# Hodge Decomposition – Harmonic Forms

Suppose ${M}$ is a compact complex manifold of dimension ${m}$ and ${V}$ is a holomorphic vector bundle of rank ${r}$ over ${M}$. From the Dolbeault isomorphism, we know that the cohomology group ${H^p(M, \mathcal{O}(V))}$ is isomorphic to the set of all ${\overline{\partial}}$-closed smooth ${V}$-valued ${(0, p)}$-forms on ${M}$ modulo the set of all ${\overline{\partial}}$-closed smooth ${V}$-valued ${(0, p)}$-forms on ${M}$. Though this gives us a more practical way of computing ${H^p(M, \mathcal{O}(V))}$ by using differential forms, it would be more convenient for computational purposes if a cohomology class is represented by a unique differential form rather than an equivalence class of differential forms.

When we have a quotient of a finite-dimensional vector space by a vector subspace, instead of a looking at an element in the quotient space, one way is to consider the orthogonal complement of the vector subspace provided we have an inner product in the ambient space. So every element in the quotient space is represented by a unique element in the orthogonal complement of the subspace instead of an equivalence class in the original ambient space. This is the same as using the element with minimum length in its equivalence class to represent the equivalence class.

In our case, the vector space of all ${\overline{\partial}}$-closed smooth ${V}$-valued ${(0, p)}$-forms on ${M}$, in general, is an infinite-dimensional vector space. Suppose we give ${M}$ a Hermitian metric and also give ${V}$ a Hermitian metric along its fibers. So there is a natural inner product on the vector space over all ${\overline{\partial}}$-closed smooth ${V}$-valued ${(0, p)}$-forms on ${M}$. If we use this method of orthogonal complement, or equivalently, the method of finding an element with minimum length in its equivalence class, we run into the trouble of closedness of the subspace, or equivalently, the convergence of a sequence of elements in an equivalence class minimizing the length function. We know that the vector space of all ${\overline{\partial}}$-closed smooth ${V}$-valued ${(0, p)}$-forms on ${M}$ is not complete in the natural inner product. In order to talk about closedness of subspaces or the convergence of a minimizing sequence of elements, we should first consider the vector space of all ${\overline{\partial}}$-closed smooth ${V}$-valued ${(0, p)}$-forms on ${M}$ to a Hilbert space. So we should consider the set of all ${\overline{\partial}}$-closed ${L^2V}$-valued ${(0, p)}$-forms on ${M}$.

1. The first question that arises in considering the set of all ${\overline{\partial}}$-closed ${L^2V}$-valued ${(0, p)}$-forms on ${M}$ is the definition of the operator ${\overline{\partial}}$ on ${L^2V}$-valued ${(0, p)}$-forms on ${M}$.
2. The second question is the closedness of the image of ${\overline{\partial}}$ after it is defined.

Let ${\mathcal{L}^2(p)}$ denote the set of all ${L^2V}$-valued ${(0, p)}$-forms on ${M}$, and let ${\mathcal{C}^\infty(p)}$ be the set of all ${V}$-valued ${(0, p)}$-forms on ${M}$. The operator ${\overline{\partial}}$ from ${\mathcal{L}^2(p)}$ to ${\mathcal{L}^2(p+1)}$ is defined on the dense subset ${C^\infty(p)}$ of ${\mathcal{L}^2(p)}$. In order to make the image of ${\overline{\partial}}$ closed, we should define it on as large a set as possible. So we consider the closure of the graph of ${\overline{\partial}}$ in the product space of ${\mathcal{L}^2(p)}$ and ${\mathcal{L}^2(p+1)}$. Given an element ${f}$ of ${\mathcal{L}^2(p)}$ and an element ${g}$ of ${\mathcal{L}^2(p+1)}$, the pair ${(f, g)}$ is in the graph if and only if there exists a sequence ${f_\nu \in \mathcal{C}^\infty(p)}$ converging in the ${L^2}$ norm of ${\mathcal{L}^2(p)}$ to ${f}$ so that ${\overline{\partial}f_\nu}$ converges in the ${L^2}$ norm of ${\mathcal{L}^2(p+1)}$ to ${g}$. We have to check that this closure is the graph of a map. The trouble is that for the closure of the graph, we may have two distinct elements of ${\mathcal{L}^2(p+1)}$ given as the image of the same element of ${\mathcal{L}^2(p+1)}$. By taking the difference of these two distinct elements of ${\mathcal{L}^2(p+1)}$, we can assume that there exists a sequence ${f_\nu \in \mathcal{C}^\infty(p)}$ converging in the ${L^2}$ norm of ${\mathcal{L}^2(p)}$ to 0 so that ${\overline{\partial}f_\nu}$ converges in the ${L^2}$ norm of ${\mathcal{L}^2(p+1)}$ to some nonzero element ${g}$ of ${\mathcal{L}^2(p+1)}$. Take an arbitrary element ${\varphi}$ of ${\mathcal{C}^\infty(p+1)}$. We have the process of integration by parts. So we get

