What property of the root system means a Lie algebra has complex structure?
$begingroup$
Given just the root system of a Lie algebra. How can we tell if the Lie algebra will or will not admit a complex representation? (e.g. a representation in terms of complex $Ntimes N$ matrices which cannot be written in terms of real $N times N$ matrices).
Is there some easy way to tell just from looking at the root system? Such as does the root system have to have a certain symmetry?
As an example from looking at $E_6$ and $E_8$ root systems, is it easy to tell that $E_6$ admits a complex representation (27D complex matrices) $E_8$ does not (fundamental representation is 248 real adjoint)?
See also this question.
representation-theory terminology lie-groups root-systems
asked May 5 '18 at 14:53
zoobyzooby
1,022716
$endgroup$
|
show 1 more comment
$begingroup$
Given just the root system of a Lie algebra. How can we tell if the Lie algebra will or will not admit a complex representation? (e.g. a representation in terms of complex $Ntimes N$ matrices which cannot be written in terms of real $N times N$ matrices).
Is there some easy way to tell just from looking at the root system? Such as does the root system have to have a certain symmetry?
As an example from looking at $E_6$ and $E_8$ root systems, is it easy to tell that $E_6$ admits a complex representation (27D complex matrices) $E_8$ does not (fundamental representation is 248 real adjoint)?
See also this question.
representation-theory terminology lie-groups root-systems
asked May 5 '18 at 14:53
zoobyzooby
1,022716
$endgroup$
$begingroup$
When you say "the root system of a Lie algebra", do you mean a (semisimple) real Lie algebra? (Because I think of complex Lie algebras first, and they obviously have complex representations.) But then, what do you mean by "the root system"? The one of the complexification? Because that one does not even determine the Lie algebra up to isomorphism.
$endgroup$
– Torsten Schoeneberg
May 8 '18 at 6:36
$begingroup$
Yes, I'm mainly thinking of simple compact Lie groups. As in the ones that can be used with Yang-Mills Theory. They have to have a complex representation (unless there are mirror bosons). Can we tell if they do just from the root system?
$endgroup$
– zooby
May 8 '18 at 14:13
$begingroup$
Every compact Lie group has a finite-dimensional faithful complex representation. This is a well-known corollary of the Peter-Weyl theorem.
$endgroup$
– Spenser
May 8 '18 at 15:35
$begingroup$
OK. Let's talk about fundamental representations then. $E_6$ has a fundamental 27 dimensional complex representation. The fundamental representation of $E_8$ is the 248 real dimensional adjoint representation. I'm sure you know what I mean even if I get some of the words wrong. Just because a group is a subgroup of GL(n,C) doesn't mean it's representation is complex. Since real numbers are a subset of complex numbers.
$endgroup$
– zooby
May 8 '18 at 16:26
$begingroup$
For mathematics people, I think the meaning of "complex representation" is a repn of of a real Lie algebra/group on a complex vector space which does not arise by tensoring with $mathbb C$ a repn on a real vector space.
$endgroup$
– paul garrett
May 8 '18 at 16:42
|
show 1 more comment
$begingroup$
Given just the root system of a Lie algebra. How can we tell if the Lie algebra will or will not admit a complex representation? (e.g. a representation in terms of complex $Ntimes N$ matrices which cannot be written in terms of real $N times N$ matrices).
Is there some easy way to tell just from looking at the root system? Such as does the root system have to have a certain symmetry?
As an example from looking at $E_6$ and $E_8$ root systems, is it easy to tell that $E_6$ admits a complex representation (27D complex matrices) $E_8$ does not (fundamental representation is 248 real adjoint)?
See also this question.
representation-theory terminology lie-groups root-systems
asked May 5 '18 at 14:53
zoobyzooby
1,022716
$endgroup$
Given just the root system of a Lie algebra. How can we tell if the Lie algebra will or will not admit a complex representation? (e.g. a representation in terms of complex $Ntimes N$ matrices which cannot be written in terms of real $N times N$ matrices).
Is there some easy way to tell just from looking at the root system? Such as does the root system have to have a certain symmetry?
