Two questions on roots of finite, simple, complex lie algebra












0














Why are there at most two root lengths for a finite, simple, complex lie algebra? I know it is from the constraint that the $2(alpha,beta)/(alpha,alpha)$ is integer, but what is the argument?



Also, if $alpha$ is a root, $kalpha$ is also a root only when k is 1 or -1? How to prove this result?






lie-groups lie-algebras root-systems







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    0














    Why are there at most two root lengths for a finite, simple, complex lie algebra? I know it is from the constraint that the $2(alpha,beta)/(alpha,alpha)$ is integer, but what is the argument?



    Also, if $alpha$ is a root, $kalpha$ is also a root only when k is 1 or -1? How to prove this result?






    lie-groups lie-algebras root-systems







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      0












      0








      0







      Why are there at most two root lengths for a finite, simple, complex lie algebra? I know it is from the constraint that the $2(alpha,beta)/(alpha,alpha)$ is integer, but what is the argument?



      Also, if $alpha$ is a root, $kalpha$ is also a root only when k is 1 or -1? How to prove this result?






      lie-groups lie-algebras root-systems







      share|cite|improve this question















      Why are there at most two root lengths for a finite, simple, complex lie algebra? I know it is from the constraint that the $2(alpha,beta)/(alpha,alpha)$ is integer, but what is the argument?



      Also, if $alpha$ is a root, $kalpha$ is also a root only when k is 1 or -1? How to prove this result?






      lie-groups lie-algebras root-systems




      lie-groups lie-algebras root-systems






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 24 '15 at 22:37

























      asked Nov 24 '15 at 19:17









      Shadumu

      1546




      1546






















          1 Answer
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          1.)The length of a root $alpha$ is given by $sqrt{(alpha,alpha)}$. The condition $2(alpha,beta)/(alpha,alpha)in mathbb{Z}$ enforces that the angle $theta$ between two roots $alpha$ and $beta$ is one of the following seven possibilities: $frac{pi}{2}, frac{pi}{3},frac{2pi}{3},frac{pi}{4},frac{3pi}{4}, frac{pi}{6},frac{5pi}{6}$. To see this study the equation $(alpha,beta)=sqrt{(alpha,alpha)}sqrt{(beta,beta)}cos (theta)$, so that
          $$
          4cos^2(theta)=2frac{(alpha,beta)}{(alpha,alpha)}cdot 2frac{(alpha,beta)}{(beta,beta)}in {0,1,2,3},
          $$
          because $0le cos^2(theta)le 4$, and $alphaneq pm beta$.
          Now for
          $theta=frac{pi}{3},frac{2pi}{3}$ we see that the roots have equal length, and for the other angles there are two possible length.
          For $theta=frac{pi}{4},frac{3pi}{4}$ the different roots have length proportional to $sqrt{2}$, and for $theta=frac{pi}{6},frac{5pi}{6}$ the factor is $sqrt{3}$.



          2.) By definition of an abstract root system, twice a root is not a root.
          So let $beta=kalpha$. We may assume that $|k|<1$, if not already
          $k=pm 1$. By definition, $2langle beta,alpharangle /langle alpha,alpharangle$ is integral (this is again an axiom). This is only possible for $k=0$ or $k=pm frac{1}{2}$. The latter is impossible, because twice a root is not a root. It follows $k=0$, a contradiction. It follows that $k=pm 1$.



          If you are not assuming that the root system of a simple complex Lie algebra satisfies the axioms of an abstract root system, you have to show that the weight spaces $L_{alpha}$ are $1$-dimensional, and $dim L_{kalpha}=0$ for all $kge 2$. This will use the Killing form of $L$, and a Diophantine argument for a dimension equation over $mathbb{Z}$ concerning the weight spaces.






          share|cite|improve this answer























          • The constraint just tells us the ratios between any two non-orthogonal roots are of ${1, sqrt(2), sqrt(3)}$, it does not rule out the possibility that there might be three roots have different lengths, right? Why twice a root is not a root in abstract root system?
            – Shadumu
            Nov 25 '15 at 15:46










          • No, the angles show that either all length are equal, or at most two different lengths exist - compute all $(alpha,alpha)$ in all seven cases. Twice a root is never a root by definition of an abstract root system (like twice a vector is always again a vector in the vector space, by definition of a vector space).
            – Dietrich Burde
            Nov 25 '15 at 16:07












