Geometry and Algebra
Let’s consider a two-element set V = {u, i}, along with its identity function 1v: V → V. This set with its identity function can be thought of as a single-morphism category; since there’s only one object (set), we can also think of it as a monoid. Actions of monoids or categories in general give rise to geometries and their algebras.
Let’s start with actions of sets. An action p of an set A on an set B is a function p: A x B → B, whose domain is a product. With A as a single-element set, the action p is an endomap or dynamical system with B as its set of states. Now let’s look at the category of actions, with functions such as p: A x B → B as its objects. The first question we need to address is: What are the morphisms of the category of actions? A morphism from an object p: A x B → B to an object q: A x C → C is a function f: B → C satisfying a commutativity condition f o p = q o (1A x f); ‘o’ denotes composition of functions, the composite ‘f o p’ is read as ‘f after p’, and 1A: A → A is the identity function of A. With A = 1 = {*}, a single-element set, the commutativity equation f o p = q o (1A x f) that a function f: B → C has to satisfy in order to be a morphism of actions reduces to f o p = q o f, which is the structure-preserving map between dynamical systems, i.e., a morphism from the dynamical system p: B → B to q: C → C. So, we think of dynamical systems as a special case of actions or of actions as a generalization of dynamical systems, with A as a set of dynamics; for example, p (a, -): B → B gives one dynamical system, while p (a’, -): B → B gives another dynamical system, i.e., we get a dynamical system for each element of the set A of a given action p: A x B → B.
Next on agenda is the composition of morphisms of the category of actions. Now we show that the composite of composable morphisms is a morphism of the category of actions (as required). Let’s consider a morphism g: C → D from the action (object) q: A x C → C to an object r: A x D → D satisfying the required commutativity equation g o q = r o (1A x g). Since the domain object q: A x C → C of the morphism g: C → D is equal to the codomain object q: A x C → C of the morphism f: B → C, we can compose these composable morphisms of actions as (g o f): B → C → D = B → D. Now we have to show that the composite (g o f): B → D is a morphism of actions, i.e., satisfies the commutativity equation:
(g o f) o p = r o (1A x (g o f))
given the two equations f o p = q o (1A x f) and g o q = r o (1A x g). Starting with the left-hand side:
(g o f) o p = g o (f o p) = g o (q o (1A x f)) = (g o q) o (1A x f) = (r o (1A x g)) o (1A x f) = r o ((1A x g) o (1A x f)) = r o (1A x (g o f)). This last step requires some explanation (for me ;)
First let’s spell out the domains and codomains of products of functions of the composite (1A x g) o (1A x f): (A x B → A x C) o (A x C → A x D).
(1A x f): A x B → A x C, whose factor functions are 1A: A → A and f: B → C
(1A x g): A x C → A x D, whose factor functions are 1A: A → A and g: C → D
Composing the two first factor functions (1A o 1A): A → A → A = 1A: A → A
Composing the two second factor functions (g o f): B → C → D = B → D
Product of these two composites is (1A x (g o f)): A x B → A x D
which is how I explained (to myself ;) the last substitution:
r o ((1A x g) o (1A x f)) = r o (1A x (g o f))
which gives us the right-hand side of the equation
(g o f) o p = r o (1A x (g o f))
thereby satisfying the commutativity condition to show that the composite of a pair of composable morphisms is a morphism of the category of actions.
If it all reads like tunneling across Himalayas to catch an ant, that’s cuz that’s exactly what I did (which is what I love doin :)
Next order of biz is get some fancy costume for our actor A so that it can play the character that we scripted: Monoid.
