text:
\begin{document}

\title{Heisenberg Hausdorff dimension \\ of Besicovitch sets}
\author{Laura Venieri}
\date{}

\maketitle

\begin{abstract}
We consider (bounded) Besicovitch sets in the Heisenberg group and prove that $L^p$ estimates for the Kakeya maximal function imply lower bounds for their Heisenberg Hausdorff dimension.
\end{abstract}

\blfootnote{\textit{Key words.} Besicovitch set, Kakeya maximal function, Heisenberg group, Hausdorff dimension.\\
\textit{Mathematics Subject Classification}. 28A75.\\
The author is supported by the Academy of Finland.}

\section{Introduction}

In this paper we investigate \textit{Besicovitch sets} (also called Kakeya sets) in the context of the Heisenberg group, examining in particular their Heisenberg Hausdorff dimension. The Kakeya sets contain a unit line segment in every direction and have Lebesgue measure zero. Their name comes from a question that Kakeya posed in 1917: what is the smallest area in which a unit line segment can be rotated of 180 degrees in the plane? Besicovitch proved that this can be done in arbitrarily small area.

In recent years much research has been done about their (Euclidean) Hausdorff dimension. The \textit{Kakeya conjecture} states that every Besicovitch set in $\mathbb{R}^n$ must have Hausdorff dimension $n$. Davies \cite{Davies1971} proved that this holds in $\mathbb{R}^2$ but it is still an open problem in $\mathbb{R}^n$ for $n \ge 3$, even if these sets have been studied extensively because of their close connection to central questions in modern Fourier analysis.

Many different methods have been used to find lower bounds for the Hausdorff dimension: the first were based on geometric observations and developed in particular by Bourgain \cite{Bourgain1991} and Wolff \cite{MR1363209}. More recently Bourgain \cite{Bourgarithm} introduced an arithmetic combinatorics method obtaining the lower bound $\frac{13}{25}n + \frac{12}{25}$, then improved by Katz and Tao \cite{Katz&Tao1} to $\frac{6}{11}n+\frac{5}{11}$. Later Katz and Tao \cite{Katz&Tao2} developed the method further and got $(2 - \sqrt{2}) (n-4)$, which is the best estimate known at the moment for $n \ge 5$.

A natural approach to the problem is via a related maximal function, the \textit{Kakeya maximal function} $f^*_\delta$ (with width $\delta >0$). This is defined for $f \in L^1_{loc }(\mathbb{R}^{n})$ as
\begin{eqnarray*}
\begin{split}
&f^*_\delta: S^{n-1} \rightarrow [0, \infty],\\
f^*_\delta(e)= \sup_{a \in \mathbb{R}^{n}} & \frac{1}{\mathcal{L}^{n}(T^\delta_e(a))} \int_{T^\delta_e(a)} |f| d \mathcal{L}^{n},
\end{split}
\end{eqnarray*}
where $S^{n-1}$ is the unit sphere and $T^\delta_e(a)$ denotes the tube with center $a$, direction $e$, length 1 and radius $\delta$:
\begin{equation*}
T^\delta_e(a)= \{ p \in \mathbb{R}^{n}: |\langle p-a, e \rangle | \le 1/2, |p-a-\langle p-a, e \rangle e| \le \delta \}.
\end{equation*} The \textit{Kakeya maximal conjecture} states that the following inequality
\begin{equation}\label{Kakeyamaxfunction}
||f^*_\delta||_{L^n(S^{n-1})} \le C_{n,\epsilon} \delta^{- \epsilon} ||f||_{L^n(\mathbb{R}^n)}
\end{equation}
holds for all $\epsilon>0$, $0 < \delta <1$ and $f \in L^n(\mathbb{R}^n)$. In particular, this conjecture implies the Kakeya conjecture.

We will show that, as in the case of the Euclidean Hausdorff dimension, estimates of type \eqref{Kakeyamaxfunction} for $L^p$ functions imply lower bounds for the Heisenberg Hausdorff dimension of Besicovitch sets. The proof we present holds only for bounded sets because of the structure of the Heisenberg group and in particular the fact that the directions of horizontal segments can get arbitrarily close to the vertical axis in unbounded sets.

We will use the following notation. The Lebesgue measure in $\mathbb{R}^n$ is denoted by $\mathcal{L}^n$ and the $s$-dimensional (euclidean) Hausdorff measure by $\mathcal{H}^s$, $s\ge0$. We recall that this is defined for any $A \subset\mathbb{R}^n$ by
\begin{equation*}
\mathcal{H}^s(A)= \lim_{\delta \rightarrow 0}\mathcal{H}_\delta^s(A),
\end{equation*}
where for $\delta >0$
\begin{equation*}
\mathcal{H}_\delta^s(A) =\inf \left\{ \sum_i d(E_i)^s : A \subset \bigcup_i E_i, d(E_i) < \delta \right\}.
\end{equation*}
Here $d(E_i)$ denotes the diameter of $E_i$ with respect to the euclidean metric.

We denote by $\sigma^{n-1}$ the surface measure on $S^{n-1}$ and by $B_E(p,r)$ the Euclidean ball with center $p$ and radius $r$. The notation $C_a$ means that the constant $C$ depends on $a$.

The Heisenberg group $\mathbb{H}^n$ is $\mathbb{R}^{2n+1}$, where we denote the points by
\begin{equation*}
p=(x,y,t)=(x_1, \dots, x_n, y_1, \dots, y_n, t)
\end{equation*}
with $x_j$, $y_j$, $t \in  \mathbb{R}$ for every $j=1, \dots, n$, with the composition law given by
\begin{equation*}
(x,y,t) \ast (x',y',t')= (x+x', y+y', t+t'+2( \langle y, x' \rangle - \langle x, y' \rangle)),
\end{equation*}
where $\langle \cdot, \cdot \rangle$ is the usual inner product in $\mathbb{R}^n$. The inverse of $p$ is $p^{-1}=(-x,-y,-t)$.

We will work with the \textit{Heisenberg metric} on $\mathbb{H}^n$, also known as the \textit{Kor\'anyi metric}. This is the left invariant metric given by
\begin{equation*}
d_H(p,q)= ||q^{-1} \ast p ||_H,
\end{equation*}
where $|| \cdot ||_H$ is the homogeneous norm defined by
\begin{equation*}
||p||_H= ((||x||^2+||y||^2)^2 +t^2)^{\frac{1}{4}}.
\end{equation*}

Here $||\cdot||$ denotes the euclidean norm. We denote by $\tau_p$ the left translation by $p$, i.e. $\tau_p(q)= p*q$. Horizontal lines in $\mathbb{H}^n$ are either lines through the origin in the $xy$-hyperplane or left translations of them.

The Heisenberg ball centred at $p$ with radius $r$ is the set
\begin{equation*}
B_H(p,r)= \{ q \in \mathbb{H}^n : d_H(p,q) <r \}.
\end{equation*}

We will estimate the Heisenberg Hausdorff dimension, which means that we consider the $s$-dimensional Hausdorff measure $\mathcal{H}^s_H$ defined with respect to the Heisenberg metric. The Heisenberg Hausdorff dimension of a set $A \subset \mathbb{H}^n$ is then defined in the usual way as
\begin{equation*}
\dim_H A= \inf \{ s: \mathcal{H}^s_H(A)=0 \} = \sup \{ s: \mathcal{H}^s_H(A)= \infty \}.
\end{equation*}
The Heisenberg Hausdorff dimension is greater or equal to the Euclidean one; for example, the Heisenberg Hausdorff dimension of $\mathbb{H}^n$ is $2n+2$.

A \textit{Kakeya set} in $\mathbb{H}^{n}$ is a Borel set $B \subset \mathbb{H}^{n} $ having zero Lebesgue measure ($\mathcal{L}^{2n+1}(B)=0$) and containing a line segment of unit length in every direction. This means that for every $e \in S^{2n}$ there exists $b \in \mathbb{R}^{2n+1}$ such that $\{ te+b: t \in [0,1] \} \subset B$.

The \textit{Kakeya maximal function} $f^*_\delta$ of $f \in L^1_{loc }(\mathbb{R}^{2n+1})$ is defined as above.

We will prove the following result.

\begin{theorem}\label{Th1}
Let $1<p<2n+1$, $\beta>0$ such that $2n+2-\beta p >0$. If
\begin{equation}\label{kn}
||f^*_\delta||_{L^p(S^{2n})} \le C_{n,p,\beta} \delta^{-\beta} ||f||_p \quad \mbox{for all} \quad f \in L^p(\mathbb{R}^{2n+1}),
\end{equation}
then the Heisenberg Hausdorff dimension of every bounded Besicovitch set in $\mathbb{H}^n$ is at least $2n+2-\beta p$. In particular, if \eqref{kn} holds for some $p$, $1 \le p < \infty$, for all $\beta >0$, then the Heisenberg Hausdorff dimension of every bounded Besicovitch set in $\mathbb{H}^n$ is 2n+2.
\end{theorem}

We recall that the analogous result in the case of Euclidean Hausdorff dimension gives the lower bound $2n+1 - \beta p$ (\cite{Mattila}, Theorem 22.9). Note that this shows in particular that Kakeya maximal conjecture implies Kakeya conjecture.

Using then inequalities of type \eqref{kn} proved by Wolff \cite{MR1363209} and Katz and Tao \cite{Katz&Tao2} for some values of $p$ and $\beta$, we will get as a corollary some lower bounds for the Heisenberg Hausdorff dimension of bounded Besicovitch sets.

\section{Bounds derived from estimates of the maximal function}

For the proof of Theorem \ref{Th1} we will use the following lemma, which states how close to the vertical axis horizontal segments can get in a bounded set.

\begin{lem}\label{anglen}
Any horizontal segment contained in a bounded set $ \subset B_E(0,R) \subset \mathbb{H}^n$ forms an angle with the $xy$-hyperplane which is at most $ \arccos \frac{1}{\sqrt{1+4R^2}} $.
\end{lem}

\begin{proof}
Let $e=(e^1,e^2,0)=(e_1, \dots, e_n, e_{n+1}, \dots, e_{2n},0) \in S^{2n}$ be a direction in the $xy$-hyperplane. A horizontal line is either a line in the $xy$-hyperplane through the origin, which has the form \begin{equation*}
\gamma(s)=(se^1,se^2,0), \qquad s \in \mathbb{R},
\end{equation*}
or a left translation of it by $p=(x,y,t) \in \mathbb{H}^n$:
\begin{equation*}
\tau_p(\gamma(s))= (se^1+x, se^2+y, t+ 2s(\langle y, e^1 \rangle - \langle x, e^2 \rangle)), \qquad s \in \mathbb{R}.
\end{equation*}

We now want to find the angle between such lines and the $xy$-hyperplane. In the first case, of course, the angle is 0.

