Simplexes, Technical Definitionn

Context

A simplex exists in the context of a topological vector space. This is rather abstract, but a simplex actually lives in a finite dimensional subset of our topological space, and such a subset is equivalent to euclidean space. So you may picture a simplex as a structure in real space, which may in turn be part of a larger topological vector space.

Geometrically Independent

Let a0 through an be a finite set of points in a topological vector space. As mentioned above, these points span a subspace that is homeomorphic to k dimensional euclidean space, for some k ≤ n+1. This set of points is geometrically independent if there is but one set of real coefficients c0 through cn such that the sum of ciai = 0, and the coefficients sum to 1.

After the points have been embedded inreal space, of dimension k, each ai is a vector of length k. The equation ∑ ciai = 0 represents k equations in parallel. To this we append the equation ∑ ci = 1. Thus there are k+1 equations in n+1 unknowns. Write this using matrix notation M*c = {0,0,0,…1}, where c is the list of coefficients, to be determined, and M is the matrix of k+1 equations. Thus M has k+1 rows and n+1 columns.

If k = n+1 then the first n+1 rows of M are linearly independent, and c is the 0 vector. The last equation, ∑ ci = 1, is not satisfied. Therefore k ≤ n.

Suppose k is less than n. The rank of M is at most n, and if there is one solution, there are many. This contradicts the definition of geometrically independent. Therefore k = n. The points a0 through an span a space of dimension n.

Barycentric Coordinates

The aforementioned matrix M is invertible. Start with a point x in n space, append a 1 to get a vector of length n+1, and apply M in reverse. The result is the unique representation of x relative to a0 through an, provided we continue to force the coefficients to sum to 1. The values of c0 through cn form the barycentric coordinates of x.

Normally, a point in n space is represented by n scalars, according to a fixed basis, but in this case x is represented by n+1 scalars, that happen to sum to 1.

Simplex

A simplex, defined by a set of geometrically independent points a0 through an, is the set of points x whose barycentric coordinates are all nonnegative. Let's illustrate with a triangle, floating in the first quadrant. The points a0 a1 and a2 form the vertices of the triangle. In a separate 3 dimensional space, picture the plane c0+c1+c2 = 1. Restrict attention to that region of the plane where the coordinates are nonnegative. In other words, keep only the portion of the plane that lies in the first octant. This is an equilateral triangle. Map this into R2 via c0a0 + c1a1 + c2a2. The image of our equilateral triangle in R3 becomes the simplex in R2.

The center of our equilateral triangle has coordinates 1/(n+1). The image of this point is the center of the simplex. In other words, add up the vectors a0 through an and divide by n+1.

Colinear Example

Don't assume any 3 points in R2 will do. Consider 1,0 ½,½ and 0,1. These points are colinear, and intuition suggests they do not form a simplex, even though the associated vectors (which include the two unit vectors) span the plane. Suppose they are geometrically independent. Write c0 + ½c1 = 0, and ½c1 + c2 = 0. Add these together to get c0+c1+c2 = 0. Yet this sum is suppose to equal 1. These points do not define a simplex.

Extending to the Entire Plane

Assuming a valid simplex in R2, let the function c0a0+c1a1+c2a2 operate across the entire domain, i.e. the plane defined by c0+c1+c2 = 1. The image is all of R2, as described earlier. In other words, every x comes from a unique set of barycentric coordinates. A plane maps onto a plane, and the map is injective and surjective. It is also continuous, hence a homeomorphism.

Transforming a Simplex

Let a0 through an define a simplex in an n dimensional space S, and let f be a linear transformation from S onto another n dimensional space T. Let f map a0 through an to b0 through bn. We will show b0 through bn forms a simplex in T. In fact, the same coefficients c0 through cn carry both sets of points to 0, while summing to 1. Suppose another set of coefficients d0 through dn satisfies ∑ dibi = 0, and ∑ di = 1. Let y be the sum over diai. Now f(y) = 0 in T, and since f is injective, y = 0. This forces di = ci, and the solution is unique. We have geometric independence, and a simplex in T.

