Definition of the Lie algebra and the Lie bracket for general vector fields
$begingroup$
I've started to go deep into the theory of Lie groups to eventually understand their representation theory. I picked up a text online and right on the first chapter something started to bother me. When the author defines a Lie algebra over a vector field, we get the standard definition:
Definition 1.2 A Lie algebra over $mathbb K = mathbb R$ or $mathbb C$ is a vector space $V$ over $mathbb K$
with a skew-symmetric $mathbb K$-bilinear form (the Lie bracket) $[,,,]:Vtimes Vrightarrow V$ which satisfies the Jacobi identity $$[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$$ for all $X,Y,Zinmathbb K$.
However, when the author first goes to use the notion of a Lie bracket to define the Lie algebra of a Lie group, he states that the pushforward of a diffeomorphism $f:Mrightarrow N$ is compatible with the Lie bracket, i.e. that
$$f_*[X,Y]=[f_* X,f_* Y].$$
As such, it seems like we completely disregarded the original abstract definition of the bracket on 1.2 and chose to use the Lie derivative because it happens to satisfy these properties and is nice to work with. This immediately seemed very arbitrary to me.
Are there any assumptions that I haven't picked up on to maybe uniquely determine that the Lie algebra of left-invariant vector fields can only have the Lie derivative as a consistent bracket (up to a scalar multiple, of course)?
differential-topology lie-groups lie-algebras
asked Jan 11 at 3:05
Gabriel GolfettiGabriel Golfetti
1847
$endgroup$
add a comment |
$begingroup$
I've started to go deep into the theory of Lie groups to eventually understand their representation theory. I picked up a text online and right on the first chapter something started to bother me. When the author defines a Lie algebra over a vector field, we get the standard definition:
Definition 1.2 A Lie algebra over $mathbb K = mathbb R$ or $mathbb C$ is a vector space $V$ over $mathbb K$
with a skew-symmetric $mathbb K$-bilinear form (the Lie bracket) $[,,,]:Vtimes Vrightarrow V$ which satisfies the Jacobi identity $$[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$$ for all $X,Y,Zinmathbb K$.
However, when the author first goes to use the notion of a Lie bracket to define the Lie algebra of a Lie group, he states that the pushforward of a diffeomorphism $f:Mrightarrow N$ is compatible with the Lie bracket, i.e. that
$$f_*[X,Y]=[f_* X,f_* Y].$$
As such, it seems like we completely disregarded the original abstract definition of the bracket on 1.2 and chose to use the Lie derivative because it happens to satisfy these properties and is nice to work with. This immediately seemed very arbitrary to me.
Are there any assumptions that I haven't picked up on to maybe uniquely determine that the Lie algebra of left-invariant vector fields can only have the Lie derivative as a consistent bracket (up to a scalar multiple, of course)?
differential-topology lie-groups lie-algebras
asked Jan 11 at 3:05
Gabriel GolfettiGabriel Golfetti
1847
$endgroup$
2
$begingroup$
This is a really weird question. It's like asking, "why is the group operation on $GL_n(mathbb{R})$ matrix multiplication, when the definition of a group doesn't say that the group operation must be matrix multiplication?".
$endgroup$
– Eric Wofsey
Jan 11 at 3:13
3
$begingroup$
Oh, looking at the text you're reading, it looks like it's not your fault and the text just does a terrible job of explaining what's going on.
$endgroup$
– Eric Wofsey
Jan 11 at 3:18
add a comment |
$begingroup$
I've started to go deep into the theory of Lie groups to eventually understand their representation theory. I picked up a text online and right on the first chapter something started to bother me. When the author defines a Lie algebra over a vector field, we get the standard definition:
Definition 1.2 A Lie algebra over $mathbb K = mathbb R$ or $mathbb C$ is a vector space $V$ over $mathbb K$
with a skew-symmetric $mathbb K$-bilinear form (the Lie bracket) $[,,,]:Vtimes Vrightarrow V$ which satisfies the Jacobi identity $$[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$$ for all $X,Y,Zinmathbb K$.
