I have been having difficulty obtaining the component expression of the codifferential of a $p$-form found on Wikipedia and in the book Riemannian Geometry and Geometric Analysis by Jürgen Jost. Let me explain.
Problem Statement
Consider an $n$-dimensional, oriented, (pseudo)-Riemannian manifold $(M,g)$. The codifferential ${\delta}:\Omega^{p}(M)\to\Omega^{p-1}(M)$ on $p$-forms is given by $$\delta = (-1)^g (-1)^{n(p-1)+1} \star \mathop{d} \star .$$
Here $(-1)^g$ is shorthand for the sign of the determinant of the metric $g$ in a positively oriented basis.
Let $A$ be an arbitrary $p$-form. Its codifferential $\delta A$ is then a $(p-1)$-form. With respect to local coordinates $\lbrace x^1,\dots,x^n\rbrace$, we write
$$
\begin{align}
A &= \frac{1}{p!}A_{i_1 \dots i_p} dx^{i_1}\wedge\dots\wedge dx^{i_{p}} \\[1.5ex]
\delta A &= \frac{1}{(p-1)!} (\delta A)_{i_{1} \dots i_{p-1}} dx^{i_1}\wedge\dots\wedge dx^{i_{p-1}}.
\end{align}
$$
Repeated indices are summed, and we use the convention
$$
dx^{i_1}\wedge\dots\wedge dx^{i_{p}} = \sum_{\sigma\in S_p} (-1)^\sigma dx^{i_{\sigma(1)}}\otimes\dots\otimes dx^{i_{\sigma(p)}}
$$
Goal: Express the components $(\delta A)_{i_{1} \dots i_{p-1}}$ in terms of the components $A_{i_1 \dots i_p}$.
Expected Result
One can express (cf. this post and Eq. (4.3.44) on page 207 in Jost) the codifferential in terms of covariant derivatives as
$$
\delta = – g^{lk} \iota(\partial_k) \nabla_l
$$
where $\iota(X)$ is the Interior Product with the vector $X$, and $\nabla_l$ is shorthand for $\nabla_{\partial_l}$. I am confident in this result.
My difficulty starts when I try to express the covariant derivatives in terms of the Christoffel symbols $\Gamma^{i}_{jk}$. Wikipedia claims that
$$
\delta A = -\frac{1}{p!}g^{lk}\left(\partial_l A_{k i_{1}\dots i_{p-1}} – \Gamma^{j}_{lk}A_{j i_{1}\dots i_{p-1}}\right) dx^{i_1}\wedge\dots\wedge dx^{i_{p-1}}.
$$
Jost in Eq. (3.3.47) on page 142 gives the same thing (modulo a factor of $\frac{1}{p}$ which I assume is due to different normalizations) with
$$
(\delta A)_{i_{1} \dots i_{p-1}} = -g^{lk}\left(\partial_l A_{k i_{1}\dots i_{p-1}} – \Gamma^{j}_{lk}A_{j i_{1}\dots i_{p-1}}\right)
$$
I am unable to attain these expressions. I get close, but I am left with an extra term that I am unable to show vanishes.
My Attempt
Before jumping in, observe the following:
Lemma: For any antisymmetric $\omega_{i_1\dots i_q}$, we have that
$$
\begin{align}
\iota(\partial_k)\left(\omega_{i_1\dots i_q} dx^{i_1}\wedge\dots\wedge dx^{i_{q}}\right) &= \sum_{j=1}^{q}(-1)^{j-1} \omega_{i_1\dots i_q} \delta^{i_j}_k dx^{i_1}\wedge\dots\wedge\widehat{dx^{i_j}}\wedge\dots\wedge dx^{i_q} \\[1.5ex]
&= \sum_{j=1}^{q}(-1)^{j-1}\left((-1)^{j-1}\omega_{i_j i_1\dots \widehat{i_j}\dots i_q}\right) \delta^{i_j}_k dx^{i_1}\wedge\dots\wedge\widehat{dx^{i_j}}\wedge\dots\wedge dx^{i_q} \\[1.5ex]
&= q\, \omega_{k i_1\dots i_{q-1}} dx^{i_1}\wedge\dots\wedge dx^{i_{q-1}}.
