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Stream: theory: category theory

Topic: ri(n)gs laws from laws of exponents

Claudio Pisani (Sep 16 2023 at 13:09):

Say that a "power algebra" $A$ is a set with three binary operations and two constants:
"product", "sum", "power", 1, and 0, such that the usual laws of exponents hold:

1) $\quad a^{b\cdot c} = (a^b)^c \quad\quad$ 1') $\quad a^1 = a$

2) $\quad a^{b+c} = a^b \cdot a^c \quad\quad$ 2') $\quad a^0 = 1$

3) $\quad (a\cdot b)^ c = a^c\cdot a^c \quad\quad$ 3') $\quad1^c = 1$

Assume furthermore that the following "Yoneda-like" axiom holds:

4) $(\forall x \quad x^a = x^b) \implies a = b$

Then, using the non-categorical analogous of Yoneda reduction techniques, one proves that

1) $(A, \cdot, 1)$ and $(A, +, 0)$ are monoids.

2) product distributes over sums on both sides.

Let us check, for instance, the associative and distributive laws:

$x^{(a\cdot b)\cdot c} = (x^{a\cdot b})^c = ((x^a)^ b)^c = (x^a)^ {b\cdot c} = x^{a\cdot( b\cdot c)}$

$x^{(a+ b)+ c} = x^{a+b}\cdot x^c = (x^a\cdot x^ b)\cdot x^c = x^a\cdot (x^ b\cdot x^c) = x^a\cdot x^{b+c} = x^{a+(b+ c)}$

$x^{a\cdot(b+c)} = (x^a)^{b+c} = (x^a)^ b\cdot (x^a)^ c = x^{a\cdot b}\cdot x^{a\cdot c} = x^{a\cdot b+a\cdot c}$

$x^{(a+b)\cdot c} = (x^{a+b})^c = (x^a\cdot x^ b)^ c = (x^a)^c\cdot (x^ b)^ c = x^{a\cdot c}\cdot x^{b\cdot c} = x^{a\cdot c+b\cdot c}$

So, any power algebra is in fact a rig, except that addition is not necessarily commutative (see also this post).
If we assume also $(a^b)^c = (a^c)^b$, then both products and sums commute.

The main instance of power algebra is the given by natural numbers.
One may wonder if there are other such natural examples, arising from isomorphism classes of a cartesian closed category $\cal C$.

If $\cal C$ is is left combinatorial (see here) then also

$(\forall X \quad X^A \cong X^B) \implies A \cong B$

so that the set of its isomorphism classes is in fact a power algebra.

Claudio Pisani (Sep 16 2023 at 13:34):

I was surprised by the effectiveness of the axiom (4) outside the realm of category theory.

Admittedly, it has a rather unusual form.

Say that a two-variable function $A\times B \to C$ is right injective if the corresponding function $B\to C^A$ is injective.
Then, axiom (4) says that the power operation is right injective.

So I ask:

do you know any other axioms or laws (in particular in algebra) that have a similar form?

dusko (Sep 16 2023 at 23:55):

Claudio Pisani said:

Say that a "power algebra" $A$ is a set with three binary operations and two constants:
"product", "sum", "power", 1, and 0, such that the usual laws of exponents hold:

1) $\quad a^{b\cdot c} = (a^b)^c \quad\quad$ 1') $\quad a^1 = a$

2) $\quad a^{b+c} = a^b \cdot a^c \quad\quad$ 2') $\quad a^0 = 1$

3) $\quad (a\cdot b)^ c = a^c\cdot a^c \quad\quad$ 3') $\quad1^c = 1$

Assume furthermore that the following "Yoneda-like" axiom holds:

4) $(\forall x \quad x^a = x^b) \implies a = b$

Then, using the non-categorical analogous of Yoneda reduction techniques, one proves that
[snip]

very nice!

it may be a matter of wording, but i wouldn't say that the reduction techniques are non-categorical, but that they are the equational fragment of closed structure. if the power algebra structure is given not over a set but over a reflexive graph and the power operation maps edges $a\to b$ to edges $x^b\to x^a$, then (4) can be refined to a bijective correspondence of each edge $a\to b$ with a distinguished family of edges $x^b\to x^a$. if there is also a bijective correspondence of the edges $a\to b$ and $1\to b^a$, it seems that reflexivity (i.e. the identity edges) and (1) give the evaluations and the composition of the edges can be derived from the mappings between the families of edges...

i am sorry that i am saying all this without checking the details, but i think it goes through because this is how the arrow parts of adjoint functors can be extracted from the adjunction correspondences.

if it does go through, then we are really looking at the equational fragment (or the skeleton) of cartesian closed structure with coproducts: (3) says that $\cdot$ is the product, (2) says that $+$ is the coproduct, (1) says that $(-)^c$ is the right adjoint to $(-)\cdot c$, and (4) is the skeletal tip of the yoneda iceberg.

there is probably a hole or two in my handwaving "construction", but i think that they can be filled and that the claim stands that we are not looking at anything non-categorical in any sense, since structures don't come together by accidents.