# quantale

A quantale $Q$ is a set with three binary operations  on it: $\wedge,\vee$, and $\cdot$, such that

1. 1.

$(Q,\wedge,\vee)$ is a complete lattice  (with $0$ as the bottom and $1$ as the top), and

2. 2.
3. 3.

$\cdot$ distributes over arbitrary joins; that is, for any $a\in Q$ and any subset $S\subseteq Q$,

 $a\cdot\big{(}\bigvee S\big{)}=\bigvee\{a\cdot s\mid s\in S\}\quad\mbox{ and }% \quad\big{(}\bigvee S\big{)}\cdot a=\bigvee\{s\cdot a\mid s\in S\}.$

It is sometimes convenient to drop the multiplication symbol, when there is no confusion. So instead of writing $a\cdot b$, we write $ab$.

The most obvious example of a quantale comes from ring theory. Let $R$ be a commutative ring with $1$. Then $L(R)$, the lattice of ideals of $R$, is a quantale.

###### Proof.
 $IJ:=\{\sum_{i=1}^{n}r_{i}s_{i}\mid r_{i}\in I\mbox{ and }s_{i}\in J\mbox{, }n% \in\mathbb{N}\},$

Now, let $S=\{I_{i}\mid i\in N\}$ be a set of ideals of $R$ and let $I=\bigvee S$. If $J$ is any ideal of $R$, we want to show that $IJ=\bigvee\{I_{i}J\mid i\in N\}$ and, since $R$ is commutative   , we would have the other equality $JI=\bigvee\{JI_{i}\mid i\in N\}$. To see this, let $a\in IJ$. Then $a=\sum r_{i}s_{i}$ with $r_{i}\in I$ and $s_{i}\in J$. Since each $r_{i}$ is a finite sum of elements of $\bigcup S$, $r_{i}s_{i}$ is a finite sum of elements of $\bigcup\{I_{i}J\mid i\in N\}$, so $a\in\bigvee\{I_{i}J\mid i\in N\}$. This shows $IJ\subseteq\bigvee\{I_{i}J\mid i\in N\}$. Conversely, if $a\in\bigvee\{I_{i}J\mid i\in N\}$, then $a$ can be written as a finite sum of elements of $\bigcup\{I_{i}J\mid i\in N\}$. In turn, each of these additive components is a finite sum of products  of the form $r_{k}s_{k}$, where $r_{k}\in I_{i}$ for some $i$, and $s_{k}\in J$. As a result, $a$ is a finite sum of elements of the form $r_{k}s_{k}$, so $a\in IJ$ and we have the other inclusion $\bigvee\{I_{i}J\mid i\in N\}\subseteq IJ$.

Finally, we observe that $R$ is the multiplicative identity  in $L(R)$, as $IR=RI=I$ for all $I\in L(R)$. This completes the proof. ∎

Remark. In the above example, notice that $IJ\leq I$ and $IJ\leq J$, and we actually have $IJ\leq I\wedge J$. In particular, $I^{2}\leq I$. With an added condition, this fact can be characterized in an arbitrary quantale (see below).

Properties. Let $Q$ be a quantale.

1. 1.

Multiplication is monotone in each argument. This means that if $a,b\in Q$, then $a\leq b$ implies that $ac\leq bc$ and $ca\leq cb$ for all $c\in Q$. This is easily verified. For example, if $a\leq b$, then $ac\vee bc=(a\vee b)c=bc$, so $ac\leq bc$. So a quantale is a partially ordered semigroup, and in fact, an l-monoid (an l-semigroup and a monoid at the same time).

2. 2.

If $1=1^{\prime}$, then $ab\leq a\wedge b$: since $a\leq 1$, then $ab\leq a1=a1^{\prime}=a$; similarly, $b\leq ab$. In particular, the bottom $0$ is also the multiplicative zero: $a0\leq a\wedge 0=0$, and $0a=0$ similarly.

3. 3.