\end{align}
$$
Onto the main task. We want to evaluate
$$
\delta A = -g^{lk}\iota(\partial_k)\nabla_l A = \frac{-g^{lk}}{p!}\iota(\partial_k)\nabla_l \left(A_{i_1\dots i_p} dx^{i_1}\wedge\dots\wedge dx^{i_p}\right).
$$
Step 1: The Covariant Derivative
Using the Leibniz rule for $\nabla_l$, we have
$$
\nabla_l \left(A_{i_1\dots i_p} dx^{i_1}\wedge\dots\wedge dx^{i_p}\right) = \partial_l A_{i_1\dots i_p} dx^{i_1}\wedge\dots\wedge dx^{i_p}+A_{i_1\dots i_p}\nabla_l\left(dx^{i_1}\wedge\dots\wedge dx^{i_p}\right)
$$
Let us focus on the right term. We recall the covariant derivative of a basis $1$-form is given by $\nabla_l dx^{i} = -\Gamma^{i}_{lm} dx^m$. Therefore once again applying the Leibniz rule we obtain
$$
\begin{align}
\nabla_l\left(dx^{i_1}\wedge\dots\wedge dx^{i_p}\right) &= \sum_{j=1}^{p}dx^{i_1}\wedge\dots\wedge\left(-\Gamma^{i_j}_{lm} dx^m\right)\wedge\dots\wedge dx^{i_p} \\[1.5ex]
&= -\sum_{j=1}^{p} (-1)^{j-1} \Gamma^{i_j}_{lm} dx^m\wedge dx^{i_1}\wedge\dots\wedge \widehat{dx^{i_j}}\wedge\dots\wedge dx^{i_p}.
\end{align}
$$
It then follows that
$$
\begin{align}
A_{i_1\dots i_p}\nabla_l\left(dx^{i_1}\wedge\dots\wedge dx^{i_p}\right) &= -\sum_{j=1}^{p} A_{i_j i_1\dots \widehat{i_j}\dots i_p} \Gamma^{i_j}_{lm} dx^m\wedge dx^{i_1}\wedge\dots\wedge \widehat{dx^{i_j}}\wedge\dots\wedge dx^{i_p} \\[1.5ex]
&= -p\,\Gamma^{j}_{lm}A_{j i_1\dots i_{p-1}} dx^m\wedge dx^{i_1}\wedge\dots\wedge dx^{i_{p-1}} \\[1.5ex]
&= -p\,\Gamma^{j}_{lm}dx^m\wedge\left(A_{j i_1\dots i_{p-1}} dx^{i_1}\wedge\dots\wedge dx^{i_{p-1}}\right)
\end{align}
$$
We now have
$$
\delta A = \frac{-g^{lk}}{p!}\iota(\partial_k)\left(\Lambda^{(1)}+\Lambda^{(2)}\right)
$$
where we have defined
$$
\begin{align}
\Lambda^{(1)} &\equiv \partial_l A_{i_1\dots i_p} dx^{i_1}\wedge\dots\wedge dx^{i_p} \\[1.5ex]
\Lambda^{(2)} &\equiv -p\,\Gamma^{j}_{lm}dx^m\wedge\left(A_{j i_1\dots i_{p-1}} dx^{i_1}\wedge\dots\wedge dx^{i_{p-1}}\right)
\end{align}
$$
Step 2: Interior Multiplication
From our lemma at the beginning, it is immediate that
$$
\iota(\partial_k)\Lambda^{(1)} = p\,\partial_l A_{k i_1 \dots i_{p-1}} dx^{i_1}\wedge\dots\wedge dx^{i_{p-1}}
$$
For the second term, we first use the fact that interior multiplication is an anti-derivation on forms.