The direction of $\tau_p(\gamma(s))$ is given by the vector $u=(e^1,e^2, 2(\langle y, e^1 \rangle - \langle x, e^2 \rangle))$, whose projection on the $xy$-hyperplane is the vector $e=(e^1,e^2,0)$. The angle between them is given by
\begin{equation*}
\theta= \arccos \frac{\langle u, e \rangle}{||u|| ||e||}= \arccos \frac{1}{\sqrt{1+4(\langle y, e^1 \rangle - \langle x, e^2 \rangle)^2}}.
\end{equation*}

Since we are interested in segments contained in $B_E(0,R)$, we consider translations by points $p=(x,y,t) \in B_E(0,R)$, that is $||(x,y,t)|| < R$. Then we try to estimate $(\langle y, e^1 \rangle - \langle x, e^2 \rangle)^2$. We can write
\begin{equation*}
(\langle y, e^1 \rangle - \langle x, e^2 \rangle)^2= |\langle (e^1, -e^2), (y,x) \rangle_{\mathbb{R}^{2n}} |^2 \le ||(e^1,-e^2)||^2_{\mathbb{R}^{2n}} ||(y,x)||^2_{\mathbb{R}^{2n}},
\end{equation*}
where the last inequality follows from Cauchy-Schwarz inequality. Since $||(e^1,-e^2)||_{\mathbb{R}^{2n}}=1$, we have
\begin{equation}\label{aybxRn}
(\langle y, e^1 \rangle - \langle x, e^2 \rangle)^2 \le ||(y,x)||^2_{\mathbb{R}^{2n}} < R^2-t^2 \le R^2.
\end{equation}

Thus
\begin{equation*}
\theta \le \arccos \frac{1}{\sqrt{1+4R^2}}.
\end{equation*}
\end{proof}

The proof of Theorem \ref{Th1} proceeds in the same way as in the case of Euclidean Hausdorff dimension, see Theorem 22.9 in \cite{Mattila}.

\begin{proof} (Theorem \ref{Th1})\\
Let $B \subset \mathbb{H}^n$ be a bounded Besicovitch set, $B \subset B_E(0,R)$ for some $R>0$. Let $0<\alpha < 2n+2-\beta p$ and for $j \in \mathbb{N}$, $j \ge1$, let $B_j= B_H(p_j, r_j)$ be Heisenberg balls such that $p_j \in B_E(0,R)$, $B \subset \bigcup_j B_j$, and $r_j \le 2^{-M}$, where $M=\left[\frac{\log(3/\tan \theta_R)}{\log 2} \right] +1$ and $\theta_R= \arccos \frac{1}{\sqrt{1+4(R+1)^2}} - \arccos \frac{1}{\sqrt{1+4R^2}} >0$. Here $[\quad]$ denotes the integer part.

Let S be the set of  $e \in S^{2n}$ such that the angle between $e$ and the $xy$-hyperplane is bigger than $\theta$, where $\theta =\arccos \frac{1}{\sqrt{1+4(R+1)^2}} > \arccos \frac{1}{\sqrt{1+4R^2}}$.
Then, by Lemma \ref{anglen}, S does not contain any horizontal direction.

Since $B$ is a Besicovitch set, it contains a unit segment parallel to $e$ for every $e \in S$. Denote by $I_e \subset B$ one of these segments. For $k=M+1,\dots$, let
\begin{equation*}
J_k = \{j: 2^{-k} \le r_j < 2^{1-k} \},
\end{equation*}
and
\begin{equation*}
S_k= \left\{e \in S : \mathcal{H}^1\left(I_e \cap \bigcup_{j \in J_k} B_j \right) \ge \frac{1}{2k^2} \right\}.
\end{equation*}

Then we have
\begin{equation*}
 \bigcup_{k=M}^\infty S_k =S.
\end{equation*}

Indeed, if there were some $e \in S \setminus \bigcup_k S_k$, then we would have $\mathcal{H}^1(I_e \cap \bigcup_{j \in J_k} B_j ) < \frac{1}{2k^2}$ for all  $k \ge M$. Since $\sum_{k=M}^\infty \frac{1}{2k^2} < \sum_{k=1}^\infty \frac{1}{2k^2}<1$, this would imply
\begin{equation*}
\sum_{k=M}^\infty \mathcal{H}^1\left(I_e \cap \bigcup_{j \in J_k} B_j \right) < \sum_{k=M}^\infty \frac{1}{2k^2} <1,
\end{equation*}
which is impossible because
\begin{equation*}
\sum_{k=M}^\infty \mathcal{H}^1\left(I_e \cap \bigcup_{j \in J_k} B_j \right) \ge \mathcal{H}^1 (I_e) \ge 1.
\end{equation*}

Let \begin{equation*}
F_k= \bigcup_{j \in J_k} B_H (p_j, R_j) \quad \mbox{and} \quad f= \chi_{F_k},
\end{equation*}

where $R_j=2 \left( 1+ \frac{1}{\sin \theta_R}\right) r_j$.

We claim that for $e \in S_k$
\begin{equation}\label{intFk}
\mathcal{L}^{2n+1} (T \cap F_k ) \ge C_{R,n}  \frac{1}{k^2} \mathcal{L}^{2n+1}(T),
\end{equation}
where $T=T^{2^{-k}}_e(b)$ with $b \in \mathbb{H}_n$ that will be specified later.

We first show how this concludes the proof and then we prove \eqref{intFk}.

By \eqref{intFk}, we have that, for $e \in S_k$, $f^*_{2^{-k}}(e) \ge C_{R,n} \frac{1}{k^2}$, which implies
\begin{equation*}
||f^*_{2^{-k}}||_p \ge C_{R,n} \frac{1}{k^2} \sigma^{2n}(S_k)^{1/p}.
\end{equation*}

Observe that $||f||_p = (\mathcal{L}^{2n+1}(F_k))^{1/p} \le (C_{R,n} \# J_k 2^{(1-k)(2n+2)})^{1/p}$, where $2^{(1-k)(2n+2)}$ is essentially the volume of a Heisenberg ball of radius $2^{1-k}$. Hence by the assumption \eqref{kn} we have
\begin{equation*}
||f^*_{2^{-k}}||_p \le C_{n,p,\beta} 2^{k\beta} ||f||_p \le  C_{n,p,\beta} 2^{k\beta} (\#J_k 2^{(1-k)(2n+2)})^{1/p}.
\end{equation*}

Combining these two inequalities, we get
\begin{eqnarray*}
\begin{split}
\sigma^{2n}(S_k) &\le C_{n,p, \beta,R} k^{2p} 2^{k \beta p} 2^{-(2n+2)k} \# J_k=C_{n,p, \beta,R} k^{2p} 2^{-k(2n+2-\beta p)} \# J_k \le \\ &\le C_{n,p, \beta,R} 2^{-k \alpha} \#J_k.
\end{split}
\end{eqnarray*}

Hence it follows
\begin{equation*}
\sum_{j=1}^\infty r_j^\alpha \ge \sum_{k=M}^\infty \# J_k 2^{-k \alpha} \ge C_{n,p, \beta,R} \sum_{k=M}^\infty \sigma^{2n}(S_k) \ge C_{n,p, \beta,R} \sigma^{2n}(S),
\end{equation*}
which concludes the proof.

We now prove \eqref{intFk}, which is trivial in the euclidean case.

First we observe that if $p \in B_H(p_j,r_j) \cap I_e$, then for any other $q \in B_H(p_j,r_j)$ we have $q \in B_H(p,2r_j)$. Hence, taking $\tilde{B}_j=B_H(p_j,2r_j)$, we can assume that $p_j \in I_e$. Then we still have for $e \in S_k$
\begin{equation*}
\mathcal{H}^1\left(I_e \cap \bigcup_{j \in J_k} \tilde{B}_j \right) \ge \frac{1}{2k^2}.
\end{equation*}

We will use the following notation: $I_e$ is given by the points $\gamma(u)=( ue_1+c_1, ue_2+c_2, ue_3 + c_3)$, with $e=(e_1,e_2,e_3) \in S$, $c=(c_1,c_2,c_3) \in \mathbb{H}^n$, $u \in [0,1]$, and $p_j=\gamma(\bar{u})=(\bar{x}, \bar{y}, \bar{t})$.

By definition, given some $j$, the Heisenberg ball $B_H(p_j,2r_j)$ is the set of points $q=(x,y,t) \in \mathbb{H}^n$ such that
\begin{equation*}
d_H(p_j,q)= ||q^{-1} \ast p_j||_H= ((||\bar{x}-x||^2 + ||\bar{y}-y||^2 )^2 + (\bar{t}-t + 2 (\langle x, \bar{y} \rangle - \langle y, \bar{x} \rangle)^2)^{1/4} <2 r_j.
\end{equation*}

If we let $H_j= \{(x,y,t) \in \mathbb{H}^n : \bar{t}-t + 2 (\langle x, \bar{y} \rangle - \langle y, \bar{x} \rangle )=0 \}$, then
\begin{equation*}
B_H(p_j,2r_j) \cap H_j= \{ (x,y,t) \in \mathbb{H}^n: (||\bar{x}-x||^2 + ||\bar{y}-y||^2 )^{1/2} < r_j, t= \bar{t} +  2 (\langle x, \bar{y} \rangle - \langle y, \bar{x} \rangle) \}.
\end{equation*}

\textbf{Step 1:} In case $p_j=(0,0,\bar{t})$, $H_j$ is parallel to the $xy$-hyperplane. If $p_j \ne (0,0,\bar{t})$, then we need to determine what angle $H_j$ makes with the $xy$-hyperplane.

A normal vector to $H_j$ is $n_j=(-2 \bar{y},  2 \bar{x}, 1)$, whereas the unit normal vector to the $xy$-hyperplane is $n=(0,0,1)$. Hence the angle between them (which is the angle between $H_j$ and the $xy$-hyperplane) is
\begin{equation*}
\theta_j= \arccos \frac{\langle n_j, n \rangle}{||n_j|| ||n||}= \arccos \frac{1}{\sqrt{1+4(||\bar{y}||^2+|| \bar{x}||^2})}.
\end{equation*}

We have $||\bar{y}||^2+ ||\bar{x}||^2 < R^2$ because $p_j \in B_E(0,R)$. Hence
$\theta_j \le \arccos \frac{1}{\sqrt{1+4R^2}}$.

Since, for $e \in S_k$, the angle that $I_e$ makes with the $xy$-hyperplane is $\bar{\theta} > \arccos \frac{1}{\sqrt{1 + 4 (R+1)^2}}$, we have that the angle between $H_j$ and $I_e$ is
\begin{equation*}
\bar{\theta} - \theta_j \ge \theta_R.
\end{equation*}

\textbf{Step 2:} Now we claim that if $a \in I_e$ belongs to a ball $B_H(p_j,2 r_j)$, then all the segments with direction parallel to $H_j$, one endpoint in $a$ and length $\frac{2^{-k}}{\sin \theta_R}$ are contained in the ball $B_H(p_j, R_j)$.

Let $a=(a_1,a_2,a_3) \in B_H(p_j,2 r_j)$. This means that
\begin{equation}\label{ainBrj}
(||\bar{x} - a_1||^2+ ||\bar{y}-a_2||^2)^2 + (\bar{t}-a_3 + 2 (\langle a_1, \bar{y} \rangle - \langle a_2, \bar{x} \rangle))^2 < (2r_j)^4.
\end{equation}

A point $q$ in any segment with direction parallel to $H_j$, one endpoint in $a$ and of length $ \frac{2^{-k}}{\sin \theta_R} $ is given by $q=\sigma^j_{e'}(s)$, with
\begin{equation}\label{segment}
\sigma^j_{e'}(s)=\left( se'_1+ a_1, s e'_2 + a_2, 2s (\langle e'_1, \bar{y} \rangle - \langle e'_2, \bar{x} \rangle) + a_3 \right),
\end{equation}
where $e'=(e'_1,e'_2)$ is in the unit sphere $S_P$ contained in the hyperplane $P-c$ and $P$ is the hyperplane orthogonal to $I_e$ passing through $c$ ($P-c$ is the translate of $P$ passing through the origin), $||e'_1||^2 + ||e'_2||^2 =1 $, and $0 \le s \le \frac{2^{-k}}{\sin \theta_R}  \le \frac{ r_j}{\sin \theta_R}$.

