Evaluate $\mathbf{u}\cdot\nabla\mathbf{u}$ (the directional derivative of $\mathbf{u}$ in the direction of $\mathbf{u}$)in cylindrical coordinates $(r, \phi,z)$, where $\bf{u}=e_{\phi}$.
The textbook that I am reading uses a purely vectorial approach and $$\mathbf{u}\cdot\nabla\mathbf{u}=-\frac{\mathbf{e_{r}}}{r} \tag{1}$$
I tried the index notation method as an alternative: $$[\mathbf{u}\cdot\nabla\mathbf{u}]_k=u^{n}\nabla_{n}u_{k}=g^{nr}u_{r}\nabla_{n}u_k$$$$=g^{nr}u_r(\partial_{n}u_{k}-\Gamma^{p}_{kn}u_p)= g^{nr}u_r\partial_n u^k- u_r u_p g^{nr} g^{pq} \Gamma_{kn,q}\tag{2}$$ $n,r,k,p=1,2,3$ and $\nabla$ is the covariant derivative.
Since $\mathbf{e_{\phi}}=\mathbf(0,1,0)$ in the cylindrical basis, the first term in the last equality of $(2)$ vanishes (i.e. $\partial_nu^k=0$), then $$[\mathbf{u}\cdot\nabla\mathbf{u}]_{k}=-g^{nr}u_r(\Gamma^p_{kn}u_p)=-u_2u_2g^{22}g^{22}\Gamma_{k2,2}=-(1)^2\frac{1}{(g_{22})^2}\Gamma_{k2,2}\tag{3}$$
Since $$\Gamma_{km,n}=\frac{1}{2}\left(g_{kn,m}+g_{mn,k}-g_{km,n} \right) \tag{4}$$
$(g_{kn,m}=\partial_{m} g_{kn})$
and $$g_{11}=g_{33}=1,g_{22}=r^2 \tag{5}$$ it follows that $$\Gamma_{12,2}=\frac{1}{2}(g_{22,1})=r, \Gamma_{2,22}=\Gamma_{3,22}=0 \tag{6}$$ so only $[\mathbf{u}\cdot\nabla\mathbf{u}]_1$ (the $r$ component) is non-zero $(6)$$\to$ $(3)$ gives $$[\mathbf{u}\cdot\nabla\mathbf{u}]=-\frac{1}{r^3} \mathbf{e_{r}} \tag{7}$$ which is clearly wrong. Can someone please explain where my conceptual errors lie?
$\bf{EDIT}$
I now understand that my errors are due to the scale factors $h_{J}$ in $\mathbf{e}_{j}=\partial_{j}/h_{J}$
But with $$(u^n\nabla_{n}u^k)\partial_k=(u^n\partial_nu^k+\frac{1}{2}u^nu^pg^{kq}(g_{pq,n}+g_{nq,p}-g_{pn,q}))\partial_{k}=\frac{1}{r}\partial_{\phi}\left(\frac{1}{r}\right)\partial_{\phi}+\frac{1}{2}\left(\frac{1}{r}\right)^2g^{\phi \phi}(g_{\phi \phi ,\phi}+g_{\phi \phi ,\phi}-g_{\phi \phi, \phi})\partial_{\phi}$$ ($u^{2}=u^{\phi}=\frac{1}{r}\partial_{\phi}$ as the only non-zero components) which has $\partial_{\phi} \to \mathbf{e}_{\phi}$ as the basis vector, not $\mathbf{e}_{r}$, which is the correct basis vector. What went wrong?
We have $\mathbf{u} = \mathbf{e}_{\phi} = \dfrac{1}{r} \dfrac{\partial}{\partial \phi}$, so the components of the vector field $\mathbf{u}$ in cylindrical coordinates basis $\left\{\dfrac{\partial}{\partial r},\dfrac{\partial}{\partial \phi}, \dfrac{\partial}{\partial z} \right\}$ are \begin{align} \begin{cases} u^r &= 0 \\ u^{\phi} &= \frac{1}{r}\\ u^z &= 0 \end{cases} \end{align} Henceforth, whenever I refer to components, I mean with respect to the above basis. The Christoffel symbols (of the Levi-Civita connection defined on $\Bbb{R}^3$ with the "standard" metric tensor field) are given as \begin{align} \Gamma^a_{bc} &= \dfrac{1}{2}g^{as}\left(\dfrac{\partial g_{sb}}{\partial x^c} + \dfrac{\partial g_{cs}}{\partial x^b} - \dfrac{\partial g_{bc}}{\partial x^s} \right), \end{align} and in cylindrical coordinates, the only non-zero components are: \begin{align} \Gamma_{\phi\phi}^r &= -r \quad \Gamma^{\phi}_{r\phi} =\Gamma^{\phi}_{\phi r} = \dfrac{1}{r} \end{align} To prove this, just use the fact that $g_{rr} = 1, g_{\phi\phi} = r^2, g_{zz} = 1$, and all other components of the metric are $0$, and then carefully resolve the Einstein summation convention.
Now, the directional derivative you're looking for is $\nabla_{\mathbf{u}}\mathbf{u}$, which we calculate as: \begin{align} \nabla_{\mathbf{u}}\mathbf{u} &= \left(u^n \dfrac{\partial u^k}{\partial x^n} + u^n u^p \Gamma_{np}^k\right)\dfrac{\partial}{\partial x^k} \end{align} Now, most of these terms vanish, because $u^r = u^z = 0$. So, in the summation, we can set $n=p = 2$ (i.e equal to the $\phi$ component) \begin{align} \nabla_{\mathbf{u}}\mathbf{u} &= \left(u^{\phi} \dfrac{\partial u^{k}}{\partial \phi} + u^{\phi} u^{\phi} \Gamma_{\phi\phi}^k\right)\dfrac{\partial}{\partial x^k} \end{align} Now, the components of $u$ do not depend on $\phi$, so the first term disappears. Finally, note that $\Gamma_{\phi\phi}^r = -r$ is the only non-zero term left. Hence, everything collapses to: \begin{align} \nabla_{\mathbf{u}}\mathbf{u} &= \left(u^{\phi}u^{\phi}\Gamma_{\phi\phi}^r\right)\dfrac{\partial}{\partial r} = \left(\dfrac{1}{r}\right)^2(-r)\dfrac{\partial}{\partial r} = -\dfrac{\mathbf{e}_r}{r}. \end{align}
I just re-read your edit, and you definitely made an error in carrying out the Einstein summation convention, because like I showed in my answer, the entire sum has to reduce to $\nabla_{\mathbf{u}}\mathbf{u} = \left(u^{\phi}u^{\phi}\Gamma_{\phi\phi}^r\right)\frac{\partial}{\partial r}$.
Also, if you carefully evaluate what you wrote on the RHS of your edit, you actually "proved" that $\nabla_{\mathbf{u}}\mathbf{u} = 0$ (all the various partial derivatives you wrote on the RHS are actually $0$). So, just carefully go over your summation and see which ones are non-zero.