In the context of arithmetic, *carrying* is part of the operation of representing addition of natural numbers by digits with respect to a base.

Given the rig of natural numbers $\mathbb{N}$, there exists a free commutative $\mathbb{N}$-algebra $\mathbb{N}[b]$ on one generator $b$ called the base. Since multiplication in a commutative algebra is power-associative, there exists a right $\mathbb{N}$-action on $\mathbb{N}[b]$ $(-)^{(-)}:\mathbb{N}[b]\times\mathbb{N}\to\mathbb{N}[b]$ called the power, and every element in $\mathbb{N}[b]$ could be written as a polynomial

$p = \sum_{n=0}^{k} a(n) b^n$

When the algebra is quotiented out by the relation $b \sim 10$, the resulting quotient algebra is isomorphic to the original rig of natural numbers $\mathbb{N}$. This means that every natural number could be expressed a polynomial with base ten,

$p = \sum_{n=0}^{k} a(n) 10^n$

There is a canonical such polynomial, one where all natural numbers in the sequence $a(n) \lt 10$ in the polynomial. Carrying arises from adding two canonical polynomials, when the sum $a_1(n) + a_2(n) \geq 10$ and the polynomial is no longer canonical; in order to make the polynomial canonical again, one would have to take the sum $a_1(n) + a_2(n)$ modulo 10 and add 1 to the sum $a_1(n+1) + a_2(n+1)$ in the next power of ten. This means there ought to be another representation of the digits in terms of integers modulo 10.

Write $\mathbb{Z}/10$ for the abelian group of addition of integers modulo 10. In the following we identify the elements as

$\mathbb{Z}/{10} = \{0,1,2, \cdots, 9\}
\,,$

as usual.

Being an abelian group, every delooping n-groupoid $\mathbf{B}^n (\mathbb{Z}/{10})$ exists.

Carrying is a 2-cocycle in the group cohomology, hence a morphism of infinity-groupoids

$c : \mathbf{B} (\mathbb{Z}/{10}) \to \mathbf{B}^2 (\mathbb{Z}/{10})
\,.$

It sends

$\array{
&& \bullet
\\
& {}^{\mathllap{a}}\nearrow
&\Downarrow^=&
\searrow^{\mathrlap{b}}
\\
\bullet &&\stackrel{a+b mod 10}{\to}&&
}
\;\;\;
\mapsto
\;\;\;
\array{
&& \bullet
\\
& {}^{\mathllap{id}}\nearrow
&\Downarrow^{c(a,b)}&
\searrow^{\mathrlap{id}}
\\
\bullet &&\stackrel{id}{\to}&& \bullet
}
\,,$

where

$c(a,b) =
\left\{
\array{
1 & a + b \geq 10
\\
0 & a + b \lt 10
\,.
}
\right.$

The central extension classified by this 2-cocycle, hence the homotopy fiber of this morphism is $\mathbb{Z}/{100}$

$\array{
\mathbf{B} (\mathbb{Z}/{100}) &\to& *
\\
\downarrow && \downarrow
\\
\mathbf{B} (\mathbb{Z}/{10}) &\stackrel{\mathbf{c}}{\to}& \mathbf{B}^2 (\mathbb{Z}/{10})
}
\,.$

That now carries a 2-cocycle

$\mathbf{B} (\mathbb{Z}/{100}) \to \mathbf{B}^2 (\mathbb{Z}/{10})
\,,$

and so on.

$\array{
\vdots
\\
\downarrow
\\
\mathbf{B} (\mathbb{Z}/{1000})
&\stackrel{c}{\to}&
\mathbf{B}^2 (\mathbb{Z}/{10})
\\
\downarrow
\\
\mathbf{B} (\mathbb{Z}/{100})
&\stackrel{c}{\to}&
\mathbf{B}^2 (\mathbb{Z}/{10})
\\
\downarrow
\\
\mathbf{B} (\mathbb{Z}/{10})
&\stackrel{c}{\to}&
\mathbf{B}^2 (\mathbb{Z}/{10})
}$

This tower can be viewed as a sort of “Postnikov tower” of $\mathbb{Z}$ (although it is of course not a Postnikov tower in the usual sense). Note that it is not “convergent”: the limit of the tower is the ring of $10$-adic integers $\mathbb{Z}_{10}$. This makes perfect sense in terms of carrying: the $10$-adic integers can be identified with “decimal numbers” that can be “infinite to the left”, with addition and multiplication defined using the usual carrying rules “on off to infinity”.

- Dan Isaksen,
*A cohomological viewpoint on elementary school arithmetic*, The American Mathematical Monthly, Vol. 109, No. 9. (Nov., 2002), pp. 796-805. (jstor)

Last revised on May 29, 2021 at 04:16:42. See the history of this page for a list of all contributions to it.