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Let $(E,\|.\|)$ be a finite dimensional normed space and take $F\subset E$ a subpace, so that we have the canonical short exact sequence $0\rightarrow F\rightarrow^\iota E\rightarrow^\pi E/F\rightarrow 0$ of normed spaces ($F$ being endowed by the restriction of $\|.\|$ and $E/F$ by the quotient norm $\|v+F\| = \inf_{f\in F} \|v-f\|$). Write $n=\dim E$ and $k = \dim F$.

We can define the Hausdorff measure on each of these spaces (say $\mu_F, \mu_E$ and $\mu_{E/F}$), deriving from each of their norms. Is this definition compatible with the short exact sequence, in the sense that for a $k$-vector $\mathfrak{f} = f_1\wedge \cdots\wedge f_k\in\bigwedge^k E$, and $\mathfrak{e} = e_{k+1}\wedge\cdots\wedge e_n\in \bigwedge^{n-k} E$, we have: $$ \mu_E(\mathfrak{f} \wedge \mathfrak{e}) = \mu_F(\mathfrak{f})\mu_{E/F}(\pi(\mathfrak{e}\wedge\mathfrak{f})). $$

Remark that this is the case when $\|.\|$ is Euclidean, as a direct consequence of the definition of the norm on the exterior algebra.

Another way of looking at the question is whether the comeasure induced by the Hausdorff measure coincidates with the measure induced by the quotient norm.

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    $\begingroup$ No it is not. You can find counterexamples in dimension two already. I think the max norm on the plane and almost any $1$-dimensional subspace will be a counterexample. I'll come back later when I have more time. $\endgroup$ Commented Mar 30, 2020 at 12:19
  • $\begingroup$ And what about with a proper renormalization of the measure, let say by a factor A_k depending only the rank k ? $\endgroup$
    – user70925
    Commented Apr 1, 2020 at 23:37
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    $\begingroup$ No that won't work either. Volumes in normed spaces are tricky things. See my paper with Thompson on A Sampler of Riemann Finsler Geometry: library.msri.org/books/Book50/contents.html $\endgroup$ Commented Apr 2, 2020 at 19:38

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