Proving $ vec{V} cdot (vec{nabla}vec{V}) = (vec{nabla}cdotvec{V})vec{V} $ using index notation

The space that these vectors live in has a metric $g_{ij}$. For example, if you're in Euclidean space and you're using Cartesian coordinates, then $g_{ij}$ is equal to the Kronecker delta $delta_{ij}$. If you're using non-Cartesian coordinates (e.g. polar coordinates) or you're working in non-Euclidean space, then your metric will be different. The index notation expression for a vector is $v^i$. The dot product between two vectors is represented by:

$$vec{a}cdotvec{b}=a^ig_{ij}b^j=a^ib_i$$

This is, in fact, the definition of a metric - it tells you the "distance" between the tips of two vectors. The "lowered index" vector $b_i$ is defined straightforwardly, as long as you know what your metric is:

$$b_i=g_{ij}b^j$$

If you're in Euclidean space and you're using Cartesian coordinates, then we conveniently have that $b_i=b^i$, since $g_{ij}=delta_{ij}$. With any other metric, this is not true. For example, in 2D polar coordinates (where $vec{b}=b^rhat{r}+b^thetahat{theta}$), our metric is defined by $g_{rr}=1$ and $g_{thetatheta}=r^2$, with the other two elements zero. In that case, we have that $b_r=g_{rr}b^r+g_{rtheta}b^{theta}$, so $b_r=b^r$, but $b_theta=g_{theta r}b^r+g_{theta theta}b^{theta}$, so $b_theta=r^2b^{theta}$. But as long as you know what your metric is, lowering the index of a vector should be straightforward.

For the rest of this discussion, let's assume that you're working in Euclidean space, since the differential geometry in non-Euclidean space gets complicated once you start taking derivatives.

The derivative operator $vec{nabla}$ is notated as $partial^i$, which is shorthand for $frac{partial}{partial x^i}$. The dyadic product $vec{nabla}vec{V}$ is therefore notated as $partial^j v^i$, which is shorthand for $frac{partial v^i}{partial x^j}$.Putting all this together, the expression you need to prove is written as:

$$v^ig_{ij}partial^kv^j=partial^ig_{ij}v^jv^k$$

or, lowering the indices:

$$v_jpartial^kv^j=partial_j v^jv^k$$

This should be enough information to start the proof.

$defvv{{bf v}} defdel{nabla} defo{cdot} defpd{partial}$Note that $$[vvo(delvv)]_j = v_i [delvv]_{ij} = v_i(pd_i v_j)$$ and
$$[(delovv)vv]_j = (pd_i v_i)v_j.$$ But that $$v_i(pd_i v_j)ne (pd_i v_i)v_j,$$ in general. (Repeated indices are to be summed over. This is Einstein's summation notation.)

For clarity let $vv$ be two dimensional. For $j=1$ the claim is that $$v_1(pd_1 v_1) + v_2(pd_2 v_1) = (pd_1 v_1+pd_2 v_2)v_1.$$ Clearly this is false. For example, if $vv=[x,y]^T$ this implies that $$x = 2x.$$

The intended claim is likely that $$vvo(delvv) = (vvodel)vv.$$ (Note that $delovv$ and $vvodel$ are completely different objects. The first is a scalar. The second is a scalar differential operator.) This result can be easily proved, $$[vvo(delvv)]_j = v_i [delvv]_{ij} = v_i(pd_i v_j) = (v_ipd_i) v_j = [(vvodel)vv]_j.$$