I am at the Shanghai Center of Mathematical Sciences at the Jiangwan campus of Fudan University in Shanghai. There are two seminars here today.
This was a talk by Lingmin Liau from the University of Wuhan. This is joint work with Cheng Liu.
In 2003, there was a paper by Jun Wu asking “How many points have the same Engel and Sylvester expansions?” Let us explain what these expansions are.
Engel’s expansion is \[\begin{equation} x= \frac{1}{d_{1}} + \frac{1}{d_{1} d_{2}} + \frac{1}{d_{1} d_{2} d_{3}} + \dots, \text{ for } 2 \leq d_{1} \leq d_{2} \leq \dots \end{equation}\]
Sylvester’s expansion is \[\begin{equation} x = \frac{1}{s_{1}} + \frac{1}{s_{2}} + \frac{1}{s_{3}} + \dots \text{ where } s_{k+1} \geq s_{j}(s_{j} - 1)+ 1 \end{equation}\]
This is related to dynamics via the following map. \[\begin{align} T(x) & = x - \tfrac{1}{ \lfloor \frac{1}{x}\rfloor + 1} \implies x \\ & = \frac{1}{ \lfloor \frac{1}{x}\rfloor + 1 } + Tx = \frac{1}{s_{1}} + \frac{1}{ \lfloor \frac{1}{ Tx} \rfloor + 1} + T^{2} x \dots \\ & = \frac{1}{s_{1}} + \frac{1}{ s_{2} } + \dots \end{align}\] There is a similar dynamics map for the first expansion which I did not understand.
Now we define the set \[\begin{equation} E = \{x : d_{1} = s_{1} , d_{2} = s_{2} , \dots\}. \end{equation}\] Then what is \(\dim_{H} E\)? The main result of Jun Wu is that this dimension is \(1/2\).
Before this Galambus showed in 1981, 1996 that the Lebesgue measure is 0.
\[\begin{equation} E(\alpha) := \{ x \in [0,1] \mid x \text{ has the same regular continued fractions and $\alpha$-CF}\}. \end{equation}\] Then, what is the \(\dim E(\alpha)\)?
We assume \(\alpha \in [0,1]\). For this, we must study the map \(T_{\alpha}\) which acts on \([\alpha - 1, \alpha)\) via \[\begin{equation} T_{\alpha}(x) = \begin{cases} |\tfrac{1}{x}| - \lfloor |\tfrac{1}{x}| + 1 - \alpha\rfloor & \text{ if }x \neq 0 \\ 0 & \text{ otherwise} \end{cases}. \end{equation}\]
For \(x \in \mathbb{R}\), let \(b_{0} \in \mathbb{Z}\) be such that \(x-b_{0} \in [\alpha-1,\alpha)\). For this
This is something similar to a phenomenon called Devil’s stairs (I don’t understand). The constant intervals of \(\alpha \mapsto S(\alpha)\) are
We have the bifurcation set (complement of the union of the above intervals) defined as: \[\begin{equation} \varepsilon = \{ x \in [0,1] \mid T^{k} x \leq x , \forall k \geq 0\}. \end{equation}\] Here \(T\) is the Gauss map \(x \mapsto \{\tfrac{1}{x}\}\). Then the Hausdorf dimension \(\dim_{H} \varepsilon = 1\).
Define \[\begin{equation} E(\alpha) : = \{x \in [0,1) : T^{n} x \leq \alpha, \forall n \geq 0\}. \end{equation}\] This is called a “survivor set”.