As an example from looking at $E_6$ and $E_8$ root systems, is it easy to tell that $E_6$ admits a complex representation (27D complex matrices) $E_8$ does not (fundamental representation is 248 real adjoint)?
See also this question.
representation-theory terminology lie-groups root-systems
representation-theory terminology lie-groups root-systems
asked May 5 '18 at 14:53
zoobyzooby
1,022716
asked May 5 '18 at 14:53
zoobyzooby
1,022716
edited May 8 '18 at 16:54
zooby
asked May 5 '18 at 14:53
zoobyzooby
1,022716
asked May 5 '18 at 14:53
zoobyzooby
1,022716
asked May 5 '18 at 14:53
zoobyzooby
1,022716
1,022716
$begingroup$
When you say "the root system of a Lie algebra", do you mean a (semisimple) real Lie algebra? (Because I think of complex Lie algebras first, and they obviously have complex representations.) But then, what do you mean by "the root system"? The one of the complexification? Because that one does not even determine the Lie algebra up to isomorphism.
$endgroup$
– Torsten Schoeneberg
May 8 '18 at 6:36
$begingroup$
Yes, I'm mainly thinking of simple compact Lie groups. As in the ones that can be used with Yang-Mills Theory. They have to have a complex representation (unless there are mirror bosons). Can we tell if they do just from the root system?
$endgroup$
– zooby
May 8 '18 at 14:13
$begingroup$
Every compact Lie group has a finite-dimensional faithful complex representation. This is a well-known corollary of the Peter-Weyl theorem.
$endgroup$
– Spenser
May 8 '18 at 15:35
$begingroup$
OK. Let's talk about fundamental representations then. $E_6$ has a fundamental 27 dimensional complex representation. The fundamental representation of $E_8$ is the 248 real dimensional adjoint representation. I'm sure you know what I mean even if I get some of the words wrong. Just because a group is a subgroup of GL(n,C) doesn't mean it's representation is complex. Since real numbers are a subset of complex numbers.
$endgroup$
– zooby
May 8 '18 at 16:26
$begingroup$
For mathematics people, I think the meaning of "complex representation" is a repn of of a real Lie algebra/group on a complex vector space which does not arise by tensoring with $mathbb C$ a repn on a real vector space.
$endgroup$
– paul garrett
May 8 '18 at 16:42
|
show 1 more comment
$begingroup$
When you say "the root system of a Lie algebra", do you mean a (semisimple) real Lie algebra? (Because I think of complex Lie algebras first, and they obviously have complex representations.) But then, what do you mean by "the root system"? The one of the complexification? Because that one does not even determine the Lie algebra up to isomorphism.
$endgroup$
– Torsten Schoeneberg
May 8 '18 at 6:36
$begingroup$
Yes, I'm mainly thinking of simple compact Lie groups. As in the ones that can be used with Yang-Mills Theory. They have to have a complex representation (unless there are mirror bosons). Can we tell if they do just from the root system?
$endgroup$
– zooby
May 8 '18 at 14:13
$begingroup$
Every compact Lie group has a finite-dimensional faithful complex representation. This is a well-known corollary of the Peter-Weyl theorem.
$endgroup$
– Spenser
May 8 '18 at 15:35
$begingroup$
OK. Let's talk about fundamental representations then. $E_6$ has a fundamental 27 dimensional complex representation. The fundamental representation of $E_8$ is the 248 real dimensional adjoint representation. I'm sure you know what I mean even if I get some of the words wrong. Just because a group is a subgroup of GL(n,C) doesn't mean it's representation is complex. Since real numbers are a subset of complex numbers.
$endgroup$
– zooby
May 8 '18 at 16:26
$begingroup$
For mathematics people, I think the meaning of "complex representation" is a repn of of a real Lie algebra/group on a complex vector space which does not arise by tensoring with $mathbb C$ a repn on a real vector space.
$endgroup$
– paul garrett
May 8 '18 at 16:42
$begingroup$
When you say "the root system of a Lie algebra", do you mean a (semisimple) real Lie algebra? (Because I think of complex Lie algebras first, and they obviously have complex representations.) But then, what do you mean by "the root system"? The one of the complexification? Because that one does not even determine the Lie algebra up to isomorphism.
