Why does Lie algebras constructed by generators and relations are usually infinite dimensional?
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I am reading Humphrey's book on Lie algebras and now the construction of Lie algebras by generators and relations was just presented.
We simply get a set $X= {hat x_i, hat y_i, hat h_i: 1leq i leq l}$ and a subset $R$ of the free lie algebra $hat L$ generated by $X$, given by
$R = {[hat h_i, hat h_j], [hat x_i, hat y_j] - delta_{ij}hat h_i, [hat h_i, hat x_j]-c_{ji}hat x_j,[hat h_i,hat y_j] + c_{ji}hat y_j$}
and then consider $hat K$ the ideal generated by $R$.
The lie algebra generated by $X$ with relations $R$ is then the quocient $L_0 = hat L/hat K.$ In the book Humphreys states that $L_0$ is usually infinite dimensional.
I wonder why is that. Why usually dim $L_0 = infty$ ?I tried to see it by finding explicitly its root system, but with no sucess.
lie-algebras
asked Nov 15 at 16:54
math.h
1,107517
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up vote
0
down vote
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I am reading Humphrey's book on Lie algebras and now the construction of Lie algebras by generators and relations was just presented.
We simply get a set $X= {hat x_i, hat y_i, hat h_i: 1leq i leq l}$ and a subset $R$ of the free lie algebra $hat L$ generated by $X$, given by
$R = {[hat h_i, hat h_j], [hat x_i, hat y_j] - delta_{ij}hat h_i, [hat h_i, hat x_j]-c_{ji}hat x_j,[hat h_i,hat y_j] + c_{ji}hat y_j$}
and then consider $hat K$ the ideal generated by $R$.
The lie algebra generated by $X$ with relations $R$ is then the quocient $L_0 = hat L/hat K.$ In the book Humphreys states that $L_0$ is usually infinite dimensional.
I wonder why is that. Why usually dim $L_0 = infty$ ?I tried to see it by finding explicitly its root system, but with no sucess.
lie-algebras
asked Nov 15 at 16:54
math.h
1,107517
2
Well, "usually" is of course vague. But the idea is that you need to be careful, or you will not be imposing enough relations to make the Lie algebra finite dimensional. I am not sure what you mean by explicitly finding its root system, as that is not something a Lie algebra will have without much stricter assumptions in the first place.
– Tobias Kildetoft
Nov 15 at 17:19
This is senseless as a mathematical statement. If I give you $10$ examples of prime numbers, $9$ of which being $le 1000$, then I can say that prime numbers are usually $le 1000$. So this means nothing more than "most presentations I can think of yield infinite-dimensional Lie algebras".
– YCor
Nov 15 at 20:51
In group theory there's a famous model by Gromov of random presentations, model depending on a parameter $tin [0,1]$, and it's a theorem that with random groups are infinite when the parameter is $<1/2$ and have at most 2 elements if the parameter $t$ is $>1/2$. There are also certainly ways to define models, of variable interest, for Lie algebras, and certainly for many of these models "with too many relations" the resulting Lie algebra will actually be trivial.
– YCor
Nov 15 at 20:54
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am reading Humphrey's book on Lie algebras and now the construction of Lie algebras by generators and relations was just presented.
We simply get a set $X= {hat x_i, hat y_i, hat h_i: 1leq i leq l}$ and a subset $R$ of the free lie algebra $hat L$ generated by $X$, given by
$R = {[hat h_i, hat h_j], [hat x_i, hat y_j] - delta_{ij}hat h_i, [hat h_i, hat x_j]-c_{ji}hat x_j,[hat h_i,hat y_j] + c_{ji}hat y_j$}
and then consider $hat K$ the ideal generated by $R$.
The lie algebra generated by $X$ with relations $R$ is then the quocient $L_0 = hat L/hat K.$ In the book Humphreys states that $L_0$ is usually infinite dimensional.
I wonder why is that. Why usually dim $L_0 = infty$ ?I tried to see it by finding explicitly its root system, but with no sucess.
lie-algebras
asked Nov 15 at 16:54
math.h
1,107517
I am reading Humphrey's book on Lie algebras and now the construction of Lie algebras by generators and relations was just presented.
We simply get a set $X= {hat x_i, hat y_i, hat h_i: 1leq i leq l}$ and a subset $R$ of the free lie algebra $hat L$ generated by $X$, given by
$R = {[hat h_i, hat h_j], [hat x_i, hat y_j] - delta_{ij}hat h_i, [hat h_i, hat x_j]-c_{ji}hat x_j,[hat h_i,hat y_j] + c_{ji}hat y_j$}
and then consider $hat K$ the ideal generated by $R$.
The lie algebra generated by $X$ with relations $R$ is then the quocient $L_0 = hat L/hat K.$ In the book Humphreys states that $L_0$ is usually infinite dimensional.
