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Let us work over $\mathbb C$. Suppose that $G$ is a semisimple algebraic group and let $H \subset G$ be a maximal torus. Consider a dominant weight $\omega$, then one can associate a unique irreducible $G$-representation $V(\omega)$.

Consider the following example. Let $G=\mathrm{SL}_3(\mathbb C)$ and let $W=\mathbb C^3$ be the standard $G$-representation. We define $$ V=\mathrm{Sym}^2W, $$ it is a $6$-dimensional irreducible representation for $G$. Moreover, we can also see $V$ as the space of symmetric $3 \times 3$ matrices. In particular, it inherits the structure of unital associative algebra using the matrix product. This is an example of Jordan algebra. On the other hand, if we define $$ \tilde V=W \otimes W^\vee, $$ it contains an irreducible $G$-representation that corresponds to the Lie algebra $\mathfrak g$.

In fact, these are all examples of $G$-representations that have also an algebra structure. My question is: given a group $G$ and a dominant weight $\omega$, can we know when $V(\omega)$ has a unital algebra structure?

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    $\begingroup$ Any algebra $V$ can be viewed as a map (the multiplication) $\mu: V \otimes V \to V$. If we don't require associativity OR any relation to $G$, then of course we can define an algebra structure on every irrep $V$ of $G$. Of course you do want some relation to $G$, I expect you want $\mu$ to be a $G$-map? Then the first step is to find out (using Weyl character formula or such) for which $V$ we have that $V$ appears as a summand in the decomposition of $V \otimes V$ into irreps. This is already a restriction. $\endgroup$
    – Vincent
    Sep 7, 2021 at 9:24
  • $\begingroup$ The second issue of the resulting algebra being unital or not is trickier (that is why my two comments are comments rather than answers) but I am also not sure if we agree on what that word means. I thought it means there must be an element $e \in V$ such that $ex = xe = x$ for all $x \in V$. But this doesn't hold in the case of Lie algebras, which seems to be one of your examples $\endgroup$
    – Vincent
    Sep 7, 2021 at 9:26
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    $\begingroup$ @Vincent: Yes, I want that $\mu$ is $G$-equivariant. I correct the misprint: I want the algebra to be unital in the sense that you wrote $\endgroup$
    – Bobech
    Sep 7, 2021 at 9:35
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    $\begingroup$ For symmetric matrices the product is not an operation (the product of two symmetric matrices is not always symmetric). You need the symmetrized product $AB+BA$, which is not associative (but indeed Jordan). $\endgroup$
    – YCor
    Sep 7, 2021 at 10:34
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    $\begingroup$ Correct, the operation on $Sym^2$ is $A.X = AXA^T$, not $A.X = AXA^{-1}$ like in the case of $W\otimes W^\ast$. The latter does commute with the Jordan algebra structure, the former does not. $\endgroup$
    – Johannes Hahn
    Sep 7, 2021 at 13:26

1 Answer 1

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The unit poses a problem here. If you require the multiplication map $\cdot: V\otimes V\to V$ to be $G$-equivariant, then this means $gv\cdot gw = g(v\cdot w)$. Letting $w=1_V$ and using that $g:V\to V$ is surjective, we see that $g(1_V) = 1_V$. In particular: $\mathbb{C} 1_V$ is a subrepresentation of $V$ isomorphic to the trivial representation. If $V$ is irreducible, this forces $V$ itself to be the trivial representation.

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