Possible error in a proof in Jost's Riemannian Geometry and Geometric Analysis

Question

I am reading Jost's Riemannian Geometry and Geometric Analysis, sixth edition. I suspect that there is an error in a proof. First let's define the following:

Let $M$ be a compact, connected, oriented Riemannian manifold. Define an inner product on the set of $p$-forms on $M$ by $$(\alpha,\beta):=\int_M\alpha\wedge\star\beta,$$ where $\star$ is the Hodge star operator.

In the book, page 116, there is the following statement:

Corollary 3.4.1. There exists a constant $c$, depending only on the Riemannian metric of $M$, with the property that for all closed forms $\beta$ that are orthogonal to the kernel of $d^*$, $$(\beta,\beta)\le c(d^*\beta,d^*\beta).$$

Note $d^*=(-1)^{n(p+1)+1}\star d\star$ when acting on $p$-form, where $n$ is the dimension of $M$.

Later on, on page 118, when trying to prove that the linear functional $\ell:d^*(\Omega^p(M))\to\Bbb R$,$\ell(d^*\varphi)=(\eta,\varphi)$ is bounded for a certain $\eta$, there is the following construction:

For $\varphi\in\Omega^p(M)$, let $\pi(\varphi)$ be the orthogonal projection onto the kernel of $d^*$, and $\psi:=\varphi-\pi(\varphi)$; in particular $d^*\varphi=d^*\psi$.

Since $\psi$ is orthogonal to the kernel of $d^*$, by Corollary 3.4.1, $\lVert \psi\rVert_{L^2}\le c\lVert d^*\psi\rVert_{L^2}=c\lVert d^*\varphi\rVert_{L^2}$.

I am not sure if the use of Corollary 3.4.1 for deducing the last inequality is correct, because I do not know if $\psi$ is closed. Can we prove that $\psi$ is actually closed? If not, how can we fix the proof?

Edit: note that the proof in the book comes before Hodge decomposition theorem, so I cannot use this theorem.

Answer 1

You should know the following elementary properties of the codifferential $d^*$:

1. the adjoint relation $(d \psi, \omega) = (\psi, d^* \omega)$, often taken as the definition of $d^*$; and
2. the fact $d^* \circ d^* = 0,$ which follows from $d \circ d = 0$.

To show $d \psi = 0$ it suffices to show $(d \psi, \omega) = 0$ for an arbitrary 2-form $\omega$. From the first fact above we know $(d \psi, \omega) = (\psi, d^* \omega).$ The second fact tells us that $d^* (d^* \omega) = 0$, so $d^* \omega$ is in the kernel of $d^*$ and thus $(\psi, d^* \omega) = 0$ since we were given $\psi \perp \ker d^*.$

Answer 2

The fact that $\psi$ is closed (and in fact exact) follows on a compact manifold from the Hodge decomposition.

This states that any form decomposes orthogonally into an exact form, a co-exact form, and a harmonic form. In other words, we can write $$\psi = d\alpha + d^*\beta + \gamma$$ for $\gamma$ harmonic.

On a compact manifold, harmonic forms are both closed and co-closed. Therefore, if $\psi$ is orthogonal to the co-closed forms, it follows that $\psi = d\alpha$, meaning $\psi$ is exact.

This might be a bit overkill, though.