# Connection under the frame change.

In the below picture ,I don't know how to compute the red line 2. But if accept the red line 2, I can compute the red line 3. But there still be a little question, that at last ,I have $$\Gamma_{ib}^aV^b=V^bF_j^a\partial_iF_b^j+...$$ $$\nabla_iV^a=V^aF_j^a\partial_iF_a^j+...$$

The $F_j^a=(F_a^j)^{-1}$, at the right part ,I always have three $a$ , it is not same as above.

I want to know how to compute the red line 2. So thanks for detail answer or hint.

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(1) Fix a coordinate vector field $E_i$ for $g(0)$ For $g(t)$, consider a frame $$F_a=F_a^i(t) E_i$$ s.t. $$g(t)(F_a(t), F_b(t))=\delta_{ab} \ \ast$$ for any $t$

And if we assume that $$\partial_t F_a^i(t)= g^{ij} R_{jk} F^k_a \ \ast\ast$$ so $$\frac{\partial }{\partial_t} [ g(t)(F_a(t), F_b(t)) ]$$ $$= -2Ric(g(t)) (F_a(t),F_b(t)) +g_{im} F_b^m g^{ij} R_{jk} F^k_a + g_{mt} F^m_a g^{tj} R_{jk} F^k_b =0$$

That is, solution in $\ast\ast$ is the desired one, which satisfies $\ast$.

(2) Define $$h:= F^\ast g,\ h_{ij}:= g_{\alpha\beta} F_i^\alpha F^\beta_j$$

That is by (3), $$\Gamma(h)_{ij}^k = \Gamma(g)_{\alpha\gamma}^\tau F^\gamma_jF^\alpha_i F_\tau^k + E_i(F^\alpha_{j}) F_\alpha^k$$ $$= \Gamma(g)_{i\gamma}^\tau F^\gamma_j F_\tau^k + E_i(F^\alpha_{j}) F_\alpha^k$$

So if $V=V^\alpha F_\alpha$, $$\nabla_i V^\gamma= E_i V^\gamma + \Gamma_{i\alpha}^\gamma V^\alpha$$

In further $$\nabla_\beta V^\gamma =F_\beta^i \nabla_i V^\gamma$$

(3) Assume that $f(x)=y$ is a diffeomorphism and $$h:=f^\ast g$$ Define $$g_{\alpha\beta }:= g(\partial_{y_\alpha},\partial_{y_\beta})$$ $$h_{ij}:= h(\partial_{x_i},\partial_{x_j})$$ $$f^\alpha_i:= \frac{\partial y^\alpha}{\partial x_i }$$

So $$\Gamma(h)_{ij}^k = \Gamma(g)_{\alpha\gamma}^\tau f^\gamma_jf^\alpha_i f_\tau^k + f^\alpha_{ij} f_\alpha^k$$