deriving the Poisson equation from Navier stokes equation using tensor algebra. I want to derive the poisson equation from the navier equations, but I'm not able to do this. 
I'm able to substitute the index form in place of the operators in the poisson equation and get back navier stokes equation but not able to do the other way round, which is indeed asked in the problem. i.e from the index form to the operator form. 

 A: Using vector symbols, incompressible Navier-Stokes equations read
\begin{align}
\frac{\partial\mathbf{v}}{\partial t}+\left(\mathbf{v}\cdot\nabla\right)\mathbf{v}&=-\frac{1}{\rho}\nabla p+\nu\Delta\mathbf{v},\tag{1}\\
\nabla\cdot\mathbf{v}&=0.\tag{2}
\end{align}
Act the divergence operator "$\nabla\cdot$" on both sides of $(1)$. Note that
\begin{align}
\nabla\cdot\frac{\partial\mathbf{v}}{\partial t}&=\frac{\partial}{\partial t}\left(\nabla\cdot\mathbf{v}\right)=0,\\
\nabla\cdot\left(\frac{1}{\rho}\nabla p\right)&=\frac{1}{\rho}\nabla\cdot\nabla p=\frac{1}{\rho}\Delta p,\\
\nabla\cdot\Delta\mathbf{v}&=\Delta\left(\nabla\cdot\mathbf{v}\right)=0,
\end{align}
where we have used $(2)$ repeatedly. Therefore, $(1)$ becomes
$$
\nabla\cdot\left(\left(\mathbf{v}\cdot\nabla\right)\mathbf{v}\right)=-\frac{1}{\rho}\Delta p.\tag{3}
$$
Now, thanks to the vector calculus identity
$$
\left(\mathbf{v}\cdot\nabla\right)\mathbf{v}=\frac{1}{2}\nabla\left\|\mathbf{v}\right\|^2-\mathbf{v}\times\left(\nabla\times\mathbf{v}\right)=\frac{1}{2}\nabla\left\|\mathbf{v}\right\|^2-\mathbf{v}\times\mathbf{w},
$$
as well as the vector calculus identity
$$
\nabla\cdot\left(\mathbf{v}\times\mathbf{w}\right)=\mathbf{w}\cdot\left(\nabla\times\mathbf{v}\right)-\mathbf{v}\cdot\left(\nabla\times\mathbf{w}\right)=\left\|\mathbf{w}\right\|^2-\mathbf{v}\cdot\left(\nabla\times\mathbf{w}\right),
$$
where we have used $\mathbf{w}=\nabla\times\mathbf{v}$ repeatedly, we have
$$
\nabla\cdot\left(\left(\mathbf{v}\cdot\nabla\right)\mathbf{v}\right)=\frac{1}{2}\Delta\left\|\mathbf{v}\right\|^2-\left\|\mathbf{w}\right\|^2+\mathbf{v}\cdot\left(\nabla\times\mathbf{w}\right).\tag{4}
$$
Finally, combine $(3)$ and $(4)$, and we eventually obtain Poisson's equation
$$
\frac{1}{\rho}\Delta\left(p+\frac{1}{2}\rho\left\|\mathbf{v}\right\|^2\right)=\left\|\mathbf{w}\right\|^2-\mathbf{v}\cdot\left(\nabla\times\mathbf{w}\right),
$$
as is expected.
