Progressive Measures
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Definition 1 (Kra et al., 2025, Definition 3.1) Let \((X,T)\) be a topological system and \(k\in\mathbb{N}\). We say that a probability measure \(\tau\in\mathcal{M}(X^{k+1})\) is progressive if, for all open sets \(U_1,...,U_k\subset X\) with \[\tau(X\times U_1\times\cdots\times U_k)>0 \] there exist infinitely many \(n\in\mathbb{N}\) such that \[\tau((X\times U_1\times\cdots\times U_{k-1}\times U_k)\cap T_{\triangle}^{-n}(U_1\times U_2\times\cdots\times U_k\times X))>0. \]
Definition 2 (cf. Kra et al., 2025, Definition 3.1) Let \((X,\Group)\) be a topological system and \(k\in\mathbb{N}\). We say that a probability measure \(\tau\in\mathcal{M}(X^{k+1})\) is progressive if, for all open sets \(U_1,...,U_k\subset X\) with \[\tau(X\times U_1\times\cdots\times U_k)>0 \] there exist infinitely many \(\GroupElement\in\Group\) such that \[\tau((X\times U_1\times\cdots\times U_{k-1}\times U_k)\cap \GroupActionPreImage{\GroupElement}{U_1\times U_2\times\cdots\times U_k\times X})>0. \]
If \((a,x_1,...,x_k)\) is a \(k+1\)-term Erdős progression where \(U_j\) is a neighbourhood of \(x_j\) for \(1\leq j\leq k\), then \[B=(X\times U_1\times\cdots\times U_{k-1}\times U_k)\cap T_{\triangle}^{-n}(U_1\times U_2\times\cdots\times U_k\times X)\] has \((a,x_1,...,x_k)\in B\) and, if further \(\tau(B)>0\), \(B\) contains infinitely many other \(k+1\)-term Erdős progressions.
Similarly, if \((a,x_{0...01},...,x_{1...1})\) is a \(k\)-dimensional Erdős cube where \(U_j\) is a neighbourhood of \(x_j\) for all \(j\in\{0,1\}^k\), \(c_i:\{0,1\}^k\rightarrow\{0,1\}^k\) such that \[c_i(j)(n)=\begin{cases}1 & j=n,\\ j(n)&\text{otherwise}, \end{cases} \] and $C_, then \[B=(X\times U_1\times\cdots\times U_{k-1}\times U_k)\cap T_{\triangle}^{-n}(U_1\times U_2\times\cdots\times U_k\times X)\]
Definition 3 (cf. Kra et al., 2025, Definition 3.1) Let \((X,\Group)\) be a topological system and \(k\in\mathbb{N}\). We say that a probability measure \(\tau\in\mathcal{M}(X^{k+1})\) is progressive if, for all open sets \(U_1,...,U_k\subset X\) with \[\tau(X\times U_1\times\cdots\times U_k)>0 \] there exist infinitely many \(\GroupElement\in\Group\) such that \[\tau((X\times U_1\times\cdots\times U_{k-1}\times U_k)\cap \GroupAction{\GroupElement}{U_1\times U_2\times\cdots\times U_k\times X})>0. \]
If \((a,x_1,...,x_k)\) is a \(k+1\)-term Erdős progression where \(U_j\) is a neighbourhood of \(x_j\) for \(1\leq j\leq k\), then \[B=(X\times U_1\times\cdots\times U_{k-1}\times U_k)\cap T_{\triangle}^{-n}(U_1\times U_2\times\cdots\times U_k\times X)\] has \((a,x_1,...,x_k)\in B\) and, if further \(\tau(B)>0\), \(B\) contains infinitely many other \(k+1\)-term Erdős progressions.
Proposition 1 (Kra et al., 2025, Proposition 3.2) Fix \(k\in\mathbb{N}\) and let \((X,T)\) be a topological system. Let \(a\in X\), \(U_1,...,U_k\subset X\) be open sets, and let \(\tau\in\mathcal{M}(X^{k+1})\) be a progressive measure satisfying \(\tau(\{a\}\times X^k)=1\).
If \(\tau(X\times U_1\times\cdots\times U_k)>0\), then there exists an Erdős progression \((a,x_1,...,x_k)\) with \(x_j\in U_j\) for all \(1\leq j\leq k\).
Proof. Let \(V=U_1\times\cdots\times U_k\). As \(\tau\) is a progressive measure, we can find \(c(1)\in\mathbb{N}\) such that \[\tau((X\times V)\cap T_{\triangle}^{-c(1)}(V\times X))>0. \]
1 Constructing Progressive Measures
- Check that the system has topological pronilfactors.
- Define a product measure in the \(k-1\)-step pronilfactor that is generic with \(\Folner\) and \(a\in\text{gen}(\mu,\Folner)\) and structured similarly to the progression.
- Project this measure back to the system by using conditional expectations.
3 Discrete Spectrum & Eigenmatrices
Definition 4 (cf. Jamneshan and Kreidler, 2025, Definition 5.1.2) For a unitary representation \(U:G\rightarrow\mathscr{U}(\mathcal{L}(X,\Borel{X}))\), we call an invariant finite-dimensional subspace \(M\subseteq\mathcal{L}(X,\Borel{X})\) irreducible if \(\{0\}\) and \(M\) are the only invariant linear subspaces contained in \(M\).
