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# Lecture 60 - Chapter 4: Closed Monoidal Posets

edited February 2020

Okay, now I've rather carefully discussed one example of $$\mathcal{V}$$-enriched profunctors, and rather sloppily discussed another. Now it's time to build the general framework that can handle both these examples.

We can define $$\mathcal{V}$$-enriched categories whenever $$\mathcal{V}$$ is a monoidal preorder: we did that way back in Lecture 29. We can also define $$\mathcal{V}$$-enriched functors whenever $$\mathcal{V}$$ is a monoidal preorder: we did that in Lecture 31. But to define $$\mathcal{V}$$-enriched profunctors, we need $$\mathcal{V}$$ to be a bit better. We can see why by comparing our examples.

Our first example involved $$\mathcal{V} = \textbf{Bool}$$. A feasibility relation

$$\Phi : X \nrightarrow Y$$ between preorders is a monotone function

$$\Phi: X^{\text{op}} \times Y\to \mathbf{Bool} .$$ We shall see that a feasibility relation is the same as a $$\textbf{Bool}$$-enriched profunctor.

Our second example involved $$\mathcal{V} = \textbf{Cost}$$. I said that a $$\textbf{Cost}$$-enriched profunctor

$$\Phi : X \nrightarrow Y$$ between $$\mathbf{Cost}$$-enriched categories is a $$\textbf{Cost}$$-enriched functor

$$\Phi: X^{\text{op}} \times Y \to \mathbf{Cost}$$ obeying some conditions. But I let you struggle to guess those conditions... without enough clues to make it easy!

To fit both our examples in a general framework, we start by considering an arbitrary monoidal preorder $$\mathcal{V}$$. $$\mathcal{V}$$-enriched profunctors will go between $$\mathcal{V}$$-enriched categories. So, let $$\mathcal{X}$$ and $$\mathcal{Y}$$ be $$\mathcal{V}$$-enriched categories. We want to make this definition:

Tentative Definition. A $$\mathcal{V}$$-enriched profunctor

$$\Phi : \mathcal{X} \nrightarrow \mathcal{Y}$$ is a $$\mathcal{V}$$-enriched functor

$$\Phi: \mathcal{X}^{\text{op}} \times \mathcal{Y} \to \mathcal{V} .$$ Notice that this handles our first example very well. But some questions appear in our second example - and indeed in general. For our tentative definition to make sense, we need three things:

1. We need $$\mathcal{V}$$ to itself be a $$\mathcal{V}$$-enriched category.

2. We need any two $$\mathcal{V}$$-enriched category to have a 'product', which is again a $$\mathcal{V}$$-enriched category.

3. We need any $$\mathcal{V}$$-enriched category to have an 'opposite', which is again a $$\mathcal{V}$$-enriched category.

Items 2 and 3 work fine whenever $$\mathcal{V}$$ is a commutative monoidal poset. We'll see why in Lecture 62.

Item 1 is trickier, and indeed it sounds rather scary. $$\mathcal{V}$$ began life as a humble monoidal preorder. Now we're wanting it to be enriched in itself! Isn't that circular somehow?

Yes! But not in a bad way. Category theory often eats its own tail, like the mythical ourobous, and this is an example.

To get $$\mathcal{V}$$ to become a $$\mathcal{V}$$-enriched category, we'll demand that it be 'closed'. For starters, let's assume it's a monoidal poset, just to avoid some technicalities.

Definition. A monoidal poset is closed if for all elements $$x,y \in \mathcal{V}$$ there is an element $$x \multimap y \in \mathcal{V}$$ such that

$$x \otimes a \le y \text{ if and only if } a \le x \multimap y$$ for all $$a \in \mathcal{V}$$.

This will let us make $$\mathcal{V}$$ into a $$\mathcal{V}$$-enriched category by setting $$\mathcal{V}(x,y) = x \multimap y$$. But first let's try to understand this concept a bit!

We can check that our friend $$\mathbf{Bool}$$ is closed. Remember, we are making it into a monoidal poset using 'and' as its binary operation: its full name is $$\lbrace \text{true},\text{false}\rbrace, \wedge, \text{true})$$. Then we can take $$x \multimap y$$ to be 'implication'. More precisely, we say $$x \multimap y = \text{true}$$ iff $$x$$ implies $$y$$. Even more precisely, we define:

$$\text{true} \multimap \text{true} = \text{true}$$ $$\text{true} \multimap \text{false} = \text{false}$$ $$\text{false} \multimap \text{true} = \text{true}$$ $$\text{false} \multimap \text{false} = \text{true} .$$ Puzzle 188. Show that with this definition of $$\multimap$$ for $$\mathbf{Bool}$$ we have

$$a \wedge x \le y \text{ if and only if } a \le x \multimap y$$ for all $$a,x,y \in \mathbf{Bool}$$.

We can also check that our friend $$\mathbf{Cost}$$ is closed! Remember, we are making it into a monoidal poset using $$+$$ as its binary operation: its full name is $$[0,\infty], \ge, +, 0)$$. Then we can define $$x \multimap y$$ to be 'subtraction'. More precisely, we define $$x \multimap y$$ to be $$y - x$$ if $$y \ge x$$, and $$0$$ otherwise.

Puzzle 189. Show that with this definition of $$\multimap$$ for $$\mathbf{Cost}$$ we have

$$a + x \le y \text{ if and only if } a \le x \multimap y .$$ But beware. We have defined the ordering on $$\mathbf{Cost}$$ to be the opposite of the usual ordering of numbers in $$[0,\infty]$$. So, $$\le$$ above means the opposite of what you might expect!

Next, two more tricky puzzles. Next time I'll show you in general how a closed monoidal poset $$\mathcal{V}$$ becomes a $$\mathcal{V}$$-enriched category. But to appreciate this, it may help to try some examples first:

Puzzle 190. What does it mean, exactly, to make $$\mathbf{Bool}$$ into a $$\mathbf{Bool}$$-enriched category? Can you see how to do this by defining

$$\mathbf{Bool}(x,y) = x \multimap y$$ for all $$x,y \in \mathbf{Bool}$$, where $$\multimap$$ is defined to be 'implication' as above?

Puzzle 191. What does it mean, exactly, to make $$\mathbf{Cost}$$ into a $$\mathbf{Cost}$$-enriched category? Can you see how to do this by defining

$$\mathbf{Cost}(x,y) = x \multimap y$$ for all $$x,y \in \mathbf{Cost}$$, where $$\multimap$$ is defined to be 'subtraction' as above?

Note: for Puzzle 190 you might be tempted to say "a $$\mathbf{Bool}$$-enriched category is just a preorder, so I'll use that fact here". However, you may learn more if you go back to the general definition of enriched category and use that! The reason is that we're trying to understand some general things by thinking about two examples.

Puzzle 192. The definition of 'closed' above is an example of a very important concept we keep seeing in this course. What is it? Restate the definition of closed monoidal poset in a more elegant, but equivalent, way using this concept.

To read other lectures go here.

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1.

Puzzle 192: A monoidal preorder is closed iff there exists a operator: $$\multimap$$ such that $(\_\otimes x) \vdash (x\multimap \_)$

Comment Source:Puzzle 192: A monoidal preorder is _closed_ iff there exists a operator: \$$\multimap\$$ such that \$(\\_\otimes x) \vdash (x\multimap \\_)\$
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2.

possible typo: in previous lectures we've defined profunctors as $$\mathcal{V}$$-enriched functors $$\mathcal{X}^\text{op}\times\mathcal{Y}\rightarrow\mathcal{V}$$ but here we're using $$\mathcal{X}\times\mathcal{Y}^\text{op}\rightarrow\mathcal{V}$$ – is this a mistake or are we swapping round for a reason?

