TITLE: When are pseudofinite theories almost sure theories?
QUESTION [3 upvotes]: Let $T$ be a complete theory (say in a countable language, though this may not be required). Assume that $T$ is complete with infinite models.
Question 1) When is the theory of $T$ an almost sure theory? What are the references for these results.
Note that I have left the term "almost sure" undefined. This is because the notion seems to change from author to author. Some authors take it to mean the limit under uniform measures, others (The Shelah-Spencer almost sure theories for example) use more complicated measures. I have no restriction in mind and would like to know what results are out there for "reasonable" notions of almost sure theories.
Question 2) What are some good sources that contain results about pseudofinite model theory and the "interaction" with almost sure theories?
Edit 1) The following should be added to the end of question 2: For example, assume that $T$ is pseudofinite. When is it an almost sure theory?

REPLY [1 votes]: In this answer, let's only think about theories in countable relational languages.
Most of the time, when people say that $T$ is an "almost-sure theory", they mean it arises as the limit theory for some sequence of probabilistic constructions which admits a zero-one law. We could formalize this by asking for a sequence of probability measures $(\mu_n)_{n\in\omega}$ on the space of all finite $L$-structures (let's say $L$-structures whose domains are initial segments of $\omega$, to make this a reasonable space), such that for every sentence $\varphi\in T$, $\lim_{n\to\infty} \mu_n([\varphi]) = 1$, where $[\varphi]$ is the set of all models of $\varphi$ in our space.
But this definition is probably too broad. Indeed, under this definition every pseudofinite theory is an almost-sure theory. Just enumerate $T$ as $\{\varphi_i\mid i\in \omega\}$, let $M_n$ be a finite model of $\bigwedge_{i\leq n}\varphi_i$, and let $\mu_n$ be the Dirac measure concentrated on the point $M_n$.
So what are some reasonable ways to refine the definition?
One very minor thing we could do is to require the measure $\mu_n$ to be a measure not on the space of all finite $L$-structures, but on the space $\mathrm{Str}_L([n])$ of $L$-structures with domain $[n] = \{0,\dots,n-1\}$. (This is the case for the usual sorts of zero-one law examples like the uniform measures and the Shelah-Spencer measures).
By a similar construction to the one above, you can show that a pseudofinite theory $T$ is almost-sure for this modified notion if and only if for every sentence $\psi$ such that $T\models \psi$, the exists some $N\in \omega$ such that $\psi$ has a model of size $n$ for all $N\leq n$. This is a somewhat reasonable notion - it rules out theories whose finite models are special enough that they only show up sporadically. But it still allows us to use Dirac measures, which feels pretty unnatural.
Here are some other possible conditions you could put on the $\mu_n$.

For any formula $\varphi(\overline{x})$ and any tuple $\overline{a}$ from $[n]$, let $[\varphi(\overline{a})] = \{M\in \mathrm{Str}_L([n])\mid M\models \varphi(\overline{a})\}$. Require that for all $\varphi(\overline{x})$ quantifier-free, all $\overline{a}$ from $[n]$,  and all $n\leq m$, $\mu_n([\varphi(\overline{a})]) = \mu_m([\varphi(\overline{a})])$. Weaker, you could only require that $\lim_{n\leq m\to \infty} \mu_m([\varphi(\overline{a})])$ converges.
The space $\mathrm{Str}_L([n])$ comes with a natural action of the symmetric group $S_n$ by permutation of the domain. Require $\mu_n$ to be invariant under this group action. This is the same as requiring that $\mu_n([\varphi(\overline{a})]) = \mu_n([\varphi(\sigma(\overline{a}))])$ for all $\sigma\in S_n$.
Require that whenever $\varphi(\overline{x})$ and $\psi(\overline{y})$ are quantifier-free formulas and $\overline{a}$ and $\overline{b}$ are disjoint tuples from $[n]$, $\mu_n([\varphi(\overline{a})\land \psi(\overline{b})]) = \mu_n([\varphi(\overline{a})])\mu_n([\psi(\overline{b})])$. Or just require that this identity is true in the limit as $n\to \infty$.

If we assume condition 1, we get that the $\mu_n$ converge in a natural way to a measure $\mu$ on $\mathrm{Str}_L(\omega)$, the space of countable labeled $L$-structures. If we assume condition 2, we get that $\mu$ is invariant under the group action of $S_\infty$ (the permutation group of $\omega$), and if we assume condition 3, it turns out that $\mu$ is ergodic for this group action. I think that these sorts of measures are very interesting, and I'm currently working on the relationship between these limit measures and the limit theory of the $\mu_n$ (under additional hypothesis, i.e. that the limit theory is the theory of a Fraïssé limit and the limit measure $\mu$ gives measure $1$ to the isomorphism class of that Fraïssé limit!).
On the other hand, I certainly don't want to suggest that a general notion of "almost sure theory" should include even condition 1, since condition 1 is not satisfied in general by the uniform measures on Fraïssé classes.


