Napoleonic Vectors

Napoleon's Theorem states that the centers of three equilateral triangles constructed on the edges of any given triangle form an equilateral triangle. In an article on Napoleon's Theorem we saw that this proposition can be expressed in terms of the three complex numbers v₁, v₂, v₃ representing the vertices of the given triangle in the complex plane. In general, three complex numbers z₁, z₂, z₃ are the vertices of an equilateral triangle if and only if

From this, given any two vertices of an equilateral triangle, we can solve for the third, choosing the appropriate root, depending on whether we want a clockwise loop or a counter-clockwise loop. The centers of the counter-clockwise equilateral triangles are then given by the averages of their vertices, so the centers are given by

The differences between these centers are

Essentially Napoleon's Theorem asserts that the sum of the squares of these three quantities vanishes for any values of v₁, v₂, v₃, and this is easily verified algebraically. Notice that the coefficients of the vertices are simply the cube roots of 1. Denoting these roots by ρ₁, ρ₂, ρ₃, we have the identities

Hence the sum of squares of the three preceding differences is

To see why this sum vanishes, note the general algebraic identity

If we set x,y,z equal to the cube roots of 1, then x² + y² + z² and xy + xz + yz both vanish, so the sum vanishes, regardless of the values of a,b,c.

This suggests an interesting algebraic structure. Let B denote the set of all ordered triples (x,y,z) of complex numbers such that xy + xz + yz = 0. An equivalent condition, for non-zero components, is 1/x + 1/y + 1/z = 0. Obviously (ρ₁, ρ₂, ρ₃) is an element of B, but there are infinitely many others, such as (–2,3,6) and (–1,2,2). In general we have the two-parameter family [x, y, –xy/(x+y)]. We define the "norm" of an element X = (x₁,x₂,x₃) of B as

Because of the trinary symmetry of the components, the norm can also be expressed in the equivalent form

Also, using the relation x₃ = −x₁x₂/(x₁+x₂) we can eliminate x₃ and write the norm as (x₁² + x₁x₂ + x₂²)/(x₁ + x₂) and similarly for the other two pairs of indices. One consequence of this reduction is that if the components of X are integers or rational, then the norm of X is an integer or rational respectively.

We can also define the "upward product" of two elements X and Y as

Denoting this product by Z = (z₁, z₂, z₃), we can evaluate the sum

Since X and Y are both elements of B, it follows that this quantity vanishes, and so the upward product of X and Y is also an element of B. Furthermore, again taking advantage of the fact that X and Y are elements of B, it's clear from identity (1) that the norm is multiplicative, i.e., we have

Likewise the "downward product" defined by

is also an element of B, and the norm is multiplicative under this operation. Naturally the corresponding "division" operations can be performed by solving the three linear equations in three unknowns. For example, if (a,b,c)↑(x,y,z) = (A,B,C), and we are given the values of x,y,z,A,B,C, we can solve for a,b,c using the relation

The determinant of the square matrix is x³ + y³ + z³ − 3xyz, which for Napoleonic triples is equivalent to (x + y + z)³. Hence division is well-defined as long as x + y + z is not zero, and we can solve the above system to give

and similarly for b and c (by rotating the coefficients A,B,C to the right). Of course, the vanishing of the denominator corresponds to the vanishing of the norm of (x,y,z), which is x² + y² + z².  In view of the identity x² + y² + z² = (x + y + z)² − 2(xy+xz+yz) we see the norm of a Napoleonic triple vanishes if and only if the sum of its components is zero.

To define an analog of addition for elements of B is slightly less straightforward, because a simple addition of corresponding components of two elements of B generally does not yield an element of B. In other words, if X and Y are elements of B, the triple of numbers x₁ + y₁, x₂ + y₂, x₃ + y₃ is generally not in B. One way of defining "addition" for elements of B is by the formula

where the scalar λ is given by

If Y is the zero element of B, i.e., Y = (0,0,0), then λ = 0 and we have

On the other hand, if the two summands are identical, the numerator of λ vanishes, so we have

with the understanding that multiplication of a Napoleonic vector by a scalar simply multiplies each component by that scalar.

The upward product is obviously not commutative, because the second and third components are transposed. In other words, if X↑Y = (z₁, z₂, z₃) then Y↑X = (z₁, z₃, z₂). The commutator can then be found by applying Napoleonic addition to X↑Y and –(Y↑X). The value of λ for this operation is

(In the last step we have made use of the fact that 2z₁z₂ + 2z₁z₃ equals −2z₂z₃.) Inserting this into the addition formula gives the commutator

Interestingly, the norm of this vector is zero, since it's just a multiple of the fundamental "equilateral" vector consisting of the cube roots of 1. It's interesting that this is formally very similar to the commutator operator in quantum mechanics. If X,Y denote the operators corresponding to two incompatible observables (such as position and momentum), then a measurement on the state φ of X followed by a measurement of Y is denoted by XY, whereas a measurement of Y followed by a measurement of X is denoted by YX, and these two compound operations do not commute. Specifically, we have

In a sense, the fundamental vector ρ = (ρ₁, ρ₂, ρ₃)/√3 and its permutations play the role of Planck's constant , i.e., they represent the minimal Napoleonic vector, having a norm of zero. The basic commutation relation of Napoleonic vectors can be written in the form

We also have

which we note is simply the signed area of the triangle whose vertices are (x₁,y₁), (x₂,y₂), and (x₃,y₃). (See our note on Net Area and Green's Theorem.) Of course, in the present case the components of X and Y are allowed to be complex, but we can still formally regard a pair of complex numbers as defining a point in the C² "plane", and define the "area" accordingly. Denoting this area by A_xy, we have

This shows that the "area" of the complex triangle defined by two given vectors plays a role analogous to that of the state vector (or wave function) in the quantum dynamical commutator relations. To carry this through, we could define X as an operator which, when applied to an "area" such as A_xy gives the Napoleonic vector (x₁,x₂,x₃), and similarly for Y. These operators, together with the definitions of multiplication and addition for Napoleonic vectors, would then enable us to write

Return to MathPages Main Menu