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I know intuitively what units are and I understand how to DO dimensional analysis fine, but it occurred to me recently that I've never really considered what units or dimensions actually ARE, that is, what properties an unit must necessarily have and what properties a dimension (in the context of dimensional analysis) must have. Specifically, I was reading several threads on the topic of how many base units are actually necessary in physics, over on the physics SE last week (didn't save the links unfortunately) and the general impression I got was that the number of base units in a unit system is analogous to the number of basis vectors in a vector space which makes the dimensions in dimensional analysis analogous to the dimensions of a vector space. That makes intuitive sense as an analogy, but unit systems in general can't actually be vector spaces since they don't satisfy the addition axioms, as addition between units of different dimensions is generally undefined (e.g. we can't add meters and kilograms). Is there some other, more general mathematical structure that unit systems are an example of, and that have a property analogous to the dimensions of a vector space that the dimensions of dimensional analysis would be an example of? If not, have any mathematicians attempted to formalize things and come up with proper definitions?

I did Google this question and found stuff on the Buckingham Pi theorem; however, looking at the Wikipedia page, if I'm understanding correctly, the theorem states that, for a unit system of n linearly independent base units, there will be an equivalent system of n dimensionless parameters. That makes it seem very much like base units are just independent variables/degrees of freedom. That sort of makes sense, except that addition between different base units is undefined and that would seem to imply that this set of dimensionless parameters that are, presumably, isometric to the original unit system, can't be either real or complex-valued, because real and complex numbers can all be added together. But if these parameters can't have real or complex values, then what type of values can they have?

And, further, it's unclear to me to what "dimensionless" even means in this context because it's unclear what "dimension" means, mathematically speaking. Even if I relate it back to the intuition I have from physics, it's still not entirely clear because different unit systems have different numbers of base units and different quantities that are used as bases. For example, in SI units, you can't add space time together without using c as a conversion factor on one of them. But if, as is actually common in some branches of physics, we simply declare that c=1, we end up with space and time having the same unit, as well as with the electric and magnetic fields having the same units, and energy, mass, and momentum all having the same unit. That presumably just means that the SI base units aren't really all linearly independent of each other and so the Buckingham Pi theorem wouldn't apply. But that brings up the question of what linear independence even means when we're not dealing with a vector space, as it doesn't seem like linear combinations are possible.

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    $\begingroup$ A measurement system is just a finite dimensional real vector space. A basis for the vector space picks out our distinguished units. For example in the cgs system centimeters, grams, and seconds are our basis. A vector might look like $cg^{0.5}$. In this vector space, addition is defined as addition of exponents (multiplication of dimensionful expressions). Scalar multiplication is multiplication of exponents by some constant (corresponding to raising the whole dimensionful part to some power). $\endgroup$ Commented Mar 27, 2023 at 2:58
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    $\begingroup$ Concretely, suppose $A_1, \ldots, A_k$ are our distinguished units. A generic vector is given by $A_1^{r_1}\cdots A_k^{r_k}$ with $r_1, \ldots, r_k \in \mathbb{R}$. We see that $(A_1^{r_1}\cdots A_k^{r_k}) \cdot (A_1^{s_1}\cdots A_k^{s_k}) = A_1^{r_1 + s_1}\cdots A_k^{r_k + s_k}$ defines an addition on the vector space. We also see that $(A_1^{r_1}\cdots A_k^{r_k})^t = A_1^{t r_1}\cdots A_k^{t r_k}$ defines scalar multiplication on the vector space. All the usual vector space properties can be checked. $\endgroup$ Commented Mar 27, 2023 at 3:02
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    $\begingroup$ It shouldn't bother you that addition looks like multiplication and multiplication looks like exponentiation. The whole point of having an abstract notion of vector space is that it enables us to make abstractions like this. The utility of the abstraction is that we can apply theorems about vector spaces to this seemingly distinct context. One theorem, that basis expansions are unique, is the whole idea behind dimensional analysis. Performing dimensional analysis is essentially just expressing a vector in a different basis as in elementary linear algebra. $\endgroup$ Commented Mar 27, 2023 at 3:06
  • $\begingroup$ @CharlesHudgins, ah got it, that makes sense. It didn't occur to me that the actual vectors would be the products of the base units all raised to some power and then adding refers adding the exponents. It makes sense now that you pointed it out though. Why not make your comments a single answer though? You're explanation is certainly thorough and clear enough to warrant being an answer. $\endgroup$ Commented Mar 27, 2023 at 3:10
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    $\begingroup$ It started out as a brief comment, but then I realized I had almost a full answer of stuff to say. $\endgroup$ Commented Mar 27, 2023 at 3:16

