TITLE: Heat kernel on the unit circle $\mathbb{R}/\mathbb{Z}$
QUESTION [1 upvotes]: The heat kernel on the circle $\mathbb S^1\cong \mathbb{R}/\mathbb{Z}$ is given by
$$(*) \qquad k_t(\theta)=\frac{1}{\sqrt{4\pi t}}\sum_{n\in \mathbb Z}e^{-\frac{(\theta-n)^2}{4 t}}, \quad \theta\in \mathbb{R}/\mathbb{Z}.$$
In a PDF, I did not understand the meaning of the following remark: $\, (*)$ is the $\mathbb{Z}$ -average of the heat kernel on $\mathbb R$ given by
$$k_t(x)=\frac{1}{\sqrt{4\pi t}} e^{-\frac{x^2}{4 t}}, \quad x\in \mathbb{R}.$$
Also, the technique of unwinding $\mathbb{Z}$-averages?
Can someone explain to me more. Thank you in advance

REPLY [1 votes]: I wouldn't call it $\mathbb{Z}$-average because "average" means dividing by the total weight, and this isn't happening here. This process is also known as "periodization" or "periodic summation" (Wikipedia). Given a function $f$ on $\mathbb{R}$, we can define a periodic function $g$ by
$$
g(x) = \sum_{n\in\mathbb{Z}}f(x+n) \tag1
$$
provided the series converges (which is does when $f$ has sufficiently fast decay at infinity).
"Unwinding $\mathbb{Z}$-average" would mean the opposite: given a periodic function $g$, find $f$ such that (1) holds.
For example, if $g(x) = \sin 2\pi x$, then one can take
$$
f(x) = \begin{cases} \sin 2\pi x,\quad &0\le x\le 1, \\ 0 \quad &\text{otherwise}\end{cases}
$$
This particular example doesn't seem important. But when $f$ solves some translation-invariant PDE on $\mathbb{R}$ (such as the heat equation), we get a solution of the same PDE on the circle $\mathbb{R}/\mathbb{Z}$ via (1), and that is nice to have.
Closely related: Poisson summation formula, which says that the Fourier series of $g$ can be obtained from the Fourier transform of $f$.


TITLE: Reference request: book on stochastic calculus (not finance)
QUESTION [0 upvotes]: I am looking at fractional Gaussian/Brownian noise from a signal theoretic and engineering point of view. In particular, I am looking at the math behind what defines these noise processes and what consequences this has on the physics, either generating them or consuming these noise signals.
As an engineer by training I am familiar with both (real/multivariate/complex) calculus and basic probability theory and  also stochastic signals. But most of what I am doing now is where fractional calculus and stochastic calculus meet (Hic sunt dracones... literally). I think I can get my way around most of the fractional calculus part, but for the stochastic calculus I am in need of better understanding of how it works.
What I am looking for is a book (or lecture notes) that not only give me an understanding and intuition how stochastic calculus works (ie. how to apply it), but I also need the proofs in order to tell what I am allowed to do with the theorems and what not. Measure theory shouldn't be much of a problem, as I have two mathematicians at hand who can explain things, if I get stuck.

REPLY [0 votes]: In my work I usually deal with Brownian motion and white noise so I am not the expert on the topic of fractional Brownian motion (fbm). But I recalled that I have seen an engineering application in the David Nualart paper on fbm (Section 4 - fbm in turbulence). Sorry, it is not on signal processing and it has a finance part (Section 5) but it is a good short intro on the topic of fbm. Also interesting paper for you could be this (fbm in a nutshell). You could check out the references in both of this papers additionaly.
But, if you type "fractional brownian motion" into the Science direct you get a lot of books that mention signal processing in engineering (fbm in engineering). And this would probably be your best shot in finding an answers to your questions.
Hope this helps you somehow.


TITLE: Statement about divisors of polynomials and their roots
QUESTION [11 upvotes]: I am a 10th grade student and there is a statement in my math book

If $a$ is a root of the polynomial $f(x)$ then $(x-a)$ is a divisor of $f(x)$

Why is $(x-a)$ a divisor of $f(x)$? Can you please tell me?

REPLY [0 votes]: Lemma $:$ For a field $F, \alpha \in F$ is a root of $a(x)$ if and only if $x-\alpha$ divides $a(x)$
Proof. $(\Longrightarrow)$ Assume that $\alpha$ is a root, i.e., $a(\alpha)=0 .$ Then we can write $a(x)$ as
$$
a(x)=(x-\alpha) q(x)+r(x)
$$
where $\operatorname{deg}(r(x))<\operatorname{deg}(x-\alpha)=1,$ i.e., $r(x)$ is a constant $r,$ where
$$
r=a(x)-(x-\alpha) q(x)
$$
Setting $x=\alpha$ in the above equation gives
$$
r=a(\alpha)-(\alpha-\alpha) q(\alpha)=0-0 \cdot q(\alpha)=0
$$
Hence $x-\alpha$ divides $a(x)$
($\Longleftarrow$) To prove the other direction, assume that $x-\alpha$ divides $a(x),$ i.e., $a(x)=$ $(x-\alpha) q(x)$ for some $q(x) .$ Then $a(\alpha)=(\alpha-\alpha) q(\alpha)=0,$ i.e., $\alpha$ is a root of $a(x)$


TITLE: Poincare-Miranda on a torus
QUESTION [3 upvotes]: The Poincare-Miranda theorem involves $d$ functions on a $d$-dimensional cube. Function $i$ is negative in one direction of dimension $i$ and positive in the opposite direction of dimension $i$. The theorem says that there must be a point in the cube in which all functions are 0.
Is there a parallel theorem on a $d$-dimensional torus, where the boundary conditions are replaced by oddity?
To be concrete, let $d=2$. A 2-dimensional torus is a cartesian product of 2 circles: $S^1 \times S^1$.
There are 2 continuous functions: $f_1(x_1,x_2), f_2(x_1,x_2)$, where $x_i$ is a point on the circle in dimension $i$. Function $i$ is odd in dimension $i$, i.e: $f_1(-x_1,x_2)=-f_1(x_1,x_2)$, and similarly $f_2(x_1,-x_2)=-f_2(x_1,x_2)$.
(Alternatively, the two functions can be defined as: $f_1(\theta_1,\theta_2), f_2(\theta_1,\theta_2)$, where $\theta_i$ is an angle on the circle in dimension $i$. Function $i$ is odd in dimension $i$, i.e: $f_1(\theta_1+\pi,\theta_2)=-f_1(\theta_1,\theta_2)$ and $f_2(\theta_1,\theta_2+\pi)=-f_2(\theta_1,\theta_2)$).
Is this true that there must be a point in the torus in which all functions are 0?
The intuition for this variant is the same as for the original variant: for each function $i$, there must be a "wall" of 0 which separates the positive side of $i$ from its negative side. Each wall is in a different dimension, so they must meet in a point. But is it true?

REPLY [1 votes]: $\newcommand{\Reals}{\mathbf{R}}$View the unit circle as the set of unit complex numbers, and define a map $f:S^{1} \times S^{1} \to \Reals^{2}$ by
$$
f(e^{i\theta_{1}}, e^{i\theta_{2}})
  = \bigl(\underbrace{\cos(\theta_{1} + \theta_{2})}_{f_{1}(\theta_{1}, \theta_{2})},
          \underbrace{\sin(\theta_{1} + \theta_{2})}_{f_{2}(\theta_{1}, \theta_{2})}\bigr).
$$
For all $\theta_{1}$ and $\theta_{2}$,
\begin{align*}
f_{1}(\theta_{1} + \pi, \theta_{2})
  &= \cos(\theta_{1} + \theta_{2} + \pi)
  = -\cos(\theta_{1} + \theta_{2}), \\
f_{2}(\theta_{1}, \theta_{2} + \pi)
  &= \sin(\theta_{1} + \theta_{2} + \pi)
  = -\sin(\theta_{1} + \theta_{2}),
\end{align*}
but $\|f(\theta_{1}, \theta_{2})\| = 1$, so the origin is not in the image of $f$; that is, $f_{1}$ and $f_{2}$ do not vanish simultaneously.
To extend this idea to an $n$-dimensional torus, define
$$
f_{1}(\theta_{1}, \dots, \theta_{n}) = \cos(\theta_{1} + \theta_{2}),\quad
f_{2}(\theta_{1}, \dots, \theta_{n}) = \sin(\theta_{1} + \theta_{2}),\quad
f_{i}(\theta_{1}, \dots, \theta_{n}) = 0,\ i > 2.
$$
The topological point seems to be that a cube is contractible, and "oddness" (even just the sign conditions on the boundary in the Poincaré-Miranda theorem) is enough to imply the image of the boundary of the cube encloses the origin, so by continuity the origin is in the image of $f$. By contrast, an $n$-torus has lots of $1$-dimensional homology, so it's "easy" to wind the image around the origin without hitting the origin, even if the mapping is odd. (At least, that's what naturally suggested the example above.)