We will prove that the \(S(\alpha)\) in our question satisfies \[\begin{equation} S(\alpha) = \{x \in [0,1] : T^{n} x < \alpha \forall n \geq 0\}. \end{equation}\] This is the same thing as \(E(\alpha)\) but with inequality turned into a strict inequality. We get \[\begin{equation} S(\alpha) \subseteq E(\alpha) \subseteq S(\alpha) \cup \mathbb{Q} \cup \bigsqcup_{n=0}^{ \infty } \{ x \in [0,1), T^{n} x = \alpha\}. \end{equation}\]
For our interest, we study \[\begin{align} S(\tfrac{1}{n}) & = \{x \mid T^{k} x < \tfrac{1}{n} \forall k \geq 0\} \\ & = \{x \mid a_{k} > n \forall k \geq 0\}. \end{align}\] We have \[\begin{equation} \lim_{n \rightarrow \infty} \dim_{H} S(\tfrac{1}{n}) = \frac{1}{2}. \end{equation}\] This is due to Good from 1941. Then \(S(0) = ?\). This is the set \(\{T^{n} x < 0\} = \{\}\)
\[\begin{equation} \forall x = [1,m,x_{3},x_{4},\dots] , m \geq 2 , x_{i} \leq m-1, \forall i \geq 3. \end{equation}\] We then have \[\begin{equation} T x = [m ,x_{3} x_{4} \dots ] < x = [1 \dots ] \geq 2 \end{equation}\] and for \(k \geq 2\) \[\begin{equation} T^{k} x = [x_{k+1}, x_{k+2} , \dots ] < x= [1,m,x_{3}, \dots ] \end{equation}\]
\[\begin{equation} \dim_{H} \bigcup_{m=2}^{\infty} \{ x = [1,m_{1},x_{3},\dots] : x_{i} \leq m - 1\}. \end{equation}\] Then we have \[\begin{equation} = \sup_{m \geq 2} \dim_{H}\{ x = [1,m,x_{3},x_{4},\dots ] : x_{i} \leq m-1, \forall i \geq 3 \} \end{equation}\] Then with this, for some reason we get our result.
We have \(F(\alpha):= S(\alpha) \cap (\text{Bounded})\). We can prove \(\alpha \mapsto \dim_{H} F(\alpha)\) is continuous on \((0,1)\). This is related to some IFS (I don’t understand).
The speaker is Nguyen-Thi Dang. She is from Université Paris-Saclay.
This is joint work with myself and J. Li. See also my notes here.
Consider \(G = \mathrm{SL}(3,\mathbb{R})\) and \(\Gamma = \mathrm{SL}(3,\mathbb{Z})\). We have \[\begin{equation} A = \{ \begin{bmatrix} a_{1} & & \\ & a_{2} & \\ & & a_{3} \end{bmatrix} \mid a_{i} > 0, a_{1}a_{2}a_{3} = 1 \}. \end{equation}\] We also have \[\begin{equation} M = \{ \begin{bmatrix} \pm 1 & & \\ & \pm 1 & \\ & & \pm 1 \end{bmatrix}, \det( \cdot ) = 1 \} \end{equation}\] We have \(\mathfrak{a} = \log A = \{ (t_{1},t_{2},t_{3}) \mid \sum_{{i}}^{} t_{i}= 0\}.\)
A periodic torus is \[\begin{equation} C(A) = \{ \text{compact } A-\text{orbit}\}. \end{equation}\]
For \(F \in C(A)\), the set of periods \[\begin{equation} P(F) = \{ Y \in \mathfrak{a} \mid e^{Y} z = z, \forall z \in F\}. \end{equation}\]
Then \(P(F) \subseteq \mathfrak{a}\) is a \(2\)-lattice in \(\mathfrak{a} \simeq \mathbb{R}^{2}\). Fix an isomorphism \(\mathfrak{a} \simeq \mathbb{R}^{2}\). We then have \({P}(F) = \sqrt{\mathrm{covol}( P(F))} g_{F} \mathbb{Z}^{2}\) for some \(g_{F} \in \mathrm{SL}(2,\mathbb{R})\)
We define a shape \([P(F)] = g_{F} \mathrm{SL}({2},\mathbb{Z}) \in \mathrm{SL}(2,\mathbb{R})/\mathrm{SL}(2,\mathbb{Z})\).
(D,G,L)
The set of shapes of periodic tori in \(M \backslash \mathrm{SL}(3,\mathbb{R})/ \mathrm{SL}(3,\mathbb{Z})\) is dense in \(\mathrm{SL}(2,\mathbb{R})/\mathrm{SL}(2,\mathbb{Z})\).
