Intuitive Argument for the Volume of the k-Dimensional Simplex
There is a simple and elegant proof to determine the volume of a (k-1)-dimensional simplex with one vertex at the origin and the remaining vertices each having one coordinate 1 and the remaining coordinates 0 in k-dimensions. This topic has been discussed in depth on the renowned online mathematics community, MathOverflow, under the heading Volume of an n-simplex. While initially complex, the argument unravels into a clear and intuitive understanding.
Understanding the k-Dimensional Simplex
A k-dimensional simplex is a geometric object that is the generalization of a triangle in 2-dimensions and a tetrahedron in 3-dimensions. Given a vertex at the origin of a k-dimensional space, the other vertices are constructed by setting one coordinate to 1 and the rest to 0. This means we have k vertices in total, each representing a standard unit vector in the k-dimensional space.
The Intuitive Argument
Let us consider the k-dimensional simplex (S_k) defined in (mathbb{R}^k) with vertices given by the origin ((0, 0, ldots, 0)) and the standard unit vectors ((1, 0, ldots, 0), (0, 1, ldots, 0), ldots, (0, 0, ldots, 1)).
The volume (V_k) of this simplex can be calculated using a simpler, more intuitive argument based on the concept of permutations.
Step-by-Step Reasoning
Consider the k-dimensional hypercube (a generalization of a square in 2D or a cube in 3D) with each side of length 1. The volume of this hypercube is clearly 1, as it is the product of its side lengths, all of which are 1.
Now, the (k-1)-dimensional simplex (S_k) can be seen as a "stack" of (k-1)-dimensional hyperplanes, where each hyperplane is a simplex of one dimension less. This stacking process can be thought of as moving from one vertex of the hypercube to another along a path that only moves to adjacent vertices in each dimension.
The number of such paths (or "lattices") from the origin to the opposite vertex is simply the number of ways to arrange k-1 ones in a sequence of length k, which is given by the binomial coefficient (binom{k}{k-1} k).
However, each of these paths corresponds to a different (k-1)-dimensional simplex (or facet) in the k-dimensional space. The volume of these facets is uniformly distributed, and the total volume of all such simplified simplices is the volume of the (k-1)-dimensional hypercube, which is 1. Since there are (k) such simplices, the volume of each (k-1)-dimensional simplex is (1/k).
Volume of the k-Dimensional Simplex
The volume of the (k-1)-dimensional simplex (S_k) is not (frac{1}{k}) but rather (frac{1}{k!}). This is because, as we move from the origin to the opposite vertex, we must consider the permutations of the coordinate axes. Each permutation of the coordinates contributes to the volume, and there are (k!) such permutations. Therefore, the volume is given by:
[frac{1}{k!}]Conclusion
The volume of the (k-1)-dimensional simplex with one vertex at the origin and the remaining vertices each having one coordinate 1 and the remaining coordinates 0 in k-dimensions is (frac{1}{k!}). This result showcases a beautiful symmetry and provides a deep connection between combinatorics and geometry.
As a concluding remark, this elegant proof not only computes the volume but also reinforces the fundamental idea that the volume of a simplex is a measure of the number of ways to order its vertices.