# Infinitesimal rotation of a vector field

My task is to show that for infinitesimal rotation $$R$$ around an axis, denoted together by d$$\boldsymbol{\phi}$$, the vector field $$\boldsymbol{A}(\boldsymbol{r})$$ transforms as

$$\boldsymbol{A}'(\boldsymbol{r}) = R\boldsymbol{A}(R^{-1}\boldsymbol{r}) = (1 - i(\boldsymbol{l} + \boldsymbol{s})\cdot \mathrm{d}\boldsymbol{\phi})\boldsymbol{A}(\boldsymbol{r}).$$

So I first shown that an arbitrary vector transforms under $$R$$ as $$v_i\mapsto v'_i = v_i + \varepsilon_{ijk}\mathrm{d}\phi_jr_k$$ and I define $$M_{ik} := \varepsilon_{ijk}\mathrm{d}\phi_j$$, which coincides with the known $$\boldsymbol{v} \mapsto \boldsymbol{v} + \mathrm{d}\boldsymbol{\phi} \times \boldsymbol{v}$$. Then $$\boldsymbol{A}$$ transforms as

$$\boldsymbol{A}'(\boldsymbol{r}) = \boldsymbol{A}(\boldsymbol{r} - d\boldsymbol{\phi}\times\boldsymbol{r}) + d\boldsymbol{\phi}\times\boldsymbol{A}(\boldsymbol{r} - d\boldsymbol{\phi}\times\boldsymbol{r}),$$

which is in first order in Taylor expansion approximately

$$\boldsymbol{A}'(\boldsymbol{r}) = \boldsymbol{A}(\boldsymbol{r}) - D_{\boldsymbol{r}}\boldsymbol{A} \cdot (d\boldsymbol{\phi}\times \boldsymbol{r}) + d\boldsymbol{\phi}\times\boldsymbol{A}(\boldsymbol{r}),$$

where $$D_{\boldsymbol{r}}\boldsymbol{A}$$ is the Jacobian at $$\boldsymbol{r}$$. And in terms of indices:

$$A'_i = A_i - \partial_j A_i (d\boldsymbol{\phi}\times \boldsymbol{r})_j + d\phi_j\varepsilon_{ijk}A_k = A_i - \partial_j A_i \varepsilon_{jlk} d\phi_l r_k + d\phi_j\varepsilon_{ijk}A_k.$$

Now I have trouble putting this result in the form $$A'_i = (\dots)_{ij}A_j$$

Assuming you're right, a little index relabelling helps:\begin{align}A_i^\prime&=A_i-\epsilon_{mlk}d\phi_l r_k\partial_mA_i+d\phi_k\epsilon_{ikj}A_j\\&=(\delta_{ij}(1-\epsilon_{mlk}d\phi_lr_k\partial_m)+d\phi_k\epsilon_{ikj})A_j.\end{align}In these three terms, the second (third) has been relabelled with $$j\leftrightarrow m$$ ($$j\leftrightarrow k$$).