$endgroup$
– Torsten Schoeneberg
May 8 '18 at 6:36
$begingroup$
When you say "the root system of a Lie algebra", do you mean a (semisimple) real Lie algebra? (Because I think of complex Lie algebras first, and they obviously have complex representations.) But then, what do you mean by "the root system"? The one of the complexification? Because that one does not even determine the Lie algebra up to isomorphism.
$endgroup$
– Torsten Schoeneberg
May 8 '18 at 6:36
$begingroup$
Yes, I'm mainly thinking of simple compact Lie groups. As in the ones that can be used with Yang-Mills Theory. They have to have a complex representation (unless there are mirror bosons). Can we tell if they do just from the root system?
$endgroup$
– zooby
May 8 '18 at 14:13
$begingroup$
Yes, I'm mainly thinking of simple compact Lie groups. As in the ones that can be used with Yang-Mills Theory. They have to have a complex representation (unless there are mirror bosons). Can we tell if they do just from the root system?
$endgroup$
– zooby
May 8 '18 at 14:13
$begingroup$
Every compact Lie group has a finite-dimensional faithful complex representation. This is a well-known corollary of the Peter-Weyl theorem.
$endgroup$
– Spenser
May 8 '18 at 15:35
$begingroup$
Every compact Lie group has a finite-dimensional faithful complex representation. This is a well-known corollary of the Peter-Weyl theorem.
$endgroup$
– Spenser
May 8 '18 at 15:35
$begingroup$
OK. Let's talk about fundamental representations then. $E_6$ has a fundamental 27 dimensional complex representation. The fundamental representation of $E_8$ is the 248 real dimensional adjoint representation. I'm sure you know what I mean even if I get some of the words wrong. Just because a group is a subgroup of GL(n,C) doesn't mean it's representation is complex. Since real numbers are a subset of complex numbers.
$endgroup$
– zooby
May 8 '18 at 16:26
$begingroup$
OK. Let's talk about fundamental representations then. $E_6$ has a fundamental 27 dimensional complex representation. The fundamental representation of $E_8$ is the 248 real dimensional adjoint representation. I'm sure you know what I mean even if I get some of the words wrong. Just because a group is a subgroup of GL(n,C) doesn't mean it's representation is complex. Since real numbers are a subset of complex numbers.
$endgroup$
– zooby
May 8 '18 at 16:26
$begingroup$
For mathematics people, I think the meaning of "complex representation" is a repn of of a real Lie algebra/group on a complex vector space which does not arise by tensoring with $mathbb C$ a repn on a real vector space.
$endgroup$
– paul garrett
May 8 '18 at 16:42
$begingroup$
For mathematics people, I think the meaning of "complex representation" is a repn of of a real Lie algebra/group on a complex vector space which does not arise by tensoring with $mathbb C$ a repn on a real vector space.
$endgroup$
– paul garrett
May 8 '18 at 16:42
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
I'm not at all an expert in the representation theory of compact Lie groups/algebras, but it seems to me that Bourbaki's Lie Groups and Algebras, chapter 9, §7 no.2 proposition 1, answers your question in principle. Bourbaki makes a distinction of complex representations of a compact group into three types: a) real, b) complex and c) quaternionic. "Real" basically means that it's just the complexification of a real representation. As I say in a comment, I assume your question is: "Which fundamental representations are not of type a)?"
Now Bourbaki says that (I paraphrase)
An irreducible representation of highest weight $lambda$ (for $lambda$ dominant w.r.t. a chosen set of simple roots) is of type b) if and only if $$-w_0(lambda) neq lambda$$ where $w_0$ is the longest element of the Weyl group (w.r.t. that set of simple roots).
Further, if it happens that $-w_0(lambda) = lambda$, then we are
in case a) if $sum_{alphain Phi^+} langle check{alpha}, lambda rangle$ is even;
in case c) if $sum_{alphain Phi^+} langle check{alpha}, lambda rangle$ is odd.
The condition on $-w_0(lambda)$ is maybe not immediately visible from the Dynkin diagram of the root system, but it's certainly known. Cf. https://math.stackexchange.com/a/59789/96384. In particular, for types $A_1, B_n, C_n, D_{2n}, E_7, E_8, F_4$ and $G_2$ it is clear that $w_0 = -id$ and hence you will certainly not be in case b) for any $lambda$.