Definition 5 (cf. Jamneshan and Kreidler, 2025, Definition 5.1.1 & Corollary 5.1.5) Let \(U:G\rightarrow\mathscr{U}(\mathcal{L}(X))\) be a unitary representation of a group \(G\). Then the closure \[ \mathcal{L}(X,\Borel{X})_\text{ds}:=\overline{\text{lin}}\bigcup\{M\subseteq \mathcal{L}(X,\Borel{X})\mid M\text{ irreducible invariant finite-dimensional subspace}\}\subseteq \mathcal{L}(X) \] is called the discrete spectrum part of \(U\). As \(M_1+M_2\) is an invariant finite-dimensional subspace for invariant finite-dimensional subspaces \(M_1,M_2\), then \(\mathcal{L}(X)\text{ds}\) is always a (closed) linear subspace of \(\mathcal{L}(X)\).
As there exists \(\ProjectionMap:X\rightarrow K/L\) where \(K\) is a compact group, \(L\subseteq K\) is a closed subgroup and \(m\) is an invariant measure on \(K/L\) such that \(\text{L}^2(X,\SigmaAlgebra{K},\Measure)\cong\text{L}^2(K/L,m)\) and, for \(f:K/L\rightarrow\mathbb{C}\), \(f\circ\ProjectionMap:X\rightarrow\mathbb{C}\).
As \(\text{L}^2(X,\SigmaAlgebra{K},\Measure)\cong\text{L}^2(K/L,m)\), we know that \(\ProjectionMap^{-1}\) exists and thus there exists \(\bar{f}_\text{c}\in\text{L}^2(K/L,m)\) such that \(\bar{f}_\text{c}=f_\text{c}\circ\ProjectionMap^{-1}\).
Theorem 1 (Farashahi, 2017, Theorem 4.6) Let \(H\) be a closed subgroup of a compact group \(G\) and \(\mu\) be the normalized \(G\)-invariant measure on \(G/H\). The Hilbert space \(\text{L}^2(G/H,\mu)\) satisfies the following orthogonality decomposition \[ \text{L}^2(G/H,\mu)=\bigoplus_{[\pi]\in\Dual{G/H}}\epsilon_\pi(G/H), \] where \(\epsilon_\pi\) denotes the linear span of the matrix elements of \(\pi\). (\(\pi\) may not be the same as the previous one)
Note:
- (Farashahi, 2017, Theorem 4.5) will be helpful in confirming that every continuous function can be approximated equicontinuously with a linear combination of “eigenfunctions”.
Let \(\{e_1,...,e_n\}\) be the (continuous) orthonormal basis of a sufficiently large invariant finite-dimensional subspace \(M\subseteq\text{L}^2(X,\SigmaAlgebra{K},\Measure)\) such that there exists \(h\in M\) where \(||f_\text{c}-h||_{\text{L}^2}<\eta/6\) and \[ h=\sum_{i=1}^n \langle h,e_i \rangle e_i. \]
By the Cauchy-Schwartz Inequality, \[ \langle h,e_i\rangle\leq ||h||_{\text{L}^2}\cdot||e_i||_{\text{L}^2}=||h||_{\text{L}^2}. \]
We also have \[\begin{align*} U_gh&=\sum_{i=1}^n\langle h,e_i \rangle \sum_{j=1}^n \langle U_ge_i,e_j\rangle e_j \\&=\sum_{j=1}^n \left(\sum_{i=1}^n\langle h,e_i \rangle \langle U_ge_i,e_j\rangle\right) e_j. \end{align*}\] As we have \[ \langle U_ge_i,e_j\rangle = \langle e_i,U_g^{-1}e_j\rangle, \] then \[\begin{align*} U_gh&=\sum_{j=1}^n \left(\sum_{i=1}^n\langle h,e_i \rangle \langle U_ge_i,e_j\rangle\right) e_j \\&=\sum_{j=1}^n \left\langle \sum_{i=1}^n\langle h,e_i \rangle e_i,U_g^{-1}e_j\right\rangle e_j \\&=\sum_{j=1}^n \langle h,U_g^{-1}e_j\rangle e_j. \end{align*}\]
Let \(\delta>0\) be sufficiently small such that, for \(1\leq i\leq n\), we have \(|e_i(y)-e_i(z)|<\eta/6n\) whenever \(d(z,y)<\delta\). Applying \(U_g\) and focusing on each \(1\leq i\leq n\), we want to show that \[ |U_ge_i(y)-U_ge_i(z)|<\eta/6n \] for all \(g\in G\).