We want to show that $q \in B_H(p_j, R_j)$. This means that
\begin{eqnarray*}
\begin{split}
&(||\bar{x}-se'_1-a_1||^2+ ||\bar{y}- s e'_2-a_2||^2)^2 + \\ &+ \left(\bar{t}- 2s (\langle e'_1, \bar{y}, \rangle - \langle e'_2, \bar{x} \rangle) - a_3 + 2 ( \langle se'_1+a_1,\bar{y} \rangle - \langle s e'_2 + a_2, \bar{x} \rangle) \right)^2 < R_j^4,
\end{split}
\end{eqnarray*}
that is
\begin{eqnarray}\label{qinBRj}
\begin{split}
(||\bar{x}-se'_1-a_1||^2+ ||\bar{y}- s e'_2-a_2||^2)^2 + (\bar{t} - a_3 + 2 (\langle a_1, \bar{y} \rangle - \langle a_2, \bar{x} \rangle))^2 < R_j^4.
\end{split}
\end{eqnarray}

By a direct calculation, the left-hand side equals
\begin{eqnarray*}
\begin{split}
& (||\bar{x} - a_1||^2+ ||\bar{y}-a_2||^2)^2 + (\bar{t}-a_3 + 2 (\langle a_1, \bar{y} \rangle - \langle a_2, \bar{x} \rangle))^2 + \\
&+ s^4
- 4 s^3 (\langle e'_1, \bar{x}-a_1 \rangle + \langle e'_2, \bar{y}- a_2 \rangle )+ \\
& + 4 s^2 (\langle e'_1, \bar{x}-a_1 \rangle + \langle e'_2, \bar{y}- a_2 \rangle )^2+ 2 s^2 (||\bar{x} - a_1||^2+ ||\bar{y}-a_2||^2) + \\
&- 4s (||\bar{x} - a_1||^2+ ||\bar{y}-a_2||^2)(\langle e'_1, \bar{x}-a_1 \rangle + \langle e'_2, \bar{y}- a_2 \rangle ).
\end{split}
\end{eqnarray*}

Using \eqref{ainBrj} and the inequalities
\begin{itemize}

\item $s \le \frac{r_j}{\sin \theta_R}$,

\item $ (\langle e'_1, \bar{x}-a_1 \rangle + \langle e'_2, \bar{y}- a_2 \rangle )^2 = (\langle (e'_1, e'_2), (\bar{x}-a_1, \bar{y}-a_2 ) \rangle )^2 \le \\ \le ||(e'_1, e'_2)||^2 ||(\bar{x}-a_1, \bar{y}-a_2 )||^2 \le ||\bar{x}-a_1||^2 + ||\bar{y}- a_2||^2 < 4 r_j^2$,

\end{itemize}
we find the following upper bound for the left-hand side of \eqref{qinBRj}:
\begin{equation*}\label{lhs}
16\left(1+ \frac{4}{\sin \theta_R}+ \frac{6}{(\sin  \theta_R)^2} + \frac{4}{(\sin \theta_R)^3} + \frac{1}{(\sin \theta_R)^4} \right) r_j^4= \left( 1 + \frac{1}{\sin \theta_R} \right)^4(2r_j)^4= R_j^4.
\end{equation*}

Hence the claim is proved.

\textbf{Step 3:} The segments $\sigma^j_{e'}(s)$ considered above are in general not parallel to each other when $p_j$ varies along $I_e$.

Consider the segment $\sigma^j_{e'}(s)$ given by \eqref{segment}, where $\bar{x}= ue_1+c_1$ and $\bar{y}=ue_2+c_2$. Its direction is given by the vector
\begin{eqnarray*}
\begin{split}
v(u)&=( e'_1, e'_2, 2(\langle e'_1, ue_2+c_2 \rangle - \langle e'_2, ue_1+c_1 \rangle))= \\&=( e'_1, e'_2, 2( u C_{e,e'} + C_{e',c})),
\end{split}
\end{eqnarray*}
where $C_{e,e'}= \langle e'_1,e_2 \rangle - \langle e'_2,e_1 \rangle$ and $C_{e',c}= \langle e'_1,c_2 \rangle - \langle e'_2,c_1 \rangle $.

The norm of $v(u)$ is then $||v(u)||= \sqrt{1+4(u C_{e,e'} + C_{e',c})^2}$. Let now
\begin{eqnarray*}
f(u)= \langle \frac{v(u)}{||v(u)||}, e \rangle = \frac{\langle e'_1, e_1\rangle + \langle e'_2,e_2 \rangle + 2 e_3 ( u C_{e,e'} + C_{e',c})}{\sqrt{1+4(u C_{e,e'} + C_{e',c})^2}}.
\end{eqnarray*}

Then $\arccos f(u)$ is the angle between $I_e$ and the segment $\sigma^j_{e'}(s)$. Observe that if $C_{e,e'}= \langle e'_1,e_2 \rangle - \langle e'_2,e_1 \rangle=0$ then $f$ is constant, hence the segments $\sigma^j_{e'}(s)$ are parallel as $p_j$ moves along $I_e$ (this happens for example if $e_1=e_2=0$, i.e. $I_e$ is vertical).

The derivative of $f$ is given by
\begin{eqnarray*}
\frac{d}{du} f(u)= \frac{2 C_{e,e'} (e_3-2C_{e',c} (\langle e'_1, e_1\rangle + \langle e'_2,e_2 \rangle) - 2u C_{e,e'} (\langle e'_1, e_1\rangle + \langle e'_2,e_2 \rangle))}{(1+4(u C_{e,e'} + C_{e',c})^2)^{3/2}}.
\end{eqnarray*}

If $\langle e'_1, e_1\rangle + \langle e'_2,e_2 \rangle=0$, then the sign of $\frac{d}{du}f$ is constant (same sign as $C_{e,e'}$), hence $f$ is monotone. If $\langle e'_1, e_1\rangle + \langle e'_2,e_2 \rangle \ne 0$ and $C_{e,e'}\ne 0$, let
\begin{equation}
C_{e,e',c}= \frac{e_3-2C_{e',c} (\langle e'_1, e_1\rangle + \langle e'_2,e_2 \rangle)}{2C_{e,e'} (\langle e'_1, e_1\rangle + \langle e'_2,e_2 \rangle)}.
\end{equation}

If $C_{e,e',c} <0$ or $C_{e,e',c} >1$, then $f$ is monotone on $[0,1]$. If $0 \le  C_{e,e',c} \le 1$, then if $f$ is increasing (respectively decreasing) for $u \in [0, C_{e,e',c})$, it is decreasing (respectively increasing) for $u \in ( C_{e,e',c},1]$.

In case $f$ is monotone for all $u \in [0,1]$, let $I_{e,e'}=I_e$. Otherwise, let $I^1_e= \gamma([0,C_{e,e',c}])$ and $I^2_e= \gamma([C_{e,e',c},1])$. Since
\begin{equation*}
\mathcal{H}^1\left(I_e \cap \bigcup_{j \in J_k} \tilde{B}_j \right) \ge \frac{1}{2k^2},
\end{equation*}
there exists $i \in \{1,2\}$ such that
\begin{equation*}
\mathcal{H}^1\left(I_e^{i} \cap \bigcup_{j \in J_k} \tilde{B}_j \right) \ge \frac{1}{4k^2}.
\end{equation*}
Let $I_{e,e'}=I_e^{i}$.

Since $f$ is monotone on $I_{e,e'}$, either the segments $\sigma^j_{e'}(s)$ or the segments $\sigma^j_{-e'}(s)$ do not intersect when $p_j \in I_{e,e'}$. Hence there is a subset $S_P'$ of $S_P$ with $\sigma^{2n-1}(S_P') = \sigma^{2n-1}(S_P)/2$ and for every $e' \in S_P'$ there exists $I_{e,e'}$ such that the segments $\sigma^j_{e'}(s)$ do not intersect for $p_j \in I_{e,e'}$ and
\begin{equation*}
\mathcal{H}^1\left(I_{e,e'} \cap \bigcup_{j \in J_k} \tilde{B}_j \right) \ge \frac{1}{4k^2}.
\end{equation*}

Let $I_1= \gamma([0,1/3])$, $I_2= \gamma([1/3,2/3])$ and $I_3=\gamma([2/3,1])$. Then there exists $i_{e'} \in \{1,2,3\}$ such that
\begin{equation}\label{Ii}
\mathcal{H}^1\left((I_{e,e'}\cap I_{i_{e'}}) \cap \bigcup_{j \in J_k} \tilde{B}_j \right) \ge  \frac{1}{12k^2}.
\end{equation}

Let $S_1=\{ e' \in S_P' : i_{e'}=1\}$, $S_2=\{ e' \in S_P' : i_{e'}=2\}$ and $S_3=\{ e' \in S_P' : i_{e'}=3\}$. Then there exists $l \in \{1,2,3\}$ such that $\sigma^{2n-1}(S_l) \ge \frac{\sigma^{2n-1}(S_P')}{3}= \frac{\sigma^{2n-1}(S_P)}{6}$.

If $l=2$ then we let $T=T^{2^{-k}}_e(\gamma(1/2))$. In case $l=1$ or $3$, we take $T=T^{2^{-k}}_e(b)$, where $b$ is the mid-point of $I_1$ or $I_3$ respectively.

For any $e' \in S_l$, consider the 2-dimensional plane $P'$ orthogonal to $P$ and containing $I_e$ and the segment joining $c$ and $e'+c$. The line containing $I_e$ divides $P'$ into two half-planes. Let $P''$ be the one that contains the segment joining $c$ and $e'+c$. Note that for $k \ge M$ (as we assumed) we have $\frac{2^{-k}}{\tan \theta_R} < \frac{1}{3} $. Thus for $p_j \in I_l$ the segments $\sigma_{e'}^j(s)$ intersect any segment $I$ parallel to $I_e$ contained in $T \cap P''$.

Moreover, by step 2 all the segments $\sigma^j_{e'}(s)$ are contained in $F_k$ and by step 3 they do not intersect for $p_j \in I_{e,e'}$. Hence by \eqref{Ii} if $I$ is any segment parallel to $I_e$, contained in $P'' \cap T$, we have
\begin{equation}\label{10}
\mathcal{H}^1\left(I \cap F_k \right) \ge C_{R,n} \frac{1}{k^2}.
\end{equation}

Hence $\mathcal{L}^{2}(P''\cap T \cap F_k) \ge C_{R,n} \frac{1}{k^2}\mathcal{L}^{2}(P'' \cap T)$. This holds for all such half-planes $P''$ containing the segment from $c$ to $e'+c$ with $e' \in S_l$. To get \eqref{intFk}, we will first integrate over 2-dimensional planes $P''$ and then over $S_l$.

