3
$\begingroup$

Consider a set $Z := \{X\in 2^\mathbb{Z}: \Vert X\Vert < \infty\}\setminus\{\emptyset\} = \{X\in 2^\mathbb{Z}: \Vert X\Vert\in \mathbb{N}_+\}$ and give it an operation of Minkowski addition, that is: $$ A + B = \{a + b: a\in A \ \wedge \ b\in B\}. $$ This structure is associative, commutative and has an identity element $\{0\}$. If I'm not mistaken, then it's an abelian monoid. What I'm interested in, is its Grothendieck group. Let us have a(n) (equivalence) relation $\sim$ on $Z\times Z$ defined as: $$ (A, B)\sim (C, D)\iff A+D+K = C + B + K $$ for some $K\in Z$.

My question is: What is the structure of the group $\ Z\times Z/\sim\ $?
Is it described in more detail somewhere?

--
Possible hints for me:

There seems to be no set $K$ that would make, for example, $ (\{1,2\}, \{2,3\})$ and $ (\{-1,1\}, \{3,7\}) $ equivalent. If I hadn't excluded infinite sets and the empty one, I could now use $\mathbb{Z}$ (or $\emptyset$, if adding it was defined the way Minkowski did) for $K$ and thus obtain a trivial group.

Now it seems that there's no set $Y\in Z$ such that for any $X\in Z$, $\ Y+X = Y$.
So, first hint I'd appreciate, is to show me, if there really is no such $K$ that makes following statement true: $$ \{4, 5, 8, 9\} + K = \{1, 2, 3, 4\} + K. $$ Or perhaps there is one...?
Probably in other words: does the cancellation law hold for this structure?

Secondly, the set of singletons $\{X\in 2^\mathbb{Z}: \Vert X\Vert = 1\} < Z$ is isomorphic with $\mathbb{Z}$. I'd say the same holds for $\{[(\{x\}, \{0\})]: x\in\mathbb{Z} \} < \ Z\times Z/\sim\ $, meaning the group really can't be trivial.

On the other hand, $(\{1,2\}, \{2,3\}) \sim (\{3,4\}, \{4,5\})$ for $K = \{0\}$, so it's not isomorphic to $Z\times Z$ either (i.e. $\sim$ isn't an identity relation). I imagine a hint would be to show me how to think about, for example, $[\{0\}, \{1,2\}]$ ("negative $\{1,2\}$ set").

$\endgroup$
1
$\begingroup$

The Grothendieck group of $Z$ is the free abelian group on 2 generators $\mathbb{Z}^2$.

To see this, first consider the equality: $$A + \mathrm{Conv}(A) = \mathrm{Conv}(A)+\mathrm{Conv}(A)$$ with $\mathrm{Conv}(A)$ the interval defined by $$\mathrm{Conv}(A) = \{ a_-,~ a_- +1,\dots,~ a_+ \}$$ with $a_-$ and $a_+$ the minimal resp. maximal element of $A$.

The equality above shows that $Z$ is not cancellative. The elements $A$ and $\mathrm{Conv}(A)$ determine the same elements in the Grothendieck group. So the Grothendieck group of $Z$ can be computed as the Grothendieck group of the submonoid $Z' \subseteq Z$ consisting of the intervals $[a,b] = \{a,~a+1,\dots,b\}$ for $a,b \in \mathbb{Z}$, $a \leq b$. It is easy to check that the Minkowski sum of two intervals $[a,b]$ and $[c,d]$ is given by $[a+c,b+d]$. This means that the map $$Z' \longrightarrow \mathbb{Z}\times\mathbb{N},\qquad [a,b] \mapsto (a,b-a)$$ is an isomorphism of monoids.

So the Grothendieck group of $Z$ agrees with the Grothendieck group of $\mathbb{Z}\times\mathbb{N}$, which is $\mathbb{Z}^2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.