Now whether you are in a) or c) in those cases, or in a), b) or c) in the remaining cases, apparently depends on which weight $lambda$ you're looking at. I admit I don't know enough to say anything about that parity distinction yet -- maybe a true expert can take it from here.
What I think is true is that e.g. for $A_n$ with even $n$, at least all fundamental weights do not get sent to their exact negative, so the corresponding representations are in case b) and hence "genuinely complex"; whereas for the cases $A_{2n+1}, D_{2n+1}$ and $E_6$, it seems that some of their fundamental representations are in b), but others are not, and then again I don't know if they would be in case a) or c).
answered May 10 '18 at 7:09
Torsten SchoenebergTorsten Schoeneberg
4,3212833
$endgroup$
$begingroup$
For type $A$ applying $-w_0$ (when the weight is written in the fundamental basis) reverses the order of the coefficients. So in this case it is easy to see which weights are self-dual. I seem to recall it being the identity for all the type $D$, but I did not actually check this
$endgroup$
– Tobias Kildetoft
May 11 '18 at 17:10
$begingroup$
@TobiasKildetoft: Bourbaki (tables at the end of ch. 6) and A. Geraschenko's answer which I linked to agree that in type $D_n$, the element $-w_0$ is the identity for even $n$, but flips the two "horns" of the Dynkin diagram for odd $n$. That's why I separated the cases $D_{2n}$ and $D_{2n+1}$.
$endgroup$
– Torsten Schoeneberg
May 11 '18 at 19:25
$begingroup$
Ahh, then I misremembered it. Thanks.
$endgroup$
– Tobias Kildetoft
May 11 '18 at 20:45
$begingroup$
@TobiasKildetoft: By the way, I came across this mnemonic to remember (after realising there is a parity distinction for $D_n$) which case is which: In case $D_4$, there is no distinct fork in the diagram (there are three), so $-w_0$ (which is always of order $le 2$) cannot choose one of them, hence has to be the identity. Now 4 is even, so ...
$endgroup$
– Torsten Schoeneberg
May 12 '18 at 22:34
add a comment |
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$begingroup$
I'm not at all an expert in the representation theory of compact Lie groups/algebras, but it seems to me that Bourbaki's Lie Groups and Algebras, chapter 9, §7 no.2 proposition 1, answers your question in principle. Bourbaki makes a distinction of complex representations of a compact group into three types: a) real, b) complex and c) quaternionic. "Real" basically means that it's just the complexification of a real representation. As I say in a comment, I assume your question is: "Which fundamental representations are not of type a)?"
Now Bourbaki says that (I paraphrase)
An irreducible representation of highest weight $lambda$ (for $lambda$ dominant w.r.t. a chosen set of simple roots) is of type b) if and only if $$-w_0(lambda) neq lambda$$ where $w_0$ is the longest element of the Weyl group (w.r.t. that set of simple roots).
Further, if it happens that $-w_0(lambda) = lambda$, then we are
in case a) if $sum_{alphain Phi^+} langle check{alpha}, lambda rangle$ is even;
in case c) if $sum_{alphain Phi^+} langle check{alpha}, lambda rangle$ is odd.
The condition on $-w_0(lambda)$ is maybe not immediately visible from the Dynkin diagram of the root system, but it's certainly known. Cf. https://math.stackexchange.com/a/59789/96384. In particular, for types $A_1, B_n, C_n, D_{2n}, E_7, E_8, F_4$ and $G_2$ it is clear that $w_0 = -id$ and hence you will certainly not be in case b) for any $lambda$.
Now whether you are in a) or c) in those cases, or in a), b) or c) in the remaining cases, apparently depends on which weight $lambda$ you're looking at. I admit I don't know enough to say anything about that parity distinction yet -- maybe a true expert can take it from here.