Before I get into monoid actions, I need coffee … no ordinary coffee but super strong malai coffee at my fav Malabar Cafe :)
Finally I’m back to the post after my coffee break as I promised which is a first: Thank you Jesus :)
I have a thing for whittling down; fortunately or unfortunately monoid is at the receiving end, so to speak. Let’s begin with a monoid M consisting of a set M = {1}; the only element 1 of the set M serves as the two-sided unit of monoid multiplication m: M x M → M, which is required to be associative. More explicitly, m (1, 1) = 1. Of course, the monoid set M can have any number of elements (in addition to the unit that we are considering), which would undoubtedly make our monoid spicy (which is exactly what I’m resisting with the express purpose of experiencing uneventfulness as the comforting familiarity that it is: my face looked purty much the same today as it did yesterday; nothing really changes between daily measures of my handsomeness and yet I changed beyond recognition over the decades; more important in the present context is that this is in the spirit of Grothendieck’s method, where nothing happens for hundreds of pages only to find an otherwise intractable theorem appear as a tautology in a vast expanse of an alien theory beyond the reach of the conceptual repertoire that sought the proof or formulated the question in the first place).
Let’s now consider an action of the monoid M = (M, 1, m) on a set T, which is a function
p: M x T → T
compatible with monoid multiplication m and its two-sided unit 1. Compatibility with the unit of monoid multiplication means
p (1, t) = t
for every element t of the set T. Compatibility with monoid multiplication m requires commutativity of the two paths from M x M x T → T, which is
p o (m x 1T) = p o (1M x p)
For every element t of the set T, we have
p o (m x 1T) (1, 1, t) = p (m (1, 1), 1T (t)) = p (1, t) = t
p o (1M x p) (1, 1, t) = p (1M (1), p (1, v)) = p (1, t) = t
A set T equipped with a monoid M-action p: M x T → T is called an M-set T.
Note that in the domain M x T of the above monoid action p: M x T → T, M is on the left-side of the set T on which it is acting. Since M = {1} is a single-element set, with the only element 1 as the unit of the monoid multiplication m: M x M → M and action p: M x T → T, we ignored the leftness. However, in general monoid actions on the left vs. right make all the difference underlying the dual: algebra vs. geometry.
Now I’m hungry … I’ll be back soon … we’ll get to inhale all the boredom there is ;)
Let us now consider a two-element monoid M = (M, 1, m), where M = {1, e} and monoid multiplication m: M x M → M defined as:
m (1, 1) = 1
m (1, e) = e
m (e, 1) = e
m (e, e) = e
A left M-action on a set T is a function a: M x T → T satisfying the commutative condition, which is equality of the two composites:
a o (m x 1T): (M x M) x T → M x T → T
a o (1M x a): M x (M x T) → M x T → T
along with
a (1, t) = t for all elements t in the set T.
Let T = {t1, t2}. First, let’s do the easy part, i.e.,
a (1, t1) = t1 and a (1, t2) = t2
Next, we have to figure out a (e, t) for each one of the two elements in the set T = {t1, t2}. There are four possibilities:
a1 (e, t1) = t1 and a1 (e, t2) = t1
a2 (e, t1) = t2 and a2 (e, t2) = t1
a3 (e, t1) = t1 and a3 (e, t2) = t2
a4 (e, t1) = t2 and a4 (e, t2) = t2
Next order of biz is to find which of above four a: M x T → T satisfy
a o (m x 1T) = a o (1M x a)
We have to evaluate the above two composites at each one of following points:
