Calculating the Formal character on the irreducible $(n+1)$ dimensional representation of $mathfrak{sl}_2$












1












$begingroup$


Let $V(n)$ be the unique, irreducible representation of $mathfrak{sl}_2$ of $(n+1)$-dimensions.



Let $rho$ be the sum of all fundamental weights.



I want to calculate the formal character $ch(V(n)) = sum_{mu in X}(dim(V(n)_mu))e^mu$



Now, I don't think I can apply the Weyl Character Formula because I don't know anything about there being a dominant weight, which makes me think I'm meant to use this formula.



I am asked to give the formal character in terms of $rho$.



My first thought is that $mathfrak{sl}_2$ has a type $A_1$ root system, and so there is one simple root, and one fundamental weight.



I know that both the roots in the $A_1$ root system are in the weight lattice, and including the fundamental weight this gives three weights to consider for the sum.



Additionally, the fact that there is only one fundamental weight, $omega$ means that $rho = omega$ and we also know that $omega = frac{1}{2}alpha$, where $alpha$ is the simple root.



I am a bit stuck now and I'm not really sure how to proceed in calculating the formal character. How do I know which weights matter?






lie-algebras root-systems







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Let $V(n)$ be the unique, irreducible representation of $mathfrak{sl}_2$ of $(n+1)$-dimensions.



    Let $rho$ be the sum of all fundamental weights.



    I want to calculate the formal character $ch(V(n)) = sum_{mu in X}(dim(V(n)_mu))e^mu$



    Now, I don't think I can apply the Weyl Character Formula because I don't know anything about there being a dominant weight, which makes me think I'm meant to use this formula.



    I am asked to give the formal character in terms of $rho$.



    My first thought is that $mathfrak{sl}_2$ has a type $A_1$ root system, and so there is one simple root, and one fundamental weight.



    I know that both the roots in the $A_1$ root system are in the weight lattice, and including the fundamental weight this gives three weights to consider for the sum.



    Additionally, the fact that there is only one fundamental weight, $omega$ means that $rho = omega$ and we also know that $omega = frac{1}{2}alpha$, where $alpha$ is the simple root.



    I am a bit stuck now and I'm not really sure how to proceed in calculating the formal character. How do I know which weights matter?






    lie-algebras root-systems







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let $V(n)$ be the unique, irreducible representation of $mathfrak{sl}_2$ of $(n+1)$-dimensions.



      Let $rho$ be the sum of all fundamental weights.



      I want to calculate the formal character $ch(V(n)) = sum_{mu in X}(dim(V(n)_mu))e^mu$



      Now, I don't think I can apply the Weyl Character Formula because I don't know anything about there being a dominant weight, which makes me think I'm meant to use this formula.



      I am asked to give the formal character in terms of $rho$.



      My first thought is that $mathfrak{sl}_2$ has a type $A_1$ root system, and so there is one simple root, and one fundamental weight.



      I know that both the roots in the $A_1$ root system are in the weight lattice, and including the fundamental weight this gives three weights to consider for the sum.



      Additionally, the fact that there is only one fundamental weight, $omega$ means that $rho = omega$ and we also know that $omega = frac{1}{2}alpha$, where $alpha$ is the simple root.



      I am a bit stuck now and I'm not really sure how to proceed in calculating the formal character. How do I know which weights matter?






      lie-algebras root-systems







      share|cite|improve this question









      $endgroup$




      Let $V(n)$ be the unique, irreducible representation of $mathfrak{sl}_2$ of $(n+1)$-dimensions.



      Let $rho$ be the sum of all fundamental weights.



      I want to calculate the formal character $ch(V(n)) = sum_{mu in X}(dim(V(n)_mu))e^mu$



      Now, I don't think I can apply the Weyl Character Formula because I don't know anything about there being a dominant weight, which makes me think I'm meant to use this formula.



      I am asked to give the formal character in terms of $rho$.



