What is a discrete invariant?
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I don´t understand the definition of discrete invariant and I wonder if someone of you would know it. The notion appears in the following sense:
Given a set $M$ and equivalence relation $sim$ on $M$, a discrete invariant is a function $f:M/simrightarrowmathbb{Z}$ that partitions $M/sim$. What does it mean?
abstract-algebra invariant-theory
asked Jan 16 at 9:09
J. OveJ. Ove
111
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add a comment |
$begingroup$
I don´t understand the definition of discrete invariant and I wonder if someone of you would know it. The notion appears in the following sense:
Given a set $M$ and equivalence relation $sim$ on $M$, a discrete invariant is a function $f:M/simrightarrowmathbb{Z}$ that partitions $M/sim$. What does it mean?
abstract-algebra invariant-theory
asked Jan 16 at 9:09
J. OveJ. Ove
111
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$begingroup$
I also don't know what "that partitions $M/sim$" means. Every function partitions its domain.
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– Qiaochu Yuan
Jan 16 at 9:11
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A function $M/sim to mathbb{Z}$ is a discrete function $M to mathbb{Z}$ which is $sim$ invariant.
$endgroup$
– reuns
Jan 16 at 9:12
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Thats what I thought @QiaochuYuan. I think that reuns have just gave us a satisfactory answer :) thanks
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– J. Ove
Jan 16 at 9:19
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Could it be that the function $f$ is defined on $M$ instead of $M/sim$? Then the condition on $f$ would make sense: It has to assign the same value to all elements of an equivalence class.
$endgroup$
– Ingix
Jan 16 at 9:58
add a comment |
$begingroup$
I don´t understand the definition of discrete invariant and I wonder if someone of you would know it. The notion appears in the following sense:
Given a set $M$ and equivalence relation $sim$ on $M$, a discrete invariant is a function $f:M/simrightarrowmathbb{Z}$ that partitions $M/sim$. What does it mean?
abstract-algebra invariant-theory
asked Jan 16 at 9:09
J. OveJ. Ove
111
$endgroup$
I don´t understand the definition of discrete invariant and I wonder if someone of you would know it. The notion appears in the following sense:
Given a set $M$ and equivalence relation $sim$ on $M$, a discrete invariant is a function $f:M/simrightarrowmathbb{Z}$ that partitions $M/sim$. What does it mean?
abstract-algebra invariant-theory
abstract-algebra invariant-theory
asked Jan 16 at 9:09
J. OveJ. Ove
111
asked Jan 16 at 9:09
J. OveJ. Ove
111
asked Jan 16 at 9:09
J. OveJ. Ove
111
asked Jan 16 at 9:09
J. OveJ. Ove
111
asked Jan 16 at 9:09
J. OveJ. Ove
111
111
$begingroup$
I also don't know what "that partitions $M/sim$" means. Every function partitions its domain.
$endgroup$
– Qiaochu Yuan
Jan 16 at 9:11
$begingroup$
A function $M/sim to mathbb{Z}$ is a discrete function $M to mathbb{Z}$ which is $sim$ invariant.
$endgroup$
– reuns
Jan 16 at 9:12
$begingroup$
Thats what I thought @QiaochuYuan. I think that reuns have just gave us a satisfactory answer :) thanks
$endgroup$
– J. Ove
Jan 16 at 9:19
$begingroup$
Could it be that the function $f$ is defined on $M$ instead of $M/sim$? Then the condition on $f$ would make sense: It has to assign the same value to all elements of an equivalence class.
$endgroup$
– Ingix
Jan 16 at 9:58
add a comment |
$begingroup$
I also don't know what "that partitions $M/sim$" means. Every function partitions its domain.
$endgroup$
– Qiaochu Yuan
Jan 16 at 9:11
$begingroup$
A function $M/sim to mathbb{Z}$ is a discrete function $M to mathbb{Z}$ which is $sim$ invariant.
$endgroup$
– reuns
Jan 16 at 9:12
$begingroup$
Thats what I thought @QiaochuYuan. I think that reuns have just gave us a satisfactory answer :) thanks
$endgroup$
– J. Ove
Jan 16 at 9:19
$begingroup$
Could it be that the function $f$ is defined on $M$ instead of $M/sim$? Then the condition on $f$ would make sense: It has to assign the same value to all elements of an equivalence class.
$endgroup$
– Ingix
Jan 16 at 9:58
$begingroup$
I also don't know what "that partitions $M/sim$" means. Every function partitions its domain.
$endgroup$
– Qiaochu Yuan
Jan 16 at 9:11
$begingroup$
I also don't know what "that partitions $M/sim$" means. Every function partitions its domain.
$endgroup$
– Qiaochu Yuan
Jan 16 at 9:11
$begingroup$
A function $M/sim to mathbb{Z}$ is a discrete function $M to mathbb{Z}$ which is $sim$ invariant.
$endgroup$
– reuns
Jan 16 at 9:12
$begingroup$
A function $M/sim to mathbb{Z}$ is a discrete function $M to mathbb{Z}$ which is $sim$ invariant.
