# The saturation of Murray von Neumann relation

Edit: According to comment of Pace Nielsen, I remove question 2 of the previous version:

Let $R$ be a unital ring. We define Murray Von Neumann relation $M$ on $R$ as follows:

We say $a M b$ iff $a=xy,\;b=yx$ for some $x,y\in R$. (This is inspired by the usual Murray Von Neumann equivalent relation in K theory, which is defined on the set of idempotents of a ring). The relation $M$ is a reflexive and symmetric relation but is not a transitive relation. So we consider its saturation. The saturation of this relation is an equivalent relation denoted by $\simeq$. In fact we say $a\simeq b$ if there are $p_{i}\in R\;$ with $p_{0} M p_{1},\;\;\;p_{1} M p_{2},\ldots p_{n-1} M p_{n}$ where $p_{0}=a,\;p_{n}=b$.

Put $R=M_{n}(\mathbb{C})$. One can show that the equivalent class containing $0$ is $$[0]=\{A\in M_{n}(\mathbb{C})\mid A^{n}=0\}$$

(In fact one can prove the following: If $A\in B(H)$ satisfy $A^{k}=0$ then there are $X,Y\in B(H)$ with $A=XY$ and $(YX)^{k-1}=0$. Here $B(H)$ is the space of bounded operators on a Hilbert space. The same is true by replacing $B(H)$ with an arbitrary Von Neumann algebra. The same also is true without any topological consideration, that is by replacing $B(H)$ with $L(V)$, the space of linear endomorphisms of a vector space $V$.)

So for $R=M_{n}(\mathbb{C}),\;\;[0]$ is an algebraic variety,i.e: the variety of nilpotent matrices $A^{n}=0$

1.Is every equivalent class of $M_{n}(\mathbb{C})$ an algebraic variety?(the zero set of polynomials on $M_{n}(\mathbb{C}) \simeq \mathbb{C}^{n^{2}}$ or the zero set of polynomials in the form $f(A)=0$ where $f\in \mathbb{C}[x]$?What is the precise description of equivalent classes?

1. Assume that $A$ is a $C^{*}$ algebra and $a\in A$ satisfies $a^{k}=0$ for some $k>1$. Are there two elements $x,y \in A$ with $a=xy$ and $(yx)^{k-1}=0$?

**Note:**Inspired by methods from K theory, I tried to construct a functor $NK$ based on the constructions above. please see A functor on the category of rings, algebras or compact Hausdorff topological space

Perhaps, it would be interesting to ask "Is there a kind of periodicity property for this functor?"

• For matrices over the complexes, $XY = A$ and $YX = B$ implies that their nonzero Jordan normal forms are the same (that is, discard the nilpotent parts of their JNFs); it follows that the equivalence relation generated by the relation is exactly equality of JNF away from zero (this is just a special case of a similar, but more complicated result when we have integer matrices). – David Handelman Feb 16 '16 at 22:53
• @DavidHandelman Thank you for your comment. Can you give a reference for that result? – Ali Taghavi Feb 17 '16 at 8:53
• $\le$ is strange notation for a relation that's symmetric. Anyway, note that your definition makes no use of the additive structure, so you might as well state it for monoids. In that setting you are asking for the (zeroth) Hochschild homology of a monoid. I don't know if this has been studied much for monoids as opposed to categories. For rings it's more natural to quotient by the subspace spanned by commutators, which gets you the usual zeroth Hochschild homology. – Qiaochu Yuan Feb 18 '16 at 6:44
• It has been pointed out at meta (meta.mathoverflow.net/questions/2749/…) that this post has undergone numerous recent edits, some of them quite minor. As you know, each edit bumps the post to the front page and pushes other questions off the front page, which can be an annoyance to others. Let's please bring the editing to a close so that users have a stable question to look at. – Todd Trimble Feb 18 '16 at 20:00
• @AliTaghavi: That's a major change to the core system; you would have to propose it in meta.stackoverflow; moderators in MO don't really have the power to make such a change. So yes, such a change would be "really difficult". Moreover, lots of people seem to be perfectly able to NOT edit their questions multiple times for minor issues, so would a minor change in your behavior towards the norm be really difficult? – Arturo Magidin Feb 19 '16 at 16:47

In answer to the second question, yes this is true. Say $x^k=0$. Let $x=v|x|$ be the polar decomposition of $x$ in $A^{**}$ (the bidual of $A$). Let $a=v|x|^{\frac 1 2}$ and $b=|x|^{\frac 1 2}$. Then clearly $x=ab$. Both $a$ and $b$ belong to $A$. In $b$'s case, by functional calculus. It is a well-known property of polar decompositions that $v|x|^{\frac 1 2}$ is also in $A$. To see this, write $p_n(|x|)\to |x|^{\frac 1 2}$, where each $p_n$ is a polynomial such that $p_n(0)=0$. Then $vp_n(|x|)\to v|x|^{\frac 1 2}$ in norm and $vp_n(|x|)\in A$ for all $n$ because we can factor out $|x|$ from $p_n(|x|)$.
Now consider $ba=|x|^{\frac 1 2}v|x|^{\frac 1 2}$ (the Aluthge transform of $x$). Then $$(ba)^{k-1}(ba)^*= (|x|^{\frac 1 2}v|x|^{\frac 1 2}\cdots |x|^{\frac 1 2}v|x|^{\frac 1 2})\cdot |x|^{\frac 1 2}v^*|x|^{\frac 1 2}= |x|^{\frac 1 2}x^{k-1} v^*|x|^{\frac 1 2}=0,$$ where we have used that $|x|^{\frac 1 2}x^{k-1}=0$ (since $|x|^{\frac 1 2}\in C^*(x^*x)$ and $(x^*x)x^{k-1}=0$). It follows that $(ba)^{k-1}((ba)^{k-1})^*=0$ which implies that $(ba)^{k-1}=0$.
• is the polar decomposition legal in the idealizer of $A$ in $A^{**}$?If i am not mistaken, it is necessary in your argument. – Ali Taghavi Mar 5 '16 at 21:31
• I really thank you very much for your very interesting and elegance answer.Is there a particular reason that you consider the polar decomposition in the bidual of A, instead of some $B(H)$ where $B(H)$ faithfully contains$A$? – Ali Taghavi Mar 7 '16 at 19:38