While investigating certain conformal blocks line bundles on $\overline{M}\_{0,n}$, I was led to what seems to be an identification between two spaces of invariants, and I am curious if there is a direct way to see this identification.


**Statement**: for any integers $n\ge 4$ and $r\ge 2$, and any integers $i\_1,\ldots,i\_n$ such that $1 \le i\_j \le r-1$ and $r=\frac{1}{2}\sum\_{j=1}^n i\_j$, I believe there is a vector space isomorphism $$(\wedge^{i\_1}\mathbb{C}^r\otimes\cdots\otimes \wedge^{i\_n}\mathbb{C}^r)^{SL(r)} \cong (S^{i\_1}\mathbb{C}^2\otimes\cdots\otimes S^{i\_n}\mathbb{C}^2)^{SL(2)},$$ where $\mathbb{C}^m$ denotes the standard representation of $SL(m)$. The invariants on the RHS are classical and well-known: a basis is given by all $2\times r$ semi-standard tableaux with entries in $\{1,\ldots,n\}$ such that $j$ occurs exactly $i\_j$ times. I wonder if the invariants on the LHS are also known, and if there's a conceptual reason why they might be in bijection with those on the RHS. 


**Background**:
This is not relevant for the question itself, but I am including it in case you are curious how this purported identity arose. It seems likely that for conformal blocks bundles on $\overline{M}\_{0,n}$ of level 1 and Lie algebra $\mathfrak{sl}(r)$, the global sections are naturally identified with a space of covariants. Specifically, the conformal blocks line bundle with weights $(\omega\_{i\_1},\ldots,\omega\_{i\_n})$, where $\omega\_i$ are fundamental weights, should have global sections $(\wedge^{i\_1}\mathbb{C}^r\otimes\cdots\otimes\wedge^{i\_n}\mathbb{C}^r)\_{\mathfrak{sl}(r)}$, since the irreducible representation associated to $\omega\_i$ is $\wedge^i\mathbb{C}^r$. This space of $\mathfrak{sl}(r)$-covariants is isomorphic to the corresponding space of $\mathfrak{sl}(r)$-invariants, which in turn is the same as the space of $SL(r)$-invariants for this representation. On the other hand, it is known (by a result of Fakhruddin) that when $\sum\_{j=1}^n i\_j = 2r$ then this conformal blocks line bundle induces the GIT morphism $\overline{M}\_{0,n} \rightarrow (\mathbb{P}^1)^n//\_{(i\_1,\ldots,i\_n)}SL(2)$, so we know that its space of global sections is $H^0((\mathbb{P}^1)^n,\mathcal{O}(i\_1,\ldots,i\_n))^{SL(2)}$.



Classical invariants involving exterior powers of standard representation
To follow up on Sasha's answer, yes there is a natural isomorphism of vector spaces which lifts the combinatorial equality. All isomorphisms in this answer will be natural.


**Schur-Weyl duality:** 


Let $\lambda$ be a partion; set $d = |\lambda|$ and let $m$ be greater than or equal to the number of parts of $\lambda$. Let $V$ be a vector space of dimension $m$, let $V\_{\lambda}$ be the irrep of $GL(V)$ with highest weight $\lambda$ an let $M\_{\lambda}$ be the $S\_d$-irrep (aka Specht module) indexed by $\lambda$. [Schur-Weyl duality](http://en.wikipedia.org/wiki/Schur%25E2%2580%2593Weyl_duality) is the isomorphism of $GL(V)$ representations:
$$\mathrm{Hom}\_{S\_d}(M\_{\lambda}, V^{\otimes d}) \cong V\_{\lambda}.$$
I'll take this to be the definition of $V\_{\lambda}$.


