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Let $V_d$ be a $d+1$-dimensional $k$-vector space with $\text{char} k = 0$. Suppose that $V_d$ is an irreducible representation of the Lie algebra $\mathfrak{sl}_2$ and let $k[V_d]^{\mathfrak{sl}_2}$ be the corresponding algebra of invariants.

Is it true that all irreducible representations of the direct sum $\mathfrak{sl}_2 \oplus \mathfrak{sl}_2$ are isomorphic to the tensor product $V_n \otimes V_m$?

If so, is it true that $k[V_n \otimes V_m]^{\mathfrak{sl}_2 \oplus \mathfrak{sl}_2} \cong k[V_m]^{\mathfrak{sl}_2} \otimes k[V_n]^{\mathfrak{sl}_2}$?

If this is the case, what is the form of this isomorphism?

The first two questions seem quite plausible, but there are some doubts.

Thank you for the answers.

Leox
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    More generally the irreps of $\mathfrak{g} \oplus \mathfrak{h}$ are always of the form $V\otimes W$ for $V,W$ irreps of $\mathfrak{g} , \mathfrak{h}$ – Callum Sep 11 '23 at 07:28
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    I'm not sure about the second question. The natural issue that occurs to me is that the tensor algebra of $V \otimes W$, $T(V\otimes W)$ is not isomorphic to $T(V) \otimes T(W)$ (the latter is much bigger, containing combinations of all degrees not just equal degrees). Clearly, if you take an element of $k[V] \otimes k[W]$, it will be invariant under the natural action of $\mathfrak{g} \oplus \mathfrak{h}$ but I don't think it is guaranteed to be a polynomial in $V\otimes W$. – Callum Sep 11 '23 at 14:15
  • @ Callum What about $k[V_n \otimes V_m]^{\mathfrak{sl}_2 \oplus \mathfrak{sl}_2} \cong k[V_m \oplus V_n]^{\mathfrak{sl}_2}$? – Leox Sep 11 '23 at 15:52
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    Just to amend on-site references for @Callum's first comment, see e.g. https://math.stackexchange.com/q/3826725/96384 and https://math.stackexchange.com/q/3350849/96384. – Torsten Schoeneberg Sep 11 '23 at 17:49
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    The easy toy example to check here is that $V_1 \otimes V_1$ should have a invariant symmetric bilinear form by the identification of $\mathfrak{sl}_2 \oplus \mathfrak{sl}_2$ with $\mathfrak{so}(4)$ that I described just recently here – Callum Sep 13 '23 at 09:12
  • Correcting my (deleted) comment: those still do not have the same tensor algebra containing them. The ones that match are $T(V\oplus W) = T(V) \otimes T(W)$ but I don't think this helps you. – Callum Sep 13 '23 at 10:48
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    In fact, you can see this toy example breaks it immediately: $V_1$ has an alternating degree 2 invariant on each factor ($\bigwedge^2 V_1$ is trivial) so $k[V_1 \oplus V_1]^{\mathfrak{sl}_2}$ contains the 2-dimensional span of these while the degree 2 invariants in $k[V_1 \otimes V_1]^{\mathfrak{sl}_2 \oplus \mathfrak{sl}_2}$ are a 1-dimensional span in the symmetric part. Clearly this rules out $k[V_1 \oplus V_1]^{\mathfrak{sl}_2} \cong k[V_1 \otimes V_1]^{\mathfrak{sl}_2 \oplus \mathfrak{sl}_2}$ as graded algebras but I think it rules out any algebra isomorphism. – Callum Sep 13 '23 at 10:54
  • @Callum By the way - what is the explicit form of the isomorphism $ \mathfrak{so}_4 \cong\mathfrak{sl}_2 \oplus \mathfrak{sl}_2$.? – Leox Sep 13 '23 at 12:25
  • I describe that isomorphism in the link I gave. If you want it even more explicit then this gives instructions on how to do that. – Callum Sep 13 '23 at 12:41

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