2
$\begingroup$

Assume that $R$ is a unital ring or a complex or real (Banach or $C^{*}$) algebra.

We define a relation $M$ on $R$ as follows: $$a\;M b \;\;\; \text{iff}\;\; a=xy,\;b=yx \;\; \text{for some}\;\; x,y\in R$$ It is a reflexive and symmetric (but not transitive ) relation. We define an equivalent relation $\simeq$ on $R$ as follows:

$a\simeq b$ if there are $p_{i}\in R,\;i=0,1,\ldots,n$ with $$\begin{cases}a=p_{0},\;\; b=p_{n},&\\ p_{i} \; M\; p_{i+1}\end{cases}$$

The space of nilpotent elements of $R$, denoted by $N(R)$, is a saturated subset of $R$, with respect to this equivalent relation, while the space of idempotent elements is not necessarily a saturated subset.

Notation: $M_{n}(R)$ is the space of $n\times n$ matrices with entries in $R$.

The natural mapping $M_{n}(R) \to M_{n+1} (R)$ with $A \mapsto A\oplus 0$ sends nilpotent elements to nilpotent elements. Moreover the above equivalent relation is preserved under this map.

We consider $\bigcup_{n=1}^{\infty} N(M_{n}(R))$, the union of all nilpotent matrices of all size. The equivalent relation $\simeq$ has a natural extension to the later space: $A\simeq B$ if there are natural numbers $k,p$ such that $A\oplus 0_{k} \simeq B\oplus 0_{p}$. The later are the zero matrices of size $k,p$, respectively.

This enable us to equip $\bigcup_{n=1}^{\infty} N(M_{n}(R))/\simeq$ to an Abelian semi group structure, via the usual matrix-sum.(Note that for two nilpotent elements $a,b$ with $ab=ba=0$, $a+b$ is again a nilpotent element. On the other hand every two elements of the above quotient space have two representation $A,B$ with $AB=BA=0$). Because, for every $a\in R$ we have $\begin{pmatrix} a&0\\0&0 \end{pmatrix} \simeq \begin{pmatrix} 0&0\\0&a \end{pmatrix}$.

The Grothendick group of this semi group is denoted by $NK(R)$.

Questions:

What is an example of a $C^{*}$ algebra $A$ for which $NK(A)$ is a non trivial group? Is there a commutative $C^{*}$ algebra $A$ with nontrivial $NK(A)$.

Note 1 The mapping $A\mapsto NK(A)$ is realy a functor on the category of rings or algebra. according to Gelfand Naimark duality this could be counted as a functor on the category of compact Hausdorff topological space

Note 2: This post is inspired by the construction in algebraic K theory and the following two posts

The saturation of Murray von Neumann relation

https://math.stackexchange.com/questions/1661660/the-reduction-of-nilpotency-order-of-nilpotent-elements-of-c-algebras

$\endgroup$
  • 1
    $\begingroup$ What does this construction give for $R={\mathbb C}$? $\endgroup$ – Yemon Choi Feb 27 '16 at 20:50
  • 1
    $\begingroup$ @YemonChoi This construction for $\mathbb{C}$ gives the trivial group(As I explained here:.mathoverflow.net/questions/231328/…) $\endgroup$ – Ali Taghavi Feb 27 '16 at 20:55
  • 1
    $\begingroup$ Let $A,B \in \cup_n M_n(R)$ be a pair of nilpotent matrices such that $AB = BA = 0$ and let $A',B' \in \cup_n M_n(R)$ be another pair of nilpotent matrices over such that $A'B' = B'A' = 0$. If $A \simeq A'$ and $B \simeq B'$, is there any reason to think that $A+B \simeq A'+B'$? This seems like a prerequisite result one would need in order to show that the desired semi-group structure on $\cup_n M_n(R)/\simeq$ is well-defined. $\endgroup$ – Yonatan Harpaz Feb 29 '16 at 19:32
  • $\begingroup$ @YonatanHarpaz Thank you for your attention to my question. I confess that I did not pay attention to this prerequisit,. The nilpotent version of proposition 4.2.4 of page 26 of Blackadar book on K theory. The reason that i did not pay attention:I was saying myself $a\simeq..a'$ and $b\simeq b'$ imply that $\begin{pmatrix} a&0\\0&b\end{pmatrix} \simeq \begin{pmatrix}a'&0\\0&b'\end {pmatrix}$,. So according to your comment, my construction collaps if that prerequisite can not be proved. $\endgroup$ – Ali Taghavi Feb 29 '16 at 20:08
  • $\begingroup$ @YonatanHarpaz Of course we can not give a counter example in matrix algebra or operator algebra or Von Neumann algebras, because all nilpotent elements are equivalent to each other. But what about other algebras? Thanks again for your very interesting comment. $\endgroup$ – Ali Taghavi Feb 29 '16 at 20:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.