MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
1  
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 $E_2(A)$.

share|cite|improve this answer
    
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
6  
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
1  
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
2  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.