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I want this :$R^{a}_{bcd}=R_{abcd}$

I tried this:$R^{a}_{bcd}=g^{ae}R_{ebcd}$ but it is not $R_{abcd}$.

How do you lower $a$ without changing it with another dummy index ?

Edit: This is what i meant: https://snag.gy/9DXwkT.jpg

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    $\begingroup$ You don't. Indices must remain balanced, so if you have $^a$ on the RHS you can't have $_a$ on the LHS. $\endgroup$
    – Chris
    Commented Jan 4, 2018 at 21:52
  • $\begingroup$ But, what you have written fundamentally is not true: $R^{a}_{bcd} \neq R_{abcd}$, it is in fact correct to say: $R^{a}_{bcd} = g^{ae} R_{ebcd}$ as you have written. $\endgroup$ Commented Jan 4, 2018 at 21:54
  • $\begingroup$ Are you trying to obtain $R^a_{bcd}$ or are you trying to obtain $R_{abcd}$? If the former, perform $g^{ae}R_{ebcd}$; if the latter, take $g_{ae}R^e_{bcd}$. $\endgroup$ Commented Jan 4, 2018 at 22:01
  • $\begingroup$ can i share a screenshot of what im trying to prove? $\endgroup$ Commented Jan 4, 2018 at 22:08
  • $\begingroup$ You can obtain that result by multiplying by the metric tensor $R_{ebcd} = g_{ea}R^a_{bcd}$; then, since the indices do not have meaning in themselves, simply change all the $~_e$'s on both sides of the equality to $~_a$'s. $\endgroup$ Commented Jan 4, 2018 at 22:20

1 Answer 1

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You can obtain that result by multiplying by the metric tensor $R_{ebcd} = g_{ea}R^a_{bcd}$; then, since the indices do not have meaning in themselves, simply change all the $~_e$'s on both sides of the equality to $~_a$'s. However, working out the matrix multiplication is more trouble than it's worth. A better way to understand the second equation in your screenshot is to note that the only $~^a$ appears on the metric tensor, so none of the other stuff is relevant to lowering that index. Therefore, you only need to consider the metric tensor with one index up and one down, which is obviously the kronecker delta.

See equation 1.25 of Markus Deserno's differential geometry notes

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