text:
\begin{document}

\begin{center}
{\LARGE\bf Rigid actions need not be strongly ergodic}

\bigskip

{\sc by Adrian Ioana$^{(1)}$ and Stefaan Vaes$^{(2)}$
\setcounter{footnote}{1}\footnotetext{Supported by a Clay Research Fellowship.\\ Mathematics Department; University of California at Los Angeles, CA 90095-1555 (United States). \\ E-mail: adiioana@math.ucla.edu}
\setcounter{footnote}{2}\footnotetext{Partially supported by ERC Starting Grant VNALG-200749, Research Programme G.0231.07 of the Research Foundation --
Flanders (FWO) and K.U.Leuven BOF research grant OT/08/032.\\ Department of Mathematics;
    K.U.Leuven; Celestijnenlaan 200B; B--3001 Leuven (Belgium).
    \\ E-mail: stefaan.vaes@wis.kuleuven.be}}

\end{center}

\begin{abstract}\noindent
A probability measure preserving action $\Gamma \actson (X,\mu)$ is called rigid if the inclusion of $\rL^\infty(X)$ into the crossed product $\rL^\infty(X) \rtimes \Gamma$ has the relative property (T) in the sense of Popa. We give examples of rigid, free, probability measure preserving actions that are ergodic but not strongly ergodic. The same examples show that rigid actions may admit non-rigid quotients.
\end{abstract}

\section*{Introduction}

Popa raised the question whether all rigid free ergodic probability measure preserving (pmp) actions are strongly ergodic. We answer this question negatively by providing a counterexample of the form $\Gamma \actson \T^4$ for some well chosen $\Gamma \subset \SL(4,\Z)$. Actually, as we recall below, the notion of rigidity makes sense for all pmp actions. In the same way as property (T) groups can have infinite center, we give examples of rigid actions that are not ergodic. By enlarging the acting group, rigidity is preserved and we obtain ergodic but non strongly ergodic rigid actions.

Popa's definition of relative property (T) for inclusions of finite von Neumann algebras is formulated in the language of bimodules. If $(M,\tau)$ is a von Neumann algebra with a faithful tracial state $\tau$ and $\cH$ is a Hilbert $M$-$M$-bimodule, we say that
\begin{itemize}
\item $\xi \in \cH$ is $M$-central if $x \xi = \xi x$ for all $x \in M$;
\item $\xi \in \cH$ is tracial if $\langle x \xi, \xi \rangle = \tau(x) = \langle \xi x, \xi \rangle$ for all $x \in M$;
\item the sequence of unit vectors $\xi_n \in \cH$ is almost $M$-central if $\|x \xi_n - \xi_n x \| \recht 0$ for all $x \in M$.
\end{itemize}

\begin{definition}[Definition 4.2.1 in \cite{Po01}]
Let $(M,\tau)$ be a von Neumann algebra with a faithful tracial state $\tau$ and $A \subset M$ a von Neumann subalgebra. We say that $A \subset M$ is \emph{rigid} if the following holds. Whenever $\cH$ is a Hilbert $M$-$M$-bimodule and $\xi_n \in \cH$ is a sequence of tracial, almost $M$-central vectors, then $\|\xi_n - P_A(\xi_n)\| \recht 0$, where $P_A$ denotes the orthogonal projection of $\cH$ onto the subspace of $A$-central vectors.
\end{definition}

For a subgroup $H \subset G$ of a countable group $G$, rigidity of $\rL H \subset \rL G$ is equivalent with the relative property (T) of $H \subset G$ in the sense of Kazhdan-Margulis, see \cite[Proposition 5.1]{Po01}. By definition, rigidity of $A \subset M$ is preserved when making $A$ smaller or $M$ larger.

Whenever $\Gamma \actson (X,\mu)$ is a pmp action of a countable group $\Gamma$, we denote by $M=\rL^\infty(X) \rtimes \Gamma$ the crossed product von Neumann algebra. We always equip $M$ with the trace given by
$$\tau(a u_g) = \begin{cases} \int_X a \; d\mu &\;\;\text{if}\;\; g = e \; , \\ 0 &\;\;\text{if}\;\; g \neq e \; .\end{cases}$$

\begin{definition}[\cite{Po01}]\label{def.rigid}
A probability measure preserving action $\Gamma \actson (X,\mu)$ is said to be rigid if the inclusion $\rL^\infty(X) \subset \rL^\infty(X) \rtimes \Gamma$ is rigid.
\end{definition}

By \cite[Theorem 0.3]{Io09} rigidity of an ergodic pmp action $\Gamma \overset{\sigma}{\actson} (X,\mu)$ is equivalent with the following ergodic theoretic property: there is no sequence of probability measures $\nu_n$ on $X \times X$ satisfying the properties
\begin{itemize}
\item $p^i_* \nu_n = \mu$ for all $n$ and $i =1,2$, where $p_i : X \times X \recht X$ denotes the projection on the $i$-th coordinate,
\item $\nu_n(\Delta) = 0$ for all $n$, where $\Delta \subset X \times X$ denotes the diagonal,
\item for all bounded measurable functions $f,g : X \recht \C$, one has
$$\int_{X \times X} f(x) g(y) \; d\nu_n(x,y) \recht \int_X f(x) g(x) \; d\mu(x) \; ,$$
\item $\|(\si_g \times \si_g)_* \nu_n - \nu_n\| \recht 0$ for all $g \in \Gamma$, where $\| \, \cdot \,\|$ denotes the norm of a bounded signed measure on $X \times X$ when viewed as a linear functional on the Banach space of bounded Borel functions $\cL^\infty(X \times X)$ equipped with the supremum norm.
\end{itemize}