TITLE: Why do we associate a graph to a ring?
QUESTION [21 upvotes]: I don't know if it is suitable for MathOverflow, if not please direct it to suitable sites.
I don't understand the following:
I find that there are many ways a graph is associated with an algebraic structure, namely Zero divisor graph (Anderson and Livingston - The zero-divisor graph of a commutative ring), Non-Commuting Graph (Abdollahi, Akbari, and Maimani - Non-commuting graph of a group) and many others.
All these papers receive hundreds of citations which means many people work in this field.
I read the papers, it basically tries to find the properties of the associated graph from the algebraic structure namely when it is connected, complete, planar, girth etc.
My questions are:

We already have a list of many unsolved problems in Abstract Algebra and Graph Theory, why do we mix the two topics in order to get more problems?

It is evident that if we just associate a graph with an algebraic structure then it is going to give us new problems like finding the structure of the graph because we just have a new graph.
Are we able to solve any existing problems in group theory or ring theory by associating a suitable graph structure? Unfortunately I could not find that in any of the papers.

Can someone show me by giving an example of a problem in group theory or ring theory which can be solved by associating a suitable graph structure?

For example suppose I take the ring $(\mathbb Z_n,+,.)$, i.e. the ring of integers modulo $n$. What unsolved problems about $\mathbb Z_n$ can we solve by associating the zero divisor graph to it?
NOTE: I got some answers/comments where people said that we study those graphs because we are curious and find them interesting.
I am not sure if this is how mathematics works.
Every subject developed because it had certain motivation.
So I don't think this reason that "Mathematicians are curious about it, so they study it" stands.
As a matter of fact, I am looking for that reason why people study this field of Algebraic Graph Theory.

REPLY [2 votes]: The Fischer graph is one of examples; see page 569 of
Suzuki, Michio. Group theory. II. Translated from the Japanese. Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], 248. Springer-Verlag, New York, 1986.
An special important case of the Fischer graph is the induced subgraph of the commuting graph of a group on a certain conjugacy class of involution so-called 3-traspositions.
The title of Section 10 of Chapter 6 of the above book of Suzuki is ``Graphs and Simple groups".


TITLE: Gauss integral asymptotics
QUESTION [0 upvotes]: I need to know the asymptotic expansion of the Gauss integral
$$\int_0^x e^{ -t^{2}} \mathrm{d}t$$
Is it known for any x as the limit? I am basically substituting the asymptotic expansion of this into another integral.

REPLY [0 votes]: As $e^{-t^2}$ is extremely quickly decaying for large $|t|$, the integral in $(-\infty,t]$ is virtually equal to the function itself.
$$t<0\to\int_{-\infty}^te^{-t^2}dt\approx e^{-t^2}.$$

And by symmetry,
$$t>0\to\int_t^{\infty}e^{-t^2}dx\approx e^{-t^2}$$
so that
$$t>0\to\int_0^te^{-t^2}dx\approx\int_0^{\infty}e^{-t^2}dx-e^{-t^2}=\frac{\sqrt\pi}2-e^{-t^2}.$$


TITLE: Why is $k[x,y,z,u]/(x^2-y^2z, uy-x) \cong k[y,u]$?
QUESTION [0 upvotes]: In the second paragraph of the accepted answer to this question, the answerer mentions that the normalization of the Whitney umbrella $k[x,y,z]/(x^2-y^2z)$ is $k[y,u]$, where $u = x/y$. I am trying to prove this claim, which I believe is equivalent to showing the isomorphism in the title, i.e. $k[x,y,z,u]/(x^2-y^2z, uy-x) \cong k[y,u]$.
From the second relation we get that the left-hand ring is isomorphic to $k[y,z,u]/(y^2u^2 - y^2z)$, but I'm not sure why this is isomorphic to $k[y,u]$. Factoring out the $y^2$, it seems to me that the relation is saying that $y^2(u^2 - z) = 0$, but I don't think I can "cancel out" the $y^2$ to conclude that $z = u^2$.
Is this misgiving correct? If so, it seems that I would need additional conditions on my ring (such as another relation) -- what might these be? If not, why is it allowed?

REPLY [3 votes]: Since $R = k[x,y,z] / (x^2 - y^2z)$ is an integral domain, it's safe to talk about fractions. The element $x/y$ is integral over $R$ because $(x/y)^2 = z$, so the integral closure includes $x/y$.
Your argument shows that this is not the same as the ring $k[x,y,z,u] / (x^2 - y^2z, uy-x)$ since the latter is not a domain. Whereas adjoining elements of the fraction field to $R$ obviously leaves it a domain.
The missing relation from that presentation is just $u^2 = z$ though, and the relations $uy=x, u^2=z$ imply $x^2 = y^2z$. You can check that $k[x,y,z,u] / (uy-x, u^2-z) \cong k[u,y] \cong R[\frac{x}{y}] \subset \mathrm{Frac}(R)$.

REPLY [2 votes]: If you attach $u=x/y$ in the fraction field of $R=k[x,y,z]/x^2-y^2z$ to $R$, you get $R[u]=S$. Since $u=x/y$, you can write $x=uy$ and then since $R$ is a domain and so is $S$ (both have the same fraction field) you can write $S=k[y.z,u]$ with the relation $u^2-z$. Thus $S=k[u,y]$.


TITLE: Mersenne number divisible by twin primes
QUESTION [2 upvotes]: I am trying to determine the smallest odd n > 1 where the Mersenne number $2^n - 1$ is divisible by twin primes p and q with $p<q$.
If n were even, by inspection, $M_4$ is divisible by 3 and 5 but the odd case is harder.
I know that p being the first twin prime implies $p \equiv_6 -1$
The textbook hints that 2 is a quadratic residue modulo p and modulo q but I don't see how this follows. I know this is the case only if p and q are congruent to +-1 modulo 8 or alternatively if $2^{p-1\over{2}}$ is congruent to 1 modulo p. I can't get there.
Can anyone give me a hint in this direction.

REPLY [2 votes]: HINT
It is known that for any natural number $n$ and for any prime $p$, $ord_p(n)$ divides $p-1$, as seen here.
Using this, it is safe to say $ord_p(2)|p-1$ and $ord_q(2)|q-1$.
In your problem, for such odd $n$ to exist,  $lcm(ord_p(2),ord_q(2))$ must divide $n$. This implies that $lcm(ord_p(2),ord_q(2))$ must be odd, or that both $ord_p(2)$ and $ord_q(2)$ are odd.
I suggest you continue from here.


TITLE: Optimization for constrained problem
QUESTION [3 upvotes]: I'm reading about Lagrange multipliers from a Pattern recognition book appendix and on one point the following is stated:

$\begin{align}
&\color{gray}{\verb+C.4+}\,\,\,\,\,\color{#08F}{\textbf{Optimization for Constrained Problems}}\\
&\color{gray}{\verb+C.4.1+}\,\,\,\,\,\text{Equality Constraints}
\end{align}$
We will first focus on linear equality constraints and then generalize to the nonlinear case. Although the philosophy for both cases is the same, it is easier to grasp the basics when linear constraints are involved . Thus the problem is cast as
$$\text{minimize}\,\,\,\, J(\theta)\\\text{subject to}\,\,\,\,A\theta=b$$
where $A$ is an $m\times l$ matrix and $b$, $\theta$ are $m\times1$ and $l\times 1$ vectors, respectively. It is assumed that the cost function $J(\theta)$ is twice continuously differentiable and it is, in general, a nonlinear function. Furthermore, we assume that the rows of $a$ are linearly independent, hence $A$ has full row rank. This assumption is known as the regularity assumption.
Let $\theta_*$ be a local minimizer of $J(\theta)$ over the set $\{\theta:\,A\theta=b\}$. Then it is not difficult to show (e.g.,[Nash 96]) that, at this points, the gradient of $J(\theta)$ is given by
$$\color{red}{\boxed{\color{black}{\displaystyle \frac{\partial}{\partial\theta}(J(\theta))|_{\theta=\theta_*}=A^T\lambda}}} \tag{C.24}$$
where $\displaystyle \lambda\equiv[\lambda_1,\cdots,\lambda_m]^T$. Taking into account that
$$\color{red}{\boxed{\color{black}{\displaystyle \frac{\partial}{\partial\theta}(A\theta)=A^T}}} \tag{C.25}$$
Eq. $\text{(C.24)}$ states that, at a constrained minimum , the gradient of the cost function is a linear combination of the gradients of the constraints. This is quite natural.