$\displaystyle (\overline{\partial}f_\nu, \varphi)_{\mathcal{L}^2(p+1)} = (f_\nu, \overline{\partial}^*, \varphi)_{\mathcal{L}^2(p)},$

where ${(\,\cdot\,,\,\cdot\,)_{\mathcal{L}^2(q)}}$ means the inner product in ${\mathcal{L}^2(q)}$ and ${\overline{\partial}^*}$ is a first-order linear partial differential operator. Letting ${\nu \rightarrow \infty}$, we conclude that

$\displaystyle (g, \varphi)_{\mathcal{L}^2(p+1)} = (0, \overline{\partial}^*\varphi)_{\mathcal{L}^2(p)} = 0$

for any ${\varphi}$, contradicting that ${g}$ is nonzero. Hence, we know that the closure of the graph of ${\overline{\partial}}$ is also a graph. This argument works in general for differential operators because we have integration by parts.

There is another way to extend the definition of ${\overline{\partial}}$ from ${\mathcal{C}^\infty}$. For any ${f \in \mathcal{L}^2(p)}$, the expression ${\overline{\partial}f}$ makes sense when differentiation is done in the sense of distributions. So in general, ${\overline{\partial}}$ would be a current. If this current can be represented by an element ${g}$ of ${\mathcal{L}^2(p+1)}$, then we say that ${f}$ belongs to the domain of ${\overline{\partial}}$ and define ${\overline{\partial}f}$ as ${g}$. The earlier extension done by using the closure of the graph of ${\overline{\partial}}$ is known as the strong extension. At first, it seems that the weak extension may have a bigger domain than the strong extension. It turns out that the two extensions are the same. This is known as the Friedrichs Extension Lemma (K.O. Friedrichs, “The identity of weak and strong extensions of differential operators”. Trans. Amer. Math. Soc. 55 (1944),  132-151). First, one observes that by using a partition of unity, we can assume that all forms involved are supported in a single coordinate chart. Then in that coordinate chart, one uses smoothing by convolution with a cutoff function. Suppose

$\displaystyle L = a(x){\partial\over{\partial x}} + b(x)$

is a first-order differential operator with smooth coefficients. Let ${\chi(x)}$ be a nonnegative function supported on the unit open ball of ${\mathbb{R}^n}$. Let

$\displaystyle \chi_\epsilon(x) = {1\over{\epsilon^n}}\chi\left({x\over{\epsilon^n}}\right).$

Suppose ${u}$ is an ${L^2}$ function. Then

$\displaystyle \chi_\epsilon * u \rightarrow u$

in ${L^2}$ norm as ${\epsilon \rightarrow 0}$. We assume that ${Lu}$ is taken in the sense of distributions in ${L^2}$. Then

$\displaystyle \chi_\epsilon * Lu \rightarrow Lu$

in ${L^2}$ norm as ${\epsilon \rightarrow 0}$. It suffices to show that

$\displaystyle \chi_\epsilon * Lu - L(\chi_\epsilon * u) \rightarrow0$

in ${L^2}$ norm as ${\epsilon \rightarrow 0}$. This is clearly true when ${u}$ belongs to the dense subset of smooth functions. So it suffices to show that

