Simple properties of the codifferential

Question

The exterior derivative $d$ has many very nice algebraic relations. For example

1. $d(\alpha\wedge\beta) = (d\alpha)\wedge \beta + (-1)^k\alpha\wedge(d\beta)$
2. $f^*(d \alpha)=d f^*(\alpha)$.
3. $d\circ d = 0$

for $\alpha, \beta$ forms on a manifold $V$ and $f:V\rightarrow W$ a smooth map.

Let $\delta = \star d \star$ the codifferential, we have $\delta\circ \delta =0$. I wonder if there are other simple and usefull properties as above. For instance can we say anything about $\delta f^*(\alpha)$?

Answer 1

To define $\star$ you need either a symplectic form or a Riemannian metric. Let us consider the first case.

In this case, this isomorphism basically depends only on a symplectic form $\omega$. That is if you have a map $f: V \to W$ such that $\Lambda^2 f(\omega_Y) = \omega_X \in \Lambda^2 W$, then this map $f$ is compatible with the Hodge star isomorphism by its construction. Therefore, you have the functoriality property $\delta f^* (\alpha) = f^* \delta (\alpha).$

The last thing is the Leibnitz rule. In the case of the codifferential one doesn't have the Leibniz rule but rather this codifferential is an operator of second order and one has what's called a Batalin-Vilkovisky structure.

Let's start with a smooth affine symplectic variety $X = \text{Spec}{R}$ of dimension $d$. In particular, you have the musical isomorphism $\Lambda^* T_X \cong \Omega^*_{X}$ with the Hodge star isomorphism $\Omega^{*}_{X} \cong \Omega^{d-*}_{X}$ that takes the de Rham differential to the codifferential. Note that $\Omega^*_{X} \cong HH_* (R)$ and $\Lambda^* T_{X} \cong HH^* (R)$ by the Hochschild-Kostant-Rosenberg theorem. Note that the de Rham differential $d_{dR}$ corresponds to the Connes differential $B: HH_*(R) \to HH_{*+1} (R)$ under the isomorphism $\Omega^*_{X} \cong HH_* (R)$. Thus, we basically transport the Connes differential on $HH_*(R)$ to the differential on $HH^* (R)$ via the Hodge star isomorphism. The algebra $HH^*(R)$ with the new differential becomes a BV-algebra. This result has a nice generalization due to Ginzburg.

Let $A$ be a Calabi-Yau algebra of dimension $d$. In particular, that means that $HH_* (A) \cong HH^{d-*} (A)$. That is one can endow $HH^{\bullet} (A)$ with the differential $\delta$ that corresponds to the Connes differential $B$. Ginzburg proves that it makes $HH^{\bullet}(A)$ a BV-algebra. As we have seen, in the case of a commutative algebra it shows that one doesn't have the Leibniz rule but rahter $\delta (abc) - \delta(ab) c + \delta(a) bc - (-1)^{|a|} a \delta(bc) - (-1)^{(|a|+1)|b|} b \delta(ac) + (-1)^{|a|} a\delta(b) c + (-1)^{|a|+|b|}ab\delta(c) = 0$.