Could perhaps someone show me easy-to-read proof of why the points in the red rectangles are true, since it's not difficult to show as the text suggests ;)
Thank you for any help :)

REPLY [1 votes]: The second point can be shown as
$$
\frac{\partial}{\partial \theta}_i (A\theta)_j = \frac{\partial}{\partial \theta}_i (\sum_kA_{jk}\theta_k) = (A)_{ji}
$$
Thus,
$$
\frac{\partial}{\partial \theta} (A\theta) = A^T
$$
Note that we are taking the gradient, not the derivative $D$ (which is the gradient transposed, technically).
The first part can be shown my setting the gradient of the relaxed, unconstrained objective (the Lagrangian)
$$J(\theta) + \lambda^T(b - A\theta)$$
equal to zero.

REPLY [1 votes]: I believe the author begins with the Lagrange multiplier formulation
$$J(\theta)-\lambda(A\theta-b)$$
of the constrainted optimization problem. Then he computes the gradient w.r.t the components $\theta_i$ of $\theta$ at the local minimizer, arriving at the expression (in compact form)
$$\nabla J(\theta)-\lambda\nabla(A\theta)=\nabla J(\theta)-\lambda A^t=0,$$
where we used the second relation in the red box.
Both relations in the boxes are checked using components, and writing the scalar product
$$\lambda A\theta :=\langle \lambda, A\theta\rangle$$
explicitly.


text:
\begin{document}

\begin{abstract}
We use Ehrenfeucht-Fra\"iss\'e games to give a local geometric criterion for elementary equivalence of II$_1$ factors.  We obtain as a corollary that two II$_1$ factors are elementarily equivalent if and only their unitary groups are elementarily equivalent as $\mathbb Z_4$-metric spaces.
 \end{abstract}

\maketitle

\section*{Introduction}

While most mathematicians are concerned in determining when two objects in their field are isomorphic, logicians tend to be concerned with the coarser notion of \emph{elementary equivalence}.  Two (classical) structures $M$ and $N$ are said to be elementarily equivalent if and only if, for any first-order sentence $\sigma$ (in the language appropriate to the study of $M$ and $N$), we have $\sigma$ is true in $M$ if and only if $\sigma$ is true in $N$.  For structures appearing in analysis, a continuous logic is used in which sentences can now take a continuum of ``truth'' values; the appropriate notion of elementary equivalence is that the truth values of all sentences are the same in both structures.

The model-theoretic study of tracial von Neumann algebras began in earnest in \cite{mtoa1}, \cite{mtoa2}, and \cite{mtoa3}.  At the moment, there are only three distinct elementary equivalence classes of II$_1$ factors known.  (This should not be so surprising as it took a while for many isomorphism classes of II$_1$ factors to be discovered and elementary equivalence is a much coarser notion.)  Indeed, it was observed in \cite{mtoa3} that Property ($\Gamma$) and the property of being McDuff are both elementary properties (for separable II$_1$ factors).  Thus, if we let $M_{\DL}$ be a separable II$_1$ factor that has Property ($\Gamma$) but is not McDuff (see \cite{DL}), then $M_{\DL}$, the hyperfinite II$_1$ factor $\R$ and the free group factor $L(\mathbb F_2)$ are mutually non-elementarily equivalent.  Amongst those studying II$_1$ factors from a model-theoretic point of view, it is widely agreed that there should be more than three elementary equivalence classes of II$_1$ factors; in fact, there should probably be continuum many elementary equivalence classes.  At the moment, we cannot even answer the question:  is $\R\otimes L(\mathbb F_2)$ elementarily equivalent to $\R$?  In order to accomplish these goals, we need more tools for understanding elementary equivalence of II$_1$ factors.

Ehrenfeucht-Fra\"iss\'e games have long been a tool in model theory for establishing that structures are elementarily equivalent.  In \cite{heinrich-henson}, the authors exhibit an Ehrenfeucht-Fra\"iss\'e-type game used to establish elementary equivalence for Banach spaces.  In this note, we adapt the game from \cite{heinrich-henson} and combine it with an argument of Kirchberg appearing in \cite{kirch} in order to characterize elementary equivalence for II$_1$ factors belonging to the class $\kop$ (to be defined below).  We should note that, currently, we do not know of a II$_1$ factor that does not belong to the class $\kop$ and the existence of such a factor would already lead to two new theories of II$_1$ factors!

Recall Dye's Theorem \cite{dye}, which states that any two factors not of type $\rm I_{2^n}$ (e.g., any two $\rm II_1$-factors) are isomorphic if and only if their unitary groups are isomorphic (even as discrete groups).  Combining Dye's Theorem with the Keisler-Shelah Theorem (which states that two structures are elementarily equivalent if and only if they have isomorphic ultrapowers) and the fact that the functors of taking ultrapowers and taking unitary groups commute, we see that two II$_1$ factors are elementarily equivalent if and only if their unitary groups are elementarily equivalent as metric groups (with respect to the $\ell_2$ metric).  Using the aforementioned Ehrenfeucht-Fra\"iss\'e games and some further arguments, our main result is that we can improve upon the previous sentence, essentially removing the group structure:

\begin{thm}
Suppose that $M$ and $N$ are II$_1$ factors belonging to the class $\kop$.  Then $M$ and $N$ are elementarily equivalent if and only if $U(M)$ and $U(N)$ are elementarily equivalent as $\mathbb Z_4$-metric \emph{spaces}.
\end{thm}

Here, by a $\mathbb Z_4$-metric space, we mean a metric space $X$ equipped with an action of $\mathbb Z_4$ on $X$ by isometries.  Unitary groups of von Neumann algebras will always be considered as $\mathbb Z_4$-metric spaces by having the generator of $\mathbb Z_4$ act by multiplication by $i$.

In this paper, we assume that the reader is familiar with some basic model theory and von Neumann algebra theory.  Good references for continuous model theory are \cite{bbhu} and \cite{mtoa2}; the latter is geared towards the model theoretic study of operator algebras.

All normed spaces are assumed to be over the complex numbers, $\bb C$. For a normed space $X$ we denote the closed unit ball $(X)_1 := \{x\in X : \nor{x}\leq 1\}$.

For the convenience of the reader, we now recall the original notion of Ehrenfeucht-Fra\"iss\'e games in the context of continuous logic.  This has not appeared in the literature but has appeared in some online lecture notes of Bradd Hart \cite{bradd}. Fix an arbitrary language $\cal L$ and atomic formulae $\varphi_1(\vec x),\ldots,\varphi_k(\vec x)$ in the variables $\vec x=(x_1,\ldots,x_n)$ and $\e>0$.  The Ehrenfeucht-Fraisse game $\fr G(\varphi_1,\ldots,\varphi_k,\e)$ is played with $\cal L$-structures $M$ and $N$ as follows:  First Player I chooses $a_1\in M$ or $b_1\in N$ respecting the sort of $x_1$.  Player II choose $b_2\in N$ or $a_2\in M$ respectively.  The players alternate in this manner until they have produced sequences $a_1,\ldots,a_n\in M$ and $b_1,\ldots,b_n\in N$.  Player II then wins the game if and only if, for each $i=1,\ldots,k$, we have $|\varphi_i(\vec a)^M-\varphi_i(\vec b)^N|\leq \e$.  It is then a theorem that $M\equiv N$ if and only Player II has a winning strategy in each $\fr G(\varphi_1,\ldots,\varphi_k,\e)$.

\section{The class $\kop$}

Given a C$^*$ algebra $A$, recall that its opposite algebra $A^{op}$ is the algebra obtained from $A$ by multiplying elements in the opposite order, that is, for $a,b\in A$, we have $a\cdot_{op}b:=b\cdot a$.  It is immediate that $A^{op}$ is once again a C$^*$ algebra.  Furthermore, if $A$ is a von Neumann algebra, then $A^{op}$ is also a von Neumann algebra.  Note also that if $(A_i \ : \ i\in I)$ is a family of C$^*$ algebras (resp. tracial von Neumann algebras) and $\uu$ is an ultrafilter on $I$, then $(\prod_\uu A_i)^{op}\cong \prod_\uu A_i^{op}$ via the identity map, where the ultraproduct is understood to be the usual C$^*$ algebra ultraproduct (resp. tracial ultraproduct).

