Difference between $\mathrm{rot}$ and $\mathrm{Rot}$ (Curl) I'm not sure whether you've heard of that difference at all and if it's of importance in English notations. In German at least you define:
$$\mathrm{rot\,\textbf{V}} = \nabla \times \textbf{V}, \quad\text{where $\textbf{V}$ is a continuously differentiable Vectorfield $\mathbb{R}^3\rightarrow\mathbb{R}^3$} $$
and at the same way
$$\mathrm{Rot}\textbf{V} = \frac{1}{2}\,\left(\textbf{V'-V'$^T$}\right),\quad \text{where $\textbf{V'}$ should equal $\mathrm{Jacobian(\textbf{V})}$}$$
I'm just aware of: $\mathrm{rot}$ describes a vector, $\mathrm{Rot}$ a matrix.  What's the matter?
 A: $\newcommand{\Vec}[1]{\mathbf{#1}}\newcommand{\Reals}{\mathbf{R}}\newcommand{\Neg}{\phantom{-}}\DeclareMathOperator{\Rot}{Rot}\DeclareMathOperator{\rot}{rot}$There's a vector space isomorphism between $\Reals^{3}$ and the set of skew-symmetric real $3 \times 3$ matrices given by
$$
\phi\left[\begin{array}{@{}c@{}} a_{1} \\ a_{2} \\ a_{3} \\ \end{array}\right]
= \left[\begin{array}{@{}ccc@{}}
    \Neg0 & -a_{3} & \Neg a_{2} \\
    \Neg a_{3} & \Neg0 & -a_{1} \\
    -a_{2} & \Neg a_{1} & \Neg 0 \\
  \end{array}\right].
$$
Under this isomorphism, the cross product corresponds to matrix multiplication in the sense that
$$
\Vec{a} \times \Vec{v} = \phi(\Vec{a})\Vec{v}
\quad\text{for all $\Vec{v}$ in $\Reals^{3}$.}
$$
It turns out that $\phi(\rot\Vec{V}) = \Rot \Vec{V}$.

To establish this claim, let $\Vec{V}$ be a continuously-differentiable vector field and expand to first order at an arbitrary point:
$$
\Vec{V}(\Vec{x}) = \Vec{V}(\Vec{x}_{0}) + D\Vec{V}(\Vec{x}_{0})(\Vec{x} - \Vec{x}_{0}) + o(\|\Vec{x} - \Vec{x}_{0}\|).
$$
If we write the Jacobian
$$
D\Vec{V}(\Vec{x}_{0})
= \left[\begin{array}{@{}ccc@{}}
    a_{1} & b_{1} & c_{1} \\
    a_{2} & b_{2} & c_{2} \\
    a_{3} & b_{3} & c_{3} \\
  \end{array}\right],
$$
then by definition
\begin{align*}
  \Rot\Vec{V}(\Vec{x}_{0})
  &= \tfrac{1}{2}\bigl[D\Vec{V}(\Vec{x}_{0}) - D\Vec{V}(\Vec{x}_{0})^{\mathsf{T}}\bigr] \\
  &= \left[\begin{array}{@{}ccc@{}}
      0 & b_{1} - a_{2} & c_{1} - a_{3} \\
      a_{2} - b_{1} & 0 & c_{2} - b_{3} \\
      a_{3} - c_{1} & b_{3} - c_{2} & 0 \\
    \end{array}\right].
\end{align*}
A short calculation gives
$$
\rot \Vec{V}(\Vec{x}_{0}) = (\nabla \times \Vec{V})(\Vec{x}_{0})
= \left[\begin{array}{@{}c@{}}
    b_{3} - c_{2} \\
    c_{1} - a_{3} \\
    a_{2} - b_{1} \\
  \end{array}\right],
$$
so we have $\phi(\rot \Vec{V})(\Vec{x}_{0}) = \Rot \Vec{V}(\Vec{x}_{0})$.
This, incidentally, is how the curl quantifies rotation in the flow of $\Vec{V}$. If we decompose $A := D\Vec{V}(\Vec{x}_{0})$ as a sum of a skew-symmetric matrix $R$, a traceless symmetric matrix $S_{0}$, and a scalar matrix $T$ (multiple of the identity), then to first order in $t$, $\exp(tA) = \exp(tR) \exp(tS_{0}) \exp(tT)$. The respective factors on the right are a rotation (the curl), a volume-preserving deformation, and a scaling (the divergence).