$\displaystyle \chi_\epsilon * Lu - L(\chi_\epsilon * u)$

is bounded in ${L^2}$ norm when ${u}$ belongs to a set bounded in ${L^2}$ norm. The zero-order part ${b(x)}$ of ${L}$ clearly has bounded contribution. So we can assume, without loss of generality, that ${b(x) = 0}$. Then

$\displaystyle \chi_\epsilon * Lu - L(\chi_\epsilon * u) = \chi_\epsilon * \left(a {{\partial u}\over{\partial x}}\right) - a{\partial\over{\partial x}}(\chi_\epsilon * u)$

$\displaystyle = \chi_\epsilon * {\partial\over{\partial x}}(au) - \chi_\epsilon * \left(u {{\partial a}\over{\partial x}}\right) - a \left({\partial\over{\partial x}} \chi_\epsilon * u\right)$

$\displaystyle = \left({\partial\over{\partial x}} \chi_\epsilon \right) * (au) - \chi_\epsilon * \left( u {{\partial a}\over{\partial x}}\right) - a\left({\partial\over{\partial x}}\chi_\epsilon - u\right).$

Clearly, the second term on the right-hand side is bounded. So we can drop it. We have

$\displaystyle \left(\left({\partial\over{\partial x}}\chi_\epsilon\right) * (au) - a\left({\partial\over{\partial x}} \chi_\epsilon * u\right)\right)(x)$

$\displaystyle \int \left(\left({\partial\over{\partial y}} \chi_\epsilon\right)(y) a(x- y) u(x - y) - a(x) \left({\partial\over{\partial x}} \chi_\epsilon * u\right)(y)u(x - y)\right) dy$

$\displaystyle =\int_{|y| < \epsilon} \left({\partial\over{\partial y}} \chi_\epsilon\right)(y)(a(x-y) - a(x))u(x - y)dy.$

In the last integral,

$\displaystyle |a(x - y) - a(x)| \le C\epsilon$

because ${|y| < \epsilon}$. We have

$\displaystyle \left|{\partial\over{\partial y}}\chi_\epsilon\right| \le {{C'}\over{\epsilon^{n+1}}}.$

Hence,

$\displaystyle \left(\left({\partial\over{\partial x}} \chi_\epsilon\right) * (au) - a\left({\partial \over{\partial x}}\chi_\epsilon * u\right)\right)(x) \le {{C''}\over{\epsilon^n}} \int_{|y| < \epsilon} |u(x - y)|dy,$

and the ${L^2}$ norm of

$\displaystyle \left({\partial\over{\partial x}} \chi_\epsilon\right) * (au) - a\left({\partial\over{\partial x}} \chi_\epsilon * u\right)$

is bounded by

$\displaystyle {{C''}\over{\epsilon^n}} \int_{|y| < \epsilon} \left( \int |u(x - y)|^2 dx\right) dy,$

which is bounded if the ${L^2}$ norm of ${u}$ is bounded.

We want to show that the image of

$\displaystyle \overline{\partial}: \mathcal{L}^2(p-1) \rightarrow \mathcal{L}^2(p)$

is closed, and the kernel of

$\displaystyle \overline{\partial}: \mathcal{L}^2(p) \rightarrow \mathcal{L}^2(p+1)$

quotiented by the image of

$\displaystyle \overline{\partial}: \mathcal{L}^2(p-1) \rightarrow \mathcal{L}^2(p)$

is isomorphic to the set of all ${\overline{\partial}}$-closed smooth ${V}$-valued ${(0, p)}$-forms on ${M}$. First, we handle the question of the closedness of the image of

$\displaystyle \overline{\partial}: \mathcal{L}^2(p-1) \rightarrow \mathcal{L}^2(p).$

This means that if we have ${f_\nu \in \mathcal{L}^2(p-1)}$ so that ${\overline{\partial}f_\nu}$ converges to some ${g_\infty}$ in ${\mathcal{L}^2(p)}$, then we want to show that ${g_\infty = \overline{\partial}f_\infty}$ for some ${f_\infty \in \mathcal{L}^2(p-1)}$. Let ${g_\nu = \overline{\partial}f_\nu}$. If we are able to solve the equation ${\overline{\partial}h_\nu = g_\nu}$ so that ${h_\nu}$ or a subsequence of it converges to some ${h_\infty}$ in ${\mathcal{L}^2(p-1)}$, then we are done.