If $f$ is a Lebesgue measurable function in $\mathbb{R}^N$, then by polar integration in translates of the $x_1, \dots , x_{N-1}$-hyperplane along the $x_N$-axis we obtain
\begin{equation*}
\int_{\mathbb{R}^N} f(x) dx = \int_{-\infty}^\infty \left( \int_0^\infty r^{N-2} \int_{S^{N-2}} f(ry , x_N ) d \sigma^{N-2}(y) dr \right) dx_N ,
\end{equation*}
where $y \in S^{N-2}$ and we have written $x=(x',x_N) \in \mathbb{R}^N$ with $ x'\in \mathbb{R}^{N-1}$. Changing the order of integration, we get
\begin{equation*}
\int_{\mathbb{R}^N} f(x) dx =\int_{S^{N-2}} \left( \int_0^\infty r^{N-2} \int_{-\infty}^\infty  f(ry , x_N ) d x_N dr \right) d \sigma^{N-2}(y),
\end{equation*}
where we first integrate over 2-dimensional planes containing the $x_N$-axis and then over the unit sphere in the $x_1, \dots , x_{N-1}$-hyperplane.

If we apply this formula to the case when $N=2n+1$, the $x_N$-axis is the line containing $I_e$ and the $x_1, \dots , x_{N-1}$-hyperplane is $P-c$, then \eqref{intFk} follows from \eqref{10}.

\end{proof}

We recall Wolff's result, which in case of a Besicovitch set in $\mathbb{R}^{2n+1}$ gives the lower bound $\frac{2n+3}{2}$ for its Euclidean Hausdorff dimension.

\begin{theorem}
Let $0 < \delta <1$. Then for $f \in L^{\frac{2n+3}{2}}(\mathbb{R}^{2n+1})$,
\begin{equation}\label{Wolff}
||f^*_\delta ||_{L^{\frac{2n+3}{2}}(S^{2n})} \le C_{n,\epsilon} \delta^{\frac{1-2n}{3+2n}- \epsilon} ||f||_{L^{\frac{2n+3}{2}}(\mathbb{R}^{2n+1})}
\end{equation}
for every $\epsilon >0$.
\end{theorem}

Later Katz and Tao improved this result for $2n+1 \ge 9$, proving the following.

\begin{theorem}
Let $0 < \delta <1$. Then for $f \in L^{\frac{8n+7}{7}}(\mathbb{R}^{2n+1})$,
\begin{equation}\label{KT}
||f^*_\delta ||_{L^{\frac{8n+7}{7}}(S^{2n})} \le C_{n,\epsilon} \delta^{-\frac{6n}{8n+7}- \epsilon} ||f||_{L^{\frac{8n+7}{7}}(\mathbb{R}^{2n+1})}
\end{equation}
for every $\epsilon >0$.
\end{theorem}

These results give the following estimates for the Heisenberg Hausdorff dimension of bounded Besicovitch sets.

\begin{cor}
Every bounded Besicovitch set in $\mathbb{H}^n$ has Heisenberg Hausdorff dimension at least $\frac{2n+5}{2}$ for $n \le 3$ and $\frac{8n+14}{7}$ for $n \ge 4$.
\end{cor}

\begin{proof}
The estimate \eqref{Wolff} corresponds to \eqref{kn} with $p=\frac{2n+3}{2}$ and $\beta = \frac{2n-1}{2n+3}+ \epsilon$. Hence it gives the lower bound $2n+2-\beta p= \frac{2n+5}{2}- \epsilon \frac{2n+3}{2}$. Taking the limit when $\epsilon \rightarrow 0$, we get the claim.

Similarly, from \eqref{KT} we get the lower bound $\frac{8n+14}{7}$.
\end{proof}

\begin{rem}
Closely related to Kakeya sets are Nikodym sets. A \textit{Nikodym set} $E \subset \mathbb{R}^n$ is a Borel set such that $\mathcal{L}^n(E)=0$ and for every $x \in \mathbb{R}^n$ there is a line $L$ through $x$ such that $E \cap L$ contains a unit line segment. The Nikodym conjecture states that Nikodym sets have full Hausdorff dimension and it is implied by Kakeya conjecture.

The \textit{Nikodym maximal function} $f^{**}_\delta$ of $f \in L^1_{loc}(\mathbb{R}^n)$ is defined as
\begin{equation*}
f^{**}_\delta(x)=\sup_{x \in T} \frac{1}{\mathcal{L}^n(T)} \int_T |f| d \mathcal{L}^n, \quad x \in \mathbb{R}^n,
\end{equation*}
where the supremum is taken over all tubes $T=T^\delta_e(a)$ that contain $x$. The Nikodym maximal conjecture asserts that the following inequality
\begin{equation*}
||f^{**}_\delta||_{L^n(\mathbb{R}^n)} \le C_{n,\epsilon} \delta^{- \epsilon} ||f||_{L^n(\mathbb{R}^n)}
\end{equation*}
holds for all $\epsilon>0$, $0 < \delta <1$ and $f \in L^n(\mathbb{R}^n)$. In \cite{Tao} Tao proved that Nikodym maximal conjecture and Kakeya maximal conjecture are equivalent.

The proof of Theorem \ref{Th1} can be easily modified (by taking the unit ball instead of the unit sphere) to treat the case of the Nikodym maximal function. The theorem still holds if we replace $f^*_\delta$ by $f^{**}_\delta$ and Besicovitch sets by Nikodym sets. In particular, Nikodym maximal conjecture implies Nikodym conjecture and the same results obtained above hold also for Nikodym sets.
\end{rem}

\textit{Acknowledgements.} The author would like to thank Pertti Mattila for constant support and helpful comments.

\bibliography{Hei_Hausdorff}{}
\bibliographystyle{plain}

Department of Mathematics and Statistics, P.O. Box 68 (Gustaf H\"allstr\"omin katu 2b), FI-00014 University of Helsinki, Finland\\
E-mail: laura.venieri@helsinki.fi

\end{document}


TITLE: How do I extract a formula for distance from a metric that employs curvature?
QUESTION [0 upvotes]: Let's say I have a metric for spacetime that looks like this:
$$
ds^2 = -c^2dt^2+a^2(t)
       \left[\frac{dr^2}{1-kr^2}+r^2d\theta^2+r^2\sin^2\theta d\phi^2\right]
$$
As I understand it, this is the Reduced Circumference version of the formula where we basically take the circumference as it's measured, but then reduce the radius such that $C=2\pi D$ where $D$ is the reduced radius.  In that way, basic trig operations can still return usable results.
I also know from basic trigonometry on a sphere that
$$D=\frac{\sin\left(r\sqrt{k}\right)}{\sqrt{k}}$$
when the curvature is positive.  Conceptually, this appears to be the same thing as the Reduced-circumference polar coordinates.  Is there a way to derive this relation from the metric?  If so, please show the steps.
I am basically trying to find the relation between $r$ and $\theta$ in a space with positive curvature, $k$ using my metric for that space.

REPLY [1 votes]: First part
It is customary to use capital letter $K$ to denote Gaussian curvature ( not $k$).  For positive Gauss curvature let $ K= 1/a^2$
I am supposing that you are trying to find a relation between $D$ and $r$.
On the question how to exploit the metric in order to find shrunk or reduced circumferential distance $r$ from arc distance  $D$ of 2D spherical surface.
For a given latitude or longitude ( same $(\phi, \theta)$ so $ d\phi=0, d\theta=0) $ these two terms vanish so we do not consider parallel circle, operate on cylindrical mode to obtain formula for geodesic distance $D$ between a pole and a geodesic parallel:
$$ dD^2= \dfrac{dr^2}{1-Kr^2}=\dfrac{dr^2}{1-r^2/a^2} \; (r< a) $$
$$dD =\dfrac{dr}{\sqrt{1-r^2/a^2}} $$
$$ D =\int \dfrac{dr}{\sqrt{1-r^2/a^2}}= a \sin^{-1}\dfrac{r}{a}$$
so that the direct Reduced Circumferentially determined radius
$$ r = a \sin \dfrac{D}{a} = a \sin D \sqrt K \;  (r<a) \tag1 $$

From trigonometry of a sphere we have
$$ \dfrac{r}{a} =\sin \dfrac{D}{a}= \sin \phi_{co-latitude}\quad r<a \tag2 $$
but not
$$ \dfrac{D}{a} =\sin \dfrac{r}{a} \tag3$$
which as an error source in your simple trigonometric calculation  is clear.. as $ r$ cannot be greater then $D$.
Also reduced circumference is
$$ 2 \pi r < 2 \pi D $$
Second part
You are in other words looking at geodesic polar coordinates for the general case with North pole as center for these coordinates. Relation between $D$ and $\theta$ in general involves metric. Even if $D$ is an uncomfortable symbol,lengths are according to the metric:
$$ ds^2= dD^2+G_{D,\theta} \;d\theta^2 $$
involving first fundamental form coefficient $G_{D,\theta} $.

It should be noted for generality that for $K<0$ circumferential dimension increases in hyperbolic geometry.


TITLE: $p^2=p\in \bar{I}$, I ideal of Banach algebra $\Rightarrow p\in I$
QUESTION [6 upvotes]: Let $I\subset A$ be a ideal of a Banach algebra $A$. Assume $p\in \overline I$ and $p^2=p$.
Show: $p\in I$.
Can someone give me a little hint how to solve this, please?

REPLY [4 votes]: This is recorded as Lemma 2.7 in

Niels Jakob Laustsen, Thomas Schlumprecht, and András Zsák, The lattice of closed ideals in the Banach algebra of operators on a certain dual Banach space, J. Operator Theory 56 (2006), no. 2, 391–402.

I assume that $A$ is unital, if not, we can pass to the unitisation.
Suppose that $(t_n)_{n=1}^\infty$ is a sequence in $I$ that converges to an idempotent $p$. Since $pAp$ is a Banach algebra with identity $p$ and $pt_np \to p^3=p$, there exists $n$ so that $pt_np$ is invertible in $pAp$. Consequently, $p = (psp)(pt_np)$ for some $s\in A$. This proves that $p\in I$.


TITLE: Why is $D(\Omega)$ continuously embeded in $H_0^1(\Omega)$?
QUESTION [1 upvotes]: let $D(\Omega)$ is the space of $C^{\infty}$ functions with compact support, with the usual notion of convergence we give in the context of distributions.
In my notes, before showing that $H^{-1} \subset D'(\Omega)$, my professor wrote that

$D(\Omega) \hookrightarrow H_0^1(\Omega)$ is a continuous embedding

I can't understand why this is true. If I consider the immersion $\mathcal{i}: D(\Omega) \rightarrow H_0^1(\Omega)$ with $i(v)=v$, I should be able to prove $$||i(v)||_{H_0^1}=||v||_{H_0^1} \leq ||v||_{D(\Omega)}$$ but I do not know what is the norm on $D(\Omega)$! I am missing some fundamental thing here, and indeed in my book I can't find what is the norm given on $D(\Omega)$. What am I missing?

REPLY [1 votes]: As per Exercise 3 in the answer I linked to
Doubt in understanding Space $D(\Omega)$
what one must show is that for every continuous seminorm $\eta$ on $H_0^1(\Omega)$, there exists a continuous seminorm $\rho$ on $\mathscr{D}(\Omega)$ such that, for all $v\in\mathscr{D}(\Omega)$,
$$
\eta(i(v))\le \rho(v)\ .
$$
Now it is enough to do this $\eta$'s which define the locally convex topology of $H_0^1(\Omega)$, namely just $\eta=\|\cdot\|_{H_0^1}$.
To construct $\rho$, just take $\rho(v):=\|i(v)\|_{H_0^1}$. To show it is a continuous seminorm, it is enough to show it is an admissible seminorm on $\mathscr{D}(\Omega)$, i.e., belongs to the set $B$ given in Example 2 of the post I mentioned.
More concretely, what you need to show is that for all compact $K\subset\Omega$, there exists $C\ge 0$ and $N\ge 0$, such that for all $v\in\mathscr{D}(\Omega)$ with support contained in $K$, we have
$$
\|v\|_{H_0^1}\le C\sup_{x\in K}\max_{|\alpha|\le N} |D^{\alpha}v(x)|
$$
which a multivariate calculus exercise.