Of course S could equal T, whence f is an automorphism.

Standard Simplex

The standard simplex is a good place to start - but is it a simplex? Let a0 be the origin, while a1 through an are the unit vectors. If the sum over ciai is to equal 0, then c1 through cn are all 0. Set c0 = 1 for the unique solution. Sure enough, the standard simplex really is a simplex.

After an appropriate linear transformation, the origin joins with any basis of Rn to build a simplex.

Interior

As shown above, the hyperplane defined by c0+c1+c2+…cn = 1 is mapped onto Rn. The function, in either direction, is accomplished by a matrix M, and is continuous. The two spaces are homeomorphic. We are merely relabeling the points. A closed set maps to a closed set, and the interior of a closed set maps to the interior of the image. With this in mind, let's talk about the inside of a simplex.

Let's illustrate with a triangle in the plane. Step back to the barycentric coordinates. The preimage of the simplex is an equilateral triangle bounded by the xy, xz, and yz planes. This triangle (interior and boundary together) has all coordinates nonnegative. The inside of this triangle has all coordinates positive. The entire triangle, and its interior, map to the simplex and its interior respectively.

Select an edge of this triangle and map it forward to one of the lines defined by our simplex. The interior of the simplex lies on one side of this line. Do this for all three lines and the interior of the simplex is a region entirely on one side of all three lines. The three lines cut the plane into 7 regions, but only one is bounded, and (between complete metric spaces) the continuous image of a closed bounded (compact) set has to be bounded. Therefore the interior of the simplex, defined by positive barycentric coordinates, is, well, the inside of the simplex.

The barycentric interior exists, i.e. it contains an open set, and the same holds for the interior of the simplex. We cannot have a flat simplex with no inside. The result would not be a simplex. We saw this earlier, when 3 colinear points in the plane were not geometrically independent.

In general, the points of an n-simplex cannot share a common hyperplane passing through n space. Such a shape has no interior, i.e. it contains no open sets. To be a simplex, the vertices must enclose a space with a finite volume.

Conversely, assume n+1 points enclose a proper space in n dimensions. Apply a linear transformation that moves n of these points onto the standard basis. The remaining point, call it a0, is still out of plane, hence its coordinates sum to something other than 1. (If a0 were the origin, we would have the standard basis.) Assign the value z to c0 and solve for c1 through cn in turn. Each ci is the opposite of z times the ith coordinate of a0. Now drop to the last equation, which constrains the sum of ci. This is z times 1 minus the coordinates of a0. The coordinates of a0 sum to something other than 1, and when this is subtracted from 1, the result is something nonzero. Set z to the reciprocal of this value to produce the one and only solution. This satisfies geometric independence, and the points form a simplex.

If a0 is "in plane", its coordinates sum to 1, and we have 0×z = 1, which is impossible. This reaffirms the fact that our simplex cannot be flat.

In summary, n+1 points form a simplex iff they cannot be contained in a hyperplane of dimension n-1.

Like a Ball

Given a simplex, apply a linear transformation, so that the simplex consists of the unit vectors and an off point a0. Slide a0, parallel to the opposite face, until it is on the main diagonal, i.e. all its coordinates are equal. Then move a0 towards or away from the opposite face, until it forms a symmetric simplex. Finally, expand the simplex from the center, like blowing up a balloon. The result is a closed ball in n space. Each step is a homotopy, and a homeomorphism, hence the n simplex is homotopic to, and homeomorphic to, the n ball.

Using the same transformations, the boundary of the simplex is homotopic to, and homeomorphic to, the boundary of the ball, which is the sphere in n-1 dimensions.

Convex

If a convex shape contains x and y, it contains the entire line segment joining x and y. That's the definition of convex.

Let x and y belong to a simplex, and pull back to the barycentric coordinates, which are all nonnegative. Move linearly from x to y by adjusting (linearly) the barycentric coordinates. They remain nonnegative at all times, hence the entire segment lies within the simplex, and the simplex is convex.