          • what about if the angle is $π/2$ then the ratio between lengths is undetermined?
            – Shadumu
            Nov 25 '15 at 17:01










          • Yes, exactly. Then $(alpha,alpha)/(beta,beta)$ is undetermined.
            – Dietrich Burde
            Nov 25 '15 at 20:31












          • @DietrichBurde How did you get the equation $(alpha,alpha)=sqrt{(alpha,alpha)}sqrt{(beta,beta)}cos (theta)$? Shouldn't it be $(alpha,beta)=sqrt{(alpha,alpha)}sqrt{(beta,beta)}cos (theta)$
            – John Doe
            Nov 30 at 3:03













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          1.)The length of a root $alpha$ is given by $sqrt{(alpha,alpha)}$. The condition $2(alpha,beta)/(alpha,alpha)in mathbb{Z}$ enforces that the angle $theta$ between two roots $alpha$ and $beta$ is one of the following seven possibilities: $frac{pi}{2}, frac{pi}{3},frac{2pi}{3},frac{pi}{4},frac{3pi}{4}, frac{pi}{6},frac{5pi}{6}$. To see this study the equation $(alpha,beta)=sqrt{(alpha,alpha)}sqrt{(beta,beta)}cos (theta)$, so that
          $$
          4cos^2(theta)=2frac{(alpha,beta)}{(alpha,alpha)}cdot 2frac{(alpha,beta)}{(beta,beta)}in {0,1,2,3},
          $$
          because $0le cos^2(theta)le 4$, and $alphaneq pm beta$.
          Now for
          $theta=frac{pi}{3},frac{2pi}{3}$ we see that the roots have equal length, and for the other angles there are two possible length.
          For $theta=frac{pi}{4},frac{3pi}{4}$ the different roots have length proportional to $sqrt{2}$, and for $theta=frac{pi}{6},frac{5pi}{6}$ the factor is $sqrt{3}$.



          2.) By definition of an abstract root system, twice a root is not a root.
          So let $beta=kalpha$. We may assume that $|k|<1$, if not already
          $k=pm 1$. By definition, $2langle beta,alpharangle /langle alpha,alpharangle$ is integral (this is again an axiom). This is only possible for $k=0$ or $k=pm frac{1}{2}$. The latter is impossible, because twice a root is not a root. It follows $k=0$, a contradiction. It follows that $k=pm 1$.



          If you are not assuming that the root system of a simple complex Lie algebra satisfies the axioms of an abstract root system, you have to show that the weight spaces $L_{alpha}$ are $1$-dimensional, and $dim L_{kalpha}=0$ for all $kge 2$. This will use the Killing form of $L$, and a Diophantine argument for a dimension equation over $mathbb{Z}$ concerning the weight spaces.






          share|cite|improve this answer























          • The constraint just tells us the ratios between any two non-orthogonal roots are of ${1, sqrt(2), sqrt(3)}$, it does not rule out the possibility that there might be three roots have different lengths, right? Why twice a root is not a root in abstract root system?
            – Shadumu
            Nov 25 '15 at 15:46










          • No, the angles show that either all length are equal, or at most two different lengths exist - compute all $(alpha,alpha)$ in all seven cases. Twice a root is never a root by definition of an abstract root system (like twice a vector is always again a vector in the vector space, by definition of a vector space).
            – Dietrich Burde
            Nov 25 '15 at 16:07












          • what about if the angle is $π/2$ then the ratio between lengths is undetermined?
            – Shadumu
            Nov 25 '15 at 17:01










          • Yes, exactly. Then $(alpha,alpha)/(beta,beta)$ is undetermined.
            – Dietrich Burde
            Nov 25 '15 at 20:31












          • @DietrichBurde How did you get the equation $(alpha,alpha)=sqrt{(alpha,alpha)}sqrt{(beta,beta)}cos (theta)$? Shouldn't it be $(alpha,beta)=sqrt{(alpha,alpha)}sqrt{(beta,beta)}cos (theta)$
            – John Doe
            Nov 30 at 3:03


