We are going to consider the operator

$\displaystyle \square = \overline{\partial}\overline{\partial}^* + \overline{\partial}^*\overline{\partial}.$

First, we would like to discuss the motivation for this operator ${\square}$. We look at the equation ${\overline{\partial}h = g}$, where ${g}$ is in ${\mathcal{L}^2(p)}$ and ${\overline{\partial}g = 0}$. We would like to solve it so that ${h}$ has a good bound when ${g}$ is bounded. The equation ${\overline{\partial}h = g}$ is equivalent to ${(\overline{\partial} h, \varphi) = (g, \varphi)}$ for all ${\varphi}$ in ${\mathcal{L}^2(p)}$, i.e. ${(h, \overline{\partial}^*\varphi) = (g, \varphi)}$. Now, ${h}$ is the unknown. To know ${h}$ is the same as knowing the linear functional ${\psi \rightarrow (h, \psi)}$. Now, we know this linear functional is ${\psi \rightarrow (g, \varphi)}$, and ${g}$ is known. The set of all ${\overline{\partial}^* \varphi}$ is a subset of ${\mathcal{L}^2(p-1)}$. If we can extend this known linear functional from the set ${\text{Im}\,\overline{\partial}^*}$ to a bounded linear functional on all of ${\mathcal{L}^2(p-1)}$, then we know ${h}$ whose norm is bounded by that of the bounded linear functional. Such an extension is possible if we have an estimate

$\displaystyle |(g, \varphi)| \le C\|\overline{\partial}^* \varphi\|.$

We write ${\varphi = \varphi_1 + \varphi_2}$ so that ${\varphi_1 \in \text{Ker}\,\overline{\partial}}$ and ${\varphi_2 \perp \text{Ker}\,\overline{\partial}}$. Since ${g \in \text{Ker}\,\overline{\partial}}$, we have ${(g, \varphi) = (g, \varphi_1)}$. From ${\varphi_2 \perp \text{Ker}\,\overline{\partial}}$, we have ${\varphi_2 \perp \text{Im}\,\overline{\partial}}$ and ${\overline{\partial}^*\varphi_2 = 0}$. Hence,

$\displaystyle |(g, \varphi)| \le C\|\overline{\partial}^* \varphi\|$

is equivalent to

$\displaystyle |(g, \varphi_1)| \le C\|\overline{\partial}^*\varphi_1\|.$

We can assume for that inequality that ${\varphi \in \text{Ker}\,\overline{\partial}}$. By the Schwarz Inequality, it suffices to have an estimate

$\displaystyle \|\varphi\| \le C\|\overline{\partial}^*\varphi\|.$

Since ${g \in \text{Ker}\,\overline{\partial}}$ to have the inequality that

$\displaystyle \|\overline{\partial}^* \varphi \|^2 + \|\overline{\partial}\varphi\|^2 \ge C\|\varphi\|^2.$

This inequality can be rewritten as

$\displaystyle (\square \varphi, \varphi) \ge C\|\varphi\|^2.$

This shows that the operator ${\square}$ is closely related to the solution ${\overline{\partial}h = g}$. If we do have the inequality

$\displaystyle (\square \varphi, \varphi) \ge C\|\varphi\|^2,$

then our preceding discussion shows that the equation ${\overline{\partial}h = g}$ admits a solution ${h}$ with

$\displaystyle \|h\| \le C'\|g\|.$

We do not expect in general to have the inequality

$\displaystyle (\square \varphi, \varphi) \ge C\|\varphi\|^2,$

because this inequality would imply that ${H^q(M, \mathcal{O}(V))}$ vanishes. However, we have a related inequality that can serve our purpose of proving the closedness of the image of ${\overline{\partial}}$. This related inequality is Gårding’s Inequality. It is says that

$\displaystyle (\square\varphi, \varphi) + (\varphi, \varphi) \ge C\|\varphi\|_1^2,$

where the norm ${\|\varphi\|_1}$ means a norm that is equivalent to the sum of the ${L^2}$-norm of ${\varphi}$ and the ${L^2}$ norm of the first derivative of ${\varphi}$.

