Changing Coefficients of Covariate derivative

Given an affine connection $$\nabla_X$$ on some vector field $$X,Y \in \mathcal{V}(M)$$ I want to compute its form with local coefficients $$\Gamma^k_{ij}$$. I.e.

$$\nabla_X(Y)|_U = \sum_{i=1}^{n} a_i \nabla_{\frac{\partial}{\partial x_i}}(\sum_{j=1}^{n}b_j \frac{\partial}{\partial x_j}) = \sum_{i,j = 1}^{n} [a_i b_j \nabla_{\frac{\partial}{\partial x_i}}(\frac{\partial}{\partial x_j})+a_i \frac{\partial b_j}{\partial x_i}\frac{\partial }{\partial x_j}]$$

I know, that $$\nabla_{\frac{\partial}{\partial x_i}}(\frac{\partial}{\partial x_j}) = \sum_{k=1}^{n} \Gamma_{ij}^k \frac{\partial}{\partial x_k}$$ on the open subspace $$(U,\varphi) \in M$$. Given another open subspace $$(V,\phi)$$, with coordinates $$y_1, \ldots y_n$$. Given the closed form of the covariate derivative on $$V$$

$$\nabla_{\frac{\partial}{\partial y_r}}(\frac{\partial}{\partial y_s}) = \sum_{t=1}^{n} \tilde\Gamma_{rs}^t \frac{\partial}{\partial y_t}$$

I want to show that I can express the local coefficients by the following expression:

$$\tilde\Gamma_{rs}^t = \sum_{i,j,k=1}^{n} \frac{\partial x_i}{\partial y_r}\frac{\partial x_j}{\partial y_s}\frac{\partial y_t}{\partial x_k} \Gamma_{ij}^k + \sum_{j=1}^{n} \frac{\partial^2 x_j}{\partial y_r \partial y_s}\frac{\partial y_t}{\partial x_j}$$

We can use the fact that $$\frac{\partial }{\partial y_r} = \sum_i \frac{\partial x_i}{\partial y_r} \frac{\partial }{\partial x_i}.$$

Then we can use your expression for $$\nabla_X(Y)$$, with $$X = \sum_i a_i \frac{\partial }{\partial x_i}, \ \ \ \ \ a_i := \frac{\partial x_i}{\partial y_r},$$ $$Y = \sum_j b_j \frac{\partial }{\partial x_j}, \ \ \ \ \ b_j := \frac{\partial x_j}{\partial y_s}.$$

This gives $$\nabla_{\frac{\partial }{\partial y_r}} \left( \frac{\partial}{\partial y_s}\right) = \sum_{i,j,k} \frac{\partial x_i}{\partial y_r}\frac{\partial x_j}{\partial y_s}\Gamma^k_{ij} \frac{\partial}{\partial x_k} + \sum_{i,j} \frac{\partial x_i}{\partial y_r} \frac{\partial}{\partial x_i} \left( \frac{\partial x_j}{\partial y_s} \right)\frac{\partial}{\partial x_j}.$$

By the chain rule, $$\sum_{i} \frac{\partial x_i}{\partial y_r} \frac{\partial}{\partial x_i} f = \frac{\partial f}{\partial y_r}$$ for any function $$f$$, so our above expression simplifies to

$$\nabla_{\frac{\partial }{\partial y_r}} \left( \frac{\partial}{\partial y_s}\right) = \sum_{i,j,k} \frac{\partial x_i}{\partial y_r}\frac{\partial x_j}{\partial y_s}\Gamma^k_{ij} \frac{\partial}{\partial x_k} + \sum_{j} \frac{\partial^2 x_j}{\partial y_s \partial y_r} \frac{\partial}{\partial x_j}.$$

Finally, we use $$\frac{\partial }{\partial x_k} = \sum_t \frac{\partial y_t}{\partial x_k} \frac{\partial }{\partial y_t}$$ to get $$\nabla_{\frac{\partial }{\partial y_r}} \left( \frac{\partial}{\partial y_s}\right) = \sum_{i,j,k,t} \frac{\partial x_i}{\partial y_r}\frac{\partial x_j}{\partial y_s}\Gamma^k_{ij} \frac{\partial y_t}{\partial x_k} \frac{\partial }{\partial y_t}+ \sum_{j, t} \frac{\partial^2 x_j}{\partial y_s \partial y_r} \frac{\partial y_t}{\partial x_j} \frac{\partial }{\partial y_t}.$$

• Thank you! I couldn't manage to get the terms right. Now I'll exercise again.
– user582360
Mar 8 '19 at 14:29