$$
\begin{align}
\iota(\partial_k)\Lambda^{(2)} =& -p\,\Gamma^{j}_{lm}\left(\iota(\partial_k)dx^m\right)\wedge\left(A_{j i_1\dots i_{p-1}} dx^{i_1}\wedge\dots\wedge dx^{i_{p-1}}\right) \\[1.5ex]
&+ p\,\Gamma^{j}_{lm} dx^m\wedge\iota(\partial_k)\left(A_{j i_1\dots i_{p-1}} dx^{i_1}\wedge\dots\wedge dx^{i_{p-1}}\right) \\[1.5ex]
=& -p\,\Gamma^{j}_{lk} A_{j i_1 \dots i_{p-1}}dx^{i_1}\wedge\dots\wedge dx^{i_{p-1}}+p(p-1)\,\Gamma^{j}_{lm}A_{j k i_1 \dots i_{p-2}} dx^m\wedge dx^{i_1}\wedge\dots\wedge dx^{i_{p-2}}
\end{align}
$$
Final Result
Putting everything together and organizing a little, we find
$$
\begin{align}
\delta A =& -\frac{1}{(p-1)!}g^{lk}\left(\partial_l A_{k i_{1}\dots i_{p-1}} – \Gamma^{j}_{lk}A_{j i_{1}\dots i_{p-1}}\right) dx^{i_1}\wedge\dots\wedge dx^{i_{p-1}} \\[2ex]
&-\frac{1}{(p-2)!} g^{lk}\Gamma^{j}_{lm}A_{j k i_1 \dots i_{p-2}} dx^m\wedge dx^{i_1}\wedge\dots\wedge dx^{i_{p-2}}
\end{align}
$$
or equivalently in components
$$
(\delta A)_{i_1\dots i_{p-1}} = -g^{lk}\left(\partial_l A_{k i_{1}\dots i_{p-1}} – \Gamma^{j}_{lk}A_{j i_{1}\dots i_{p-1}}\right) – (p-1)g^{lk}\Gamma^{j}_{li_1}A_{j k i_1 \dots i_{p-2}}
$$
Discussion
My result agrees with what is expected if it can be shown that
$$
-\frac{1}{(p-2)!} g^{lk}\Gamma^{j}_{lm}A_{j k i_1 \dots i_{p-2}} dx^m\wedge dx^{i_1}\wedge\dots\wedge dx^{i_{p-2}} = 0.
$$
It is not clear to me that this is obvious just from symmetry alone. For instance, since $A_{jk i_1 \dots i_{p-2}}$ is antisymmetric, it would be sufficient to show that
$$
g^{lk}\Gamma^{j}_{lm} = g^{lj}\Gamma^{k}_{lm}.
$$
But a short calculation gives
$$
\begin{align}
g^{lk}\Gamma^{j}_{lm} – g^{lj}\Gamma^{k}_{lm} &= \frac{1}{2}g^{lk}g^{jr}(\partial_m g_{rl} +\partial_l g_{rm} – \partial_r g_{lm}) -\frac{1}{2}g^{lj}g^{kr}(\partial_m g_{rl} +\partial_l g_{rm} – \partial_r g_{lm}) \\[1.5ex]
&= \frac{1}{2}g^{lk}g^{jr}(\partial_m g_{rl} +\partial_l g_{rm} – \partial_r g_{lm}) -\frac{1}{2}g^{rj}g^{kl}(\partial_m g_{lr} +\partial_r g_{lm} – \partial_l g_{rm}) \\[1.5ex]
&= g^{lk}g^{jr}(\partial_l g_{rm} – \partial_r g_{lm})
\end{align}
$$
which I don't believe vanishes in general.
The situation would be somewhat different if $A$ was of opposite valence (i.e. $A = A^{i_1\dots i_p} \partial_{i_1}\wedge\dots\wedge \partial_{i_p}$) in which case the extraneous term would be of the form (cf. the answer to this post)
$$
\Gamma^{m}_{lj}A^{j l i_1 \dots i_{p-2}} \partial_m\wedge \partial_{i_1}\wedge\dots\wedge \partial_{i_{p-2}}
$$
which does vanish due to the symmetry of $\Gamma^{m}_{lj}$ on $l$ and $j$.
