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A Riemann tensor equation

  1. Jul 25, 2016 #1
    The Riemann-Christoffel Tensor (##R^{k}_{\cdot n i j}##) is defined as:

    $$
    R^{k}_{\cdot n i j}= \frac{\delta \Gamma^{k}_{j n}}{\delta Z^{i}} - \frac{\delta \Gamma^{k}_{i n}}{\delta Z^{j}}+ \Gamma^{k}_{i l} \Gamma^{l}_{j n}- \Gamma^{k}_{j l} \Gamma^{l}_{i n}
    $$

    My question is that it seems that the equation can be simplified as follows, and I'm wondering if my understanding is correct or not.

    Given following equation for the Christoffel symbol (##\Gamma^{k}_{i j}##):

    $$
    \Gamma^{k}_{i j} = \textbf{Z}^{k} \frac{\delta \textbf{Z}_{i}}{\delta Z^{j}}
    $$


    Based on this equation, we consider the following term in the Riemann curvature tensor equation

    $$

    \begin{align}

    \Gamma^{k}_{il}\Gamma^{l}_{jn} &= \textbf{Z}^{k} \frac{\delta \textbf{Z}_i}{\delta Z^{l}} \textbf{Z}^{l} \frac{\delta \textbf{Z}_j}{\delta Z^{n}}

    \\

    &=\textbf{Z}^{k l} \frac{\delta \textbf{Z}_i}{\delta Z^{l}} \frac{\delta \textbf{Z}_j}{\delta Z^{n}}

    \end{align}

    $$


    Similarly:

    $$

    \begin{align}

    \Gamma^{k}_{j l}\Gamma^{l}_{i n} &= \textbf{Z}^{k} \frac{\delta \textbf{Z}_j}{\delta Z^{l}} \textbf{Z}^{l} \frac{\delta \textbf{Z}_i}{\delta Z^{n}}

    \\

    &=\textbf{Z}^{k l} \frac{\delta \textbf{Z}_j}{\delta Z^{l}} \frac{\delta \textbf{Z}_i}{\delta Z^{n}}

    \\

    &=\textbf{Z}^{k l} \frac{\delta \textbf{Z}_i}{\delta Z^{l}} \frac{\delta \textbf{Z}_j}{\delta Z^{n}}

    \end{align}

    $$


    Thus:

    $$
    \Gamma^{k}_{i l} \Gamma^{l}_{j n}- \Gamma^{k}_{j l} \Gamma^{l}_{i n}=0
    $$


    If this is true, the Riemann curvature tensor can be simply written as follows:

    $$
    R^{k}_{\cdot n i j}= \frac{\delta \Gamma^{k}_{j n}}{\delta Z^{i}} - \frac{\delta \Gamma^{k}_{i n}}{\delta Z^{j}}
    $$

    Where is my mistake? I'm not sure.
     
  2. jcsd
  3. Jul 25, 2016 #2

    haushofer

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    I'm not familiar with your notation, but it seems you use a basis in which the connection vanishes while its derivative does not. This is always possible, but it does not result into a tensor equation since the connections are not tensors.
     
  4. Jul 25, 2016 #3

    Orodruin

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    What you are doing from (4) to (5) does not seem correct. It is difficult to tell since, as pointed out in #2, you are not following the typical notation.
     
  5. Jul 26, 2016 #4
    Notation is from Pavel Grinfeld: Introduction to Tensor Analysis and the Calculus of Moving Surfaces
    I'm happy to put in different notation; if you could refer a page to me in the notation you prefer, I'm happy to change.
    Steps (4) & (5) are really the key point. Why can't the terms be switched?

    ##\frac{\delta x}{\delta y} \frac{\delta z}{\delta t} = \frac{\delta z}{\delta y} \frac{\delta x}{\delta t}##?

    Or similarly:

    ## \frac{\delta x}{\delta z} \frac{\delta z}{\delta y} = \frac{\delta x}{\delta y} \frac{\delta z}{\delta z} ##?
     
  6. Jul 26, 2016 #5

    Orodruin

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    Why would they be interchangable? They represent different things.
     
  7. Jul 26, 2016 #6
    I see the mistake; thanks!
     
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