Before we prove Gårding’s Inequality, let us look at its consequences. Rewrite Gårding’s Inequality in the form

$\displaystyle ((\square + 1)\varphi, \varphi) \ge C\|\varphi\|_1^2.$

We know that the operator ${\square + 1}$ admits an inverse whose norm is ${\le 1}$, because the operator ${\square}$ is nonnegative. Replacing ${\varphi}$ by ${(\square + 1)^{-1}\varphi}$ in Gårding’s Inequality, we get

$\displaystyle ((\square + 1)^{-1}\varphi, \varphi) \ge C\|(\square + 1)^{-1}\varphi\|_1^2.$

Since

$\displaystyle ((\square + 1)^{-1}\varphi, \varphi) \le \|(\square + 1)^{-1}\varphi\|^2 \|\varphi\|^2 \le \|(\square + 1)^{-1}\varphi\|_1^2 \|\varphi\|^2,$

it follows that

$\displaystyle C\|(\square + 1)^{-1}\varphi\|_1 \le \|\varphi\|.$

This means that the map ${(\square + 1)^{-1}}$ from ${\mathcal{L}^2(p)}$ to ${\mathcal{L}_1^2(p)}$ is continuous, where ${\mathcal{L}_1^2}$ is the Hilbert space of all ${L^2V}$-valued ${(0, p)}$-forms whose first derivatives are also ${L^2}$. Since by the Rellich-Kondrachov Theorem the inclusion map

$\displaystyle \mathcal{L}_1^2(p) \rightarrow \mathcal{L}^2(p)$

is a compact operator, it follows that the map ${(\square + 1)^{-1}}$ from ${\mathcal{L}^2(p)}$ to ${\mathcal{L}^2(p)}$ is a compact operator. A compact operator means that it maps a bounded set of the domain space to a relatively compact subset of the target space. We will prove the Rellich-Kondrachov Theorem later. By the spectral theorem for self-adjoint compact operators, we know that the eigenvalues ${\mu_i}$ of the operator ${(\square + 1)^{-1}}$ from ${\mathcal{L}^2(p)}$ to ${\mathcal{L}^2(p)}$ are in ${(0, 1]}$, the only possible limit point of ${\{\mu_i\}}$ is 0, the eigenspace for each ${\mu_i}$ is finite-dimensional, and the eigenfunctions of ${(\square + 1)^{-1}}$ span ${\mathcal{L}^2(p)}$. The eigenvalues ${\mu_i}$ are limited to ${(0, 1]}$, because ${(\square + 1)^{-1}}$ is positive and its norm is ${\le 1}$. An equation

$\displaystyle (\square + 1)^{-1}f_i = \mu_i f_i$

is equivalent to

$\displaystyle \square f_i = \lambda_i f_i,\text{ with }\lambda_i = {{1 - \mu_i}\over{\mu_i}}.$

So we conclude that the eigenvalues ${\lambda_i}$ of ${\square}$ are in ${[0, \infty)}$, the eigenspace for each ${\lambda_i}$ is finite-dimensional, and the eigenfunction ${f_i}$ of ${\square}$ span ${\mathcal{L}^2(p)}$. If 0 is an eigenvalue, we call it ${\lambda_0}$. By allowing some positive eigenvalues ${\lambda_i}$ (${1 \le i < \infty}$) to be counted more than once, we can assume that the eigenspace for each positive eigenvalue ${\lambda_i}$ is only ${1}$-dimensional, and the totality of the eigenfunction ${f_i}$ for ${\lambda_i}$ (${1 \le i < \infty}$) is an orthonormal basis of the orthonal complement ${(\text{Ker}\,\square)^\perp}$ of ${\text{Ker}\,\square}$ in ${\mathcal{L}^2(p)}$. We know that ${\text{Ker}\,\square}$ is finite-dimensional. On ${(\text{Ker}\,\square)^\perp}$, we can define the inverse of ${\square}$ by