          1














          1.)The length of a root $alpha$ is given by $sqrt{(alpha,alpha)}$. The condition $2(alpha,beta)/(alpha,alpha)in mathbb{Z}$ enforces that the angle $theta$ between two roots $alpha$ and $beta$ is one of the following seven possibilities: $frac{pi}{2}, frac{pi}{3},frac{2pi}{3},frac{pi}{4},frac{3pi}{4}, frac{pi}{6},frac{5pi}{6}$. To see this study the equation $(alpha,beta)=sqrt{(alpha,alpha)}sqrt{(beta,beta)}cos (theta)$, so that
          $$
          4cos^2(theta)=2frac{(alpha,beta)}{(alpha,alpha)}cdot 2frac{(alpha,beta)}{(beta,beta)}in {0,1,2,3},
          $$
          because $0le cos^2(theta)le 4$, and $alphaneq pm beta$.
          Now for
          $theta=frac{pi}{3},frac{2pi}{3}$ we see that the roots have equal length, and for the other angles there are two possible length.
          For $theta=frac{pi}{4},frac{3pi}{4}$ the different roots have length proportional to $sqrt{2}$, and for $theta=frac{pi}{6},frac{5pi}{6}$ the factor is $sqrt{3}$.



          2.) By definition of an abstract root system, twice a root is not a root.
          So let $beta=kalpha$. We may assume that $|k|<1$, if not already
          $k=pm 1$. By definition, $2langle beta,alpharangle /langle alpha,alpharangle$ is integral (this is again an axiom). This is only possible for $k=0$ or $k=pm frac{1}{2}$. The latter is impossible, because twice a root is not a root. It follows $k=0$, a contradiction. It follows that $k=pm 1$.



          If you are not assuming that the root system of a simple complex Lie algebra satisfies the axioms of an abstract root system, you have to show that the weight spaces $L_{alpha}$ are $1$-dimensional, and $dim L_{kalpha}=0$ for all $kge 2$. This will use the Killing form of $L$, and a Diophantine argument for a dimension equation over $mathbb{Z}$ concerning the weight spaces.






          share|cite|improve this answer























          • The constraint just tells us the ratios between any two non-orthogonal roots are of ${1, sqrt(2), sqrt(3)}$, it does not rule out the possibility that there might be three roots have different lengths, right? Why twice a root is not a root in abstract root system?
            – Shadumu
            Nov 25 '15 at 15:46










          • No, the angles show that either all length are equal, or at most two different lengths exist - compute all $(alpha,alpha)$ in all seven cases. Twice a root is never a root by definition of an abstract root system (like twice a vector is always again a vector in the vector space, by definition of a vector space).
            – Dietrich Burde
            Nov 25 '15 at 16:07












          • what about if the angle is $π/2$ then the ratio between lengths is undetermined?
            – Shadumu
            Nov 25 '15 at 17:01










          • Yes, exactly. Then $(alpha,alpha)/(beta,beta)$ is undetermined.
            – Dietrich Burde
            Nov 25 '15 at 20:31












          • @DietrichBurde How did you get the equation $(alpha,alpha)=sqrt{(alpha,alpha)}sqrt{(beta,beta)}cos (theta)$? Shouldn't it be $(alpha,beta)=sqrt{(alpha,alpha)}sqrt{(beta,beta)}cos (theta)$
            – John Doe
            Nov 30 at 3:03
















          1












          1








          1






          1.)The length of a root $alpha$ is given by $sqrt{(alpha,alpha)}$. The condition $2(alpha,beta)/(alpha,alpha)in mathbb{Z}$ enforces that the angle $theta$ between two roots $alpha$ and $beta$ is one of the following seven possibilities: $frac{pi}{2}, frac{pi}{3},frac{2pi}{3},frac{pi}{4},frac{3pi}{4}, frac{pi}{6},frac{5pi}{6}$. To see this study the equation $(alpha,beta)=sqrt{(alpha,alpha)}sqrt{(beta,beta)}cos (theta)$, so that
          $$
          4cos^2(theta)=2frac{(alpha,beta)}{(alpha,alpha)}cdot 2frac{(alpha,beta)}{(beta,beta)}in {0,1,2,3},
          $$
          because $0le cos^2(theta)le 4$, and $alphaneq pm beta$.
          Now for
          $theta=frac{pi}{3},frac{2pi}{3}$ we see that the roots have equal length, and for the other angles there are two possible length.
          For $theta=frac{pi}{4},frac{3pi}{4}$ the different roots have length proportional to $sqrt{2}$, and for $theta=frac{pi}{6},frac{5pi}{6}$ the factor is $sqrt{3}$.



