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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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    $\begingroup$ 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? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Mar 27, 2014 at 2:14
  • $\begingroup$ @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 $n>2$?" $\endgroup$
    – Ali Taghavi
    Commented Mar 27, 2014 at 2:37
  • $\begingroup$ @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? $\endgroup$
    – Ali Taghavi
    Commented Mar 27, 2014 at 2:42
  • $\begingroup$ @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 $\endgroup$
    – Ali Taghavi
    Commented Mar 27, 2014 at 4:07

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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 $E_2(A)$.

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  • $\begingroup$ 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 $\endgroup$
    – Ali Taghavi
    Commented Feb 1, 2014 at 18:26
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    $\begingroup$ 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 $p_i(u_j)=\delta_{ij}$. $\endgroup$
    – Neil Strickland
    Commented Feb 1, 2014 at 19:15
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    $\begingroup$ Somewhat strange as for example $A = \mathbf{C}$ even does not have mutually orthogonal Elements. $\endgroup$
    – hänsel
    Commented Mar 20, 2016 at 18:55
  • $\begingroup$ @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. $\endgroup$
    – Ali Taghavi
    Commented Mar 21, 2016 at 7:34
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    $\begingroup$ @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. $\endgroup$
    – Simon Henry
    Commented Mar 23, 2016 at 14:32

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