Lobachevskii Journal of Mathematics Vol. 9, 2001, 47 – 53 \begin{matrix}c \\ ◯ \end{matrix}D.Kh. Mushtari

D.Kh. Mushtari

SOME REMARKS ON MEASURES ON ORTHOGONAL RATIONAL PROJECTIONS AND THE RATIONAL SPHERE

Abstract. We examine measures on the quantum logic of all projections with rational n × n-matrices and on its sublogic generated by all projections onto one-dimensional subspaces in ℚ^{n} passing through elements of the unit sphere in ℚ^{n}.

Introduction

We consider two logics of projections acting in the space ℚ^{n} which is the n-dimensional linear space over the field ℚ of rationals endowed with the standard scalar product (·, ·). Let Π(ℚ^{n}) be a quantum logic of all orthogonal idempotent operators with rational n × n-matrices, S(ℚ^{n}) be the unit sphere in ℚ^{n}, Π_{0}(ℚ^{n}) be a sublogic of Π(ℚ^{n}) generated by all projections onto onedimensional subspaces in ℚ^{n} passing through elements of S(ℚ^{n}).

We consider measures on Π(ℚ^{n}) (or on Π_{0}(ℚ^{n})), i.e., [0, ∞)-valued functions μ satisfying the relation μ(P + Q) = μ(P) + μ(Q) for all pairs of orthogonal (P Q = QP = 0) projections.

Problem. Does a measure μ defined on Π(ℚ^{n}) or on Π_{0}(ℚ^{n}) admit the well-known Gleason [1] representation:

μ(P) = tr(T P)  (1)

(see also Problem G in [2])?

Obviously, the measure μ is uniquely defined by its values on the onedimensional projections. So, we can consider instead of a measure a function on unit vectors (not necessarily in S(ℚ^{n})). Such a function (we denote it μ, too) is defined by the relation

μ(x) := μ(P), where P (y) = (y, x)x for every S(ℚ^{n}).

Supported by the Russian FBR grant, 99-01-00441.

Key words and phrases. Gleason theorem, quantum logic, projection, measure, rational sphere.

Here x ∈ S(ℚ^{n}) in case of the logic Π_{0}(ℚ^{n}) and x ∈ {x ∈ ℝ^{n} : ∥x∥ = 1, Lin(x) ∩ ℚ^{n} ≠ ∅} in case of the logic Π(ℚ^{n}). The function μ has to satisfy the following additivity condition:

μ(x_{1}) + μ(x_{2}) + ... + μ(x_{n}) = Const

for all orthonormal bases (x_{i}) in S(ℚ^{n}) or in {x ∈ ℝ^{n} : ∥x∥ = 1, Lin(x) ∩ ℚ^{n} ≠ ∅}.

This paper is initiated by [3] where the authors constructed a colouring, c, of S(ℚ^{3}) by three colours (we denote them {1, 2, 3}) with the following property: any two orthogonal elements x, y ∈ S(ℚ^{n}) have different colours c(x) ≠ c(y). We call such a colouring regular. The colouring is constructed in the following way: each element x ∈ S(ℚ^{3}) is presented in the form p^{−1}(n_{1}, n_{2}, n_{3}) where n_{i} are coprime integers and p ∈ ℤ. It is easy to verify (by dividing n_{i}^{2} on 4 and comparing the remainders) that only one n_{i} is odd (we denote it i(x)), and i(x), i(y) differ in case of orthogonal vectors x, y. The colouring is defined by the relation: c(x) = i(x). Such a colouring immediately produces a {0, 1}- valued measure μ on Π_{0}(ℚ^{3}). In fact, μ may be constructed being equal to 1 on every one-dimensional projection of colour 1 and equal to 0 on projections of colours 2 and 3.

It is easy to see that such μ does not admit a Gleason-like representation. At the first glance this solves Problem G. But this counterexample does not touch the measures on Π(ℚ^{n}) and the measures on Π_{0}(ℚ^{n}) for n > 3 or much larger. So, this is not a counterexample for Problem G in [2]. Even the case n = 4 differs from n = 3 because we have two situations for p^{−1}(n_{1}, n_{2}, n_{3}, n_{4}): (odd, even, even, even) and (odd, odd, odd, odd) (for example, (1/2, 1/2, 1/2, 1/2)). The question of the colouring of S(ℚ^{4}) in four colours became especially interesting for P.Ovchinnikov who organized a e-mail round table of colouring problems (2000, March – May). It turned out that a study of remainders on dividing n_{i}^{2} by 8 or 16 proves the existence of a regular colouring by 4 colours of S(ℚ^{4}) (O.Okunev) and by 5 colours of S(ℚ^{5}) (P.Ovchinnikov).

