What is the anticommutator of the interior product and codifferential (adjoint of exterior derivative)?

Question

What is $\eta=i_X \delta + \delta i_X$ acting on differential forms? Here $\delta$ is the usual Hodge adjoint of the exterior derivative and $i_X$ is contraction of a form with the vector field $X$. Clearly $\eta=0$ trivially on 0-forms and 1-forms, but does this hold for forms of arbitrary degree?

Answer 1

Clifford algebra might make this manipulation easier: its geometric product of vectors (or forms) is associative like the wedge product, but still has the metrical terms needed for the codifferential.

In clifford notation, we have,

$$\eta(\alpha) = X \rfloor (\nabla \rfloor \alpha) + \nabla \rfloor (X \rfloor \alpha)$$

You can see here that the contraction operator $\rfloor$ plays a similar role in relating $\nabla \rfloor$ to $\delta$ as it does in relating $X \rfloor$ to $i_X$.

Now we can use associativity of the geometric product to our advantage, considering the grade $k-2$ terms of a general product:

$$\langle \nabla (X \alpha) \rangle_{k-2} = (\nabla \wedge X) \rfloor \alpha - (X \wedge \nabla) \rfloor \alpha = \nabla \rfloor (X \rfloor \alpha)$$

Expand the $(X \wedge \nabla) \rfloor \alpha$ term as

$$\langle X \nabla \alpha \rangle_{k-2} = (X \wedge \nabla) \rfloor \alpha = X\rfloor (\nabla \rfloor \alpha)$$

So we can conclude that the anticommutator is

$$\eta(\alpha) = (\nabla \wedge X) \rfloor \alpha$$

Hm, I'm not familiar with how this should be denoted in pure differential forms. $\nabla \wedge X$ ought to be something like $dX$, but since $X$ isn't a form (it's a vector field), that's a little weird. Similarly, $(\nabla \wedge X) \rfloor \alpha$ ought to read like an interior product into $\alpha$, but with 2-vector (or a 2-form, or a mixed 1-vector/1-form) into $\alpha$.

Answer 2

You can easily verify that $$\nabla_Y \iota_X \omega = \iota_X \nabla_Y \omega + \iota_{\nabla_YX} \omega$$ for vector fields $X$ and $Y$. For the codifferential, we have the formula $$\delta = - \sum_{j=1}^n \iota_{e_j} \nabla_{e_j} \omega$$ which holds for any orthonormal frame $e_1, \dots, e_n$. Hence $$\delta \iota_X \omega = - \sum_{j=1}^n \iota_{e_j} \nabla_{e_j}\iota_X \omega = - \sum_{j=1}^n \iota_{e_i}\bigl(\iota_X \nabla_{e_i}\omega + \iota_{\nabla_{e_i}X} \omega \bigr) = \sum_{j=1}^n \iota_X \iota_{e_i} \nabla_{e_i}\omega - \sum_{j=1}^n\iota_{e_i}\iota_{\nabla_{e_i}X} \omega$$ where we used that $\iota_X \iota_Y = -\iota_Y \iota_X$ for all vector fields $X$ and $Y$. This can be written as $$\delta \iota_X \omega + \iota_X \delta \omega = -\iota_{\nabla \wedge X} \omega$$ where $$\nabla \wedge X := e_i \wedge \nabla_{e_i} X$$ is an anti-symmetric bi-vector.