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Stream: theory: mathematics

Topic: Quantale of intervals

Ralph Sarkis (Aug 22 2025 at 13:02):

Any two elements $a \leq b$ inside a quantale $V$ define an interval $[a,b] = \{x \in V \mid a \leq x \leq b \} \subseteq V$. The set of all intervals in $V$ plus the empty set is a complete lattice with supremum and infimum defined as follows:

$$\bigvee_i [a_i,b_i] = [\wedge_i a_i, \vee_i b_i] \text{ and } \bigwedge_i [a_i,b_i] = [\vee_i a_i, \wedge_i b_i].$$

Thus, you can see the set of intervals as the complete lattice $V^{\mathrm{op}} \times V$ after identifying all the intervals $[a,b]$ where $a \geq b$ with the empty set.

If you define addition $[a,b] \oplus [a',b'] = [a+a', b+b']$ (where $+$ is the operation of the quantale $V$), you do not always get a quantale because $+$ might not preserve infimums.

In the specific case of the quantale of extended real numbers $([-\infty,\infty], \leq, +, 0)$, $+$ does not preserve infimums, but you can extend addition in a different way that will preserve infimums but not supremums, i.e. it yields a quantale $([-\infty, \infty], \geq, +, 0)$---the result of adding positive infinity to negative infinity is the only thing that distinguishes these operations. Then, you obtain the quantale of intervals of extended real numbers by defining addition with $[a,b] \oplus [a',b'] = [a+a', b+b']$ where the $+$ on the left is the one that preserves infimums, and the $+$ on the right is the one that preserves supremum (again they only differ on the value of $-\infty+\infty$).

More generally, you can define the quantale of intervals inside a complete lattice $V$ if you have a quantale structure on $V$ and one on $V^{\mathrm{op}}$. Has that construction been studied? Maybe, the specific one for extended reals at least? (I have seen a couple of papers discussing the complete lattice of intervals inside a complete lattice.)

There is a bit more details in this note.

Matteo Capucci (he/him) (Aug 24 2025 at 17:08):

Cool!

Matteo Capucci (he/him) (Aug 24 2025 at 17:14):

It seems you are looking at defining the twisted arrow construction in some flavours of quantales. Since you need a duality on the 2-category of such objects, it seems the minimal context is a 'linearly distributive monoidal poset with infima', these being an infinitary extension of Cockett and Seeley's notion of 'LDC with products', from 'Linearly distributive functors' '99.

As an aside, in https://arxiv.org/abs/2406.04936 I, like you, observed that the 'two additions', distinguished by how they resolve the indeterminate form $-\infty+\infty$, come together to make the extended reals an (isomix) $*$-autonomous quantale. Indeed, having a duality is a quick way to define an LDC structure!

Ralph Sarkis (Aug 24 2025 at 18:03):

I recognized the twisted arrow category but 1) it does not contain the empty set (hence it does not have all (co)products) and 2) you get stuck really quickly if you want to define (co)products in the twisted arrow category of an arbitrary category with (co)products, so even before considering the monoidal product, this construction seems restricted to orders.

Ralph Sarkis (Aug 24 2025 at 18:03):

Ah... Sorry I should have cited your paper. I knew about it but forgot you had the same tables.

Matteo Capucci (he/him) (Sep 01 2025 at 07:15):

No problem! I wasn't fishing for a citation!

Matteo Capucci (he/him) (Sep 01 2025 at 07:16):

Ralph Sarkis said:

I recognized the twisted arrow category but 1) it does not contain the empty set (hence it does not have all (co)products) and 2) you get stuck really quickly if you want to define (co)products in the twisted arrow category of an arbitrary category with (co)products, so even before considering the monoidal product, this construction seems restricted to orders.

You're right, there is no empty set there :thinking: that's quite intriguing

Matteo Capucci (he/him) (Sep 01 2025 at 07:17):

I'm not following your remarks about coproducts. Is this the analogue to joins of the quantale?

Ralph Sarkis (Sep 01 2025 at 08:13):

Yes, when constructing the joins of intervals in a complete lattice, we use the fact that the category is thin, so there is only one choice of morphism $\wedge a_i \to \vee b_i$. In an arbitrary category with (co)products, if meets and joins are interpreted as products and coproducts respectively, then given morphisms $f_i : A_i \to B_i$, there are many choices of morphisms $\prod_i A_i \rightarrow \coprod_i B_i$. For example, for each $i$, we have

$$\prod_i A_i \xrightarrow{\pi_i} A_i \xrightarrow{f_i} B_i \xrightarrow{\kappa_i} \coprod_i B_i.$$

Thus, it is not obvious how one should define products in the twisted arrow category.