Comment Source:possible typo: in previous lectures we've defined profunctors as \$$\mathcal{V}\$$-enriched functors \$$\mathcal{X}^\text{op}\times\mathcal{Y}\rightarrow\mathcal{V}\$$ but here we're using \$$\mathcal{X}\times\mathcal{Y}^\text{op}\rightarrow\mathcal{V}\$$ – is this a mistake or are we swapping round for a reason?
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3.
edited July 2018

re Puzzle 188:

$$a\wedge x \leq y$$ is false iff $$a\wedge x = \text{true}$$ and $$y = \text{false}$$ iff $$a = \text{true}$$, $$x = \text{true}$$, $$y = \text{false}$$

$$a \leq x\multimap y$$ is false iff $$a = \text{true}$$ and $$x\multimap y = \text{false}$$ iff $$a = \text{true}$$, $$x = \text{true}$$, $$y = \text{false}$$

so $$a\wedge x \leq y$$ iff $$a \leq x\multimap y$$

Comment Source:re **Puzzle 188**: \$$a\wedge x \leq y\$$ is false iff \$$a\wedge x = \text{true}\$$ and \$$y = \text{false}\$$ iff \$$a = \text{true}\$$, \$$x = \text{true}\$$, \$$y = \text{false}\$$ \$$a \leq x\multimap y\$$ is false iff \$$a = \text{true}\$$ and \$$x\multimap y = \text{false}\$$ iff \$$a = \text{true}\$$, \$$x = \text{true}\$$, \$$y = \text{false}\$$ so \$$a\wedge x \leq y\$$ iff \$$a \leq x\multimap y\$$
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4.

Puzzle 190. What does it mean, exactly, to make $$\mathbf{Bool}$$ into a $$\mathbf{Bool}$$-enriched category? Can you see how to do this by defining

$$\mathbf{Bool}(x,y) = x \multimap y$$ for all $$x,y \in \mathbf{Bool}$$, where $$\multimap$$ is defined to be 'implication' as above?

Let's recall the definition John gave in Lecture 29:

Definition. A $$\mathcal{V}$$-enriched category $$\mathcal{X}$$ consists of two parts, satisfying two properties. First:

1. one specifies a set $$\mathrm{Ob}(\mathcal{X})$$, elements of which are called objects;

2. for every two objects $$x,y$$, one specifies an element $$\mathcal{X}(x,y)$$ of $$\mathcal{V}$$.

Then:

a) for every object $$x\in\text{Ob}(\mathcal{X})$$ we require that

$$I\leq\mathcal{X}(x,x) .$$ b) for every three objects $$x,y,z\in\mathrm{Ob}(\mathcal{X})$$, we require that $$\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z).$$

We want

$$\mathcal{X}(a,b) := a \multimap_{\mathbf{Bool}} b$$ Let's go through the parts of the definition.

1. The objects are specified by $$\mathrm{Ob}(\multimap_{\mathbf{Bool}}) = \{ \text{true}, \text{false} \}$$
2. The behavior of $$a \multimap_{\mathbf{Bool}} b$$ is specified by its truth table below.

$$\begin{array}{|c|c|c|} \hline a & b &a \multimap b\\ \hline \text{true}&\text{true}&\text{true}\\ \hline \mathbf{true}&\mathbf{false} &\mathbf{false}\\ \hline \text{false}&\text{true} &\text{true}\\ \hline \text{false}&\text{false} &\text{true}\\ \hline \end{array}$$ Now we need to verify the two laws, (a) and (b), hold.

(a) We need $$I \le \mathcal{X}(x,x)$$, ie $$\text{true} \leq x \multimap x$$, for all x. But as we can see from the truth table, $$x \multimap x = \text{true}$$ holds for $$x = \text{true}$$ and $$x = \text{false}$$.

(b) We need to check $$x \multimap y \wedge y \multimap z \leq x \multimap z$$. We can show this using another truth table, enumerating all of the possible values. We just need to check that the last two columns, and verify that $$x \multimap y \wedge y \multimap z \leq x \multimap z$$, remembering that $$\text{false} \le \text{true}$$.

$$\begin{array}{|c|c|c|c|c|c|c|c|} \hline x & y & z & x \multimap y & y \multimap z & x \multimap y \wedge y \multimap z & x \multimap z \\ \hline \text{true}&\text{true}&\text{true} &\text{true} &\text{true} &\mathbf{true} & \mathbf{true} \\ \hline \text{true}&\text{true}&\text{false}&\text{true} &\text{false}&\mathbf{false}&\mathbf{false}\\ \hline \text{true}&\text{false}&\text{true}&\text{false} &\text{true}&\mathbf{false}&\mathbf{true}\\ \hline \text{true}&\text{false}&\text{false}&\text{false} &\text{true}&\mathbf{false}&\mathbf{false}\\ \hline \text{false}&\text{true}&\text{true} &\text{true} &\text{true} &\mathbf{true} & \mathbf{true} \\ \hline \text{false}&\text{true}&\text{false}&\text{true} &\text{false}&\mathbf{false}&\mathbf{true}\\ \hline \text{false}&\text{false}&\text{true}&\text{true} &\text{true}&\mathbf{true}&\mathbf{true}\\ \hline \text{false}&\text{false}&\text{false}&\text{true} &\text{true}&\mathbf{true}&\mathbf{true}\\ \hline \end{array}$$

Comment Source:> **Puzzle 190.** What does it mean, exactly, to make \$$\mathbf{Bool}\$$ into a \$$\mathbf{Bool}\$$-enriched category? Can you see how to do this by defining > > $\mathbf{Bool}(x,y) = x \multimap y$ > > for all \$$x,y \in \mathbf{Bool}\$$, where \$$\multimap\$$ is defined to be 'implication' as above? Let's recall the definition John gave in [Lecture 29](https://forum.azimuthproject.org/discussion/2121/lecture-29-chapter-2-enriched-categories): > **Definition.** A **\$$\mathcal{V}\$$-enriched category** \$$\mathcal{X}\$$ consists of two parts, satisfying two properties. First: > > 1. one specifies a set \$$\mathrm{Ob}(\mathcal{X})\$$, elements of which are called **objects**; > > 2. for every two objects \$$x,y\$$, one specifies an element \$$\mathcal{X}(x,y)\$$ of \$$\mathcal{V}\$$. > > Then: > > a) for every object \$$x\in\text{Ob}(\mathcal{X})\$$ we require that > > $I\leq\mathcal{X}(x,x) .$ > > b) for every three objects \$$x,y,z\in\mathrm{Ob}(\mathcal{X})\$$, we require that > > $\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z).$ We want $\mathcal{X}(a,b) := a \multimap_{\mathbf{Bool}} b$ Let's go through the parts of the definition. 1. The objects are specified by \$$\mathrm{Ob}(\multimap_{\mathbf{Bool}}) = \\{ \text{true}, \text{false} \\} \$$ 2. The behavior of \$$a \multimap_{\mathbf{Bool}} b\$$ is specified by its [*truth table*](https://en.wikipedia.org/wiki/Material_conditional#Truth_table) below. $\begin{array}{|c|c|c|} \hline a & b &a \multimap b\\\\ \hline \text{true}&\text{true}&\text{true}\\\\ \hline \mathbf{true}&\mathbf{false} &\mathbf{false}\\\\ \hline \text{false}&\text{true} &\text{true}\\\\ \hline \text{false}&\text{false} &\text{true}\\\\ \hline \end{array}$ Now we need to verify the two laws, (a) and (b), hold. (a) We need \$$I \le \mathcal{X}(x,x)\$$, ie \$$\text{true} \leq x \multimap x\$$, for all x. But as we can see from the truth table, \$$x \multimap x = \text{true}\$$ holds for \$$x = \text{true}\$$ and \$$x = \text{false}\$$. (b) We need to check \$$x \multimap y \wedge y \multimap z \leq x \multimap z\$$. We can show this using another truth table, enumerating all of the possible values. We just need to check that the last two columns, and verify that \$$x \multimap y \wedge y \multimap z \leq x \multimap z\$$, remembering that \$$\text{false} \le \text{true}\$$. $\begin{array}{|c|c|c|c|c|c|c|c|} \hline x & y & z & x \multimap y & y \multimap z & x \multimap y \wedge y \multimap z & x \multimap z \\\\ \hline \text{true}&\text{true}&\text{true} &\text{true} &\text{true} &\mathbf{true} & \mathbf{true} \\\\ \hline \text{true}&\text{true}&\text{false}&\text{true} &\text{false}&\mathbf{false}&\mathbf{false}\\\\ \hline \text{true}&\text{false}&\text{true}&\text{false} &\text{true}&\mathbf{false}&\mathbf{true}\\\\ \hline \text{true}&\text{false}&\text{false}&\text{false} &\text{true}&\mathbf{false}&\mathbf{false}\\\\ \hline \text{false}&\text{true}&\text{true} &\text{true} &\text{true} &\mathbf{true} & \mathbf{true} \\\\ \hline \text{false}&\text{true}&\text{false}&\text{true} &\text{false}&\mathbf{false}&\mathbf{true}\\\\ \hline \text{false}&\text{false}&\text{true}&\text{true} &\text{true}&\mathbf{true}&\mathbf{true}\\\\ \hline \text{false}&\text{false}&\text{false}&\text{true} &\text{true}&\mathbf{true}&\mathbf{true}\\\\ \hline \end{array}$ 
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5.
edited July 2018

