## Compositional Thermostatics (Part 3)

guest post by Owen Lynch

This is the third part (Part 1, Part 2) of a blog series on a paper that we wrote recently:

• John Baez, Owen Lynch and Joe Moeller, Compositional thermostatics.

In the previous two posts we talked about what a thermostatic system was, and how we think about composing them. In this post, we are going to back up from thermostatic systems a little bit, and talk about operads: a general framework for composing things! But we will not yet discuss how thermostatic systems use this framework—we’ll do that in the next post.

The basic idea behind this framework is the following. Suppose that we have a bunch of widgets, and we want to compose these widgets. If we are lucky, given two widgets there is a natural way of composing them. This is the case if the widgets are elements of a monoid; we simply use the monoid operation. This is also the case if the widgets are morphisms in a category; if the domain and codomain of two widgets match up, then they can be composed. More generally, n-morphisms in a higher category also have natural ways of composition.

However, there is not always a canonical way of composing widgets. For instance, let $R$ be a commutative ring, and let $a$ and $b$ be elements of $R.$ Then there are many ways to compose them: we could add them, subtract them, multiply them, etc. In fact any element of the free commutative ring $\mathbb{Z}[x,y]$ gives a way of composing a pair of elements in a commutative ring. For instance, $x^2 + xy - y^2,$ when applied to $a$ and $b,$ gives $a^2 + ab - b^2.$ Note that there is nothing special here about the fact that we started with two elements of $R;$ we could start with as many elements of $R$ as we liked, say, $a_1,\ldots,a_n,$ and any element of $\mathbb{Z}[x_1,\ldots,x_n]$ would give a ‘way of composing’ $a_1,\ldots,a_n.$

The reader familiar with universal algebra should recognize that this situation is very general: we could do the exact same thing with vector spaces, modules, groups, algebras, or any more exotic structures that support a notion of ‘free algebra on $n$ variables’.

Let’s also discuss a less algebraic example. A point process $X$ on a subset of Euclidean space $A \subseteq \mathbb{R}^n$ can be described as an assignment of a $\mathbb{N}$-valued random variable $X_U$ to each measurable set $U \subseteq A$ that is countably additive under disjoint union of measurable sets.

The interpretation is that a point process gives a random collection of points in $A,$ and $X_U$ counts how many points fall in $U.$ Moreover, this collection of points cannot have a limit point; there cannot be infinitely many points in any compact subset of $A.$

Now suppose that $f \colon B \to A$ and $g \colon C \to A$ are rigid embeddings such that $f(B) \cap g(C) = \emptyset,$ and that $X$ is a point process on $B$ and $Y$ is a point process on $C.$ Then we can define a new point process $Z$ on $A$ (assuming that $X$ and $Y$ are independent) by letting

$Z_U = X_{f^{-1}(U)}+ Y_{g^{-1}(U)}$

This is the union of the point process $X$ running in $f(B)$ and the point process $Y$ running in $g(C).$

The precise details here are not so important: what I want to display is the intuition that we are geometrically composing things that ‘live on’ a space. The embeddings $f$ and $g$ give us a way of gluing together a point process on $B$ and a point process on $C$ to get a point process on $A.$ We could have picked something else that lives on a space, like a scalar/vector field, but I chose point processes because they are easy to visualize and composing them is fairly simple (when composing vector fields one has to be careful that they ‘match’ at the edges).

In all of the examples in the previous section, we have things that we want to compose, and ways of composing them. This situation is formalized by operads and operad algebras (which we will define very shortly). However, the confusing part is that the operad part corresponds to'”ways of composing them’, and the operad algebra part corresponds to ‘things we want to compose’. Thus, the mathematics is somewhat ‘flipped’ from the way of thinking that comes naturally; we first think about the ways of composing things, and then we think about what things we want to compose, rather than first thinking about the things we want to compose and only later thinking about the ways of composing them!

Unfortunately, this is the logical way of presenting operads and operad algebras; we must define what an operad is before we can talk about their algebras, even if what we really care about is the algebras. Thus, without further ado, let us define what an operad is.

