Skip to main content
8 of 16
added 9 characters in body
Ali Taghavi
  • 356
  • 8
  • 31
  • 123

The saturation of Murray von Neumann relation

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

We say $a\leq b$ 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). It 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 $p_{0}=a\simeq p_{1}\simeq p_{2}\ldots\simeq p_{n}=b$

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 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. What is an example of a unital ring with an element $a\simeq 0$ but $a$ is not a nilpotent element?
Ali Taghavi
  • 356
  • 8
  • 31
  • 123