TITLE: Operator norm and continuity
QUESTION [0 upvotes]: I've read in the solution of an exercise: "$T$ has a finite norm, thus $T$ is continuous".
We are in a normed vector space $(V,||.||)$ and $T$ is a linear selfmap over the vector space $V$. The purpose of the exercise is to show that the set $L_w=\{v\in V:T(v)=w\}$ is closed. The first statement in the solution is the one above, and once showed continuity of $T$, the exercise is pretty straightforward.
Since I am working with linear operators in normed vector spaces, my idea is that it is linked with the equivalence of boundedness and continuity of an operator, but I am not sure. Is my idea right? If not, why the statement is true?

REPLY [1 votes]: The norm of a linear operator $L:V\to V$ is the smallest $M$ such that:
$$
||Lv||\le M||v|| \quad \forall v \in V
$$
So, your idea is correct:  a linear operator $L$ in a  normed spaces $V$ is bounded if and only if it is a continuous linear operator (here a simple proof).


TITLE: Find the number of integers $n$ of the form $30a0b03$ that are divisible by $13$.
QUESTION [0 upvotes]: Find the number of integers $n$ of the form $30a0b03$ that are divisible by $13$.

What I Tried: This problem looks simple but I cannot find it's solution. I recalled all the divisible rules of $13$ and tried to put them here, but none of them really worked. One thing which I got was :-
$$\rightarrow 13 | bo3 - 0a0 + 3$$
I only got this by applying one divisibility rule of $13$.
I might try and test for each values of $a$ and $b$ but there will be $100$ configurations, and I don't want to do that.
Can anyone help me?
Edit: After the hints @Daniel gave me, I figured out the solution and found $7$ solutions, can anyone tell me if this is correct or not?

REPLY [0 votes]: This can alternately be written as 30,000,003+10,000a+100b
1.30,000,003≡6(mod 13)
Thus, 10,000a+100b≡7(mod 13)

3a+9b≡7(mod 13)
3a+9b=7: No possibilities
3a+9b=20:No possibilities
Now you have to realize that the RHS has to divide 3 by Bezouts lemma. So, 13x+7=3y
3y-13x=7
2x(mod 3)≡1(mod 3). Thus, x≡2(mod 3) and can be represented as 3a+2.
3y-13(3a+2)=7 so 3y-39a=33 and y-13a=11. y=13a+11. y also has to be less than or equal to 9. However, no solutions for y satisfy this.


TITLE: Uniform Convergence and Continuous Functions at Boundary
QUESTION [0 upvotes]: I'm stumped on this question:
Show that a if series of continuous functions $f_n$ in the interval $[a,b]$ uniformly converges on $(a,b]$, then the limit $lim_{n \to \infty}f(a)$ exists.
I tried using Cauchy's Criterion for convergence and the fact the series is continuous but my proof doesn't seem right.

REPLY [1 votes]: Was it this proof? Fix $\epsilon>0$. By the Cauchy criterion for uniform convergence, there is $N$ such that for all $n,m>N$ we have
$$\vert f_n(x)-f_m(x) \vert<\epsilon$$
for all $x\in (a,b]$. Fix any $n,m>N$. Since $f_n$ and $f_m$ are continuous on $[a,b]$, there is $\delta_n>0$ and $\delta_m>0$ such that
$$\vert f_n(a)-f_n(x)\vert<\epsilon$$
and
$$\vert f_m(a)-f_m(x)\vert<\epsilon$$
for $\vert x-a\vert<{\rm min}\{\delta_n,\delta_m\}$. Then for any such $x$
\begin{eqnarray}
\vert f_n(a)-f_m(a)\vert &=& \vert f_n(a)-f_m(a)+f_n(x)-f_n(x)+f_m(x)-f_m(x) \vert
\\
 &\leq& \vert f_n(a)-f_n(x)\vert
 +\vert f_m(x)-f_m(a)\vert + \vert f_n(x)-f_m(x)\vert <3\epsilon
\end{eqnarray}
Since $n,m\geq N$ were arbitrary, it follows that $f_n(a)$ is a Cauchy sequence in $\mathbb{R}$ and thus converges.


text:
\section{Introduction}
\label{sec:intro}

There has been a tremendous surge in the development and application of machine learning models in recent years due to their flexibility and capability to represent trends in complex systems \citep{hastie2016elements}.
The parameters of a machine learning model can often be calibrated, with sufficient data, to give high fidelity representations of the underlying process
\citep{raissi2017,jones2018machine,frankel2019predicting,frankel2019prediction}.
It is now feasible to construct deep learning models over datasets of tens of thousands to millions of data points with modern computational resources  \citep{dean2012large}. In many scientific applications, however, there may not be large amount\REVISION{s} of data available for training. Unlike data from internet or text searches, computational and physical experiments are typically extremely expensive. Moreover, even if ample data exists, the machine learning model may yield behaviors that are inconsistent with what is expected physically when queried in an extrapolatory regime.

To aid and improve the process of building machine learning models for scientific applications, it is desirable to have a framework that allows the incorporation of physical principles and other a priori information to supplement the limited data and regularize the behavior of the model.  Such a framework is often referred to as ``physics-constrained'' machine learning within the scientific computing community \citep{raissi2018, pan2018data, lusch2018deep, brunton2016discovering, LeeCarlberg, ling2016machine,jones2018machine}.
\citet{karpatne2017theory} provide a taxonomy for theory-guided data science,
with the goal of incorporating scientific
consistency in the learning of generalizable models.  The information used to constrain models can be simple, such as known range or positivity constraints, shape constraints, or monotonicity constraints that the machine learning model must satisfy.  The constraints can also be more complex; for example, they can encode knowledge of the underlying data-generating process in the form of a partial differential equation.

Several recent conferences highlight the interest in ``physics-informed'' machine learning~\citep{conference1,conference2,conference3,conference4,conference5}.

Much of the existing research in physics-informed machine learning has focused on incorporating constraints in neural networks \citep{ling2016machine,jones2018machine}, often through the use of objective/loss functions which penalize constraint violation \citep{raissi2018deep,raissi2019physics,magiera2019constraint,cyr2019robust,mao2020physics}. Other works have focused on incorporating prior knowledge using Bayesian inference that expresses the data-generating process as dependent on a set of parameters, the initial distribution of which is determined by the available information, e.g., functional constraints \citep{wangBerger, jidling2017linearly}. Unlike deterministic learning approaches, the predictions made using approximations trained with Bayesian inference are accompanied with probabilistic estimates of uncertainty/error.

Within the Bayesian regression framework, Gaussian processes (GPs) are popular for constructing ``surrogates'' or ``emulators'' of data sources that are very expensive to query.  The use of GPs in a
regression framework to predict a set of function values is called Gaussian Process Regression (GPR).
An accurate GPR can often be constructed using only a relatively small number of training data (e.g. tens to hundreds), which consists of pairs of input
parameters and corresponding response values.  Once constructed, the GPR can be thought of as a machine-learned metamodel and used to provide
fast, cheap function evaluations for the purposes of prediction, sensitivity
analysis, uncertainty quantification, calibration, and optimization.
GP regression models are constructed with data obtained from computational simulation~\citep{gramacy2020surrogates} or field data; in geostatistics, the process of applying Gaussian processes to field data has been used for decades and is frequently referred to as \emph{kriging} \citep{chiles2018fifty}.

In this survey we focus on the use of constrained GPRs

that honor or incorporate a wide variety of physical constraints~\citep{riihimaki, daveiga, jensen, solak, raissi2017, yang, lopez2018, bachoc2019}.
Specifically,

we focus on the following topics, after a short review of Gaussian process regression in Section \ref{gpr}.
Section \ref{overview_of_constraints} presents an overview and a classification of constraints according to how the constraint is enforced during the construction of a GP.
Section \ref{sec:bound_constraints} discusses bound constraints, in which the GP prediction may be required to be positive, for example, or the prediction may be required to fall between upper and lower bounds.
Section \ref{sec:monotonicity_constraints} discusses monotonicity and related convexity constraints.
Constraints may also be more
tightly integrated with the underlying physics:  the GP can be constrained to satisfy
linear operator constraints which represent physical laws expressed as partial different equations (PDE).
This is discussed in Section~\ref{sec:pde_constraints}.
Section \ref{sec:boundary_constraints} discusses intrinsic boundary condition constraints.
We review several different approaches for enforcing each of these constraint types.
Finally, Section \ref{sec:computation_considerations} is a compendium of computational details for implementing the constraints of Sections \ref{sec:bound_constraints} -- \ref{sec:boundary_constraints}, together with a summary of computational strategies for improving GPR and brief commentary about the challenges of applying these strategies for the constrained GPs considered here.

The taxonomy we present is formulated to enable practitioners to easily query this overview for information on the specific constraint(s) they may be interested in.
For approaches that enforce different constraints but have significant overlap in methodology, references are made between sections to the prerequisite subsection where the technical basis of an approach is first discussed in detail. This is done, for example, when discussing spline-based approaches which are used for both bound constraints in Section \ref{sec:bound_constraints} and monotonicity constraints in Section \ref{sec:monotonicity_constraints}.

Not all physical constraints can be neatly divided into the categories that we focus on in Sections \ref{sec:bound_constraints} -- \ref{sec:boundary_constraints}. For example, with a view toward computer vision, \citet{salzmann2010implicitly} considered GPR for pose estimation under rigid (constant angle and length) and non-rigid (constant length) constraints between points. They proved that linear equality constraints of the form $A\mathbf{y} = \mathbf{b}$, if satisfied by all the data vectors $\mathbf{y}$, are satisfied by the posterior mean predictor of a GP. Then, at the cost of squaring the input dimension, they translated quadratic length constraints into such linear constraints for pairwise products of the input variables.
In another example, \citet{frankel2019tensor} applied GPR to predict the behavior of hyperelastic materials, in which the stress-stretch constitutive relation naturally exhibits rotational invariance. The rotational invariance was enforced by deriving a finite expansion of the Cauchy stress tensor in powers of the Finger tensor that satisfies the rotational invariance by virtue of its structure, and GPR was performed for the coefficients of the expansion.
We mention these examples to illustrate that physical constraints are varied, and in some cases the method to enforce them can depend highly on the specific nature of the constraint.

Even within the selected categories represented by Sections \ref{sec:bound_constraints} -- \ref{sec:boundary_constraints}, the literature on constrained Gaussian processes is extensive and expanding rapidly. Consequently, we cannot provide a complete survey of every instance of constrained GPR.
Rather, we strive to discuss main areas of research within the field. The goal is to aid readers in selecting methods appropriate for their applications and enable further exploration of the literature. We present selected implementation details and numerical examples, giving references to the original works for further details. Many of the authors of these works have developed codebases and released them publicly. Finally, we remark that we have adopted consistent notation (established in Section \ref{gpr}) for GPR that does not always follow the notation of the original works exactly.

config: arxiv


TITLE: On the meaning of the complex measure $\int_{\mathbb{C}} d z d \bar{z}$
QUESTION [3 upvotes]: I have a problem to understand the meaning of a complex measure; i.e., when someone writes ($i \equiv \sqrt{-1}$)
$$
\int_{\mathbb{R}^2} d\mathrm{Re}z \, d\mathrm{Im}z  \equiv \int_{\mathbb{C}} \frac{dz d\bar{z}}{2i} \quad (\ast)
$$
The lefthand-side will yield a real number (after performing the integration over a real-valued function), while it is not obvious that the righthand-side yields a real number.
Furthermore, how can one obtain the equivalence relation? Is the factor $\frac{1}{2i}$ the Jacobian of some transformation like
$$
z = \mathrm{Re} z + i \, \mathrm{Im} z ,\\
\bar{z} = \mathrm{Re} z - i \, \mathrm{Im} z ,\\
$$
So, the complex measure $dz d\bar{z}$ does not have the same meaning as a ‘simple’ complex integration which in complex calculus (integration over a path in the complex plane).
Please provide an explanation for the complex measure and the equivalence relation ($\ast$) above.