(1, 1, t1)
(1, 1, t2)
(1, e, t1)
(1, e, t2)
(e, 1, t1)
(e, 1, t2)
(e, e, t1)
(e, e, t2)
Let’s evaluate beginning with (1, 1, t1) and all the way to (e, e, t2)
a o (m x 1T) (1, 1, t1) = a (m (1, 1), 1T (t1)) = a (1, t1) = t1
a o (1M x a) (1, 1, t1) = a (1M (1), a (1, t1)) = a (1, t1) = t1
a o (m x 1T) (1, 1, t2) = a (m (1, 1), 1T (t2)) = a (1, t2) = t2
a o (1M x a) (1, 1, t2) = a (1M (1), a (1, t2)) = a (1, t2) = t2
Next with a1 (e, t1) = t1 and a1 (e, t2) = t1
a1 o (m x 1T) (1, e, t1) = a1 (m (1, e), 1T (t1)) = a1 (e, t1) = t1
a1 o (1M x a1) (1, e, t1) = a1 (1M (1), a1 (e, t1)) = a1 (1, t1) = t1
a1 o (m x 1T) (1, e, t2) = a1 (m (1, e), 1T (t2)) = a1 (e, t2) = t1
a1 o (1M x a1) (1, e, t2) = a1 (1M (1), a1 (e, t2)) = a1 (1, t1) = t1
a1 o (m x 1T) (e, 1, t1) = a1 (m (e, 1), 1T (t1)) = a1 (e, t1) = t1
a1 o (1M x a1) (e, 1, t1) = a1 (1M (e), a1 (1, t1)) = a1 (e, t1) = t1
a1 o (m x 1T) (e, 1, t2) = a1 (m (e, 1), 1T (t2)) = a1 (e, t2) = t1
a1 o (1M x a1) (e, 1, t2) = a1 (1M (e), a1 (1, t2)) = a1 (e, t2) = t1
a1 o (m x 1T) (e, e, t1) = a1 (m (e, e), 1T (t1)) = a1 (e, t1) = t1
a1 o (1M x a1) (e, e, t1) = a1 (1M (e), a1 (e, t1)) = a1 (e, t1) = t1
a1 o (m x 1T) (e, e, t2) = a1 (m (e, e), 1T (t2)) = a1 (e, t2) = t1
a1 o (1M x a1) (e, e, t2) = a1 (1M (e), a1 (e, t2)) = a1 (e, t1) = t1
Thus a1: M x T → T defined as
a1 (1, t1) = t1 and a1 (1, t2) = t2
a1 (e, t1) = t1 and a1 (e, t2) = t1
is a left M-action on the set T. The set T with the left M-action a1: M x T → T is a left M-set.
I need to go get sugar-high ;)
For now, except for the a2 (e, t1) = t2 and a2 (e, t2) = t1, all other three (a1, a3, and a4) are left M-actions. I’ll back soon to verify. Once we verify and have all three corresponding left M-sets, we define morphisms and eventually characterize the category of left M-sets.
See you soon with my good ol’ friend Cayley (who’s on the way to see what this medium is all about ;)
Summing all that’s been said above, a left M-set T is a set T equipped with a left M-action, which is a function a: M x T → T, where M is a set of elements one of which is a two-sided unit 1 with respect to the action a: M x T → T and the associative monoid operation m: M x M → M. Rephrasing it all, the function a: M x T → T satisfies:
a (1, t) = t for all t in T
and
a o (m x 1T) = a o (1M x a)
Now let’s see if a2: M x T → T, with a2 (e, t1) = t2 and a2 (e, t2) = t1, is a left M-action, which involves verification like that we did for a1: M x T → T. Here we go checking the equality of above composites point-wise.
a2 o (m x 1T) (1, e, t1) = a2 (m (1, e), 1T (t1)) = a2 (e, t1) = t2
a2 o (1M x a2) (1, e, t1) = a2 (1M (1), a2 (e, t1)) = a2 (1, t2) = t2
a2 o (m x 1T) (1, e, t2) = a2 (m (1, e), 1T (t2)) = a2 (e, t2) = t1
a2 o (1M x a2) (1, e, t2) = a2 (1M (1), a2 (e, t2)) = a2 (1, t1) = t1
a2 o (m x 1T) (e, 1, t1) = a2 (m (e, 1), 1T (t1)) = a2 (e, t1) = t2
a2 o (1M x a2) (e, 1, t1) = a2 (1M (e), a2 (1, t1)) = a2 (e, t1) = t2
a2 o (m x 1T) (e, 1, t2) = a2 (m (e, 1), 1T (t2)) = a2 (e, t2) = t1
a2 o (1M x a2) (e, 1, t2) = a2 (1M (e), a2 (1, t2)) = a2 (e, t2) = t1
a2 o (m x 1T) (e, e, t1) = a2 (m (e, e), 1T (t1)) = a2 (e, t1) = t2
a2 o (1M x a2) (e, e, t1) = a2 (1M (e), a2 (e, t1)) = a2 (e, t2) = t1
Now that we have a point (e, e, t1) at which the values of the two composites are not equal, we conclude that the two composite functions are not equal. As such, the function a2: M x T → T is not a left M-action on the set T, which is to say that the set T with the function a2: M x T → T is not a left M-set.
