Very nice! Also, is your passage from power series to polynomials really necessary? To my eye it looks as if you could run the same argument, but using dimension theory for power series rings instead. As you say, that would imply that the series in the question does not converge.

Atiyah-Hirzebruch is unnecessary. The point is that Hodge is about the image of Chow -> cohomology, and so translates immediately via these direct sum decompositions (which are compatible with the algebraic $K_0$ to topological $K_0$ map.) Explicitly, the topological Chern isomorphism gives you a Hodge structure on rationalized topological K-theory (or rather, a direct sum of Hodge structures of different weights), and the middle piece is again the algebraic part.

The relevant piece of technology (at least rationally) is the Chern character isomorphism between topological/algebraic K-theory (tensor Q) and a direct sum of cohomology/Chow groups (tensor Q). Perhaps there is more that can be said integrally, but I am kind of skeptical...

My impression is that this duality is supposed to be substantially more subtle than anything coming from Serre or Poincare duality (but I am very far from an expert.) A related paper is arxiv.org/abs/1701.07329v3 - it does not answer your question but it may be enlightening nonetheless, explaining how the properties of this duality can be derived via the wonderful compactification.

Yep - that decomposition is fine (and works on the level of vector bundles).

Incidentally, Anderson-Jantzen is mainly about the positive characteristic case. If you are interested only in the characteristic zero case, I think this theorem goes back to Kostant. In fact it also follows from the Grauert-Riemenschneider theorem applied to the Springer resolution (I am not sure who first observed this.)

The answer to your question is no, as the vector bundle corresponding to $S^2(\mathfrak{n}^*)$ does not split into line bundles. This is because $S^2(\mathfrak{n}^*)$ does not split into a sum of one-dimensional representations as a $B$-representation ($B$ is not semisimple.) Instead, what you get is a filtration of your vector bundle by line bundles.

Under your conditions, $f$ must be an isomorphism. Apply Zariski's main theorem to factor $f$ into an open immersion and a finite mapping. This finite mapping is birational, hence an isomorphism because $Y$ is normal (this is all you need for conditions on $Y$), so $f$ is an open immersion. The only bijective open immersions are isomorphisms, giving you the result.

What version of inhomogeneous Koszul duality are you using? The only version I know of requires $A$ to be a filtered deformation of a quadratic algebra, in which case what is your filtration on $A$? The only one that comes to mind for me makes $A$ a homogeneous quadratic algebra. Also I was under the impression that in this case you still necessarily have the identity that the power series of dimensions of grade components of $A$ and $B$ multiplied to $1$, which is evidently false here.

I don't think there is any relationship between $A$ and $B$. The Koszul dual of $B$ is the symmetric algebra, which looks completely different from $A$ (in particular, it is not finite dimensional.) $B$ is also very canonical in a sense, while $A$ is not (it depends highly on an exceptional collection.)

I think it's misleading to say that geometric Langlands ala Gaitsgory is a geometrization/categorification of complex archimedian Langlands. Rather, it's an analogue of nonarchimedian Langlands over the complex numbers. On the other hand, as I understand it, Ben-Zvi & Nadler's "Loop Spaces and Langlands Parameters" (which I assume is the paper Marty was thinking of) tries to explain archimedian Langlands as a S^1-equivariant localization of geometric Langlands, both in the real and complex cases. I'm secretly hoping David Ben-Zvi sees this and writes an answer...

@Qfwfq: Well, there is such a functional equation, (A.22) in Drinfeld-Wang, which gives a formula for the decategorification of $\operatorname{Eis}_!$. But I don't know any way to express the decategorification of $\operatorname{Eis}_!$ in terms of standard automorphic form operations, so I'm not sure this is really useful for anything...

@Qfwfq: I added some more details to my answer. Does this answer your question satisfactorily?