I Christoffel symbol ("undotting")

kiuhnm

I hope you can understand my notation. The Christoffel symbol can be defined through the relation$$\frac{\partial \pmb{Z}_i} {\partial Z^k} = \Gamma_{ik}^j \pmb{Z}_j$$ I can solve for the Christoffel symbol this way: $$\frac{\partial \pmb{Z}_i} {\partial Z^k} \cdot \pmb{Z}^m = \Gamma_{ik}^j \pmb{Z}_j \cdot \pmb{Z}^m = \Gamma_{ik}^j \delta^m_j = \Gamma_{ik}^m$$
This might be a stupid question, but how can I go from the last relation back to the first one? Probably by multiplying by $\pmb{Z}_m$: $$\left( \frac{\partial \pmb{Z}_i} {\partial Z^k} \cdot \pmb{Z}^m \right) \pmb{Z}_m = \ldots\$$ Now, either I need to rewrite the dot product more explicitly or there's some propriety I don't know of.

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Orodruin

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For any vector $\vec v$, it holds that if you expand it in terms of $\vec Z_m$, you have the components $v^m$ as
$$\vec v = v^m \vec Z_m.$$
Multiplying by $\vec Z^i$ gives you
$$\vec Z^i \cdot \vec v = v^m \vec Z^i \cdot \vec Z_ m = v^m \delta^i_m = v^i$$
and therefore
$$\vec v = (\vec Z^m \cdot \vec v) \vec Z_m.$$
So yes, you regain the first equation by multiplying the second with $\vec Z_m$.

kiuhnm

Ah, yes. After all that's exactly what the Christoffel symbol means: it "contains" the components of all the n^2 vectors wrt the covariant basis.
In hindsight it really was a dumb question! Thank you.

META: how should I thank people on this forum? Should I just press the "like" button?

Orodruin

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META: how should I thank people on this forum? Should I just press the "like" button?
That is one way that is giving some sort of official record, but a simple "thanks!" in the thread is also always appreciated. (Or you can do both )

"Christoffel symbol ("undotting")"

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