Of late, I’m finding inequality quite fascinating more so than equality (all blame should be credited to Grassmann’s pure math ;) So, we check to see if the two composites take different values at the remaining point:
a2 o (m x 1T) (e, e, t2) = a2 (m (e, e), 1T (t2)) = a2 (e, t2) = t1
a2 o (1M x a2) (e, e, t2) = a2 (1M (e), a2 (e, t2)) = a2 (e, t1) = t2
So the values of two composites a2 o (m x 1T) and a2 o (1M x a2) are not equal at two points: (e, e, t1) and (e, e, t2).
Comparing a1: M x T → T, which is a left M-action, with a2: M x T → T that isn’t reminds of two i-words: idempotent vs. involution.
Now that we are in a mission-mode, we move on to define morphisms between left M-sets. Since we have one left M-set a1: M x T → T, let’s content ourselves with an endomorphism from a1 to a1, which is function l: T → T satisfying l o a1 = a1 o (1M x l). With T = {t1, t2} we have four endomaps:
l1 (t1) = t1 and l1 (t2) = t1
l2 (t1) = t2 and l2 (t2) = t1
l3 (t1) = t1 and l3 (t2) = t2
l4 (t1) = t2 and l4 (t2) = t2
Now we check for the equality l o a1 = a1 o (1M x l) for each one of the above four functions at each one of four points: (1, t1), (1, t2), (e, t1), (e, t2).
l1 o a1 (1, t1) = l1 (a1 (1, t1)) = l1 (t1) = t1
a1 o (1M x l1) (1, t1) = a1 (1M (1), l1 (t1)) = a1 (1, t1) = t1
l1 o a1 (1, t2) = l1 (a1 (1, t2)) = l1 (t2) = t1
a1 o (1M x l1) (1, t2) = a1 (1M (1), l1 (t2)) = a1 (1, t1) = t1
l1 o a1 (e, t1) = l1 (a1 (e, t1)) = l1 (t1) = t1
a1 o (1M x l1) (e, t1) = a1 (1M (e), l1 (t1)) = a1 (e, t1) = t1
l1 o a1 (e, t2) = l1 (a1 (e, t2)) = l1 (t1) = t1
a1 o (1M x l1) (e, t2) = a1 (1M (e), l1 (t2)) = a1 (e, t1) = t1
As such, we have the equality of the two composite functions:
l1 o a1 = a1 o (1M x l1)
which means l1: T → T is an endomorphism on the left M-set T equipped with a1: M x T → T. A morphism from a left M-set T with a1: M x T → T to a left M-set T’ with a1': M x T’ → T’ is a function l’: T → T’ satisfying:
l’ o a1 = a1' o (1M x l’)
These left M-sets along with the morphisms constitute a category.