An operad $\mathcal{O}$ consists of a collection $\mathcal{O}_0$ of types (which are abstract, just like the ‘objects’ in a category are abstract), and for every list of types $X_1,\ldots,X_n,Y \in \mathcal{O}_0,$ a collection of operations $\mathcal{O}(X_1,\ldots,X_n;Y).$

These operations are the ‘ways of composing things’, but they themselves can be composed by ‘feeding into’ each other, in the following way.

Suppose that $g \in \mathcal{O}(Y_1,\ldots,Y_n;Z)$ and for each $i = 1,\ldots,n,$ $f_i \in \mathcal{O}(X_{i,1},\ldots,X_{i,k_i};Y_i).$ Then we can make an operation

$g(f_1,\ldots,f_n) \in \mathcal{O}(X_{1,1},\ldots,X_{1,k_1},\ldots,X_{n,1},\ldots,X_{n,k_n};Z)$

We visualize operads by letting an operation be a circle that can take several inputs and produces a single output. Then composition of operations is given by attaching the output of circles to the input of other circles. Pictured below is the composition of a unary operator $f_1,$ a nullary operator $f_2,$ and a binary operator $f_3$ with a ternary operator $g$ to create a ternary operator $g(f_1,f_2,f_3).$

Additionally, for every type $X \in \mathcal{O}_0,$ there is an ‘identity operation’ $1_X \in \mathcal{O}(X;X)$ that satisfies for any $g \in \mathcal{O}(X_1,\ldots,X_n;Y)$

$g(1_{X_1},\ldots,1_{X_n}) = g$

and for any $f \in \mathcal{O}(X;Y)$

$1_Y(f) = f$

There is also an associativity law for composition that is a massive pain to write out explicitly, but is more or less exactly as one would expect. For unary operators $f,g,h,$ it states

$f(g(h)) = f(g)(h)$

The last condition for being an operad is that if $f \in \mathcal{O}(X_1,\ldots,X_n;Y)$ and $\sigma \in S(n),$ the symmetric group on $n$ elements, then we can apply $\sigma$ to $f$ to get

$\sigma^\ast(f) \in \mathcal{O}(X_{\sigma(1)},\ldots,X_{\sigma(n)};Y).$

We require that $(\sigma \tau)^\ast(f) = \tau^\ast(\sigma^\ast(f))$ if $\sigma,\tau \in S(n),$ and there are also some conditions for how $\sigma^\ast$ interacts with composition, which can be straightforwardly derived from the intuition that $\sigma^\ast$ permutes the arguments of an operation.

Note that our definition of an operad is what might typically be known as a ‘symmetric, colored operad’, but as we will always be using symmetric, colored operads, we choose to simply drop the modifiers.

That was a long definition, so it is time for an example. This example corresponds to the first situation in the first section, where we wanted to compose ring elements.

Define $\mathcal{R}$ to be an operad with one type, which we will call $R \in \mathcal{R}_0,$ and let $\mathcal{R}(R^n;R) = \mathbb{Z}[x_1,\ldots,x_n],$ where $\mathcal{R}(R^n;R)$ is $\mathcal{R}(R,\ldots,R;R)$ with $R$ repeated $n$ times.

Composition is simply polynomial substitution. That is, if

$q(y_1,\ldots,y_n) \in \mathbb{Z}[y_1,\ldots,y_n] \cong \mathcal{R}(R^n;R)$

and

$p_i(x_{i,1},\ldots,x_{i,k_i}) \in \mathbb{Z}[x_{i,1},\ldots,x_{i,k_i}] \cong \mathcal{R}(R^{k_i};R)$

then

$q(p_1(x_{1,1},\ldots,x_{1,k_1}),\ldots,p_n(x_{n,1},\ldots,x_{n,k_n})) \in \mathcal{R}(R^{\sum_{i=1}^n k_i};R)$

is the composite of $p_1,\ldots,p_n,q.$ For instance, composing

$x^2 \in \mathbb{Z}[x] \cong \mathcal{R}(R;R)$

and

$y+z \in \mathbb{Z}[y,z] \cong \mathcal{R}(R,R;R)$

results in

$(y+z)^2 \in \mathbb{Z}[y,z] \cong \mathcal{R}(R,R;R)$

The reader is invited to supply details for identities and the symmetry operators.

