1
$\begingroup$

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:

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ Thank you! I couldn't manage to get the terms right. Now I'll exercise again. $\endgroup$
    – user582360
    Mar 8 '19 at 14:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy