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• CommentRowNumber1.
• CommentAuthorFosco
• CommentTimeJul 14th 2015
• (edited May 7th 2019)

In an attempt to find a mathematically convicing definition of a “unit of measurement” I came up with this construction; it’s nice, but not perfect. I am interested in your opinion.

In elementary physics units of measurement are something you attach to scalars (identified with “pure quantities”). You can multiply “meters” and “kilograms”, and you can divide the result by “seconds”^2. But you can’t add “inhomogeneous” quantities like meters and kilograms, this has no physical meaning.

Ok then, I want to propose the following interpretation for this situation:

• “you can’t add inhomogeneous quantities” means that you can’t compose non-contiguous arrows
• “you can multiply freely” means that you can freely tensor objects and arrows.

This led me to the following definition(s): let $\mathbb{F}$ be a ring, and $A$ a set of symbols (to fix ideas, let $A=\{ meters, kilograms, seconds\}$, whatever these three words mean). I denote by $\mathbf{A}$ the abelian group obtaining inverting all elements in the free monoid over $A$. I define the following category $\mathcal{M}_{\mathbb{F}}(A)$

• objects are elements of $\mathbf{A}$
• there are no arrows which are not endomorphisms, and $\hom_{\mathcal{M}_{\mathbb{F}}(A)}([v], [v]) = \mathbb{F}$ for each $[v]\in \mathcal{M}_{\mathbb{F}}(A)$.

Composition in $\mathcal{M}_{\mathbb{F}}(A)$ is sum of elements of $\mathbb{F}$; this turns $\mathcal{M}_{\mathbb{F}}(A)$ into a groupoid (there are several other equivalent ways to define it).

Moreover, $\mathcal{M}_{\mathbb{F}}(A)$ has a monoidal structure: let $(a, [v])\otimes (b, [w]) = (ab, [v.w])$, where $ab$ is the product of $a,b$ in $\mathbb{F}$ and “dot” is the concatenation of words in $\mathbf{A}$. (I denote an arrow in $\mathcal{M}_{\mathbb{F}}(A)$ with the pair of its domain-codomain and the name of the arrow $a\colon [v]\to [v]$ in $\mathbb{F}$).

The fact that this turns $\mathcal{M}_{\mathbb{F}}(A)$ into a monoidal category is a consequence of the ring structure on $\mathbb{F}$.

Plusses of this construction:

• If $\mathbb{F}$ is a differential ring, you can define endofunctors $\mathbf{D}_{[v]}\colon \mathcal{M}_{\mathbb{F}}(A)\to \mathcal{M}_{\mathbb{F}}(A)$ as

where $a\mapsto a'$ is the derivation of $\mathbb{F}$. Funny thing, this is “linear and Leibniz”, in the sense that it is (a functor and) such that

$\mathbf{D}_{[v]}\big( \alpha \otimes \beta \big) = (\mathbf{D}_{[v]}\alpha \otimes \beta)\circ (\alpha\otimes \mathbf{D}_{[v]}\beta) = (a'b + a b' , [u.w.v^{-1}])$

for each $\alpha = (a, [u])$ e $\beta = (b, [w])$.

• Nothing forces you to have $\mathbb{F}$ as “ring/field of scalars” for each $[v]\in\mathbf{A}$: some physical quantities are quantized, some are not, so for some $[v]\in\mathbf{A}$ I can choose $\mathbb{R}$, for some others I can choose $\mathbb{Q}$ or even $\mathbb{Z}$ or even $\mathbb{Z}/3\mathbb{Z}$.

That’s all. I find it beautiful; help me to turn it into something more beautiful!

• CommentRowNumber2.
• CommentAuthorTodd_Trimble
• CommentTimeJul 14th 2015

Fosco, do you know about Jim Dolan’s lectures on Algebraic Geometry for Category Theorists?

• CommentRowNumber3.
• CommentAuthorDavidRoberts
• CommentTimeJul 14th 2015

First thing I thought of, too.

Speaking of that, I wish we could get notes from Jim’s recent Australian Category Seminar talks.

• CommentRowNumber4.
• CommentAuthorFosco
• CommentTimeJul 14th 2015

The links to Jim Dolan’s videos are all broken; is there another source to see them?

• CommentRowNumber5.
• CommentAuthorFosco
• CommentTimeJul 14th 2015

Ah, now I remember about “dimensional categories”!

• CommentRowNumber6.
• CommentAuthorDavidRoberts
• CommentTimeJul 15th 2015
• CommentRowNumber7.
• CommentAuthorFosco
• CommentTimeJul 15th 2015

This is strange, now I can download the videos but they have no sound (also, the first video is of rather poor quality, to the point that I can’t read on the blackboard…)

• CommentRowNumber8.
• CommentAuthorUrs
• CommentTimeJul 15th 2015

By the way, there is the beginning of something in the $n$Lab entry physical unit. Certainly deserves to be expanded…

• CommentRowNumber9.
• CommentAuthorNikolajK
• CommentTimeJul 15th 2015

I remember this

https://terrytao.wordpress.com/2012/12/29/a-mathematical-formalisation-of-dimensional-analysis/