Notes

1. A similar question is asked here; yet no clear proof or justification is provided.

2. An example of the relation appears, for instance, in Altland, A. and B. D. Simons. “Condensed Matter Field Theory” (2nd ed., 2010), p. 102:

REPLY [3 votes]: You have to  be aware that there are intagration of measures and integration of differiential forms. You cab see the difference when using the Change of variables formula.
For example if $V,W$ are vector spaces of same dimension, $\lambda,\mu$ Lebesgue measures on $V$ and $W$ (or rather a Haar measure), and $f : V \rightarrow W$ smooth and compactely supported, then :
$$\int_{x \in V} d\mu(f(x)) = \int_{x \in V} |\mathrm{Jac}(f(x))| ~ d\lambda(x).$$
In your context it is integration of differential form, and you need the following formula.
If $V,W$ are oriented vector spaces not necessarily of the same dimension, $f : V \rightarrow W$ smooth, $v$ an invertible differential form on $V$ (i.e. volume form), $w$ a differential form on $W$ of degree $\dim V$ then
$$\int_{x \in V} f^*v~(x) = \int_{x \in V} \mathrm{Jac}(f(x)) ~ v(x),$$
where $\mathrm{Jac}(f(x))$ is the unique number such that $f^*v~(x) = \mathrm{Jac}(f(x)) ~ v(x)$ (recall that $v$ is invertible, and it coincides with the determinant when $V$ and $W$ are of the same dimension).
In your situation $V=\mathbb{C}$, $W = \mathbb{C} \times \mathbb{C}$, $f : z=x+iy \mapsto (z,\bar{z})$, $v= dx \wedge dy$, and $w = dz_1 \wedge dz_2$, which yields :
$$\int_{z \in \mathbb{C}} dz \wedge d\bar{z} = \int_{z \in \mathbb{C}} -2i ~ dx \wedge dy.$$
Note : $dz$, $d\bar{z}$, $dz.d\bar{z}$ are not measures nor complex measures. They are complex differential forms of degree 1 or 2.


TITLE: Irreducible representations of $S_3$
QUESTION [7 upvotes]: I am trying to do a problem from artin's algebra 2nd ed (Chapter 10, Exercise 2.3) but having trouble:
Let $(\rho , V)$ be a representation of the symmetric group $S_3$. Let $x=(123), y=(12)$ be the usual generators of $S_3$.
(a) Let $u\neq 0$ be a vector in $V$. Let $v=u+xu+x^2u$ and $w=u+yu$. By analyzing the $S_3$-orbits of $v,w$, show that $V$ contains a nonzero invariant subspace of dimension $\leq 2$.
(b) Prove that all 2-dimensional irreducible representations of $S_3$ are isomorphic, and determine them all.
After some calculations, i found that $xv=v$ and $yw=w$. But I dont know what to do next. Also, we are not allowed to use character theory, thanks in advance.

REPLY [7 votes]: We have $$x^3 = y^2 = (xy)^2 = 1,$$ so $$yx = (yx)^{-1} = x^2y,\text{ }yx^2 = x^2yx = x^4y = xy.$$
a.
Fix a nonzero $u\in V$, and let $$v = u+xu+x^2u = (1+x+x^2)u,\text{ }w=u+yu = (1+y)u.$$ Since $\{1,x,x^2\},\{1,y\}$ are subgroups of $G = S_3$, we have $xv = x^2v = v$ and $yw = w$.
We have the operator identities $$y(1+x+x^2) = y+yx+yx^2 = y+x^2y+xy = (1+x+x^2)y,$$ $$x(1+y) = x+xy = x+yx^2,$$$$x^2(1+y) = x^2+x^2y = x^2+yx.$$ It follows that $yv = (1+x+x^2)(yu)$, so $x,x^2$ fix $yv$ $($just as they fix $v$$)$. Of course, $y(yv) = y^2v = v$, so $\text{span}(\text{Orb}(v)) = \text{span}(v,yv)$ is a $G$-invariant subspace of dimension at most $2$. This gives us a nonzero invariant subspace of dimension at most $2$ if $v\ne0$.
Otherwise, if $v=0$, we analyze the $G$-orbit of $w$ instead. Since $$xw = (x+yx^2)u, \text{ }x^2w = (x^2+yx)u,$$ we have $$y(xw) = (yx+x^2)u = x^2w,\text{ }y(x^2w) = (yx^2+x)u = xw.$$
But $1+y$ commutes with $1+x+x^2$, so $$w+xw+x^2w = (1+x+x^2)yu = y(1+x+x^2)u = yv = 0$$ $($from $v=0$$)$. Thus $$\text{span}(\text{Orb}(w)) = \text{span}(w,xw,x^2w) = \text{span}(w,xw)$$ is $G$-invariant with dimension at most $2$, and we are fine as long as $w\ne0$.
Thus we have reduced the problem to the case in which $$(1+x+x^2)u = (1+y)u = 0$$ for all $($nonzero$)$ vectors $u\in V$. In this case, we must have $$1+\rho_y = 1+\rho_x+(\rho_x)^2 = 0$$ identically $($as operators$)$, so in particular $\rho_y$ must fix all one-dimensional eigenspaces. Now set $u$ to be an arbitrary eigenvector of $\rho_x$; then $\rho_x$, $\rho_y$ both fix the nonzero subspace $\text{span}(u)$ of dimension $1$, and we are done.
We remark that a simpler approach altogether is just to start with an eigenvector of $\rho_x$.
b.
First we classify all $1$-dimensional $($irreducible$)$ representations of $G$ $($over some $1$-dimensional $V = (v)$, $v\ne0$$)$. Write $xv = \alpha v$ and $yv = \beta v$ $($for complex $\alpha$, $\beta$$)$; then $$x^3v = y^2v = (xy)^2v = v,$$ so $$\alpha^3 = \beta^2 = (\alpha\beta)^2 = 1,$$ which forces $$\alpha^3 = \alpha^2 = 1\implies \alpha = 1,$$ and $\beta = \pm1$. Thus the only $1$-dimensional $($irreducible$)$ representations of $G$ are the trivial $($$\beta=1$$)$ and sign $($$\beta=-1$$)$ representations. $($In terms of matrices, $R_x = [\alpha]$ and $R_y = [\beta]$ for both cases.$)$
Now consider an irreducible representation of $G$ over a $2$-dimensional $V$. Take an eigenvector $v\ne0$ of $x$ with eigenvalue $\lambda$, so $xv=\lambda v$ and $\lambda^3 = 1$. Define $w = yv$, so $$yw = y^2v = v.$$ Then $$xw = xyv = yx^2v = y\lambda^2 v = \lambda^2 w.$$
Of course, $w\ne v$, or else $(v)$ would be a $1$-dimensional $G$-invariant subspace of $V$. Similarly, $\lambda\ne1$, or else we would have $xv=v$, $xw=w$, $yv+yw = w+v$, and $(v+w)$ would be a $1$-dimensional $G$-invariant subspace.
Also, without loss of generality, we may assume $\lambda = e^{2\pi i/3}$, or else we could have started with the eigenvector $w\ne0$ of $x$ $($with eigenvalue $\lambda^2$$)$ instead of $v$. It follows that for any irreducible representation of $G$ over a $2$-dimensional $V$, there exist basis vectors $v$, $w\in V$ such that the matrices $R_x$, $R_y$ of $\rho_x$, $\rho_y$ with respect to the basis $B = (v,w)$ are $$\begin{pmatrix} e^{2\pi i/3} & 0 \\ 0 & e^{-2\pi i/3} \end{pmatrix},\, \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},$$ respectively, so we are done.
We remark that it is easy to check that the three representations presented above are in fact irreducible. The one-dimensional representations are clearly irreducible; for the two-dimensional one, we use the matrix interpretation $R_x$, $R_y$ and note that a $G$-invariant one-dimensional subspace $(v)$ $($with $v = (p,q)^\text{T}$$)$ would have to have $(q,p) = yv\in (v)$ and thus $p^2 = q^2 \ne0$. But it is easy to check that $R_x$ sends $(p,p)$ to $(\lambda p, \lambda^{-1} p)\notin (v)$ and $(p,-p)$ to $(\lambda p,-\lambda^{-1} p)\notin (v)$.


TITLE: Deep research in probability theory
QUESTION [2 upvotes]: Recently I came across the concept of probability theory in my statistics class. I found the concept fascinating and plan to pursue it in depth.
Problem is, I'm an economics major and due to the the rigid education system in my country I can't pick up another subject along with it. I've decided to give it a go by myself. After searching for it I found out that measure theory and functional analysis is used for in depth research in probability.
What I want to know is what topics do I have to cover before I get there and if you could name the books/sources you used, that would be much appreciated. If there's any autodidact out there do tell what sources you used as I understand this is on the very right side of the spectrum when it comes to difficulty.

REPLY [2 votes]: What you really need is Real Analysis, measure theory, and some comfort with elementary set theory.
I would recommend Rudin's book on Real and Complex Analysis.  He also wrote one on functional analysis as well (all when he was under something like 25).  They are sort of the standard texts many people use.
I have never found a great text on the probability theory (the actual name of the theory of probability is often referred to as Mathematical Statistics).  I find that if you google it there are lots of professors who have written their own notes, essentially, you can download.  When I learned it the professor lectured it all from his head, as he said there weren't any good books.
It is a a bit of a balancing act, trying to find something that you know enough math to make progress on the probability, then learning more math that underlies it, then back to probability, etc.
In fact, probability was not even on a firm mathematical footing until the Russian Mathematician Kolmogorov put it there well into the 1900s.


TITLE: When does the capacitor start to get charged in opposite polarity in a LC circuit
QUESTION [0 upvotes]: I have been reading in pretty much all the books that in a LC circuit when you switch on the circuit, Capacitor starts to discharge. It will take some time for current to be maximum in an inductor which is understandable. And then they say, after the current has reached maximum, it starts to charge the capacitor ( with opposite polarity).
Why doesn't capacitor start to get charged (in opposite polarity) when current is building up in inductor.? why it has to wait till current reaches maximum in inductor.?

