Mixing
Connection between a Lie algebra's root system and it's Lie bracket
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I have for some time been trying to understand what the root systems of Lie algebras "mean". I understand that vaguely speaking, the Lie algebra is the derivative of the corresponding Lie group at the identity, and the Lie bracket corresponds to the derivative of the conjugation of two elements close to the identity.
I understand that root systems are lattices that are preserved by reflections through any of the roots. But I don't understand what these reflections have to do with the the Lie bracket. I've read a lot of different explanations, but none of them have made much sense to me so far. Is there a simple relationship?
lie-algebras
edited Jan 14 at 21:59
Bernard
121k740116
asked Jan 14 at 21:55
user142857user142857
32
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add a comment |
$begingroup$
I have for some time been trying to understand what the root systems of Lie algebras "mean". I understand that vaguely speaking, the Lie algebra is the derivative of the corresponding Lie group at the identity, and the Lie bracket corresponds to the derivative of the conjugation of two elements close to the identity.
I understand that root systems are lattices that are preserved by reflections through any of the roots. But I don't understand what these reflections have to do with the the Lie bracket. I've read a lot of different explanations, but none of them have made much sense to me so far. Is there a simple relationship?
lie-algebras
edited Jan 14 at 21:59
Bernard
121k740116
asked Jan 14 at 21:55
user142857user142857
32
$endgroup$
add a comment |
$begingroup$
I have for some time been trying to understand what the root systems of Lie algebras "mean". I understand that vaguely speaking, the Lie algebra is the derivative of the corresponding Lie group at the identity, and the Lie bracket corresponds to the derivative of the conjugation of two elements close to the identity.
I understand that root systems are lattices that are preserved by reflections through any of the roots. But I don't understand what these reflections have to do with the the Lie bracket. I've read a lot of different explanations, but none of them have made much sense to me so far. Is there a simple relationship?
lie-algebras
edited Jan 14 at 21:59
Bernard
121k740116
asked Jan 14 at 21:55
user142857user142857
32
$endgroup$
I have for some time been trying to understand what the root systems of Lie algebras "mean". I understand that vaguely speaking, the Lie algebra is the derivative of the corresponding Lie group at the identity, and the Lie bracket corresponds to the derivative of the conjugation of two elements close to the identity.
I understand that root systems are lattices that are preserved by reflections through any of the roots. But I don't understand what these reflections have to do with the the Lie bracket. I've read a lot of different explanations, but none of them have made much sense to me so far. Is there a simple relationship?
lie-algebras
lie-algebras
edited Jan 14 at 21:59
Bernard
121k740116
asked Jan 14 at 21:55
user142857user142857
32
edited Jan 14 at 21:59
Bernard
121k740116
asked Jan 14 at 21:55
user142857user142857
32
edited Jan 14 at 21:59
Bernard
121k740116
edited Jan 14 at 21:59
Bernard
121k740116
edited Jan 14 at 21:59
Bernard
121k740116
121k740116
asked Jan 14 at 21:55
user142857user142857
32
asked Jan 14 at 21:55
user142857user142857
32
asked Jan 14 at 21:55
user142857user142857
32
32
add a comment |
add a comment |
1 Answer
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I would answer your final question with "no".
Note first of all that only reductive Lie groups/algebras, and in a way only their (split) semisimple part, "have" a root system. This shows that the root system lies on a different structural level than the mere existence of a Lie group resp. Lie algebra structure: only a special subclass has them.
Now, the prototypical reductive Lie group resp. algebra is $GL_n(Bbb C)$ resp. $mathfrak{gl}_n(Bbb C)$. If you work with them, you'll figure out very soon that computing matrix products $Acdot B$ resp. commutators $[A, B]$ by hand each time is not the way you want to spend your hours. So one looks deeper into the structure of matrices, and one figures out that the diagonal matrices on the one hand, and specific unipotent (resp. nilpotent) matrices $I_n + E_{i,j}$ (resp. just $E_{i,j}$) on the other, are at the basis of everything: If one understands products resp. commutators of these, one understands the entire group resp. algebra.
Then, for (split) semisimple Lie groups/algebras, I personally think of root systems as a vast generalisation of that. Cf. my answer here. The analogue of diagonal matrices that you want to find in your group resp. algebra is a torus resp. Cartan subalgebra, the analogue of those elementary matrices are the root spaces etc. Then, as after a while you figure out that in the matrices you don't really need all $E_{i,j}$, but only the $E_{i,i+1}$, you will find that all the information of the root system is already in its so-called simple roots, etc.