One associates to any ergodic pmp action $\Gamma \actson (X,\mu)$ the type II$_1$ orbit equivalence relation $\cR := \cR(\Gamma \actson X)$, giving rise to a II$_1$ factor $\rL \cR$ with Cartan subalgebra $\rL^\infty(X)$. When the action $\Gamma \actson (X,\mu)$ is essentially free, we have $\rL \cR = \rL^\infty(X) \rtimes \Gamma$. In the non-free case this equality does not hold. Moreover, it is an open problem whether in general the rigidity of a non-free ergodic pmp action $\Gamma \actson (X,\mu)$ in the sense of Definition \ref{def.rigid} (i.e.\ the relative property (T) of $\rL^\infty(X) \subset \rL^\infty(X) \rtimes \Gamma$ or the above ergodic theoretic criterion) is equivalent or not with the rigidity of the orbit equivalence relation, i.e.\ the relative property (T) of $\rL^\infty(X) \subset \rL \cR$. We refer to \cite[Remark 4.6]{Io09} for more details.

We finally recall the notion of strong ergodicity. A pmp action $\Gamma \actson (X,\mu)$ is said to be strongly ergodic if there are no non-trivial asymptotically $\Gamma$-invariant subsets in $X$. More precisely, whenever $X_n \subset X$ is a sequence of measurable subsets such that the measure of the symmetric difference $g \cdot X_n \bigtriangleup X_n$ tends to zero for all $g \in \Gamma$, then $\mu(X_n) (1-\mu(X_n)) \recht 0$. Equivalently, there is no bounded sequence of functions $a_n \in \rL^\infty(X)$ such that $\|g \cdot a_n - a_n \|_2 \recht 0$ for all $g \in \Gamma$ and $\int_X a_n \; d\mu = 0$ for all $n$.

In \cite[Theorem 2.1]{JS85} it is shown that an ergodic pmp action $\Gamma \actson (X,\mu)$ is non strongly ergodic if and only if there exists an ergodic pmp action $\Z \actson (Y,\eta)$ and a non-singular quotient map $\pi : X \recht Y$ satisfying $\pi(\Gamma \cdot x) = \Z \cdot \pi(x)$ for a.e.\ $x \in X$. Since actions of amenable groups are never rigid, in an attempt to establish that rigidity implies strong ergodicity, we tried to prove that quotients of rigid actions remain rigid. As we observe in Remark 3 at the end of the article, quotients of rigid actions actually need not be rigid.

A pmp action $\Gamma \actson (X,\mu)$ is strongly ergodic whenever it has spectral gap, i.e.\ if the unitary representation $\Gamma \recht \cU(\rL^2(X) \ominus \C 1)$ does not weakly contain the trivial representation. Since $\SL(2,\Z)$ is non-amenable while all stabilizers for the action of $\SL(2,\Z)$ on $\Z^2 - \{0\}$ are amenable, it follows that the natural action of $\SL(2,\Z)$ on $\R^2/\Z^2 = \T^2$ has spectral gap and in particular is strongly ergodic, see \cite{Sc79}. From this it follows immediately that the action $\SL(n,\Z) \actson \R^n/\Z^n = \T^n$ has spectral gap whenever $n \geq 3$.

If $\Gamma$ has property (T) then all ergodic pmp actions are strongly ergodic. An amenable group $\Gamma$ cannot act strongly ergodically on a standard non-atomic probability space.

\section*{Examples of rigid actions that are ergodic but not strongly ergodic}

Let $k \geq 0$ and $n \geq 2$ be integers. Fix an arbitrary subgroup $\Lambda \subset \SL(k,\Z)$ and define the group $G \subset \SL(k+n+1,\Z)$ given by
$$G = \begin{pmatrix} \Lambda & \M_{k,n}(\Z) & \Z^k \\ 0 & \SL(n,\Z) & \Z^n \\ 0 & 0 & 1 \end{pmatrix} \; .$$
Denote by $I_s$ the $s \times s$ identity matrix and define the following subgroups of $G$.
\begin{align*}
\Gamma = \begin{pmatrix} \Lambda & \M_{k,n}(\Z) & 0 \\ 0 & \SL(n,\Z) & 0 \\ 0 & 0 & 1 \end{pmatrix} \qquad & , \qquad
H = \begin{pmatrix} I_k & 0 & \Z^k \\ 0 & I_n & \Z^n \\ 0 & 0 & 1 \end{pmatrix} \; , \\
H_1 = \begin{pmatrix} I_k & \M_{k,n}(\Z) & \Z^k \\ 0 & I_n & \Z^n \\ 0 & 0 & 1 \end{pmatrix} \qquad & , \qquad
H_2 = \begin{pmatrix} I_k & \M_{k,n}(\Z) & \Z^k \\ 0 & \SL(n,\Z) & \Z^n \\ 0 & 0 & 1 \end{pmatrix} \; .
\end{align*}