For the other example, define an operad $\mathcal{P}$ by letting $\mathcal{P}_0$ be the set of compact subsets of $\mathbb{R}^2$ (we could consider something more exciting, but this works fine and is easy to visualize). An operation $f \in \mathcal{P}(X_1,\ldots,X_n;Y)$ consists of disjoint embeddings $f_1,\ldots,f_n,$ where $f_i \colon X_i \to Y.$

We can visualize such an operation as simply a shape with holes in it.

Composition of such operations is just given by nesting the holes.

The outcome of the above composition is given by simply taking away the intermediate shapes (i.e. the big circle and the triangle).

Another source of examples for operads comes from the following construction. Suppose that $(C,\otimes,1)$ is a symmetric monoidal category. Define $\mathrm{Op}(C,\otimes,1) = \mathrm{Op}(C)$ by letting

$\mathrm{Op}(C)_0 = C_0$

where $C_0$ is the collection of objects in $C,$ and

$\mathrm{Op}(C)(X_1,\ldots,X_n;Y) = \mathrm{Hom}_C(X_1 \otimes \cdots \otimes X_n, Y)$

To compose operations $f_1,\ldots,f_n$ and $g$ (assuming that the types are such that these are composable), we simply take $g \circ (f_1 \otimes \ldots \otimes f_n).$ Moreover, the identity operation is simply the identity morphism, and the action of $\sigma \in S(n)$ is given by the symmetric monoidal structure.

In fact, the second example that we talked about is an example of this construction! If we let $C$ be the category where the objects are compact subsets of $\mathbb{R}^2,$ with embeddings as the morphisms, and let the symmetric monoidal product be disjoint union, then it is not too hard to show that the operad we end up with is the same as the one we described above.

Perhaps the most important example of this construction is when it is applied to $(\mathsf{Set}, \times, 1),$ because this is important in the next section! This operad has as types, sets, and an operation

$f \in \mathrm{Op}(\mathsf{Set})(X_1,\ldots,X_n;Y)$

is simply a function

$f \colon X_1 \times \cdots \times X_n \to Y$

Although ‘operad algebra’ is the name that has stuck in the literature, I think a better term would be ‘operad action’, because the analogy to keep in mind is that of a group action. A group action allows a group to ‘act on’ elements of a set; an operad algebra similarly allows an operad to ‘act on’ elements of a set.

Moreover, a group action can be described as a functor from the 1-element category representing that group to $\mathsf{Set},$ and as we will see, an operad algebra can also be described as an ‘operad morphism’ from the operad to $\mathrm{Op}(\mathsf{Set}),$ the operad just described in the last section.

In fact, this is how we will define an operad algebra; first we will define what an operad morphism is, and then we will define an operad algebra as an operad morphism to $\mathrm{Op}(\mathsf{Set}).$

An operad morphism $F$ from an operad $\mathcal{O}$ to an operad $\mathcal{P}$ is exactly what one would expect: it consists of

• For every $X_1,\ldots,X_n,Y \in \mathcal{O}_0,$ a map

$F \colon \mathcal{O}(X_1,\ldots,X_n;Y) \to \mathcal{P}(F(X_1),\ldots,F(X_n);F(Y))$

such that $F$ commutes with all of the things an operad does, i.e. composition, identities, and the action of $\sigma \in S(n).$

Thus an operad morphism $F$ from $\mathcal{O}$ to $\mathrm{Op}(\mathsf{Set}),$ also known as an operad algebra, consists of

• A set $F(X)$ for every $X \in \mathcal{O}_0$
• A function $F(f) \colon F(X_1) \times \cdots \times F(X_n) \to F(Y)$ for every operation $f \in \mathcal{O}(X_1,\ldots,X_n;Y)$

such that the assignment of sets and functions preserves identities, composition, and the action of $\sigma \in S(n).$

Without further ado, let’s look at the examples. From any ring $A$ we can produce an algebra $F_A$ of $\mathcal{R}.$ We let $F_A(R) = A$ (considered as a set), and for