Many of the naturally occurring tracial von Neumann algebras are isomorphic to their opposites, e.g. $\R$ and $L(G)$ ($G$ any group).  There are examples of tracial von Neumann algebras that are not isomorphic to their opposites (see \cite{Connes}).  During a seminar talk given by the first author at Vanderbilt University, Jesse Peterson asked whether or not the class of all tracial von Neumann algebras isomorphic to their opposites is an axiomatizable class.  While we do not know the answer to this question (although we suspect the answer is negative), the answer is positive if one replaces the word ``isomorphism'' by ``elementary equivalence'' as we show in the following:

\begin{prop}\label{kop}
The class of all tracial von Neumann algebras that are elementarily equivalent to their opposites is an elementary class.
\end{prop}

\begin{defn}
We let $\kop$ denote the class of all tracial von Neumann algebras elementarily equivalent to their opposites.
\end{defn}

\begin{proof}[Proof of Proposition \ref{kop}] We present a proof suggested to us by Todor Tsankov as well as independently by the anonymous referee. There are a collection of axioms for the class $\kop$:  for every term $t$, recursively define the term $t^{op}$ by   defining $(t_1\cdot t_2)^{op}:=t_2^{op}\cdot t_1^{op}$.  Then one can recursively define, for any formula $\varphi$, the formula $\varphi^{op}$, the key clause being the atomic formulae, where one replaces every occurrence of a term $t$ by the term $t^{op}$.  Then the conditions $|\sigma-\sigma^{op}|=0$, as $\sigma$ ranges over all sentences, axiomatizes the class $\kop$.
\end{proof}

We remark in passing that alternately by \cite[Proposition 5.14]{bbhu}, it suffices to show that $\kop$ is closed under isomorphisms, ultraproducts, and ultraroots. We leave it as an exercise to the reader to verify these properties for $\kop$.

Since $\R$ and $L(\mathbb F_2)$ are isomorphic to their opposites, they belong to $\kop$.  Moreover, the example $M_{\DL}$ of a II$_1$ factor with Property ($\Gamma$) that is not McDuff given by Lance and Dixmier  in \cite{DL} is also isomorphic to its opposite.  Thus, we have:

\begin{cor}
If there is a II$_1$ factor that does not belong to $\kop$, then there are at least five theories of II$_1$ factors.
\end{cor}

\begin{proof}
If $N$ is a II$_1$ factor that does not belong to $\kop$, then the theories of $N$ and $N^{op}$ differ from each other and from the three known theories of II$_1$ factors.

\end{proof}

\begin{question}
Are there more ``explicit'' axioms for the class $\kop$?  Can one use typical model-theoretic preservation theorems to show that $\kop$ is universally axiomatizable or $\forall\exists$-axiomatizable?
\end{question}

\begin{question}
Is there a single sentence $\sigma$ such that, adding the condition ``$\sigma=0$'' to the axioms for II$_1$ factors gives an axiomatization of $\kop$?
\end{question}

A negative answer to the last question implies that there must be infinitely many elementary equivalence classes of II$_1$ factors not belonging $\kop$.  Indeed, if there are only finitely many elementary equivalence classes of II$_1$ factors not belonging to $\kop$, then the class of II$_1$ factors not belonging to $\kop$ is readily verified to be elementary as well, whence a typical compactness argument is used to show that the last question has a positive answer.

\section{Model theory of Banach pairs} In order to frame the main results of the paper in the next section on the model theory of $\rm II_1$-factors, we introduce a class of linear (unbounded) metric structures (``Banach pairs'') for which $\rm II_1$-factors will be the primary set of examples. The important fact which we will see is that the theory of a $\rm II_1$-factor regarded as a Banach pair will determine its theory as $\rm II_1$-factor. For this reason we feel it is justified to introduce this treatment, despite several existing approaches in the literature for dealing with linear metric structures, e.g., \cite{ben, bbhu, henson-moore}, with at least one treatment \cite{mtoa2} being devoted to C$^*$-algebras and tracial von Neumann algebras.

\begin{defn} A \emph{Banach pair} $(X,\cal C)$ consist of a normed space $X$ and a distinguished subset $\cal C\subset (X)_1$ which is:
\begin{itemize}
\item complete;
\item roundly convex, i.e., $\lambda x  + \mu y\in \cal C$ for all $x,y\in \cal C$ and $\lambda,\mu\in \bb C$ with $\abs{\lambda} + \abs{\mu}\leq 1$; and
\item generating, i.e., $\bigcup_n n\cdot \cal C = X$.
\end{itemize}
\end{defn}

\noindent The main examples of Banach pairs we will be interested in are where $X = M$, a tracial von Neumann algebra equipped with the $2$-norm $\nor{x}_2 := \tr(x^*x)^{1/2}$, and $\cal C= (M)_1$, the (norm) closed unit ball.

A Banach pair $(X,\cal C)$ can be intepreted as an structure for the following language $\cal L_{\BP}$:

\begin{itemize}
\item There is one sort each for $\bb C$ and $X$.
\item There is a sequence of domains of quantification $\cal C_n$ for $X$.
\item There are function symbols $\imath_{m,n}: \cal C_m\to \cal C_n$ for $m\leq n$ to be interpreted as the usual inclusion maps.
\item $X$ is given the usual complex normed space axioms.
\item Axioms which show $0_X\in \cal C_1\subset (X)_1$.
\item Axioms to show each $\cal C_n$ is roundly convex.
\end{itemize}

For a Banach pair $(X,\cal C)$, for $x\in X$ we define $\nor{x}_{\cal C} := \inf\{t>0 : x\in t\cdot\cal C\}$ which can be checked to be a Banach norm on $X$. However, note that $\nor{\,\cdot\,}_{\cal C}$ is a definable predicate if and only if it is uniformly continuous with respect to the usual norm. (In the case that $X$ is a tracial von Neumann algebra this will be the case if and only if $X$ is finite-dimensional.)

As an $\cal L_{\BP}$-structure the ultrapower $(X,\cal C)^\uu$ can be identified with the Banach pair $(X^\uu, \cal C^\uu)$ where $X^\uu$ is the quotient space of $\{(x_i) : \lim_\uu \nor{x_i}_{\cal C}<\infty\}$ modulo the subspace $\{(z_i) : \lim_\uu\nor{z_i}_{\cal C}<\infty,\ \lim_\uu\nor{z_i}=0\}$ and $\cal C^\uu\subset X^\uu$ is defined in the obvious way.

We say that two Banach pairs $(X,\cal C)$ and $(Y,\cal D)$ are isomorphic (written $(X,\cal C)\cong (X,\cal D)$) if they are isomorphic as $\cal L_{\BP}$-structures, that is, if there is an isometry $T: X\to Y$ so that $T(\cal C) = \cal D$. By definition, the aforementioned Banach pairs are elementarily equivalent (written $(X,\cal C)\equiv (Y,\cal D))$ if $\Th(X,\cal C) = \Th(Y, \cal D)$. As a consequence of the Keisler--Shelah theorem in continuous logic, we have that $(X,\cal C)\equiv (Y,\cal D)$ if and only if there is an ultrafilter so that $(X,\cal C)^\uu \cong (Y,\cal D)^\uu$. See \S10 in \cite{henson-iovino} for a proof of this fact in the context of normed spaces or \S3 in \cite{heinrich-henson} for a more explicit construction for Banach spaces.

Our main observation in this section is that for Banach pairs $(X,\cal C)$ and $(Y,\cal D)$ elementary equivalence can be characterized in terms of the pairs ``having the same local geometric structure'' by the use of Ehrenfeucht--Fra\"iss\'e games. For the very similar case of Banach spaces, this was done by Heinrich and Henson \cite[Theorem 4]{heinrich-henson} and the case of normed spaces is largely similar (see Remark 10.10 in \cite{henson-iovino}).

We now describe precisely what we mean when we say that two Banach pairs $(X,\cal C)$ and $(Y,\cal D)$ have the same local geometric structure. For $E$ a subspace of $X$ and $F$ a subspace of $Y$ we say that a linear bijection $T: E\to F$ is an \emph{$\e$-almost isometry} if $\nor{T},\nor{T^{-1}}\leq 1+\e$ and $T(E\cap \cal C) \subset_\e F\cap \cal D$ and $T^{-1}(F\cap \cal D)\subset_\e E\cap \cal C$. (We write $A\subset_\e B$ if $\sup_{x\in A}\inf_{y\in B}\nor{x - y}\leq \e$.)