I wonder why is that. Why usually dim $L_0 = infty$ ?I tried to see it by finding explicitly its root system, but with no sucess.
lie-algebras
lie-algebras
asked Nov 15 at 16:54
math.h
1,107517
asked Nov 15 at 16:54
math.h
1,107517
asked Nov 15 at 16:54
math.h
1,107517
asked Nov 15 at 16:54
math.h
1,107517
asked Nov 15 at 16:54
math.h
1,107517
1,107517
2
Well, "usually" is of course vague. But the idea is that you need to be careful, or you will not be imposing enough relations to make the Lie algebra finite dimensional. I am not sure what you mean by explicitly finding its root system, as that is not something a Lie algebra will have without much stricter assumptions in the first place.
– Tobias Kildetoft
Nov 15 at 17:19
This is senseless as a mathematical statement. If I give you $10$ examples of prime numbers, $9$ of which being $le 1000$, then I can say that prime numbers are usually $le 1000$. So this means nothing more than "most presentations I can think of yield infinite-dimensional Lie algebras".
– YCor
Nov 15 at 20:51
In group theory there's a famous model by Gromov of random presentations, model depending on a parameter $tin [0,1]$, and it's a theorem that with random groups are infinite when the parameter is $<1/2$ and have at most 2 elements if the parameter $t$ is $>1/2$. There are also certainly ways to define models, of variable interest, for Lie algebras, and certainly for many of these models "with too many relations" the resulting Lie algebra will actually be trivial.
– YCor
Nov 15 at 20:54
add a comment |
2
Well, "usually" is of course vague. But the idea is that you need to be careful, or you will not be imposing enough relations to make the Lie algebra finite dimensional. I am not sure what you mean by explicitly finding its root system, as that is not something a Lie algebra will have without much stricter assumptions in the first place.
– Tobias Kildetoft
Nov 15 at 17:19
This is senseless as a mathematical statement. If I give you $10$ examples of prime numbers, $9$ of which being $le 1000$, then I can say that prime numbers are usually $le 1000$. So this means nothing more than "most presentations I can think of yield infinite-dimensional Lie algebras".
– YCor
Nov 15 at 20:51
In group theory there's a famous model by Gromov of random presentations, model depending on a parameter $tin [0,1]$, and it's a theorem that with random groups are infinite when the parameter is $<1/2$ and have at most 2 elements if the parameter $t$ is $>1/2$. There are also certainly ways to define models, of variable interest, for Lie algebras, and certainly for many of these models "with too many relations" the resulting Lie algebra will actually be trivial.
– YCor
Nov 15 at 20:54
2
2
Well, "usually" is of course vague. But the idea is that you need to be careful, or you will not be imposing enough relations to make the Lie algebra finite dimensional. I am not sure what you mean by explicitly finding its root system, as that is not something a Lie algebra will have without much stricter assumptions in the first place.
– Tobias Kildetoft
Nov 15 at 17:19
Well, "usually" is of course vague. But the idea is that you need to be careful, or you will not be imposing enough relations to make the Lie algebra finite dimensional. I am not sure what you mean by explicitly finding its root system, as that is not something a Lie algebra will have without much stricter assumptions in the first place.
– Tobias Kildetoft
Nov 15 at 17:19
This is senseless as a mathematical statement. If I give you $10$ examples of prime numbers, $9$ of which being $le 1000$, then I can say that prime numbers are usually $le 1000$. So this means nothing more than "most presentations I can think of yield infinite-dimensional Lie algebras".
– YCor
Nov 15 at 20:51
This is senseless as a mathematical statement. If I give you $10$ examples of prime numbers, $9$ of which being $le 1000$, then I can say that prime numbers are usually $le 1000$. So this means nothing more than "most presentations I can think of yield infinite-dimensional Lie algebras".
– YCor
Nov 15 at 20:51
In group theory there's a famous model by Gromov of random presentations, model depending on a parameter $tin [0,1]$, and it's a theorem that with random groups are infinite when the parameter is $<1/2$ and have at most 2 elements if the parameter $t$ is $>1/2$. There are also certainly ways to define models, of variable interest, for Lie algebras, and certainly for many of these models "with too many relations" the resulting Lie algebra will actually be trivial.
– YCor
Nov 15 at 20:54
In group theory there's a famous model by Gromov of random presentations, model depending on a parameter $tin [0,1]$, and it's a theorem that with random groups are infinite when the parameter is $<1/2$ and have at most 2 elements if the parameter $t$ is $>1/2$. There are also certainly ways to define models, of variable interest, for Lie algebras, and certainly for many of these models "with too many relations" the resulting Lie algebra will actually be trivial.