      My first thought is that $mathfrak{sl}_2$ has a type $A_1$ root system, and so there is one simple root, and one fundamental weight.



      I know that both the roots in the $A_1$ root system are in the weight lattice, and including the fundamental weight this gives three weights to consider for the sum.



      Additionally, the fact that there is only one fundamental weight, $omega$ means that $rho = omega$ and we also know that $omega = frac{1}{2}alpha$, where $alpha$ is the simple root.



      I am a bit stuck now and I'm not really sure how to proceed in calculating the formal character. How do I know which weights matter?






      lie-algebras root-systems




      lie-algebras root-systems






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 5 at 18:34









      user366818user366818

      938410




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          1 Answer
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          $begingroup$

          I believe I have solved this problem:



          Let $e,h,f in mathfrak{sl}_2$ be the usual basis with commutator relations:



          $[ef] = h, ; [he] = 2e, ; [hf] = -2f$



          Then we know that $V(n)$ has a basis $v_0, dots, v_n$ such that:



          $e cdot v_0 = 0$



          $h cdot v_0 = nv_0$



          $f cdot v_i = v_i+1, ; forall i<n$



          $f cdot v_n = 0$



          Also, we know that $mathfrak{sl}_2$ has root system of type $A_1$ and thus has only one simple root $alpha$. This means that if $X$ is the weight lattice, then $X = {frac{m}{2}alpha mid m in mathbb Z}$.



          Using the fact that $alpha(h) = 2$, we see that:



          $V(n)_{frac{m}{2}alpha} = {v in V(n) mid h cdot v = mv} $



          Additionally, we may see that $h cdot v_i = (n-2i)v_i, ; forall 1leq ileq n$



          By comparing dimensions, we may see that:



          $$bigoplus_{i = 0}^{n} V(n)_{frac{(n-2i)}{2}alpha} = V(n)$$



          Knowing all this, it is trivial to use Weyl Character Formula to compute that:



          $$ch(V(n)) = {e^ {-nrho}}frac{e^{2(n+1)rho} - 1}{e^{2rho} - 1}$$






          share|cite|improve this answer









          $endgroup$













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            1 Answer
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            1 Answer
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            active

            oldest

            votes









            active

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            active

            oldest

            votes









            0












            $begingroup$

            I believe I have solved this problem:



            Let $e,h,f in mathfrak{sl}_2$ be the usual basis with commutator relations:



            $[ef] = h, ; [he] = 2e, ; [hf] = -2f$



            Then we know that $V(n)$ has a basis $v_0, dots, v_n$ such that:



            $e cdot v_0 = 0$



            $h cdot v_0 = nv_0$



            $f cdot v_i = v_i+1, ; forall i<n$



            $f cdot v_n = 0$



            Also, we know that $mathfrak{sl}_2$ has root system of type $A_1$ and thus has only one simple root $alpha$. This means that if $X$ is the weight lattice, then $X = {frac{m}{2}alpha mid m in mathbb Z}$.



            Using the fact that $alpha(h) = 2$, we see that:



            $V(n)_{frac{m}{2}alpha} = {v in V(n) mid h cdot v = mv} $



            Additionally, we may see that $h cdot v_i = (n-2i)v_i, ; forall 1leq ileq n$



            By comparing dimensions, we may see that:



            $$bigoplus_{i = 0}^{n} V(n)_{frac{(n-2i)}{2}alpha} = V(n)$$



            Knowing all this, it is trivial to use Weyl Character Formula to compute that:



            $$ch(V(n)) = {e^ {-nrho}}frac{e^{2(n+1)rho} - 1}{e^{2rho} - 1}$$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              I believe I have solved this problem:



              Let $e,h,f in mathfrak{sl}_2$ be the usual basis with commutator relations:



              $[ef] = h, ; [he] = 2e, ; [hf] = -2f$



              Then we know that $V(n)$ has a basis $v_0, dots, v_n$ such that:



              $e cdot v_0 = 0$



              $h cdot v_0 = nv_0$



              $f cdot v_i = v_i+1, ; forall i<n$



              $f cdot v_n = 0$



              Also, we know that $mathfrak{sl}_2$ has root system of type $A_1$ and thus has only one simple root $alpha$. This means that if $X$ is the weight lattice, then $X = {frac{m}{2}alpha mid m in mathbb Z}$.



