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Are all Lie groups with a (linear) representation a matrix Lie group?
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I am self studying quantum field theories and as many of you know this can not be properly done without at least some knowledge of Lie groups and the corresponding algebra.
I am reading the QFT text by Michele Maggiore “A modern introduction to quantum field theoryâ€Â. In his chapter on Lie groups (I know, a single chapter is far from enough) he defines a representation of the Lie group as an operation that assigns a linear operator to each element within the Lie group.
My question is, does all Lie groups has such a (linear) representation? Or is Michele Maggiore restricting his discussion on Lie groups to that of matrix Lie groups without explicitly stating so?
This might be trivial question, but I am new to Lie groups and is just seeking some clarification on the topic.
lie-groups
asked Aug 7 at 22:34
Wuberdall
716
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up vote
2
down vote
favorite
I am self studying quantum field theories and as many of you know this can not be properly done without at least some knowledge of Lie groups and the corresponding algebra.
I am reading the QFT text by Michele Maggiore “A modern introduction to quantum field theoryâ€Â. In his chapter on Lie groups (I know, a single chapter is far from enough) he defines a representation of the Lie group as an operation that assigns a linear operator to each element within the Lie group.
My question is, does all Lie groups has such a (linear) representation? Or is Michele Maggiore restricting his discussion on Lie groups to that of matrix Lie groups without explicitly stating so?
This might be trivial question, but I am new to Lie groups and is just seeking some clarification on the topic.
lie-groups
asked Aug 7 at 22:34
Wuberdall
716
1
You can always define the adjoint representation on the associated Lie algebra.
– David Hill
Aug 7 at 22:43
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am self studying quantum field theories and as many of you know this can not be properly done without at least some knowledge of Lie groups and the corresponding algebra.
I am reading the QFT text by Michele Maggiore “A modern introduction to quantum field theoryâ€Â. In his chapter on Lie groups (I know, a single chapter is far from enough) he defines a representation of the Lie group as an operation that assigns a linear operator to each element within the Lie group.
My question is, does all Lie groups has such a (linear) representation? Or is Michele Maggiore restricting his discussion on Lie groups to that of matrix Lie groups without explicitly stating so?
This might be trivial question, but I am new to Lie groups and is just seeking some clarification on the topic.
lie-groups
asked Aug 7 at 22:34
Wuberdall
716
I am self studying quantum field theories and as many of you know this can not be properly done without at least some knowledge of Lie groups and the corresponding algebra.
I am reading the QFT text by Michele Maggiore “A modern introduction to quantum field theoryâ€Â. In his chapter on Lie groups (I know, a single chapter is far from enough) he defines a representation of the Lie group as an operation that assigns a linear operator to each element within the Lie group.
My question is, does all Lie groups has such a (linear) representation? Or is Michele Maggiore restricting his discussion on Lie groups to that of matrix Lie groups without explicitly stating so?
This might be trivial question, but I am new to Lie groups and is just seeking some clarification on the topic.
lie-groups
asked Aug 7 at 22:34
Wuberdall
716
asked Aug 7 at 22:34
Wuberdall
716
asked Aug 7 at 22:34
Wuberdall
716
asked Aug 7 at 22:34
Wuberdall
716
716
1
You can always define the adjoint representation on the associated Lie algebra.
– David Hill
Aug 7 at 22:43
add a comment |Â
1
You can always define the adjoint representation on the associated Lie algebra.
– David Hill
Aug 7 at 22:43
1
1
You can always define the adjoint representation on the associated Lie algebra.
– David Hill
Aug 7 at 22:43
You can always define the adjoint representation on the associated Lie algebra.
– David Hill
Aug 7 at 22:43
add a comment |Â
3 Answers
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For any Lie group $G$ and any vector space $V$, there is a representation of $G$ which sends every element of $G$ to the identity operator on $V$. So, every Lie group has a representation.
In any case, though, I think you are kind of missing the point. We aren't interested in "groups which have a representation". Rather, we are interested in studying the many different representations which a single group may have. So even though a matrix Lie group "comes with" a canonical representation (since every element already is a matrix), we are still interested in other representations of such a group.
answered Aug 7 at 22:55
Eric Wofsey
165k12192306
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Presumably you are wondering whether every Lie group has a faithful finite-dimensional linear representation. Or in other words: is every Lie group realizable as a closed subgroup of $mathrmGL(V)$ for some finite dimensional vector space $V$?