**Tensor product and restriction:** 


Let $\mu$, $\lambda\_1$, ..., $\lambda\_k$ be partitions, with $|\mu| = \sum |\lambda\_i|$. Let $m$ be greater than or equal to the number of parts of $\mu$. Set $d\_i = |\lambda\_i|$. Then the above shows that
\begin{align\*}
&\mathrm{Hom}\_{GL(V)}\left( V\_{\mu}, V\_{\lambda\_1} \otimes \cdots \otimes V\_{\lambda\_k} \right)\\
&\cong \mathrm{Hom}\_{GL(V)} \left( \mathrm{Hom}\_{S\_{\sum d\_i}} \left( M\_{\mu}, V^{\otimes \sum d\_i} \right), \mathrm{Hom}\_{S\_{d\_1} \times \cdots \times S\_{d\_k}}\left( M\_{\lambda\_1} \boxtimes \cdots \boxtimes M\_{\lambda\_k}, V^{\otimes \sum d\_i} \right) \right) \\
&\cong \mathrm{Hom}\_{S\_{d\_1} \times \cdots \times S\_{d\_k}} \left( M\_{\lambda\_1} \boxtimes \cdots \boxtimes M\_{\lambda\_k}, \left( M\_{\mu} \right)|\_{S\_{d\_1} \times \cdots \times S\_{d\_k}} \right).
\end{align\*}
In other words, every Hom in the second expression is induced from composition with one of the Hom's in the third expression. Here, if $U$ and $V$ are $G$ and $H$-representations, then $U \boxtimes V$ denotes $U \otimes V$ with $G$ and $H$ acting on the first and second factors respectively.


**Relation to transpose:** Now let $\lambda\_i$ and $\mu$ be as above. Let $\epsilon(d)$ be the sign representation of $S\_d$, and let $\lambda^T$ be the transpose of $\lambda$. Then $M\_{\lambda} \cong M\_{\lambda^T} \otimes \epsilon(|\lambda|)$. How to make this natural depends on exactly how you define $M\_{\lambda}$. For example, if you use the [Vershik-Okounkov](http://arxiv.org/abs/math.RT/0503040) approach, the bases they construct for $M\_{\lambda}$ and $M\_{\lambda^T}$ correspond to each other.




Now, $\epsilon(d\_1) \boxtimes \cdots \boxtimes \epsilon(d\_k) \cong \epsilon(\sum d\_i)|\_{S\_{d\_1} \times \cdots \times S\_{d\_k}}$, and is irreducible. So we deduce that $$\mathrm{Hom}\_{S\_{d\_1} \times \cdots \times S\_{d\_k}} \left( M\_{\lambda\_1} \boxtimes \cdots \boxtimes M\_{\lambda\_k}, \left( M\_{\mu} \right)|\_{S\_{d\_1} \times \cdots \times S\_{d\_k}} \right) \cong \mathrm{Hom}\_{S\_{d\_1} \times \cdots \times S\_{d\_k}} \left( M\_{\lambda\_1}^T \boxtimes \cdots \boxtimes M\_{\lambda\_k}^T, \left( M\_{\mu}^T \right)|\_{S\_{d\_1} \times \cdots \times S\_{d\_k}} \right)$$
and this isomorphism is natural if we define $M\_{\lambda} \cong M\_{\lambda^T} \otimes \epsilon(|\lambda|)$ in a natural way.


Putting it all together, if $\dim V$ is greater than or equal to the number of parts in $\mu$, and $\dim W$ is greater than or equal to the number of parts in $\mu^T$, then we have a natural isomorphism:
$$\mathrm{Hom}\_{GL(V)}\left( V\_{\mu}, V\_{\lambda\_1} \otimes \cdots \otimes V\_{\lambda\_k} \right) \cong \mathrm{Hom}\_{GL(W)}\left( W\_{\mu^T}, W\_{\lambda^T\_1} \otimes \cdots \otimes W\_{\lambda^T\_k} \right).$$


**Your question:**


Take $\lambda\_j = 1^{i\_j}$ and $\mu = 2^r$. Let $\dim V = r$ and $\dim W=2$. 


Since $M\_{\lambda\_j}$ is the sign rep, we have $V\_{\lambda\_j} \cong \bigwedge^{i\_j} V$ naturally. Similarly, since $M\_{\lambda\_j^T}$ is the trivial rep, we have $W\_{\lambda\_j^T} \cong \mathrm{Sym}^{i\_j}(W)$.


Finally, one needs to use the fact that $V\_{\mu}$ and $W\_{\mu^T}$ are trivial one-dimensional $SL(V)$ and $SL(W)$ reps. (More precisely, they are $\det^2$ and $\det^r$, as $GL$-reps.) I'm not sure what the easiest proof of this is, but it can only introduce a scalar factor to your isomorphisms.