Note that $H,H_1,H_2$ are all normal subgroups of $G$ and that we can view $G$ as the semidirect product
$$G = H \rtimes \Gamma = H_1 \rtimes (\Lambda \times \SL(n,\Z)) = H_2 \rtimes \Lambda \; .$$
We identify the group von Neumann algebra $\rL H$ with $\rL^\infty(\Hh)$, where $\Hh \cong \T^{k+n}$ is equipped with the Lebesgue measure. In this way, we get a natural probability measure preserving action $\Gamma \actson \Hh$ which has the property that $\rL^\infty(\Hh) \rtimes \Gamma = \rL G$.

\begin{theorem}\label{ourthm1}
Let $k \geq 0$ and $n \geq 2$ be integers. Fix an arbitrary subgroup $\Lambda \subset \SL(k,\Z)$ and define the groups $G, \Gamma, H, H_1, H_2$ as above.
\begin{enumerate}
\item The pair $H_1 < H_2$ has the relative property (T). In particular, $H < G$ has the relative property (T) and $\Gamma \actson \Hh$ is a rigid action.
\item The action $\Gamma \actson \Hh$ is ergodic if and only if $\{A x \mid A \in \Lambda\}$ is infinite for all $x \in \Z^k - \{0\}$. In particular, if $k \geq 1$ and $\Lambda = \{I_k\}$ is the trivial group, the action $\Gamma \actson \Hh$ is rigid but not ergodic.
\item The action $\Gamma \actson \Hh$ is strongly ergodic if and only if the natural action $\Lambda \actson \widehat{\Z^k} = \T^k$ is strongly ergodic.
\item The action $\Gamma \actson \Hh$ is essentially free.
\end{enumerate}
In particular, if $k,n \geq 2$, if $A \in \SL(k,\Z)$ is a matrix with no eigenvalues being a root of unity and if $\Lambda = A^\Z$, then the action $\Gamma \actson \Hh$ is rigid, free, probability measure preserving and ergodic, but not strongly ergodic.
\end{theorem}

\begin{proof}
1.\ Define the following subgroups of $H_1$.
$$K = \begin{pmatrix} I_k & \M_{k,n}(\Z) & 0 \\ 0 & I_n & 0 \\ 0 & 0 & 1 \end{pmatrix} \qquad , \qquad
L = \begin{pmatrix} I_k & 0 & 0 \\ 0 & I_n & \Z^n \\ 0 & 0 & 1 \end{pmatrix} \; .$$
Embedding $\Z^n$ as the $i$-th row into $\M_{k,n}(\Z)$, we define the subgroup $K_i < K$. Since $n \geq 2$, we know that $\Z^n \subset \Z^n \rtimes \SL(n,\Z)$ has the relative property (T). Therefore, $L < H_2$ and $K_i < H_2$ have the relative property (T). The group $H_1$ is boundedly generated by $L,K_1,\ldots,K_k$. More precisely one checks easily that $K=K_1 \cdots K_k$ and $H_1 = KLKL$. Hence $H_1 < H_2$ has the relative property (T).

2.\ The action $\Gamma \actson \Hh$ is ergodic if and only if $\{g x \mid g \in \Gamma\}$ is infinite for all $x \in \Z^{k+n} - \{0\}$. It is easy to see that the latter is equivalent with $\{A x \mid A \in \Lambda\}$ being infinite for all $x \in \Z^k - \{0\}$.

3.\ Denote the action $\Gamma \actson \rL(\Z^{k+n}) = \rL^\infty(\T^{k+n})$ by $(\si_g)_{g \in \Gamma}$. Denote by $\tau$ the natural tracial state on $\rL(\Z^{k+n})$, corresponding to integration with respect to the Lebesgue measure on $\rL^\infty(\T^{k+n})$. Let $\|\, \cdot \,\|_2$ be the associated $2$-norm. View $\rL(\Z^k) \subset \rL(\Z^{k+n})$ and denote by $E : \rL(\Z^{k+n}) \recht \rL(\Z^k)$ the unique trace preserving conditional expectation. It suffices to prove the following claim: if
$a_i \in \rL(\Z^{k+n})$ is a bounded sequence satisfying $\|a_i - \si_g(a_i)\|_2 \recht 0$ for all $g \in \Gamma$, then $\|a_i - E(a_i)\|_2 \recht 0$. To prove this claim, it suffices to show the stronger statement that $(\si_g)_{g \in \Gamma}$, viewed as a unitary representation on $\ell^2(\Z^{k+n})$, is such that its restriction to $\ell^2(\Z^{k+n}) \ominus \ell^2(\Z^k)$ does not weakly contain the trivial representation. The latter follows because, for $n \geq 2$, the unitary representation $\SL(n,\Z) \recht \cU(\ell^2(\Z^n - \{0\}))$ does not weakly contain the trivial representation.

