. acknowledges the support of the Spanish MINECO ``Centro de Excelencia Severo Ochoa'' Programme under grant SEV-2012-0249.  P.W.P.
thanks DMR19-19143 for partial funding of this project.
\bibliography{hydrobib}
\onecolumngrid
\appendix
\clearpage
\section*{Appendix: Proof of Theorem III.1}
\begin{proof}
The proof is a simple generalization of the standard fact that holds for the normal Laplacian on forms.
For simplicity (and without loss of generality) we may assume that $M_1=\mathbb R^{n}$ and $M_2=\mathbb R^{d-n}$ with Euclidean metrics. We then calculate
\begin{equation}
D_\gamma A= \sum _{\substack{k=1\\j_k\in\{ 1, \cdots, n\} }}^{d-p} \frac{\partial}{\partial x^{j_k}}A_{i_1,\cdot , i_p}\, dx^{j_k}\wedge dx^{i_1}\wedge \cdots \wedge dx^{i_p}+\sum _{\substack{k=1\\j_k\in\{ n+1, \cdots, d-n\} }}^{d-p} \frac{\partial}{\partial x^{j_k}}(\Delta _2^{\frac{\gamma-1}{2}} A_{i_1,\cdot , i_p})\, dx^{j_k}\wedge dx^{i_1}\wedge \cdots \wedge dx^{i_p}
\end{equation} whence
\begin{equation}
\star D_\gamma A=\sum _{\substack{k=1\\j_k\in\{ 1, \cdots, n\} }}^{d-p} (-1) ^{p+k-1}\frac{\partial}{\partial x^{j_k}}A_{i_1,\cdot , i_p}\, dx^{j_1}\wedge \cdots \widehat{dx^{j_k}}\wedge  \cdots \wedge dx^{j_{d-p}}+\sum _{\substack{k=1\\j_k\in\{ n+1, \cdots, d-n\} }}^{d-p} (-1) ^{p+k-1}\frac{\partial}{\partial x^{j_k}}(\Delta _2^{\frac{\gamma-1}{2}} A_{i_1,\cdot , i_p})\, dx^{j_1}\wedge \cdots \widehat{dx^{j_k}}\wedge  \cdots \wedge dx^{j_{d-p}}
\end{equation}
which yields
\begin{equation}
\begin{aligned}
&D_\gamma \star D_\gamma A= \sum _{\substack{k=1\\j_k\in\{ 1, \cdots, n\} }}^{d-p} (-1) ^{p+k-1}\frac{\partial^2}{(\partial x^{j_k})^2 }A_{i_1,\cdot , i_p}\,dx^{j_k}\wedge dx^{j_1}\wedge \cdots \widehat{dx^{j_k}}\wedge  \cdots \wedge dx^{j_{d-p}}+\\&\sum _{\substack{k=1\\j_k\in\{ 1, \cdots, n\} }}^{d-p} (-1) ^{p+k-1}\frac{\partial^2}{\partial x^{j_k}\partial x^{i_\ell}}A_{i_1,\cdot , i_p}\,dx^{i_\ell}\wedge dx^{j_1}\wedge \cdots \widehat{dx^{j_k}}\wedge  \cdots \wedge dx^{j_{d-p}}+\\&\sum _{\substack{k=1\\j_k\in\{ n+1, \cdots, d-n\} }}^{d-p} (-1) ^{p+k-1}\frac{\partial^2}{(\partial x^{j_k})^2 }(\Delta _2^{\frac{\gamma-1}{2}} A_{i_1,\cdot , i_p})\, dx^{j_k}\wedge dx^{j_1}\wedge \cdots \widehat{dx^{j_k}}\wedge  \cdots \wedge dx^{j_{d-p}}+\\&\sum _{\substack{k=1\\j_k\in\{ n+1, \cdots, d-n\} }}^{d-p} (-1) ^{p+k-1}\frac{\partial^2}{\partial x^{j_k}\partial x^{i_\ell}}(\Delta _2^{\frac{\gamma-1}{2}} A_{i_1,\cdot , i_p})\, dx^{i_\ell}\wedge dx^{j_1}\wedge \cdots \widehat{dx^{j_k}}\wedge  \cdots \wedge dx^{j_{d-p}}
\end{aligned}
\end{equation}

