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Consider a system given by the action:

$$S = \int \sum_{i,j} G_{ij}(q) \dot{q}^i \dot{q}^j dt$$

Now, consider the quantity $Q_v = \sum_{i,j} G_{ij} v^i \dot{q}^j $ with $v^i$ the Killing vector.

I want to show that $Q_v$ is conserved so that $\frac{dQ_v}{dt} = 0 $.

How can one show this by using this relation of the Killing vector: $$\sum_{i} (\partial_i G_{jk} v^i + G_{ij} \partial_k v^i + G_{ki} \partial_j v^i) = 0 ~?$$

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It is a bland plugin, provided you recall to use the E-L equations of motion, which you skipped, $\frac{d}{dt} (2G_{ij}\dot{q} ^i)= \partial_j G_{ik} \dot q ^i \dot q ^k $. I'm using repeated index summation convation and utilizing the index-symmetry of the relevant metric.

The Killing symmetry condition then yields $$ \dot q ^i \dot q ^k (\partial_j G_{ik} v^j + 2G_{ij} \partial_k v^j) = 0 . $$

It thus follows that the Killing symmetry charge is conserved, $$\frac{dQ_v}{dt}=\frac{d}{dt}\!(G_{ij}\dot q ^i) ~~v^j + G_{ij}\dot q ^i \dot v ^j \\ = \tfrac{1}{2} \partial_j G_{ik} \dot q ^i \dot q ^k v^j + G_{ij}\dot q ^i \dot q ^k \partial_k v ^j= \frac{\dot q ^i \dot q ^k}{2} ( \partial_j G_{ik} v^j + 2G_{ij} \partial_k v ^j) = 0 ~.$$

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