However, Wikipedia and Jost are very clearly working with forms and not vectors.
I would be very grateful if someone could explain how to see that the extra term I arrive at vanishes for a general coordinate frame, or point a mistake a step in my derivation and how to correctly arrive at what is expected.
It might be helpful to examine the derivation given on page 207 of Jost. He does not include many details, and there is no mention of the extra term I have in any point of his derivation. I cannot understand how he arrives at his result.
Best Answer
The expression given by Wikipedia and by the book is incorrect. Here is a counterexample.
Consider the open unit cube $U = \left\lbrace(x,y,z)\;\vert\; 0<x,y,z<1\right\rbrace\subset \mathbb{R}^{3}$ with standard orientation. We define a metric (inspired from https://math.stackexchange.com/a/402684/649763) on $U$ $$ \begin{align} g_{ij} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \sin^{2}(x) & 0 \\ 0 & 0 & \sin^{2}(x)\,\sin^{2}(y) \end{pmatrix} \qquad && \qquad g^{ij} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \csc^{2}(x) & 0 \\ 0 & 0 & \csc^{2}(x)\,\csc^{2}(y) \end{pmatrix} \end{align} $$
The Christoffel symbols are computed as $$ \begin{align} \Gamma^{1}_{ij} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & -\cos(x)\sin(x) & 0 \\ 0 & 0 & -\cos(x)\sin(x)\sin^{2}(y) \end{pmatrix} \qquad && \qquad \Gamma^{2}_{ij} = \begin{pmatrix} 0 & \cot(x) & 0 \\ \cot(x) & 0 & 0 \\ 0 & 0 & -\cos(y)\sin(y) \end{pmatrix} \end{align} $$ $$ \Gamma^{3}_{ij} = \begin{pmatrix} 0 & 0 & \cot(x) \\ 0 & 0 & \cot(y) \\ \cot(x) & \cot(y) & 0 \end{pmatrix} $$
We take $A$ to be the 2-form $$ A = e^{y}\,dx\wedge dy $$ which has components $$ A_{ij} = \begin{pmatrix} 0 & e^y & 0 \\ -e^y & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} $$
We compute $\delta A$ first directly and then compare to what is obtained by the expressions listed in the original post.
Direct Calculation
In the situation under consideration $(-1)^{g}(-1)^{n(p-1)+1} = (-1)^{3(2-1)+1} = 1$, so that $\delta A = \star \mathop{d}\star$.
Observe that the 1-forms $dx$, $\sin(x)dy$ and $\sin(x)\sin(y)dz$ form an orthonormal basis for $g^{ij}$, and therefore the Riemannian volume element is given by $$ \omega = dx\wedge(\sin(x)dy)\wedge(\sin(x)\sin(y)dz) $$ This allows us to easily take Hodge duals. $$ \begin{align} \star A &= \frac{e^{y}}{\sin(x)}\star(dx\wedge\sin(x)dy) \\[1.5ex] &= \frac{e^{y}}{\sin(x)}\sin(x)\sin(y)dz \\[1.5ex] &= e^{y}\sin(y) dz \end{align} $$ The exterior derivative gives $$ \begin{align} d\star A &= e^{y}(\cos(y)+\sin(y))dy\wedge dz \\[1.5ex] &= e^{y}(1+\cot(y))\csc^2(x)\,(\sin(x)dy\wedge\sin(x)\sin(y)dz) \end{align} $$ Taking the dual again gives the codifferential as $$ \delta A = e^{y}(1+\cot(y))\csc^2(x)dx $$ or in components $$ (\delta A)_{i} = \begin{pmatrix} e^{y}(1+\cot(y))\csc^2(x) & 0 & 0 \end{pmatrix} $$