Finally note that for non-split semisimple Lie algebras / algebraic groups, which exist over non-algebraically closed fields, the story is more complicated and in general, root systems alone do not help. Tits and Satake did a lot of work there. While over $Bbb R$, the Cartan classification and properties of compact forms remedy this, over number fields ike $Bbb Q$ the situation becomes much trickier.
Finally, to get back to the first point, even over $Bbb C$ and $Bbb R$, there is a huge class of Lie groups resp. algebras, namely the solvable ones, where the theory of root systems is quite useless.
answered Jan 15 at 18:31
Torsten SchoenebergTorsten Schoeneberg
4,1212833
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1 Answer
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$begingroup$
I would answer your final question with "no".
Note first of all that only reductive Lie groups/algebras, and in a way only their (split) semisimple part, "have" a root system. This shows that the root system lies on a different structural level than the mere existence of a Lie group resp. Lie algebra structure: only a special subclass has them.
Now, the prototypical reductive Lie group resp. algebra is $GL_n(Bbb C)$ resp. $mathfrak{gl}_n(Bbb C)$. If you work with them, you'll figure out very soon that computing matrix products $Acdot B$ resp. commutators $[A, B]$ by hand each time is not the way you want to spend your hours. So one looks deeper into the structure of matrices, and one figures out that the diagonal matrices on the one hand, and specific unipotent (resp. nilpotent) matrices $I_n + E_{i,j}$ (resp. just $E_{i,j}$) on the other, are at the basis of everything: If one understands products resp. commutators of these, one understands the entire group resp. algebra.
Then, for (split) semisimple Lie groups/algebras, I personally think of root systems as a vast generalisation of that. Cf. my answer here. The analogue of diagonal matrices that you want to find in your group resp. algebra is a torus resp. Cartan subalgebra, the analogue of those elementary matrices are the root spaces etc. Then, as after a while you figure out that in the matrices you don't really need all $E_{i,j}$, but only the $E_{i,i+1}$, you will find that all the information of the root system is already in its so-called simple roots, etc.
Finally note that for non-split semisimple Lie algebras / algebraic groups, which exist over non-algebraically closed fields, the story is more complicated and in general, root systems alone do not help. Tits and Satake did a lot of work there. While over $Bbb R$, the Cartan classification and properties of compact forms remedy this, over number fields ike $Bbb Q$ the situation becomes much trickier.
Finally, to get back to the first point, even over $Bbb C$ and $Bbb R$, there is a huge class of Lie groups resp. algebras, namely the solvable ones, where the theory of root systems is quite useless.
answered Jan 15 at 18:31
Torsten SchoenebergTorsten Schoeneberg
4,1212833
$endgroup$
add a comment |
$begingroup$
I would answer your final question with "no".
Note first of all that only reductive Lie groups/algebras, and in a way only their (split) semisimple part, "have" a root system. This shows that the root system lies on a different structural level than the mere existence of a Lie group resp. Lie algebra structure: only a special subclass has them.
Now, the prototypical reductive Lie group resp. algebra is $GL_n(Bbb C)$ resp. $mathfrak{gl}_n(Bbb C)$. If you work with them, you'll figure out very soon that computing matrix products $Acdot B$ resp. commutators $[A, B]$ by hand each time is not the way you want to spend your hours. So one looks deeper into the structure of matrices, and one figures out that the diagonal matrices on the one hand, and specific unipotent (resp. nilpotent) matrices $I_n + E_{i,j}$ (resp. just $E_{i,j}$) on the other, are at the basis of everything: If one understands products resp. commutators of these, one understands the entire group resp. algebra.
Then, for (split) semisimple Lie groups/algebras, I personally think of root systems as a vast generalisation of that. Cf. my answer here. The analogue of diagonal matrices that you want to find in your group resp. algebra is a torus resp. Cartan subalgebra, the analogue of those elementary matrices are the root spaces etc. Then, as after a while you figure out that in the matrices you don't really need all $E_{i,j}$, but only the $E_{i,i+1}$, you will find that all the information of the root system is already in its so-called simple roots, etc.
Finally note that for non-split semisimple Lie algebras / algebraic groups, which exist over non-algebraically closed fields, the story is more complicated and in general, root systems alone do not help. Tits and Satake did a lot of work there. While over $Bbb R$, the Cartan classification and properties of compact forms remedy this, over number fields ike $Bbb Q$ the situation becomes much trickier.
