Sum of Three Idempotents Equals A Constant

The problem is to classify k idempotents in an algebra whose sum is equal to a constant I, where I is the identity. As we shall see, the case of three idempotents can be completely resolved using algebra, indeed using NCAlgebra. It is known in various cases that if k is equal to four, and if is a Banach algebra, the conclusion is greatly simplified. This, of course, can not be analyzed using NCAlgebra. Now we turn to the case of three idempotents and derive the following theorem.

Theorem Let be an algebra and A, B and C be elements of . If A, B and C are idempotents and A + B + C = , then satisfies a 5th degree polynomial whose roots are 0, 1, 2, 3 and ³/2. Furthermore, A, B, and C can be classified, and A, B and C commute in all cases except = ³/2.

Now we turn to proving the theorem and deriving the classification claimed by the theorem. The first main step in the proof of the theorem is the following proposition.

Proposition 1 Let be an algebra and A, B and C be elements of . If A, B and C are idempotents and A + B + C = , then satisfies a 5th degree polynomial - + 20 ³ - ² + 9 , whose roots are 0, 1, 2, 3 and ³/2.

Proof Let I be the ideal of K[A,B,C, ] generated by

{A²-A, B²-B, C²-C, A+B+C- , A - A , B - B , C - C }

The spreadsheet shows that is a root of a 5th degree polynomial - + 20 ³ - ² + 9 . The Solve command in Mathematica shows that the roots of this polynomial are 0, 1, 2, 3 and ³/2.

Now we turn to the classification of A, B and C for each of the five cases, which will show that A, B and C commute in the cases where lambda does not equal 3/2.
Case = 0: Let be an algebra. If A, B and C are elements of which are idempotents and A + B + C = 0, then A = 0, B = 0 and C = 0.

Proof See the sum of three idempotents equals zero problem.

Case = 1: There exists an algebra and idempotents a, b and c of such that a + b + c = 1. Moreover, one can pick a, b and c to be nonzero.

Proof

• Example 1: Let equal the complex numbers and a = 1, b = 0 and c = 0.
• Example 2: Take equal to the collection of 3 × 3 matrices over the complex numbers and let

 1 0 0 a = 0 0 0 0 0 0
 0 0 0 b = 0 1 0 0 0 0
 0 0 0 c = 0 0 0 0 0 1

• A Gröbner Basis approach is to consider the ideal I generated by {A² - A, B² - B, C² - C, A+B+C-1}. If we set = Q[A,B,C]/I and a = A + I, b = B + I and c = C + I, then a, b and c are idempotents which satisfy a + b + c = 1. The Gröbner Basis is shown here. Note that it is not necessary for A, B, and C to be zero, but A and B commute, and their product is zero. From the given relation C = 1 - A - B, we can see that C also commutes with A and B. Also, C multiplied by either A or B is zero.

Case = 2 There exists an algebra and idempotents a, b and c of such that a + b + c = 2. Moreover, one can pick a, b and c to be nonzero.

Proof The proof is analogous to the one above where the Gröbner Basis is here. Notice that A and B commute again, and therefore C commutes with both of them.

Case = 3 For every algebra , there exists idempotents a, b and c of such that a + b + c = 3. Moreover, a, b and c must be equal to the identity element of .

Proof The proof is analogous to the one above where the Gröbner Basis is here.

Case = ³/2 For every algebra , there exists idempotents a, b and c of such that a + b + c = ³/2.

Proof The proof is analogous to the one above where the Gröbner Basis is here. In this case, A and B need not commute. The equation to take note of is

B A + A B - A - B = ³/4.