Mixing
Grothendieck group of integer sets with Minkowski addition
$begingroup$
Consider a set $Z := {Xin 2^mathbb{Z}: Vert XVert < infty}setminus{emptyset} = {Xin 2^mathbb{Z}: Vert XVertin mathbb{N}_+}$ and give it an operation of Minkowski addition, that is:
$$ A + B = {a + b: ain A wedge bin B}. $$
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 $Ztimes Z$ defined as:
$$ (A, B)sim (C, D)iff A+D+K = C + B + K $$
for some $Kin Z$.
My question is: What is the structure of the group $ Ztimes 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 $Yin Z$ such that for any $Xin Z$, $ 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 ${Xin 2^mathbb{Z}: Vert XVert = 1} < Z$ is isomorphic with $mathbb{Z}$. I'd say the same holds for ${[({x}, {0})]: xinmathbb{Z} } < Ztimes 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 $Ztimes Z$ 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").
group-theory monoid grothendieck-construction
asked Mar 1 '18 at 11:51
AdHocAdHoc
335
$endgroup$
add a comment |
$begingroup$
Consider a set $Z := {Xin 2^mathbb{Z}: Vert XVert < infty}setminus{emptyset} = {Xin 2^mathbb{Z}: Vert XVertin mathbb{N}_+}$ and give it an operation of Minkowski addition, that is:
$$ A + B = {a + b: ain A wedge bin B}. $$
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 $Ztimes Z$ defined as:
$$ (A, B)sim (C, D)iff A+D+K = C + B + K $$
for some $Kin Z$.
My question is: What is the structure of the group $ Ztimes 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 $Yin Z$ such that for any $Xin Z$, $ 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 ${Xin 2^mathbb{Z}: Vert XVert = 1} < Z$ is isomorphic with $mathbb{Z}$. I'd say the same holds for ${[({x}, {0})]: xinmathbb{Z} } < Ztimes 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 $Ztimes Z$ 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").
group-theory monoid grothendieck-construction
asked Mar 1 '18 at 11:51
AdHocAdHoc
335
$endgroup$
add a comment |
$begingroup$
Consider a set $Z := {Xin 2^mathbb{Z}: Vert XVert < infty}setminus{emptyset} = {Xin 2^mathbb{Z}: Vert XVertin mathbb{N}_+}$ and give it an operation of Minkowski addition, that is:
$$ A + B = {a + b: ain A wedge bin B}. $$
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 $Ztimes Z$ defined as:
$$ (A, B)sim (C, D)iff A+D+K = C + B + K $$
for some $Kin Z$.
My question is: What is the structure of the group $ Ztimes 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 $Yin Z$ such that for any $Xin Z$, $ 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 ${Xin 2^mathbb{Z}: Vert XVert = 1} < Z$ is isomorphic with $mathbb{Z}$. I'd say the same holds for ${[({x}, {0})]: xinmathbb{Z} } < Ztimes 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 $Ztimes Z$ 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").
group-theory monoid grothendieck-construction
asked Mar 1 '18 at 11:51
AdHocAdHoc
335
$endgroup$
Consider a set $Z := {Xin 2^mathbb{Z}: Vert XVert < infty}setminus{emptyset} = {Xin 2^mathbb{Z}: Vert XVertin mathbb{N}_+}$ and give it an operation of Minkowski addition, that is:
$$ A + B = {a + b: ain A wedge bin B}. $$
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 $Ztimes Z$ defined as:
$$ (A, B)sim (C, D)iff A+D+K = C + B + K $$
for some $Kin Z$.
My question is: What is the structure of the group $ Ztimes 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 $Yin Z$ such that for any $Xin Z$, $ 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 ${Xin 2^mathbb{Z}: Vert XVert = 1} < Z$ is isomorphic with $mathbb{Z}$. I'd say the same holds for ${[({x}, {0})]: xinmathbb{Z} } < Ztimes 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 $Ztimes Z$ 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").
group-theory monoid grothendieck-construction
group-theory monoid grothendieck-construction
asked Mar 1 '18 at 11:51
AdHocAdHoc
335
asked Mar 1 '18 at 11:51
AdHocAdHoc
335
asked Mar 1 '18 at 11:51
AdHocAdHoc
335
asked Mar 1 '18 at 11:51
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asked Mar 1 '18 at 11:51
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1 Answer
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$begingroup$
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
$$Z' longrightarrow mathbb{Z}timesmathbb{N},qquad [a,b] mapsto (a,b-a)$$
is an isomorphism of monoids.
So the Grothendieck group of $Z$ agrees with the Grothendieck group of $mathbb{Z}timesmathbb{N}$, which is $mathbb{Z}^2$.
answered Jan 9 at 15:55
Jens HemelaerJens Hemelaer
476
$endgroup$
add a comment |
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$begingroup$
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
$$Z' longrightarrow mathbb{Z}timesmathbb{N},qquad [a,b] mapsto (a,b-a)$$
is an isomorphism of monoids.
So the Grothendieck group of $Z$ agrees with the Grothendieck group of $mathbb{Z}timesmathbb{N}$, which is $mathbb{Z}^2$.
answered Jan 9 at 15:55
Jens HemelaerJens Hemelaer
476
$endgroup$
add a comment |
$begingroup$
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
$$Z' longrightarrow mathbb{Z}timesmathbb{N},qquad [a,b] mapsto (a,b-a)$$
is an isomorphism of monoids.
So the Grothendieck group of $Z$ agrees with the Grothendieck group of $mathbb{Z}timesmathbb{N}$, which is $mathbb{Z}^2$.
answered Jan 9 at 15:55
Jens HemelaerJens Hemelaer
476
$endgroup$
add a comment |
$begingroup$
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
$$Z' longrightarrow mathbb{Z}timesmathbb{N},qquad [a,b] mapsto (a,b-a)$$
is an isomorphism of monoids.
So the Grothendieck group of $Z$ agrees with the Grothendieck group of $mathbb{Z}timesmathbb{N}$, which is $mathbb{Z}^2$.
answered Jan 9 at 15:55
Jens HemelaerJens Hemelaer
476
$endgroup$
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
$$Z' longrightarrow mathbb{Z}timesmathbb{N},qquad [a,b] mapsto (a,b-a)$$
is an isomorphism of monoids.
So the Grothendieck group of $Z$ agrees with the Grothendieck group of $mathbb{Z}timesmathbb{N}$, which is $mathbb{Z}^2$.
answered Jan 9 at 15:55
Jens HemelaerJens Hemelaer
476
answered Jan 9 at 15:55
Jens HemelaerJens Hemelaer
476
answered Jan 9 at 15:55
Jens HemelaerJens Hemelaer
476
answered Jan 9 at 15:55
Jens HemelaerJens Hemelaer
476
476
add a comment |
add a comment |
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