This is true for compact Lie groups (by the Peter-Weyl theorem) but false in general: for instance, the universal cover of $mathrmSL_2(mathbfR)$ has no faithful finite-dimensional representation.
answered Aug 7 at 23:05
Stephen
10.2k12237
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Your confusion is reasonable. First, as @DavidHill noted, we can always let a Lie group act by Adjoint on its Lie algebra, so there is at least one quite non-trivial linear representation. But, yes, this representation may map much of the Lie group to the trivial operator, if the center is large.
A question that may be underlying your question is about faithful linear representations, that is, where the group maps injectively/one-to-one to the linear transformations on the vector space. The question of whether a given abstract Lie group admits a faithful (finite-dimensional) linear representation is subtler, and the answer is not always "yes".
But compact Lie groups, for example, do always admit faithful linear representations, as was known already to Peter and Weyl and others almost 100 years ago.
And, yes, matrix Lie algebras behave somewhat more predictably that more general families.
answered Aug 7 at 23:06
paul garrett
30.9k360116
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
For any Lie group $G$ and any vector space $V$, there is a representation of $G$ which sends every element of $G$ to the identity operator on $V$. So, every Lie group has a representation.
In any case, though, I think you are kind of missing the point. We aren't interested in "groups which have a representation". Rather, we are interested in studying the many different representations which a single group may have. So even though a matrix Lie group "comes with" a canonical representation (since every element already is a matrix), we are still interested in other representations of such a group.
answered Aug 7 at 22:55
Eric Wofsey
165k12192306
add a comment |Â
up vote
3
down vote
For any Lie group $G$ and any vector space $V$, there is a representation of $G$ which sends every element of $G$ to the identity operator on $V$. So, every Lie group has a representation.
In any case, though, I think you are kind of missing the point. We aren't interested in "groups which have a representation". Rather, we are interested in studying the many different representations which a single group may have. So even though a matrix Lie group "comes with" a canonical representation (since every element already is a matrix), we are still interested in other representations of such a group.
answered Aug 7 at 22:55
Eric Wofsey
165k12192306
add a comment |Â
up vote
3
down vote
up vote
3
down vote
For any Lie group $G$ and any vector space $V$, there is a representation of $G$ which sends every element of $G$ to the identity operator on $V$. So, every Lie group has a representation.
In any case, though, I think you are kind of missing the point. We aren't interested in "groups which have a representation". Rather, we are interested in studying the many different representations which a single group may have. So even though a matrix Lie group "comes with" a canonical representation (since every element already is a matrix), we are still interested in other representations of such a group.
answered Aug 7 at 22:55
Eric Wofsey
165k12192306
For any Lie group $G$ and any vector space $V$, there is a representation of $G$ which sends every element of $G$ to the identity operator on $V$. So, every Lie group has a representation.
In any case, though, I think you are kind of missing the point. We aren't interested in "groups which have a representation". Rather, we are interested in studying the many different representations which a single group may have. So even though a matrix Lie group "comes with" a canonical representation (since every element already is a matrix), we are still interested in other representations of such a group.
answered Aug 7 at 22:55
Eric Wofsey
165k12192306
answered Aug 7 at 22:55
Eric Wofsey
165k12192306
answered Aug 7 at 22:55
Eric Wofsey
165k12192306
answered Aug 7 at 22:55
Eric Wofsey
165k12192306
165k12192306
add a comment |Â
add a comment |Â
up vote
3
down vote
Presumably you are wondering whether every Lie group has a faithful finite-dimensional linear representation. Or in other words: is every Lie group realizable as a closed subgroup of $mathrmGL(V)$ for some finite dimensional vector space $V$?
This is true for compact Lie groups (by the Peter-Weyl theorem) but false in general: for instance, the universal cover of $mathrmSL_2(mathbfR)$ has no faithful finite-dimensional representation.
answered Aug 7 at 23:05
Stephen
10.2k12237
add a comment |Â
up vote
3
down vote
Presumably you are wondering whether every Lie group has a faithful finite-dimensional linear representation. Or in other words: is every Lie group realizable as a closed subgroup of $mathrmGL(V)$ for some finite dimensional vector space $V$?
This is true for compact Lie groups (by the Peter-Weyl theorem) but false in general: for instance, the universal cover of $mathrmSL_2(mathbfR)$ has no faithful finite-dimensional representation.
answered Aug 7 at 23:05
Stephen
10.2k12237
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Presumably you are wondering whether every Lie group has a faithful finite-dimensional linear representation. Or in other words: is every Lie group realizable as a closed subgroup of $mathrmGL(V)$ for some finite dimensional vector space $V$?