and also
\begin{equation}
\begin{aligned}
&\star D_\gamma \star D_\gamma A=\\& \sum _{\substack{k=1\\j_k\in\{ 1, \cdots, n\} }}^{d-p} (-1) ^{p+p(d-p)}\frac{\partial^2}{(\partial x^{j_k})^2 }A_{i_1,\cdot , i_p}\,dx^{i_1}\wedge \cdots \wedge dx^{i_p}+\\&\sum _{\substack{k=1\\j_k\in\{ 1, \cdots, n\} }}^{d-p} (-1) ^{pd+\ell}\frac{\partial^2}{\partial x^{j_k}\partial x^{i_\ell}}A_{i_1,\cdot , i_p}\,dx^{j_k} \wedge dx^{1_1}\wedge \cdots \wedge \widehat{ dx^{i_\ell}}\wedge \cdots\wedge dx^{i_p}+\\&\sum _{\substack{k=1\\j_k\in\{ n+1, \cdots, d-n\} }}^{d-p} (-1) ^{p+p(d-p)}\frac{\partial^2}{(\partial x^{j_k})^2 }(\Delta _2^{\frac{\gamma-1}{2}} A_{i_1,\cdot , i_p})\, dx^{i_1}\wedge \cdots \wedge dx^{i_p}+\\&\sum _{\substack{k=1\\j_k\in\{ n+1, \cdots, d-n\} }}^{d-p} (-1) ^{pd+\ell}\frac{\partial^2}{\partial x^{j_k}\partial x^{i_\ell}}(\Delta _2^{\frac{\gamma-1}{2}} A_{i_1,\cdot , i_p})\, dx^{j_k} \wedge dx^{1_1}\wedge \cdots \wedge \widehat{ dx^{i_\ell}}\wedge \cdots\wedge dx^{i_p}.
\end{aligned}
\end{equation}
A similar calculation shows that
\begin{equation}
\begin{aligned}
&D_\gamma \star D_\gamma \star A=\\& \sum _{\substack{k=1\\j_k\in\{ 1, \cdots, n\} }}^{d-p} (-1) ^{p(d-p)+d-p+\ell-1}\frac{\partial^2}{(\partial x^{j_k})^2 }A_{i_1,\cdot , i_p}\,dx^{i_1}\wedge \cdots \wedge dx^{i_p}+\\&\sum _{\substack{k=1\\j_k\in\{ 1, \cdots, n\} }}^{d-p} (-1) ^{p(d-p)+d-p+\ell-1}\frac{\partial^2}{\partial x^{j_k}\partial x^{i_\ell}}A_{i_1,\cdot , i_p}\,dx^{j_k} \wedge dx^{1_1}\wedge \cdots \wedge \widehat{ dx^{i_\ell}}\wedge \cdots\wedge dx^{i_p}+\\&\sum _{\substack{k=1\\j_k\in\{ n+1, \cdots, d-n\} }}^{d-p} (-1) ^{p(d-p)+d-p+\ell-1}\frac{\partial^2}{(\partial x^{j_k})^2 }(\Delta _2^{\frac{\gamma-1}{2}} A_{i_1,\cdot , i_p})\, dx^{i_1}\wedge \cdots \wedge dx^{i_p}+\\&\sum _{\substack{k=1\\j_k\in\{ n+1, \cdots, d-n\} }}^{d-p} (-1) ^{p(d-p)+d-p+\ell-1}\frac{\partial^2}{\partial x^{j_k}\partial x^{i_\ell}}(\Delta _2^{\frac{\gamma-1}{2}} A_{i_1,\cdot , i_p})\, dx^{j_k} \wedge dx^{1_1}\wedge \cdots \wedge \widehat{ dx^{i_\ell}}\wedge \cdots\wedge dx^{i_p}
\end{aligned}
\end{equation} and therefore
\begin{equation}
(D_\gamma ^* D_\gamma + D_\gamma D_\gamma ^* ) A= (\Delta _1 A_I )d^I - (\Delta _2 ^\gamma A_I) dx^I
\end{equation}
as we wanted to prove.