$p(x_1,\ldots,x_n) \in \mathbb{Z}[x_1,\ldots,x_n] = \mathcal{R}(X_1,\ldots,X_n;Y)$

we let

$F(p)(a_1,\ldots,a_n) = p(a_1,\ldots,a_n)$

We can also make an operad algebra of point processes, $\mathrm{PP},$ for $\mathcal{P}.$ For $A \in \mathcal{P}_0,$ we let $\mathrm{PP}(A)$ be the set of point processes on $A.$ If $f \colon A_1 \sqcup \cdots \sqcup A_n \to B$ is an embedding, then we let $\mathrm{PP}(f)$ be the map that sends point processes $X_1,\ldots,X_n$ on $A_1,\ldots,A_n$ respectively to the point process $Y$ defined by

$Y_U = X_{f^{-1}(U) \cap A_1} + \cdots + X_{f^{-1}(U) \cap A_n}$

Finally, if $(C,\otimes,1)$ is a symmetric monoidal category, there is a way to make an operad algebra of $\mathrm{Op}(C)$ from a special type of functor $F \colon C \to \mathsf{Set}.$ This is convenient, because it is often easier to prove that the functor satisfies the necessary properties than it is to prove that the algebra is in fact well-formed.

The special kind of functor we need is a lax symmetric monoidal functor. This is a functor $F$ equipped with a natural transformation $\tau_{A,B} \colon F(A) \times F(B) \to F(A \otimes B)$ that is well-behaved with respect to the associator, identity, and symmetric structure of $(C, \otimes, 1).$ We call $\tau$ the laxator, and formally speaking, a lax symmetric monoidal functor consists of a functor along with a laxator. I won’t go into detail about the whole construction that makes an operad algebra out of a lax symmetric monoidal functor, but the basic idea is that given an operation $f \in \mathrm{Op}(C)(X,Y;Z)$ (which is a morphism $f \colon X \otimes Y \to Z$), we can construct a function $F(X) \times F(Y) \to F(Z)$ by composing

$\tau_{X,Y} \colon F(X) \times F(Y) \to F(X \otimes Y)$

with

$F(f) \colon F(X \otimes Y) \to F(Z)$

This basic idea can be extended using associativity to produce a function $X_1 \times \cdots \times X_n \to Y$ from an operation $f \colon X_1 \otimes \cdots \otimes X_n \to Y.$

As an example of this construction, consider point processes again. We can make a lax symmetric monoidal functor $\mathrm{PP}$ by sending a set $A$ to $\mathrm{PP}(A),$ the set of point processes on $A,$ and an embedding $f \colon A \to B$ to the map $F(f)$ that sends a point process $X$ to a point process $Y$ defined by

$Y_U = X_{f^{-1}(U)}$

The laxator $\tau_{A,B} \colon F(A) \times F(B) \to F(A \sqcup B)$ sends a point process $X$ on $A$ and a point process $Y$ on $B$ to a point process $Z$ on a $A \sqcup B$ defined by

$Z_{U} = X_{U \cap A} + Y_{U \cap B}$

The reader should inspect this definition and think about why it is equivalent to the earlier definition for the operad algebra of point processes.

### Summary

This was a long post, so I’m going to try and go over the main points so that you can organize what you just learned in some sort of coherent fashion.

First I talked about how there frequently arises situations in which there isn’t a canonical way of ‘composing’ two things. The two examples that I gave were elements of a ring, and structures on spaces, specifically point processes.

I then talked about the formal way that we think about these situations. Namely, we organize the ‘ways of composing things’ into an operad, and then we organize the ‘things that we want to compose’ into an operad algebra. Along the way, I discussed a convenient way of making an operad out of a symmetric monoidal category, and an operad algebra out of a lax symmetric monoidal functor.

This construction will be important in the next post, when we make an operad of ‘ways of composing thermostatic systems’ and an operad algebra of thermostatic systems to go along with it.

See all four parts of this series:

Part 1: thermostatic systems and convex sets.

Part 2: composing thermostatic systems.

Part 3: operads and their algebras.

