The Hodge dual (or formal adjoint) to the exterior derivative $d: \Omega^k(M) \to \Omega^{k+1}(M)$ on a smooth manifold $M$ is the codifferential $ d^* $, a linear map $$ d^*: \Omega^k(M) \to \Omega^{k-1}(M),$$ defined by
$ d^* = (-1)^{n(k+1)+1} * d * $
Could someone kindly provide a detailed explanation of the operation denoted by $ * d * $ ?
Hodge Dual: The Hodge dual, denoted by $(*)$, is an operation that maps a $( k )$-form to an $( (n-k) )$-form on an $( n )$-dimensional manifold. It’s defined using the metric of the manifold and the orientation. If $( \omega )$ is a $( k )$-form, then its Hodge dual $( *\omega )$ is an $( (n-k) )$-form such that for any ( k )-form ( \eta ), we have the inner product $( \langle \omega, \eta \rangle = \int_M \omega \wedge *\eta )$. Exterior Derivative: The exterior derivative ( d ) is a map $( d: \Omega^k(M) \rightarrow \Omega^{k+1}(M) )$ that takes a $( k )$-form to a ( (k+1) )-form. It generalizes the concept of taking derivatives of functions to higher-dimensional analogs. Codifferential $( d^ )*$: The codifferential is the formal adjoint of the exterior derivative with respect to the inner product on forms. It’s a map $( d^: \Omega^k(M) \rightarrow \Omega^{k-1}(M) )$ that takes a ( k )-form to a $( (k-1) )$-form. The codifferential is defined using the Hodge dual and the exterior derivative as follows: $[ d^ = (-1)^{n(k+1)+1} * d * ]$ This means that to compute $( d^* )$ of a $( k )$-form, you first apply the Hodge dual $(*)$, then the exterior derivative ( d ), and then the Hodge dual again. The operation (d) is essentially the first and last part of computing the codifferential $( d^* )$. It involves applying the Hodge dual to a ( k )-form, then the exterior derivative, and then the Hodge dual again. This process transforms a $( k )$-form into an $( $(n-k-1)$ )$-form.