The following is adapted from \S2 of \cite{heinrich-henson}: see also \S8 of \cite{henson-moore}. We describe a game $\fr G(n,\e)$ played by two players with Banach pairs $(X,\cal C)$ and $(Y,\cal D)$, where $\e>0$ and $n$ are fixed parameters.

\noindent {\bf Step 1.} Player I chooses a one-dimensional subspace, either $E_1\subset X$ or $F_1\subset Y$. Player II then chooses a subspace, respectively $F_1\subset Y$ or $E_1\subset X$ and a linear bijection $T_1: E_1\to F_1$.

\noindent{\bf Step i.} Player I chooses an at most one-dimensional extension, either $E_i \supset E_{i-1}$ or $F_i\supset F_{i-1}$. Player II then chooses a subspace, respectively $F_i\subset Y$ or $E_i\subset X$ and a linear bijection $T_i: E_i\to F_i$ which extends $T_{i-1}$.

\noindent{\bf Step n.} The players make their choices, and the game terminates. Player II wins if $T_n: E_n\to F_n$ is an $\e$-almost isometry; otherwise, Player I wins.

During the course of proofs, we may speak of Player I playing $x_i\in X$, in which case we mean that Player I plays $\sp(E_{i-1}\cup \{x_i\})$.  We may then also say that Player II responds with $y_i\in Y$, in which case we mean that Player II plays the linear bijection $T_i$ extending $T_{i-1}$ that sends $x_i$ to $y_i$.

\begin{defn} We say that Banach pairs $(X,\cal C)$ and $(Y,\cal D)$ are \emph{locally equivalent} (written $(X,\cal C)\cong_{\loc} (Y,\cal D)$) if for every $\e>0$ and every $n$, Player II has a winning strategy for the game $\fr G(n,\e)$.
\end{defn}

\begin{rem}\label{alternate} Since $\e$ is arbitrary, and we need only deal with at most one-dimensional extensions, we see that local isomorphism remains the same under an alternate version of $\e$-almost isometry, namely, the existence of linear bijections $T: E\to F$, $S:F\to E$ with strict containment $T(E\cap \cal C)\subseteq F\cap \cal D$ and $S(F\cap \cal D)\subseteq E\cap \cal C$ so that $\nor{ST - \id_E}$, $\nor{TS - \id_F}<\e$ and $\|T\|,\|S\|<1+\e$.
\end{rem}

\begin{prop}\label{EFBP} The following statements are equivalent:
\begin{enumerate}
\item $(X,\cal C)\equiv (Y,\cal D)$;
\item there exists an ultrafilter so that $(X,\cal C)^\uu\cong (Y,\cal D)^\uu$ as Banach pairs;
\item $(X,\cal C)\cong_{\loc} (Y,\cal D)$.
\end{enumerate}
\end{prop}

\noindent As noted above (1) $\Leftrightarrow$ (2) is the Keisler--Shelah theorem applied to the language of Banach pairs.  The proof of (2) $\Rightarrow$ (3) is straightforward using representing sequences. Therefore we only need to prove (3) $\Rightarrow$ (1).  The proof is more or less identical to the Banach space version as in \cite{heinrich-henson}.  However, since we are working in a different logic, we sketch a (nearly complete) proof here for the convenience of the reader.

\begin{proof}[Sketch of (3) $\Rightarrow$ (1)]
First, we work with the notion of $\epsilon$-almost isometry as described in Remark \ref{alternate}.  Let $\sigma$ be a sentence of the form $\inf_{v_1}\sup_{v_2}\cdots Q_{v_n} \rho(v_1,\ldots,v_n)$, where $Q$ is $\inf$ if $n$ is odd and $\sup$ if $n$ is even and where $\rho$ is quantifier-free.  (We suppress mention of the sorts $\cal C_i$ corresponding to each $v_i$.)  Fix $\epsilon>0$.  It suffices to show that $\sigma^{(Y,\cal D)}\leq \sigma^{(X,\cal C)}+\epsilon$ for all $\epsilon>0$.  Indeed, by symmetry of the relation of local equivalence, this shows that all sentences of the above form have the same truth values in $(X,\cal C)$ and $(Y,\cal D)$.  Since any sentence in prenex normal form is equivalent to one of the above form (by adding dummy variables) and since the set of  sentences in prenex normal form is dense in the set of all sentences (see \cite[\S6]{bbhu}), we obtain that $(X,\cal C)\equiv (Y,\cal D)$.

Fix sufficiently small $\delta>0$.  (We will see exactly how small $\delta$ needs to be in a moment.)  Fix a winning strategy $\cal S$ for Player II in $\fr G(n,\delta)$.  Call a play of the game $\fr G(n,\delta)$ \emph{regular} if:
\begin{itemize}
\item For odd $i$, Player I plays $x_i\in X$, while for even $i$, Player I plays $y_i\in Y$;
\item For each $i$, Player I's move at Round $i$ is always in the sort corresponding to the variable $v_i$;
\item Player II always plays according to $\cal S$.
\end{itemize}
We say that sequences $x_1,\ldots,x_k\in X$ and $y_1,\ldots,y_k\in Y$ \emph{correspond} if they are the results of the first $k$ rounds of a regular play of $\fr G(n,\delta)$.

For $0\leq l \leq n$, let $\sigma_l(v_1,\ldots,v_{n-l})$ denote the formula obtained from $\sigma$ by removing the first $n-l$ quantifiers.  One now proves, by induction on $l$ ($0\leq l\leq n$), that if $x_1,\ldots,x_{n-l}\in X$ and $y_1,\ldots,y_{n-l}\in Y$ correspond, then
$$\sigma_l(y_1,\ldots,y_{n-1})^{(Y,\cal D)}\leq \sigma_l(x_1,\ldots,x_{n-1})^{(X,\cal C)}+\e.$$  The base case $l=0$ follows from the fact that $T_n:\sp(x_1,\ldots,x_n)\to \sp(y_1,\ldots,y_n)$ is a $\delta$-almost isometry if $\delta$ is chosen sufficiently small.  We now prove the induction step.  Suppose that the claim holds for $l$ and that $x_1,\ldots,x_{n-l-1}\in X$ and $y_1,\ldots,y_{n-l-1}\in Y$ correspond.  Let $r:=\sigma_{l+1}(x_1,\ldots,x_{n-l-1})^{(X,\cal C)}$.  First suppose that $n-l$ is odd, so that $\sigma_{l+1}(v_1,\ldots,v_{n-l-1})=\inf_{v_{n-l}}\sigma_l(v_1,\ldots,v_{n-l})$.  Fix $\eta>0$ and let $x_{n_l}\in X$ be of the same sort as $v_{n-l}$ so that $\sigma_l(x_1,\ldots,x_{n-l})^{(X,\cal C)}\leq r+\eta$.  Let $y_{n-l}\in Y$ be Player II's response to $x_{n-l}$ according to the strategy $\cal S$.  Then, by induction,
$$\sigma_l(y_1,\ldots,y_{n-l})^{(Y,\cal D)}\leq \sigma_l(x_1,\ldots,x_{n-l})^{(X,\cal C)}+\e\leq r+\e+\eta.$$  Letting $\eta$ go to $0$ yields the desired result.  The case that $n-l$ is even is similar and is left to the reader.
\end{proof}

\section{Elementary equivalence of $\rm II_1$-factors}

We say that two tracial von Neumann algebras $M$ and $N$ are locally equivalent if the associated Banach pairs $(M,(M)_1)$ and $(N, (N)_1)$ are locally equivalent.  Somewhat miraculously, it turns out that for II$_1$ factors belonging to $\kop$, local equivalence is the same as elementary equivalence.  This essentially follows from an argument of Kirchberg in \cite{kirch}.  First, we need to recall a fact about \emph{Jordan morphisms} between von Neumann algebras.

Given a C$^*$ algebra $A$, the \emph{special Jordan product} on $A$ is the operation $\circ$ defined by $a\circ b:=\frac{1}{2}(ab+ba)$ for all $a,b\in A$.  If $B$ is also a C$^*$ algebra, then a linear map $T:A\to B$ is a \emph{Jordan morphism} if it preserves the special Jordan product and the involution.  We need the following:

\begin{fact}(See \cite[Corollary 7.4.9]{jordan})
If $M$ and $N$ are von Neumann algebras and $T:M\to N$ is a normal Jordan homomorphism, then $T$ is the sum of a $*$-homomorphism and a $*$-antihomomorphism.
\end{fact}

Recall that a map $A\to B$ between C$^*$ algebras is a $*$-antihomomorphism if and only if it is a $*$-homorphism $A\to B^{op}$.