4.\ Because $\SL(k+n,\Z) \actson \T^{k+n}$ is essentially free and $\Gamma \subset \SL(k+n,\Z)$, also $\Gamma \actson \Hh$ is essentially free.

Concrete example.\ Take $A$ and $\Lambda = A^\Z$ as in the theorem. Since $\Lambda \cong \Z$ is amenable, the action $\Lambda \actson \T^k$ is not strongly ergodic. By 3, also $\Gamma \actson \Hh$ is not strongly ergodic. All eigenvalues of all matrices in $\Lambda - \{I_k\}$ differ from $1$. Hence, $B x \neq x$ whenever $x \in \Z^k - \{0\}$ and $B \in \Lambda - \{I_k\}$. Hence, $\{B x \mid B \in \Lambda\}$ is infinite for all $x \in \Z^k - \{0\}$. By 2, $\Gamma \actson \Hh$ is ergodic.
\end{proof}

\section*{Concluding remarks}

{\bf 1)\ } If $k \geq 0$ and $n \geq 3$, the group $H_2$ defined above has property (T) and its center is isomorphic with $\Z^k$.

{\bf 2)\ } If $n =2$ and if $\Lambda \subset \SL(k,\Z)$ has the Haagerup property, then $G/H_1 \cong \Lambda \times \SL(2,\Z)$ has the Haagerup property, so that $\rL H_1 \subset \rL G$ is an HT inclusion in the sense of \cite[Definition 6.1]{Po01}.

{\bf 3)\ } Let $\Lambda \subset \SL(k,\Z)$ be a subgroup such that $\Lambda x$ is infinite for all $x \in \Z^k \setminus \{0\}$ and $\Lambda \actson \widehat{\Z^k}$ is non-rigid, e.g.\ because $\Lambda$ is amenable. Denote by $\pi : \Gamma \recht \Lambda$ the natural quotient homomorphism. The action $\Gamma \actson \widehat{\Z^k}$ given by $g \cdot \om = \pi(g) \cdot \om$ is a \emph{non-rigid quotient} of the rigid action $\Gamma \actson \widehat{H}$. Since $\Gamma$ acts on $\widehat{\Z^k}$ through the quotient homomorphism $\pi$, the action $\Gamma \actson \widehat{\Z^k}$ is not free. The following simple modification also provides free counterexamples: the diagonal action $\Gamma \actson \widehat{H} \times \widehat{H}$ given by $g \cdot (\om,\om') = (g \cdot \om,g\cdot \om')$ is free, ergodic, rigid and admits the diagonal action $\Gamma \actson \widehat{\Z^k} \times \widehat{H}$ given by $g \cdot (\om,\om') = (\pi(g) \cdot \om,g \cdot \om')$ as a quotient. This quotient action is free and ergodic. To see that it is non-rigid, it suffices to observe that $\Z^k \times H \subset (\Z^k \times H) \rtimes \Gamma$ does not have the relative property (T). The latter holds because
$$\Z^k \times H \subset (\Z^k \times H) \rtimes \Gamma \subset (\Z^k \times \Lambda) \times (H \rtimes \Gamma)$$
and because $\Z^k \times \Lambda$ is amenable.

{\bf 4)\ }
By Theorem \ref{ourthm1}, for $n \geq 3$, the natural action of
$$\Gamma_1 = \begin{pmatrix} 1 & \M_{1,n-1}(\Z) \\ 0 & \SL(n-1,\Z)\end{pmatrix}$$
by automorphisms of $\Z^{n}$ yields the rigid embedding $\rL \Z^{n} \subset \rL \Z^{n} \rtimes \Gamma_1$ and hence a rigid pmp action $\Gamma_1 \actson \T^{n}$. Under the natural isomorphism $\rL \Z^{n} \cong \rL^\infty(\T^{n})$, the automorphism of $\rL \Z^{n}$ given by $A \in \SL(n,\Z)$ gets transformed into the pmp transformation of $\T^{n} = \R^{n} / \Z^{n}$ given by $(A^{-1})^t$. So, the natural action of
$$\Gamma_2 = \begin{pmatrix} 1 & 0 \\ \M_{n-1,1}(\Z) & \SL(n-1,\Z) \end{pmatrix}$$
on $\R^n / \Z^n = \T^{n}$ is rigid. The $\Gamma_2$-invariant functions in $\rL^\infty(\T^{n})$ are precisely the functions that only depend on the first variable. Whenever $\al \in \R - \Q$, multiplication by $(\exp(2\pi i \al x_1),\ldots,\exp(2\pi i \al x_{n}))$ provides an ergodic pmp action of $\Z^{n}$ on $\T^{n}$. This action is in semidirect product position with the action of $\Gamma_2$ that we considered so far. Together they yield an action of
$$\Gamma = \left( \mbox{}\hspace{-6ex}\begin{matrix} 1 & 0 & 0 \\ \Z & 1 & 0 \\ \hspace{7ex}\M_{n-1,2}(\Z)\hspace{-3ex}\mbox{} & & \SL(n-1,\Z) \end{matrix}\right)$$
on $\T^{n}$. This provides a different family of examples of rigid, free, pmp actions that are ergodic but not strongly ergodic.

