Consider the inverse Fourier transform
$\mathbf f(x)=\int\frac{dk}{2\pi}\mathbf{\hat f}(k)e^{ik\cdot x}$
where $\{x,k\}\subset\mathbb{R}$, and $\mathbf{\hat f}$ (a (0, -1) Fourier transform) is valued in degree-1 elements of the exterior algebra of the complex free vector space of $\mathbb R$, $\mathbf{\hat f}:\mathbb R\to \Lambda^1 \mathbb{C}^\mathbb{R}$, with generating elements of $\Lambda\,\mathbb{C}^\mathbb{R}$ are notated $\theta_{k}$. Importantly, restrict $\mathbf{\hat f}$ to the form $\mathbf{\hat f}(k)=\hat{f}(k)\theta_{k}$ for any $\hat{f}:\mathbb{R}\to\mathbb{C}$.
Now, the product of two independent, single-mode functions $\mathbf{f}'(x)$ and $\mathbf{f}''(x)$ (k-space deltas, where $\mathbf{\hat f}'(k)=\delta(k-k')\theta_{k}$ and $\mathbf{\hat f}''(k)=\delta(k-k'')\theta_{k}$ for some $k'\ne k''$) is a non-zero 2-vector proportional to $\theta_{k'}\theta_{k''}$.
Question: What about the product of two x-space deltas (where $\mathbf{\hat f}'(k)=e^{ik\cdot x'}\theta_{k}$ and $\mathbf{\hat f}''(k)=e^{ik\cdot x''}\theta_{k}$)? Does the Fourier transform change basis to yield a non-zero 2-vector proportional to $\theta_{x'}\theta_{x''}$? (Taking a linear combination of $\theta_{k}$ to a linear combination of $\theta_{x}$?) A change of basis seems possible for a finite-dimensional exterior algebra (e.g. $\Lambda\,\mathbb{R}^3$), at the very least.
Or do $\mathbf{f}'$ & $\mathbf{f}''$ pick up enough $\theta_k$ such that $\mathbf{f}'(x)\cdot\mathbf{f}''(x)=0$ everywhere?