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Tag: noncommutative

some skew Smullyan stumpers

Published September 21, 2023 by lieven

Raymond Smullyan‘s logic puzzles are legendary. Among his best known are his Knights (who always tell the truth) and Knaves (who always lie) puzzles. Here’s a classic example.

“On the day of his arrival, the anthropologist Edgar Abercrombie came across three inhabitants, whom we will call $A$, $B$ and $C$. He asked $A$: “Are you a Knight or a Knave?” $A$ answered, but so indistinctly that Abercrombie could not understand what he said.

He then asked $B$: “What did he say?” $B$ replies: “He said that he is a knave.” At this point, $C$ piped up and said: “Don’t believe that; it’s a lie!”

Was $C$ a Knight or a Knave?”

If you are stumped by this, try to figure out what kind of inhabitant can say “I am a Knave”.

Some years ago, my friend and co-author Karin Cvetko-Vah wrote about a much stranger island, the island of two truths.

“The island was ruled by a queen and a king. It is important to stress that the queen was neither inferior nor superior to the king. Rather than as a married couple one should think of the queen and the king as two parallel powers, somewhat like the Queen of the Night and the King Sarastro in Mozart’s famous opera The Magic Flute. The queen and the king had their own castle each, each of them had their own court, their own advisers and servants, and most importantly each of them even had their own truth value.

On the island, a proposition p is either FALSE, Q-TRUE or K-TRUE; in each of the cases we say that p has value 0, Q or K, respectively. The queen finds the truth value Q to be superior, while the king values the most the value K. The queen and the king have their opinions on all issues, while other residents typically have their opinions on some issues but not all.”

The logic of the island of two truths is the easiest example of what Karin and I called a non-commutative frame or skew Heyting algebra (see here), a notion we then used, jointly with Jens Hemelaer, to define the notion of a non-commutative topos.

If you take our general definitions, and take Q as the distinguished top-element, then the truth tables for the island of two truths are these ones (value of first term on the left, that of the second on top):

\[
\begin{array}{c|ccc}
\wedge & 0 & Q & K \\
\hline
0 & 0 & 0 & 0 \\
Q & 0 & Q & Q \\
K & 0 & K & K
\end{array} \quad
\begin{array}{c|ccc}
\vee & 0 & Q & K \\
\hline
0 & 0 & Q & K \\
Q & Q & Q & K \\
K & K & Q & K
\end{array} \quad
\begin{array}{c|ccc}
\rightarrow & 0 & Q & K \\
\hline
0 & Q & Q & K \\
Q & 0 & Q & K \\
K & 0 & Q & K
\end{array} \quad
\begin{array}{c|c}
& \neg \\
\hline
0 & Q \\
Q & 0 \\
K & 0
\end{array}
\]

Note that on this island the order of statements is important! That is, the truth value of $p \wedge q$ may differ from that of $q \wedge p$ (and similarly for $\vee$).

Let’s reconsider Smullyan’s puzzle at the beginning of this post, but now on an island of two truths, where every inhabitant is either of Knave, or a Q-Knight (uttering only Q-valued statements), or a K-Knight (saying only K-valued statements).

Again, can you determine what type $C$ is?

Well, if you forget about the distinction between Q- and K-valued sentences, then we’re back to classical logic (or more generally, if you divide out Green’s equivalence relation from any skew Heyting algebra you obtain an ordinary Heyting algebra), and we have seen that then $B$ must be a Knave and $C$ a Knight, so in our new setting we know that $C$ is either a Q-Knight or a K-Knight, but which of the two?

Now, $C$ claims the negation of what $B$ said, so the truth value is $\neg 0 = Q$, and therefore $C$ must be a Q-Knight.

Recall that in Karin Cvetko-Vah‘s island of two truths all sentences have a unique value which can be either $0$ (false) or one of the non-false values Q or K, and the value of combined statements is given by the truth tables above. The Queen and King both have an opinion on all statements, which may or may not coincide with the actual value of that statement. However, if the Queen assigns value $0$ to a statement, then so does the King, and conversely.

Other inhabitants of the island have only their opinion about a subset of all statements (which may be empty). Two inhabitants agree on a statement if they both have an opinion on it and assign the same value to it.

Now, each inhabitant is either loyal to the Queen or to the King (or both), meaning that they agree with the Queen (resp. King) on all statements they have an opinion of. An inhabitant loyal to the Queen is said to believe a sentence when she assigns value $Q$ to it (and symmetric for those loyal to the King), and knows the statement if she believes it and that value coincides with the actual value of that statement.

Further, if A is loyal to the Queen, then the value of the statement ‘A is loyal to the Queen’ is Q, and if A is not loyal to the Queen, then the value of the sentence ‘A is loyal to the Queen’ is $0$ (and similarly for statements about loyalty to the King).

These notions are enough for the first batch of ten puzzles in Karin’s posts

Just one example:

Show that if anybody on the island knows that A is not loyal to the Queen, then everybody that has an opinion about the sentence ‘A is loyal to the Queen’ knows that.

After these two posts, Karin decided that it was more fun to blog about the use of non-commutative frames in data analysis.

But, she once gave me a text containing many more puzzles (as well as all the answers), so perhaps I’ll share these in a follow-up post.

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A noncommutative moduli space

Published July 19, 2014 by lieven

Supernatural numbers also appear in noncommutative geometry via James Glimm’s characterisation of a class of simple $C^*$-algebras, the UHF-algebras.

