$\newcommand{\al}{\alpha}$ $\newcommand{\euc}{\mathcal{e}}$ $\newcommand{\Cof}{\operatorname{Cof}}$ $\newcommand{\Det}{\operatorname{Det}}$

Let $M,N$ be smooth $n$-dimensional Riemannian manifolds (perhaps with smooth boundary), and let $\, f:M \to N$ be a smooth **immersion**. Let $\Omega^k(M,f^*TN)$ be the space of $f^*TN$-valued $k$-forms.

Let $d:\Omega^k(M,f^*TN) \to \Omega^{k+1}(M,f^*TN)$ be the covariant exterior derivative associated with the pullback connection of the Levi-Civita connection on $N$ (via $f$), and let $\delta$ be its adjoint.

Denote $\delta_{k}:\Omega^k(M,f^*TN) \to \Omega^{k-1}(M,f^*TN)$.

**Question:** Is $\ker \delta_1$ is infinite dimensional?

Since $\delta=\star d \star$ (up to sign), $\dim(\ker \delta_1)=\dim(\ker d_{n-1})$. So, this is a question about $d$.

**Edit:**

In the special case where $N$ is flat, $\delta_{k} \circ \delta_{k+1}=0$, so $\operatorname{Image}(\delta_{k+1}) \subseteq \ker(\delta_k)$.

In particulr, $\operatorname{Image}(\delta_{2}) \subseteq \ker(\delta_1)$.

Is $\operatorname{Image}(\delta_{2})$ is infinite-dimensional? Can we at least something when $N=\mathbb{R}^n$? or even when $M=N=\mathbb{R}^n$?

Let's see what happens when $M=N=\mathbb{R}^n,f=\operatorname{Id}$:

$$ d_{n-1}(\sigma)(e_1,...,e_n)=\sum_{j=1}^{d} (-1)^{j+1} \nabla^{T\mathbb{R}^n}_{e_j} \big( \sigma(e_1,...,\hat{e_j},...,e_d) \big)$$

Write $\sigma=(-1)^{j+1}a_j^idx^1 \wedge \dots \wedge \hat{dx^j} \wedge \dots \wedge dx^d \otimes e_i, a_j^i \in C^{\infty}(\mathbb{R}^n)$.

Then $ \sigma(e_1,...,\hat{e_j},...,e_d)=(-1)^{j+1}a_j^i e_i $, so

$$ \nabla^{T\mathbb{R}^n}_{e_j} \big( \sigma(e_1,...,\hat{e_j},...,e_d) \big)= \nabla^{T\mathbb{R}^n}_{e_j}((-1)^{j+1}a_j^i e_i)=(-1)^{j+1}\frac{\partial a_j^i}{\partial x_j}e_i.$$

Thus, $$\sigma \in \ker d_{n-1} \iff \frac{\partial a_j^i}{\partial x_j}e_i=0 $$ (in the last term there is a double summation, on $i,j$).

Since $e_i$ are independent, this is equivalent to $$\sum_{j=1}^d \frac{\partial a_j^i}{\partial x_j}=0 \, \text{ for all } \, i=1,...,d $$

Denoting $\bar a^i=(a_1^i,...,a_d^i)$, we get that $\operatorname{div} (\bar a^i)=0$.

I guess it shouldn't be too hard to see now that $\dim(\ker d_{n-1}) = \infty$. (By taking the $a_j^i$ to be constants, one immediately gets $\dim(\ker d_{n-1}) \ge n^2$. Since the condition on the divergence do not touch $\frac{\partial a_j^i}{\partial x_k}$ for $k \neq j$, it seems plausible that the dimension is indeed infinite. Perhaps someone can come with a slick argument to show this.)

The next case we should try is $M=N=\mathbb{R}^d$, and $f$ an arbitrary mapping....

**Comment:**

I know that $\operatorname{Cof}(df) \in \ker \delta_1$, where $\operatorname{Cof}(df)$ is the corresponding *cofactor map* of $df$:
$$ \Cof df= (-1)^{d-1} \star_{f^*TN}^{n-1} (\wedge^{n-1} df) \star_{TM}^1. $$

Since $f$ is an immersion, $\operatorname{Cof}(df) \neq 0$, so $\dim (\ker \delta_1 ) \ge 1$.

For my purposes, It would suffice to know that $\ker \delta_1$ always contains elements which are linearly independent of $\Cof df$.