Suppose that $M$ and $N$ are von Neumann algebras and $T:M\to N$ is a unital, bijective, normal Jordan homomorphism.  Write $T=T_1+T_2$, where $T_1:M\to N$ and $T_2:M\to N^{op}$ are $*$-homomorphisms.  Since $T_i(1)$ is a projection for $i=1,2$ and $T_1(1)+T_2(1)=1$, $T_1(1)$ and $T_2(1)$ are orthogonal projections.  Since $T(M)=N$, it follows that each $T_i(1)$ is a central projection.  Thus, if $N$ is a factor, it follows that $\{T_1(1),T_2(1)\}=\{0,1\}$, whence $T$ is either an isomorphism or an anti-isomorphism.

The following is basically Proposition 4.6 in \cite{kirch}.

\begin{prop}[Kirchberg]\label{jordan} Suppose that $M$ and $N$ are II$_1$ factors. If there is an isometry $T: L^2(M,\tr_M)\to L^2(N,\tr_N)$ so that $T$ maps $M$ onto $N$ contractively, then $M\cong N$ or $M\cong N^{op}$.
\end{prop}

\begin{proof} We first show that $T$ maps unitaries to unitaries. If $u\in M$ is a unitary, we have $1 = \nor{u}_2^2 = \nor{T(u)}_2^2 = \ip{T(u)}{T(u)} = \ip{T(u)^*T(u)}{1}$. On the other hand $\nor{T(u)^*T(u)}_2\leq \nor{T(u)}\cdot \nor{T(u)}_2\leq 1$.  It follows that $T(u)^*T(u) = 1$. We thus have that $T'(x) := T(1)^*T(x)$ is unital, contractive, trace-preserving, and takes unitaries to unitaries. By the same reasoning as in the proof of \cite[Proposition 4.6]{kirch}, $T'$ is a weakly-continuous Jordan morphism and the result follows from the discussion preceding this proposition.
\end{proof}

\begin{cor}
Suppose that $M$ and $N$ are II$_1$ factors.  Then $M$ is locally equivalent to $N$ if and only if $M$ is elementarily equivalent to $N$ or to $N^{op}$.  In particular, if $M$ and $N$ are II$_1$ factors belonging to the class $\kop$, then $M$ is locally equivalent to $N$ if and only if $M$ is elementarily equivalent to $N$.
\end{cor}

\begin{proof}
By the Downward L\"owenheim-Skolem Theorem (see \cite[Section 4.2]{mtoa2}), we may suppose that $M$ and $N$ are separable.  Suppose that $M$ is locally equivalent to $N$.  Then by Proposition \ref{EFBP}, there is an isometry $L^2(M^\uu)\to L^2(N^\uu)$ that maps $M^\uu$ into $N^\uu$ contractively.  By Proposition \ref{jordan}, $M^\uu$ is isomorphic to either $N^\uu$ or $(N^\uu)^{op}$.  It follows that $M$ is elementarily equivalent to either $N$ or $N^{op}$.  The converse is trivial.
\end{proof}

We now introduce a more useful test for determining elementary equivalence which works in the more specific case of Banach pairs $(M, (M)_1)$ where $M$ is a $\rm II_1$-factor (or more generally a tracial von Neumann algebra) equipped with the $2$-norm and $(M)_1$ is the (operator norm) unit ball of $M$.

We define the game $\fr G_{\vN}(n,\e)$ in parameters $n$ and $\e>0$ which is played by two players with $\rm II_1$-factors $M$ and $N$ as follows.

\noindent{\bf Step i.} Player I chooses a unitary either $u_i\in U(M)$ or $v_i\in U(N)$. Player II then chooses a unitary, respectively $v_i\in U(N)$ or $u_i\in U(M)$ in the same manner.

\noindent{\bf Step n.} The players make their choices, and the game terminates. Player II wins if $\abs{\ip{u_i}{u_j} - \ip{v_i}{v_j}}<\e$ for all $1\leq i,j\leq n$; otherwise, Player I wins.

\begin{thm}\label{local-ii} The $\rm II_1$-factors $M$ and $N$ are locally equivalent if and only if Player II has a winning strategy for the game $\fr G_{\vN}(n,\e)$ for all parameters $(n,\e)$.
\end{thm}

In order to prove this result we will first need one lemma.

\begin{lem}\label{unitary} Let $M$ and $N$ be $\rm II_1$-factors, $E\subset M$ and $F\subset N$ be subspaces, and $T: (E, E\cap (M)_1)\to (F,F\cap (N)_1)$ is an $\e$-almost isometry. If $u\in E$ is a unitary, then there exists a unitary $v\in N$ so that $\nor{T(u) - v}_2\leq 4\sqrt{\e}$.
\end{lem}

\begin{proof} In a $\rm II_1$-factor a $u$ is a unitary if and only if it is a contraction with $\nor{u}_2=1$. By definition, we see that there exists a contraction $y\in N$ with $\nor{y -T(u)}_2\leq \e$. In particular $\nor{y}_2\geq 1-2\e$. By a standard estimate we have that $$\nor{1 - \abs{y}}_2^2\leq  1 + \nor{\abs{y}}_2^2 - 2\tr(\abs{y}) = 1 + \tr(\abs{y}^2) -2\tr(\abs{y})\leq 1 -\tr(\abs{y})\leq 1 - \nor{y}_2\leq 2\e,$$ whence writing $y = v\abs{y}$ for $v\in U(N)$ we have that $\nor{T(u) -v}_2\leq 4\sqrt{\e}$.
\end{proof}

\begin{proof} [Proof of Theorem \ref{local-ii}]

First suppose that $M$ and $N$ are locally equivalent.  Fix $n$ and $\e>0$; we describe a winning strategy for Player II in the game $\frak G_{\vN}(n,\e)$.  For simplicity, we suppose that $n=2$ and describe a winning strategy for Player II; the general case is no more difficult, only the notation is more cumbersome.  Fix $\delta$ sufficiently small (to be specified later) and fix a winning strategy $\cal S$ for Player II in the game $\frak G(2,\delta)$.    Suppose that Player I first plays $u_1\in U(M)$.  (The case that Player I first plays a unitary in $N$ is similar.)  Let $y_1\in N$ be Player II's response to $u_1$ in the game $\frak G(2,\delta)$ according to $\cal S$.  Since $u_1\mapsto y_1$ determines a $\delta$-almost isometry, by Lemma \ref{unitary}, there is $v_1\in U(N)$ such that $\|y_1-v_1\|_2\leq 4\sqrt{\delta}$.  Now suppose that Player II responds with $v_2\in U(N)$.  (The case that Player II responds with a unitary in $M$ is similar.)  Let $x_2\in M$ be Player II's response to $(u_1,y_1,v_2)$ in the game $\frak G(2,\delta)$ according to $\cal S$.  Since $u_1\mapsto y_1$, $x_2\mapsto v_2$ determines a $\delta$-almost isometry, we once again have $u_2\in U(M)$ such that $\|x_2-u_2\|_2\leq 4\sqrt{\delta}$.

We need to verify that $|\langle u_i,u_j\rangle -\langle v_i,v_j\rangle |<\e$ for $i,j=1,2$.  If $\delta$ is chosen small enough so that a $\delta$-almost isometry preserves inner products within an error of $\frac{\e}{3}$ (use, for example, the Polarization Identity) and such that perturbing entries of an inner product by a distance of no more than $4\sqrt{\delta}$ changes the inner product by an amount not exceeding $\frac{\e}{3}$, then the desired estimates hold.  For example:
$$\langle u_1,u_2\rangle \sim_{\frac{\e}{3}} \langle u_1,x_2\rangle \sim_{\frac{\e}{3}} \langle y_1,v_2\rangle \sim_{\frac{\e}{3}} \langle v_1,v_2\rangle.$$

We now prove the converse.  Suppose that Player II has a winning strategy in all of the games $\frak G_{\vN}(n,\e)$; we show that $M$ and $N$ are elementarily equivalent as Banach pairs.  By symmetry, it is enough to show that $\sigma^{(M, (M)_1)}\leq r$ implies that $\sigma^{(N,(N)_1)}\leq r$ for any positive real number $r$ and any prenex normal form sentence $\sigma$.  Since $\sigma\dotminus r$ is equivalent to a prenex normal form sentence, it is enough to prove that $\sigma^{(M,(M)_1)}=0$ implies $\sigma^{(N,(N)_1)}=0$ for any prenex normal form sentence $\sigma$.