There is a general proof of (b) above.

For any commutative closed monoidal preorder we have

$$x \multimap y \otimes y \multimap z \le x \multimap z$$ We don't really care if we are in $$\mathbf{Bool}$$ or $$\mathbf{Cost}$$ or whatever.

First observe

$$a \multimap b \le a \multimap b \text{ for all } a \text{ and } b$$ This follows from reflexivity of $$\le$$. By the adjointness of $$\otimes \dashv\, \multimap$$ we know

$$a \multimap b \otimes a \le b \text{ for all } a \text{ and } b$$ In particular, we have

\begin{align} x \multimap y \otimes x & \le y \tag{A}\\ y \multimap z \otimes y & \le z \tag{B} \end{align} If we take $$(A)$$, we can tensor both sides by applying $$y \multimap z\, \otimes$$ to get

$$y \multimap z \otimes x \multimap y \otimes x \le y \multimap z \otimes y$$ This follows since $$\otimes$$ is monotone. Along with $$(B)$$ and transitivity of $$\le$$ we have

$$y \multimap z \otimes x \multimap y \otimes x \le z$$ applying $$\otimes \dashv\, \multimap$$ once again to this gives

$$y \multimap z \otimes x \multimap y \le x \multimap z \tag{★}$$ Finally, since we are assuming $$\otimes$$ is commutative, we know

$$y \multimap z \otimes x \multimap y = x \multimap y \otimes y \multimap z$$ taking this along with $$(★)$$ finally yields:

$$x \multimap y \otimes y \multimap z \le x \multimap z$$

Comment Source:There is a general proof of (b) above. For *any* commutative closed monoidal preorder we have $x \multimap y \otimes y \multimap z \le x \multimap z$ We don't really care if we are in \$$\mathbf{Bool}\$$ or \$$\mathbf{Cost}\$$ or whatever. First observe $a \multimap b \le a \multimap b \text{ for all } a \text{ and } b$ This follows from reflexivity of \$$\le\$$. By the adjointness of \$$\otimes \dashv\, \multimap\$$ we know $a \multimap b \otimes a \le b \text{ for all } a \text{ and } b$ In particular, we have \begin{align} x \multimap y \otimes x & \le y \tag{A}\\\\ y \multimap z \otimes y & \le z \tag{B} \end{align} If we take \$$(A)\$$, we can *tensor* both sides by applying \$$y \multimap z\, \otimes \$$ to get $y \multimap z \otimes x \multimap y \otimes x \le y \multimap z \otimes y$ This follows since \$$\otimes\$$ is monotone. Along with \$$(B)\$$ and transitivity of \$$\le\$$ we have $y \multimap z \otimes x \multimap y \otimes x \le z$ applying \$$\otimes \dashv\, \multimap\$$ once again to this gives $y \multimap z \otimes x \multimap y \le x \multimap z \tag{★}$ Finally, since we are assuming \$$\otimes\$$ is commutative, we know $y \multimap z \otimes x \multimap y = x \multimap y \otimes y \multimap z$ taking this along with \$$(★)\$$ finally yields: $x \multimap y \otimes y \multimap z \le x \multimap z$
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6.
edited July 2018

I think we can simplify that slightly @Matthew

We have your (A) and (B), ie $$x\otimes (x\multimap y) \leq y$$ and $$y \otimes (y\multimap z) \leq z$$.

Therefore $$x\otimes (x\multimap y) \otimes (y\multimap z) \leq y \otimes (y\multimap z) \leq z$$.

So by adjunction $$(x\multimap y) \otimes (y\multimap z) \leq (x\multimap z)$$.

(EDIT: Oh I see this is the same proof, but you spelled it out in a bit more detail!)

Comment Source:I think we can simplify that slightly @Matthew We have your (A) and (B), ie \$$x\otimes (x\multimap y) \leq y\$$ and \$$y \otimes (y\multimap z) \leq z\$$. Therefore \$$x\otimes (x\multimap y) \otimes (y\multimap z) \leq y \otimes (y\multimap z) \leq z\$$. So by adjunction \$$(x\multimap y) \otimes (y\multimap z) \leq (x\multimap z)\$$. (EDIT: Oh I see this is the same proof, but you spelled it out in a bit more detail!)
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7.
edited July 2018

I think we can simplify that slightly @Matthew

We have your (A) and (B), ie $$x\otimes (x\multimap y) \leq y$$ and $$y \otimes (y\multimap z) \leq z$$.

Sure, but it's important to note that we are using commutativity to flip (B).

Therefore $$x\otimes (x\multimap y) \otimes (y\multimap z) \leq y \otimes (y\multimap z) \leq z$$.

So by adjunction $$(x\multimap y) \otimes (y\multimap z) \leq (x\multimap z)$$.

Again, you are using commutativity...

That's pedantry, I know. The proof is ugly and long if we don't just use commutativity implicitly.

However, I am not sure if this theorem is true in a closed non-commutative monoidal preorder.

Comment Source:> I think we can simplify that slightly @Matthew > > We have your (A) and (B), ie \$$x\otimes (x\multimap y) \leq y\$$ and \$$y \otimes (y\multimap z) \leq z\$$. Sure, but it's important to note that we are using commutativity to flip (B). > Therefore \$$x\otimes (x\multimap y) \otimes (y\multimap z) \leq y \otimes (y\multimap z) \leq z\$$. > > So by adjunction \$$(x\multimap y) \otimes (y\multimap z) \leq (x\multimap z)\$$. Again, you are using commutativity... That's pedantry, I know. The proof is ugly and long if we don't just use commutativity implicitly. However, I am not sure if this theorem is true in a closed non-commutative monoidal preorder.
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8.
edited July 2018

One detail that puzzles me about the adjunction: the maps $$(x\otimes -)$$ and $$(x\multimap -)$$ are monotone but they don't seem to (necessarily) be monoidal monotone. So this is an adjunction between preorders, not an adjunction between monoidal preorders. That seems a bit unexpected to me.

