Prove the dual root system is a root system
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Let $mathfrak{g}$ be a semisimple Lie Algebra, $mathfrak{t}$ a Cartan Subalgebra, $Phi$ the roots with respect to $mathfrak{t}$, and $E = text{Span}_{mathbb R}(Phi)$ a Euclidean space with the Euclidean inner product $(cdot, cdot)_E$
For $alpha in E$ define $checkalpha : E rightarrow mathbb R$ by: $checkalpha(lambda) = frac{2(alpha,lambda)_E}{(alpha,alpha)_E}$.
Then $(Phi, E)$ is a root system, i.e. it satisfies:
$0 notin Phi, ; E = text{Span}_{mathbb R}(Phi)$
If $alpha, beta in Phi$ then $checkbeta(alpha) in mathbb Z$
Define $omega_alpha : E rightarrow E $ by $omega_alpha(lambda) = lambda - checkalpha(lambda)alpha$.
Then $alpha in Phi Rightarrow omega_alpha(Phi) = Phi$
If $alpha in Phi$ and $calpha in Phi$ then $c = pm 1$
Consider now $checkPhi = {checkalpha mid alpha in Phi}$ and $E^*$ the dual space of $E$. I am asked to prove that $(checkPhi, E^*)$, the dual root system, is indeed a root system.
I have shown that properties 1 and 4 hold for $(checkPhi, E^*)$ but I am struggling with 2 and 3.
Specifically, I am unsure of what $(cdot, cdot)_{E^*}$ is. I suspect that I will want to define the functions:
$check{checkalpha} : E^* rightarrow mathbb R$ and $omega_{checkalpha} : E^* rightarrow E^*$ for a given $checkalpha$ such that for $checklambda in E^*$ we have:
$check{check{alpha}}(checklambda) = frac{(checkalpha, checklambda)_{E^*}}{(checkalpha, checkalpha)_{E^*}}$, and $omega_{checkalpha}(checklambda) = checklambda - check{checkalpha}(checklambda)checkalpha$
But I don't really understand what this means. Specifically, I suspect there ought to be some sort of relationship between $(cdot, cdot)_E$ and $(cdot, cdot)_{E^*}$ but I'm not sure what it is.
Should I instead endow $E^*$ with a specific inner product that I construct so that everything works? If I do that, how should I go about constructing one?
lie-algebras
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1
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Let $mathfrak{g}$ be a semisimple Lie Algebra, $mathfrak{t}$ a Cartan Subalgebra, $Phi$ the roots with respect to $mathfrak{t}$, and $E = text{Span}_{mathbb R}(Phi)$ a Euclidean space with the Euclidean inner product $(cdot, cdot)_E$
For $alpha in E$ define $checkalpha : E rightarrow mathbb R$ by: $checkalpha(lambda) = frac{2(alpha,lambda)_E}{(alpha,alpha)_E}$.
Then $(Phi, E)$ is a root system, i.e. it satisfies:
$0 notin Phi, ; E = text{Span}_{mathbb R}(Phi)$
If $alpha, beta in Phi$ then $checkbeta(alpha) in mathbb Z$
Define $omega_alpha : E rightarrow E $ by $omega_alpha(lambda) = lambda - checkalpha(lambda)alpha$.
Then $alpha in Phi Rightarrow omega_alpha(Phi) = Phi$
If $alpha in Phi$ and $calpha in Phi$ then $c = pm 1$
Consider now $checkPhi = {checkalpha mid alpha in Phi}$ and $E^*$ the dual space of $E$. I am asked to prove that $(checkPhi, E^*)$, the dual root system, is indeed a root system.
I have shown that properties 1 and 4 hold for $(checkPhi, E^*)$ but I am struggling with 2 and 3.
Specifically, I am unsure of what $(cdot, cdot)_{E^*}$ is. I suspect that I will want to define the functions:
$check{checkalpha} : E^* rightarrow mathbb R$ and $omega_{checkalpha} : E^* rightarrow E^*$ for a given $checkalpha$ such that for $checklambda in E^*$ we have:
$check{check{alpha}}(checklambda) = frac{(checkalpha, checklambda)_{E^*}}{(checkalpha, checkalpha)_{E^*}}$, and $omega_{checkalpha}(checklambda) = checklambda - check{checkalpha}(checklambda)checkalpha$
But I don't really understand what this means. Specifically, I suspect there ought to be some sort of relationship between $(cdot, cdot)_E$ and $(cdot, cdot)_{E^*}$ but I'm not sure what it is.
Should I instead endow $E^*$ with a specific inner product that I construct so that everything works? If I do that, how should I go about constructing one?
lie-algebras
1
There is a canonical isomorphism $fcolon E to E^*$ given by $f(v)(w) := (v,w)_E$. If you equip $E^*$ with the unique inner product that makes $f$ an isometry, can you work it out?
– Claudius
Nov 14 at 15:40
@Claudius So then $f(frac{2alpha}{(alpha, alpha)_E}) = checkalpha$ so we want $(checkalpha, checkbeta)_{E^*} = (frac{2alpha}{(alpha,alpha)_E}, frac{2beta}{(beta,beta)_E})_E = frac{4}{(alpha,alpha)_E(beta,beta)_E}(alpha,beta)_E$ ?
