1
$\begingroup$

I asked this in MSE, it flashed and disappeared.

Let $V_n$ be the volume on the set of polytopes in $\mathbb R^n$, defined axiomatically, i.e. a functional, that assigns to each polytope $P\subseteq\mathbb R^n$ a real number $V_n(P)\ge 0$ in such a way that the following conditions are fulfilled:

  1. For the unit hypercube $C\subseteq\mathbb R^n$ $$ V_n(C)=1, $$

  2. If $P\cap Q=\varnothing$, then $$ V_n(P\cup Q)=V_n(P)+V_n(Q), $$

  3. If $P$ is made from $Q$ by a motion, then $$ V_n(P)=V_n(Q). $$ Question:

    Is there a simple way to prove that the $n$-th volume of a parallelepiped $P_{a_1,...,a_n}$ spanned by $n$ vectors $a_1,...,a_n$ is the poduct of the $n-1$-th volume of $P_{a_1,...,a_{n-1}}$ and the height of $P_{a_1,...,a_n}$ with respect to the base $P_{a_1,...,a_{n-1}}$, $$ V_n(P_{a_1,...,a_n})=V_{n-1}(P_{a_1,...,a_{n-1}})\cdot \left|\text{pr}_{\{a_1,...,a_{n-1}\}^\perp}(a_n)\right| $$ ?

(Here $\text{pr}_{\{a_1,...,a_{n-1}\}^\perp}(a_n)$ is the orthogonal projection of $a_n$ to the orthogonal complement $\{a_1,...,a_{n-1}\}^\perp$ of $\{a_1,...,a_{n-1}\}$, i.e. the height of $P_{a_1,...,a_n}$.)

P.S. This is not for math reasearch, this is for teaching.

Improve this question
$\endgroup$
10
  • 1
    $\begingroup$ What does flashed and disappeared mean... ? $\endgroup$
    – Zev Chonoles
    Jun 20, 2015 at 7:31
  • $\begingroup$ Perhaps, this is my bad English. What is it called when something appears and disappears immediately? $\endgroup$
    – Sergei Akbarov
    Jun 20, 2015 at 7:34
  • $\begingroup$ Well, that's indeed what I thought you meant, but my confusion is that your math.SE question is still visible to me. (If you felt it didn't get enough attention on math.SE, editing-to-bump or a bounty are the standard approaches.) $\endgroup$
    – Zev Chonoles
    Jun 20, 2015 at 7:38
  • $\begingroup$ I meant, disappeared from the list of the questions on the main page. I thought, this means that nobody is interested. $\endgroup$
    – Sergei Akbarov
    Jun 20, 2015 at 7:43
  • 1
    $\begingroup$ It seems like this should be straightforward by breaking the parallelopipeds into tiny congruent pieces and then using these pieces to approximately tile each other. $\endgroup$
    – Eric Wofsey
    Jun 20, 2015 at 7:43

1 Answer 1

Reset to default
1
$\begingroup$

If you replace your 2nd axiom by the more natural one, namely the inclusion-exclusion principle,

$$\mu(P\cup Q)=\mu(P)+\mu(Q)-\mu(P\cap Q), $$

then a lot can be said. (Functions on the set of polytopes satisfying the above conditions are called valuations.) I will assume this holds the sequel.

$\newcommand{\Pix}{Pix}$ Denote by $\Pix(n)$ the set of pixelations of $\newcommand{\bR}{\mathbb{R}}$ $\bR^n$, i.e., sets that are unions of paralelipipeds with faces parallel to the coordinate face.

Theorem 2.19 from these very nice notes imply that for any box $B=[a_1,b_1]\times \cdots \times [a_n,b_n]$ we have

$$\mu(B)=\prod_k(b_k-a_k). $$

If you replace the nonnegativity assumption with a stronger continuity assumption, then the result is a special case of Hadwiger's characterization theorem; see Chapter 4 of the same notes.

Improve this answer
$\endgroup$
2
  • 4
    $\begingroup$ Aren't the two axioms equivalent? One can extend (2) to three disjoint sets and then apply it to $P\cap Q$, $P\setminus Q$ and $Q\setminus P$ $\endgroup$
    – Federico Poloni
    Jun 20, 2015 at 12:43
  • 1
    $\begingroup$ Federico, as far as I understand, the problem with the equivalence of these two axioms arises when we define polytopes as compact sets (spanned by... etc), then $P\setminus Q$ needs not to be a polytope, and the functional $V_n$ is not defined on $P\setminus Q$. $\endgroup$
    – Sergei Akbarov
    Jun 20, 2015 at 19:55

Not the answer you're looking for? Browse other questions tagged or ask your own question.