Help proving that signed volume of n-parallelepiped is multilinear

Question

Overview

I am trying to build some intuition about the volumes of parallelepipeds and determinants. I would like to define the determinant as the unique function of $N$ vectors in $\mathbb{R}^N$ which is multilinear, anti-symmetric, and normalized and then show that the signed volume of the $N$-parallelepiped satisfies these conditions without using any facts about determinants. The challenging one here for me is multi-linearity. Furthermore, I would like the signed volume to be defined in terms of Lebesgue integrals.

Problem Setup

Let $v_1, \ldots, v_N \in \mathbb{R}^N$. A parallelepiped $P = P(v_1, \ldots, v_N) \subset \mathbb{R}^N$ is the set

$$ P = P(v_1, \ldots, v_N) = \left\{\sum_{i=1}^N t_i v_i \mid 0 \le t_i \le 1 \text{ for $i$ from 1 to $N$} \right\} $$

My goal is to define a function $\text{svol}(P) = \text{svol}(v_1,\ldots, v_N)$ which is the signed volume of the parallelepiped $P$. The volume $\text{vol}$ of the parallelepiped is defined as follows. Let $1_P$ be the indicator function on the set $P$.

$$ \text{vol}(P) = \text{vol}(v_1, \ldots, v_N) = \int_{\mathbb{R}^N} 1_P dV $$

Where the integral is the Lebesgue integral. We should have

$$ |\text{svol}(P)| = \text{vol}(P) $$

We must additionally define, or provide an algorithm, to set the sign of $\text{svol}(P)$.

It is apparently the case, once everything above has been set up correctly, that for $a, b \in \mathbb{R}$ and $w \in \mathbb{R}^N$

\begin{equation} \tag{*} \text{svol}(av_1 + bw, v_2, \ldots v_N) = a\cdot \text{svol}(v_1, v_2, \ldots, v_N) + b\cdot \text{svol}(w, v_2, \ldots v_N) \end{equation}

The Direct Questions

My questions are

• How should $\text{svol}(v_1,\ldots, v_N)$ be defined such that (1) it is defined in terms of an integral over $\mathbb{R}^N$ and (2) we are able to prove $(*)$.
• Given the definition that answers the previous question, how do we prove $(*)$?

An Illustrative Image

The answer to this question includes a beautiful illustration the depicts exactly what I am trying to prove. It is intuitively clear to me from this diagram and Cavalieri's principle that $(*)$ should hold. But of course, for me, the illustration does not constitute a fully rigorous proof following the conditions I laid out above.

An Almost Solution

The closest I have come to a satisfactory proof is as follows. It is possible to define the volume of the $n$-parallelepiped inductively. We say the volume of the 1-parallelepiped $\overline{\text{vol}}(P) = ||v||$. The volume of the $n$-parallelepiped is then defined by $\overline{\text{vol}}_N(v_1, \ldots, v_N) = \overline{\text{vol}}_{N-1}(v_1, \ldots, v_{N-1}) \cdot ||v_{N, \perp}||$ where $v_{N, \perp}$ is the component of $v_N$ which is orthogonal to the span of $\{v_1, \ldots, v_{N-1}\}$. This component may be called the $N^{\text{th}}$ altitude of the parallelepiped. Under this definition the proof follows because, from $(*)$, $||v_{1, \perp}|| + ||w_{\perp}|| = ||(v_1 + w)_{\perp}||$. This proof is essentially a conversion of the image above into a more rigorous definition and proof. I think the quick proof I've given here lacks something defining the sign of $\overline{\text{vol}_N}$.

The problems with this proof are the lack of control of the sign of the volume and that the definition of $\overline{\text{vol}}_N$ is not based on an integral. One appropriate answer to the question I am asking in this thread would be a way to control the sign of $\overline{\text{vol}_N}$ and relate $\overline{\text{vol}_N}$ to the integration based definition of volume. This would essentially be an equation relating $\text{svol}$ and $\overline{\text{vol}_N}$.

Comment About Sign

I realize that to define $\text{sgn}(\text{svol}(v_1,\ldots, v_N))$ that we must make some convention choice. This convention choice should be consistent with $\text{sgn}(\text{svol}(e_1,\ldots,e_N)) = +1$ where $\{e_1, \ldots, e_N\}$ is the standard basis for $\mathbb{R}^N$. Beyond that, I'm not sure how to define the sign in a way that does not use any facts about determinants. I don't know if there's a way to determine if two sets of $N$ vectors, $\{v_1, \ldots, v_N\}$ and $\{w_1, \ldots, w_N\}$ have the same orientation other than computing the sign of the determinant. If someone has an answer here I would appreciate it. If not I would be comfortable with the following.