Since we can’t keep good ol’ Cayley waiting for too long, we rush to self-actions of the monoid M = (M, 1, m), which is a function m: M x M → M satisfying the commutative condition, i.e., equality of the two composites:
m o (m x 1M): (M x M) x M → M x M → M
m o (1M x m): M x (M x M) → M x M → M
Lest you’re thinking I’m making it all on the fly, rest assured I have Professor F. William Lawvere’s Perugia Notes with me all along :)
Since it’s already past 5 AM, I’m taking Cayley for Madhur Chai … hopefully he’ll like it as much as I do :)
Now we have to check for the equality of the following two composites:
m o (m x 1M) = m o (1M x m)
With M = {1, e}, we have a total of eight points at which we have to evaluate the above functions. Let’s start:
m o (m x 1M) (1, 1, 1) = m (m (1, 1), 1M (1)) = m (1, 1) = 1
m o (1M x m) (1, 1, 1) = m (1M (1), m (1, 1)) = m (1, 1) = 1
m o (m x 1M) (1, 1, e) = m (m (1, 1), 1M (e)) = m (1, e) = e
m o (1M x m) (1, 1, e) = m (1M (1), m (1, e)) = m (1, e) = e
m o (m x 1M) (1, e, 1) = m (m (1, e), 1M (1)) = m (e, 1) = e
m o (1M x m) (1, e, 1) = m (1M (1), m (e, 1)) = m (1, e) = e
m o (m x 1M) (1, e, e) = m (m (1, e), 1M (e)) = m (e, e) = e
m o (1M x m) (1, e, e) = m (1M (1), m (e, e)) = m (1, e) = e
m o (m x 1M) (e, 1, 1) = m (m (e, 1), 1M (1)) = m (e, 1) = e
m o (1M x m) (e, 1, 1) = m (1M (e), m (1, 1)) = m (e, 1) = e
m o (m x 1M) (e, 1, e) = m (m (e, 1), 1M (e)) = m (e, e) = e
m o (1M x m) (e, 1, e) = m (1M (e), m (1, e)) = m (e, e) = e
m o (m x 1M) (e, e, 1) = m (m (e, e), 1M (1)) = m (e, 1) = e
m o (1M x m) (e, e, 1) = m (1M (e), m (e, 1)) = m (e, e) = e
m o (m x 1M) (e, e, e) = m (m (e, e), 1M (e)) = m (e, e) = e
m o (1M x m) (e, e, e) = m (1M (e), m (e, e)) = m (e, e) = e
Hence m o (m x 1M) = m o (1M x m), which means
m: M x M → M is a self-action of the monoid M = (M, 1, m) on M.
Now, what’s so special about this monoid self-action?
Given any left M-set T, there is a 1–1 correspondence between its elements t: 1 → T and morphisms from the representable set M (resulting from the self-action of the monoid M on M) to the given left M-set T. A morphism from the representable m: M x M → M to a left M-set a: M x T → T is a function l: M → T satisfying l o m = a o (1M x l). So, corresponding to the two elements of the set T = {t1, t2}, we need two functions
l’: M → T satisfying l’ o m = a o (1M x l’)
l’’: M → T satisfying l’’ o m = a o (1M x l’’)
With a: M x T → T defined as
a1 (1, t1) = t1 and a1 (1, t2) = t2
a1 (e, t1) = t1 and a1 (e, t2) = t1
we have to find out which, if any, of the four possible functions l: M → T satisfy the above commutative conditions.
l1 (1) = t1 and l1 (e) = t1
l2 (1) = t2 and l2 (e) = t1
l3 (1) = t1 and l3 (e) = t2
l4 (1) = t2 and l4 (e) = t2
Let’s see if l1: M → T corresponds to the element t1 of the set T. First we have check if l1 o m = a o (1M x l1).
l1 o m (1, 1) = l1 (1) = t1
a o (1M x l1) (1, 1) = a o (1M (1), l1 (1)) = a (1, t1) = t1
l1 o m (1, e) = l1 (e) = t1
a o (1M x l1) (1, e) = a o (1M (1), l1 (e)) = a (1, t1) = t1
l1 o m (e, 1) = l1 (e) = t1
a o (1M x l1) (e, 1) = a o (1M (e), l1 (1)) = a (e, t1) = t1
l1 o m (e, e) = l1 (e) = t1
a o (1M x l1) (e, e) = a o (1M (e), l1 (e)) = a (e, t1) = t1
Hence we have a morphism l1 (corresponding to the element t1) from the representable (self-action of the monoid M) m: M x M → M to the left M-set a: M x T → T.