{\bf 5)\ }
For all $n \geq 4$, Theorem \ref{ourthm1} provides examples of subgroups $\Gamma \subset \SL(n,\Z)$ such that the corresponding action $\Gamma \actson \T^n$ is rigid and ergodic, but not strongly ergodic. We prove that for $n \leq 3$ such examples do not exist. So, let $n \leq 3$ and let $\Gamma \subset \SL(n,\Z)$ be a subgroup such that $\Gamma \actson \R^n/\Z^n = \T^n$ is rigid, but not strongly ergodic. We prove that $\Gamma \actson \T^n$ is not ergodic. Since $\Gamma \actson \T^n$ is rigid, the group $\Gamma$ is non-amenable. In particular, $\Gamma$ has no abelian finite index subgroup. From the strong ergodicity criterion in \cite[Theorem 6.5(ii)]{FS96} it then follows that the linear action of $\Gamma$ on $\R^n$ has a non-trivial invariant subspace. Since $n \leq 3$, this invariant subspace has dimension 1 or co-dimension 1. We find a non-zero vector $v \in \R^n$ such that $v$ is either a common eigenvector for the matrices in $\Gamma$ or a common eigenvector for the matrices in the transpose $\Gamma^t$ of $\Gamma$.

We make use of the following observation that we prove below: if $n \in \{2,3\}$ and if $v \in \R^n - \{0\}$ is a common eigenvector for all the matrices in a non-amenable subgroup $\Gamma \subset \SL(n,\Z)$, then there exists $A \in \SL(n,\Z)$ such that $A^{-1} \Gamma A \subset K$ where
$$K = \begin{pmatrix} \pm 1 & \Z^{n-1} \\ 0 & \GL(n-1,\Z) \end{pmatrix} \; .$$
Using this observation, we may assume that either $\Gamma$ or $\Gamma^t$ is a subgroup of $K$. The projection $\Z^n \recht \Z$ on the first coordinate, extends to a group homomorphism $\Z^n \rtimes K^t \recht \Z \rtimes \Z/2\Z$, implying that $\Z^n \subset \Z^n \rtimes K^t$ does not have the relative property (T). Taking the Fourier transform, it follows that $K \actson \R^n/\Z^n$ is not rigid. So, it is impossible that $\Gamma \subset K$. Hence, we may assume that $\Gamma \subset K^t$. Since $K^t \actson \R^n/\Z^n$ is not ergodic, it follows that $\Gamma \actson \R^n/\Z^n$ is not ergodic.

It remains to prove the observation in the previous paragraph. Let $\om : \Gamma \recht \R^*$ be a group morphism and $v \in \R^n$ a non-zero vector such that $B v = \om(B) v$ for all $B \in \Gamma$. Let $\al_1,\ldots,\al_k \in \R$ be a basis over $\Q$ for the $\Q$-linear span of the coefficients of $v$. Take $v_1,\ldots,v_k \in \Q^n$ such that $v = \al_1 v_1 + \cdots \al_k v_k$. It follows that for all $B \in \Ker \om$ and all $i = 1,\ldots,k$, we have $B v_i = v_i$.
Since $\Ker \om$ is non-amenable and $n \leq 3$, it follows that $\lspan\{v_1,\ldots,v_k\}$ is one-dimensional. We conclude that $v$ is a multiple of a vector in $\Q^n$. Hence, we find a non-zero vector $w \in \Z^n$ such that the greatest common divisor of all coefficients of $w$ equals $1$ and $B w = \om(B) w$ for all $B \in \Gamma$. Denote by $e_1 \in \Q^n$ the first standard basis vector. Take $A \in \SL(n,\Z)$ such that $A e_1 = w$. It follows that
$$A^{-1} \Gamma A \subset \begin{pmatrix} * & \R^{n-1} \\ 0 & \GL(n-1,\R) \end{pmatrix} \; .$$
Since $A^{-1} \Gamma A \subset \SL(n,\Z)$, it follows that actually $A^{-1} \Gamma A \subset K$.

config: arxiv


TITLE: How do I show this relationship of 2 differentials?
QUESTION [1 upvotes]: If I have some $H = U + pV$, where $dU = TdS - pdV$,  how do I show for some varying $H$, $S$, $p$ that $$\left(\frac{\partial V}{\partial S}\right)_p = \left(\frac{\partial T}{\partial p}\right)_S$$
which will then further show that
$$\left(\frac{\partial S}{\partial V}\right)_p \left(\frac{\partial T}{\partial p}\right)_V - \left(\frac{\partial S}{\partial p}\right)_V \left(\frac{\partial T}{\partial V}\right)_p = 1$$
by considering $U$ as a function of $p$ and $V$ and considering two expressions for $\partial^2U/\partial p \partial V$.
What I have done:

Looking at the question at first, I'm thinking that I should first integrate $dU$ to get $U = Ts - pV$, which I then plug back into the first equation concerning $H$.
Then I would have $H = TS$. But this doesn't contain $p$ in it, so it is hard to find the partial derivatives as required.
Looking at $dU = TdS - pdV$, I can make an expression of $\frac{dV}{dS}$ but this is not a partial derivative.

