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I just read Terence Tao's blog post about mathematical formulations of physical units (also referenced in this SE post). Toward the end, he introduces three one-dimensional real vector spaces $V^M$, $V^L$, and $V^T$, whose elements represent units of mass, length, and time, respectively. Other physical quantities are then represented as tensor products of these three spaces and their duals. E.g., velocities are elements of the space $V^{L^1T^{-1}} = V^L \otimes (V^T)^*$. Under this scheme, a "choice of physical units" amounts to choosing a basis (i.e., a distinguished element) for each of $V^M$, $V^L$, and $V^T$.

This approach successfully describes the space of physical quantities without reference to a distinguished choice of base unit sizes (e.g., meter vs. yard for length). But one downside is that it still seems to single out a distinguished choice of physical quantities as base units, namely mass, length, and time. One might argue that such a choice is undesirable, especially since different systems of units single out different fundamental base units. (For example, one could regard "natural units" used in particle physics as taking energy, action, and velocity to be fundamental.$^*$)

Now, given the set $\mathcal S$ of tensor powers of $V^{L,M,T}$ and their duals, one could also consider (e.g.) the three spaces $W^E = V^{ML^2T^{-2}}$, $W^A = V^{ML^2T^{-1}}$, and $W^V = V^{LT^{-1}}$ (these spaces correspond to energy, action, and velocity), and then express $\mathcal S$ as instead being isomorphic to the set of tensor powers of $W^{E,A,V}$ and their duals. So it seems that this framework still offers the possibility of describing $\mathcal S$ as some sort of "graded set of vector spaces on three generators," without reference to which particular generators. How can this idea be made precise?

$^*$Physicists think of these units as "setting $\hbar = c=1$," but this can also be thought of as expressing quantities in base units of MeV (for energy), $\hbar$ (for action), and $c$ (for velocity), and then suppressing the powers of $\hbar$ and $c$ for convenience.

J. W. Tanner
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WillG
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1 Answers1

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There's already no need to decide which physical quantities are the basic ones. The key point is that $V_1\otimes V_2$ is naturally isomorphic to $\mathbb{R}$ if and only if $V_2$ is naturally isomorphic to $V_1^*$.

Each physical quantity has its own vector space $V$. The set $S$ that is the union of all physical quantities and $\mathbb{R}$ is closed under tensor product and dualization. In other words, if $V_1, V_2 \in S$, then $V_1\otimes V_2 \in S$ and if $V \in S$, then $V^*\in S$. Also, $V\otimes\mathbb{R} = \mathbb{R}\otimes V = V$. So $S$ is a free abelian group, where the group product is the tensor product, the inverse is the dual, and the identity element is $\mathbb{R}$.

A subset of $S$ can be called fundamental if it generates $S$.

Deane
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    I agree with your first sentence, but my point is that the construction of the space itself (as done by Tao) singles out certain quantities. It is like defining a 3D vector space as the free vector space on ${e_1, e_2, e_3}$, and then noting that other bases can be constructed. This is doable, but arguably not as pure as a definition that avoids a particular basis altogether. I'm aiming for something like the latter for the set $\mathcal S$. But how does one describe $\mathcal S$ without reference to the generators? A set of vector spaces that could be generated in such a way? – WillG Jan 08 '25 at 04:10
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    @WillG, could you say more about why you find the second paragraph of my answer unsatisfactory? A physical quantity is a 1-dimensional vector space. The set $S$ of all physical quantities is a set of 1-dimensional vector spaces that is closed under tensor product and dualization. It therefore is an abelian group. Since there are no nontrivial relations, it is free. There's no mention of generators in this description. – Deane Jan 08 '25 at 04:40
  • Aha, I missed that point in your second paragraph, namely that $S$ can be defined without reference to a fundamental generating set. – WillG Jan 08 '25 at 06:52
  • One other point of confusion is that a "literal" realization of the tensor product space $V_1 \otimes V_2$ (e.g., as a quotient of the free vector space on $V_1 \times V_2$) may actually treat $V_1$ and $V_2$ as fundamental: $V_1 \otimes V_2$ is "really" constructed from $V_1$ and $V_2$, whereas $V_1$ only happens to be isomorphic to $V_1 \otimes V_2 \otimes V_2^*$. So I suppose we should think of these as abstract tensor products, with naturally isomorphic versions identified? – WillG Jan 08 '25 at 06:52
  • "So I suppose we should think of these as abstract tensor products, with naturally isomorphic versions identified?" Yes, that's right. – Deane Jan 08 '25 at 15:47