Comment Source:One detail that puzzles me about the adjunction: the maps \$$(x\otimes -)\$$ and \$$(x\multimap -)\$$ are monotone but they don't seem to (necessarily) be *monoidal* monotone. So this is an adjunction between preorders, not an adjunction between monoidal preorders. That seems a bit unexpected to me.
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9.
edited July 2018

Anindya wrote:

One detail that puzzles me about the adjunction: the maps $$(x\otimes -)$$ and $$(x\multimap -)$$ are monotone but ...

You know, there might be a typo in the lecture notes.

I usually see $$(x\otimes -) \dashv (x\multimap -)$$. For instance, this is how the Haskell adjunctions library defines them (see the hackage docs).

But in the lecture notes, John wrote

$$a \otimes x \le y \text{ if and only if } a \le x \multimap y \tag{✴}$$

So technically $$(- \otimes x) \dashv (x\multimap -)$$ if we don't assume commutativity.

$$x \otimes a \le y \text{ if and only if } a \le x \multimap y \tag{✴'}$$ Then your quick proof in #6 works, the theorem we proved holds without demanding commutativity, and $$(x\otimes -) \dashv (x\multimap -)$$ like you say.

Comment Source:[Anindya](https://forum.azimuthproject.org/discussion/comment/20082/#Comment_20082) wrote: > One detail that puzzles me about the adjunction: the maps \$$(x\otimes -)\$$ and \$$(x\multimap -)\$$ are monotone but ... You know, there might be a typo in the lecture notes. I usually see \$$(x\otimes -) \dashv (x\multimap -)\$$. For instance, this is how the Haskell adjunctions library defines them (see [the hackage docs](https://hackage.haskell.org/package/adjunctions-4.0.2/docs/Data-Functor-Adjunction.html#t:Adjunction)). But in the lecture notes, John wrote > $a \otimes x \le y \text{ if and only if } a \le x \multimap y \tag{✴}$ So technically \$$(- \otimes x) \dashv (x\multimap -)\$$ if we don't assume commutativity. If we go instead with $x \otimes a \le y \text{ if and only if } a \le x \multimap y \tag{✴'}$ Then your quick proof in [#6](https://forum.azimuthproject.org/discussion/comment/20080/#Comment_20080) works, the theorem we proved holds without demanding commutativity, and \$$(x\otimes -) \dashv (x\multimap -)\$$ like you say.
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10.
edited July 2018

I've been thinking about the case when $$\otimes$$ is not symmetric.

As @Matthew says, we can turn $$\mathcal{V}$$ into a $$\mathcal{V}$$-enriched preorder if $$(x\otimes -)$$ has a right adjoint.

If on the other hand it's $$(-\otimes x)$$ that has a right adjoint, the proof doesn't work. However in this case we could slightly vary the definition of a $$\mathcal{V}$$-enriched category to use right-to-left composition instead of left-to-right composition.

This amount to replacing the $$\mathcal{V}(x, y)\otimes\mathcal{V}(y, z)\leq\mathcal{V}(x, z)$$ condition with $$\mathcal{V}(y, z)\otimes\mathcal{V}(x, y)\leq\mathcal{V}(x, z)$$ – similarly to how we define $$\text{hom}$$ in a $$\textbf{Set}$$-category. Then the proof would go through.

So whether to go with $$(x\otimes -)$$ or $$(-\otimes x)$$ depends on which way round we write composites.

A speculative interpretation of this: we could read the non-symmetric $$A \otimes B$$ as "A and then B". In this case a right adjoint of $$A \otimes -$$ would be like a permit to turn any A that you've already got into a B, because $$A \otimes (A \multimap B) \leq B$$, while a right adjoint of $$- \otimes A$$ would be a permit to turn any A that you get in the future into a B, because $$(A \multimap B)\otimes A \leq B$$. Maybe there's some "time-conscious" logic sitting in here that might be useful in some situations.

Comment Source:I've been thinking about the case when \$$\otimes\$$ is not symmetric. As @Matthew says, we can turn \$$\mathcal{V}\$$ into a \$$\mathcal{V}\$$-enriched preorder if \$$(x\otimes -)\$$ has a right adjoint. If on the other hand it's \$$(-\otimes x)\$$ that has a right adjoint, the proof doesn't work. However in this case we could slightly vary the definition of a \$$\mathcal{V}\$$-enriched category to use right-to-left composition instead of left-to-right composition. This amount to replacing the \$$\mathcal{V}(x, y)\otimes\mathcal{V}(y, z)\leq\mathcal{V}(x, z)\$$ condition with \$$\mathcal{V}(y, z)\otimes\mathcal{V}(x, y)\leq\mathcal{V}(x, z)\$$ – similarly to how we define \$$\text{hom}\$$ in a \$$\textbf{Set}\$$-category. Then the proof would go through. So whether to go with \$$(x\otimes -)\$$ or \$$(-\otimes x)\$$ depends on which way round we write composites. A speculative interpretation of this: we could read the non-symmetric \$$A \otimes B\$$ as "A and then B". In this case a right adjoint of \$$A \otimes -\$$ would be like a permit to turn any A *that you've already got* into a B, because \$$A \otimes (A \multimap B) \leq B\$$, while a right adjoint of \$$- \otimes A\$$ would be a permit to turn any A *that you get in the future* into a B, because \$$(A \multimap B)\otimes A \leq B\$$. Maybe there's some "time-conscious" logic sitting in here that might be useful in some situations.
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11.
edited July 2018

Just thinking out loud here...

If we define a function called $$\mathrm{when}$$,

$\mathrm{when}(b,x) :\mathbf{Bool}\otimes \mathcal{X} \to \mathcal{X} \\ := \text{if } b \text{ then } x \text{ else } \varnothing,$

then $$\mathrm{when}$$ is adjoint (in fact equivalent to) to taking the Cartesian product with a Boolean variable,

$(b \times -) \vdash \mathrm{when}(b,-).$

So then a $$\mathbf{Cost}$$-morphism is something like $$\mathrm{when}(x\multimap y,r)$$, ie are $$\mathbf{Bool}$$-morphisms that are labeled by a real number.

Comment Source:Just thinking out loud here... If we define a function called \$$\mathrm{when}\$$, \$\mathrm{when}(b,x) :\mathbf{Bool}\otimes \mathcal{X} \to \mathcal{X} \\\\ := \text{if } b \text{ then } x \text{ else } \varnothing, \$ then \$$\mathrm{when}\$$ is adjoint (in fact equivalent to) to taking the Cartesian product with a Boolean variable, \$(b \times -) \vdash \mathrm{when}(b,-). \$ So then a \$$\mathbf{Cost}\$$-morphism is something like \$$\mathrm{when}(x\multimap y,r)\$$, ie are \$$\mathbf{Bool}\$$-morphisms that are labeled by a real number.
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12.
edited July 2018

Anindya wrote:

possible typo: in previous lectures we've defined profunctors as $$\mathcal{V}$$-enriched functors $$\mathcal{X}^\text{op}\times\mathcal{Y}\rightarrow\mathcal{V}$$ but here we're using $$\mathcal{X}\times\mathcal{Y}^\text{op}\rightarrow\mathcal{V}$$ – is this a mistake or are we swapping round for a reason?

This is just a mistake - I'll fix it, thanks!

