Simplicial Generalization of Pythagoras

Question

I recently heard about a claim that

For a triangle in 3-space, its area squared equals the sum of squares of areas of its projections onto three pairwise orthogonal planes.

I currently don't have any proof for that at hand. However, if it were true, in combination with the well-known Pythagoras theorem, where the line segments simply are to be considered as 1D simplices, it seems that it might become extrapolatable for higher dimensional simplices too:

Then it ought state like this:

For an nD simplex within (n+1)D space, its nD hyper-volume squared equals the sum of the n+1 squares of the nD hyper-volumes of its projections onto the (mutually orthogonal) coordinate hyperplanes.

If that theorem could be proven in general, then Pythagoras would simply become the case n=1 and the heard of claim would just be case n=2.

Thus, any proof would be welcome.

--- rk

Answer 1

This is true. It's called de Gua's theorem (from 1783!), and the higher-dimensional generalization is due to Conant and Beyer (from 1974!). The cleanest proof I can think of involves using either exterior algebra or geometric algebra; they are basically equivalent but I'll stick to the exterior algebra since I think it is a little more familiar.

We need the following facts. Let $V$ be a finite-dimensional real inner product space of dimension $n$ and let $e_1, \dots e_n$ be any orthonormal basis of it.

1. From $V$ we can form another vector space called the $k$-th exterior power $\Lambda^k(V)$, which consists of sums of symbols of the form $v_1 \wedge v_2 \wedge \dots v_k$ (wedge products or exterior products) subject to the relations that $\wedge$ is associative, antisymmetric ($v \wedge w = - w \wedge v$, or equivalently $v \wedge v = 0$), and bilinear. This vector space has dimension ${n \choose k}$, and given a basis of $V$ as above, a basis of $\Lambda^k(V)$ is given by wedge products $e_{i_1} \wedge e_{i_2} \wedge \dots \wedge e_{i_k}$ of $k$ distinct elements of a basis of $V$ in increasing order.

2. $\Lambda^k(V)$ has an inner product induced from the inner product on $V$ given by the Gram determinant $(v_1 \wedge \dots v_k) \cdot (w_1 \wedge \dots w_k) = \det (v_i \cdot w_j)$. It has the property that if $e_i$ is an orthonormal basis of $V$ then the basis of wedge products as defined above is an orthonormal basis of $\Lambda^k(V)$.

3. This inner product induces a norm on $\Lambda^k(V)$, and the squared norm $\| v_1 \wedge \dots \wedge v_k \|^2 = \det(v_i \cdot v_j)$ is the squared volume of the oriented parallelepiped determined by $v_1, \dots v_k$; equivalently, it is $k!^2$ times the squared volume of the $k$-simplex with vertices $0, v_1, \dots v_k$.

4. If we expand a wedge product $v_1 \wedge \dots \wedge v_k = \sum c_{i_1, \dots i_k} e_{i_1} \wedge \dots \wedge e_{i_k}$ in terms of the orthonormal basis from above, then the coefficients $c_{i_1, \dots i_k}$ are the oriented volumes of the projection of the parallelepiped determined by $v_1, \dots v_k$ onto the coordinate hyperplane determined by $e_{i_1}, \dots e_{i_k}$.

These are all "well-known" although I wouldn't know where to direct you for proofs from scratch. The most interesting parts are the identification of all these formal calculations with concrete volumes. And with these the result is a simple calculation analogous to the calculation that proves the Pythagorean theorem in an inner product space: now we just have to write

$$v_1 \wedge \dots \wedge v_k = \sum c_{i_1, \dots i_k} e_{i_1} \wedge \dots \wedge e_{i_k}$$

and then take the squared Euclidean norm of both sides, arriving at

$$\| v_1 \wedge \dots \wedge v_k \|^2 = \sum c_{i_1, \dots i_k}^2.$$

These are the volumes of parallelepipeds but we can get the volumes of simplices by dividing both sides by $k!^2$. Of course this means the hard part of the proof is in the establishment of all the facts above, especially fact 4 (fact 3 is a corollary). Note that the statement you asked for is the case $k = n - 1$ but the argument works for any $k$.

Answer 2

Consider ${ \mathbb{R} ^n . }$

Consider a box

$${ B := p + \text{box}(v _1, \ldots, v _k) }$$

in ${ \mathbb{R} ^n . }$

Note that the ${ k }$ volume of the box is

$${ \text{vol} _k (B) = \sqrt{\det( P ^T P ) } }$$

where

$${ P := [v _1, \ldots, v _k ] . }$$

Note that by Cauchy Binet theorem

$${ {\begin{aligned} &\, \det(P ^T P) \\ = &\, \sum _{\substack{I \subseteq [n] \\ \vert I \vert = k}} \det((P ^T) _{[k], I} ) \det(P _{I, [k]}) \\ = &\, \sum _{\substack{I \subseteq [n] \\ \vert I \vert = k}} \det((P _{I, [k]}) ^T) \det(P _{I, [k]}) \\ = &\, \sum _{\substack{I \subseteq [n] \\ \vert I \vert = k}} \det(P _{I, [k]} ^T P _{I, [k]} ) . \end{aligned}} }$$

For ${ I \subseteq [n] }$ and ${ \vert I \vert = k , }$ let ${ \pi _I }$ denote the orthogonal projection from ${ \mathbb{R} ^n }$ onto ${ \text{span}\lbrace e _i : i \in I \rbrace . }$

Note that for ${ I \subseteq [n] }$ and ${ \vert I \vert = k , }$

$${ \text{vol} _k (\pi _I (B)) = \sqrt{\det(P _{I, [k]} ^T P _{I, [k]} )} . }$$

Hence

$${ \boxed{ \text{vol} _k (B) ^2 = \sum _{\substack{ I \subseteq [n] \\ \vert I \vert = k } } \text{vol} _k (\pi _I (B)) ^2 } . }$$

Since the above result is true for boxes, we have (Link to Wikipedia page: Link)

Obs: Let ${ U }$ be a measurable subset of a ${ k }$ dimensional affine subspace of ${ \mathbb{R} ^n . }$ Then

$${ \text{vol} _k (U) ^2 = \sum _{\substack{ I \subseteq [n] \\ \vert I \vert = k } } \text{vol} _k (\pi _I (U)) ^2 . }$$