Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 102 / 4

Choose positive roots ${\Phi }^{+}$ and letbe the Weyl vector. For a dominant integral weight $\lambda$, the Weyl dimension formula isHere ${\alpha }^{\vee }$ is the coroot and $⟨,⟩$ is the natural weight-coroot pairing.

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Write the nonzero highest weight of $V$ asChoose $i$ with $m_i>0$. Then $\lambda -{\omega }_i$ is dominant. Every positive coroot is a nonnegative combination of simple coroots, so every numerator in the Weyl dimension formula for $\lambda$ is at least the corresponding numerator for ${\omega }_i$. ThusMinimality of $\dim V$ forces equality. If $\lambda \ne {\omega }_i$, some simple-coroot factor is strictly larger, making the product strict. Hence $\lambda ={\omega }_i$ andso $V$ is a fundamental representation.

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For an irreducible representation $V(\lambda )$, writeFormThe tensor product of highest-weight vectors is killed by all positive-root spaces and has weight $\lambda$. It therefore generates a highest-weight constituent isomorphic to $V(\lambda )$. By Complete reducibility of semisimple Lie algebra representations, this constituent is a subrepresentation of $T$. This proves the generation by fundamental representations statement.

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For ${\mathfrak{sl}}_3(\mathbb{C})$ there are two fundamental weights. The corresponding Fundamental representations of sl3 are

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Let $V=V({\omega }_1)$, so $V^{\ast }=V({\omega }_2)$. First,The highest-weight tensor-product rule givesandThe dimensions $15+3+6+3=27$ check the decomposition. Therefore the Triple tensor decomposition for the defining sl3 representation isThus one may take

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