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When doing an exercise about linear representations of finite groups I stumbled upon this Isomorphism in the comments of another post which I was not aware of.

In this context $V$ and $W$ are finite dimensional Vector complex vector spaces. I was already able to show that the dimension of the two Vector spaces or rather Tensors are the same. As such it should be sufficient to construct an injective or surjective map between them.

However I tried to construct one such as: $$ \phi: (v_1 \oplus w_1) \cdot (v_2\oplus w_2) \in Sym^2(V \oplus W) \mapsto (v_1 \cdot v_2 ) \oplus ((v_1+v_2) \otimes (w_1 + w_2)) \oplus (w_1 \cdot w_2) \in Sym^2(V)\oplus (V\otimes W)\oplus Sym^2(W) $$ However it seems to me as if this map is not injective nor surjective since $\phi((v_1 \oplus w_1) \cdot (v_2\oplus w_2)) = \phi((v_1 \oplus w_2) \cdot (v_2\oplus w_1))$.

Any tips for constructing such a map or should I be trying a different approach?

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    $\begingroup$ I'd just like to point out that the title says "isomorphism of algebras", but those spaces are not algebras, just vector spaces. $\endgroup$ Commented Sep 10, 2022 at 12:14

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Here is an abstract approach. There is no need to assume that $V$ is finite-dimensional or over $\mathbb{C}$: the discussion below applies to $V$ an arbitrary module over an arbitrary commutative ring $R$. The symmetric powers $S^n(V) = V^{\otimes n} / S_n$ organize into a single object called the symmetric algebra

$$S(V) = \bigoplus_{n \ge 0} S^n(V).$$

The symmetric algebra is the free commutative algebra on $V$; formally, it's the left adjoint of the forgetful functor from commutative $R$-algebras to $R$-modules. As a left adjoint, it preserves colimits, so in particular it sends coproducts to coproducts. The coproduct of commutative $R$-algebras is the tensor product, so this gives the "exponential law"

$$S(V \oplus W) \cong S(V) \otimes S(W)$$

and writing this isomorphism down in each degree we get a family of isomorphisms

$$S^n(V \oplus W) \cong \bigoplus_{i+j=n} S^i(V) \otimes S^i(W).$$

The desired isomorphism is the special case $n = 2$. A nice exercise here is to check that when $V, W$ are finite-dimensional vector spaces the dimensions of the two sides agree, which gives a nice identity; more ambitiously you can check that the characters of the two sides agree as representations of $GL(V) \times GL(W)$.

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  • $\begingroup$ This is a nice solution. Can you do the same with the alternating algebra? $\endgroup$
    – Alex Byard
    Commented Oct 19, 2023 at 0:36
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    $\begingroup$ @Alex: yes, the exterior algebra is the free graded-commutative algebra on a module regarded as living in degree $1$. We get the same isomorphism although to promote it to an isomorphism of algebras requires using Koszul signs. $\endgroup$ Commented Oct 19, 2023 at 5:31
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Here is a tip: $S^2(V\oplus W)$ has a basis consisting of elements of the form $v_1v_2$ or $vw$ or $w_1w_2$ in the obvious notation (prove it). Those things slot nicely into $S^2(V) \oplus (V\otimes W) \oplus S^2(W)$.

If you want to write down a linear map, it's easier the other way round: there are "obvious" maps $S^2(W), V\otimes W, S^2(V) \to S^2(V\oplus W)$ and you can form their direct sum and show it is an isomorphism.

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    $\begingroup$ That's a very intuitive approach. I think I now get the idea idea behind the isomorphism. I will try to formulate that idea out and post an answer myself. Thank you. $\endgroup$ Commented Sep 10, 2022 at 12:23

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