What is the weight system for these $suleft(5right)$ representations?












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I need to work out the weight systems for the fundamental representation $mathbf{5}$ and the conjugate representation $overline{mathbf{5}}$. I'm not clear what this means. The 5 representation is of course just the representation of $suleft(5right)$ by itself. After picking a Cartan subalgebra as the diagonal matrices with zero trace, we can of course see that the roots are $L_i−L_j$ where $L_i$ picks out the ith element on the diagonal, and the weights are simply $L_i$ in this case. So what is the 'weight system'?



It is supposed to be the case that I can use the weight systems of representations to show for instance that $mathbf{5}otimes mathbf{5}=mathbf{5}oplus mathbf{15}$.










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  • The representation labelled by $5$ is not the algebra itself, since that is way more than $5$-dimensional. It is instead the "defining" representation coming from this being a subalgebra of $mathfrak{gl}(5)$. In general, labelling the irreducible representations by their dimensions here is going to cause a lot of issues, since there will potentially be many representations of a given dimension.
    – Tobias Kildetoft
    Nov 28 '18 at 9:47










  • 55 = 1015.
    – Cosmas Zachos
    Dec 18 '18 at 1:38


















0














I need to work out the weight systems for the fundamental representation $mathbf{5}$ and the conjugate representation $overline{mathbf{5}}$. I'm not clear what this means. The 5 representation is of course just the representation of $suleft(5right)$ by itself. After picking a Cartan subalgebra as the diagonal matrices with zero trace, we can of course see that the roots are $L_i−L_j$ where $L_i$ picks out the ith element on the diagonal, and the weights are simply $L_i$ in this case. So what is the 'weight system'?



It is supposed to be the case that I can use the weight systems of representations to show for instance that $mathbf{5}otimes mathbf{5}=mathbf{5}oplus mathbf{15}$.










share|cite|improve this question






















  • The representation labelled by $5$ is not the algebra itself, since that is way more than $5$-dimensional. It is instead the "defining" representation coming from this being a subalgebra of $mathfrak{gl}(5)$. In general, labelling the irreducible representations by their dimensions here is going to cause a lot of issues, since there will potentially be many representations of a given dimension.
    – Tobias Kildetoft
    Nov 28 '18 at 9:47










  • 55 = 1015.
    – Cosmas Zachos
    Dec 18 '18 at 1:38
















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0








0







I need to work out the weight systems for the fundamental representation $mathbf{5}$ and the conjugate representation $overline{mathbf{5}}$. I'm not clear what this means. The 5 representation is of course just the representation of $suleft(5right)$ by itself. After picking a Cartan subalgebra as the diagonal matrices with zero trace, we can of course see that the roots are $L_i−L_j$ where $L_i$ picks out the ith element on the diagonal, and the weights are simply $L_i$ in this case. So what is the 'weight system'?



It is supposed to be the case that I can use the weight systems of representations to show for instance that $mathbf{5}otimes mathbf{5}=mathbf{5}oplus mathbf{15}$.










share|cite|improve this question













I need to work out the weight systems for the fundamental representation $mathbf{5}$ and the conjugate representation $overline{mathbf{5}}$. I'm not clear what this means. The 5 representation is of course just the representation of $suleft(5right)$ by itself. After picking a Cartan subalgebra as the diagonal matrices with zero trace, we can of course see that the roots are $L_i−L_j$ where $L_i$ picks out the ith element on the diagonal, and the weights are simply $L_i$ in this case. So what is the 'weight system'?



It is supposed to be the case that I can use the weight systems of representations to show for instance that $mathbf{5}otimes mathbf{5}=mathbf{5}oplus mathbf{15}$.







representation-theory lie-groups lie-algebras






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asked Nov 28 '18 at 3:06









Joshua Tilley

549313




549313












  • The representation labelled by $5$ is not the algebra itself, since that is way more than $5$-dimensional. It is instead the "defining" representation coming from this being a subalgebra of $mathfrak{gl}(5)$. In general, labelling the irreducible representations by their dimensions here is going to cause a lot of issues, since there will potentially be many representations of a given dimension.
    – Tobias Kildetoft
    Nov 28 '18 at 9:47










  • 55 = 1015.
    – Cosmas Zachos
    Dec 18 '18 at 1:38




















  • The representation labelled by $5$ is not the algebra itself, since that is way more than $5$-dimensional. It is instead the "defining" representation coming from this being a subalgebra of $mathfrak{gl}(5)$. In general, labelling the irreducible representations by their dimensions here is going to cause a lot of issues, since there will potentially be many representations of a given dimension.
    – Tobias Kildetoft
    Nov 28 '18 at 9:47










  • 55 = 1015.
    – Cosmas Zachos
    Dec 18 '18 at 1:38


















The representation labelled by $5$ is not the algebra itself, since that is way more than $5$-dimensional. It is instead the "defining" representation coming from this being a subalgebra of $mathfrak{gl}(5)$. In general, labelling the irreducible representations by their dimensions here is going to cause a lot of issues, since there will potentially be many representations of a given dimension.
– Tobias Kildetoft
Nov 28 '18 at 9:47




The representation labelled by $5$ is not the algebra itself, since that is way more than $5$-dimensional. It is instead the "defining" representation coming from this being a subalgebra of $mathfrak{gl}(5)$. In general, labelling the irreducible representations by their dimensions here is going to cause a lot of issues, since there will potentially be many representations of a given dimension.
– Tobias Kildetoft
Nov 28 '18 at 9:47












55 = 1015.
– Cosmas Zachos
Dec 18 '18 at 1:38






55 = 1015.
– Cosmas Zachos
Dec 18 '18 at 1:38












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