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Recall that there is a bijection between irreducible representations of a compact real Lie group $G$ and the cocharacters (homomorphisms $U(1) \to G$, modulo conjugation) of the Langlands dual group $^LG$.

The irreducible representations of $G$ have additional structure related to tensoring representations: Given representations $\alpha, \beta, \gamma$ of $G$, we have the invariant subspace $V_{\alpha\beta\gamma}$ of the tensor product $\alpha\otimes\beta\otimes\gamma$. (Or, if you prefer, the space $V_{\alpha\beta}^{\gamma^*}$ of homomorphisms from $\gamma^*$ tp $\alpha\otimes\beta$.)

Is there an intrinsic way to define a Langlands dual structure on the cocharacters of $^LG$? In other words, in a natural way (and without using Langlands duality) associate to cocharacters $a,b,c$ of $^LG$ a vector space $V_{abc}$?

One possibility would be to think of the cocharacters as (equivalence classes of) geodesics in $^LG$ and replace the tensoring of $G$ representations with the splicing of $^LG$ geodesics. The resulting families of broken geodesics $a \cdot b$ could flow to actual geodesics $\{c_i\}$. Before pursuing this idea I wanted to check whether there are already known answers to the main question above.


ADDED LATER:

The motivation for the above question was geometric Langlands TQFTs applied to the operation of gluing two disks together to obtain another disk. I had forgotten that on the Rep($G$) side a 2-sphere gives the same monoidal category as a disk, and that therefore the geometric Satake isomorphism answers my question. I still wonder whether thinking in terms of disks instead of spheres would give a different (but presumably equivalent) construction of the monoidal product on the cocharacter side.

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This answer will be cheating: you can look at the affine grassmanian of $^LG$, $\mathcal{G}r(^LG)=^LG(\mathcal{K})/^LG(\mathcal{O})$ where $\mathcal{O}=\mathbb{C}[ [t]]$ and $\mathcal{K}$ its field of fractions. The Geometric Satake isomorphism states that the category of $^LG(\mathcal{O})$-equivariant perverse sheaves on $\mathcal{G}r$ is equivalent as a tensor category to the category of finite dimensional representations of $G$. For each cocharacter $a$ of $^LG$ there is a canonical way of constructing such a perverse sheaf, namely one looks at the $^LG(\mathcal{O})$ orbit of $a$ and extend from it the trivial local system to a perverse sheaf $\lambda_a$ on $\mathcal{G}r$. Now for $a,b,c$ we have $Hom(\lambda_a, \lambda_b \otimes \lambda_c)$ is the vector space you are looking for. By the above mentioned theorem, this corresponds to the respective vector space of morphisms between the $G$-modules.

Edit: I should have mentioned that I switched from compact Lie groups over $\mathbb{R}$ to the corresponding algebraic groups and I'm thinking on cocharacters as $\mathbb{G}_m \rightarrow G$ since this does not affect the question.

Of course you can argue that the tensor product "$\lambda_b \otimes \lambda_c$" that I used is very hard to construct and that the geometric Satake equivalence can be considered as the definition of $^LG$ and that's the cheating part, but since this answer is so natural and so far there is no other answer.

Standard references for the tensor product and the Satake equivalence are:

[1] I. Mirkovic and K. Vilonen, Geometric Langlands duality and representations of algebraic groups over commutative rings, Ann. of Math. (2) 166 (2007), no. 1, 95–143.
[2] V. Ginzburg, Perverse sheaves on a Loop group and Langlands’ duality (1995), available at http://arxiv.org/abs/alg-geom/9511007.
[3] A. Beilinson and V. Drinfeld, Quantization of Hitchin’s integrable system and Hecke eigensheaves, available at http://www.math.uchicago.edu/~mitya/langlands/hitchin/BD-hitchin.pdf.

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  • $\begingroup$ Thanks for the answer. Do you have a reference for the tensor product on $^LG(\mathcal{O})$-equivariant perverse sheaves on the affine grassmanian? $\endgroup$ – Kevin Walker Nov 21 '14 at 2:11
  • $\begingroup$ Edited to add references $\endgroup$ – Reimundo Heluani Nov 21 '14 at 10:45
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This is another version of Reimundo Heluani's answer, avoiding perverse sheaves.

In general, the $H$-orbits on $H/P \times H/Q$ are indexed by $W_P \backslash W_H/W_Q$. If $H = {}^L G(\mathcal K)$, so $W_H$ is the affine Weyl group $W_G \ltimes $[the weight lattice], and $P=Q={}^LG(\mathcal O)$, then $W_P \backslash W_H/W_Q =$ dominant weights. For $L,L' \in \mathcal Gr$, write $L \stackrel{\lambda}{\to} L'$ if $(L,L')$ is in the $\lambda$-orbit, for $\lambda$ a dominant weight.

Now consider the space of "triangles" $$\mathcal V_{\lambda\mu\nu} := \{(L,L') : L,L' \in \mathcal Gr, base \stackrel{\lambda}{\to} L \stackrel{\mu}{\to} L' \stackrel{\nu}{\to} base \}$$ and I believe the statement is that $V_{\lambda\mu\nu} \cong H_{top}(\mathcal V_{\lambda\mu\nu})$.

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  • $\begingroup$ Thanks, that's very helpful. I'm giving the green checkmark to the other answer since it came sooner and gives references, but I think your answer is checkmark-worthy. $\endgroup$ – Kevin Walker Nov 21 '14 at 14:30

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