This is true for compact Lie groups (by the Peter-Weyl theorem) but false in general: for instance, the universal cover of $mathrmSL_2(mathbfR)$ has no faithful finite-dimensional representation.
answered Aug 7 at 23:05
Stephen
10.2k12237
Presumably you are wondering whether every Lie group has a faithful finite-dimensional linear representation. Or in other words: is every Lie group realizable as a closed subgroup of $mathrmGL(V)$ for some finite dimensional vector space $V$?
This is true for compact Lie groups (by the Peter-Weyl theorem) but false in general: for instance, the universal cover of $mathrmSL_2(mathbfR)$ has no faithful finite-dimensional representation.
answered Aug 7 at 23:05
Stephen
10.2k12237
answered Aug 7 at 23:05
Stephen
10.2k12237
answered Aug 7 at 23:05
Stephen
10.2k12237
answered Aug 7 at 23:05
Stephen
10.2k12237
10.2k12237
add a comment |Â
add a comment |Â
up vote
2
down vote
Your confusion is reasonable. First, as @DavidHill noted, we can always let a Lie group act by Adjoint on its Lie algebra, so there is at least one quite non-trivial linear representation. But, yes, this representation may map much of the Lie group to the trivial operator, if the center is large.
A question that may be underlying your question is about faithful linear representations, that is, where the group maps injectively/one-to-one to the linear transformations on the vector space. The question of whether a given abstract Lie group admits a faithful (finite-dimensional) linear representation is subtler, and the answer is not always "yes".
But compact Lie groups, for example, do always admit faithful linear representations, as was known already to Peter and Weyl and others almost 100 years ago.
And, yes, matrix Lie algebras behave somewhat more predictably that more general families.
answered Aug 7 at 23:06
paul garrett
30.9k360116
add a comment |Â
up vote
2
down vote
Your confusion is reasonable. First, as @DavidHill noted, we can always let a Lie group act by Adjoint on its Lie algebra, so there is at least one quite non-trivial linear representation. But, yes, this representation may map much of the Lie group to the trivial operator, if the center is large.
A question that may be underlying your question is about faithful linear representations, that is, where the group maps injectively/one-to-one to the linear transformations on the vector space. The question of whether a given abstract Lie group admits a faithful (finite-dimensional) linear representation is subtler, and the answer is not always "yes".
But compact Lie groups, for example, do always admit faithful linear representations, as was known already to Peter and Weyl and others almost 100 years ago.
And, yes, matrix Lie algebras behave somewhat more predictably that more general families.
answered Aug 7 at 23:06
paul garrett
30.9k360116
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Your confusion is reasonable. First, as @DavidHill noted, we can always let a Lie group act by Adjoint on its Lie algebra, so there is at least one quite non-trivial linear representation. But, yes, this representation may map much of the Lie group to the trivial operator, if the center is large.
A question that may be underlying your question is about faithful linear representations, that is, where the group maps injectively/one-to-one to the linear transformations on the vector space. The question of whether a given abstract Lie group admits a faithful (finite-dimensional) linear representation is subtler, and the answer is not always "yes".
But compact Lie groups, for example, do always admit faithful linear representations, as was known already to Peter and Weyl and others almost 100 years ago.
And, yes, matrix Lie algebras behave somewhat more predictably that more general families.
answered Aug 7 at 23:06
paul garrett
30.9k360116
Your confusion is reasonable. First, as @DavidHill noted, we can always let a Lie group act by Adjoint on its Lie algebra, so there is at least one quite non-trivial linear representation. But, yes, this representation may map much of the Lie group to the trivial operator, if the center is large.
A question that may be underlying your question is about faithful linear representations, that is, where the group maps injectively/one-to-one to the linear transformations on the vector space. The question of whether a given abstract Lie group admits a faithful (finite-dimensional) linear representation is subtler, and the answer is not always "yes".
But compact Lie groups, for example, do always admit faithful linear representations, as was known already to Peter and Weyl and others almost 100 years ago.
And, yes, matrix Lie algebras behave somewhat more predictably that more general families.
answered Aug 7 at 23:06
paul garrett
30.9k360116
answered Aug 7 at 23:06
paul garrett
30.9k360116
answered Aug 7 at 23:06
paul garrett
30.9k360116
answered Aug 7 at 23:06
paul garrett
30.9k360116
30.9k360116
add a comment |Â
add a comment |Â
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1
You can always define the adjoint representation on the associated Lie algebra.
– David Hill
Aug 7 at 22:43