Expressing the codifferential of a $p$-form in terms of covariant derivatives

Question

Let $\alpha$ be a $p$-form on an $n$-manifold $M$. I read somewhere that the codifferential $\delta\alpha$ of $\alpha$ can be expressed in terms of covariant derivatives as follow: \begin{align} (\delta\alpha)_{i_1\cdots i_{p-1}}=-g^{jk}\nabla_j\alpha_{ki_1\cdots i_{p-1}} \end{align} I would like to know how to prove this. Since only the definition of $\delta$: \begin{align} \delta\alpha=(-1)^{np+n+1}*d*\alpha \end{align} (where $*$ is the Hodge star operator) was given to me (I'm aware that $\delta$ is usually defined as the adjoint of $d$, but this is the definition I was given; nevertheless, both can be assumed here), and the computation of $*\alpha$ usually involves the factor $\sqrt{\det g_{ij}}$, I'm not sure how to proceed to obtain the expression above in terms of only the covariant derivatives.

Thanks in advance for any comment, hint, and answer.

Answer 1

In general a differential form can be written as

$$ \sum_{i_1< \cdots< i_p} \alpha_{i_1\cdots i_p} dx^{i_1}\wedge\cdots \wedge dx^{i_p}.$$

The equality that you want to show is linear in $\alpha$, so it suffices to check it for

$$\alpha = f dx^1\wedge \cdots \wedge dx^p.$$

Moreover, since the above equality is first order (involve only $g, \partial g$), it suffices to check it at the center of a normal coordinates. That is, one may assume that $g_{ij} = \delta_{ij}$. Then

\begin{align} *\alpha &= f dx^{p+1}\wedge \cdots\wedge dx^n \\ \Rightarrow d*\alpha &= \sum_{i=1}^p \frac{\partial f}{\partial x^i} dx^i \wedge dx^{p+1}\wedge \cdots\wedge dx^n \\ \Rightarrow *d*\alpha &= \sum_{i=1}^p \frac{\partial f}{\partial x^i} (-1)^{(p-1)(n-p)+i-1} dx^1\wedge \cdots\widehat {dx^i}\wedge \cdots \wedge dx^p\\ &=(-1)^{np+n+1}\sum_{i=1}^p\frac{\partial f}{\partial x^i} (-1)^{i} dx^1\wedge \cdots\widehat {dx^i}\wedge\cdots \wedge dx^p.\\ \Rightarrow \delta\alpha &= \sum_{i=1}^p\frac{\partial f}{\partial x^i} (-1)^{i} dx^1\wedge \cdots\widehat {dx^i}\wedge\cdots \wedge dx^p. \end{align} On the other hand, at a normal coordinates,

$$-g^{jk}\nabla_j\alpha_{ki_1\cdots i_{p-1}} = - \sum_{j=1}^n \frac{\partial \alpha_{ji_1\cdots i_{p-1}}}{\partial x^j}.$$

So if $\alpha = fdx^1 \wedge\cdots\wedge dx^p$,

$$\alpha_{i_1\cdots i_p} = \begin{cases} f & \text{if } i_1 = 1,\cdots i_p = p, \\ 0 &\text{otherwise.}\end{cases}$$

and thus

$$ - \sum_{j=1}^n \frac{\partial \alpha_{ji_1\cdots i_{p-1}}}{\partial x^j} = \begin{cases} (-1)^i\frac{\partial f}{\partial x^i} & \text{if } (i_1, \cdots, i, \cdots, i_p) = (1, \cdots, p), \\ 0 &\text{otherwise.}\end{cases}$$

This is what you want.

Answer 2

Notes

1. I use the notation $\operatorname{d}^{\dagger}$ rather than $\delta$ for the codifferential.
2. Where did you get the definition $\operatorname{d}^{\dagger}=(−1)^{nk+n+1}\star\operatorname{d}\star$? Doesn't agree with mine, which is $\operatorname{d}^{\dagger} = (-1)^{kn}\operatorname{sgn}(g)\star\operatorname{d}\star$, even if we forget about the sign.

3. The computation of $\starα$ usually involves the factor $\sqrt{\det g_{ij}}$

   Well, it only does if you choose a coordinate system from the beggining. For the abstract calculations you don't really need to know $\star\alpha$, just its properties. In this particular calculation you'll see it suffices to know its definition.