– user366818
Nov 14 at 17:00
1
Since you were asked to prove this, the next question here might be about the base of the dual root system, see this question.
– Dietrich Burde
Nov 14 at 22:29
1
Yes, now use this to compute $check{checkalpha} (check beta)$ and $omega_{check alpha}$. You might have to use that $omega_alpha$ is an isometry.
– Claudius
Nov 15 at 7:28
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $mathfrak{g}$ be a semisimple Lie Algebra, $mathfrak{t}$ a Cartan Subalgebra, $Phi$ the roots with respect to $mathfrak{t}$, and $E = text{Span}_{mathbb R}(Phi)$ a Euclidean space with the Euclidean inner product $(cdot, cdot)_E$
For $alpha in E$ define $checkalpha : E rightarrow mathbb R$ by: $checkalpha(lambda) = frac{2(alpha,lambda)_E}{(alpha,alpha)_E}$.
Then $(Phi, E)$ is a root system, i.e. it satisfies:
$0 notin Phi, ; E = text{Span}_{mathbb R}(Phi)$
If $alpha, beta in Phi$ then $checkbeta(alpha) in mathbb Z$
Define $omega_alpha : E rightarrow E $ by $omega_alpha(lambda) = lambda - checkalpha(lambda)alpha$.
Then $alpha in Phi Rightarrow omega_alpha(Phi) = Phi$
If $alpha in Phi$ and $calpha in Phi$ then $c = pm 1$
Consider now $checkPhi = {checkalpha mid alpha in Phi}$ and $E^*$ the dual space of $E$. I am asked to prove that $(checkPhi, E^*)$, the dual root system, is indeed a root system.
I have shown that properties 1 and 4 hold for $(checkPhi, E^*)$ but I am struggling with 2 and 3.
Specifically, I am unsure of what $(cdot, cdot)_{E^*}$ is. I suspect that I will want to define the functions:
$check{checkalpha} : E^* rightarrow mathbb R$ and $omega_{checkalpha} : E^* rightarrow E^*$ for a given $checkalpha$ such that for $checklambda in E^*$ we have:
$check{check{alpha}}(checklambda) = frac{(checkalpha, checklambda)_{E^*}}{(checkalpha, checkalpha)_{E^*}}$, and $omega_{checkalpha}(checklambda) = checklambda - check{checkalpha}(checklambda)checkalpha$
But I don't really understand what this means. Specifically, I suspect there ought to be some sort of relationship between $(cdot, cdot)_E$ and $(cdot, cdot)_{E^*}$ but I'm not sure what it is.
Should I instead endow $E^*$ with a specific inner product that I construct so that everything works? If I do that, how should I go about constructing one?
lie-algebras
Let $mathfrak{g}$ be a semisimple Lie Algebra, $mathfrak{t}$ a Cartan Subalgebra, $Phi$ the roots with respect to $mathfrak{t}$, and $E = text{Span}_{mathbb R}(Phi)$ a Euclidean space with the Euclidean inner product $(cdot, cdot)_E$
For $alpha in E$ define $checkalpha : E rightarrow mathbb R$ by: $checkalpha(lambda) = frac{2(alpha,lambda)_E}{(alpha,alpha)_E}$.
Then $(Phi, E)$ is a root system, i.e. it satisfies:
$0 notin Phi, ; E = text{Span}_{mathbb R}(Phi)$
If $alpha, beta in Phi$ then $checkbeta(alpha) in mathbb Z$
Define $omega_alpha : E rightarrow E $ by $omega_alpha(lambda) = lambda - checkalpha(lambda)alpha$.
Then $alpha in Phi Rightarrow omega_alpha(Phi) = Phi$
If $alpha in Phi$ and $calpha in Phi$ then $c = pm 1$
Consider now $checkPhi = {checkalpha mid alpha in Phi}$ and $E^*$ the dual space of $E$. I am asked to prove that $(checkPhi, E^*)$, the dual root system, is indeed a root system.
I have shown that properties 1 and 4 hold for $(checkPhi, E^*)$ but I am struggling with 2 and 3.
Specifically, I am unsure of what $(cdot, cdot)_{E^*}$ is. I suspect that I will want to define the functions:
$check{checkalpha} : E^* rightarrow mathbb R$ and $omega_{checkalpha} : E^* rightarrow E^*$ for a given $checkalpha$ such that for $checklambda in E^*$ we have:
$check{check{alpha}}(checklambda) = frac{(checkalpha, checklambda)_{E^*}}{(checkalpha, checkalpha)_{E^*}}$, and $omega_{checkalpha}(checklambda) = checklambda - check{checkalpha}(checklambda)checkalpha$
But I don't really understand what this means. Specifically, I suspect there ought to be some sort of relationship between $(cdot, cdot)_E$ and $(cdot, cdot)_{E^*}$ but I'm not sure what it is.