Below we describe (Proposition 4) the structure of S(ℚ^{4}), this immediately implies the existence of a regular colouring for S(ℚ^{4}). We also give the proofs of some results announced at the round table (and also in [4]). The Proposition 1 was proved by the author during the conference ”Quantum Structures V. March 31 – April 5, 2001” and was announced at the conference. But after examining some Los-Alamos preprints the author addressed to A.Peres and received a e-mail containing a picture from [5] with a construction just similar to that of the author. A.Peres attributed this construction to S.Kochen, but remarked that its proof is not published. So, we think that the proof of Proposition 1 in this paper could be helpful.

We close the Introduction by the remark that all the proofs presented in the paper are simple. But the subject contains also problems which seem to be rather difficult. These problems wait attention of specialists in the Number Theory.

Notation. e_{i} = (0, ..., 0, 1, 0, ...) where 1 takes the i-th place.

When {0, 1}-valued measures do not exist

Proposition 1. (S.Kochen, see the picture on p. 197 in [5]) The logic Π(ℚ^{3}) does not admit nontrivial {0, 1}-valued measures.

Proof. Here we denote a one-dimensional projection on Lin(p, q, r) as (p, q, r), where p, q, r are coprime numbers in ℤ.

Let μ be a {0, 1}-valued measure on Π(ℚ^{3}). Since {(1, 0, 0), (0, 1, 0), (0, 0, 1)} is an orthonormal basis, we may assume without loss of generality that

μ(1, 0, 0) = 1, μ(0, 1, 0) = 0, μ(0, 0, 1) = 0.  (2)

It follows from (2) that

μ(0, a, b) = 0 ∀a, b ∈ ℤ.  (3)

The projections (1, 1, 0), (1, −1, 0), (0, 0, 1) also form an orthonormal basis. Therefore, μ(1, 1, 0) = 1 or μ(1, −1, 0) = 1, the same is true for a pair ((1, 0, 1), (1, 0, −1)). Without loss of generality we assume that

μ(1, 1, 0) = 1, μ(1, 0, 1) = 1,  (4a)

μ(1, −1, 0) = 0, μ(1, 0, −1) = 0.  (4b)

Since μ is {0, 1}-valued, the measure of any projection orthogonal to one of projections in (4a) is equal to 0.

Thus, we obtain:

μ(1, 1, −1) = 0, μ(1, −1, −1) = 0, μ(1, −1, 1) = 0.  (5)

Let us consider the values of μ on elements constructed by one number ±2 and two numbers ±1:

μ(2, 1, −1) = 1, μ(1, 1, 2) = 1.  (6)

We explain this: (2, 1, −1) is orthogonal to two mutually orthogonal projections (0, 1, 1) (see (3)) and (1, −1, 1) (see (5)) with the measure 0, (1, 1, 2) is orthogonal to (1, 1, −1) and (1, −1, 0).

Now we finish the proof: μ(1, 0, 2) = 1 or μ(2, 0, −1) = 1 (since μ(0, 1, 0) = 0), μ(1, 0, 2) = 1 contradicts μ(2, 1, −1) = 1, μ(2, 0, −1) = 1 contradicts μ(1, 1, 2) = 1.

Proposition 2. In case n ≥ 6 the logic Π_{0}(ℚ^{n}) does not admit nontrivial {0, 1}-valued measures.

Proof. Suppose the contrary. Without loss of generality we may assume that μ(e_{1}) = 1, and μ(e_{i}) = 0 for all i ≠ 1.

We also use another orthonormal basis

f_{1} = (1/2, 1/2, 1/2, 1/2, 0, 0), f_{2} = (1/2, 1/2, −1/2, −1/2, 0, 0),

f_{3} =(1/2, −1/2, 1/2, −1/2, 0, 0), f_{4} =(1/2, −1/2, −1/2, 1/2, 0, 0), f_{5} =e_{5}, f_{6} =e_{6}. One of numbers μ(f_{i}) (i ≤ 4) is equal to 1. Without loss of generality we assume that μ(f_{1}) = 1.