Comparison with Index Expressions
We calculate the components via $$ (\delta A)^{(\text{wiki})}_{i} = -g^{lk}\left(\partial_{l}A_{k i}- \Gamma^{j}_{lk}\, A_{ji}\right) $$
For $i=1$, noting that $g^{ij}$ is diagonal, we find $$ \begin{align} (\delta A)^{(\text{wiki})}_{1} &= -g^{22}\partial_{y}A_{2 1}+ g^{lk}\Gamma^{2}_{lk}\, A_{21} \\[1.5ex] &= -\csc^2(x)(-e^{y})+(\csc^2(x)\csc^2(y))(-\cos(y)\sin(y))(-e^{y})\\[1.5ex] &= e^{y}(1+\cot(y))\csc^2(x) \end{align} $$
For $i=2$, we have $$ \begin{align} (\delta A)^{(\text{wiki})}_{2} &= -g^{11}\partial_{x}A_{1 2}+ g^{lk}\Gamma^{1}_{lk}\, A_{12} \\[1.5ex] &= 0+\bigg(\big(\csc^{2}(x)\times-\cos(x)\sin(x)\big)+\big(\csc^{2}(x)\csc^{2}(y)\times -\cos(x)\sin(x)\sin^{2}(y)\big)\bigg)(e^{y})\\[1.5ex] &= -2e^{y}\cot(x) \end{align} $$ It is obvious that $(\delta A)^{(\text{wiki})}_{3}=0$, so we find $$ (\delta A)^{(\text{wiki})}_{i} = \begin{pmatrix} e^{y}(1+\cot(y))\csc^2(x) & -2e^{y}\cot(x) & 0 \end{pmatrix} $$ This clearly does not coincide with $(\delta A)_{i}$.
If we examine the extra term which I was unable to show vanish (setting $p=2$) $$ (\delta A)^{(\text{ex})}_{i} = -(p-1)\,g^{lk}\,\Gamma^{j}_{li} A_{jk} = -\,\left(g^{22}\,\Gamma^{1}_{2i} A_{12} + g^{11}\,\Gamma^{2}_{1i} A_{21}\right) $$ We have for $i=1$, $$ (\delta A)^{(\text{ex})}_{1} = -(\csc^{2}(x)e^{y}\times0-e^{y}\times 0) = 0. $$ For $i=2$, $$ (\delta A)_{2}^{(\text{ex})} = -(\csc^{2}(x)e^{y}\times(-\cos(x)\sin(x))-e^{y}\times\cot(x)) = 2e^{y}\cot(x) $$ For $i=3$, $$ (\delta A)_{3}^{(\text{ex})} = -(\csc^{2}(x)e^{y}\times 0 -e^{y}\times 0) = 0. $$
So clearly it is not possible to show that $(\delta A)^{(\text{ex})}_{i}$ vanishes in general. Furthermore, we have that $$ \begin{align} (\delta A)^{(\text{wiki})}_{i} + (\delta A)^{(\text{ex})}_{i} &= \begin{pmatrix} e^{y}(1+\cot(y))\csc^2(x) & -2e^{y}\cot(x) & 0 \end{pmatrix} \quad+ \quad\begin{pmatrix} 0 & 2e^{y}\cot(x) & 0 \end{pmatrix} \\[2ex] &= \begin{pmatrix} e^{y}(1+\cot(y))\csc^2(x) & 0 & 0 \end{pmatrix}\\[2ex] &=(\delta A)_i \end{align} $$ So I think $$ (\delta A)_{i_{1}\dots i_{p-2}}=-g^{lk}\left(\partial_{l}A_{k i_{1}\dots i_{p-1}}- \Gamma^{j}_{lk}\, A_{ji_{1}\dots i_{p-1}} + (p-1)\,\Gamma^{j}_{l i_{1}}A_{jk i_{2}\dots i_{p-1}}\right) $$ is about the most condensed you're gonna get with explicit Christoffel symbols.