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Let $E$ be a real normed space, and let $v_1,...,v_n\in E$ be linearly independent. The parallelepiped defined by these vectors is $P=\{\sum_{i=1}^{n}\alpha_i v_i|~0\le\alpha_i\le 1\}$. Since $E$ is a metric space, we can define the $n$-dimensional Hausdorff measure $\mu_n$ on it.

Is there a method to calculate $\mu_n(P)$ if we know "everything" about $v_1,...,v_n$ and $E$?

If the norm of $E$ is induced by an inner product, then the volume can be calculated by taking the square root of the Gram matrix of $v_1,...,v_n$. I am hoping there is something of the kind if $E$ is a concrete space of functions or sequences. Since the $(i,j)$ entry of the Gram matrix is $<v_i,v_j>=\frac{1}{2}\left(\|v_i+v_j\|^2-\|v_i\|^2-\|v_j\|^2\right)$, maybe one can get $\mu_n(P)$ from the distances inside the parallelepiped somehow (also Cayley-Menger determinant comes to mind).

Remark. If we identify $\mathbb{R}^n$ with $span \{v_1,...,v_n\}$ via a linear isomorphism, the pull-back of $\mu_{n}$ will be an invariant measure, and so a multiple of the usual Lebesgue measure. The only unknown thing is the constant factor.

Remark 2. As I understand Liviu Nicolaescu's suggestion in comments, let $\varphi:\mathbb{R}^n\to E$ be the linear map that sends every standard base element $e_k$ into $v_k$. Let $\hat{\mu}_n$ be the pull-back of $\mu_n$ with respect to $\varphi$. Let $B=\varphi^{-1}(B_{E})$, and let $\Omega_n$ be the volume of the standard Euclidean ball in $\mathbb{R}^n$. Then, $\hat{\mu}_n(B)=\Omega_n$ (as was proved by Busemann$^1$), and so the constant factor from the first remark can be obtained by comparing $\Omega_n$ with the actual volume of $B$. Moreover, since the pull-back of the parallelepiped is the unit cube, we get that $$\mu_n(P)=\hat{\mu}_n([0,1]^n)=\lambda_n([0,1]^n)\frac{\hat{\mu}_n(B)}{\lambda_n(B)}=1\cdot\frac{\hat{\mu}_n(B)}{\Omega_n}=\frac{\hat{\mu}_n(B)}{\Omega_n},$$ where $\lambda_n$ is the Lebesgue measure on $\mathbb{R}^n$. Thus, the question is reduced to the following:

Let $E$ be a real normed space, and let $v_1,...,v_n\in E$ be linearly independent. Let $\lambda_n$ be the Lebesgue measure on $\mathbb{R}^n$, and let $\varphi:\mathbb{R}^n\to E$ be the linear map with $\varphi(e_k)=v_k$, $k\in\overline{1,n}$. How to calculate $\lambda_n[\varphi^{-1}( B_{E})]$?

1 - see Theorem 3.7.5 in Minkowski Geometry by A.C.Thompson.

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  • $\begingroup$ Denote by $\bar{\mu}_n$ the standard Lebesgue measure on $\mathbb{R}^n$ and by $B$ the unit ball with respect to the norm $\Vert-\Vert$ and by $\Omega_n$ the Euclidean volume of the Euclidean unit ball in $\mathbb{R}^n$. Then $\mu_n(B)=\Omega_n$ so $\mu_n=c\bar{\mu}_n$, $c=\frac{\Omega_n}{\bar{\mu}_n(B)}$. $\endgroup$ Commented May 24, 2019 at 9:49
  • $\begingroup$ @LiviuNicolaescu but how knowing $v_1,...,v_n$ find $\overline{\mu}_n(B)$? Also, it is not obvious to me, why $\mu_n(B)=\Omega_n$.. $\endgroup$
    – erz
    Commented May 24, 2019 at 10:10
  • $\begingroup$ $B$ is exactly covered by single unit ball (in the norm $\Vert-\Vert$. So its Hausdorff measure is the Euclidean measure $\Omega_n$ of the Euclidean unit ball. Identifying the span of $\{v_1,\dotsc, v_n\}$ with $\mathbb{R}^n$ we identify the paralellepiped with the unit cube $C$ in $\mathbb{R}^n$ so that $\mu_n(C)=c\bar{\mu}_n(C)=c$. $\endgroup$ Commented May 24, 2019 at 12:09
  • $\begingroup$ I need to rethink the statement $\mu_n(B)=\Omega_n$. $\endgroup$ Commented May 24, 2019 at 12:18
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    $\begingroup$ Have a look at Theorem 3.23 in this paper library.msri.org/books/Book50/files/02AT.pdf $\endgroup$ Commented May 24, 2019 at 12:25

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