A factor $R$ is called stable if $M\_n(R)\cong R$ for all $n>0$. For the sake of this question, we call a factor *backwards stable* if $R\cong M\_n(S)$ implies $S\cong R$ where $S$ is allowed to be any other factor.
If $R$ is stable then it is backwards stable if $M\_n(R)\cong M\_n(S) \implies R\cong S$.


My question is which factors are backwards stable.


I'm particularly interested in hyperfinite type III factors. For hyperfinite type II factors I can show backwards stability using the uniqueness of the II$\_1$ factor.


As a side question, are there examples of factors that are not stable other than those of finite type I? For infinite factors, one can always find Cuntz isometries so that they should be stable and, by uniqueness, the hyperfinite II$\_1$ factor is stable. This leaves open the case of non-hyperfinite type II$\_1$ factors.



Backwards stable factors
For $\textrm{II}\_1$ factors, your definitions of backwards stable and stable are the same. The point here is that for a $\textrm{II}\_1$ factor $R$ you can talk about $M\_n(R)$ for all positive number $n$ instead of just positive integers. This is called amplification and is usually written as $R^n$. This is done by choosing an integer $m \geq n$ and a projection $p$ of trace $\frac{n}{m}$ in $M\_m(R)$, then defining $R^n = pM\_m(R)p$. The fact that two projections of the same trace are equivalent in a $\textrm{II}\_1$ factor ensures that this is well-defined and does not depend on the choice of $m$ or $p$. With this in mind, as $R = M\_n(R)^{\frac{1}{n}}$, $M\_n(R) \simeq M\_n(S) \Rightarrow R \simeq S$ always holds for $\textrm{II}\_1$ factors $R$ and $S$. As you have observed, this means stability implies backwards stability. The converse holds by observing that, if $R$ is backwards stable, as $R \simeq M\_n(R^{\frac{1}{n}})$ (this is a general fact about amplifications), $R \simeq R^{\frac{1}{n}}$. Then $R \simeq M\_n(R^{\frac{1}{n}}) \simeq M\_n(R)$, which shows $R$ is stable.


There are plenty of non-stable $\textrm{II}\_1$ factors. In fact, for a $\textrm{II}\_1$ factor $R$, the collection of positive real numbers $r$ for which $R \simeq R^r$ form a group under multiplication, which is known as the fundamental group of $R$. Plenty of $\textrm{II}\_1$ factors have fundamental groups that are not $\mathbb{R}\_+$. For example, all property (T) $\textrm{II}\_1$ factors have countable fundamental groups. They could even have trivial fundamental groups. (See, for example, chapter 18 in [https://www.math.ucla.edu/~popa/Books/IIunV15.pdf](https://www.math.ucla.edu/%7Epopa/Books/IIunV15.pdf) .) It is a famous open question as to what is the fundamental group of the free group factor $L(\mathbb{F}\_2)$, which is currently only known to be either trivial or $\mathbb{R}\_+$.