Towards this end, we introduce the ``unitary transform'' of a sentence in prenex normal form.  Suppose that $\sigma$ is a sentence in prenex normal form, say $$\sigma=Q_1x_1\cdots Q_n x_n \varphi(\vec x),$$ where $\varphi(\vec x)$ is quantifier-free.  We form the new sentence $\sigma^u$ as follows:
\begin{itemize}
\item If $Q_i=\inf$ and $x_i$ is of sort $n_i$, replace each occurrence of the variable $x_i$ by the term $t_i(u_i,v_i):=n_i\cdot (\frac{u_i+v_i}{2})$, where $u_i$ and $v_i$ are variables of sort $\cal C_1$, and replace the quantifier $Q_i x_i$ by the quantifiers $Q_i u_i Q_i v_i$.
\item The quantifier-free part of $\sigma^u$ should now be $$\max(\varphi,\max_i(\max(1\dotminus \|u_i\|_2,1\dotminus \|v_i\|_2))).$$
\end{itemize}

For example, if $\sigma=\sup_{x_1}\inf_{x_2} \varphi(x_1,x_2)$ where $x_2$ is of sort $\mathcal C_1$ (for simplicity), then $\sigma^u=\sup_{x_1}\inf_{u_2}\inf_{v_2} \varphi(x_1,\frac{u_2+v_2}{2})$.

Also, we let $\sigma^{uu}$ be the ``formula'' defined in the exact same way as $\sigma^u$ except that we only allow quantifiers over the unitary groups rather than the entire unit ball.  (Formally, $\sigma^{uu}$ is not a formula in the sense of continuous logic, but it will be useful in the remainder of the proof.)

\

\noindent \textbf{Claim 1:}  We have $\sigma^{(M,(M)_1)}=0$ if and only if $(\sigma^u)^{(M,(M)_1)}=0$ (and the corresponding statement for $(N,(N)_1)$).

\

\noindent Claim 1 follows from the fact that, in a \emph{finite} von Neumann algebra, any contraction is an average of two unitaries. Indeed if $x$ is a contraction in a finite von Neumann algebra, then it has polar decomposition $x = u |x|$, where $u$ is a unitary. As $|x|$ is a self-adjoint contraction, by functional calculus it may be written as the average of two unitaries.

\

\noindent \textbf{Claim 2:}  $(\sigma^u)^{(M,(M)_1)}=0$ if and only if $(\sigma^{uu})^{(M,(M)_1)}=0$ (and the corresponding statement for $(N,(N)_1)$).

\

\noindent The backwards direction of Claim 2 is trivial; the forwards direction follows from the fact that if $x$ is a contraction in a finite factor and $\nor{x}_2\geq 1-\e$, then there is a unitary $u$ so that $\nor{u - x}_2\leq 2\sqrt \e$.

\

\noindent Finally, suppose that $\sigma$ is a sentence in prenex normal form and $\sigma^{(M,(M)_1)}=0$.  Then by Claims 1 and 2, we have $(\sigma^{uu})^{(M,(M)_1)}=0$.  Since atomic formulae are of the form $\nor{\lambda_1x_1 + \dotsb + \lambda_nx_n}_2$ and arbitrary quantifier-free formulae are continuous combinations of atomic formulae, it follows from a winning strategy for Player II in $\frak G_{\vN}(n,\e)$ (for suitably small $\e$) that $(\sigma^{uu})^{(N,(N)_1)}=0$, whence $\sigma^{(N,(N)_1)}=0$ by Claims 1 and 2 again.

\end{proof}

Suppose now that $\cal L_i=\{\phi\}$, where $\phi$ is a unary function symbol with modulus of uniform continuity $\Delta_\phi(\e)=\e$.  If $M$ is a tracial von Neumann algebra, we view $U(M)$ as an $\cal L_i$-structure by interpreting $\phi$ as multiplication by $i$.  We then have:

\begin{cor} Let $M$ and $N$ be $\rm II_1$-factors.  Then $M$ and $N$ are locally equivalent if and only if $U(M)$ and $U(N)$ are elementarily equivalent as $\cal L_i$-structures.
\end{cor}

\begin{proof}
If $M$ and $N$ are locally equivalent, then $M$ is elementarily equivalent to either $N$ or $N^{op}$.  It follows that there is an ultrafilter $\uu$ such that $M^\uu$ is isomorphic to $N^\uu$ or $(N^{op})^\uu$.  In either case, $(U(M))^\uu=U(M^\uu)$ is isomorphic to $(U(N))^\uu=U(N^\uu)$ as $\cal L_i$-structures, whence $U(M)$ and $U(N)$ are elementarily equivalent as $\cal L_i$-structures.

Conversely, assume that $U(M)$ and $U(N)$ are elementarily equivalent as $\cal L_i$-structures.  Then Player II has a winning strategy for the EF-games for $U(M)$ and $U(N)$ as $\cal L_i$-structures.  It then follows that Player II has a winning strategy in the games $\frak G_{\vN}$ for $M$ and $N$.  Indeed, this follows from the fact that
$$\Re \langle u_i,u_j\rangle =1-\frac{1}{2}d(u_i,u_j)^2, \quad \Im \langle u_i,u_j\rangle=1-\frac{1}{2}d(u_i,i\cdot u_j)^2.$$
\end{proof}

\begin{rem}
Notice that the proof of the previous corollary gives an alternative proof of the forward direction of Theorem \ref{local-ii}.

\end{rem}

\begin{cor}\label{kop-ee}
Let $M$ and $N$ be $\rm II_1$ factors in the class $\kop$.  Then $M$ and $N$ are elementarily equivalent if and only if $U(M)$ and $U(N)$ are elementarily equivalent as $\cal L_i$-structures.
\end{cor}

\begin{cor} Let $M$ and $N$ be $\rm II_1$-factors. Suppose for every $\e$ that there is an $(1+\e)$-Lipschitz homeomorphism $f: U(M)\to U(N)$, that is, $f$ is bijective with $(1+\e)^{-1}\nor{u - v}_2\leq \nor{f(u) - f(v)}_2\leq (1+\e)\nor{u - v}_2$, that is further assumed to preserve the action by $\mathbb Z_4$. Then $M$ and $N$ are locally isomorphic.
\end{cor}

We will say that $M$ and $N$ are \emph{approximately Lipschitz isometric} if the condition of the previous corollary is satisfied. Although this relation ought to be in principle much stronger than elementary equivalence, to the best of our knowledge the results of \cite{mtoa3} heretofore furnish the only know examples of properties invariant under this relation namely, the McDuff property and property ($\Gamma$). It is, however, tempting to speculate that approximate Lipschitz isometry ought to be equivalent to isomorphism (up to opposites).

In lieu of this, it would be highly interesting to determine whether hyperfiniteness is an invariant of approximate Lipschitz isometry. If true, this would be in contrast with \cite[Theorem 4.3]{mtoa3} which shows in particular that hyperfiniteness is not an invariant of elementary equivalence. Though one can show, essentially by Fact 3.1 and Proposition 3.2 (see also \cite[chapter XIV.2]{tak}), that for every $n$ there exists $\e>0$ so that any $\e$-approximate Lipschitz embedding $\theta$ of $M_n$ into a $\rm II_1$-factor $N$ there is a $\ast$-homomorphism $\theta': M_n\to N$ so that the image of the unit ball under $\theta$ is $\e$-contained in $2$-norm in the image unit ball under $\theta'$ of $M_n$, this still does not seem sufficient, unless $\e$ could be taken independent of $n$.

\section{Further remarks and open problems}

Of course, Corollary \ref{kop-ee} raises the question:  which $\mathbb Z_4$-metric spaces arise as unitary groups of II$_1$ factors? Even more importantly, what are the theories of such $\mathbb Z_4$-metric spaces?  Ignoring the extra structure for a moment, an important example of a complete theory of (noncompact) metric spaces is the theory of the \emph{Urysohn metric space}.  (See, for example, \cite{ealygold}.)  Recall that the Urysohn metric space is the unique (up to isometry) complete, separable metric space that is universal (that is, every separable metric space isometrically embeds) and ultrahomogeneous (every isometry between finite--even compact--subspaces extends to an isometry of the entire space).  However, the Urysohn space (or rather, its bounded counterpart, the Urysohn sphere) could never be isometric to the unitary group of a II$_1$ factor as the latter's metric is always negative definite.

Note that for $M$ with separable predual, $U(M)$ isometrically embeds naturally in $\mathbb S^\infty$, the Hilbert sphere in $\ell^2$.  The space $\mathbb S^\infty$ is the ``Hilbertian Urysohn sphere" in the sense described in \cite{nvt}, section 1.4.2.