Let

$$ D(v_1, \ldots, v_N) = \sum_{\sigma \in S_N} \text{sgn}(\sigma) \prod_{i=1}^N v_{i, \sigma(i)} $$

where $S_N$ is the symmetric group of size $N$ and if $j=\sigma(i)$ then $v_{i, j}$ indicates the $j^{\text{th}}$ component of vector $v_i$. Then let

$$ \text{sgn}(\text{svol}(v_1, \ldots, v_N)) = \text{sgn}(D(v_1, \ldots, v_N)) $$

so that

$$ \text{svol}(v_1, \ldots, v_N) = \text{sgn}(D(v_1, \ldots, v_N)) \cdot \text{vol}(v_1, \ldots, v_N) $$

This last section essentially serves as an answer to my first question above. The question of how to prove $(*)$ from this definition still remains.

Answer 1

DISCLAIMER: Partial answer, to be continued. I'm posting it because I wrote down the formulas and want them to be saved.

After a lot of work I was not able to come up with a 100% answer to my question, but I was able to learn a lot. This question was motivated by trying to understand the relationship between volumes and alternating multilinear forms. What I've come to understand, however, is that (1) volumes are not directly related to alternating multilinear forms per-se. (2) Rather,

In particular, I think it is not possible to characterize the signed volume of a parallelepiped without reference to determinants

Let $v_1, \ldots, v_N \in \mathbb{R}^N$ be linearly independent and

$$ P_N = P(v_1, \ldots, v_N) = \left\{\sum_{i=1}^N t_i v_i : 0 \le t_1, \ldots, t_N \le 1\right\} $$

Perform a Gram-Schmidt orthogonalization on $\{v_1, \ldots, v_N\}$ to get an orthonormal basis $\{e_1, \ldots, e_N\}$. These two bases are related by

$$ v_i = \sum_{j=1}^N e_j \langle e_j, v_i\rangle = \sum_{j=1}^i e_j\langle e_j, v_i\rangle $$

The summation in the last expression is truncated because the matrix relating the two bases that arises from the Gram-Schmidt procedure is upper triangular. The components of this matrix $U$ are given by

$$ U_{ji} = \langle e_j, v_i\rangle $$

We also define $R = U^{-1}$.

We rewrite

$$ \sum_{i=1}^N t_i v_i = \sum_{i=1}^N\sum_{j=1}^i t_i e_j U_{ji} = \sum_{i=1}^N\sum_{j=i}^N t_j e_i U_{ij} $$

Let

$$ x_i = \sum_{j=i}^N U_{ij} t_j = \sum_{j=1}^N U_{ij} t_j $$

If we define vectors $X$ and $T$ with components $x_i$ and $t_i$ then

\begin{align} X =& UT\\ T =& RX \end{align}

We get

$$ t_j = \sum_{k=j}^N R_{jk} x_k $$

We then manipulate

$$ \sum_{i=1}^N e_i \sum_{j=i}^N U_{ij} t_j = \sum_{i=1}^N e_i \left(U_{ii} t_i + \sum_{j=i+1}^N \sum_{k=j}^N U_{ij}R_{jk} x_k \right) $$

We can write the parallelepiped as

$$ P_N = \left\{\sum_{i=1}^N e_i \left(U_{ii} t_i + \sum_{j=i+1}^N \sum_{k=j}^N U_{ij}R_{jk} x_k \right): 0\le t_1, \ldots, t_n \le 1 \right\} $$

The key intuition here is that the "width" of the parallelepiped in the direction $e_i$ is always given by $U_{ii} = \langle e_i, v_i\rangle = ||v_i^{\perp}||$, even though that slice may be offset by a distance given by the second term depending on all of the $x_k$ with $k>i$. The translation invariance of the integral will allow us to neglect this offset in each dimension. with $x_k$ defined above. We are now ready to perform an integration to find the volume of this set.

$$ \text{vol}(P_N) = \int_{x_N=0}^{||v_N^{\perp}||}\ldots\int_{x_1=\sum_{j=2}^N\sum_{k=j}^NU_{ij}R_{jk}x_k}^{||v_1^{\perp}|| + \sum_{j=2}^N\sum_{k=j}^NU_{ij}R_{jk}x_k} 1 dx_1\ldots dx_N $$

Because the offset on the bounds in each integral only depend on coordinates integrated over in integrals with are "further out" in the nesting, we can always shift the integration bounds by that shift:

$$ \text{vol}(P_N) = \int_{x_N=0}^{||v_N^{\perp}||}\ldots\int_{x_1=0}^{||v_1^{\perp}||} 1 dx_1\ldots dx_N = ||v_1^{\perp}||\ldots ||v_N^{\perp}|| $$