Index notation: associative property

Question

I have given the following term, that I can write in index notation as following:
$$ (\mathbf{a} \cdot \nabla)\mathbf{b} = a_j \partial_j b_i$$
Now I can exchange the order and get
$$ \partial_j b_i a_j = (\partial_j b_i) a_j $$
which equals to
$$ (\partial_j b_i) a_j = (\nabla \mathbf{b})\cdot\mathbf{a}$$
Is this enough to show that
$$ (\mathbf{a} \cdot \nabla)\mathbf{b} = (\nabla \mathbf{b})\cdot\mathbf{a}$$?

Answer 1

In 7. f) of your source there is a remark noting that $B\cdot a \neq a\cdot B$, when $B$ is a 2-tensor.

This is because $a\cdot B$ is the contraction of $a$ with the first index of $B$, while $B\cdot a$ is to be interpreted as contraction of $a$ with the second index of $B$.

Hence in the present case we have $$(\nabla b \cdot a)_i = (\nabla b)_{ij}a_j = \partial_ib_ja_j$$ c.f. with your equation $(\nabla b\cdot a)_i=\partial_jb_ia_j$, which should actually be $(a\cdot\nabla b)_i=a_j\partial_jb_i=\partial_jb_ia_j$.

Thus, the identity you actually derived is $$(a\cdot \nabla)b = a\cdot (\nabla b)$$