Polygons Again
Jul. 31st, 2012 02:06 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
stoutfellowI want to make another run at explaining my research on polygons. What's under the cut isn't dependent on the stuff I posted before. Be warned: there will be vectors and matrices.
For simplicity, I'm going to talk specifically about quadrilaterals here. I'll point out the necessary modifications for polygons with other numbers of sides as they come up.
First: what is a quadrilateral? We usually describe one by giving its vertices A,B,C,D, in order. The thing to notice is that there's a "next vertex" concept: the next vertex after A is B, after B is C, after C is D, and - and this is the important bit - after D is A again. Because "next vertex" cycles back to the beginning after four steps, it's natural (to a mathematician) to think of the vertices as indexed by the integers mod 4, {0,1,2,3}. "Next vertex" means "add 1 to the index"; mod 4, 3+1=0, so the next vertex after #3 is #0 again.
Okay, so what are these vertices? They're points in the Euclidean plane; but we've known since the time of Descartes and Fermat that we can think of a point as a pair of real numbers. Even better, thanks to Messrs. Wessel, Argand, and Gauss, we can think of it as a single complex number. So a quadrilateral is given by four complex numbers, in order, representing its vertices. For instance, the quadruple (0, 2, 2+i, i) describes a rectangle with length 2 and height 1.
Once we've done this, it's natural (again, to a mathematician) to think of vectors: a quadrilateral can be thought of as a vector in complex 4-space. This means, in particular, that we can multiply a quadrilateral by a complex number z; this has the effect of scaling the quadrilateral by the modulus |z| and rotating it by the angle Arg(z). We can also, less intuitively, add two quadrilaterals, which will be significant in a little bit.
Since we're using vectors to describe something geometric, it's important to ask: are we using the best basis? That is, given what we're trying to look at, is there a better coordinate system we can use? If Q=(a,b,c,d), we can write Q=a(1,0,0,0)+b(0,1,0,0)+c(0,0,1,0)+d(0,0,0,1), using the four vectors on the right as our basis. But what quadrilateral does, say, (1,0,0,0) represent? Its zeroth vertex is at 1, and the other three at 0. For (0,1,0,0), the zeroth vertex is 0, the first 1, and the other two 0; and so on. These are not very interesting or appealing quadrilaterals, and I'm going to suggest a different set of four to serve as our basis.
Since we're working with four-sided figures, we're going to do something involving the complex numbers whose fourth power is 1: that is, 1, i, -1, -i. These are, respectively, the zeroth, first, second, and third powers of i. (Note, again, that the fourth power of i brings us back to 1.) So, for j=0,1,2,3, we'll define the quadrilateral Qj to be given by the powers of ij. So Q0 is given by the quadruple (1,1,1,1) - i.e., all four vertices coincide at 1; Q1 has vertices (1,i,-1,-i) - a square, oriented counterclockwise; Q2 is given by (1,-1,1,-1) - its vertices alternate between 1 and -1; and finally Q3 has vertices (1,-i,-1,i) - it's the same square as Q1, but oriented clockwise.
Let's look at these four. Q0 is the prototype of punctual quadrilaterals, with all vertices coinciding. The other three quadrilaterals are all centered at 0; so, if Q=aQ0+bQ1+cQ2+dQ3, it is the coefficient a that determines where the center of Q is. That's the only role played by Q0. What does Q2 contribute? Well, the midpoints of the diagonals of the other three are each located at the center of the quadrilateral, which is not true for Q2; therefore, it's the coefficient c that determines where the midpoints of the diagonals are, relative to the center of the quadrilateral. Note that, for these two, reversing the order of the vertices has no effect. The two squares, Q1 and Q3, are interchanged by this reversal. We'll consider them together; it turns out that any sum of multiples of these squares is a parallelogram.
In other words, we can write any quadrilateral as z0Q0+z1Q1+z2Q2+z3Q3. The coefficient z0 identifies the center of the quadrilateral; z1 and z3 describe a parallelogram with its center at 0, and z2 describes the "perturbation". (Take the quadrilateral described by z1 and z3, and then shift vertices 0 and 2 by z2 and vertices 1 and 3 by -z2. Then move the whole thing bodily by z0.)
Why is this basis better? That'll be the next topic.