However, when the author first goes to use the notion of a Lie bracket to define the Lie algebra of a Lie group, he states that the pushforward of a diffeomorphism $f:Mrightarrow N$ is compatible with the Lie bracket, i.e. that
$$f_*[X,Y]=[f_* X,f_* Y].$$
As such, it seems like we completely disregarded the original abstract definition of the bracket on 1.2 and chose to use the Lie derivative because it happens to satisfy these properties and is nice to work with. This immediately seemed very arbitrary to me.
Are there any assumptions that I haven't picked up on to maybe uniquely determine that the Lie algebra of left-invariant vector fields can only have the Lie derivative as a consistent bracket (up to a scalar multiple, of course)?
differential-topology lie-groups lie-algebras
asked Jan 11 at 3:05
Gabriel GolfettiGabriel Golfetti
1847
$endgroup$
I've started to go deep into the theory of Lie groups to eventually understand their representation theory. I picked up a text online and right on the first chapter something started to bother me. When the author defines a Lie algebra over a vector field, we get the standard definition:
Definition 1.2 A Lie algebra over $mathbb K = mathbb R$ or $mathbb C$ is a vector space $V$ over $mathbb K$
with a skew-symmetric $mathbb K$-bilinear form (the Lie bracket) $[,,,]:Vtimes Vrightarrow V$ which satisfies the Jacobi identity $$[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$$ for all $X,Y,Zinmathbb K$.
However, when the author first goes to use the notion of a Lie bracket to define the Lie algebra of a Lie group, he states that the pushforward of a diffeomorphism $f:Mrightarrow N$ is compatible with the Lie bracket, i.e. that
$$f_*[X,Y]=[f_* X,f_* Y].$$
As such, it seems like we completely disregarded the original abstract definition of the bracket on 1.2 and chose to use the Lie derivative because it happens to satisfy these properties and is nice to work with. This immediately seemed very arbitrary to me.
Are there any assumptions that I haven't picked up on to maybe uniquely determine that the Lie algebra of left-invariant vector fields can only have the Lie derivative as a consistent bracket (up to a scalar multiple, of course)?
differential-topology lie-groups lie-algebras
differential-topology lie-groups lie-algebras
asked Jan 11 at 3:05
Gabriel GolfettiGabriel Golfetti
1847
asked Jan 11 at 3:05
Gabriel GolfettiGabriel Golfetti
1847
asked Jan 11 at 3:05
Gabriel GolfettiGabriel Golfetti
1847
asked Jan 11 at 3:05
Gabriel GolfettiGabriel Golfetti
1847
asked Jan 11 at 3:05
Gabriel GolfettiGabriel Golfetti
1847
1847
2
$begingroup$
This is a really weird question. It's like asking, "why is the group operation on $GL_n(mathbb{R})$ matrix multiplication, when the definition of a group doesn't say that the group operation must be matrix multiplication?".
$endgroup$
– Eric Wofsey
Jan 11 at 3:13
3
$begingroup$
Oh, looking at the text you're reading, it looks like it's not your fault and the text just does a terrible job of explaining what's going on.
$endgroup$
– Eric Wofsey
Jan 11 at 3:18
add a comment |
2
$begingroup$
This is a really weird question. It's like asking, "why is the group operation on $GL_n(mathbb{R})$ matrix multiplication, when the definition of a group doesn't say that the group operation must be matrix multiplication?".
$endgroup$
– Eric Wofsey
Jan 11 at 3:13
3
$begingroup$
Oh, looking at the text you're reading, it looks like it's not your fault and the text just does a terrible job of explaining what's going on.
$endgroup$
– Eric Wofsey
Jan 11 at 3:18
2
2
$begingroup$
This is a really weird question. It's like asking, "why is the group operation on $GL_n(mathbb{R})$ matrix multiplication, when the definition of a group doesn't say that the group operation must be matrix multiplication?".
$endgroup$
– Eric Wofsey
Jan 11 at 3:13
$begingroup$
This is a really weird question. It's like asking, "why is the group operation on $GL_n(mathbb{R})$ matrix multiplication, when the definition of a group doesn't say that the group operation must be matrix multiplication?".
$endgroup$
– Eric Wofsey
Jan 11 at 3:13
3
3
$begingroup$
Oh, looking at the text you're reading, it looks like it's not your fault and the text just does a terrible job of explaining what's going on.