Actually, $a0=0a=0$ is true even without $1=1^{\prime}$: since $a\varnothing=\{ab\mid b\in\varnothing\}=\varnothing$ and $0:=\bigvee\varnothing$, we have $a0=a\bigvee\varnothing=\bigvee a\varnothing=\bigvee\varnothing=0$. Similarly $0a=0$. So a quantale is a semiring  , if $\vee$ is identified as $+$ (with $0$ as the additive identity), and $\cdot$ is again $\cdot$ (with $1^{\prime}$ the multiplicative identity).

4. 4.

Viewing quantale $Q$ now as a semiring, we see in fact that $Q$ is an idempotent semiring, since $a+a=a\vee a=a$.

5. 5.

Now, view $Q$ as an i-semiring. For each $a\in Q$, let $S=\{1^{\prime},a,a^{2},\ldots\}$ and define $a^{*}=\bigvee S$. We observe some basic properties

• $1^{\prime}+aa^{*}=a^{*}$: since $1^{\prime}\vee(a\bigvee S)=1^{\prime}\vee(\bigvee\{a1^{\prime},aa,aa^{2},% \ldots\})=\bigvee\{1^{\prime},a,a^{2},\ldots\}=\bigvee S=a^{*}$

• $1^{\prime}+a^{*}a=a^{*}$ as well

• if $ab\leq b$, then $a^{*}b\leq b$: by induction  on $n$, we have $a^{n}b\leq b$ whenever $a\leq b$, so that $a^{*}b=\bigvee\{a^{n}b\mid n\in\mathbb{N}\cup\{0\}\}\leq b$.

• similarly, if $ba\leq b$, then $ba^{*}\leq b$

All of the above properties satisfy the conditions for an i-semiring to be a Kleene algebra. For this reason, a quantale is sometimes called a standard Kleene algebra.

6. 6.

Call the multiplication idempotent   if each element is an idempotent with respect to the multiplication: $aa=a$ for any $a\in Q$. If $\cdot$ is idempotent and $1=1^{\prime}$, then $\cdot=\wedge$. In other words, $ab=a\wedge b$.

###### Proof.

As we have seen, $ab\leq a\wedge b$ in the 2 above. Now, suppose $c\leq a\wedge b$. Then $c\leq a$ and $c\leq b$, so $c=c^{2}\leq cb\leq ab$. So $ab$ is the greatest lower bound  of $a$ and $b$, i.e., $ab=a\wedge b$. This also means that $ba=b\wedge a=a\wedge b=ab$. ∎

7. 7.

In fact, a locale is a quantale if we define $\cdot:=\wedge$. Conversely, a quantale where $\cdot$ is idempotent and $1=1^{\prime}$ is a locale.

###### Proof.

If $Q$ is a locale with $\cdot=\wedge$, then $aa=a\wedge a=a$ and $a1=a\wedge 1=a=1\wedge a=1a$, implying $1=1^{\prime}$. The infinite  distributivity of $\cdot$ over $\vee$ is just a restatement of the infinite distributivity of $\wedge$ over $\vee$ in a locale. Conversely, if $\cdot$ is idempotent and $1=1^{\prime}$, then $\cdot=\wedge$ as shown previously, so $a\wedge(\bigvee S)=a(\bigvee S)=\bigvee\{as\mid s\in S\}=\bigvee\{a\wedge s% \mid s\in S\}$. Similarly $(\bigvee S)\wedge a=\bigvee\{s\wedge a\mid s\in S\}$. Therefore, $Q$ is a locale. ∎

Remark. A quantale homomorphism between two quantales is a complete lattice homomorphism and a monoid homomorphism at the same time.

## References

• 1 S. Vickers, Topology via Logic, Cambridge University Press, Cambridge (1989).
Title quantale Quantale 2013-03-22 17:00:08 2013-03-22 17:00:08 CWoo (3771) CWoo (3771) 12 CWoo (3771) Definition msc 06F07 standard Kleene algebra quantale homomorphism