We will look at \(0 < a_{1} < a_{2}\) in \(\mathbb{N}\) such that \(a_{1} \gg 1\) and \(a_{2} - a_{1} \gg 1\). Then \(f(X) = X(X-a_{1})(X-a_{2}) - 1\) is irreducible over \(\mathbb{Q}\).
We consider the field \(K = \mathbb{Q}[X]/ \langle f(X)\rangle\) This is isomorphic to \(\mathbb{Q}^{3}\) additively.
For \(\sigma = (\sigma_{1},\sigma_{2},\sigma_{3}) : K \rightarrow \mathbb{R}\) are given by \(X \mapsto \alpha_{i}\) where \(\alpha_{1},\alpha_{2},\alpha_{3}\) are the roots of the polynomial.
(DUT)
For all \([\mathcal{O}: \mathcal{O}_{K}] < \infty\) and \(\mathcal{O}\) is a subring \[\begin{equation} \mathcal{O}^{\times} \simeq \langle \pm 1 , \beta_{1}, \beta_{2} \rangle \simeq \mathbb{Z}^{2}. \end{equation}\] That is, multiplicatively \(\mathcal{O}^{\times} \simeq \mathbb{Z}^{2}\).
Using this, one can construct a periodic torus. We consider the action of \(K^{\times}\) on \(K\) and get \[\begin{align} K & \xrightarrow[]{} \mathbb{R}^{3} \\ x & \mapsto \mathrm{diag}( \sigma_{1}(x),\sigma_{2}(x), \sigma_{3}(x) ). \end{align}\] Then we look at \(\mathcal{O}^{\times}\) acting on \(\mathcal{O}\) and write \(\mathcal{O} = h_{\mathcal{O}} \mathbb{Z}^{3} \subseteq \mathbb{R}^{3}\).
For \(\mathcal{O} = \mathbb{Z}[X]/ \langle f(X)\rangle = \mathbb{Z}[\alpha]\), then \(\mathcal{O}^{\times} = \langle \pm 1, \alpha, \alpha- a_{1}\rangle\).
Indeed, observe that \(\alpha, \alpha-a_{1},\alpha - a_{2} \in \mathcal{O}^{\times}\). This lemma is due to an optimal inequality for cubic fields by Cusick.
But we want density of shapes!
What we will try is \[\begin{align} (a_{1},a_{2}) & \simeq (0,1)\pmod{2^{c}}, \text{ for }c \in \mathbb{Z}_{\geq 0} \\ (a_{1},a_{2}) & \simeq (1,0)\pmod{3^{d}}, \text{ for }d \in \mathbb{Z}_{\geq 0} \\ (a_{1},a_{2}) & \simeq (1,1)\pmod{5^{r}}, \text{ for }r \in \mathbb{Z}_{\geq 0} \\ \end{align}\] Then we have that \(\mathcal{O}_{(c,d,r)} = \mathbb{Z} + 2^{c} 3^{d} 5^{r} \mathcal{O}\) satisfies \[\begin{equation} \mathcal{O}_{(c,d,r)}^{\times} = \{ \pm \alpha^{m} (\alpha - a_{1})^{n} \mid \underbrace{ \begin{cases} \varphi_{(0,1)}(m,n) \simeq 0 \pmod{ 7\cdot 2^{c-1}} \\ \varphi_{(1,0)}(m,n) \simeq 0 \pmod{ 8\cdot 3^{d-1}} \\ \varphi_{(1,1)}(m,n) \simeq 0 \pmod{ 24\cdot 5^{r-1}} \\ \end{cases}}_{\text{ defines a lattice }\Lambda_{(c,d,r)} \subseteq \mathbb{Z}^{2}} \} \end{equation}\] where \(\varphi_{(0,1)}, \varphi_{(1,0)}, \varphi_{(1,1)} \in (\mathbb{Z}^{2})^{*}\).
(D,G,L)
\[\overline{\{ [\Lambda_{(c,d,r)}]\} }= \mathrm{SL}(2,\mathbb{R})/\mathrm{SL}(2,\mathbb{Z})\]
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