I Christoffel symbol ("undotting")

64
1
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.
 

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##.
 
64
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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 :wink:)
 

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