Skip to main content
2 of 2
Replaced \bullet by \cdot, as mentioned in the comments
David Roberts
  • 35.4k
  • 11
  • 124
  • 349

Let ${\bf K}$ be a vector field in the neighbourhood of ${\bf p}\in{\mathbb R}^n$, and let ${\bf X}$ and ${\bf Y}$ be two tangent vectors at ${\bf p}$. These two vectors span a parallelogram $P$ with one vertex at ${\bf p}$. The "circulation" of ${\bf K}$ around $P$ computes to $$ \int_{\partial P}{\bf K}\cdot \mathrm{d}{\bf x}= (L\,{\bf X})\cdot{\bf Y}- (L\,{\bf Y})\cdot{\bf X} + o(|P|^2) $$ with $L:=\mathrm{d}{\bf K}({\bf p})$ and $|P|:= \mathrm{diam}(P)$. It follows that there is a certain skew bilinear function ${\rm Rot}{\bf K}({\bf p}):T_{\bf p}\times T_{\bf p}\to{\mathbb R}$ with $$ \int_{\partial P}{\bf K}\cdot \mathrm{d}{\bf x}={\rm Rot}{\bf K}({\bf p})({\bf X},{\bf Y})+ o(|P|^2) \quad (|P|\to 0). $$ In the case $n=3$ the bilinear form ${\rm Rot}$ can be represented by the vector ${\rm curl}{\bf K}$ in the form $$ {\rm Rot}{\bf K}({\bf p})({\bf X},{\bf Y}) = {\rm curl}{\bf K}({\bf p})\cdot({\bf X}\times{\bf Y}). $$