$endgroup$
– Eric Wofsey
Jan 11 at 3:18
$begingroup$
Oh, looking at the text you're reading, it looks like it's not your fault and the text just does a terrible job of explaining what's going on.
$endgroup$
– Eric Wofsey
Jan 11 at 3:18
add a comment |
2 Answers
2
active
oldest
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$begingroup$
All that's going on here is that the author is using the term "Lie bracket" with two different meanings and not explaining that they are separate. On the one hand, there is the term "Lie bracket" which refers to the operation in a general Lie algebra as in Definition 1.2. On the other hand, there is the term "Lie bracket" which specifically refers to the Lie derivative operation on vector fields on a manifold. When the author talks about pushforwards under diffeomorphism, he is using "Lie bracket" in this second sense, not in the sense of Definition 1.2.
The author is in no way asserting that the Lie derivative is the only possible Lie bracket (in the sense of Definition 1.2) that can be defined on vector fields. Rather, he is just (rather confusingly in context) using the term "Lie bracket" as a name for the Lie derivative operation on vector fields. Of course, this is connected to Definition 1.2 in that vector fields with the Lie derivative operation are an example of a Lie algebra. But they're just an example; no assertion is being made about the meaning of "Lie bracket" in an arbitrary Lie algebra.
answered Jan 11 at 3:27
Eric WofseyEric Wofsey
192k14220352
$endgroup$
add a comment |
$begingroup$
There can be many other consistent brackets.
Suppose that there is a basis $v_1,v_2,cdots,v_n$, you can define a consistent bracket just through the commutation relation which is completely determined by some constants.
However, only the Lie derivative is useful to get the needed Lie algebra which corresponds the Lie group.
answered Jan 11 at 3:23
W. muW. mu
755310
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
All that's going on here is that the author is using the term "Lie bracket" with two different meanings and not explaining that they are separate. On the one hand, there is the term "Lie bracket" which refers to the operation in a general Lie algebra as in Definition 1.2. On the other hand, there is the term "Lie bracket" which specifically refers to the Lie derivative operation on vector fields on a manifold. When the author talks about pushforwards under diffeomorphism, he is using "Lie bracket" in this second sense, not in the sense of Definition 1.2.
The author is in no way asserting that the Lie derivative is the only possible Lie bracket (in the sense of Definition 1.2) that can be defined on vector fields. Rather, he is just (rather confusingly in context) using the term "Lie bracket" as a name for the Lie derivative operation on vector fields. Of course, this is connected to Definition 1.2 in that vector fields with the Lie derivative operation are an example of a Lie algebra. But they're just an example; no assertion is being made about the meaning of "Lie bracket" in an arbitrary Lie algebra.
answered Jan 11 at 3:27
Eric WofseyEric Wofsey
192k14220352
$endgroup$
add a comment |
$begingroup$
All that's going on here is that the author is using the term "Lie bracket" with two different meanings and not explaining that they are separate. On the one hand, there is the term "Lie bracket" which refers to the operation in a general Lie algebra as in Definition 1.2. On the other hand, there is the term "Lie bracket" which specifically refers to the Lie derivative operation on vector fields on a manifold. When the author talks about pushforwards under diffeomorphism, he is using "Lie bracket" in this second sense, not in the sense of Definition 1.2.
The author is in no way asserting that the Lie derivative is the only possible Lie bracket (in the sense of Definition 1.2) that can be defined on vector fields. Rather, he is just (rather confusingly in context) using the term "Lie bracket" as a name for the Lie derivative operation on vector fields. Of course, this is connected to Definition 1.2 in that vector fields with the Lie derivative operation are an example of a Lie algebra. But they're just an example; no assertion is being made about the meaning of "Lie bracket" in an arbitrary Lie algebra.
answered Jan 11 at 3:27
Eric WofseyEric Wofsey
192k14220352
$endgroup$
add a comment |
$begingroup$
All that's going on here is that the author is using the term "Lie bracket" with two different meanings and not explaining that they are separate. On the one hand, there is the term "Lie bracket" which refers to the operation in a general Lie algebra as in Definition 1.2. On the other hand, there is the term "Lie bracket" which specifically refers to the Lie derivative operation on vector fields on a manifold. When the author talks about pushforwards under diffeomorphism, he is using "Lie bracket" in this second sense, not in the sense of Definition 1.2.