It is well worth pointing out the following proposition, which is an immediate consequence of Ozawa's \cite{ozawa} fundamental result on the non-existence of a universal, separable $\rm II_1$ factor.

\begin{prop}\label{Z4universal} For any separable II$_1$ factor $M$, $U(M)$ is not universal among all $\bb Z_4$-metric spaces which embed (as $\mathbb Z_4$-metric spaces) in $\mathbb S^\infty$.
\end{prop}

\begin{proof} Suppose, towards a contradiction, that there is a II$_1$ factor $M$ for which $U(M)$ is universal among all $\bb Z_4$-metric spaces which embed in $\bb S^\infty$.  In particular, for any $\rm II_1$-factor $N$ with separable predual, $U(N)$ isometrically embeds in $U(M)$ in a way which commutes with the action of $i$. Since this embedding respects the inner product, it is not hard to see it must extend to an isometric embedding $L^2(N)\to L^2(M)$ which takes $N$ into $M$ contractively. Thus, as above, there is a unital injective $\ast$-homomorphism $N\hookrightarrow pMp\oplus ((1-p) M (1-p))^{op}$, whence $N$ embeds in either $M$ or $M^{op}$ since $N$ is a factor. However, this would contradict the fact \cite{ozawa} that there is no separable universal $\rm II_1$-factor (pick $M\star M^{op}$).
\end{proof}

\begin{question} Can $U(M)$ ever be universal among all metric spaces which embed in $\cal S^\infty$?
\end{question}

Proposition \ref{Z4universal} is good evidence that the answer to the previous question is no. We remark that a positive answer to the previous question would be equivalent to demonstrating the existence of a separable $\rm II_1$-factor for which there is an isometric embedding $\bb S^\infty\hookrightarrow U(M)$. We currently do not know whether $\bb S^\infty$ embeds isometrically in the unitary group of \emph{any} $\rm II_1$-factor. The existence of such an embedding ought to have striking consequences as the following proposition, which is similar in  spirit, demonstrates.

\begin{prop}
Suppose that $M$ is a separable II$_1$  factor belonging to the class $\kop$.  Further suppose that, for each $n$, the $n$-dimensional complex spheres $\bb S^{n}$ isometrically embed in $U(M)$ with respect to the natural $\bb Z_4$-actions.  Then $M$ is a locally universal $\rm II_1$-factor, that is, every separable II$_1$ factor embeds into an ultrapower of $M$. In particular, if, for each $n$, the $n$-dimensional complex spheres $\bb S^{n}$ isometrically embed in $U(\R)$ with respect to the natural $\bb Z_4$-actions, then Connes' embedding problem has a positive answer.
 \end{prop}

 \begin{proof}
Suppose that $M$ satisfies the assumption of the proposition and let $N$ be an II$_1$ factor.  Let $F$ be any finite subset of $U(N)$. Then choosing an orthogonal projection $P$ onto a suitably large finite-dimensional subspace so that $\|P(u)\|> 1- \e$ for all $u\in F \cup iF$, we can correct to an (effective in) $\e$-almost $\bb Z_4$-embedding of $F$ into some $\bb S^n$, and therefore also in $U(M)$. But $\bb Z_4$-embeddings preserve inner products, whence pairs of inner products in $F$ can be modeled arbitrarily well in $U(M)$.  As above, Kirchberg's argument shows that $N$ embeds in $M^\uu$.
 \end{proof}

We now remark how our main result recasts Kirchberg's characterization of $\R^\om$-embeddability in a game-theoretical light. Let $(A,\tr)$ be an arbitrary tracial C$^*$-algebra which we view as a normed space with respect to the $2$-norm. To introduce a bit of terminology, we say that a subspace $E\subset A$ is \emph{$\e$-almost representable} in $\R$ if there exists a subspace $F\subset \R$ and a linear bijection $T: E\to F$ so that $\nor{T}, \nor{T^{-1}}\leq 1+\e$ and $T(E\cap (A)_1)\subset_\e F\cap (\R)_1$. Then by Proposition 4.6 in \cite{kirch}, $A$ is $\R^\om$-embeddable if and only if for every $\e>0$, every finite-dimensional subspace of $A$ is $\e$-representable in $\R$.

Let us introduce the following ``one-sided, one-round game'' $\fr G_{\R}(n,\e)$ for which the winning condition is that, for all $u_1,\dotsc,u_n\in U(A)$ which are linearly independent, there exist $n$ unitaries $v_1,\dotsc,v_n\in U(\R)$ so that the map $$T: \operatorname{span}\{u_1,\dotsc,u_n\}\to \operatorname{span}\{v_1,\dotsc,v_n\}$$ defined by $T(u_i) = v_i$ satisfies $\nor{T},\nor{T^{-1}}\leq 1+\e$.

\begin{prop} There is a constant $N = N(n,\e)$ so that every $n$-dimensional subspace $E$ of any tracial $C^*$-algebra $(A,\tr)$ is $\e$-almost representable in $\R$ if $\fr G_{\R}(N,\e/4)$ is winnable.
\end{prop}

\begin{proof} We first claim that there is a uniform constant $K(n,\e)$ so that for every $n$-dimensional subspace $E\subset A$ of any tracial $C^*$-algebra $(A,\tr)$ there exists a set of unitaries ${\bar u} =\{u_1,\dotsc,u_l\}\subset U(A)$ with $l\leq K$ so that every element of $E\cap (A)_1$ is $\e$-approximated in $2$-norm by a convex combination of elements of $\bar u$.

Indeed, choose an $\e/2$-net $x_1, \dotsc, x_m\in E\cap (A)_1$. The cardinality of such a net is bounded in particular by the $\e/4$-covering number of the unit ball in $\ell^2_n$. We may perturb each $x_i$ so that $\nor{x_i}<1 - \e/4$ and still have an $\e$-net for $E\cap (A)_1$. By the main result of \cite{popa-rd} there is a constant $C$ depending only on $\e$ so that each $x_i$ is a convex combination of at most $C$ unitaries in $U(A)$, whence the claim follows.

We next claim that if $A$ is infinite-dimensional and if $E\subset A$ is a finite-dimensional subspace, then for every $\e>0$ and $u\in U(A)$ there exists $u'\in U(A)$ with $\nor{u - u'}_2<\e$ and so that $u'$ is linearly independent from $E$.  To see this, let $P_E: L^2(A)\to E$ be the orthogonal projection onto $E$. By the Kaplansky density theorem, we have that $U(A)$ is $2$-norm dense in $U(A'')$. Since $M := A''\subset \cal B(L^2(A,\tr))$ is infinite-dimensional, it contains a diffuse abelian subalgebra. Therefore, there is a projection $p\in M$ with trace $\tr(p) = 1-\e^2/2$ and a sequence of unitaries $v_n\in U(M)$ so that $v_n\to p$ weakly. Since $P_E$ is a finite-rank operator, we thus have that $P_E(uv_n)\to P_E(up)$ strongly, whence $\nor{P_E(uv_n)}_2 \to \nor{P_E(up)}_2\leq \nor{p}_2 = \sqrt{1 - \e^2/2}$. It is now easy to see that choosing $n$ sufficiently large and $u'\in U(A)$ sufficiently close to $uv_n$ works.

We now can proceed with the proof of the proposition. Let $E = \operatorname{span}\{u_1,\dotsc,u_n\}\subset A$. (Every $n$-dimensional subspace of a $C^*$-algebra is a subspace of a space spanned by at most $4n$ unitaries, so we may assume this is the case without loss of generality.) By the previous claims, we can extend $u_1, \dotsc, u_n$ to $u_1,\ldots,u_n, u_{n+1}, \dotsc, u_s$ ($s\leq n + K(n,\e)$) to a complete collection of linearly independent unitaries so that all elements in $E\cap (A)_1$ are $2\e$-approximated in $2$-norm by a convex combination of unitaries in the collection. If $\fr G_R(s,\e/4)$ is winnable, then it is easy to check that for $S = T|_E$ we have that $S(E\cap (A)_1) \subset_{\e} S(E)\cap (\R)_1$, and we are done.
\end{proof}

\begin{prob} Let $\cal C\subset \ell^2_n$ be a convex subset of the unit ball in $n$-dimensional Hilbert space. For every $\e>0$ does there exists a $\rm II_1$-factor $M$ so that $(\ell^2_n,\cal C)$ is $\e$-represented in $M$? Can one always choose a locally universal $\rm II_1$-factor (in the sense of \cite{mtoa3}) or even $\R$?
\end{prob}

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