What I think is true is that e.g. for $A_n$ with even $n$, at least all fundamental weights do not get sent to their exact negative, so the corresponding representations are in case b) and hence "genuinely complex"; whereas for the cases $A_{2n+1}, D_{2n+1}$ and $E_6$, it seems that some of their fundamental representations are in b), but others are not, and then again I don't know if they would be in case a) or c).
answered May 10 '18 at 7:09
Torsten SchoenebergTorsten Schoeneberg
4,3212833
$endgroup$
$begingroup$
For type $A$ applying $-w_0$ (when the weight is written in the fundamental basis) reverses the order of the coefficients. So in this case it is easy to see which weights are self-dual. I seem to recall it being the identity for all the type $D$, but I did not actually check this
$endgroup$
– Tobias Kildetoft
May 11 '18 at 17:10
$begingroup$
@TobiasKildetoft: Bourbaki (tables at the end of ch. 6) and A. Geraschenko's answer which I linked to agree that in type $D_n$, the element $-w_0$ is the identity for even $n$, but flips the two "horns" of the Dynkin diagram for odd $n$. That's why I separated the cases $D_{2n}$ and $D_{2n+1}$.
$endgroup$
– Torsten Schoeneberg
May 11 '18 at 19:25
$begingroup$
Ahh, then I misremembered it. Thanks.
$endgroup$
– Tobias Kildetoft
May 11 '18 at 20:45
$begingroup$
@TobiasKildetoft: By the way, I came across this mnemonic to remember (after realising there is a parity distinction for $D_n$) which case is which: In case $D_4$, there is no distinct fork in the diagram (there are three), so $-w_0$ (which is always of order $le 2$) cannot choose one of them, hence has to be the identity. Now 4 is even, so ...
$endgroup$
– Torsten Schoeneberg
May 12 '18 at 22:34
add a comment |
$begingroup$
I'm not at all an expert in the representation theory of compact Lie groups/algebras, but it seems to me that Bourbaki's Lie Groups and Algebras, chapter 9, §7 no.2 proposition 1, answers your question in principle. Bourbaki makes a distinction of complex representations of a compact group into three types: a) real, b) complex and c) quaternionic. "Real" basically means that it's just the complexification of a real representation. As I say in a comment, I assume your question is: "Which fundamental representations are not of type a)?"
Now Bourbaki says that (I paraphrase)
An irreducible representation of highest weight $lambda$ (for $lambda$ dominant w.r.t. a chosen set of simple roots) is of type b) if and only if $$-w_0(lambda) neq lambda$$ where $w_0$ is the longest element of the Weyl group (w.r.t. that set of simple roots).
Further, if it happens that $-w_0(lambda) = lambda$, then we are
in case a) if $sum_{alphain Phi^+} langle check{alpha}, lambda rangle$ is even;
in case c) if $sum_{alphain Phi^+} langle check{alpha}, lambda rangle$ is odd.
The condition on $-w_0(lambda)$ is maybe not immediately visible from the Dynkin diagram of the root system, but it's certainly known. Cf. https://math.stackexchange.com/a/59789/96384. In particular, for types $A_1, B_n, C_n, D_{2n}, E_7, E_8, F_4$ and $G_2$ it is clear that $w_0 = -id$ and hence you will certainly not be in case b) for any $lambda$.
Now whether you are in a) or c) in those cases, or in a), b) or c) in the remaining cases, apparently depends on which weight $lambda$ you're looking at. I admit I don't know enough to say anything about that parity distinction yet -- maybe a true expert can take it from here.
What I think is true is that e.g. for $A_n$ with even $n$, at least all fundamental weights do not get sent to their exact negative, so the corresponding representations are in case b) and hence "genuinely complex"; whereas for the cases $A_{2n+1}, D_{2n+1}$ and $E_6$, it seems that some of their fundamental representations are in b), but others are not, and then again I don't know if they would be in case a) or c).
answered May 10 '18 at 7:09
Torsten SchoenebergTorsten Schoeneberg
4,3212833
$endgroup$
$begingroup$
For type $A$ applying $-w_0$ (when the weight is written in the fundamental basis) reverses the order of the coefficients. So in this case it is easy to see which weights are self-dual. I seem to recall it being the identity for all the type $D$, but I did not actually check this
$endgroup$
– Tobias Kildetoft
May 11 '18 at 17:10
$begingroup$
@TobiasKildetoft: Bourbaki (tables at the end of ch. 6) and A. Geraschenko's answer which I linked to agree that in type $D_n$, the element $-w_0$ is the identity for even $n$, but flips the two "horns" of the Dynkin diagram for odd $n$. That's why I separated the cases $D_{2n}$ and $D_{2n+1}$.