But, oh well. Here you go. A coordinate free proof in good old Penrose graphical notation. At the end I repeat the calculation using abstract indices, so you can compare.

We begin with some explanations about the notation. Now to the proof.

And that's it. It was very lengthy in part because I was very explicit in the steps. Now the calculation is very easy to write in abstract index notation.

\begin{align} (\star\omega)_{a_{1}\dots a_{n-k}} &= \frac{1}{k!}\omega^{m_1\dots m_k}\epsilon_{m_1\dots m_ka_{1}\dots a_{n-k}}\\ \implies (\operatorname{d}\star\omega)_{b_1\dots b_{n-k+1}} &= \frac{1}{k!(n-k)!}\nabla_{[b_1}(\omega^{a_1\dots a_k}\epsilon_{|a_1\dots a_k|b_2\dots b_{n-k+1}]})\\ &=\frac{1}{k!(n-k)!}\nabla_{[b_1}\omega^{a_1\dots a_k}\epsilon_{|a_1\dots a_k|b_{2}\dots b_{n-k+1}]}\\ \implies (\star\operatorname{d}\star\omega)_{c_1\dots c_{k-1}} &=\frac{1}{k!(n-k)!} \nabla^{[b_1}\omega_{a_1\dots a_k} \epsilon^{|a_1\dots a_k|b_{2}\dots b_{n-k+1}]} \epsilon_{b_1\dots b_{n-k+1}c_1\dots c_{k-1}}\\ &=\frac{1}{k!(n-k)!} \nabla^{b_1}\omega_{a_1\dots a_k} \epsilon^{a_1\dots a_kb_{2}\dots b_{n-k+1}} \epsilon_{b_1\dots b_{n-k+1}c_1\dots c_{k-1}}\\ &=\frac{\operatorname{sgn}(g)}{k!(n-k)!} \nabla^{b_1}\omega_{a_1\dots a_k} \delta^{a_1\dots a_kb_{2}\dots b_{n-k+1}}_{b_1\dots b_{n-k+1}c_1\dots c_{k-1}}\\ &=\frac{\operatorname{sgn}(g)(-1)^{(k-1)(n-k)}}{k!(n-k)!} \nabla^{b_1}\omega_{a_1\dots a_k} \delta^{a_1\dots a_kb_{2}\dots b_{n-k+1}}_{b_1c_1\dots c_{k-1}b_2\dots b_{n-k+1}}\\ &=\frac{\operatorname{sgn}(g)(-1)^{(k-1)(n-k)}}{k!} \nabla^{b_1}\omega_{a_1\dots a_k} \delta^{a_1\dots a_k}_{b_1c_1\dots c_{k-1}}\\ &=\operatorname{sgn}(g)(-1)^{(k-1)(n-k)} \nabla^{b_1}\omega_{[b_1c_1\dots c_{k-1}]}\\ &=\operatorname{sgn}(g)(-1)^{(k-1)(n-k)} \nabla^{a}\omega_{ac_1\dots c_{k-1}} \end{align}

Since $(-1)^{(k-1)(n-k)} = (-1)^{nk+n}$, we have $$(\star\operatorname{d}\star\omega)_{c_1\dots c_{k-1}}=\operatorname{sgn}(g)(-1)^{nk+n}\nabla^{a}\omega_{ac_1\dots c_{k-1}}$$ And coming back to the definition of the codifferential $$\operatorname{d}^{\dagger} = (-1)^{kn}\operatorname{sgn}\star\operatorname{d}\star$$ we obtain \begin{align} (\operatorname{d}^{\dagger}\omega)_{c_1\dots c_{k-1}} &=(-1)^{kn}\operatorname{sgn}(g)\operatorname{sgn}(g)(-1)^{nk+n}\nabla^{a}\omega_{ac_1\dots c_{k-1}}\\ &=(-1)^{2kn+n}\nabla^{a}\omega_{ac_1\dots c_{k-1}}\\ &=(-1)^{n}\nabla^{a}\omega_{ac_1\dots c_{k-1}} \end{align}

Answer 3

First, note that in a torsion-less manifold, one can use the covariant derivative instead of the partial derivative to define the exterior derivative. This is most easily seen using this expression for the covariant derivative (I am using the suggestive notation $\partial/\partial dx^a$ for $\iota_{\frac{\partial}{\partial x^a}}$) $$\nabla_a=\partial_a-{\Gamma^b}_{ac}dx^c\frac{\partial}{\partial dx^b},$$ which is valid when working with differential forms. We then have (I am supressing all wedge products) $$d x^a\nabla_a=dx^a\partial_a-{\Gamma^b}_{ac}dx^adx^c\frac{\partial}{\partial dx^b}=dx^a\partial_a=d.$$ The Christoffel symbol part vanishes because of the lack of torsion ${\Gamma^b}_{ac}={\Gamma^b}_{ca}$.

