Is Taylor expansion related to Helmholtz decomposition? The Taylor expansion of a vector field $f(x)$ to the order of one is
$$f(x)=f(x_0)+Jf(x_0)\cdot\Delta x+o(\Delta x)$$
where $Jf$ is Jacobian of the vector field and $\Delta x=x-x_0$. 
Suppose we decompose it into symmetric and skew-symmetric part such that
$$S=\frac{Jf+(Jf)^T}{2}$$
$$A=\frac{Jf-(Jf)^T}{2}$$
Then
$$f(x)\approx f(x_0)+S\Delta x+A\Delta x$$
Could we regard $f(x_0)$ as a translation, $S\Delta x$ as an expansion or a contradiction along axes of eigen directions for it resembles strain tensor and $A\Delta x$ a rotation since skew-symmetric matrix is infinitesimal of rotation matrix?
If it is correct, I'm a bit confused with, say operator $A$, what object it acts on?
Otherwise, does this decomposition have any practical meaning in terms of vector field or velocity field?
 A: you ask whether the decomposition of the Jacobi tensor $J{f}$ into symmetric and antisymmetric parts $S$ and $A$ produces a Helmholtz decomposition of $\Delta f=J{f}\cdot\Delta{x}$ into a curl-free part $\chi=S\cdot\Delta x$ and a divergence-free part $\omega=A\cdot\Delta x$.
this is indeed correct, I will try to write it out explicitly (trying to avoid the errors I made in my first answer)
notation: the subscript $0$ indicates the dependence on $x_0$ and derivatives with respect to $x_0$; I abbreviate $\xi=\Delta x=x-x_0$ and denote by $\nabla_\xi$ the derivative with respect to $\xi$ (or $x$).
$${\omega}(\xi)={A}(x_0)\cdot\xi=\tfrac{1}{2}\xi\cdot(\nabla_0 f_0)-\tfrac{1}{2}(\nabla_0 f_0)\cdot\xi=\tfrac{1}{2}(\nabla_0\times{f}_0)\times\xi$$
$$\Rightarrow\nabla_\xi\cdot{\omega}(\xi)=-\tfrac{1}{2}(\nabla_0\times f_0)\cdot(\nabla_\xi\times\xi)=0$$
$$\chi(\xi)=S(x_0)\cdot\xi=\tfrac{1}{2}(\nabla_0 f_0)\cdot\xi+\tfrac{1}{2}\xi\cdot(\nabla_0 f_0)$$
$$\Rightarrow\nabla_\xi\times\chi(\xi)=-\tfrac{1}{2}\nabla_0\times f_0+\tfrac{1}{2}\nabla_0\times f_0=0$$

vector identities used:
$$a\times(b\times c)=b(a\cdot c)-c(a\cdot b),\;\;
a\cdot(b\times c)=c\cdot(a\times b)$$