$endgroup$
– Torsten Schoeneberg
May 11 '18 at 19:25
$begingroup$
Ahh, then I misremembered it. Thanks.
$endgroup$
– Tobias Kildetoft
May 11 '18 at 20:45
$begingroup$
@TobiasKildetoft: By the way, I came across this mnemonic to remember (after realising there is a parity distinction for $D_n$) which case is which: In case $D_4$, there is no distinct fork in the diagram (there are three), so $-w_0$ (which is always of order $le 2$) cannot choose one of them, hence has to be the identity. Now 4 is even, so ...
$endgroup$
– Torsten Schoeneberg
May 12 '18 at 22:34
add a comment |
$begingroup$
I'm not at all an expert in the representation theory of compact Lie groups/algebras, but it seems to me that Bourbaki's Lie Groups and Algebras, chapter 9, §7 no.2 proposition 1, answers your question in principle. Bourbaki makes a distinction of complex representations of a compact group into three types: a) real, b) complex and c) quaternionic. "Real" basically means that it's just the complexification of a real representation. As I say in a comment, I assume your question is: "Which fundamental representations are not of type a)?"
Now Bourbaki says that (I paraphrase)
An irreducible representation of highest weight $lambda$ (for $lambda$ dominant w.r.t. a chosen set of simple roots) is of type b) if and only if $$-w_0(lambda) neq lambda$$ where $w_0$ is the longest element of the Weyl group (w.r.t. that set of simple roots).
Further, if it happens that $-w_0(lambda) = lambda$, then we are
in case a) if $sum_{alphain Phi^+} langle check{alpha}, lambda rangle$ is even;
in case c) if $sum_{alphain Phi^+} langle check{alpha}, lambda rangle$ is odd.
The condition on $-w_0(lambda)$ is maybe not immediately visible from the Dynkin diagram of the root system, but it's certainly known. Cf. https://math.stackexchange.com/a/59789/96384. In particular, for types $A_1, B_n, C_n, D_{2n}, E_7, E_8, F_4$ and $G_2$ it is clear that $w_0 = -id$ and hence you will certainly not be in case b) for any $lambda$.
Now whether you are in a) or c) in those cases, or in a), b) or c) in the remaining cases, apparently depends on which weight $lambda$ you're looking at. I admit I don't know enough to say anything about that parity distinction yet -- maybe a true expert can take it from here.
What I think is true is that e.g. for $A_n$ with even $n$, at least all fundamental weights do not get sent to their exact negative, so the corresponding representations are in case b) and hence "genuinely complex"; whereas for the cases $A_{2n+1}, D_{2n+1}$ and $E_6$, it seems that some of their fundamental representations are in b), but others are not, and then again I don't know if they would be in case a) or c).
answered May 10 '18 at 7:09
Torsten SchoenebergTorsten Schoeneberg
4,3212833
$endgroup$
I'm not at all an expert in the representation theory of compact Lie groups/algebras, but it seems to me that Bourbaki's Lie Groups and Algebras, chapter 9, §7 no.2 proposition 1, answers your question in principle. Bourbaki makes a distinction of complex representations of a compact group into three types: a) real, b) complex and c) quaternionic. "Real" basically means that it's just the complexification of a real representation. As I say in a comment, I assume your question is: "Which fundamental representations are not of type a)?"
Now Bourbaki says that (I paraphrase)
An irreducible representation of highest weight $lambda$ (for $lambda$ dominant w.r.t. a chosen set of simple roots) is of type b) if and only if $$-w_0(lambda) neq lambda$$ where $w_0$ is the longest element of the Weyl group (w.r.t. that set of simple roots).
Further, if it happens that $-w_0(lambda) = lambda$, then we are
in case a) if $sum_{alphain Phi^+} langle check{alpha}, lambda rangle$ is even;
in case c) if $sum_{alphain Phi^+} langle check{alpha}, lambda rangle$ is odd.