\end{proof}
\end{document}


TITLE: Why is the largest eigenvalue Lipschitz continuous and not differentiable?
QUESTION [3 upvotes]: Let
$$
A:\mathbb R^n\to \mathbb R^{n\times n}
$$
where $A(x)$ is symmetric for any $x=(x_1,..,x_n)$. $$A(x) = A_0+x_1A_1+x_2A_2+...x_nA_n$$ and all $A$ is positive semidefinite.
Consider
$$
\underset{x\in\mathbb R^n}{\rm minimize}~ \lambda_{\rm max}(A(x))
$$
I would like to show that $\lambda_{\rm max}(A(x))$ is Lipschitz continuous but not differentiable.
Attempt:
It is Lipschitz since $\lambda_{\rm max}(A(x))$ is equal to $\|A(x)\|_2$ (is this true?).
Does the non-differentiability also follow from this fact?

REPLY [1 votes]: I was looking for a reference that I could cite. There is a theorem of Hoffman and Wielandt that says (the norm is the Frobenius norm):

Theorem 1. If $A$ and $B$ are normal matrices with eigenvalues $\alpha_1,\ldots,\alpha_n$ and $\beta_1,\ldots,\beta_n$ respectively, then there exists a suitable numbering of the eigenvalues such that $\sum_i \lvert \alpha_i - \beta_i \rvert^2 \leq \lVert A - B \rVert_F^2$.

For a Hermitian matrix $A$, the “suitable numbering” is such that the eigenvalues are ordered monotonously:

Remark 2. Although the arrangement of the eigenvalues mentioned in Theorem 1 is difficult, in general, to describe more explicitly, it is easy in the special case that $A$ is Hermitian. Then a “best” arrangement is $\alpha_1 \geq \ldots \geq \alpha_n$ and $\operatorname{Re}\beta_1 \geq \ldots \geq \operatorname{Re}\beta_n$.

See also The Wielandt-Hoffman theorem in Wilkinson (paragraph 48, page 104). There the version for real symmetric matrices is given.
Let $\lambda_k(\cdot)$ map a Hermitian matrix to its $k$-th largest eigenvalue and let $A$, $B$ be Hermitian. We obtain from the above, for every $1 \leq k \leq n$,
$$
    \lvert \lambda_k(A) - \lambda_k(B) \rvert^2 \leq\sum_i \lvert \lambda_i(A) - \lambda_i(B) \rvert^2 \leq  \lVert A - B \rVert_F^2,
$$
from which Lipschitz continuity with Lipschitz constant 1 follows. Note that this does not contradict the “Added…” remark by orangeskid, because different norms are used in the two answers.
Reference

Hoffman, A. J.; Wielandt, H. W.: The variation of the spectrum of a normal matrix. Duke Math. J. 20 (1953), no. 1, 37–39. doi:10.1215/S0012-7094-53-02004-3. https://projecteuclid.org/euclid.dmj/1077465062
Wilkinson, J. H.: The Algebraic Eigenvalue Problem (1988)


TITLE: Is this seemingly novel irrational constant also transcendental?
QUESTION [24 upvotes]: I recently discovered a constant that is constructed as follows:
$\chi=\sum_{n=1}^\infty (\frac{\cos{n}}{2|\cos{n}|}+\frac{1}{2}) 2^{-n}$
Furthermore I can prove that it is an irrational number whose decimal approximation is .555609809015...
I conjecture that it is also a transcendental number but have not been able to prove this.
Anyone else want to take a crack at it? I am not a professional mathematician so if this number is not novel, then I appologize in advance.
One of the reasons I want to know this is that if this prototype number is proven transcendental than that would imply that an infinite set of similar transcendental numbers exists on the open interval (0,1) that are given by
$\chi_f=\sum_{n=1}^\infty (\frac{\cos{f(n)}}{2|\cos{f(n)}|}+\frac{1}{2}) 2^{-n}$
where f(n) is any algebraic function of n.

