My attempt at computing transformation of Christoffel symbols under frame transformation Let $\triangledown$ be a connection on $M$. Suppose we are given two local frames $\{E_i\}$ and $\{\tilde{E}_i\}$ on an open subset $U \subset M$, related by $\tilde{E}_i=A_i^jE_j$ for some matrix of functions $(A_i^j)$. Let $\Gamma_{i,j}^k$ and $\tilde{\Gamma}_{i,j}^k$ denote the Christoffel symbols of $\triangledown$ with respect to these two frames. Compute the law expressing $\Gamma_{i,j}^k$ in terms of $\tilde{\Gamma}_{i,j}^k$ and $A_i^j$.

My attempt: I just looked for some relationship through evaluating the connection on basis elements on each frame.
$\Sigma_{k=1}^n\tilde{\Gamma}_{i,j}^k=\triangledown_{\tilde{E}_i}\tilde{E}_j=\triangledown_{A_i^zE_z}A_j^wE_w$ where we are using Einstein summation notation.
Carrying on, we have:
$\triangledown_{A_i^zE_z}A_j^wE_w=A_i^z\triangledown_{E_z}A_j^wE_w=A_i^z(E_zA_j^w)E_w+A_j^w\triangledown_{E_z}E_w=\tilde{E_i}\tilde{E_j}+A_j^w\triangledown_{E_z}E_w=\tilde{E_i}\tilde{E_j}+A_j^w(\Sigma_{k=1}^n\Gamma_{z,w}^k)$
Thus, by relabeling, we have:
$\Sigma_{k=1}^n\tilde{\Gamma}_{i,j}^k=\tilde{E_i}\tilde{E_j}+A_j^w(\Sigma_{k=1}^n\Gamma_{z,w}^k)$
This isn't right.. I must have messed up my summations somewhere.. Does anybody see where my error is?
Also, this probably a very inefficient method anyway, as I'm not relating one Christoffel symbol to another, but rather a summatino of them in one frame to a summation in the other frame.
Help appreciated!! Thanks!
 A: It seems that the problem is that you don't use the product rule correctly, and also what you claim to be computing isn't exactly what you actually compute. 
Using the definition of the Christoffel symbols and the relation between the frames, we have
$$ \nabla_{\tilde{E}_i} \tilde{E}_j = \tilde{\Gamma}_{ij}^k \tilde{E}_k = \tilde{\Gamma}_{ij}^k A_k^l E_l  $$
(which is not what you wrote in the question body). 
On the other hand, using the relation between the frames and the product rule, we have
$$ \nabla_{\tilde{E}_i} \tilde{E}_j = \nabla_{A^z_i E_z} \left( A^w_j E_w \right) = A_i^z \left( \nabla_{E_z} \left( A^w_j E_w \right) \right) = A_i^z \left( \left( E_z A^w_j \right)E_w + A^w_j \nabla_{E_z} E_w \right) \\
=  A_i^z \left( \left( E_z A^w_j \right)E_w + A^w_j \Gamma_{zw}^k E_k\right).  $$
Relabeling the indices and cleaning up, we have
$$ \nabla_{\tilde{E}_i} \tilde{E}_j = \left( A_i^z \left( E_z A^l_j \right) +  A_i^z A^w_j \Gamma^l_{zw} \right) E_l. $$
Comparing both expressions, we get
$$ \tilde{\Gamma}_{ij}^k A_k^l = \left( A_i^z \left( E_z A^l_j \right) +  A_i^z A^w_j \Gamma^l_{zw} \right). $$
Multiplying both sides by $\left( A^{-1} \right)^r_l$ the left side becomes
$$ \tilde{\Gamma}_{ij}^k A_k^l \left( A^{-1} \right)^r_l = \tilde{\Gamma}_{ij}^k \delta_k^r = \tilde{\Gamma}_{ij}^r $$
while the right hand side is
$$ \left( A_i^z \left( E_z A^l_j \right) +  A_i^z A^w_j \Gamma^l_{zw} \right)\left( A^{-1} \right)^r_l$$
and so you get
$$ \tilde{\Gamma}_{ij}^r = \left( A_i^z \left( E_z A^l_j \right) +  A_i^z A^w_j \Gamma^l_{zw} \right)\left( A^{-1} \right)^r_l. $$
You can clean up the right hand side a little but this already gives you the transformation rule between the Christoffel symbols.