The condition on $-w_0(lambda)$ is maybe not immediately visible from the Dynkin diagram of the root system, but it's certainly known. Cf. https://math.stackexchange.com/a/59789/96384. In particular, for types $A_1, B_n, C_n, D_{2n}, E_7, E_8, F_4$ and $G_2$ it is clear that $w_0 = -id$ and hence you will certainly not be in case b) for any $lambda$.
Now whether you are in a) or c) in those cases, or in a), b) or c) in the remaining cases, apparently depends on which weight $lambda$ you're looking at. I admit I don't know enough to say anything about that parity distinction yet -- maybe a true expert can take it from here.
What I think is true is that e.g. for $A_n$ with even $n$, at least all fundamental weights do not get sent to their exact negative, so the corresponding representations are in case b) and hence "genuinely complex"; whereas for the cases $A_{2n+1}, D_{2n+1}$ and $E_6$, it seems that some of their fundamental representations are in b), but others are not, and then again I don't know if they would be in case a) or c).
answered May 10 '18 at 7:09
Torsten SchoenebergTorsten Schoeneberg
4,3212833
edited Jan 25 at 0:27
answered May 10 '18 at 7:09
Torsten SchoenebergTorsten Schoeneberg
4,3212833
answered May 10 '18 at 7:09
Torsten SchoenebergTorsten Schoeneberg
4,3212833
answered May 10 '18 at 7:09
Torsten SchoenebergTorsten Schoeneberg
4,3212833
4,3212833
$begingroup$
For type $A$ applying $-w_0$ (when the weight is written in the fundamental basis) reverses the order of the coefficients. So in this case it is easy to see which weights are self-dual. I seem to recall it being the identity for all the type $D$, but I did not actually check this
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– Tobias Kildetoft
May 11 '18 at 17:10
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@TobiasKildetoft: Bourbaki (tables at the end of ch. 6) and A. Geraschenko's answer which I linked to agree that in type $D_n$, the element $-w_0$ is the identity for even $n$, but flips the two "horns" of the Dynkin diagram for odd $n$. That's why I separated the cases $D_{2n}$ and $D_{2n+1}$.
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– Torsten Schoeneberg
May 11 '18 at 19:25
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Ahh, then I misremembered it. Thanks.
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– Tobias Kildetoft
May 11 '18 at 20:45
$begingroup$
@TobiasKildetoft: By the way, I came across this mnemonic to remember (after realising there is a parity distinction for $D_n$) which case is which: In case $D_4$, there is no distinct fork in the diagram (there are three), so $-w_0$ (which is always of order $le 2$) cannot choose one of them, hence has to be the identity. Now 4 is even, so ...
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– Torsten Schoeneberg
May 12 '18 at 22:34
add a comment |
$begingroup$
For type $A$ applying $-w_0$ (when the weight is written in the fundamental basis) reverses the order of the coefficients. So in this case it is easy to see which weights are self-dual. I seem to recall it being the identity for all the type $D$, but I did not actually check this
$endgroup$
– Tobias Kildetoft
May 11 '18 at 17:10
$begingroup$
@TobiasKildetoft: Bourbaki (tables at the end of ch. 6) and A. Geraschenko's answer which I linked to agree that in type $D_n$, the element $-w_0$ is the identity for even $n$, but flips the two "horns" of the Dynkin diagram for odd $n$. That's why I separated the cases $D_{2n}$ and $D_{2n+1}$.
$endgroup$
– Torsten Schoeneberg
May 11 '18 at 19:25
$begingroup$
Ahh, then I misremembered it. Thanks.
$endgroup$
– Tobias Kildetoft
May 11 '18 at 20:45
$begingroup$
@TobiasKildetoft: By the way, I came across this mnemonic to remember (after realising there is a parity distinction for $D_n$) which case is which: In case $D_4$, there is no distinct fork in the diagram (there are three), so $-w_0$ (which is always of order $le 2$) cannot choose one of them, hence has to be the identity. Now 4 is even, so ...