          2.) By definition of an abstract root system, twice a root is not a root.
          So let $beta=kalpha$. We may assume that $|k|<1$, if not already
          $k=pm 1$. By definition, $2langle beta,alpharangle /langle alpha,alpharangle$ is integral (this is again an axiom). This is only possible for $k=0$ or $k=pm frac{1}{2}$. The latter is impossible, because twice a root is not a root. It follows $k=0$, a contradiction. It follows that $k=pm 1$.



          If you are not assuming that the root system of a simple complex Lie algebra satisfies the axioms of an abstract root system, you have to show that the weight spaces $L_{alpha}$ are $1$-dimensional, and $dim L_{kalpha}=0$ for all $kge 2$. This will use the Killing form of $L$, and a Diophantine argument for a dimension equation over $mathbb{Z}$ concerning the weight spaces.






          share|cite|improve this answer














          1.)The length of a root $alpha$ is given by $sqrt{(alpha,alpha)}$. The condition $2(alpha,beta)/(alpha,alpha)in mathbb{Z}$ enforces that the angle $theta$ between two roots $alpha$ and $beta$ is one of the following seven possibilities: $frac{pi}{2}, frac{pi}{3},frac{2pi}{3},frac{pi}{4},frac{3pi}{4}, frac{pi}{6},frac{5pi}{6}$. To see this study the equation $(alpha,beta)=sqrt{(alpha,alpha)}sqrt{(beta,beta)}cos (theta)$, so that
          $$
          4cos^2(theta)=2frac{(alpha,beta)}{(alpha,alpha)}cdot 2frac{(alpha,beta)}{(beta,beta)}in {0,1,2,3},
          $$
          because $0le cos^2(theta)le 4$, and $alphaneq pm beta$.
          Now for
          $theta=frac{pi}{3},frac{2pi}{3}$ we see that the roots have equal length, and for the other angles there are two possible length.
          For $theta=frac{pi}{4},frac{3pi}{4}$ the different roots have length proportional to $sqrt{2}$, and for $theta=frac{pi}{6},frac{5pi}{6}$ the factor is $sqrt{3}$.



          2.) By definition of an abstract root system, twice a root is not a root.
          So let $beta=kalpha$. We may assume that $|k|<1$, if not already
          $k=pm 1$. By definition, $2langle beta,alpharangle /langle alpha,alpharangle$ is integral (this is again an axiom). This is only possible for $k=0$ or $k=pm frac{1}{2}$. The latter is impossible, because twice a root is not a root. It follows $k=0$, a contradiction. It follows that $k=pm 1$.



          If you are not assuming that the root system of a simple complex Lie algebra satisfies the axioms of an abstract root system, you have to show that the weight spaces $L_{alpha}$ are $1$-dimensional, and $dim L_{kalpha}=0$ for all $kge 2$. This will use the Killing form of $L$, and a Diophantine argument for a dimension equation over $mathbb{Z}$ concerning the weight spaces.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 30 at 9:08

























          answered Nov 25 '15 at 11:57









          Dietrich Burde

          77.3k64386




          77.3k64386












          • The constraint just tells us the ratios between any two non-orthogonal roots are of ${1, sqrt(2), sqrt(3)}$, it does not rule out the possibility that there might be three roots have different lengths, right? Why twice a root is not a root in abstract root system?
            – Shadumu
            Nov 25 '15 at 15:46










          • No, the angles show that either all length are equal, or at most two different lengths exist - compute all $(alpha,alpha)$ in all seven cases. Twice a root is never a root by definition of an abstract root system (like twice a vector is always again a vector in the vector space, by definition of a vector space).
            – Dietrich Burde
            Nov 25 '15 at 16:07












          • what about if the angle is $π/2$ then the ratio between lengths is undetermined?
            – Shadumu
            Nov 25 '15 at 17:01