We have \[ U_ge_i=\sum_{j=1}^n \langle U_ge_i,e_j\rangle e_j. \] For each \(e_j\), we have that
- Use continuity as \(e_j\) is assumed to be continuous
- Orthogonal invariance
- Kronecker factor is uniquely ergodic so every point has dense orbit
Below is incorrect at the moment:
, as \(e_1,...,e_n\) is an orthonormal basis and \(|e_j(y)-e_j(z)|<\eta/6n\) for all \(1\leq j\leq n\), we get \[\begin{align*} |U_ge_i(y)-U_ge_i(z)|^2&= \sum_{j=1}^n |\langle U_ge_i,e_j\rangle (e_j(y)-e_j(z))|^2 \\&=\left(\frac{\eta}{6n}\right)^2\sum_{j=1}^n |\langle U_ge_i,e_j\rangle|^2 \\&=\left(\frac{\eta}{6n}\right)^2. \end{align*}\] Thus, \[ |U_ge_i(y)-U_ge_i(z)|<\frac{\eta}{6n}. \]
3.1 Rough work
By definition of the dual representation, we have \[\begin{align*} \overline{h}(f)&=\langle f,h\rangle \\&=\langle f,\sum_{i=1}^n \langle h,e_i \rangle e_i\rangle \\&=\sum_{i=1}^n \overline{\langle h,e_i \rangle} \langle f,e_i\rangle \\&=\sum_{i=1}^n \overline{\langle h,e_i \rangle} \overline{e_i}(f), \end{align*}\] for any \(f\in\mathcal{L}(X)\). Thus, \[ \overline{h}=\sum_{i=1}^n \overline{\langle h,e_i \rangle}\overline{e_i}. \]
With our tensor product \(U\otimes\overline{U}:G\rightarrow\mathscr{U}(\mathcal{L}(X)\otimes\mathcal{L}(\mathcal{L}(X)))\) with the dual representation, we get \[\begin{align*} (U_g\otimes\overline{U_g})(h\otimes\overline{h})&=(U_gh)\otimes(\overline{U_gh}) \\&=\sum_{i=1}^n\left( \langle h,e_i \rangle U_ge_i\right)\otimes\left(\overline{\langle h,e_i \rangle}\overline{U_ge_i}\right) \\&=\sum_{i=1}^n\left(\langle h,e_i \rangle \sum_{j=1}^n \langle U_ge_i,e_j\rangle e_j\right)\otimes\left(\overline{\langle h,e_i \rangle}\sum_{k=1}^n\overline{\langle U_ge_i,e_k\rangle}\overline{e_k}\right) \end{align*}\] for all \(g\in G\).
Continuing on, we have \[\begin{align*} (U_g\otimes\overline{U_g})(h\otimes\overline{h})&=\sum_{i=1}^n\left(\langle h,e_i \rangle \sum_{j=1}^n \langle U_ge_i,e_j\rangle e_j\right)\otimes\left(\overline{\langle h,e_i \rangle}\sum_{k=1}^n\overline{\langle U_ge_i,e_k\rangle}\overline{e_k}\right) \\&= \sum_{j=1}^n \sum_{k=1}^n\left(\sum_{i=1}^n\langle h,e_i \rangle \overline{\langle h,e_i \rangle} \langle U_ge_i,e_j\rangle \overline{\langle U_ge_i,e_k\rangle}\right) e_j\otimes\overline{e_k}. \end{align*}\]
As \(U_x\) is unitary, then \(U_ge_1,...,U_ge_n\) also forms an orthonormal basis and we get \[ e_k=\sum_{i=1}^n\langle e_k,U_ge_i\rangle U_ge_i. \]
We also have \[ \sum_{i=1}^n\langle h,e_i \rangle \overline{\langle h,e_i \rangle} \langle U_ge_i,e_j\rangle \overline{\langle U_ge_i,e_k\rangle}=\sum_{i=1}^n\langle\langle e_i,h \rangle h,e_i \rangle \langle\langle e_k,U_ge_i\rangle U_ge_i,e_j\rangle \]
As \[ U_gh=\sum_{i=1}^n \langle h,e_i\rangle U_ge_i=\sum_{j=1}^n \langle h,U_g^{-1}e_j\rangle e_j \] and \(U_ge_1,...,U_ge_n\) are orthonormal, we get \[ \langle h,e_i\rangle U_ge_i=\sum_{j=1}^n\langle h,U_g^{-1}e_j\rangle \langle e_j,U_ge_i\rangle U_ge_i \]
Using the Cauchy-Schwartz Inequality, we get \[\begin{align*} |\langle h,U_g^{-1}e_i\rangle|&\leq ||h||\cdot||U_g^{-1}e_i|| \\& = ||h||\cdot ||e_i|| \\& = ||h|| \end{align*}\]
We have \[ U_ge_i=\sum_{j=1}^n \langle U_ge_i,e_j\rangle e_j, \] and so \[ U_ge_i(y)-U_ge_i(z)=\sum_{j=1}^n \langle U_ge_i,e_j\rangle (e_j(y)-e_j(z)). \] As \(|e_j(y)-e_j(z)|<\eta/6n\) for all \(1\leq j\leq n\), we get \[\begin{align*} |U_ge_i(y)-U_ge_i(z)|&\leq \sum_{j=1}^n \langle U_ge_i,e_j\rangle |(e_j(y)-e_j(z))| \\&<\frac{\eta}{6n}\sum_{j=1}^n \langle U_ge_i,e_j\rangle \end{align*}\]