REPLY [66 votes]: Yes. It's known to be transcendental. The sequence of coefficients of your number is a variant of a Sturmian sequence. It has very low complexity. The definition of this: let the digit sequence be $a_1,a_2,a_3\ldots$ taking values in $ \lbrace 0,1\ldots,d-1 \rbrace ^{\mathbb N}$. A subword of length $k$ is a string $a_ia_{i+1}a_{i+2}\ldots a_{i+k-1}$. The complexity, $p(k)$, is a function from $\mathbb N$ to $\mathbb N$ taking $k$ to the number of subwords of the sequence of length $k$.
In a 2007 paper in the Annals of Mathematics (vol 165, p547--565), Adamczewski and Bugeaud (On the complexity of algebraic numbers. I. Expansions in integer bases) showed that if a number is algebraic, then its digit sequence in base $b$ has complexity satisfying $p(k)/k\to\infty$.
In your case, the complexity of the sequence of base 2 digits satisfies $p(k)=2k$. How to see this?
Define a map $f$ from $[0,1)$ to $\lbrace 0,1\rbrace$ by $f(x)=1$ if $x\in [0,1/2)$ and 0 otherwise.
The $n$th term of your sequence is $f(\alpha n\bmod 1)$, where $\alpha=1/(2\pi)$. Write $T$ for the transformation from $[0,1)$ to itself given by $T(x)=x+\alpha\bmod 1$. Then the $n$th term is just $f(T^n0)$. The sub-block of the digit sequence of length $k$ starting at the $j$th term is $f(T^j0)\ldots f(T^{j+k-1}0)$. Since the $T^j0$ are dense in $[0,1)$, we need to ask how many blocks $f(x)\ldots f(T^{k-1}x)$ are possible.
Consider taking $x=0$ and moving it around the circle (=$[0,1)$) once. As you move it, the $T^ix$ also each move around the circle one time. The sequence changes each time one of the $(T^ix)_{0\le i< k}$ crosses 0 or 1/2. This is a total of $2k$ changes. Hence the sequence takes on $2k$ values as $x$ moves around the circle, hence the estimate for the complexity.


TITLE: Find minimal $x\in\Bbb N$ that solves the linear congruence
QUESTION [1 upvotes]: I need to find minimal $x\in\Bbb N$ that solves the linear congruences:
$6x \equiv 2 \pmod {\!4}$
$3x \equiv 6 \pmod {\!9}$
$x  \equiv 15 \pmod {\! 17}$
I divided the first congruence by $2$ and the second congruence by $3$, then used the Chinese remainder theorem.
I got $x\equiv 393 \pmod {2\cdot 3\cdot 17} $
I checked that this solution actually solves the 3 equations.
But is it the smallest one? If so, then why ?
Thanks.

REPLY [1 votes]: From the comments you're interested in using Chinese Remainder Theorem in general. Here is one algorithm you can use.
Inbetween, instead of using clever manipulations, you can use EEA, etc. as explained below.
$x\equiv 1\pmod{\! 2}\iff (x=2k+1$ for some $k\in\Bbb Z)$.
$\,\bmod{3}\!:\ \ \ x\equiv 2\iff \color{#00F}2k+1\equiv 2\iff \color{#00F}{-1}k\equiv 1\iff k\equiv -1\equiv 2$
$\iff (k=3n+2$ for some $n\in\Bbb Z)\iff x=2(3n+2)+1=6n+5$.
$\,\bmod{17}\!:\ \ \ x\equiv 15\equiv -2\iff 6n+5\equiv -2\iff 6n\equiv -7\equiv -24$
$\stackrel{:6}\iff n\equiv -4\equiv 13\iff n=17m+13$ for some $m\in\Bbb Z$.
$\iff x=6(17m+13)+5=102m+83$.
We've used clever manipulations here. If you want to solve it algorithmically in a short time, the numbers are huge and you don't see a simple solution to $ax\equiv b\pmod{\! m}$ using manipulations like those above, then use either Extended Euclidean algorithm or what Bill Dubuque calls Gauss's algorithm. Here is how you apply EEA:
$$ax\equiv b\pmod{\! m}\iff ax+mk=b$$ for some $k\in\Bbb Z$. This has a solution $(x,k)$ iff $(a,m)\mid b=:(a,m)h$ (see here).
Using EEA you find $x_1,k_1\in\Bbb Z$ such that $$ax_1+mk_1=(a,m)\iff a(x_1h)+m(k_1h)=b$$
So $x\equiv x_1h\pmod{\! m}$ is your solution.