Let’s now see if l4: M → T corresponds to the element t2 of the set T. First we have check if l4 o m = a o (1M x l4).
l4 o m (1, 1) = l4 (1) = t2
a o (1M x l4) (1, 1) = a o (1M (1), l4 (1)) = a (1, t2) = t2
l4 o m (1, e) = l4 (e) = t2
a o (1M x l4) (1, e) = a o (1M (1), l4 (e)) = a (1, t2) = t2
l4 o m (e, 1) = l4 (e) = t2
a o (1M x l4) (e, 1) = a o (1M (e), l4 (1)) = a (e, t2) = t2
l4 o m (e, e) = l4 (e) = t2
a o (1M x l4) (e, e) = a o (1M (e), l4 (e)) = a (e, t2) = t2
Hence we have a morphism l4 (corresponding to the element t2) from the representable (self-action of the monoid M) m: M x M → M to the left M-set a: M x T → T.
I don’t know about you but I have no idea how to go from morphisms l1 and l4 to elements t1 and t2, respectively.
Thus concludes the action-packed Bengaluru marathon ;)
Given the feverish-mode I seem to be in, I’m back to see what the remaining two functions l2, l3: M → T correspond to. Since there are only two elements t1 and t2 in the set T to which we (we some trepidation, but not completely devoid of any reason) found that the two morphisms from a self-acting representing monoid M to a left M-set T, i.e., l1, l4: M → T correspond.
Now let’s see if the function l2: M → T satisfies l2 o m = a o (1M x l2), which is required for the function l2 to be a morphism from the self-acting monoid m: M x M → M to the left M-set a: M x T → T. In other words, we check to see if
a o (1M x l2) = a o (1M x l2)
is satisfied.
l2 o m (1, 1) = l2 (1) = t1
a o (1M x l2) (1, 1) = a o (1M (1), l2 (1)) = a (1, t2) = t2
Since l2 o m (1, 1) is not equal to a o (1M x l2) (1, 1), we conclude that function l2: M → T is not a morphism and in turn doesn’t correspond to any element of the set T.
Next we check to see if l3: M x T satisfies (l3 o m) = a o (1M x l3).
(l3 o m) (1, 1) = l3 (m (1, 1)) = l3 (1) = t1
a o (1M x l3) (1, 1) = a (1M (1), l3 (1)) = a (1, t1) = t1
(l3 o m) (1, e) = l3 (m (1, e)) = l3 (e) = t2
a o (1M x l3) (1, e) = a (1M (1), l3 (e)) = a (1, t2) = t2
(l3 o m) (e, 1) = l3 (m (e, 1)) = l3 (e) = t2
a o (1M x l3) (e, 1) = a (1M (e), l3 (1)) = a (e, t1) = t1
Since the values of the two functions (l3 o m) and a o (1M x l3) are not equal at the point (e, 1), the two functions are not equal, which means l3: M → T is not a morphism from the self-acting M-set M to the M-set T. So, at least there are exactly two morphisms from the self-acting M-set M to the M-set T, which can be thought of as corresponding to the two elements of T.
But what is getting on my nerves is that when we say there is a 1–1 correspondence between two sets, we should be able to start with an element in a set and get the corresponding element in the other set and vice verse; it’s this process that I can’t figure out (fear not, there’s Perugia Notes ;)
I’ll be back as soon I get a feel for what’s going on here … for this is a simple instance of more involved Yoneda actions of categories, Professor F. William Lawvere’s functorial semantics, and the holy grail of Isbell conjugacy between geometry of figures and its algebra of properties that we are after.