$\displaystyle G\left(\sum_{i=1}^\infty \alpha_if_i\right) = \sum_{i=1}^\infty {{\alpha_i}\over{\lambda_i}}f_i.$

We extend ${G}$ to all of ${\mathcal{L}^2(p)}$ by defining ${G}$ to be zero on ${\text{Ker}\,\square}$. We call ${G}$ Green’s operator. The reason for the name is that ${G}$ is the inverse of the Laplace operator ${\square}$. Let ${H}$ be the orthogonal projection from ${\mathcal{L}^2(p)}$ onto ${\text{Ker}\,\square}$. Then we have the identity

$\displaystyle 1 - H = G\square = \square G.$

Now, we are ready to prove the closedness of ${\text{Im}\,\overline{\partial}}$. We go back to our earlier notation of ${g_\nu \rightarrow g_\infty}$ in ${\mathcal{L}^2(p)}$ so that ${g_\nu \in \text{Im}\,\overline{\partial}}$. Suppose ${g \in \mathcal{L}^2(p)}$ with ${\overline{\partial}g = 0}$. We have

$\displaystyle g - Hg = (\overline{\partial}\overline{\partial}^* + \overline{\partial}^* \overline{\partial})Gg.$

By applying ${\overline{\partial}}$ to both sides, we get

$\displaystyle 0 = \overline{\partial}\overline{\partial}^*\overline{\partial}Gg.$

Taking the inner product with ${Gg}$, we obtain

$\displaystyle 0 = (\overline{\partial}\overline{\partial}^*\overline{\partial} Gg, \overline{\partial} Gg) = \|\overline{\partial}^*\overline{\partial} Gg\|^2.$

Hence,

$\displaystyle \overline{\partial}Gg = 0,\text{ and }g - Hg = \overline{\partial}\overline{\partial}^* Gg.$

This shows that though in general we can not solve the equation ${g = \overline{\partial} h}$ because ${H^p(M, \mathcal{O}(V))}$ may not vanish, we can still always solve the equation

$\displaystyle g = Hg = \overline{\partial}h\text{ with }h = \overline{\partial}^* Gg.$

From

$\displaystyle \overline{\partial}h = g - Hg,\text{ }\overline{\partial}^*h = 0,$

and Gårding’s Inequality, we conclude that

$\displaystyle \|g - Hg\|^2 \ge C\|h\|_1^2,\text{ }\|g\|^2 \ge C'\|h\|_1^2,$

because

$\displaystyle \|g - Hg\| \le C'' \|g\|.$

We go back to our earlier notation of ${g_\nu \rightarrow g_\infty}$ in ${\mathcal{L}^2(p)}$ so that ${g_\nu \in \text{Im}\,\overline{\partial}}$. Since ${g_\nu \in \text{Ker}\,\overline{\partial}}$, we have

$\displaystyle g-\nu - Hg_\nu = \overline{\partial}h_\nu \text{ with }\|g_\nu\|^2 \ge C'\|h_\nu\|_1^2.$

An element ${f}$ of ${\mathcal{L}^2(p)}$ belongs to ${\text{Ker}\,\square}$ if and only if

$\displaystyle \overline{\partial}f = 0,\text{ }\overline{\partial}^*f = 0,$

because

$\displaystyle \square = \overline{\partial}\overline{\partial}^* + \overline{\partial}^*\overline{\partial} \text{ and }(\square f, f) = \|\overline{\partial}f\|^2 + \|\overline{\partial}^*f\|^2.$

This implies every element of ${\text{Ker}\,\square}$ is perpendicular to both ${\text{Im}\,\overline{\partial}}$ and ${\text{Im}\,\overline{\partial}^*}$. Since