REPLY [1 votes]: For the first part.
We have that the enthalpy $H = U + pV$, so $$\mathrm dH=\mathrm dU+p\,\mathrm dV+V\,\mathrm dp\quad\Longrightarrow\quad\mathrm d H= T\,\mathrm d S+V\,\mathrm d p$$
Now $H=H(S,p)$ so that
$$\mathrm d H=\left(\frac{\partial H}{\partial S}\right)_p\,\mathrm d S+\left(\frac{\partial H}{\partial p}\right)_S\,\mathrm d p\quad\Longrightarrow\quad \left(\frac{\partial H}{\partial S}\right)_p=T\quad\text{and}\quad  \left(\frac{\partial H}{\partial p}\right)_S=V$$
For the differential form of $H$ to be exact the second derivatives of both
terms in the differential are equal, i.e.
$$\left(\frac{\partial }{\partial S}\left(\frac{\partial H}{\partial p}\right)_S\right)_p=\left(\frac{\partial }{\partial p}\left(\frac{\partial H}{\partial S}\right)_p\right)_S $$
that is
$$\left(\frac{\partial V}{\partial S}\right)_p = \left(\frac{\partial T}{\partial p}\right)_S$$
Now you can prove the second part.


TITLE: Determine an invertible matrix $S \in \operatorname{Mat}_3(\mathbb{R})$ such that $S^{-1}AS$ is a diagonal matrix
QUESTION [1 upvotes]: Consider the matrix
$$
A =
\begin{pmatrix}
1 & 0 & -1 \\
2 & 2 & 2 \\
-1 & 0 & 1
\end{pmatrix}
\in \text{Mat}_3(\mathbb{R})
$$
Then I have to determine an invertible matrix $S \in \operatorname{Mat}_3(\mathbb{R})$ such that $S^{-1}AS$ is a diagonal matrix.
My attempt
If I have understood what I have to do correctly I need to find the eigenvectors for the matrix $A$. In an earlier question, I calculated the eigenvalues which gave me that
$$
t_1 = t_2 = 0 \ \text{and} \ t_3 =2
$$
and afterwards the eigenspaces in order to calculate the geometric multiplicity which gave me the following basis
$$\operatorname{Span}(1,-2,1)^T
$$
and
$$
\operatorname{Span}(-1,1,1)^T
$$
but where will the last basis come from that I need to have in order to construct my matrix $S$? I thought that as $t_1 = t_2 = 0$ they would have the same basis but if I construct
$$
S =
\begin{pmatrix}
1 & 1 & -1 \\
-2 & -2 & 1 \\
1 & 1 & 1
\end{pmatrix}
$$
and afterwards find the inverse of $S$ and calculate $S^{-1}AS$ I don't get the desired result as it should be
$$
\begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 2
\end{pmatrix}
$$
if I am not mistaken. I am not sure where my mistake is. Can you help me in the right direction?

REPLY [1 votes]: I have examined the matrix, it has eigenvalues $2,2, 0$, rather than $0,0,2$.
Let's compute $A-2I=\begin{bmatrix} -1 & 0 & -1 \\ 2 & 0 & 2\\ -1 & 0 & -1\end{bmatrix}$. Rank of $A-2I$ is $1$, hence, there are two independent eigenvectors corresponding to $2$, that is the geometric multiplicty is $2$. Obviously $\begin{bmatrix} 0 \\ 1 \\ 0\end{bmatrix}$ is an eigenvector for eigenvalue 2. Use the eigenvectors to construct $S$.
I will let you take it from here.


TITLE: Maxima $|1+z|+|1-z+z^2|$ when $|z| = 1$
QUESTION [2 upvotes]: I was solving this problem 5 from book namely Complex Analysis A to Z , here we have to find range of
$$ y=|1+z|+|1-z+z^2|$$ with $|z|=1$ .

First of all i didn't get the step
$|1-z+z^2|=\sqrt{|7-2t^2|}$  .
Then i tried to solve this question by substituting $z= \cos\theta+i\sin\theta$
After simplification and plotting the graph i got
$$y=|t|+\sqrt{t+3}$$ where $ t=2\cos\theta-1$ and $$\sqrt3\leq y\leq {13\over4}$$ .
So i actually got complex numbers $z=-{7\over8}\pm \frac{\sqrt15}{8}$  which gives $y={13\over4} > 3\sqrt{{7\over 6}}$ maxima we needed to prove .
So which answer is right did i make a mistake ?

