Weights of $SUleft(5right)$ representation












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Consider the representation $Lambda^2V$ of $suleft(5right)$ where $V$ is the fundamental representation. How can I work out the Dynkin labels of its weights?



Are these the correct Dynkin labels for the weights of $V$:
$$left(1,0,0,0right),left(-1,1,0,0right),left(0,-1,1,0right),left(0,0,-1,1right),left(0,0,0,-1right)$$










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    1














    Consider the representation $Lambda^2V$ of $suleft(5right)$ where $V$ is the fundamental representation. How can I work out the Dynkin labels of its weights?



    Are these the correct Dynkin labels for the weights of $V$:
    $$left(1,0,0,0right),left(-1,1,0,0right),left(0,-1,1,0right),left(0,0,-1,1right),left(0,0,0,-1right)$$










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      Consider the representation $Lambda^2V$ of $suleft(5right)$ where $V$ is the fundamental representation. How can I work out the Dynkin labels of its weights?



      Are these the correct Dynkin labels for the weights of $V$:
      $$left(1,0,0,0right),left(-1,1,0,0right),left(0,-1,1,0right),left(0,0,-1,1right),left(0,0,0,-1right)$$










      share|cite|improve this question













      Consider the representation $Lambda^2V$ of $suleft(5right)$ where $V$ is the fundamental representation. How can I work out the Dynkin labels of its weights?



      Are these the correct Dynkin labels for the weights of $V$:
      $$left(1,0,0,0right),left(-1,1,0,0right),left(0,-1,1,0right),left(0,0,-1,1right),left(0,0,0,-1right)$$







      representation-theory lie-groups lie-algebras






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      asked Nov 28 '18 at 11:23









      Joshua Tilley

      549313




      549313






















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          Pick a basis $v_1, ldots, v_5$ for $V$, where $g in SU(5)$ acts by matrices. Then the maximal torus $T$ of diagonal matrices in $SU(5)$ acts on $V$ as well, and this representation is completely reducible, and decomposes as $mathbb{C}v_1 oplus cdots oplus mathbb{C}v_5$. The action of $(t_1, ldots, t_5)$ on $v_1$ is $(t_1, ldots, t_5) cdot v_1 = t_1 v_1$, which we could call the character $(1, 0, 0, 0, 0)$. All the other characters are similar, for example the action on $v_3$ is by the character $(0, 0, 1, 0, 0)$. Note also that since $t_1 cdots t_5 = 1 in T$, these characters are only given up to shift by $(1, 1, 1, 1, 1)$.



          Now, for $bigwedge^2 V$, you can do exactly the same thing. It has a basis $v_1 wedge v_2, v_1 wedge v_3, ldots, v_4 wedge v_5$, and the action of $T$ on $v_1 wedge v_2$ is by $(t_1, ldots, t_5) cdot (v_1 wedge v_2) = t_1 v_1 wedge t_2 v_2$, which is the character $(1, 1, 0, 0, 0)$. You can work out the others.






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            1 Answer
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            active

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            1














            Pick a basis $v_1, ldots, v_5$ for $V$, where $g in SU(5)$ acts by matrices. Then the maximal torus $T$ of diagonal matrices in $SU(5)$ acts on $V$ as well, and this representation is completely reducible, and decomposes as $mathbb{C}v_1 oplus cdots oplus mathbb{C}v_5$. The action of $(t_1, ldots, t_5)$ on $v_1$ is $(t_1, ldots, t_5) cdot v_1 = t_1 v_1$, which we could call the character $(1, 0, 0, 0, 0)$. All the other characters are similar, for example the action on $v_3$ is by the character $(0, 0, 1, 0, 0)$. Note also that since $t_1 cdots t_5 = 1 in T$, these characters are only given up to shift by $(1, 1, 1, 1, 1)$.