I repeatedly amaze myself by my inability to stick to arbitrary conventions. In my defense I'll mention that there's a heated and pointless debate among category theorists about which convention is better. It doesn't really matter which you use as long as you stick with it, but people who get used to one find the other counterintuitive. I'm trying to use Fong and Spivak's convention $$\mathcal{X}^\text{op}\times\mathcal{Y}\rightarrow\mathcal{V}$$, and this is actually my favorite convention, mainly because it makes the hom-functor

$$\text{hom} : \mathcal{C}^{\text{op}} \times \mathcal{C} \to \mathbf{Set}$$ into a profunctor.

(We are currently just studying enriched profunctors between enriched categories, and for a $$\mathcal{V}$$-enriched category $$\mathcal{C}$$ will have an enriched profunctor

$$\text{hom} : \mathcal{C}^{\text{op}} \times \mathcal{C} \to \mathcal{V},$$ but the idea is just the same for ordinary categories and ordinary profunctors.)

Comment Source:Anindya wrote: > possible typo: in previous lectures we've defined profunctors as \$$\mathcal{V}\$$-enriched functors \$$\mathcal{X}^\text{op}\times\mathcal{Y}\rightarrow\mathcal{V}\$$ but here we're using \$$\mathcal{X}\times\mathcal{Y}^\text{op}\rightarrow\mathcal{V}\$$ – is this a mistake or are we swapping round for a reason? This is just a mistake - I'll fix it, thanks! I repeatedly amaze myself by my inability to stick to arbitrary conventions. In my defense I'll mention that there's a heated and pointless debate among category theorists about which convention is better. It doesn't really matter which you use as long as you stick with it, but people who get used to one find the other counterintuitive. I'm _trying_ to use Fong and Spivak's convention \$$\mathcal{X}^\text{op}\times\mathcal{Y}\rightarrow\mathcal{V}\$$, and this is actually my favorite convention, mainly because it makes the hom-functor $\text{hom} : \mathcal{C}^{\text{op}} \times \mathcal{C} \to \mathbf{Set}$ into a profunctor. (We are currently just studying _enriched_ profunctors between _enriched_ categories, and for a \$$\mathcal{V}\$$-enriched category \$$\mathcal{C}\$$ will have an enriched profunctor $\text{hom} : \mathcal{C}^{\text{op}} \times \mathcal{C} \to \mathcal{V},$ but the idea is just the same for ordinary categories and ordinary profunctors.)
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13.
edited July 2018

Matthew wrote:

$$x \otimes a \le y \text{ if and only if } a \le x \multimap y \tag{✴'}$$ Then your quick proof in
#6 works, the theorem we proved holds without demanding commutativity, and $$(x\otimes -) \dashv (x\multimap -)$$ like you say.

This is an interesting issue. I was using the convention in Fong and Spivak's book (see Section 2.5.1), and I should stick with that to avoid confusion. It could be that in the noncommutative case the convention you mention here works better. I will avoid trouble by the simple expedient of assuming $$\mathcal{V}$$ is commutative. But there's more to say...

The book mainly focuses on categories enriched over commutative monoidal posets - or symmetric monoidal preorders, which work more or less the same way. Things get more fussy if we drop the assumption of commutativity, or symmetry, and certain things just don't work at all.

For example, a $$\mathcal{V}$$-enriched category has an 'opposite' when $$\mathcal{V}$$ is symmetric monoidal but not, I believe, when $$\mathcal{V}$$ is merely monoidal! To define composition in the would-be opposite you need to switch two things in $$\mathcal{V}\past each other. Similarly, we can take the product of two \(\mathcal{V}$$-enriched categories when $$\mathcal{V}$$ is symmetric monoidal but not, I believe, when $$\mathcal{V}$$ is merely monoidal. Again, to define composition we need to switch two things in $$\mathcal{V}$$ past each other.

We will see these things soon.

For these two reasons it's hopeless (I believe) to develop the theory of enriched profunctors when $$\mathcal{V}$$ is merely monoidal. Thus, I could have spared myself some grief by assuming $$\mathcal{V}$$ is a commutative monoidal poset throughout this lecture. But in fact closed monoidal categories, and closed monoidal posets, are interesting even in the noncommutative case! So I couldn't resist getting into trouble.

I will rewrite the lecture to avoid this issue by assuming commutativity. The reason is that in the noncommutative case one must really distinguish between left closed and right closed monoidal posets. In one, the operation $$a \otimes -$$ has a right adjoint for each $$a$$. In the other, the operation $$- \otimes a$$ has a right adjoint for each $$a$$.

(When we want to distinguish these two, we do it by turning around the lollipop in $$\multimap$$. I don't know how to do that in this TeX installation!)

Puzzle. You can think of a monoid as a 'discrete' monoidal poset, meaning one where $$x \le y$$ iff $$x = y$$. Show that a monoid which is group gives a monoidal poset that is both left closed and right closed.

A monoidal poset (or monoidal category) that's both left and right closed is called biclosed.

You're raising the issue of whether it's left closed or right closed monoidal posets that become enriched over themselves... or both... or neither. You may have already settled this, but my instinct is that if one works, the other must too! The reason is that I don't believe the concept of "enriched category" has a left/right asymmetry built into it: I don't think there are "left" and "right" enriched categories.

Comment Source:Matthew wrote: > If we go instead with > $x \otimes a \le y \text{ if and only if } a \le x \multimap y \tag{✴'}$ > Then your quick proof in [#6](https://forum.azimuthproject.org/discussion/comment/20080/#Comment_20080) works, the theorem we proved holds without demanding commutativity, and \$$(x\otimes -) \dashv (x\multimap -)\$$ like you say. This is an interesting issue. I was using the convention in Fong and Spivak's book (see Section 2.5.1), and I should stick with that to avoid confusion. It could be that in the noncommutative case the convention you mention here works better. I will avoid trouble by the simple expedient of assuming \$$\mathcal{V}\$$ is commutative. But there's more to say... <img width = "100" src = "http://math.ucr.edu/home/baez/mathematical/warning_sign.jpg"> The book mainly focuses on categories enriched over commutative monoidal posets - or symmetric monoidal preorders, which work more or less the same way. Things get more fussy if we drop the assumption of commutativity, or symmetry, and certain things just don't work at all. For example, a \$$\mathcal{V}\$$-enriched category has an 'opposite' when \$$\mathcal{V}\$$ is symmetric monoidal but not, I believe, when \$$\mathcal{V}\$$ is merely monoidal! To define composition in the would-be opposite you need to switch two things in \$$\mathcal{V}\\past each other. Similarly, we can take the product of two \\(\mathcal{V}\$$-enriched categories when \$$\mathcal{V}\$$ is symmetric monoidal but not, I believe, when \$$\mathcal{V}\$$ is merely monoidal. Again, to define composition we need to switch two things in \$$\mathcal{V}\$$ past each other. We will see these things soon. For these two reasons it's hopeless (I believe) to develop the theory of enriched profunctors when \$$\mathcal{V}\$$ is merely monoidal. Thus, I could have spared myself some grief by assuming \$$\mathcal{V}\$$ is a commutative monoidal poset throughout this lecture. But in fact closed monoidal categories, and closed monoidal posets, are interesting even in the noncommutative case! So I couldn't resist getting into trouble. I will rewrite the lecture to avoid this issue by assuming commutativity. The reason is that in the noncommutative case one must really distinguish between **left closed** and **right closed** monoidal posets. In one, the operation \$$a \otimes -\$$ has a right adjoint for each \$$a\$$. In the other, the operation \$$- \otimes a\$$ has a right adjoint for each \$$a\$$. (When we want to distinguish these two, we do it by turning around the lollipop in \$$\multimap\$$. I don't know how to do that in this TeX installation!) **Puzzle.** You can think of a monoid as a 'discrete' monoidal poset, meaning one where \$$x \le y\$$ iff \$$x = y\$$. Show that a monoid which is _group_ gives a monoidal poset that is both left closed and right closed. A monoidal poset (or monoidal category) that's both left and right closed is called **biclosed**. You're raising the issue of whether it's _left closed_ or _right closed_ monoidal posets that become enriched over themselves... or both... or _neither_. You may have already settled this, but my instinct is that if one works, the other must too! The reason is that I don't believe the concept of "enriched category" has a left/right asymmetry built into it: I don't think there are "left" and "right" enriched categories. 
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14.