The author is in no way asserting that the Lie derivative is the only possible Lie bracket (in the sense of Definition 1.2) that can be defined on vector fields. Rather, he is just (rather confusingly in context) using the term "Lie bracket" as a name for the Lie derivative operation on vector fields. Of course, this is connected to Definition 1.2 in that vector fields with the Lie derivative operation are an example of a Lie algebra. But they're just an example; no assertion is being made about the meaning of "Lie bracket" in an arbitrary Lie algebra.
answered Jan 11 at 3:27
Eric WofseyEric Wofsey
192k14220352
$endgroup$
All that's going on here is that the author is using the term "Lie bracket" with two different meanings and not explaining that they are separate. On the one hand, there is the term "Lie bracket" which refers to the operation in a general Lie algebra as in Definition 1.2. On the other hand, there is the term "Lie bracket" which specifically refers to the Lie derivative operation on vector fields on a manifold. When the author talks about pushforwards under diffeomorphism, he is using "Lie bracket" in this second sense, not in the sense of Definition 1.2.
The author is in no way asserting that the Lie derivative is the only possible Lie bracket (in the sense of Definition 1.2) that can be defined on vector fields. Rather, he is just (rather confusingly in context) using the term "Lie bracket" as a name for the Lie derivative operation on vector fields. Of course, this is connected to Definition 1.2 in that vector fields with the Lie derivative operation are an example of a Lie algebra. But they're just an example; no assertion is being made about the meaning of "Lie bracket" in an arbitrary Lie algebra.
answered Jan 11 at 3:27
Eric WofseyEric Wofsey
192k14220352
answered Jan 11 at 3:27
Eric WofseyEric Wofsey
192k14220352
answered Jan 11 at 3:27
Eric WofseyEric Wofsey
192k14220352
answered Jan 11 at 3:27
Eric WofseyEric Wofsey
192k14220352
192k14220352
add a comment |
add a comment |
$begingroup$
There can be many other consistent brackets.
Suppose that there is a basis $v_1,v_2,cdots,v_n$, you can define a consistent bracket just through the commutation relation which is completely determined by some constants.
However, only the Lie derivative is useful to get the needed Lie algebra which corresponds the Lie group.
answered Jan 11 at 3:23
W. muW. mu
755310
$endgroup$
add a comment |
$begingroup$
There can be many other consistent brackets.
Suppose that there is a basis $v_1,v_2,cdots,v_n$, you can define a consistent bracket just through the commutation relation which is completely determined by some constants.
However, only the Lie derivative is useful to get the needed Lie algebra which corresponds the Lie group.
answered Jan 11 at 3:23
W. muW. mu
755310
$endgroup$
add a comment |
$begingroup$
There can be many other consistent brackets.
Suppose that there is a basis $v_1,v_2,cdots,v_n$, you can define a consistent bracket just through the commutation relation which is completely determined by some constants.
However, only the Lie derivative is useful to get the needed Lie algebra which corresponds the Lie group.
answered Jan 11 at 3:23
W. muW. mu
755310
$endgroup$
There can be many other consistent brackets.
Suppose that there is a basis $v_1,v_2,cdots,v_n$, you can define a consistent bracket just through the commutation relation which is completely determined by some constants.
However, only the Lie derivative is useful to get the needed Lie algebra which corresponds the Lie group.
answered Jan 11 at 3:23
W. muW. mu
755310
answered Jan 11 at 3:23
W. muW. mu
755310
answered Jan 11 at 3:23
W. muW. mu
755310
answered Jan 11 at 3:23
W. muW. mu
755310
755310
add a comment |
add a comment |
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$begingroup$
This is a really weird question. It's like asking, "why is the group operation on $GL_n(mathbb{R})$ matrix multiplication, when the definition of a group doesn't say that the group operation must be matrix multiplication?".
$endgroup$
– Eric Wofsey
Jan 11 at 3:13
3
$begingroup$
Oh, looking at the text you're reading, it looks like it's not your fault and the text just does a terrible job of explaining what's going on.
$endgroup$
– Eric Wofsey
Jan 11 at 3:18