Finally, to get back to the first point, even over $Bbb C$ and $Bbb R$, there is a huge class of Lie groups resp. algebras, namely the solvable ones, where the theory of root systems is quite useless.
answered Jan 15 at 18:31
Torsten SchoenebergTorsten Schoeneberg
4,1212833
$endgroup$
add a comment |
$begingroup$
I would answer your final question with "no".
Note first of all that only reductive Lie groups/algebras, and in a way only their (split) semisimple part, "have" a root system. This shows that the root system lies on a different structural level than the mere existence of a Lie group resp. Lie algebra structure: only a special subclass has them.
Now, the prototypical reductive Lie group resp. algebra is $GL_n(Bbb C)$ resp. $mathfrak{gl}_n(Bbb C)$. If you work with them, you'll figure out very soon that computing matrix products $Acdot B$ resp. commutators $[A, B]$ by hand each time is not the way you want to spend your hours. So one looks deeper into the structure of matrices, and one figures out that the diagonal matrices on the one hand, and specific unipotent (resp. nilpotent) matrices $I_n + E_{i,j}$ (resp. just $E_{i,j}$) on the other, are at the basis of everything: If one understands products resp. commutators of these, one understands the entire group resp. algebra.
Then, for (split) semisimple Lie groups/algebras, I personally think of root systems as a vast generalisation of that. Cf. my answer here. The analogue of diagonal matrices that you want to find in your group resp. algebra is a torus resp. Cartan subalgebra, the analogue of those elementary matrices are the root spaces etc. Then, as after a while you figure out that in the matrices you don't really need all $E_{i,j}$, but only the $E_{i,i+1}$, you will find that all the information of the root system is already in its so-called simple roots, etc.
Finally note that for non-split semisimple Lie algebras / algebraic groups, which exist over non-algebraically closed fields, the story is more complicated and in general, root systems alone do not help. Tits and Satake did a lot of work there. While over $Bbb R$, the Cartan classification and properties of compact forms remedy this, over number fields ike $Bbb Q$ the situation becomes much trickier.
Finally, to get back to the first point, even over $Bbb C$ and $Bbb R$, there is a huge class of Lie groups resp. algebras, namely the solvable ones, where the theory of root systems is quite useless.
answered Jan 15 at 18:31
Torsten SchoenebergTorsten Schoeneberg
4,1212833
$endgroup$
I would answer your final question with "no".
Note first of all that only reductive Lie groups/algebras, and in a way only their (split) semisimple part, "have" a root system. This shows that the root system lies on a different structural level than the mere existence of a Lie group resp. Lie algebra structure: only a special subclass has them.
Now, the prototypical reductive Lie group resp. algebra is $GL_n(Bbb C)$ resp. $mathfrak{gl}_n(Bbb C)$. If you work with them, you'll figure out very soon that computing matrix products $Acdot B$ resp. commutators $[A, B]$ by hand each time is not the way you want to spend your hours. So one looks deeper into the structure of matrices, and one figures out that the diagonal matrices on the one hand, and specific unipotent (resp. nilpotent) matrices $I_n + E_{i,j}$ (resp. just $E_{i,j}$) on the other, are at the basis of everything: If one understands products resp. commutators of these, one understands the entire group resp. algebra.
Then, for (split) semisimple Lie groups/algebras, I personally think of root systems as a vast generalisation of that. Cf. my answer here. The analogue of diagonal matrices that you want to find in your group resp. algebra is a torus resp. Cartan subalgebra, the analogue of those elementary matrices are the root spaces etc. Then, as after a while you figure out that in the matrices you don't really need all $E_{i,j}$, but only the $E_{i,i+1}$, you will find that all the information of the root system is already in its so-called simple roots, etc.
Finally note that for non-split semisimple Lie algebras / algebraic groups, which exist over non-algebraically closed fields, the story is more complicated and in general, root systems alone do not help. Tits and Satake did a lot of work there. While over $Bbb R$, the Cartan classification and properties of compact forms remedy this, over number fields ike $Bbb Q$ the situation becomes much trickier.
Finally, to get back to the first point, even over $Bbb C$ and $Bbb R$, there is a huge class of Lie groups resp. algebras, namely the solvable ones, where the theory of root systems is quite useless.
answered Jan 15 at 18:31
Torsten SchoenebergTorsten Schoeneberg
4,1212833
edited Jan 19 at 22:15
answered Jan 15 at 18:31
Torsten SchoenebergTorsten Schoeneberg
4,1212833
answered Jan 15 at 18:31
Torsten SchoenebergTorsten Schoeneberg
4,1212833
answered Jan 15 at 18:31
Torsten SchoenebergTorsten Schoeneberg
4,1212833
4,1212833
add a comment |
add a comment |
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