Anindya wrote:

However in this case we could slightly vary the definition of a $$\mathcal{V}$$-enriched category to use right-to-left composition instead of left-to-right composition.

Hmm, as mentioned in my previous comment I don't see a left/right asymmetry built into the definition of enriched category:

Definition. A $$\mathcal{V}$$-enriched category $$\mathcal{X}$$ consists of two parts, satisfying two properties. First:

1. one specifies a set $$\mathrm{Ob}(\mathcal{X})$$, elements of which are called objects;

2. for every two objects $$x,y$$, one specifies an element $$\mathcal{X}(x,y)$$ of $$\mathcal{V}$$.

Then:

a) for every object $$x\in\text{Ob}(\mathcal{X})$$ we require that

$$I\leq\mathcal{X}(x,x) .$$ b) for every three objects $$x,y,z\in\mathrm{Ob}(\mathcal{X})$$, we require that $$\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z).$$

The interesting thing is that you can interpret $$\mathcal{X}(x,y)$$ as 'morphisms from $$x$$ to $$y$$' or 'morphisms from $$y$$ to $$x$$' and the definition makes sense either way!

Are you suggesting that we could switch to using

$$\mathcal{X}(y,z)\otimes \mathcal{X}(x,y) \leq\mathcal{X}(x,z)?$$ We could, but you'll notice this isn't just a left/right reflection of the usual definition: I could explain the difference to someone who can't tell the difference between left and right! In the usual definition, the two $$y$$'s here are next to each other:

$$\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)$$ while in the modified definition they aren't.

Comment Source:<img width = "100" src = "http://math.ucr.edu/home/baez/mathematical/warning_sign.jpg"> Anindya wrote: > However in this case we could slightly vary the definition of a \$$\mathcal{V}\$$-enriched category to use right-to-left composition instead of left-to-right composition. Hmm, as mentioned in my previous comment I don't see a left/right asymmetry built into the definition of enriched category: > **Definition.** A **\$$\mathcal{V}\$$-enriched category** \$$\mathcal{X}\$$ consists of two parts, satisfying two properties. First: > 1. one specifies a set \$$\mathrm{Ob}(\mathcal{X})\$$, elements of which are called **objects**; > 2. for every two objects \$$x,y\$$, one specifies an element \$$\mathcal{X}(x,y)\$$ of \$$\mathcal{V}\$$. > Then: > a) for every object \$$x\in\text{Ob}(\mathcal{X})\$$ we require that > $I\leq\mathcal{X}(x,x) .$ > b) for every three objects \$$x,y,z\in\mathrm{Ob}(\mathcal{X})\$$, we require that > $\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z).$ The interesting thing is that you can interpret \$$\mathcal{X}(x,y)\$$ as 'morphisms from \$$x\$$ to \$$y\$$' or 'morphisms from \$$y\$$ to \$$x\$$' and the definition makes sense either way! Are you suggesting that we could switch to using $\mathcal{X}(y,z)\otimes \mathcal{X}(x,y) \leq\mathcal{X}(x,z)?$ We could, but you'll notice this isn't just a left/right reflection of the usual definition: I could explain the difference to someone who can't tell the difference between left and right! In the usual definition, the two \$$y\$$'s here are next to each other: $\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)$ while in the modified definition they aren't. 
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15.
edited July 2018

Okay, you actually are suggesting that:

This amount to replacing the $$\mathcal{V}(x, y)\otimes\mathcal{V}(y, z)\leq\mathcal{V}(x, z)$$ condition with $$\mathcal{V}(y, z)\otimes\mathcal{V}(x, y)\leq\mathcal{V}(x, z)$$ – similarly to how we define $$\text{hom}$$ in a $$\textbf{Set}$$-category. Then the proof would go through.

Sorry, I'm not always good at reading what people write!

Comment Source:Okay, you actually are suggesting that: > This amount to replacing the \$$\mathcal{V}(x, y)\otimes\mathcal{V}(y, z)\leq\mathcal{V}(x, z)\$$ condition with \$$\mathcal{V}(y, z)\otimes\mathcal{V}(x, y)\leq\mathcal{V}(x, z)\$$ – similarly to how we define \$$\text{hom}\$$ in a \$$\textbf{Set}\$$-category. Then the proof would go through. Sorry, I'm not always good at reading what people write! 
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16.
edited July 2018

Anindya wrote:

One detail that puzzles me about the adjunction: the maps $$(x\otimes -)$$ and $$(x\multimap -)$$ are monotone but they don't seem to (necessarily) be monoidal monotone. So this is an adjunction between preorders, not an adjunction between monoidal preorders. That seems a bit unexpected to me.

It's actually familiar. One fundamental example of a monoidal preorder is a group, with $$x \le y$$ iff $$x = y$$, with $$x \otimes y$$ defined to be the product $$xy$$. This is closed if we define $$x \multimap y = yx^{-1}$$, since the definition of 'closed'

$$a \otimes x \le y \text{ if and only if } a \le x \multimap y$$ turns into

$$ax = y \text{ if and only if } a = yx^{-1}$$ The moral is:

The monoidal structure is to multiplication as the closed structure is to division.

But back to your point: multiplying by $$x$$ is not a group homomorphism, nor is dividing by $$x$$:

$$(aa')x \ne (ax)(a'x)$$ and

$$(yy')x^{-1} \ne (yx^{-1}) (y' {x}^{-1}) .$$ So, in the general case, we shouldn't want the maps $$(x\otimes -)$$ and $$(x\multimap -)$$ to be monoidal.

Comment Source:Anindya wrote: > One detail that puzzles me about the adjunction: the maps \$$(x\otimes -)\$$ and \$$(x\multimap -)\$$ are monotone but they don't seem to (necessarily) be *monoidal* monotone. So this is an adjunction between preorders, not an adjunction between monoidal preorders. That seems a bit unexpected to me. It's actually familiar. One fundamental example of a monoidal preorder is a _group_, with \$$x \le y\$$ iff \$$x = y\$$, with \$$x \otimes y\$$ defined to be the product \$$xy\$$. This is closed if we define \$$x \multimap y = yx^{-1} \$$, since the definition of 'closed' $a \otimes x \le y \text{ if and only if } a \le x \multimap y$ turns into $ax = y \text{ if and only if } a = yx^{-1}$ The moral is: > **The monoidal structure is to multiplication as the closed structure is to division.** But back to your point: multiplying by \$$x\$$ is not a group homomorphism, nor is dividing by \$$x\$$: $(aa')x \ne (ax)(a'x)$ and $(yy')x^{-1} \ne (yx^{-1}) (y' {x}^{-1}) .$ So, in the general case, we shouldn't want the maps \$$(x\otimes -)\$$ and \$$(x\multimap -)\$$ to be monoidal.
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17.