          • Yes, exactly. Then $(alpha,alpha)/(beta,beta)$ is undetermined.
            – Dietrich Burde
            Nov 25 '15 at 20:31












          • @DietrichBurde How did you get the equation $(alpha,alpha)=sqrt{(alpha,alpha)}sqrt{(beta,beta)}cos (theta)$? Shouldn't it be $(alpha,beta)=sqrt{(alpha,alpha)}sqrt{(beta,beta)}cos (theta)$
            – John Doe
            Nov 30 at 3:03




















          • The constraint just tells us the ratios between any two non-orthogonal roots are of ${1, sqrt(2), sqrt(3)}$, it does not rule out the possibility that there might be three roots have different lengths, right? Why twice a root is not a root in abstract root system?
            – Shadumu
            Nov 25 '15 at 15:46










          • No, the angles show that either all length are equal, or at most two different lengths exist - compute all $(alpha,alpha)$ in all seven cases. Twice a root is never a root by definition of an abstract root system (like twice a vector is always again a vector in the vector space, by definition of a vector space).
            – Dietrich Burde
            Nov 25 '15 at 16:07












          • what about if the angle is $π/2$ then the ratio between lengths is undetermined?
            – Shadumu
            Nov 25 '15 at 17:01










          • Yes, exactly. Then $(alpha,alpha)/(beta,beta)$ is undetermined.
            – Dietrich Burde
            Nov 25 '15 at 20:31












          • @DietrichBurde How did you get the equation $(alpha,alpha)=sqrt{(alpha,alpha)}sqrt{(beta,beta)}cos (theta)$? Shouldn't it be $(alpha,beta)=sqrt{(alpha,alpha)}sqrt{(beta,beta)}cos (theta)$
            – John Doe
            Nov 30 at 3:03


















          The constraint just tells us the ratios between any two non-orthogonal roots are of ${1, sqrt(2), sqrt(3)}$, it does not rule out the possibility that there might be three roots have different lengths, right? Why twice a root is not a root in abstract root system?
          – Shadumu
          Nov 25 '15 at 15:46




          The constraint just tells us the ratios between any two non-orthogonal roots are of ${1, sqrt(2), sqrt(3)}$, it does not rule out the possibility that there might be three roots have different lengths, right? Why twice a root is not a root in abstract root system?
          – Shadumu
          Nov 25 '15 at 15:46












          No, the angles show that either all length are equal, or at most two different lengths exist - compute all $(alpha,alpha)$ in all seven cases. Twice a root is never a root by definition of an abstract root system (like twice a vector is always again a vector in the vector space, by definition of a vector space).
          – Dietrich Burde
          Nov 25 '15 at 16:07






          No, the angles show that either all length are equal, or at most two different lengths exist - compute all $(alpha,alpha)$ in all seven cases. Twice a root is never a root by definition of an abstract root system (like twice a vector is always again a vector in the vector space, by definition of a vector space).
          – Dietrich Burde
          Nov 25 '15 at 16:07














          what about if the angle is $π/2$ then the ratio between lengths is undetermined?
          – Shadumu
          Nov 25 '15 at 17:01




          what about if the angle is $π/2$ then the ratio between lengths is undetermined?
          – Shadumu
          Nov 25 '15 at 17:01












          Yes, exactly. Then $(alpha,alpha)/(beta,beta)$ is undetermined.
          – Dietrich Burde
          Nov 25 '15 at 20:31






          Yes, exactly. Then $(alpha,alpha)/(beta,beta)$ is undetermined.
          – Dietrich Burde
          Nov 25 '15 at 20:31














          @DietrichBurde How did you get the equation $(alpha,alpha)=sqrt{(alpha,alpha)}sqrt{(beta,beta)}cos (theta)$? Shouldn't it be $(alpha,beta)=sqrt{(alpha,alpha)}sqrt{(beta,beta)}cos (theta)$
          – John Doe
          Nov 30 at 3:03






          @DietrichBurde How did you get the equation $(alpha,alpha)=sqrt{(alpha,alpha)}sqrt{(beta,beta)}cos (theta)$? Shouldn't it be $(alpha,beta)=sqrt{(alpha,alpha)}sqrt{(beta,beta)}cos (theta)$
          – John Doe
          Nov 30 at 3:03




















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