Should I instead endow $E^*$ with a specific inner product that I construct so that everything works? If I do that, how should I go about constructing one?
lie-algebras
lie-algebras
asked Nov 14 at 15:27
user366818
808410
808410
1
There is a canonical isomorphism $fcolon E to E^*$ given by $f(v)(w) := (v,w)_E$. If you equip $E^*$ with the unique inner product that makes $f$ an isometry, can you work it out?
– Claudius
Nov 14 at 15:40
@Claudius So then $f(frac{2alpha}{(alpha, alpha)_E}) = checkalpha$ so we want $(checkalpha, checkbeta)_{E^*} = (frac{2alpha}{(alpha,alpha)_E}, frac{2beta}{(beta,beta)_E})_E = frac{4}{(alpha,alpha)_E(beta,beta)_E}(alpha,beta)_E$ ?
– user366818
Nov 14 at 17:00
1
Since you were asked to prove this, the next question here might be about the base of the dual root system, see this question.
– Dietrich Burde
Nov 14 at 22:29
1
Yes, now use this to compute $check{checkalpha} (check beta)$ and $omega_{check alpha}$. You might have to use that $omega_alpha$ is an isometry.
– Claudius
Nov 15 at 7:28
add a comment |
1
There is a canonical isomorphism $fcolon E to E^*$ given by $f(v)(w) := (v,w)_E$. If you equip $E^*$ with the unique inner product that makes $f$ an isometry, can you work it out?
– Claudius
Nov 14 at 15:40
@Claudius So then $f(frac{2alpha}{(alpha, alpha)_E}) = checkalpha$ so we want $(checkalpha, checkbeta)_{E^*} = (frac{2alpha}{(alpha,alpha)_E}, frac{2beta}{(beta,beta)_E})_E = frac{4}{(alpha,alpha)_E(beta,beta)_E}(alpha,beta)_E$ ?
– user366818
Nov 14 at 17:00
1
Since you were asked to prove this, the next question here might be about the base of the dual root system, see this question.
– Dietrich Burde
Nov 14 at 22:29
1
Yes, now use this to compute $check{checkalpha} (check beta)$ and $omega_{check alpha}$. You might have to use that $omega_alpha$ is an isometry.
– Claudius
Nov 15 at 7:28
1
1
There is a canonical isomorphism $fcolon E to E^*$ given by $f(v)(w) := (v,w)_E$. If you equip $E^*$ with the unique inner product that makes $f$ an isometry, can you work it out?
– Claudius
Nov 14 at 15:40
There is a canonical isomorphism $fcolon E to E^*$ given by $f(v)(w) := (v,w)_E$. If you equip $E^*$ with the unique inner product that makes $f$ an isometry, can you work it out?
– Claudius
Nov 14 at 15:40
@Claudius So then $f(frac{2alpha}{(alpha, alpha)_E}) = checkalpha$ so we want $(checkalpha, checkbeta)_{E^*} = (frac{2alpha}{(alpha,alpha)_E}, frac{2beta}{(beta,beta)_E})_E = frac{4}{(alpha,alpha)_E(beta,beta)_E}(alpha,beta)_E$ ?
– user366818
Nov 14 at 17:00
@Claudius So then $f(frac{2alpha}{(alpha, alpha)_E}) = checkalpha$ so we want $(checkalpha, checkbeta)_{E^*} = (frac{2alpha}{(alpha,alpha)_E}, frac{2beta}{(beta,beta)_E})_E = frac{4}{(alpha,alpha)_E(beta,beta)_E}(alpha,beta)_E$ ?
– user366818
Nov 14 at 17:00
1
1
Since you were asked to prove this, the next question here might be about the base of the dual root system, see this question.
– Dietrich Burde
Nov 14 at 22:29
Since you were asked to prove this, the next question here might be about the base of the dual root system, see this question.
– Dietrich Burde
Nov 14 at 22:29
1
1
Yes, now use this to compute $check{checkalpha} (check beta)$ and $omega_{check alpha}$. You might have to use that $omega_alpha$ is an isometry.
– Claudius
Nov 15 at 7:28
Yes, now use this to compute $check{checkalpha} (check beta)$ and $omega_{check alpha}$. You might have to use that $omega_alpha$ is an isometry.
– Claudius
Nov 15 at 7:28
add a comment |
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There is a canonical isomorphism $fcolon E to E^*$ given by $f(v)(w) := (v,w)_E$. If you equip $E^*$ with the unique inner product that makes $f$ an isometry, can you work it out?
– Claudius
Nov 14 at 15:40
@Claudius So then $f(frac{2alpha}{(alpha, alpha)_E}) = checkalpha$ so we want $(checkalpha, checkbeta)_{E^*} = (frac{2alpha}{(alpha,alpha)_E}, frac{2beta}{(beta,beta)_E})_E = frac{4}{(alpha,alpha)_E(beta,beta)_E}(alpha,beta)_E$ ?
– user366818
Nov 14 at 17:00
1
Since you were asked to prove this, the next question here might be about the base of the dual root system, see this question.
– Dietrich Burde
Nov 14 at 22:29
1
Yes, now use this to compute $check{checkalpha} (check beta)$ and $omega_{check alpha}$. You might have to use that $omega_alpha$ is an isometry.
– Claudius
Nov 15 at 7:28