              Using the fact that $alpha(h) = 2$, we see that:



              $V(n)_{frac{m}{2}alpha} = {v in V(n) mid h cdot v = mv} $



              Additionally, we may see that $h cdot v_i = (n-2i)v_i, ; forall 1leq ileq n$



              By comparing dimensions, we may see that:



              $$bigoplus_{i = 0}^{n} V(n)_{frac{(n-2i)}{2}alpha} = V(n)$$



              Knowing all this, it is trivial to use Weyl Character Formula to compute that:



              $$ch(V(n)) = {e^ {-nrho}}frac{e^{2(n+1)rho} - 1}{e^{2rho} - 1}$$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                I believe I have solved this problem:



                Let $e,h,f in mathfrak{sl}_2$ be the usual basis with commutator relations:



                $[ef] = h, ; [he] = 2e, ; [hf] = -2f$



                Then we know that $V(n)$ has a basis $v_0, dots, v_n$ such that:



                $e cdot v_0 = 0$



                $h cdot v_0 = nv_0$



                $f cdot v_i = v_i+1, ; forall i<n$



                $f cdot v_n = 0$



                Also, we know that $mathfrak{sl}_2$ has root system of type $A_1$ and thus has only one simple root $alpha$. This means that if $X$ is the weight lattice, then $X = {frac{m}{2}alpha mid m in mathbb Z}$.



                Using the fact that $alpha(h) = 2$, we see that:



                $V(n)_{frac{m}{2}alpha} = {v in V(n) mid h cdot v = mv} $



                Additionally, we may see that $h cdot v_i = (n-2i)v_i, ; forall 1leq ileq n$



                By comparing dimensions, we may see that:



                $$bigoplus_{i = 0}^{n} V(n)_{frac{(n-2i)}{2}alpha} = V(n)$$



                Knowing all this, it is trivial to use Weyl Character Formula to compute that:



                $$ch(V(n)) = {e^ {-nrho}}frac{e^{2(n+1)rho} - 1}{e^{2rho} - 1}$$






                share|cite|improve this answer









                $endgroup$



                I believe I have solved this problem:



                Let $e,h,f in mathfrak{sl}_2$ be the usual basis with commutator relations:



                $[ef] = h, ; [he] = 2e, ; [hf] = -2f$



                Then we know that $V(n)$ has a basis $v_0, dots, v_n$ such that:



                $e cdot v_0 = 0$



                $h cdot v_0 = nv_0$



                $f cdot v_i = v_i+1, ; forall i<n$



                $f cdot v_n = 0$



                Also, we know that $mathfrak{sl}_2$ has root system of type $A_1$ and thus has only one simple root $alpha$. This means that if $X$ is the weight lattice, then $X = {frac{m}{2}alpha mid m in mathbb Z}$.



                Using the fact that $alpha(h) = 2$, we see that:



                $V(n)_{frac{m}{2}alpha} = {v in V(n) mid h cdot v = mv} $



                Additionally, we may see that $h cdot v_i = (n-2i)v_i, ; forall 1leq ileq n$



                By comparing dimensions, we may see that:



                $$bigoplus_{i = 0}^{n} V(n)_{frac{(n-2i)}{2}alpha} = V(n)$$



                Knowing all this, it is trivial to use Weyl Character Formula to compute that:



                $$ch(V(n)) = {e^ {-nrho}}frac{e^{2(n+1)rho} - 1}{e^{2rho} - 1}$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 9 at 19:20









                user366818user366818

                938410




                938410






























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