John wrote:

For these two reasons it's hopeless (I believe) to develop the theory of enriched profunctors when $$\mathcal{V}$$ is merely monoidal. Thus, I could have spared myself some grief by assuming $$\mathcal{V}$$ is a commutative monoidal poset throughout this lecture. But in fact closed monoidal categories, and closed monoidal posets, are interesting even in the noncommutative case! So I couldn't resist getting into trouble.

Okay, that makes things clearer.

The reason is that in the noncommutative case one must really distinguish between left closed and right closed monoidal posets. In one, the operation $$a \otimes -$$ has a right adjoint for each $$a$$. In the other, the operation $$- \otimes a$$ has a right adjoint for each $$a$$.

(When we want to distinguish these two, we do it by turning around the lollipop in $$\multimap$$. I don't know how to do that in this TeX installation!)

I think the usual symbol is \multimapinv, but it doesn't work here. However, if you don't mind copy-pasting unicode, you can use $$⟜$$ which works fine.

Comment Source:John wrote: > For these two reasons it's hopeless (I believe) to develop the theory of enriched profunctors when \$$\mathcal{V}\$$ is merely monoidal. Thus, I could have spared myself some grief by assuming \$$\mathcal{V}\$$ is a commutative monoidal poset throughout this lecture. But in fact closed monoidal categories, and closed monoidal posets, are interesting even in the noncommutative case! So I couldn't resist getting into trouble. Okay, that makes things clearer. > The reason is that in the noncommutative case one must really distinguish between **left closed** and **right closed** monoidal posets. In one, the operation \$$a \otimes -\$$ has a right adjoint for each \$$a\$$. In the other, the operation \$$- \otimes a\$$ has a right adjoint for each \$$a\$$. > > (When we want to distinguish these two, we do it by turning around the lollipop in \$$\multimap\$$. I don't know how to do that in this TeX installation!) I think the usual symbol is \multimapinv, but it doesn't work here. However, if you don't mind copy-pasting unicode, you can use \$$⟜\$$ which works fine.
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18.

I've decided to flip the definition of 'closed monoidal category' around a bit to make it nicer in the noncommutative case, following the suggestions of Anindya and Matthew. In the commutative case, which is the main case we care about, this doesn't change anything. But it will make Lecture 61 a bit smoother!

Comment Source:I've decided to flip the definition of 'closed monoidal category' around a bit to make it nicer in the noncommutative case, following the suggestions of Anindya and Matthew. In the commutative case, which is the main case we care about, this doesn't change anything. But it will make Lecture 61 a bit smoother!
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19.
edited July 2018

Interesting, in a class on sub-structural type theory they used / and \ for ordered implication/closure.

Comment Source:Interesting, in a class on sub-structural type theory they used / and \ for ordered implication/closure.
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20.
edited July 2018

Interesting, in a class on sub-structural they used / and \ for ordered implication/closure.

A closed monoidal poset that happens to be lattice is called residuated lattice.

Some textbooks (and Wikipedia) denote $$x \multimap z$$ with $$x / z$$ and call it the right residual of $$z$$ by $$x$$. Dually they denote $$z ⟜ y$$ as the left residual of $$z$$ by $$y$$ and it is denoted with $$z \backslash y$$.

The "lollipop" $$\multimap$$ was, AFAIK, introduced by Jean-Yves Girard in Linear Logic (1987). Just as fragments of intuitionistic logic can be given semantics in Cartesian Closed Categories (CCC)s, the $$\otimes,\multimap$$-fragement of linear logic can be given semantics in Symmetric Monoidal Closed Categories (SMCCs). There is a nice discussion of this in Introduction to Categories and Categorical Logic (2011) by Abramsky and Tzevelekos.

Comment Source:> Interesting, in a class on sub-structural they used / and \ for ordered implication/closure. A closed monoidal poset that happens to be lattice is called [*residuated lattice*](https://en.wikipedia.org/wiki/Residuated_lattice#Definition). Some textbooks (and Wikipedia) denote \$$x \multimap z\$$ with \$$x / z\$$ and call it the *right residual of \$$z\$$ by \$$x\$$*. Dually they denote \$$z ⟜ y\$$ as the *left residual of \$$z\$$ by \$$y\$$* and it is denoted with \$$z \backslash y\$$. The "lollipop" \$$\multimap\$$ was, AFAIK, introduced by Jean-Yves Girard in [*Linear Logic* (1987)](https://www.sciencedirect.com/science/article/pii/0304397587900454). Just as fragments of intuitionistic logic can be given semantics in *Cartesian Closed Categories* (CCC)s, the \$$\otimes,\multimap\$$-fragement of linear logic can be given semantics in *Symmetric Monoidal Closed Categories* (SMCCs). There is a nice discussion of this in [*Introduction to Categories and Categorical Logic (2011)*](https://arxiv.org/pdf/1102.1313.pdf) by Abramsky and Tzevelekos.
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21.
edited July 2018

Yes, I think Girard introduced the lollipop along with some other strange symbols like the upside-down and.

In logic and physics we mainly use braided and symmetric monoidal closed categories, and I wrote an introduction to those here:

The first time I saw really interesting applications of closed monoidal categories that weren't symmetric or even braided was in Lambek's work on linguistics. When you don't have a braiding isomorphism $$x \otimes y \cong y \otimes x$$ you need to distinguish between left and right closed monoidal categories, which have two operations x⊸y and y⟜x, or in other words x\y and y/x. Lambek uses these to describe grammatical operations! His pregroups are monoidal preorders that are both left and right closed. My student Jade and another student at ACT2018 wrote a pretty good introduction to this idea, which is pretty simple:

Comment Source:Yes, I think Girard introduced the lollipop along with some other strange symbols like the upside-down and. In logic and physics we mainly use _braided_ and _symmetric_ monoidal closed categories, and I wrote an introduction to those here: * John Baez and Mike Stay, [Physics, topology, logic and computation: a Rosetta Stone](http://arxiv.org/abs/0903.0340) in _New Structures for Physics_, ed. Bob Coecke, Lecture Notes in Physics vol. 813, Springer, Berlin, 2011, pp. 95-172. The first time I saw really interesting applications of closed monoidal categories that weren't symmetric or even braided was in Lambek's work on linguistics. When you don't have a braiding isomorphism \$$x \otimes y \cong y \otimes x\$$ you need to distinguish between left and right closed monoidal categories, which have two operations x⊸y and y⟜x, or in other words x\y and y/x. Lambek uses these to describe grammatical operations! His **pregroups** are monoidal preorders that are both left and right closed. My student Jade and another student at ACT2018 wrote a pretty good introduction to this idea, which is pretty simple: * Jade Master and Cory Griffith, [Linguistics using category theory](https://golem.ph.utexas.edu/category/2018/02/linguistics_using_category_the.html), _The n-Category Caf&eacute;_, 6 February 2018.
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22.
edited July 2018

I did not know about the application to linguistics. This is really cool John! Thanks for the reference!

Comment Source:I did not know about the application to linguistics. This is really cool John! Thanks for the reference!
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23.

Sure! Lambek's original papers are also very clear.

Do you know any other good applications of left and right residuated monoidal posets that aren't commutative?

Comment Source:Sure! Lambek's original papers are also very clear. Do you know any other good applications of left and right residuated monoidal posets that aren't commutative?
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24.

I found the string diagram for lollipops from the paper above and redrew them showing $$a \otimes x \le y \text{ if and only if } a \le x \multimap y$$. It's a pretty dynamic symbol propping up the string on the other side.

