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Recall the celebrated rank-level strange duality isomorphism:

$$H^0(SU_C(r),L^l) \to H^0(U_C(l), \mathcal{O}(r\Theta))$$

between level $l$ generalized theta functions on the moduli space of rank $r$ vector bundles with trivial determinant on a smooth projective curve $C$, and the global sections of the level $r$ generalized theta functions on the moduli of VB of rank $l$ and degree $l(g-1)$ on the same curve $C$.

Moreover, by results of Narasimhan-Ramadas and Su, factorization formulas exist for generalized theta functions. Roughly speaking, this means that when a curve $X_0$ is reducible (for instance with a separating node, and two components $X_1$ and $X_2$) then we have an isomorphism

$$ H^0(U_{X_0},\Theta_{X_0}) \cong \sum H^0(U^\mu_{X_1},\Theta_{X_1})\otimes H^0(U^\mu_{X_2},\Theta_{X_2}),$$

where $U^\mu$ means the moduli space of parabolic bundles with some parabolic structure at the point corresponding to the node, and the summation is taken over the possible values of $\mu$.

My question is: to what extent the strange duality isomorphism is compatible with this factorization? That is, over nodal curves does strange duality factorize to strange dualities (for parabolic bundles) on the irreducible components of the nodal curves? Is strange duality for nodal curves the tensor product of strange dualities for the irreducible components? I am not completely aware of the literature about strange duality for nodal curves hence there may be gaps in my writing, which admittedly is still a bit rough.

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    $\begingroup$ I think you need to provide more details to your question. $\endgroup$ Commented Jan 23 at 15:20
  • $\begingroup$ yeah, I agree, I will do this as soon as I have a moment to sit down $\endgroup$ Commented Jan 24 at 7:53
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    $\begingroup$ Various papers of Takeshi Abe, e.g., "Degeneration of the strange duality map for symplectic bundles" might be useful. I have not looked at any of them carefully, but I think he proves what you want in the symplectic setting and I assume that the same statements can be proved (by similar, perhaps more elementary methods) in the linear case. $\endgroup$ Commented Jan 27 at 9:04
  • $\begingroup$ @naf: thank you, would yoou want to make an answer out of your comment so that I can award you the bounty? $\endgroup$ Commented Jan 30 at 7:58
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    $\begingroup$ It's not really an answer since I have not checked that Abe's papers do actually contain what you asked... $\endgroup$ Commented Jan 30 at 10:39

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