Now, the space of differential forms $\Omega^k(M)$ has an inner product $$\langle\mu,\nu\rangle=\int\mu\star\nu.$$ The codifferential $\delta$ is the formal adjoint of $d$ under this inner product for, if $\mu\in\Omega^{k-1}(M)$ and $\nu\in\Omega^k(M)$, we have $$\langle d\mu,\nu\rangle=\int d\mu\star\nu=\int\left(d(\mu\star\nu)+(-1)^{k-1}\mu d\star\nu\right)=(-1)^{k-1+(d-k+1)(k-1)}\int\mu\star\star d\star\nu=\langle\mu,\delta\nu\rangle.$$ In here we used the fact that $\star\star=(-1)^{k(d-k)}$ when acting on $k$-forms.

On the other hand, we can compute the adjoint using the expression in terms of the covariant derivative. First, it is clear that the adjoint of the covariant derivative is minus itself. This is a simple consequence of integration by parts and the fact that the metric is covariantly constant $$\langle\nabla_a\mu,\nu\rangle=\int\nabla_a\mu\star\nu=\int d^dx \sqrt{g}\nabla_a\mu_{a_1\cdots a_k}\nu^{a_1\cdots a_k}=-\int d^dx \sqrt{g}\mu^{a_1\cdots a_k}\nabla_a\nu_{a_1\cdots a_k}=-\langle\mu,\nabla_a\nu\rangle.$$ On the other hand, as shown in this post, the adjoint of $dx^a$ is $g^{ab}\partial/\partial dx^b$. We then conclude that $$\delta=-\nabla_a g^{ab}\frac{\partial}{\partial dx^b},$$ which is what you wanted to show.

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EDIT: This is wrong! The problem is that $\nabla_a$ is not anti-self-adjoint. The problem is that what I thought was a boundary term is not a boundary term. It is $$\int d^dx\sqrt{g}\nabla_a(\mu_{a_1\cdots a_k}\nu^{a_1\cdots a_k})=\int d^dx\sqrt{g}\partial_a(\mu_{a_1\cdots a_k}\nu^{a_1\cdots a_k})=-\int d^dx \sqrt{g}\partial_a\ln\sqrt{g}\mu_{a_1\cdots a_k}\nu^{a_1\cdots a_k}.$$

Therefore, in fact $$(\nabla_a)^\dagger=-\nabla_a-{\Gamma^b}_{ab}.$$ Therefore, $$\delta=-\nabla_a\left(g^{ab}\frac{\partial}{\partial dx^b}\right)-g^{ab}{\Gamma^c}_{ac}\frac{\partial}{\partial dx^b}=-\partial_a g^{ab}\frac{\partial}{\partial dx^b}-g^{ab}\nabla_a\frac{\partial}{\partial dx^b}-g^{ab}{\Gamma^c}_{ac}\frac{\partial}{\partial dx^b}.$$

We can compute $$\left[\nabla_a,\frac{\partial}{\partial dx^b}\right]=-{\Gamma^d}_{ac}\left\{dx^c,\frac{\partial}{\partial dx^b}\right\}\frac{\partial}{\partial dx^d}=-{\Gamma^d}_{ab}\frac{\partial}{\partial dx^d}.$$ Therefore $$\delta=-\left(\partial_a g^{ab}+g^{ab}{\Gamma^c}_{ac}+g^{ac}{\Gamma^b}_{ad}\right)\frac{\partial}{\partial dx^b}-g^{ab}\frac{\partial}{\partial dx^b}\nabla_a.$$ The term in parenthesis vanishes because of the metricity condition of the Levi-Civita tensor. We conclude that $$\delta=-g^{ab}\frac{\partial}{\partial dx^b}\nabla_a.$$ This is now the correct equation!