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 $-n$.