# A generalized K- theory via generalized idempotents

Edit After the answer by Neil Strickland, I add the word "a ring" in this new version.

In the literature, there is a concept of generalized idempotent: an n- idempotent is an element $a$ of a Banach algebra or a ring with $a^{n}=a$.

Can the 3 equivalent relations, Murray-Von Neumann, similarity and homotopy on 2-idempotents be generalized to n-idempotents,for arbitrary $n>2$? Does this processes gives us a useful and new type of K theory?

We know that "Vector bundles" are the topological analogy of 2-idempotents. Now what is a topological analogy for generalized idempotents?

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Do you have any example and/or application and/or concrete motivation thqt justifies changing the question in a way that renders a perfectly correct answer irrelevant? – Mariano Suárez-Alvarez Mar 27 '14 at 2:14
@MarianoSuárez-Alvarez the answer was not irrelevant . It was very interesting. But does it contains an obviouse answer to this part of my question(which was presented in the first version of my post)?"Can the 3 equivalent relations, Murray-Von Neumann, similarity and homotopy on 2-idempotents be generalized to n-idempotents,for arbitrary $n>2$?" – Ali Taghavi Mar 27 '14 at 2:37
@MarianoSuárez-Alvarez Moreover please see my comment on his answer(My question about "ring case" In the new version I add only this case. I do not think that this new version render his interesting answer irrelelevant. however I think my question on ring case is stile nonobviouse. Do you mean that I should present this question(ring) in a new post and I should not change the first version of the current post? – Ali Taghavi Mar 27 '14 at 2:42
@MarianoSuárez-Alvarez Any way I explained in the head of this new version about this change, so I do not think that it is an unusual conduct – Ali Taghavi Mar 27 '14 at 4:07

Let $E_n(A)$ be the set of $n$-idempotents in $A$, and let $u_1,\dotsc,u_n$ be the elements of $E_n(\mathbb{C})$. Let $E'_n(A)$ be the set of $n$-tuples $e_1,\dotsc,e_n\in E_2(A)$ with $e_ie_j=0$ for $i\neq j$, and $\sum_ie_i=1$. Define $f\colon E'_n(A)\to E_n(A)$ by $f(e_1,\dotsc,e_n)=\sum_iu_ie_i$. Then it is not hard to see that $f$ is bijective. Thus, $E_n(A)$ does not really tell you anything that is not already determined by $E_2(A)$.
Some question on your answer:1)Is your statement true for a complex algebra( without any topological consideration, so without holomorphic functional calculus)? 2)what can we say if we are interested in pure algebraic generalized K theory(for rings) 3)According to your answer, do you belive that the three realations on generalized idempotents in banach algebras(Mouray Von.similarity, homotopy...) is well defined? Does it leads to triviality? $)could you please write the argument in your answer for banach algebras , explicitly? Thanks – Ali Taghavi Feb 1 '14 at 18:26 You don't need any fancy functional calculus. The inverse is just$f^{-1}(a)=(p_1(a),\dotsc,p_n(a))$, where$p_i(t)\in\mathbb{C}[t]$is the unique polynomial of degree$n$such that$p_i(u_j)=\delta_{ij}$. – Neil Strickland Feb 1 '14 at 19:15 Somewhat strange as for example$A = \mathbf{C}$even does not have mutually orthogonal Elements. – hänsel Mar 20 at 18:55 @hänsel thank you very much for your very helpfull comment. When I have accepted this answer, I did not pay attention to the point which you mentioned. – Ali Taghavi Mar 21 at 7:34 @hänsel : That is not true : for each$n$, there is exactly$n$famillies of$n$mutually orthogonal projections in$\mathbb{C}$(a one and$(n-1)$zeros in the various possible order). As there is exactly$n$elements in$E_n(\mathbb{C})\$ I don't see the problem. – Simon Henry Mar 23 at 14:32