$\displaystyle Hg_\nu = g_\nu - \overline{\partial}h_\nu$

belongs to ${\text{Im}\,\overline{\partial}}$, it follows from ${Hg_\nu \in \text{Ker}\,\square}$ that ${Hg_\nu = 0}$. Hence, ${g_\nu = \overline{\partial}h_\nu}$. Since ${\|h_\nu\|_1^2}$ is bounded independent of ${\nu}$, by the Rellich-Kondrachov Theorem, we can select a subsequence ${h_{\nu_j}}$ of ${h_\nu}$ converging to some ${h_\infty}$ in ${\mathcal{L}^2(p)}$. Then ${\overline{\partial}h_\infty = g_\infty}$, because

$\displaystyle h_{\nu_j} \rightarrow h_\infty,\text{ }\overline{\partial}h_{\nu_j} = g_{\nu_j} \rightarrow g_\infty.$

Hence, the image of

$\displaystyle \overline{\partial}: \mathcal{L}^2(p-1) \rightarrow \mathcal{L}^2(p).$

Now, we want to show that the cohomology group ${H^p(M, \mathcal{O}(V))}$ can be calculated by using ${L^2V}$-valued forms instead of ${V}$-valued forms. For this we need a regularity result for the equation ${\square f = g}$ in ${\mathcal{L}^2(p)}$. More precisely, we want to show that if the norm ${\|g\|_k}$, which is equivalent to the sum of all the ${L^2}$ norms of derivatives of ${g}$ of order ${\le k}$ is finite, then ${\|f\|_{k+2}}$ is finite. Note that we assume already that the ${L^2}$ norms of both ${f}$ and ${g}$ are finite. First, we observe that if ${\varphi \in \mathcal{L}^2(p)}$ and ${\overline{\partial}\varphi \in \mathcal{L}_k^2(p+1)}$ and ${\overline{\partial}\varphi \in \mathcal{L}_k^2(p-1)}$ for some ${k \ge 1}$, then ${\varphi \in \mathcal{L}_{k+1}^2(p)}$. The case ${k = 0}$ is given right away by Gårding’s Inequality. For the general case, we take any vector field ${X}$ and apply the argument to ${\nabla_X \varphi}$ and use induction on ${k}$. Now, we come back to the equation ${\square f = g}$ in ${\mathcal{L}^2(p)}$. Again, we look at first the case ${k = 0}$. By Gårding’s Inequality,

$\displaystyle \|\overline{\partial}f\|_1^2 \le C(\square\overline{\partial}f, \overline{\partial}f) + C\|\overline{\partial}f\|^2$

$\displaystyle = C(\overline{\partial}^*\partial f, \overline{\partial}^* \overline{\partial} f) + C\|\overline{\partial}f \|^2$

$\displaystyle = C(\square f, \square f) + C\|\overline{\partial}f\|^2 \text{ (because }(\overline{\partial}^*\partial f, \overline{\partial}\overline{\partial}^* f) = 0\text{)}$

$\displaystyle \le C(\square f, \square f) + C(\square f, f)^2 = C\|g\|^2 + C(g, f).$

Likewise,

$\displaystyle \|\overline{\partial}^* f\|_1^2 \le C\|g\|^2 + C(g, f).$

Hence, by the above observation, ${\|f\|_2}$ is finite. For the case of a general ${k}$, again we take any vector field ${X}$ and apply the argument to ${\nabla_X \varphi}$ and use induction on ${k}$. In these arguments, we have been quite sloppy above justifying such things as integration by parts even though our forms are not smooth. The rigorous way to justify such arguments is to smooth out the forms first in the graph norm of the first-order operators by using the result of Friedrichs and get estimates on the approximating smooth forms using constants dependent only on our original nonsmooth form and then take limits at the end. A form is in ${\mathcal{L}_k^2(p)}$ if and only if it can be approximated by smooth forms in the norm of ${\mathcal{L}_k^2(p)}$.