A uniformly hyperfine (or, UHF) algebra $A$ is a $C^*$-algebra that can be written as the closure, in the norm topology, of an increasing union of finite-dimensional full matrix algebras

$M_{c_1}(\mathbb{C}) \subset M_{c_2}(\mathbb{C}) \subset … \quad \subset A$

Such embedding are only possible if the matrix-sizes divide each other, that is $c_1 | c_2 | c_3 | … $, and we can assign to $A$ the supernatural number $s=\prod_i c_i$ and denote $A=A(s)$.

In his paper On a certain class of operator algebras, Glimm proved that two UHF-algebras $A(s)$ and $B(t)$ are isomorphic as $C^*$-algebras if and only if $s=t$. That is, the supernatural numbers $\mathbb{S}$ are precisely the isomorphism classes of UHF-algebras.

An important invariant, the Grothendieck group $K_0$ of $A(s)$, can be described as the additive subgroup $\mathbb{Q}(s)$ of $\mathbb{Q}$ generated by all fractions of the form $\frac{1}{n}$ where $n$ is a positive integer dividing $s$.

A “noncommutative space” is a Morita class of $C^*$-algebras, so we want to know when two $UHF$-algebras $A(s)$ and $B(t)$ are Morita-equivalent. This turns out to be the case when there are positive integers $n$ and $m$ such that $n.s = m.t$, or equivalently when the $K_0$’s $\mathbb{Q}(s)$ and $\mathbb{Q}(t)$ are isomorphic as additive subgroups of $\mathbb{Q}$.

Thus Morita-equivalence defines an equivalence relation on $\mathbb{S}$ as follows: if $s=\prod p^{s_p}$ and $t= \prod p^{t_p}$ then $s \sim t$ if and only if the following two properties are satisfied:

(1): $s_p = \infty$ iff $t_p= \infty$, and

(2): $s_p=t_p$ for all but finitely many primes $p$.

That is, we can view the equivalence classes $\mathbb{S}/\sim$ as the moduli space of noncommutative spaces associated to UHF-algebras!

Now, the equivalence relation is described in terms of isomorphism classes of additive subgroups of the rationals, which was precisely the characterisation of isomorphism classes of points in the arithmetic site, that is, the finite adรจle classes

$\mathbb{S}/\sim~\simeq~\mathbb{Q}^* \backslash \mathbb{A}^f_{\mathbb{Q}} / \widehat{\mathbb{Z}}^*$

and as the induced topology of $\mathbb{A}^f_{\mathbb{Q}}$ on it is trivial, this “space” is usually thought of as a noncommutative space.

That is, $\mathbb{S}/\sim$ is a noncommutative moduli space of noncommutative spaces defined by UHF-algebras.

The finite integers form one equivalence class, corresponding to the fact that the finite dimensional UHF-algebras $M_n(\mathbb{C})$ are all Morita-equivalent to $\mathbb{C}$, or a bit more pompous, that the Brauer group $Br(\mathbb{C})$ is trivial.

Multiplication of supernaturals induces a well defined multiplication on equivalence classes, and, with that multiplication we can view $\mathbb{S}/\sim$ as the ‘Brauer-monoid’ $Br_{\infty}(\mathbb{C})$ of simple UHF-algebras…

(Btw. the photo of James Glimm above was taken by George Bergman in 1972)

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3 related new math-sites

Published September 14, 2011 by lieven

F_un Mathematics

Hardly a ‘new’ blog, but one that is getting a new life! On its old homepage you’ll find a diagonal banner stating ‘This site has moved’ and clicking on it will guide you to its new location : cage.ugent.be/~kthas/Fun.

From now on, this site will be hosted at the University of Ghent and maintained by Koen Thas. So, please update your bookmarks and point your RSS-aggregator to the new feed.

Everyone interested in contributing to this blog dedicated to the mathematics of the field with one element should contact Koen by email.

angst

Though I may occasionally (cross)post at F_un mathematics, my own blog-life will center round a new blog to accompany the master-course ‘seminar noncommutative geometry’ I’m running at Antwerp University this semester. Its URL is noncommutative.org and it is called :

Here, angs is short for Antwerp Noncommutative Geometry Seminar and the additions @t resp. + are there to indicate we will experiment a bit trying to find useful interactions between the IRL seminar, its blog and social media such as twitter and Google+.

The seminar (and blog) are scheduled to start in earnest september 30th, but I may post some prep-notes already. This semester the seminar will try to decode Smirnov’s old idea to prove the ABC-conjecture in number theory via geometry over the field with one element and connect it with new ideas such as Borger’s $\mathbb{F}_1$-geometry using $\lambda$-rings and noncommutative ideas proposed by Connes, Consani and Marcolli.

Again, anyone willing to contribute actively is invited to send me an email or to comment on ‘angst’, tweet about it using the hashtag #angs (all such tweets will appear on the frontpage) or share its posts on Google+.

Noncommutative Arithmetic Geometry Media Library

Via the noncommutative geometry blog a new initiative maintained by Alain Connes and Katia Consani was announced : the Noncommutative Arithmetic Geometry Media Library.

This site is dedicated to maintain articles, videos, and news about meetings and activities related to noncommutative arithmetic geometry. The website is still `under construction’ and the plan is to gradually add more videos (also from past conferences and meetings), as well as papers and slides.

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