Comment Source:* John Baez and Mike Stay, [Physics, topology, logic and computation: a Rosetta Stone](http://arxiv.org/abs/0903.0340) in _New Structures for Physics_, ed. Bob Coecke, Lecture Notes in Physics vol. 813, Springer, Berlin, 2011, pp. 95-172. I found the string diagram for lollipops from the paper above and redrew them showing \$$a \otimes x \le y \text{ if and only if } a \le x \multimap y \$$. It's a pretty dynamic symbol propping up the string on the other side. ![lollipop string diagram](http:\\aether.co.kr/images/lollipop_string_diagram.svg) 
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25.
edited July 2018

Hey John,

Do you know any other good applications of left and right residuated monoidal posets that aren't commutative?

The ones I've encountered are action algebras.

These extend regular expressions, which are insanely common in computer programming. For instance, if you try to sign up for a website, programmers do not want you to give an invalid email. The programmer will check that your email matches a pattern like

^[a-zA-Z0-9_.+-]+@[a-zA-Z0-9-]+.[a-zA-Z0-9-.]+$ (I lifted that from this stack overflow) Suppose I had three regular expression: • one called $$\mathtt{email}$$ for checking emails • one called $$\mathtt{comma}$$ for checking there's a comma • one called $$\mathtt{phone}$$ for checking phone numbers I can combine these to match a line of text where there's an email, a comma, and a phone number (lots of files are like this). This is done using a monoidal tensor. I've seen this tensor denoted $$;$$ in Propositional Dynamic Logic and programming language design courses so I'll use that notation. So we can make a new regular expression like this: $$\mathtt{email} ; \mathtt{comma} ; \mathtt{phone}$$ Composition here is not commutative so this will not match a phone followed by a comma follows by an email. There's also a regular expression $$I$$ that just matches an empty string, and plays the role of identity $$I ; R = R ; I = R$$. We can pre-order regular expressions too: $$A \leq B$$ if any text $$A$$ matches then $$B$$ matches it. Action algebras explore what programming is like if $$(;)$$ had left and right adjoints such that $$(- \leftarrow A) \dashv (A ; -) \dashv (A \to -)$$ Real world regular expressions never implement this, but these residuals are still useful for understanding regular expressions. For instance, action algebras are finitely axiomatizable and form a variety. Regular expressions correspond to a fragment of this variety. Regular expressions are remarkably not finitely axiomatizable! One proof of this comes from none other than John Conway of surreal-numbers and game-of-life fame. Comment Source:Hey John, > Do you know any other good applications of left and right residuated monoidal posets that aren't commutative? The ones I've encountered are [*action algebras*](https://en.wikipedia.org/wiki/Action_algebra). These extend [*regular expressions*](https://en.wikipedia.org/wiki/Regular_expression), which are insanely common in computer programming. For instance, if you try to sign up for a website, programmers do not want you to give an invalid email. The programmer will check that your email matches a pattern like <center><code>^[a-zA-Z0-9_.+-]+@[a-zA-Z0-9-]+\.[a-zA-Z0-9-.]+$</code></center> (I lifted that from this [stack overflow](https://stackoverflow.com/a/719543)) Suppose I had three regular expression: - one called \$$\mathtt{email}\$$ for checking emails - one called \$$\mathtt{comma}\$$ for checking there's a comma - one called \$$\mathtt{phone}\$$ for checking phone numbers I can combine these to match a line of text where there's an email, a comma, and a phone number (lots of files are like this). This is done using a *monoidal tensor*. I've seen this tensor denoted \$$;\$$ in [Propositional Dynamic Logic](https://plato.stanford.edu/entries/logic-dynamic/) and programming language design courses so I'll use that notation. So we can make a new regular expression like this: $\mathtt{email} ; \mathtt{comma} ; \mathtt{phone}$ Composition here is *not* commutative so this will not match a phone followed by a comma follows by an email. There's also a regular expression \$$I\$$ that just matches an empty string, and plays the role of identity \$$I ; R = R ; I = R\$$. We can pre-order regular expressions too: \$$A \leq B\$$ if any text \$$A\$$ matches then \$$B\$$ matches it. Action algebras explore what programming is like if \$$(;)\$$ had left and right adjoints such that $(- \leftarrow A) \dashv (A ; -) \dashv (A \to -)$ Real world regular expressions never implement this, but these residuals are still useful for understanding regular expressions. For instance, action algebras are finitely axiomatizable and form a variety. Regular expressions correspond to a fragment of this variety. Regular expressions are remarkably *not* finitely axiomatizable! One proof of this comes from none other than [John Conway](https://en.wikipedia.org/wiki/Kleene_algebra#History) of surreal-numbers and game-of-life fame.
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26.

Ordered types can be used to describe elements of a (possibly bi-directional) pipeline running asynchronously. (If its synchronous I think all you need is subsignleton types.) See notes for lectures 5-10 (sep 13-sep 29) of http://www.cs.cmu.edu/~fp/courses/15816-f16/schedule.html

Comment Source:Ordered types can be used to describe elements of a (possibly bi-directional) pipeline running asynchronously. (If its synchronous I think all you need is subsignleton types.) See notes for lectures 5-10 (sep 13-sep 29) of http://www.cs.cmu.edu/~fp/courses/15816-f16/schedule.html
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27.

Ordered types can be used to describe elements of a (possibly bi-directional) pipeline running asynchronously. (If its synchronous I think all you need is subsignleton types.) See notes for lectures 5-10 (sep 13-sep 29) of http://www.cs.cmu.edu/~fp/courses/15816-f16/schedule.html

I am confused, what does this have to do with residuated lattices over non-commutative monoids?

Comment Source:> Ordered types can be used to describe elements of a (possibly bi-directional) pipeline running asynchronously. (If its synchronous I think all you need is subsignleton types.) See notes for lectures 5-10 (sep 13-sep 29) of http://www.cs.cmu.edu/~fp/courses/15816-f16/schedule.html I am confused, what does this have to do with residuated lattices over non-commutative monoids?
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28.
edited July 2018

Because $$(\bullet, /, \\ )$$ is one if I understand correctly.

Comment Source:Because \$$(\bullet, /, \\\ ) \$$ is one if I understand correctly.
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29.
edited July 2018

Okay, the Frank Pfenning lecture notes you linked do have a bunch of substructural logics with things that behave like residuals. These aren't exactly residuated lattices, since Prof. Pfenning holds on to proof terms and has multiple assumptions. This is because he's trying to give denotational semantics for programs given some background context.

If you drop the proof terms, and restrict yourself to the single-premise and single conclusion fragment of Prof. Pfenning's sequent calculi, you can recover a preorder using the turnstyle $$\vdash$$ he defines. It looks like some work to show the meta-theorem $$X \bullet Y \vdash Z \iff Y \vdash X / Z$$.

Comment Source:Okay, the Frank Pfenning lecture notes you linked do have a bunch of substructural logics with things that behave like residuals. These aren't exactly residuated lattices, since Prof. Pfenning holds on to proof terms and has multiple assumptions. This is because he's trying to give denotational semantics for programs given some background context. If you drop the proof terms, and restrict yourself to the single-premise and single conclusion fragment of Prof. Pfenning's sequent calculi, you can recover a preorder using the turnstyle \$$\vdash\$$ he defines. It looks like some work to show the meta-theorem \$$X \bullet Y \vdash Z \iff Y \vdash X / Z\$$.
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30.

Yeah, I kind of forgot about the lattice part. Though I think if you are willing to work with preorders instead of posets you can lift the lattice on the types into a lattice-like-preorder on terms.

Comment Source:Yeah, I kind of forgot about the _lattice_ part. Though I think if you are willing to work with preorders instead of posets you can lift the lattice on the types into a lattice-like-preorder on terms.