TITLE: Why in a directional derivative it has to be a unit vector
QUESTION [5 upvotes]: Why when we compute the directional derivative we have to use the unit vector $\vec{\bf{u}}$? I know that using $2\vec{\bf{u}}$ would change the directional derivative as the second point is further apart. But why not use $\frac{\vec{\bf{u}}}{2}$ then? Wouldn't that gradient be "more precise"?

REPLY [2 votes]: At the class we were given the definition of a directional derivative when the direction vector is a unit vector. Because of that I got confused. The real definition can be found here: https://en.wikipedia.org/wiki/Directional_derivative (Variation using only direction of vector for euclidean space). Here we can see that using the unit vector just simplifies things and is not necessary. It is very similar to 2D derivative as the distance between the two point $f(x+hv)-f(x)$ is the length of vector $\vec{\bf{v}}$.


TITLE: Example of a group $G$, and prime $p$, such that every $p$-nilpotent subquotient of $G$ has normal Sylow-$p$-subgroup, but $G$ does not
QUESTION [2 upvotes]: A few years ago I proved a theorem (invariant-theoretic) which applies to all groups $G$ and primes $p$, such that $G$ contains a subquotient which is $p$-nilpotent with non-normal Sylow-$p$-subgroup.
The theorem is not true if $G$ contains a normal Sylow-$p$-subgroup.
I wondered if anyone could give me an example of a group such that neither of these holds?

REPLY [2 votes]: Going by the problem as stated in the title, there is no $p$-solvable example of what you want. This is because if $G$ is a $p$-solvable group in which the automizer of every $p^{\prime}$-subgroup of $G$ is a $p^{\prime}$-group, then $G$ has a normal Sylow $p$-subgroup ( sketch proof: by induction, we may suppose that $O_{p}(G) = 1$, and then the Hall-Higman centralizer lemma implies that $O_{p^{\prime}}(G)$ contains its centralizer, but the hypotheses imply that it is centralized by a Sylow $p$-subgroup of $G$, which must then be trivial).
However, I think an example of what you want (if I am reading it correctly) is provided by the alternating group $A_{5}= G$ for $p = 5.$ Then every proper section of $G$ has a normal Sylow $5$-subgroup, but $G$ clearly does not have a normal Sylow $5$-subgroup ( recall that a section of a group $G$ is a group of the form $X/Y$ where $Y \lhd X$ and $Y,X$ are subgroups of $G$. The section is said to be proper if either $X \neq G$ or $Y \neq 1$).
Later edit: In fact, for every prime $p>3,$ the simple group $G = {\rm PSL}(2,p)$ has the property that every proper section of $G$ has a normal Sylow $p$-subgroup, but $G$ itself does not have a normal Sylow $p$-subgroup.
Even later edit: We have now established that there examples of the type you seek for every prime $p >3.$ Now we note that there is no example of the type you want when $p=2.$ For if $G$ is a finite group such that all $2$-nilpotent sections of $G$ are $2$-closed, then every odd order subgroup of $G$ has automizer of odd order. We claim that this implies that $G$ has a normal Sylow $2$-subgroup. As above, by induction, we may suppose that $O_{2}(G) = 1.$ But now, if $G$ has even order, $G$ contains a involution $t$ and $t$ can invert no non-identity element of odd order. Hence $\langle t,t^{g} \rangle$ is a $2$-group for each $g \in G$  and hence (by the Baer-Suzuki theorem) $t \in O_{2}(G),$ a contradiction.
To complete the picture, set $G = {\rm SL}(2,8).$ Then all proper sections of $G$ have a normal Sylow $3$-subgroup, but $G$ itself does not.
Hence there is an example of the kind you want for every odd prime $p,$ but there is no such example for $p=2.$


TITLE: derivative of a parametric function at a point
QUESTION [0 upvotes]: I need to find the derivative of
$$l\begin{cases}
y=t\cdot \cos(t)\\
x=e^t-2t-1
\end{cases}
$$
at $(0,0)$.
How do I approach this?
Thank you.