For simplicity, I'm going to talk specifically about quadrilaterals here. I'll point out the necessary modifications for polygons with other numbers of sides as they come up.
First: what is a quadrilateral? We usually describe one by giving its vertices A,B,C,D, in order. The thing to notice is that there's a "next vertex" concept: the next vertex after A is B, after B is C, after C is D, and - and this is the important bit - after D is A again. Because "next vertex" cycles back to the beginning after four steps, it's natural (to a mathematician) to think of the vertices as indexed by the integers mod 4, {0,1,2,3}. "Next vertex" means "add 1 to the index"; mod 4, 3+1=0, so the next vertex after #3 is #0 again.
Okay, so what are these vertices? They're points in the Euclidean plane; but we've known since the time of Descartes and Fermat that we can think of a point as a pair of real numbers. Even better, thanks to Messrs. Wessel, Argand, and Gauss, we can think of it as a single complex number. So a quadrilateral is given by four complex numbers, in order, representing its vertices. For instance, the quadruple (0, 2, 2+i, i) describes a rectangle with length 2 and height 1.
If we were looking at pentagons, we'd index the vertices using the integers mod 5, and end up with a quintuple of complex numbers; and similarly for any number of vertices.
Once we've done this, it's natural (again, to a mathematician) to think of vectors: a quadrilateral can be thought of as a vector in complex 4-space. This means, in particular, that we can multiply a quadrilateral by a complex number z; this has the effect of scaling the quadrilateral by the modulus |z| and rotating it by the angle Arg(z). We can also, less intuitively, add two quadrilaterals, which will be significant in a little bit.
Since we're using vectors to describe something geometric, it's important to ask: are we using the best basis? That is, given what we're trying to look at, is there a better coordinate system we can use? If Q=(a,b,c,d), we can write Q=a(1,0,0,0)+b(0,1,0,0)+c(0,0,1,0)+d(0,0,0,1), using the four vectors on the right as our basis. But what quadrilateral does, say, (1,0,0,0) represent? Its zeroth vertex is at 1, and the other three at 0. For (0,1,0,0), the zeroth vertex is 0, the first 1, and the other two 0; and so on. These are not very interesting or appealing quadrilaterals, and I'm going to suggest a different set of four to serve as our basis.
Since we're working with four-sided figures, we're going to do something involving the complex numbers whose fourth power is 1: that is, 1, i, -1, -i. These are, respectively, the zeroth, first, second, and third powers of i. (Note, again, that the fourth power of i brings us back to 1.) So, for j=0,1,2,3, we'll define the quadrilateral Qj to be given by the powers of ij. So Q0 is given by the quadruple (1,1,1,1) - i.e., all four vertices coincide at 1; Q1 has vertices (1,i,-1,-i) - a square, oriented counterclockwise; Q2 is given by (1,-1,1,-1) - its vertices alternate between 1 and -1; and finally Q3 has vertices (1,-i,-1,i) - it's the same square as Q1, but oriented clockwise.
For pentagons, we'd rely on the complex numbers with fifth power 1, i.e. exp(2 Pi j/5 with j=0,1,2,3,4.
Let's look at these four. Q0 is the prototype of punctual quadrilaterals, with all vertices coinciding. The other three quadrilaterals are all centered at 0; so, if Q=aQ0+bQ1+cQ2+dQ3, it is the coefficient a that determines where the center of Q is. That's the only role played by Q0. What does Q2 contribute? Well, the midpoints of the diagonals of the other three are each located at the center of the quadrilateral, which is not true for Q2; therefore, it's the coefficient c that determines where the midpoints of the diagonals are, relative to the center of the quadrilateral. Note that, for these two, reversing the order of the vertices has no effect. The two squares, Q1 and Q3, are interchanged by this reversal. We'll consider them together; it turns out that any sum of multiples of these squares is a parallelogram.
In other words, we can write any quadrilateral as z0Q0+z1Q1+z2Q2+z3Q3. The coefficient z0 identifies the center of the quadrilateral; z1 and z3 describe a parallelogram with its center at 0, and z2 describes the "perturbation". (Take the quadrilateral described by z1 and z3, and then shift vertices 0 and 2 by z2 and vertices 1 and 3 by -z2. Then move the whole thing bodily by z0.)
Why is this basis better? That'll be the next topic.
