Determinant of adjoint representation
$begingroup$
Let $G$ be a semisimple Lie group with Iwasawa decomposition $G=KAN$ and consider the determinant of the adjoint representation $operatorname{Ad}$ of $AN$.
I want to determine what the derived representation looks like on $mathfrak{a}$ (on $mathfrak{n}$ it is obviously zero). I suspect that one can calculate this values using the root space decomposition w.r.t the root system $(mathfrak{g},mathfrak{a})$.
I know that the derivative of the determinant is the trace. Does this help?
Thanks for any hints.
lie-groups lie-algebras root-systems lie-derivative
asked Jan 13 at 17:33
Luke MathwalkerLuke Mathwalker
361213
$endgroup$
add a comment |
$begingroup$
Let $G$ be a semisimple Lie group with Iwasawa decomposition $G=KAN$ and consider the determinant of the adjoint representation $operatorname{Ad}$ of $AN$.
I want to determine what the derived representation looks like on $mathfrak{a}$ (on $mathfrak{n}$ it is obviously zero). I suspect that one can calculate this values using the root space decomposition w.r.t the root system $(mathfrak{g},mathfrak{a})$.
I know that the derivative of the determinant is the trace. Does this help?
Thanks for any hints.
lie-groups lie-algebras root-systems lie-derivative
asked Jan 13 at 17:33
Luke MathwalkerLuke Mathwalker
361213
$endgroup$
$begingroup$
What is $AN$? A Borel subgroup? Or a parabolic subgroup?
$endgroup$
– D_S
Jan 13 at 23:00
$begingroup$
Also, are you assuming that $G$ is split?
$endgroup$
– D_S
Jan 13 at 23:00
add a comment |
$begingroup$
Let $G$ be a semisimple Lie group with Iwasawa decomposition $G=KAN$ and consider the determinant of the adjoint representation $operatorname{Ad}$ of $AN$.
I want to determine what the derived representation looks like on $mathfrak{a}$ (on $mathfrak{n}$ it is obviously zero). I suspect that one can calculate this values using the root space decomposition w.r.t the root system $(mathfrak{g},mathfrak{a})$.
I know that the derivative of the determinant is the trace. Does this help?
Thanks for any hints.
lie-groups lie-algebras root-systems lie-derivative
asked Jan 13 at 17:33
Luke MathwalkerLuke Mathwalker
361213
$endgroup$
Let $G$ be a semisimple Lie group with Iwasawa decomposition $G=KAN$ and consider the determinant of the adjoint representation $operatorname{Ad}$ of $AN$.
I want to determine what the derived representation looks like on $mathfrak{a}$ (on $mathfrak{n}$ it is obviously zero). I suspect that one can calculate this values using the root space decomposition w.r.t the root system $(mathfrak{g},mathfrak{a})$.
I know that the derivative of the determinant is the trace. Does this help?
Thanks for any hints.
lie-groups lie-algebras root-systems lie-derivative
lie-groups lie-algebras root-systems lie-derivative
asked Jan 13 at 17:33
Luke MathwalkerLuke Mathwalker
361213
asked Jan 13 at 17:33
Luke MathwalkerLuke Mathwalker
361213
edited Jan 13 at 21:40
Luke Mathwalker
asked Jan 13 at 17:33
Luke MathwalkerLuke Mathwalker
361213
asked Jan 13 at 17:33
Luke MathwalkerLuke Mathwalker
361213
asked Jan 13 at 17:33
Luke MathwalkerLuke Mathwalker
361213
361213
$begingroup$
What is $AN$? A Borel subgroup? Or a parabolic subgroup?
$endgroup$
– D_S
Jan 13 at 23:00
$begingroup$
Also, are you assuming that $G$ is split?
$endgroup$
– D_S
Jan 13 at 23:00
add a comment |
$begingroup$
What is $AN$? A Borel subgroup? Or a parabolic subgroup?
$endgroup$
– D_S
Jan 13 at 23:00
$begingroup$
Also, are you assuming that $G$ is split?
$endgroup$
– D_S
Jan 13 at 23:00
$begingroup$
What is $AN$? A Borel subgroup? Or a parabolic subgroup?
$endgroup$
– D_S
Jan 13 at 23:00
$begingroup$
What is $AN$? A Borel subgroup? Or a parabolic subgroup?
$endgroup$
– D_S
Jan 13 at 23:00
$begingroup$
Also, are you assuming that $G$ is split?
$endgroup$
– D_S
Jan 13 at 23:00
$begingroup$
Also, are you assuming that $G$ is split?
$endgroup$
– D_S
Jan 13 at 23:00
add a comment |
1 Answer
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$begingroup$
My question is answered in Knapps book Lie groups beyond an introduction on page 472 (Integration, Application to Reductive Lie Groups). The answer is $2rholog(a)$, where $rho$ is the half sum of positive roots (counted with multiplicities).
answered Jan 13 at 23:07
Luke MathwalkerLuke Mathwalker
361213
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
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$begingroup$
My question is answered in Knapps book Lie groups beyond an introduction on page 472 (Integration, Application to Reductive Lie Groups). The answer is $2rholog(a)$, where $rho$ is the half sum of positive roots (counted with multiplicities).
answered Jan 13 at 23:07
Luke MathwalkerLuke Mathwalker
361213
$endgroup$
add a comment |
$begingroup$
My question is answered in Knapps book Lie groups beyond an introduction on page 472 (Integration, Application to Reductive Lie Groups). The answer is $2rholog(a)$, where $rho$ is the half sum of positive roots (counted with multiplicities).
answered Jan 13 at 23:07
Luke MathwalkerLuke Mathwalker
361213
$endgroup$
add a comment |
$begingroup$
My question is answered in Knapps book Lie groups beyond an introduction on page 472 (Integration, Application to Reductive Lie Groups). The answer is $2rholog(a)$, where $rho$ is the half sum of positive roots (counted with multiplicities).
answered Jan 13 at 23:07
Luke MathwalkerLuke Mathwalker
361213
$endgroup$
My question is answered in Knapps book Lie groups beyond an introduction on page 472 (Integration, Application to Reductive Lie Groups). The answer is $2rholog(a)$, where $rho$ is the half sum of positive roots (counted with multiplicities).
answered Jan 13 at 23:07
Luke MathwalkerLuke Mathwalker
361213
answered Jan 13 at 23:07
Luke MathwalkerLuke Mathwalker
361213
answered Jan 13 at 23:07
Luke MathwalkerLuke Mathwalker
361213
answered Jan 13 at 23:07
Luke MathwalkerLuke Mathwalker
361213
361213
add a comment |
add a comment |
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$begingroup$
What is $AN$? A Borel subgroup? Or a parabolic subgroup?
$endgroup$
– D_S
Jan 13 at 23:00
$begingroup$
Also, are you assuming that $G$ is split?
$endgroup$
– D_S
Jan 13 at 23:00