There's a similar question asked so just confirm me if book is wrong and question can be merged with that ?

Also the step i mentioned earlier how author got that ?

REPLY [1 votes]: You are correct. The mistake in the book comes from the fact that one has
\begin{align}
\vert 1- z + z^2 \vert^2 &= 3-2z^* - 2z +z^2 +(z^*)^2 \\
& = (1 - z^* -z)^2,
\end{align}
with $z^*$ denoting complex conjugate of $z$. Hence, using the notation of the book,
$$
\vert 1- z + z^2 \vert = \vert 3 - t^2 \vert \ .
$$
Indeed $f(t) = t + \vert 3 - t^2 \vert$ satisfies
$$
\sqrt{3}\le f(t) \le \frac{13}{4}
$$
when $t\in[0,2]$. I have no idea where the mistake may have originated. It also seems that this way of thought is not the simplest either.


TITLE: Levi's book on Leibnizian calculus
QUESTION [3 upvotes]: Raphael Levi learned from Leibniz at a late stage in Leibniz's career. This might be a definite advantage for understanding Leibniz. Leibniz did not elaborate some of the philosophical principles behind the calculus until quite late in his career. If he conveyed them to Levi the latter might have included some interesting explanations in this text. Some examples:

Does Levi comment on the nature of infinitesimals: fictions? ideal entities?
Does Levi mention the idea that equality is a generalized relation of equality "up to", in line with the Leibnizian "transcendental law of homogeneity" which he did not elaborate explicitly until 1710?
Does Levi say anything about the law of continuity along the lines of Leibniz's fairly late text "Cum prodiisset" (1701)?

Two books on Leibnizian calculus were published by his secretary Rafael Levi, see http://opac.tib.uni-hannover.de/DB=1/LNG=EN/CHARSET=iso-8859-1/CMD?ACT=SRCHA&IKT=1016&SRT=YOP&TRM=Raphael+Levi&Submit=go Does anyone know anything about those books? What's in them?
The main title of the two books seems to be:
Calculus differentialis oder Rechnung des Unendlichen des Herrn von Leibnitz
The books date from 1747 and 1776. Thus they appear to be different from the ones mentioned in Beenakker's answer in the name of Schwarzschild. I haven't been able to get a pdf to see what the introduction says. If anyone has the pdf I would appreciate it.
The 1776 entry: http://opac.tib.uni-hannover.de/DB=1/XMLPRS=N/PPN?PPN=031333311 is in manuscript. Whoever has access please let me know.

REPLY [5 votes]: This is what Steven and Henry Schwarzschild write here about these works of Rafael Levi, which indeed seem to be his own research, based no doubt on what he had learned from Leibniz:

In his scholarly capacity Levi published several books over the
  decades. They are essentially pedestrian in content and reflect his
  two areas of expertise - in German on commercial arithmetic, and in
  Hebrew on Jewish astronomical calendration. By thus putting into
  literary practice his commercial philosophico-mathematical training as
  well as his Jewish learning, Levi served in a double-barreled way the
  Jewish mercantile class and the religious leadership (typically
  combined in personal union).

The mathematical works of Rafael Levi include:

Vorbericht vom Gebrauch der neuerfundenen Logarithmischen
Wechseltafeln + Supplement (Frankfurt-Hannover 1747/1748/1749)
Vorbericht zum Gebrauch der neuerfundenen logarithmische Wechsel-Tabellen, wodurch ein Cours aus 2. 3. oder mehr gegebene Coursen bloß durch addiren und subtrahiren kan gefunden werden (Frankfurt-Leipzig 1749).

This has been scanned and can be accessed from here.

Rechnungsmethode (Hannover 1783, edited posthumously by Meyer Aaron)


TITLE: Sum of the first $n$ terms of polynome
QUESTION [3 upvotes]: There are several related questions here, but none seem to have helped me so far. There is an exercise in my book that goes like this:
Determine the sum of the first $n$ elements of $$x^3 + x^5 + x^7 + ...$$
However the theoretic/general approach in my book is very confusing.
So far I've got $$x_k = x^{2k+1}$$
Could someone explain to mo how I can find the sum of this expression? I know the sum of a geometric series is $$ s_k = a\frac{1 - x^{k+1}}{1-x}$$ but this does not help my very much.
I think if I understand this example, this general approach becomes much more clear.

REPLY [2 votes]: What you want is
$$x^3+x^5+x^7+\cdots+x^{2n+1}.$$
Let it be $S$. Then, consider
$$x^2S=\color{red}{x^5+x^7+\cdots+x^{2n+1}}+x^{2n+3}$$
$$S=x^3+\color{red}{x^5+x^7+\cdots+x^{2n+1}}$$
Subtracting the latter from the former gives you
$$(x^2-1)S=x^{2n+3}-x^3.$$
Hence, if $x^2-1\not= 0$, i.e. $x\not=\pm 1$, then we get$$S=\frac{x^{2n+3}-x^3}{x^2-1}.$$
The cases $x=\pm 1$ are trivial.