REPLY [1 votes]: Electrical energy oscillates between the capacitor and the inductor.
The energy stored in the capacitor is a maximum when the charge stored on it is a maximum, and zero when it is discharged. The energy stored in the inductor is a maximum when the current flowing through it is a maximum and zero when there is no current flowing. At other times there is energy stored in both the capacitor and the inductor. The polarity of the charge and the direction of the current have no effect on how much energy is stored.
There is a fixed total amount of energy stored in the circuit. When the energy stored in the capacitor is decreasing the energy stored in the inductor must be increasing. And vice versa.
If the current through the inductor is increasing while the charge on the capacitor is also increasing then the total energy stored in the circuit is increasing. This cannot happen : there is nowhere for the extra energy to come from.
See also
Where does the extra energy come from in an LC circuit?
Why does the current increase in an LC circuit while the capacitor is being discharged?


text:
\begin{document}

\newcommand{\arXivNumber}{1406.1072}

\allowdisplaybreaks

\renewcommand{\thefootnote}{$\star$}

\renewcommand{\PaperNumber}{089}

\FirstPageHeading

\ShortArticleName{Maximal Green Sequences of Exceptional Finite Mutation Type Quivers}

\ArticleName{Maximal Green Sequences\\
of Exceptional Finite Mutation Type Quivers\footnote{This paper is a~contribution to the Special Issue on New Directions
in Lie Theory.
The full collection is available at
\href{http://www.emis.de/journals/SIGMA/LieTheory2014.html}{http://www.emis.de/journals/SIGMA/LieTheory2014.html}}}

\Author{Ahmet I.~SEVEN}

\AuthorNameForHeading{A.I.~Seven}

\Address{Middle East Technical University, Department of Mathematics, 06800, Ankara, Turkey}
\Email{\href{mailto:aseven@metu.edu.tr}{aseven@metu.edu.tr}}

\ArticleDates{Received June 18, 2014, in f\/inal form August 15, 2014; Published online August 19, 2014}

\Abstract{Maximal green sequences are particular sequences of mutations of quivers which were introduced by Keller in
the context of quantum dilogarithm identities and independently by Cecotti--C\'ordova--Vafa in the context of
supersymmetric gauge theory.
The existen\-ce of maximal green sequences for exceptional f\/inite mutation type quivers has been shown~by
Alim--Cecotti--C\'ordova--Espahbodi--Rastogi--Vafa except for the quiver $X_7$.
In this paper we show that the quiver $X_7$ does not have any maximal green sequences.
We also generalize the idea of the proof to give suf\/f\/icient conditions for the non-existence of maximal green sequences
for an arbitrary quiver.}

\Keywords{skew-symmetrizable matrices; maximal green sequences; mutation classes}

\Classification{15B36; 05C50}

\renewcommand{\thefootnote}{\arabic{footnote}}
\setcounter{footnote}{0}

\section{Introduction and main results}

Maximal green sequences are particular sequences of mutations of quivers.
They were used in~\cite{Ke2} to study the ref\/ined Donaldson--Thomas invariants and quantum dilogarithm identities.
Moreover, the same sequences appeared in theoretical physics where they yield the complete spectrum of a~BPS particle,
see~\cite[Section~4.2]{CCV}.
The existence of maximal green sequences for exceptional f\/inite mutation type quivers has been shown in~\cite{ACCERV}
except for the quiver~$X_7$.
In this paper, we show that the quiver~$X_7$ does not have any maximal green sequences.
We also give some general suf\/f\/icient conditions for the non-existence of maximal green sequences for an arbitrary
quiver.

To be more specif\/ic, we need some terminology.
Formally, a~quiver is a~pair $Q=(Q_0,Q_1)$ where $Q_0$ is a~f\/inite set of vertices and $Q_1$ is a~set of arrows between
them.
It is represented as a~directed graph with the set of vertices $Q_0$ and a~directed edge for each arrow.
We consider quivers with no loops or 2-cycles and represent a~quiver~$Q$ with vertices $1,\dots,n$,
by the uniquely associated skew-symmetric matrix $B=B^Q$ def\/ined as follows: the entry $B_{i,j}>0$ if and only if there are~$B_{i,j}$
many arrows from~$j$ to~$i$; if~$i$ and~$j$ are not connected to each other by an edge then $B_{i,j}=0$.
We will also consider more general skew-symmetrizable matrices: recall that an $n\times n$ integer matrix~$B$ is
skew-symmetrizable if there is a~diagonal matrix~$D$ with positive diagonal entries such that $DB$ is skew-symmetric.
To def\/ine the notion of a~green sequence, we consider pairs $(\mathbf{c}, B)$, where~$B$ is a~skew-symmetrizable integer
matrix and $\mathbf{c}=(\mathbf{c}_1,\dots,\mathbf{c}_n)$
such that each $\mathbf{c}_i=(c_1,\dots,c_n) \in \mathbb{Z}^n$ is non-zero.
Motivated by the structural theory of cluster algebras, we call such a~pair $(\mathbf{c}, B)$ a~$Y$-seed.
Then, for $k = 1, \dots, n$ and any~$Y$-seed $(\mathbf{c}, B)$ such that all entries of $ \mathbf{c}_k $ are
non-negative or all are non-positive, the \emph{$Y$-seed mutation} $\mu_k$ transforms $(\mathbf{c}, B)$ into
the~$Y$-seed $\mu_k(\mathbf{c}, B)=(\mathbf{c}', B')$ def\/ined as follows~\cite[equation~(5.9)]{CAIV}, where we use the
notation $[b]_+ = \max(b,0)$:
\begin{itemize}\itemsep=0pt
\item the entries of the exchange matrix $B'=(B'_{ij})$ are given~by
\begin{gather}
\label{eq:matrix-mutation}
B'_{ij} =
\begin{cases}
-B_{ij} & \text{if $i=k$ or $j=k$,}
\\
B_{ij} + [B_{ik}]_+ [B_{kj}]_+ - [-B_{ik}]_+ [-B_{kj}]_+ & \text{otherwise;}
\end{cases}
\end{gather}
\item the tuple $\mathbf{c}'=(\mathbf{c}_1',\dots,\mathbf{c}_n')$ is given~by
\begin{gather}
\label{eq:y-mutation}
\mathbf{c}'_i =
\begin{cases}
-\mathbf{c}_{i} & \text{if} \  i = k,
\\
\mathbf{c}_i+[\sgn(\mathbf{c}_k)B_{k,i}]_+\mathbf{c}_k & \text{if} \  i \neq k.
\end{cases}
\end{gather}
\end{itemize}
The matrix $B'$ is skew-symmetrizable with the same choice of~$D$.
We also use the notation $B' = \mu_k(B)$ (in~\eqref{eq:matrix-mutation}) and call the transformation $B \mapsto B'$ the
\emph{matrix mutation}.
This operation is involutive, so it def\/ines a~\emph{mutation-equivalence} relation on skew-symmetrizable matrices.

We use the~$Y$-seeds in association with the vertices of a~regular tree.
To be more precise, let~$\mathbb{T}_n$ be an \emph{$n$-regular tree} whose edges are labeled by the numbers $1, \dots,
n$, so that the~$n$ edges emanating from each vertex receive dif\/ferent labels.
We write $t \stackrel{k}{-} t'$ to indicate that vertices $t,t'\in\mathbb{T}_n$ are joined by an edge labeled by~$k$.
Let us f\/ix an initial seed at a~vertex $t_0$ in $\mathbb{T}_n$ and assign the (initial)~$Y$-seed $(\mathbf{c}_0,B_0)$,
where $\mathbf{c}_0$ is the tuple of standard basis.
This def\/ines a~\emph{$Y$-seed pattern} on $\mathbb{T}_n$, i.e.~an assignment of a~$Y$-seed $(\mathbf{c}_t, B_t)$ to
every vertex $t \in \mathbb{T}_n$, such that the seeds assigned to the endpoints of any edge $t \stackrel{k}{-} t'$ are
obtained from each other by the seed mutation~$\mu_k$; we call $(\mathbf{c}_t, B_t)$ a~\emph{$Y$-seed with respect to
the initial~$Y$-seed $(\mathbf{c}_0,B_0)$}.
We write:
\begin{gather*}

\mathbf{c}_t =\mathbf{c}= (\mathbf{c}_{1},\dots,\mathbf{c}_{n}),
\qquad
B_t=B = (B_{ij}).
\end{gather*}
We refer to~$B$ as the \emph{exchange matrix} and $\mathbf{c}$ as the \emph{$\mathbf{c}$-vector} tuple of the~$Y$-seed.
These vectors have the following \emph{sign coherence property}~\cite{DWZ2}:
\begin{gather}
\label{eq:C-sign-coherence}
\text{each vector $\mathbf{c}_{j}$ has either all entries nonnegative or all entries nonpositive.}
\end{gather}
Note that this property is conjectural if $B$ is a~general non-skew-symmetric (but skew-sym\-met\-ri\-zable) matrix.
It implies, in particular, that the {$Y$-seed mutation} in~\eqref{eq:y-mutation} is def\/ined for any $Y$-seed
$(\mathbf{c}_t, B_t)$, furthermore $\mathbf{c}_{t}$ is a~basis of $ \mathbb{Z}^n$~\cite[Proposition~1.3]{NZ}.
We also write $\mathbf{c}_{j}>0$ (resp.\
$\mathbf{c}_{j}<0$) if all entries are non-negative (resp.\
non-positive).

Now we can recall the notion of a~green sequence~\cite{BDP}:
\begin{definition}

Let $B_0$ be a~skew-symmetrizable $n\times n$ matrix.
A~\emph{green sequence for $B_0$} is a~sequence $\mathbf{i} = (i_1, \ldots, i_l)$ such that,   for any $1 \leq k \leq l$
with $(\mathbf{c},B)=\mu_{i_{k-1}} \circ \dots \circ \mu_{i_1}(\mathbf{c}_0,B_0)$, we have $\mathbf{c}_{i_{k}}>0$,
i.e.~each coordinate of $\mathbf{c}_{i_{k}}$ is greater than or equal to~$0$; here if $k=1$, then we take
$(\mathbf{c},B)=(\mathbf{c}_0,B_0)$.
\emph{A green sequence for a~quiver} is a~green sequence for the associated skew-symmetric matrix.

A green sequence $\mathbf{i} = (i_1, \ldots, i_l)$ is maximal if, for $(\mathbf{c},B)=\mu_{i_{l}} \circ \dots \circ
\mu_{i_1}(\mathbf{c}_0,B_0)$, we have $\mathbf{c}_k<0$ for all $k=1,\dots,n$.
\end{definition}

In this paper, we study the maximal green sequences for the quivers which are mutation-equivalent to the quiver $X_7$
(Fig.~\ref{fig:X7}).
Our result is the following:

\begin{theorem}
\label{th:cyclic}
Suppose that~$Q$ is mutation-equivalent to the quiver $X_7$ $($so $Q$ is one of the quivers in Fig.~{\rm \ref{fig:X7})}.
Then~$Q$ does not have any maximal green sequences.
\end{theorem}

\begin{figure}[ht]
\centering
$
\begin{array}{cc}
\xymatrix{& \bullet\ar[rd] & & \bullet\ar@<.5ex>[rd]\ar@<-.5ex>[rd]
\\
\bullet\ar@<.5ex>[ru]\ar@<-.5ex>[ru] & & \bullet\ar[ll]\ar[ru]\ar[rd] & & \bullet\ar[ll]
\\
& \bullet\ar[ru] && \bullet\ar@<.5ex>[ll]\ar@<-.5ex>[ll] &} & \xymatrix{& \bullet\ar@/^1pc/[rrdd]\ar[ld] &&
\bullet\ar[ll]\ar[ld] &
\\
\bullet\ar[rd]\ar[rr] & & \bullet \ar[lu]\ar[rr]\ar[ld] & & \bullet \ar[lu]\ar@/^1pc/[llll]
\\
& \bullet\ar@/^1pc/[rruu]\ar[rr] & & \bullet\ar[lu]\ar[ru]}
\end{array}
$