Now we consider the third orthonormal basis

g_{1} = (−1/2, 1/2, 1/2, 1/2, 0, 0), g_{2} = (1/2, −1/2, 1/2, 1/2, 0, 0),

g_{3} = (1/2, 1/2, −1/2, 1/2, 0, 0), g_{4} = (1/2, 1/2, 1/2, −1/2, 0, 0), g_{5} = e_{5}, g_{6} = e_{6}. One of the numbers μ(g_{i}) (i ≤ 4) is equal to 1. Without loss of generality we assume that μ(g_{2}) = 1.

At the end we consider a fourth orthonormal basis

h_{1} = (1/2, 1/2, 0, 0, 1/2, 1/2), h_{2} = (1/2, 1/2, 0, 0, −1/2, −1/2),

h_{3} =(1/2, −1/2, 0, 0, 1/2, −1/2), h_{4} =(1/2, −1/2, 0, 0, −1/2, 1/2), h_{5} =e_{3}, h_{6} =e_{4}. One of the numbers μ(h_{i}) (i ≤ 4) is equal to 1. It is easy to see that h_{i} ⊥ f_{1} or h_{i} ⊥ g_{2}. We have a contradiction.

The case n = 3 in Proposition 1 covers the case n = 4, but below we give the following simple proof in this case.

Proposition 3. In case n ≥ 4 the logic Π(ℚ^{n}) does not admit nontrivial {0, 1}-valued measures.

Proof. We denote by {e′_{i}}, {f′_{i}}, {g′_{i}} the restrictions to {1, 2, 3, 4} of elements of bases {e_{i}},{f_{i}},{g_{i}} in the proof of Proposition 2 (for example, e′_{1} = (1, 0, 0, 0), f′_{1} = (1/2, 1/2, 1/2, 1/2) and so on). We suppose that μ(e′_{1}) = 1. One of the μ(f′_{i}) is equal to 1, without loss of generality we suppose that μ(f′_{1}) = 1. One of the μ(g′_{i}) is equal to 1, we suppose that μ(g′_{2}) = 1 or μ(g′_{1}) = 1 (the cases μ(g′_{3}) = 1 and μ(g′_{3}) = 1 are similar to μ(g′_{2}) = 1).

Let now consider an other orthonormal basis

h′_{1} = (1/\sqrt{2}, 1/\sqrt{2}, 0, 0), h′_{2} = (1/\sqrt{2}, −1/\sqrt{2}, 0, 0), h′_{3} = e′_{3}, h′_{4} = e_{4}.

Obviously, μ(h′_{1}) = 1 or μ(h′_{2}) = 1 (because μ(e′_{3}) = μ(e′_{4}) = 0). But h′_{1} is orthogonal to g′_{1} or g′_{2}, h′_{2} ⊥ f′_{1}, we have a contradiction.

Corollary. The logics Π_{0}(ℚ^{(∞)}) and Π(ℚ^{(∞)}) do not admit nontrivial finite set valued measures.

The unit sphere in ℚ^{4}

Let us consider three orthonormal bases B^{(i)} = {e_{1}^{(i)}, e_{2}^{(i)}, e_{3}^{(i)}, e_{4}^{(i)}}, i ≤ 3, in S(ℚ^{4}), where

e_{1}^{(1)} = e_{1}, e_{2}^{(1)} = e_{2}, e_{3}^{(1)} = e_{3}, e_{4}^{(1)} = e_{4};
e_{1}^{(2)} = (1/2, 1/2, 1/2, 1/2), e_{2}^{(2)} = (1/2, 1/2, −1/2, −1/2),
e_{3}^{(2)} = (1/2, −1/2, 1/2, −1/2), e_{4}^{(2)} = (1/2, −1/2, −1/2, 1/2);
e_{1}^{(3)} = (−1/2, 1/2, 1/2, 1/2), e_{2}^{(3)} = (1/2, −1/2, 1/2, 1/2),
e_{3}^{(3)} = (1/2, 1/2, −1/2, 1/2), e_{4}^{(3)} = (1/2, 1/2, 1/2, −1/2).

