# BRST cohomology and vanishing cycles

Consider the $$\mathbb{C}$$-variety $$\mathbb{A}^{1}$$, equipped with the potential (ie global function) $$P:=\frac{z^{n+1}}{n+1}$$. We can form the twisted de Rham complex $$H_{dR}(\mathbb{A}^{1},P)$$ which by definition is the complex of forms on $$\mathbb{A}^{1}$$ equipped with the differential $$d+dP$$. Now a (very) special case of a theorem of Sabbah implies that this computes (shifted) vanishing cycles for the pair $$(\mathbb{A}^{1},P)$$. Note that this is very easy in this case, the cohomology is $$n$$ dimensional in degree $$1$$ and vanishes elsewhere. Further theorems of Sabbah (and I believe many others) imply that the cohomology is also isomorphic to that of forms with the differential $$dP$$. This is a sort of Hodge to de Rham type result.

Consider now the chiral version of the above. Namely, we replace $$\Omega_{\mathbb{A}^{1}}$$ with $$\Omega^{ch}_{\mathbb{A}^{1}}$$ and the differntial $$dP$$ with $$Res_{z=0}(dP(z))$$. We get some bi-graded family of vector spaces generalizing the above vanishing cohomology. Let us denote this $$H^{ch}(\mathbb{A}^{1},P)$$.

Remark. If one takes a twisted de Rham type chiral differential (ie $$d^{ch}+(dP)_{(0)}$$), one gets nothing new. This is not hard to show and the $$P=0$$ case is a well known fact about the chiral de Rham complex. This is why I use the Hodge type differential above.

Question. Is anything known about the space $$H^{ch}(\mathbb{A}^{1},P)$$? If we denote conformal weight with a variable $$q$$ and cohomological degree with a variable $$y$$, I can see that the bi-graded Euler characteristic is $$ny+(n-1)(1+y)^{2}q +\mathcal{O}(q^{2})$$. Further it is easy enough to see that the $$y=-1$$ specialisation is just $$-n$$.