\caption{Quivers which are mutation-equivalent to $X_7$; the f\/irst one is the quiver $X_7$, see~\cite{DO}.}
\label{fig:X7}
\end{figure}

We prove the theorem using the following general statement, which can be easily checked to give a~suf\/f\/icient condition
for the non-existence of maximal green sequences:
\begin{proposition}
\label{prop:positive}
Let $B_0$ be a~skew-symmetrizable initial exchange matrix.
Suppose that there is a~vector $u>0$ such that, for any~$Y$-seed $(\mathbf{c},B)$ with respect to the initial seed
$(\mathbf{c}_0,B_0)$, the coordinates of~$u$ with respect to $\mathbf{c}$ are non-negative.
Then, under assumption~\eqref{eq:C-sign-coherence}, the matrix $B_0$ does not have any maximal green sequences.
\end{proposition}

We establish such a~vector for the quiver $X_7$:

\begin{proposition}
\label{prop:radical}
Suppose that $Q_0$ is a~quiver which is mutation-equivalent to $X_7$, so $Q_0$ is one of the quivers in
Fig.~{\rm \ref{fig:X7}}, and $B_0$ is the corresponding skew-symmetric matrix.
Let $u=(a_1,a_2,\dots,a_7)$ be the vector defined as follows:

\begin{quotation}
$(*)$~if $Q_0$ is the quiver $X_7$ $($so $Q$ is the f\/irst quiver in Fig.~{\rm \ref{fig:X7})}, then the coordinate
corresponding to the ``center'' is equal $2$, and the rest is equal to $1$; if $Q_0$ is not the quiver $X_7$ $($so $Q$ is the second quiver in Fig.~{\rm \ref{fig:X7})}, then all coordinates are equal to~$1$.
\end{quotation}

Then, for any~$Y$-seed $(\mathbf{c},B)$ with respect to the initial seed $(\mathbf{c}_0,B_0)$, the coordinates of~$u$
with respect to $\mathbf{c}$ is of the same form as in $(*)$.
In particular, the coordinates of~$u$ with respect to $\mathbf{c}$ are positive.

$($The vector $u$ is a~radical vector for $B_0$, i.e.~$B_0u=0$.
In fact, any radical vector for $B_0$ is a~multiple of $u.)$
\end{proposition}

We generalize this statement to an arbitrary quiver as follows:

\begin{theorem}
\label{th:radical}
Let $B_0$ be a~skew-symmetrizable initial exchange matrix and suppose that $u_0>0$ is a~radical vector for $B_0$,
i.e.~$ B_0u_0=0 $.
Suppose also that, for any~$Y$-seed $(\mathbf{c},B)$ with respect to the initial seed $(\mathbf{c}_0,B_0)$, the
coordinates of $u_0$ with respect to $\mathbf{c}$ are non-negative.
Then, under assumption~\eqref{eq:C-sign-coherence}, for any $B$ which is mutation-equivalent to $B_0$, the matrix $B$ does not have any maximal green sequences.
\end{theorem}

We prove our results in Section~\ref{sec:proof}. For related applications of maximal green sequences, we refer the reader to~\cite{CLS} and~\cite{S6}.

\section{Proofs of main results}
\label{sec:proof}

Let us f\/irst note how the coordinates of a~vector change under the mutation operation, which can be easily checked using
the def\/initions (assuming~\eqref{eq:C-sign-coherence}):

\begin{proposition}
\label{prop:coord_change}
Suppose that $(\mathbf{c},B)$ is a~$Y$-seed with respect to an initial~$Y$-seed.
Suppose also that the coordinate vector of~$u$ with respect to $\mathbf{c}$ is $(a_1,\dots,a_n)$.
Let $(\mathbf{c}', B')=\mu_k(\mathbf{c}, B)$ and $(a'_1,\dots,a'_n)$ be the coordinates of~$u$ with respect to
$\mathbf{c}'$.
Then $a_i=a'_i$ if $i \ne k$ and $a'_k=-a_k+\sum a_i[\sgn(\mathbf{c}_k)B_{k,i}]_+$, where the sum is over all $i \ne k$.
\end{proposition}

As we mentioned, in view of Proposition~\ref{prop:positive}, Theorem~\ref{th:cyclic} follows from
Proposition~\ref{prop:radical}.
To prove Proposition~\ref{prop:radical}, it is enough to show that the coordinates of the vector $u$ change as stated,
i.e.~show that if the coordinates of $u$ with respect to $\mathbf{c}$ are as in $(*)$, then for the~$Y$-seed
$(\mathbf{c}',B')=\mu_k(\mathbf{c},B)$, the coordinates with respect to $\mathbf{c}'$ are also of the form in $(*)$.
This can be checked easily using the formula in Proposition~\ref{prop:coord_change}.

To prove Theorem~\ref{th:radical}, let us f\/irst note the following property of the coordinates of the radical vectors:

\begin{lemma}
\label{lem:rad_coord_change}
Suppose that $(\mathbf{c},B)$ is a~$Y$-seed with respect to an initial~$Y$-seed $(\mathbf{c}_0,B_0)$ and $u_0$ is
a~radical vector for $B_0$.
Suppose that the coordinate vector of $u_0$ with respect to $\mathbf{c}$ is $(a_1,\dots,a_n)$.
Then, for any index $k$, we have the following:
\begin{gather*}
\sum a_i[\sgn(\mathbf{c}_k)B_{k,i}]_+=\sum a_i[-\sgn(\mathbf{c}_k)B_{k,i}]_+,
\end{gather*}
where the sum is over all $i \ne k$.

In particular, for radical vectors, the formula in Proposition~{\rm \ref{prop:coord_change}} that describe the change of
coordinates under mutation depends only on the exchange matrix, not on the $\mathbf{c}$-vectors.
\end{lemma}

To prove the lemma, suppose that $ D=\operatorname{diag}(d_1,\dots,d_n) $ is a~skew-symmetrizing matrix for~$B_0$, so it is also
skew-symmetrizing for~$B$, so $ DB=C $ is skew-symmetric, i.e.~$ C_{i,k}=d_iB_{i,k}=-d_kB_{k,i}=-C_{k,i} $ for all~$i$,~$k$.
Let $ u=(a_1,\dots,a_n) $.
Then $u$ is a~radical vector for $B$, so it is also a~radical vector for $C=DB$, i.e.~$ Cu=0 $, which means that
for any index $k$, we have $ \sum a_i[\sgn(\mathbf{c}_k)C_{k,i}]_+=\sum a_i[\sgn(\mathbf{c}_k)C_{i,k}]_+$, which is
equal to $\sum a_i[-\sgn(\mathbf{c}_k)C_{k,i}]_+$, where the sum is over all $i \ne k$.
Then, writing $ C_{k,i}= d_kB_{k,i}$, we have
\begin{gather*}
\sum a_i[\sgn(\mathbf{c}_k)d_kB_{k,i}]_+=\sum a_i[-\sgn(\mathbf{c}_k)d_kB_{k,i}]_+.
\end{gather*}
Dividing both sides by $d_k>0$, we obtain the lemma.

 We will also need the following property of the radical vectors:

\begin{lemma}
\label{lem:rad_coord_change2}
In the set-up of Theorem~{\rm \ref{th:radical}}, let $u$ denote the vector which represents $u_0$ with respect to the
basis $\mathbf{c}$.
Then $u$ is a~radical vector for $B$, i.e.~$Bu=0$.
\end{lemma}

To prove the lemma, let us note that $u$ can be obtained from $u_0$ by applying the formula in
Proposition~\ref{prop:coord_change} along with the mutations.
Thus, to prove the lemma, it is enough to show that, for any $k=1,\dots,n$, we have the following:
\begin{quotation}
$(**)$~if $u=(a_1,\dots,a_n)$ is a~radical vector for $B$, then $u'$ is a~radical vector for $B'=\mu_k(B)$, i.e.~$B'u'=0$,
where $u'=(a'_1,\dots,a'_n)$ is the vector as in Proposition~\ref{prop:coord_change}.
\end{quotation}

To show $(**)$, we write $B'$ in matrix notation as follows~\cite[Lemma~3.2]{CAIII}: for $\epsilon=\sgn(\mathbf{c}_k)$,  we have
\begin{gather*}

B' = (J_{n,k} + E_k)  B  (J_{n,k} + F_k),
\end{gather*}
where
\begin{itemize}\itemsep=0pt
\item $J_{n,k}$ denotes the diagonal $n\times n$ matrix whose diagonal entries are all~$1$'s, except for $-1$ in
the~$k$th position;
\item $E_k$   is the $n\times n$ matrix whose only nonzero entries are   $e_{ik} = [-\varepsilon b_{ik}]_+$;
\item $F_k$   is the $n\times n$ matrix whose only nonzero entries are $f_{kj} = [\varepsilon b_{kj}]_+$.
\end{itemize}
It follows from a~direct check that $(J_{n,k} + F_k)u'=u$.
Then $ B'u'= (J_{n,k} + E_k) B (J_{n,k} + F_k)u'= (J_{n,k} + E_k) Bu=(J_{n,k} + E_k) 0=0$.
This completes the proof of the lemma.

Let us now prove Theorem~\ref{th:radical}.
For this, we f\/irst consider the $Y$-seed pattern def\/ined by the initial $Y$-seed $(\mathbf{c}_0,B_0)$ at the initial
vertex $t_0$.
Let us suppose that $t_1$ is a~vertex such that the corresponding $Y$ seed $(\mathbf{c},B)$ has the exchange matrix~$B$.
Then we can consider the $Y$-seed pattern def\/ined by the initial $Y$-seed $(\mathbf{c}_0,B)$ at the initial vertex
$t_1$ (where $\mathbf{c}_0$ is the tuple of standard basis).
Then we have the following: for any f\/ixed vertex $t$ of the $n$-regular tree $\mathbb{T}_n$, the exchange matrices
of the $Y$-seeds assigned by these patterns coincide because the pattern is formed by mutating at the labels
of the $n$-regular tree $\mathbb{T}_n$ and mutation is an involutive operation on matrices; let us denote these seeds~by
$(\mathbf{c}',B')$ and $(\mathbf{c}'',B')$ respectively.

On the other hand, let $u$ denote the vector which represents~$u_0$ with respect to the basis~$\mathbf{c}$, which
can be obtained by applying the formula in Proposition~\ref{prop:coord_change} along with the mutations.
Then~$u$ is a~radical vector for~$B$, i.e.~$Bu=0$ (Lemma~\ref{lem:rad_coord_change2}).
Furthermore, the coordinates of the vectors $u_0$ and $u$ with respect to the bases~$\mathbf{c}'$ and~$\mathbf{c}''$
respectively will coincide by Lemma~\ref{lem:rad_coord_change} (which says that for radical vectors the formula in
Proposition~\ref{prop:coord_change} depends only on the exchange matrices, not on the $ \mathbf{c}$-vectors).
In particular, the coordinates of~$u$ with respect to any basis of~$ \mathbf{c} $-vectors are non-negative.
Thus, by Proposition~\ref{prop:positive}, the matrix $B$ does not have any maximal green sequences.
This completes the proof.

\subsection*{Acknowledgements}

The author's research was supported in part by the Scientif\/ic and Technological Research Council of Turkey (TUBITAK)
grant \#~113F138.
The author also thanks Christof\/f Geiss for drawing his attention to the paper~\cite{ACCERV} by presenting its results at
the Workshop on Hall and Cluster Algebras in CRM, University of Montreal.
He also thanks the organizers for organizing the conference.

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