            Now, for $bigwedge^2 V$, you can do exactly the same thing. It has a basis $v_1 wedge v_2, v_1 wedge v_3, ldots, v_4 wedge v_5$, and the action of $T$ on $v_1 wedge v_2$ is by $(t_1, ldots, t_5) cdot (v_1 wedge v_2) = t_1 v_1 wedge t_2 v_2$, which is the character $(1, 1, 0, 0, 0)$. You can work out the others.






            share|cite|improve this answer


























              1














              Pick a basis $v_1, ldots, v_5$ for $V$, where $g in SU(5)$ acts by matrices. Then the maximal torus $T$ of diagonal matrices in $SU(5)$ acts on $V$ as well, and this representation is completely reducible, and decomposes as $mathbb{C}v_1 oplus cdots oplus mathbb{C}v_5$. The action of $(t_1, ldots, t_5)$ on $v_1$ is $(t_1, ldots, t_5) cdot v_1 = t_1 v_1$, which we could call the character $(1, 0, 0, 0, 0)$. All the other characters are similar, for example the action on $v_3$ is by the character $(0, 0, 1, 0, 0)$. Note also that since $t_1 cdots t_5 = 1 in T$, these characters are only given up to shift by $(1, 1, 1, 1, 1)$.



              Now, for $bigwedge^2 V$, you can do exactly the same thing. It has a basis $v_1 wedge v_2, v_1 wedge v_3, ldots, v_4 wedge v_5$, and the action of $T$ on $v_1 wedge v_2$ is by $(t_1, ldots, t_5) cdot (v_1 wedge v_2) = t_1 v_1 wedge t_2 v_2$, which is the character $(1, 1, 0, 0, 0)$. You can work out the others.






              share|cite|improve this answer
























                1












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                1






                Pick a basis $v_1, ldots, v_5$ for $V$, where $g in SU(5)$ acts by matrices. Then the maximal torus $T$ of diagonal matrices in $SU(5)$ acts on $V$ as well, and this representation is completely reducible, and decomposes as $mathbb{C}v_1 oplus cdots oplus mathbb{C}v_5$. The action of $(t_1, ldots, t_5)$ on $v_1$ is $(t_1, ldots, t_5) cdot v_1 = t_1 v_1$, which we could call the character $(1, 0, 0, 0, 0)$. All the other characters are similar, for example the action on $v_3$ is by the character $(0, 0, 1, 0, 0)$. Note also that since $t_1 cdots t_5 = 1 in T$, these characters are only given up to shift by $(1, 1, 1, 1, 1)$.



                Now, for $bigwedge^2 V$, you can do exactly the same thing. It has a basis $v_1 wedge v_2, v_1 wedge v_3, ldots, v_4 wedge v_5$, and the action of $T$ on $v_1 wedge v_2$ is by $(t_1, ldots, t_5) cdot (v_1 wedge v_2) = t_1 v_1 wedge t_2 v_2$, which is the character $(1, 1, 0, 0, 0)$. You can work out the others.






                share|cite|improve this answer












                Pick a basis $v_1, ldots, v_5$ for $V$, where $g in SU(5)$ acts by matrices. Then the maximal torus $T$ of diagonal matrices in $SU(5)$ acts on $V$ as well, and this representation is completely reducible, and decomposes as $mathbb{C}v_1 oplus cdots oplus mathbb{C}v_5$. The action of $(t_1, ldots, t_5)$ on $v_1$ is $(t_1, ldots, t_5) cdot v_1 = t_1 v_1$, which we could call the character $(1, 0, 0, 0, 0)$. All the other characters are similar, for example the action on $v_3$ is by the character $(0, 0, 1, 0, 0)$. Note also that since $t_1 cdots t_5 = 1 in T$, these characters are only given up to shift by $(1, 1, 1, 1, 1)$.



                Now, for $bigwedge^2 V$, you can do exactly the same thing. It has a basis $v_1 wedge v_2, v_1 wedge v_3, ldots, v_4 wedge v_5$, and the action of $T$ on $v_1 wedge v_2$ is by $(t_1, ldots, t_5) cdot (v_1 wedge v_2) = t_1 v_1 wedge t_2 v_2$, which is the character $(1, 1, 0, 0, 0)$. You can work out the others.







                share|cite|improve this answer












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                answered Nov 28 '18 at 22:16









                Joppy

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