If $\Omega\subseteq\mathbb R^3$ is open and $f:\Omega\to\mathbb R^3$ is differentiable, there is the identity $$(f\cdot\nabla)f=\frac12\nabla|f|^2-f\times\operatorname{curl}f\tag1,$$ where $$\operatorname{curl}f:=\nabla\times f.$$
Is there an analogue of $(1)$ in the two-dimensional case?
I think we could rewrite $(1)$ using the alternating tensor product $\nabla\otimes f-f\otimes\nabla$ in place of $\operatorname{curl}f$ and obtain a natural analogue in this way. In any case, I'd really like to know whether $(1)$ can be written in this way.