$endgroup$
– Torsten Schoeneberg
May 12 '18 at 22:34
$begingroup$
For type $A$ applying $-w_0$ (when the weight is written in the fundamental basis) reverses the order of the coefficients. So in this case it is easy to see which weights are self-dual. I seem to recall it being the identity for all the type $D$, but I did not actually check this
$endgroup$
– Tobias Kildetoft
May 11 '18 at 17:10
$begingroup$
For type $A$ applying $-w_0$ (when the weight is written in the fundamental basis) reverses the order of the coefficients. So in this case it is easy to see which weights are self-dual. I seem to recall it being the identity for all the type $D$, but I did not actually check this
$endgroup$
– Tobias Kildetoft
May 11 '18 at 17:10
$begingroup$
@TobiasKildetoft: Bourbaki (tables at the end of ch. 6) and A. Geraschenko's answer which I linked to agree that in type $D_n$, the element $-w_0$ is the identity for even $n$, but flips the two "horns" of the Dynkin diagram for odd $n$. That's why I separated the cases $D_{2n}$ and $D_{2n+1}$.
$endgroup$
– Torsten Schoeneberg
May 11 '18 at 19:25
$begingroup$
@TobiasKildetoft: Bourbaki (tables at the end of ch. 6) and A. Geraschenko's answer which I linked to agree that in type $D_n$, the element $-w_0$ is the identity for even $n$, but flips the two "horns" of the Dynkin diagram for odd $n$. That's why I separated the cases $D_{2n}$ and $D_{2n+1}$.
$endgroup$
– Torsten Schoeneberg
May 11 '18 at 19:25
$begingroup$
Ahh, then I misremembered it. Thanks.
$endgroup$
– Tobias Kildetoft
May 11 '18 at 20:45
$begingroup$
Ahh, then I misremembered it. Thanks.
$endgroup$
– Tobias Kildetoft
May 11 '18 at 20:45
$begingroup$
@TobiasKildetoft: By the way, I came across this mnemonic to remember (after realising there is a parity distinction for $D_n$) which case is which: In case $D_4$, there is no distinct fork in the diagram (there are three), so $-w_0$ (which is always of order $le 2$) cannot choose one of them, hence has to be the identity. Now 4 is even, so ...
$endgroup$
– Torsten Schoeneberg
May 12 '18 at 22:34
$begingroup$
@TobiasKildetoft: By the way, I came across this mnemonic to remember (after realising there is a parity distinction for $D_n$) which case is which: In case $D_4$, there is no distinct fork in the diagram (there are three), so $-w_0$ (which is always of order $le 2$) cannot choose one of them, hence has to be the identity. Now 4 is even, so ...
$endgroup$
– Torsten Schoeneberg
May 12 '18 at 22:34
add a comment |
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When you say "the root system of a Lie algebra", do you mean a (semisimple) real Lie algebra? (Because I think of complex Lie algebras first, and they obviously have complex representations.) But then, what do you mean by "the root system"? The one of the complexification? Because that one does not even determine the Lie algebra up to isomorphism.
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– Torsten Schoeneberg
May 8 '18 at 6:36
$begingroup$
Yes, I'm mainly thinking of simple compact Lie groups. As in the ones that can be used with Yang-Mills Theory. They have to have a complex representation (unless there are mirror bosons). Can we tell if they do just from the root system?
$endgroup$
– zooby
May 8 '18 at 14:13
$begingroup$
Every compact Lie group has a finite-dimensional faithful complex representation. This is a well-known corollary of the Peter-Weyl theorem.
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– Spenser
May 8 '18 at 15:35
$begingroup$
OK. Let's talk about fundamental representations then. $E_6$ has a fundamental 27 dimensional complex representation. The fundamental representation of $E_8$ is the 248 real dimensional adjoint representation. I'm sure you know what I mean even if I get some of the words wrong. Just because a group is a subgroup of GL(n,C) doesn't mean it's representation is complex. Since real numbers are a subset of complex numbers.
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– zooby
May 8 '18 at 16:26
$begingroup$
For mathematics people, I think the meaning of "complex representation" is a repn of of a real Lie algebra/group on a complex vector space which does not arise by tensoring with $mathbb C$ a repn on a real vector space.
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– paul garrett
May 8 '18 at 16:42