Part 4: the operad for composing thermostatic systems.

### 4 Responses to Compositional Thermostatics (Part 3)

1. John Baez says:

Since there’s no operad (or more precisely, no Set-operad, the only kind discussed here) whose algebras are exactly commutative rings, the reader is invited to figure out what are the algebras of Owen’s operad called $\mathcal{R}.$

• Toby Bartels says:

I give up. I've looked at this back and forth, and as long as Owen means what I mean by ‘polynomial’ (an element of the free commutative ring on the set of variables), then it seems to me that the algebras are commutative rings. Although I should look at what the algebra homomorphisms are (which was not defined here); maybe the true answer is Morita equivalence classes of rings or something like that.

• John Baez says:

There can’t possibly be an operad whose algebras in Set are commutative rings, since operads can only describe sets equipped with operations obeying equations with no deletion or duplication of variables. This is why there’s an operad for monoids (that is, an operad whose algebras in Set are monoids) but no operad for groups: the axiom

$g g^{-1} = 1$

duplicates a variable at left and deletes it at right! There’s also no operad for rings, or commutative rings, due to the axiom

$a(b+c) = ab + ac$

There’s an adjunction between Lawvere theories and operads, so you can take the Lawvere theory for groups or rings or commutative rings and then take its underlying operad—but the resulting operad has something else as its algebras in Set.

I believe this is what Owen is implicitly doing here: taking the Lawvere theory for commutative rings, whose n-ary operations are integer-coefficient polynomials in n variables, and then taking the underlying operad.

I’m not sure exactly what the algebras of this operad are like: one possibility is that they’re things like rings but not obeying the axioms where variables are duplicated or deleted, like

$a(b+c) = ab + ac$

and

$a 0 = 0$

• Toby Bartels says:

I thought of that, but then Owen also said that he's using a fancy version of operads, so I decided to focus on the details. And in the details, the set of $n$-ary operations is the set of polynomials in $n$ variables, and $\mathrm x_1(\mathrm x_2+\mathrm x_3)$ is equal to $\mathrm x_1\mathrm x_2+\mathrm x_1\mathrm x_3$ in the set of polynomials in (say) $3$ variables.

Maybe the problem is with the composition of operators. If I compose the $2$-variable polynomial $\mathrm x_1\mathrm x_2$ with the $1$-variable polynomial $\mathrm x_1$ (which is the identity) and the $2$-variable polynomial $\mathrm x_1+\mathrm x_2$, then I get the $1+2=3$-variable polynomial $\mathrm x_1(\mathrm x_2+\mathrm x_3)=\mathrm x_1\mathrm x_2+\mathrm x_1\mathrm x_3$. But if I compose the $2$-variable polynomial $\mathrm x_1+\mathrm x_2$ with the $2$-variable polynomial $\mathrm x_1\mathrm x_2$ twice, then I get the $2+2=4$-variable polynomial $\mathrm x_1\mathrm x_2+\mathrm x_3\mathrm x_4$. And I could apply a symmetry to this to turn it into $\mathrm x_1\mathrm x_3+\mathrm x_2\mathrm x_4$, but nothing will turn it into $\mathrm x_1\mathrm x_2+\mathrm x_1\mathrm x_3$, which is an operation with a different arity. So in an algebra of the operad, applying $\mathrm x_1\mathrm x_2+\mathrm x_1\mathrm x_3$ to the $3$ algebra elements $a$, $b$, and $c$, will typically give a different result from applying $\mathrm x_1\mathrm x_3+\mathrm x_2\mathrm x_4$ to the $4$ elements $a$, $a$, $b$, and $c$, even though we'd be tempted to write both of them as $ab+ac$.

So it's not that we don't have $a(b+c)=ab+ac$; we should be so lucky! No, instead, we have two different interpretations of $ab+ac$, one of which equals $a(b+c)$, and one of which doesn't. Or rather, we have infinitely many interpretations of $ab+ac$, one of which equals one of the infinitely many interpretations of $a(b+c)$, and another of which doesn't equal any of them. In fact, every expression has infinitely many distinct interpretations, even integer constants!

So the algebras are actually extremely complicated.

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