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Here's one way to do it with $1$-dimensional physical quantities for which you can have both positive and negative amounts. The amount of a physical quantity can be represented without units by an oriented $1$-dimensional abstract vector space $V$. The properties of a vector space mirror the assumption that you can add amounts of the same quantity, as well as multiply the amount by any scale factor. The orientation is needed, because presumably there is an observable difference between positive and negative amounts.

Choosing a unit is simply choosing a nonzero vector $u \in V$ and declaring it to be a unit amount. Then any other amount $v \in V$ can be written uniquely as $v = cu$, and therefore consists of $c$ units of the quantity.

If you have a physical quantity that is the product of two other quantities, e.g., area, the compound quantity is represented by a tensor product. If $V$ represents length, then $V\otimes V$ represents length squared, i.e., area. If you choose a unit $u \in V$ for length, then $u\otimes u$ represents a square unit of area.

Then the concept of "per quantity" is modeled by the dual vector space $V^*$. If $u \in V$ a unit quantity, then there is a unique $u^* \in V$ such that $\langle u^*,u\rangle = 1$. $u^*$ represents "per unit quantity"

For example, if $V$ represents distance and $T$ represents time, then speed is represented by $V\otimes T^*$. If you choose units $u \in V$ and $t \in T$, then you get a unit for speed, $u\otimes\tau$, where $\langle \tau,t\rangle = 1$.

A general framework for higher dimensional physical measurements seems more complicated to me. However, if you restrict to space, then things are already well understood. For example, flat space is just an abstract vector space with an inner product, and choosing units is just choosing a basis. Normally, we want the units to be isotropic (use same units in all directions), so the basis is chosen to be orthonormal.

You also sometimes want to use fractional powers of units. This is no problem, because an oriented $1$-dimensional vector space can be raised to any fractional power.

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  • $\begingroup$ Fascinating! It never occurred to me that we could have a separate vector space for each unit, and then use the tensor product to combine them to make compound units. I'm a little unclear on the purpose of the dual space here though. The dual space V* is just the vector space of all linear functions that map V to F, yes? But division isn't a linear operation, so why should there be a member of the dual space corresponding to the reciprocal function? $\endgroup$ Commented Mar 27, 2023 at 3:26
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    $\begingroup$ @MikaylaEckelCifrese An element $\alpha \in V^*$ can be multiplied with an element of $v \in V$ by means of $\alpha \cdot v = \alpha(v)$. Now since $V$ is one-dimensional, every $v \in V$ is of the form $v = \lambda e$ for some chosen non-zero $e \in V$. Now, define $v^{-1} \in V^*$ by $v^{-1} (\mu e) = \mu \cdot \frac{1}{\lambda}$, which is clearly linear. Then $v^{-1} \cdot v = v^{-1} (\lambda e) = \frac{\lambda}{\lambda} = 1$, which justifies the notation $v^{-1}$. Also, clearly $v^{-1}$ is the unique element of $V^*$ with this property. $\endgroup$ Commented Mar 27, 2023 at 11:15
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Units with a fixed dimension (e.g. units of length) do not form a vector space, but rather a positive space - i.e. given a pair of units, the "sum" is another unit and we can multiply a unit with a stricly positive number to obtain another unit. For the formal definition, see this paper. Tensor products are also defined in the context of positive spaces and may be used to construct units of area from units of length etc.

If we consider units with different dimensions, then we get the follwing structure: Suppose that units form a set $X$. We have an equivalence relation on $X$, where each equivalence class consist of all the units with a fixed dimension. The equivalence classes are closed under multiplication with positive real numbers and for each equivalence class we are given a function "addition" such that it comes with the structure of a positive space. Furthermore we can multiply any two units (not necessarely with the same dimension) and consider rational powers of units. This defines a $\mathbb Q$-vector space structure.

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