We denote by

x = \frac{m_{1}^{(i)}(x)}{r^{(i)}(x)}e_{1}^{(i)} + \frac{m_{2}^{(i)}(x)}{r^{(i)}(x)}e_{2}^{(i)} + \frac{m_{3}^{(i)}(x)}{r^{(i)}(x)}e_{3}^{(i)} + \frac{m_{4}^{(i)}(x)}{r^{(i)}(x)}e_{4}^{(i)}  (8)

the decomposition of x ∈ S(ℚ^{4}) with respect to the basis B^{(i)} where {m_{1}^{(i)}(x), m_{2}^{(i)}(x), m_{3}^{(i)}(x), m_{4}^{(i)}(x)} is a coprime collection of integers and r^{(i)}(x) ∈ ℤ. Here

(m_{1}^{(i)}(x))^{2} + (m_{2}^{(i)}(x))^{2} + (m_{3}^{(i)}(x))^{2} + (m_{4}^{(i)}(x))^{2} = (r^{(i)}(x))^{2}.

Considering the remainders by dividing (m_{k}^{(i)}(x))^{2}, 1 ≤ k ≤ 4, and (r^{(i)}(x))^{2} on 4 we see that there are only two possibilities:

α) only one of numbers m_{k}^{(i)}(x) is odd, three other ones are even, r^{(i)}(x) is odd;

β) all numbers m_{k}^{(i)}(x) are odd, r^{(i)}(x) is even.

We introduce the set S^{(i)} ⊂ S(ℚ^{4}) as the set of all x ∈ S(ℚ^{4}) such that numbers m_{k}^{(i)}(x), 1 ≤ k ≤ 4, satisfy α).

Proposition 4. S(ℚ^{4}) admits the following decomposition:

S(ℚ^{4}) = S^{(1)} ∪ S^{(2)} ∪ S^{(3)} where S^{(i)} ∩ S^{(j)} = ∅ for i ≠ j, .  (9)

Proof. We are going to formulate several obvious statements.

i) If x ∈ S^{(i)} and x ⊥ y then y ∈ S^{(i)}. Otherwise,

0 = m_{1}^{(i)}(x)m_{1}^{(i)}(y) + m_{2}^{(i)}(x)m_{2}^{(i)}(y) + m_{3}^{(i)}(x)m_{3}^{(i)}(y) + m_{4}^{(i)}(x)m_{4}^{(i)}(y)

is not an even number.

ii) If {x_{1}, x_{2}, x_{3}, x_{4}} ⊂ S^{(i)} is an orthonormal basis and (α_{1}, α_{2}, α_{3}, α_{4}) ∈ S^{(1)} then

α_{1}x_{1} + α_{2}x_{2} + α_{3}x_{3} + α_{4}x_{4} ∈ S^{(i)}.

iii) There is readily verified that

e_{1}^{(i)}, e_{2}^{(i)}, e_{3}^{(i)}, e_{4}^{(i)} ∉ S^{(j)} for j ≠ i.

i), ii), iii) give us

S^{(i)} ∩ S^{(j)} = ∅ for i ≠ j.

In order to prove this, we embed, by [6, ch. 2, Lemma 4.2], x ∈ S^{(i)} into an orthonormal basis {x_{1} = x, x_{2}, x_{3}, x_{4}} in S(ℚ^{4}). By i), {x_{1}, x_{2}, x_{3}, x_{4}} ⊂ S^{(i)}. We denote m_{kj}^{(i)} = m_{j}^{(i)}(x_{k}), r_{k}^{(i)} = r^{(i)}(x_{k}). Obviously, two vectors x_{k}, x_{l} with different k and l have different odd coefficients in the decomposition (8). This permits us to prove that the determinant D = |[m_{kj}^{(i)}]_{k≤4,j≤4}| is an odd number. In fact, it is easy to see that D = 1 (mod 2). In the same way we prove that a minor D_{kj} of D is odd if and only for any l ≠ k the number n of the odd coefficient m_{ln}^{(i)} does not coincide with j. This implies that for any e_{j}^{(i)} the coefficients of its decomposition by (x_{k}) satisfy the condition α). Thus, by ii) and iii),

x_{1} ∈ S^{(j)} ⇒ x_{1}, x_{2}, x_{3}, x_{4} ∈ S^{(j)} ⇒ e_{j}^{(i)} ∈ S^{(j)} ⇒ j = i.

In order to finish the proof we need only to show that

S(ℚ^{4}) ⊃ S^{(1)} ∪ S^{(2)} ∪ S^{(3)}.