Now that we have the regularity result for ${\square}$, we consider the map ${\Psi}$ from the space

$\displaystyle \text{Ker}(\overline{\partial}: \mathcal{C}^\infty(p) \rightarrow \mathcal{C}^\infty(p+1))/\text{Im}(\overline{\partial}: \mathcal{C}^\infty(p-1) \rightarrow \mathcal{C}^\infty(p))$

to the space

$\displaystyle \text{Ker}(\overline{\partial}: \mathcal{L}^\infty(p) \rightarrow \mathcal{L}^\infty(p+1))/\text{Im}(\overline{\partial}: \mathcal{L}^\infty(p-1) \rightarrow \mathcal{L}^\infty(p))$

We want to show that ${\Psi}$ is an isomorphism. First, we show that it is surjective. Take ${\varphi \in \mathcal{L}^2(p)}$ with ${\overline{\partial}\varphi = 0}$. We have

$\displaystyle \varphi = H\varphi = \overline{\partial}^*\overline{\partial} G \varphi + \overline{\partial} \overline{\partial}^* G\varphi.$

As we argued before, when we apply ${\overline{\partial}}$ to both sides and take the inner product with ${G\varphi}$, we conclude that

$\displaystyle \overline{\partial}^*\overline{\partial}G\varphi = 0.$

From ${\square H \varphi = 0}$ and the regularity of ${\square}$, we know that ${H\varphi}$ is smooth. Since ${\varphi}$ differs from the ${\overline{\partial}}$-closed smooth form ${H\varphi}$ by the ${\overline{\partial}}$-exact form ${\overline{\partial}^* G\varphi}$, we conclude that ${\Psi}$ is surjective. Now, we look at the injectivity of ${\Psi}$. Suppose ${\varphi \in \mathcal{C}^\infty(p)}$ and ${\varphi = \overline{\partial}\psi}$ for some ${\psi \in \mathcal{L}^2(p-1)}$. Take the decomposition

$\displaystyle \varphi = H\varphi + \overline{\partial}^*\overline{\partial}G\varphi + \overline{\partial}\overline{\partial}^* G\varphi.$

It is clear that the three spaces ${\text{Ker}\,\square}$, ${\text{Im}\,\overline{\partial}}$, ${\text{Im}\,\overline{\partial}^*}$ are mutually orthogonal, because

$\displaystyle \text{Ker}\,\square = (\text{Ker}\,\overline{\partial}) \cap (\text{Ker}\,\overline{\partial}^*).$

From

$\displaystyle 0 = H\varphi + \overline{\partial}^*\overline{\partial} G \varphi + \overline{\partial}(\overline{\partial}^* G\varphi - \psi),$

we conclude that

$\displaystyle H\varphi = 0,\text{ }\overline{\partial}^*\overline{\partial} G\varphi = 0.$

So

$\displaystyle \varphi = \overline{\partial}\overline{\partial}^* G\varphi.$

Since ${\varphi}$ is smooth, by the regularity of ${\square}$ and

$\displaystyle \square G \varphi = \varphi - H\varphi,$

we know that ${G\varphi}$ is smooth and ${\overline{\partial}^* G\varphi}$ is smooth. Since ${\varphi}$ is the ${\overline{\partial}}$ of the smooth form ${\overline{\partial}^*G\varphi}$, we conclude that ${\Psi}$ is injective. From the above argument, we also see that both spaces

$\displaystyle \text{Ker}(\overline{\partial}: \mathcal{C}^\infty(p) \rightarrow \mathcal{C}^\infty(p+1))/\text{Im}(\overline{\partial}: \mathcal{C}^\infty(p-1) \rightarrow \mathcal{C}^\infty(p))$

and

$\displaystyle \text{Ker}(\overline{\partial}: \mathcal{L}^\infty(p) \rightarrow \mathcal{L}^\infty(p+1))/\text{Im}(\overline{\partial}: \mathcal{L}^\infty(p-1) \rightarrow \mathcal{L}^\infty(p))$

are isomorphic to ${\text{Ker}\,\square}$. An element of ${\text{Ker}\,\square}$ is called a harmonic form. So we have proved that the cohomology group ${H^p(M, \mathcal{O}(V))}$ is isomorphic to the space of all harmonic ${V}$-valued ${(0, p)}$-forms. Since ${\text{Ker}\,\square}$ is finite-dimensional from the spectral theorem, we know that ${H^p(M, \mathcal{O}(V))}$ is always finite-dimensional for a compact manifold ${M}$.