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:




  1. $0 notin Phi, ; E = text{Span}_{mathbb R}(Phi)$


  2. If $alpha, beta in Phi$ then $checkbeta(alpha) in mathbb Z$



  3. Define $omega_alpha : E rightarrow E $ by $omega_alpha(lambda) = lambda - checkalpha(lambda)alpha$.



    Then $alpha in Phi Rightarrow omega_alpha(Phi) = Phi$



  4. 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?










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  • 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















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:




  1. $0 notin Phi, ; E = text{Span}_{mathbb R}(Phi)$


  2. If $alpha, beta in Phi$ then $checkbeta(alpha) in mathbb Z$



  3. Define $omega_alpha : E rightarrow E $ by $omega_alpha(lambda) = lambda - checkalpha(lambda)alpha$.



    Then $alpha in Phi Rightarrow omega_alpha(Phi) = Phi$



  4. 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?










share|cite|improve this question


















  • 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













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:




  1. $0 notin Phi, ; E = text{Span}_{mathbb R}(Phi)$


  2. If $alpha, beta in Phi$ then $checkbeta(alpha) in mathbb Z$



  3. Define $omega_alpha : E rightarrow E $ by $omega_alpha(lambda) = lambda - checkalpha(lambda)alpha$.



    Then $alpha in Phi Rightarrow omega_alpha(Phi) = Phi$



  4. 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?










share|cite|improve this question













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:




  1. $0 notin Phi, ; E = text{Span}_{mathbb R}(Phi)$


  2. If $alpha, beta in Phi$ then $checkbeta(alpha) in mathbb Z$



  3. Define $omega_alpha : E rightarrow E $ by $omega_alpha(lambda) = lambda - checkalpha(lambda)alpha$.



    Then $alpha in Phi Rightarrow omega_alpha(Phi) = Phi$



  4. 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?







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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














  • 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















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