Suppose that m_{1}^{(1)}(x), m_{2}^{(1)}(x), m_{3}^{(1)}(x), m_{4}^{(1)}(x) are all odd. Then the remainders by dividing (m_{1}^{(1)}(x))^{2}, (m_{2}^{(1)}(x))^{2}, (m_{3}^{(1)}(x))^{2}, (m_{4}^{(1)}(x))^{2} on 16 are equal to 1 or 9, and numbers of remainders 1 and 9 are even. This implies that r^{(1)}(x) = 2s where s is odd. We denote m_{1}^{(1)}(x) = 2m_{1} + 1, m_{2}^{(1)}(x) = 2m_{2} + 1, m_{3}^{(1)}(x) = 2m_{3} + 1, m_{4}^{(1)}(x) = 2m_{4} + 1.

It is readily verified that if a) null, b) two, or c) four of the numbers m_{1}, m_{2}, m_{3}, m_{4} are odd, then x ∈ S^{(2)}, i.e., in the decomposition (8) for i = 2 only one of numbers m_{j}^{(2)}(x) (j ≤ 4) is odd. For proving this we must enumerate all the variants of remainders by dividing m_{j}^{(2)}(x) (j ≤ 4) on 8. We consider the remainders by dividing

a_{1} = m_{1}^{(1)}(x) + m_{2}^{(1)}(x) + m_{3}^{(1)}(x) + m_{4}^{(1)}(x),

a_{2} = m_{1}^{(1)}(x) + m_{2}^{(1)}(x) − m_{3}^{(1)}(x) − m_{4}^{(1)}(x),
a_{3} = m_{1}^{(1)}(x) − m_{2}^{(1)}(x) + m_{3}^{(1)}(x) − m_{4}^{(1)}(x),
a_{4} = m_{1}^{(1)}(x) − m_{2}^{(1)}(x) − m_{3}^{(1)}(x) + m_{4}^{(1)}(x)

on 8 and verify that there are two of these numbers with different number of divisors 2. This signifies that m_{j}^{(2)}(x) do not satisfy β).

For example, in the case a) the remainders of m_{j}^{(1)}(x) may be equal to 1 or 5. If all of them are equal to 1 then a_{2}, a_{3}, a_{4} are divisible on 8, but a_{1} is divisible only on 4, If the remainders are 1, 5, 5, 5 then a_{1} is divisible on 8 but all other a_{i} are not divisible on 8 (in fact, the alternative α) or β) shows that this variant is not possible) and so on. Similarly, we may enumerate the cases b) or c).

If three or one of the numbers m_{1}, m_{2}, m_{3} , m_{4} are odd, then we prove in the same way that x ∈ S^{(3)}.

Corollary. The unit sphere in ℚ^{4} admits a regular colouring.

Proof. Every set S^{(i)} admits a regular colouring by four colours 1, 2, 3, 4. The x ∈ S^{(i)} has the colour j when m_{j}^{(i)}(x) is odd.

Remark. During the e-mail round table E.Specker proposed to make from S(ℚ^{4}) a graph connecting all the orthogonal pairs of points in S(ℚ^{4}). He posed the question: ”Is such a graph connected?” P.Ovchinnikov showed that the graph S(ℚ^{3}) is not connected. It follows from the proof of Proposition 4 that the graph S(ℚ^{4}) is not connected, too.

Question. Are the sets S^{(i)} connected subgraphs?

References

1. A.M.Gleason. Measures on the closed subspaces of Hilbert space. J.Rat. Mech. Anal., 1957, vol. 6, pp. 885–894.

2. D.H.Mushtari. Gleason type theorems for signed measures on orthomodular posets of projections on linear spaces. Int. J. Theor. Phys., 1995, v. 34, no. 8, pp. 1627–1635.

3. H.Havlicek, G.Krenn, J.Summhammer, K.Svozil. On coloring the rational quantum sphere. http:/XXX.lanl.gow/abs/quant-ph/, 9911040.

4. D.H.Mushtari. Remarks on measures on orthogonal rational projections. Quantum Structures V. Book of abstracts. Cesena-Cesenatico, Italy, March 31 — April 5, 2001, pp. 65–66.

5. A.Peres. Quantum theory: Concepts and Methods. Kluwer, 1995.

6. J.W.S. Cassels. Rational quadratic forms. Academic Press, 1978.

Research Institute of Mathematics and Mechanics, Kazan State University, Universitetskaya Str. 17, Kazan, 420008 Russia E-mail address: Daniar.Mushtari@ksu.ru

Received November, 30, 2001