I Riemann curvature tensor derivation

[cozycoz]

Riemann tensor is defined mathematically like this:
$∇_k∇_jv_i-∇_j∇_kv_i={R^l}_{ijk}v_l$

Using covariant derivative formula for covariant tensors and covariant vectors. which are

$∇_av_b=∂_av_b-{Γ^c}_{ab}v_c$
$∇_aT_{bc}=∂_av_{bc}-{Γ^d}_{ac}v_{db}-{Γ^d}_{ab}v_{dc}$,

I got these:

$∇_k∇_jv_i=∂_k∂_jv_i-∂_k{Γ^m}_{ji}v_m-{Γ^m}_{ji}∂_kv_m-{Γ^l}_{ki}∂_lv_j+{Γ^l}_{ki}{Γ^m}_{lj}v_m-{Γ^l}_{kj}∂_lv_i+{Γ^l}_{kj}{Γ^m}_{li}v_m$(I'll call each term by its number : $∂_k∂_jv_i$ is "1" because it's the first term)

$∇_j∇_kv_i=∂_j∂_kv_i-∂_j{Γ^m}_{ki}v_m-{Γ^m}_{ki}∂_jv_m-{Γ^l}_{ji}∂_lv_k+{Γ^l}_{ji}{Γ^m}_{lk}v_m-{Γ^l}_{jk}∂_lv_i+{Γ^l}_{jk}{Γ^m}_{li}v_m$

Then 1, 3, 4, 6, 7 terms all vanish when we substract the lower from the upper' according to my professor.

Especially he noted that 3rd term vanishes because it's basically the same when you just exchange the indices j⇔k and add up all m's.
But then isn't it also the case for the second term?In 2nd

$∂_j{Γ^m}_{ki}v_m$
$∂_k{Γ^m}_{ji}v_m$ ,

​

And in 3rd

${Γ^m}_{ki}∂_jv_m$
${Γ^m}_{ji}∂_kv_m$
(They are the same)​

I see no reason why I can't apply the logic for 3rd to 2nd...could you help me please

 

[Mentz114]

Riemann tensor is defined mathematically like this:
$∇_k∇_jv_i-∇_j∇_kv_i={R^l}_{ijk}v_l$

Using covariant derivative formula for covariant tensors and covariant vectors. which are

$∇_av_b=∂_av_b-{Γ^c}_{ab}v_c$
$∇_aT_{bc}=∂_av_{bc}-{Γ^d}_{ac}v_{db}-{Γ^d}_{ab}v_{dc}$,
[..]
In 2nd

$∂_j{Γ^m}_{ki}v_m$
$∂_k{Γ^m}_{ji}v_m$ ,

​

And in 3rd

${Γ^m}_{ki}∂_jv_m$
${Γ^m}_{ji}∂_kv_m$
(They are the same)​

I see no reason why I can't apply the logic for 3rd to 2nd...could you help me please

This $∂_j{Γ^m}_{ki}v_m-∂_k{Γ^m}_{ji}v_m$ can be written $∂_{[j}{Γ^{|m|}}_{k]i}v_m$ and the result depends on the swapping-symmetry of indexes $k,j$. I can't see if $j,k$ are symmetric or not so I would write out all the components to check.

The square barcket "[" is an anti-symmetrization bracket (Bach bracket)

$V_{[\alpha \beta ]}=\frac{1}{2}\left ( V_{\alpha \beta }-V_{\beta \alpha } \right )$

see
https://physics.stackexchange.com/questions/95133/bracket-notation-on-tensor-indices

 

[sweet springs]

I think term 4 of upper equation be \Gamma^l_{\ ki}\partial_j v_l. So term 3 + term 4 are symmetric for exchange of j and k. Thus when we subtract the lower from the upper they cancel. Please check it out and let me know please.

 

[Pencilvester]

$∇_aT_{bc}=∂_av_{bc}-{Γ^d}_{ac}v_{db}-{Γ^d}_{ab}v_{dc}$

I assume you are intending that $T = v$ here. If that’s the case, this formula would only hold if $T_{bc}$ is symmetric on $b$ and $c$. You can see on the RHS the second term has the $b$ in v’s second spot, whereas on the LHS the $b$ is in the first spot. Since in this case, $T_{bc} = \nabla_b v_c$, $~$ $T$ will in general not be symmetric, so your formula is incorrect. This error was transcribed into your equations, which is what @sweet springs pointed out.

 

[cozycoz]

thanks it helped a lot..