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The category of (pseudo-)Euclidean vector spaces (vector spaces with a nondegenerate but not necessarily positive-definite quadratic form) is not closed under products because $R^n$ over $R$ and $Z_2^m$ over $Z_2$ are vector spaces, but their module product $R^n \times Z_2^m$ over the ring $R \times Z_2$ is not because the scalar ring is not a field. Similar structures, including related products like the module $R^n \times Z_2^m$ over $R^n \times Z_2^m$, are useful in studying hybrid automata and hybrid dynamical systems.

[Note added in response to McLaury comment: W. Noll (Euclidean geometry and Minkowskian chronometry, American Mathematical Monthly, V. 71, pp. 129-144, 1964, see pp. 134-136, especially Remark 1) observes that the underlying axioms and uniqueness theorem for inner product spaces, along with the representation they induce for automorphisms as rigid displacements, remain valid for commutative rings of characteristic other than 2, attributing these observations to Bourbaki.]

1. How little must one weaken the notion of pseudo-Euclidean vector space to obtain a structure that has the properties central to the use of vector spaces in classical and relativistic mechanics, at minimum including a well-defined notion of dimension or rank and supporting the decomposition of isometries into translations and orthogonal transformations, yet is preserved under taking of products?

Call structures satisfying these conditions "vectal" spaces.

2. Is there an established name for this type of structure?

Free modules over commutative rings with nondegenerate symmetric bilinear forms compatible with pseudo-Euclidean metrics (FMOCRNSBFCPEM for short) have the desired vector-like properties and form a category closed under products.

3. Are vectal spaces correctly characterized as FMOCRNSBFCPEMs?

4. Is every FMOCRNSBFCPEM isomorphic to a product of pseudo-Euclidean vector spaces?

5. Is every vectal space isomorphic to a product of pseudo-Euclidean vector spaces?

6. [A simpler question added in response to Lspice and McLaury comments] Are there answers to these problems if one drops the pseudo-Euclidean requirement, but keeping the dimensionality/rank requirement? Free modules over commutative rings?

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  • $\begingroup$ What is a pseudo-Euclidean space? What is a mechnical property? Aside from the fact that the ground ring $\mathbb R \times \mathbb F_2$ is not a field, what properties is $\mathbb R^n \times \mathbb F_2^m$ missing that $\mathbb R^n$ and $\mathbb F_2^m$ individually both have? Would you expect $\mathbb R^2$ as an $\mathbb R^2$-module to be an example of your kind of structure? Why is PEFMOCR and not FMOCRNSBFCPEM the abbreviation for your property? $\endgroup$
    – LSpice
    Commented May 10, 2019 at 15:27
  • $\begingroup$ 1. According to the definitions I see online, the field of scalars of a pseudo Euclidean vector space is taken to be the reals by definition. 2. What to you mean by a pseudo Euclidean metric here other than the nondegenerate bilinear form itself? $\endgroup$
    – Daniel McLaury
    Commented May 13, 2019 at 12:47
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    $\begingroup$ $\mathbb R^m \times \mathbb Z_2^n$ is not a free $(\mathbb R \times \mathbb Z_2)$-module unless $m = n$. $\endgroup$
    – lambda
    Commented May 13, 2019 at 13:58
  • $\begingroup$ I edited the question to clarify the points raised by Lspice and McLaury, and to provide some additional information. $\endgroup$
    – Jon Doyle
    Commented May 13, 2019 at 14:47
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    $\begingroup$ @JonDoyle No, it's not a free module. Your family of vectors is not linearly independent. For example $(1,0) \cdot (0,0,1) = (0,0,0)$. (Also, as all French people I love acronyms, but FMOCRNSBFCPEM is just too much.) $\endgroup$
    – Najib Idrissi
    Commented May 13, 2019 at 14:56

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