Is the space of exact forms modulo exact forms whose anti-derivatives vanish on the boundary finite dimensional?

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Let $M$ be a smooth, connected, compact manifold with boundary (of dimension $d \ge 2$). Let $E$ be the space of exact $k$-forms on $M$. Define $$E_0=\{ \eta \in E \, | \, \eta=d\theta, \theta|_{\partial M}=0\}.$$

Is it true that $ E/E_0$ infinite dimensional?

The answer is positive for $k=1$:

$$ E/E_0=\{ df \, | \, f \in C^{\infty}(M) \, \}\, / \, \{ dg \, | \, g \in C^{\infty}(M) \, \, \text{and} \,g|_{\partial M}=0\}.$$

Let $f_i$ be an infinite set of functions on $\partial M$, such that $\{1,f_1,f_2\dots\}$ are linearly independent in $C^{\infty}(\partial M) $. Let $F_i$ be smooth extensions of $f_i$ to all of $M$. Then $[dF_i]$ are linearly independent in $E/E_0$. Indeed, suppose that $a^i[dF_i]=0$. Then $0=[a^idF_i]=[d(a^iF_i)]$, so $d(a^iF_i)=dg$ for some $g \in C^{\infty}(M)$ vanishing on $\partial M$. Since $M$ is connected, this implies $a^iF_i-g$ is some constant $c$, so in particular $a^if_i=c$, which implies $a_i=0$.

What about $k>1$?

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Yes.

Let $V$ denote the quotient space $$\frac{\{k-1\mathrm{-forms\;on\;}\partial M\}}{\{\mathrm{closed\;}k-1\mathrm{-forms\;on\;}\partial M\}}.$$ As $\partial M$ is a smooth manifold of positive dimension, the space $V$ is infinite dimensional (assuming $k\leq d$). Let $\alpha_1,\alpha_2,\ldots$ be an infinite sequence of $k-1$-forms on $\partial M$ whose image in $V$ is linearly independent. As $\partial M\subset M$ is closed as a subset, every $\alpha_i$ can be extended to a $k-1$-form on $M$, which we call $\overline{\alpha}_i$. Suppose we have $$\sum_{i=1}^n c_id\overline{\alpha}_i=d\beta,\quad\beta|_{\partial M}=0,\quad n\in\mathbb{N}.$$Then it follows that $$d\left(\sum_1^nc_i\overline{\alpha}_i-\beta\right)=0,$$and restriction to the boundary yields $$d\left(\sum c_i\alpha_i\right)=0.$$By construction, this means that all the $c_i$'s vanish.