REPLY [1 votes]: $$\frac{dy}{dx}\bigg|_{(0,0)}=\biggl(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\biggr)_{t=0}$$
Therefore $\frac{dy}{dx}\bigg|_{(0,0)}=\biggr(\frac{y'(t)}{x'(t)}\biggl)_{t=0}=\biggl(\frac{cos(t)-tsin(t)}{e^t-2}\biggr)_{t=0}=-1$


TITLE: Casimir effect as an entropic force
QUESTION [13 upvotes]: When I first learned about the depletion interaction, my initial reaction was that it looks very similar to the Casimir effect. On making this remark to the professor, he replied somewhat mystically: "It is the Casimir effect." No further detail was supplied, however.
Nevertheless, it really looks like both problems can be understood in terms of degrees of freedom maximising their positional entropy by reducing the volume of some "forbidden region" between two objects. For the depletion interaction, the fluctuating degrees of freedom are small particles, while in the Casimir effect, these degrees of freedom are long-wavelength modes of the free radiation field. However, a key difference is that in the first case the fluctuations arise thermally, while in the second case they are unavoidable quantum fluctuations of the vacuum.

Is it possible to derive the Casimir force between two conducting plates from entropic considerations alone?

REPLY [2 votes]: Partial answer:
Because the distance between the plates is finite, you have a natural correspondence between the quantum Casimir problem, and a thermodynamic problem.
In fact, in a $d$-dimensional space, if you calculate the Helmholtz free energy, in a grand-canonical formalism, for a perfect photon gaz (so with zero chemical potential), with the correspondence $\beta =  \frac{2a}{\hbar c}, V = aL^{d-1}$, for plates with characteristic length $L$ separated by a distance $a$ , you will find exactly the Casimir energy for a photon field.
And it works for fermions too, where you have to consider a perfect gaz of relativist fermions (with zero chemical potential).


TITLE: Find the Coordinates of a Unkown Point of a Triangle
QUESTION [1 upvotes]: The situation is as follows: I am creating a game in this game I have a line($A$ to $B$) and a mouse position $C$. Now I want to calculate point $D$ on line $A$$B$.
I know the coordinates of: $A$, $B$ and $C$ and the angle of point $B$. It should be possible to calculate point $D$ but I have no clue where to start. Can any one point me in a direction?
Note: it is possible that point $A$ and $B$ are on the same axis. For example: $A(1,1)$ and $B(1,3)$.

REPLY [0 votes]: Do it with vectors!
Let A, B and C be position vectors of points A, B, C, so AB ( = A - B) is the vector from A to B, and DC (= D - C) is from D to C. Now AB and DC are perpendicular, so their dot product is zero. Let t be the ratio AD/AB (so it's 1 if D is at B, and 0 if it's at A).
These requirements give us:
$$ D = A + tAB $$
$$ (D-C).AB = 0 $$
Solve for t:
$$ (A + tAB - C).AB = 0 $$
$$ (A - C).AB - t(AB.AB) = 0 $$
$$ t = \frac{(A-C).AB}{AB.AB} = \frac{(A-C).AB}{\|AB\|^2}$$
Herewith some Python:
class Vector:
    def __init__(self, x, y):
        (self.x, self.y) = (x,y)

    def sub(self, other):
        return Vector( self.x - other.x, self.y - other.y )

    def add(self, other):
        return Vector( self.x - other.x, self.y - other.y )

    def dot(self, other):
        return self.x*other.x + self.y*other.y

    def scalarMul(self, s):
        return Vector( self.x * s, self.y * s )

def intersect(A,B,C):
    AB = A.sub(B)
    AC = A.sub(C)
    t = (AC.dot(AB)) / (AB.dot(AB))
    return A.add(AB.scalarMul(t))

A=Vector(4.5, 0.5)
B=Vector(0.5, 1.5)
C=Vector(1.5, 0.5)

D=intersect(A,B,C)
print D.x, D.y