$endgroup$
– reuns
Jan 16 at 9:12
$begingroup$
Thats what I thought @QiaochuYuan. I think that reuns have just gave us a satisfactory answer :) thanks
$endgroup$
– J. Ove
Jan 16 at 9:19
$begingroup$
Thats what I thought @QiaochuYuan. I think that reuns have just gave us a satisfactory answer :) thanks
$endgroup$
– J. Ove
Jan 16 at 9:19
$begingroup$
Could it be that the function $f$ is defined on $M$ instead of $M/sim$? Then the condition on $f$ would make sense: It has to assign the same value to all elements of an equivalence class.
$endgroup$
– Ingix
Jan 16 at 9:58
$begingroup$
Could it be that the function $f$ is defined on $M$ instead of $M/sim$? Then the condition on $f$ would make sense: It has to assign the same value to all elements of an equivalence class.
$endgroup$
– Ingix
Jan 16 at 9:58
add a comment |
1 Answer
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votes
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Well, the function $f:M/equiv ;rightarrow {Bbb Z}$ assigns to each equivalence class $bar a ={bin Mmid aequiv b}$ an integer $f(bar a)$.
For instance, take the set of words $M={text{car}, text{auto},text{face},text{book},text{app}}$. If the equivalence relation is ''same number of letters as'', then
$$M/equiv = {{text{car},text{app}},{text{auto},text{face},text{book}}}$$
and the invariant $f$ could assign the ''length information'' to each class:
$f({text{car},text{app}}) = 3$ and $f({text{auto},text{face},text{book}})=4$.
answered Jan 16 at 14:05
WuestenfuxWuestenfux
4,5121413
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$begingroup$
Well, the function $f:M/equiv ;rightarrow {Bbb Z}$ assigns to each equivalence class $bar a ={bin Mmid aequiv b}$ an integer $f(bar a)$.
For instance, take the set of words $M={text{car}, text{auto},text{face},text{book},text{app}}$. If the equivalence relation is ''same number of letters as'', then
$$M/equiv = {{text{car},text{app}},{text{auto},text{face},text{book}}}$$
and the invariant $f$ could assign the ''length information'' to each class:
$f({text{car},text{app}}) = 3$ and $f({text{auto},text{face},text{book}})=4$.
answered Jan 16 at 14:05
WuestenfuxWuestenfux
4,5121413
$endgroup$
add a comment |
$begingroup$
Well, the function $f:M/equiv ;rightarrow {Bbb Z}$ assigns to each equivalence class $bar a ={bin Mmid aequiv b}$ an integer $f(bar a)$.
For instance, take the set of words $M={text{car}, text{auto},text{face},text{book},text{app}}$. If the equivalence relation is ''same number of letters as'', then
$$M/equiv = {{text{car},text{app}},{text{auto},text{face},text{book}}}$$
and the invariant $f$ could assign the ''length information'' to each class:
$f({text{car},text{app}}) = 3$ and $f({text{auto},text{face},text{book}})=4$.
answered Jan 16 at 14:05
WuestenfuxWuestenfux
4,5121413
$endgroup$
add a comment |
$begingroup$
Well, the function $f:M/equiv ;rightarrow {Bbb Z}$ assigns to each equivalence class $bar a ={bin Mmid aequiv b}$ an integer $f(bar a)$.
For instance, take the set of words $M={text{car}, text{auto},text{face},text{book},text{app}}$. If the equivalence relation is ''same number of letters as'', then
$$M/equiv = {{text{car},text{app}},{text{auto},text{face},text{book}}}$$
and the invariant $f$ could assign the ''length information'' to each class:
$f({text{car},text{app}}) = 3$ and $f({text{auto},text{face},text{book}})=4$.
answered Jan 16 at 14:05
WuestenfuxWuestenfux
4,5121413
$endgroup$
Well, the function $f:M/equiv ;rightarrow {Bbb Z}$ assigns to each equivalence class $bar a ={bin Mmid aequiv b}$ an integer $f(bar a)$.
For instance, take the set of words $M={text{car}, text{auto},text{face},text{book},text{app}}$. If the equivalence relation is ''same number of letters as'', then
$$M/equiv = {{text{car},text{app}},{text{auto},text{face},text{book}}}$$
and the invariant $f$ could assign the ''length information'' to each class:
$f({text{car},text{app}}) = 3$ and $f({text{auto},text{face},text{book}})=4$.
answered Jan 16 at 14:05
WuestenfuxWuestenfux
4,5121413
answered Jan 16 at 14:05
WuestenfuxWuestenfux
4,5121413
answered Jan 16 at 14:05
WuestenfuxWuestenfux
4,5121413
answered Jan 16 at 14:05
WuestenfuxWuestenfux
4,5121413
4,5121413
add a comment |
add a comment |
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$begingroup$
I also don't know what "that partitions $M/sim$" means. Every function partitions its domain.
$endgroup$
– Qiaochu Yuan
Jan 16 at 9:11
$begingroup$
A function $M/sim to mathbb{Z}$ is a discrete function $M to mathbb{Z}$ which is $sim$ invariant.
$endgroup$
– reuns
Jan 16 at 9:12
$begingroup$
Thats what I thought @QiaochuYuan. I think that reuns have just gave us a satisfactory answer :) thanks
$endgroup$
– J. Ove
Jan 16 at 9:19
$begingroup$
Could it be that the function $f$ is defined on $M$ instead of $M/sim$? Then the condition on $f$ would make sense: It has to assign the same value to all elements of an equivalence class.
$endgroup$
– Ingix
Jan 16 at 9:58