– YCor
Nov 15 at 20:54
add a comment |
1 Answer
1
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up vote
1
down vote
As said, "usually" is too vague. It really depends on the relations. For example, the Lie algebra generated by $e_1$ and $e_2$, and Lie brackets
$$
[e_1,e_i]=e_{i+1}, forall; ige 1,; [e_2,e_3]=e_5,; [e_2,e_5]=re_7
$$
is infinite-dimensional for $r=1$ and $r=9/10$, but $11$-dimensional for $rneq 1,9/10$, i.e., with $e_i=0$ for $ige 12$ ( it is a nilpotent Lie algebra). So it is "usually" a finite-dimensional Lie algebra, and not an infinite-dimensional one.
answered Nov 15 at 20:11
Dietrich Burde
76.7k64286
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
As said, "usually" is too vague. It really depends on the relations. For example, the Lie algebra generated by $e_1$ and $e_2$, and Lie brackets
$$
[e_1,e_i]=e_{i+1}, forall; ige 1,; [e_2,e_3]=e_5,; [e_2,e_5]=re_7
$$
is infinite-dimensional for $r=1$ and $r=9/10$, but $11$-dimensional for $rneq 1,9/10$, i.e., with $e_i=0$ for $ige 12$ ( it is a nilpotent Lie algebra). So it is "usually" a finite-dimensional Lie algebra, and not an infinite-dimensional one.
answered Nov 15 at 20:11
Dietrich Burde
76.7k64286
add a comment |
up vote
1
down vote
As said, "usually" is too vague. It really depends on the relations. For example, the Lie algebra generated by $e_1$ and $e_2$, and Lie brackets
$$
[e_1,e_i]=e_{i+1}, forall; ige 1,; [e_2,e_3]=e_5,; [e_2,e_5]=re_7
$$
is infinite-dimensional for $r=1$ and $r=9/10$, but $11$-dimensional for $rneq 1,9/10$, i.e., with $e_i=0$ for $ige 12$ ( it is a nilpotent Lie algebra). So it is "usually" a finite-dimensional Lie algebra, and not an infinite-dimensional one.
answered Nov 15 at 20:11
Dietrich Burde
76.7k64286
add a comment |
up vote
1
down vote
up vote
1
down vote
As said, "usually" is too vague. It really depends on the relations. For example, the Lie algebra generated by $e_1$ and $e_2$, and Lie brackets
$$
[e_1,e_i]=e_{i+1}, forall; ige 1,; [e_2,e_3]=e_5,; [e_2,e_5]=re_7
$$
is infinite-dimensional for $r=1$ and $r=9/10$, but $11$-dimensional for $rneq 1,9/10$, i.e., with $e_i=0$ for $ige 12$ ( it is a nilpotent Lie algebra). So it is "usually" a finite-dimensional Lie algebra, and not an infinite-dimensional one.
answered Nov 15 at 20:11
Dietrich Burde
76.7k64286
As said, "usually" is too vague. It really depends on the relations. For example, the Lie algebra generated by $e_1$ and $e_2$, and Lie brackets
$$
[e_1,e_i]=e_{i+1}, forall; ige 1,; [e_2,e_3]=e_5,; [e_2,e_5]=re_7
$$
is infinite-dimensional for $r=1$ and $r=9/10$, but $11$-dimensional for $rneq 1,9/10$, i.e., with $e_i=0$ for $ige 12$ ( it is a nilpotent Lie algebra). So it is "usually" a finite-dimensional Lie algebra, and not an infinite-dimensional one.
answered Nov 15 at 20:11
Dietrich Burde
76.7k64286
edited Nov 16 at 19:13
answered Nov 15 at 20:11
Dietrich Burde
76.7k64286
answered Nov 15 at 20:11
Dietrich Burde
76.7k64286
answered Nov 15 at 20:11
Dietrich Burde
76.7k64286
76.7k64286
add a comment |
add a comment |
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Well, "usually" is of course vague. But the idea is that you need to be careful, or you will not be imposing enough relations to make the Lie algebra finite dimensional. I am not sure what you mean by explicitly finding its root system, as that is not something a Lie algebra will have without much stricter assumptions in the first place.
– Tobias Kildetoft
Nov 15 at 17:19
This is senseless as a mathematical statement. If I give you $10$ examples of prime numbers, $9$ of which being $le 1000$, then I can say that prime numbers are usually $le 1000$. So this means nothing more than "most presentations I can think of yield infinite-dimensional Lie algebras".
– YCor
Nov 15 at 20:51
In group theory there's a famous model by Gromov of random presentations, model depending on a parameter $tin [0,1]$, and it's a theorem that with random groups are infinite when the parameter is $<1/2$ and have at most 2 elements if the parameter $t$ is $>1/2$. There are also certainly ways to define models, of variable interest, for Lie algebras, and certainly for many of these models "with too many relations" the resulting Lie algebra will actually be trivial.
– YCor
Nov 15 at 20:54