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TODO show (and put this back in worksheet #whatever for metric-cancelling connections):
${\hat\Gamma^b}_{ab} = \partial_a log( \sqrt{|g|} )$ for Levi-Civita connection $\hat\Gamma$
Ricci curvature:
$R_{ab} = {R^u}_{aub}$
$=
e_u ({\Gamma^u}_{ba})
- e_b ({\Gamma^u}_{ua})
+ {\Gamma^u}_{uv} {\Gamma^v}_{ba}
- {\Gamma^u}_{bv} {\Gamma^v}_{ua}
- {\Gamma^u}_{va} {c_{ub}}^v$
Antisymmetric components of Ricci curvature:
$R_{[ab]} =
e_u ({\Gamma^u}_{[ba]})
- e_{[b} ({\Gamma^u}_{u|a]})
+ {\Gamma^u}_{uv} {\Gamma^v}_{[ba]}
- {\Gamma^u}_{[b|v} {\Gamma^v}_{u|a]}
- {\Gamma^u}_{v[a} {c_{u|b]}}^v$
$ =
\frac{1}{2} e_u ( {T^u}_{ba} + {c_{ba}}^u )
- e_{[b} ({\Gamma^u}_{u|a]})
+ \frac{1}{2} {\Gamma^u}_{uv} {T^v}_{ba}
+ \frac{1}{2} {\Gamma^u}_{av} {T^v}_{ub}
- \frac{1}{2} {\Gamma^u}_{bv} {T^v}_{ua}
+ \frac{1}{2} {\Gamma^u}_{uv} {c_{ba}}^v
+ \frac{1}{2} {\Gamma^u}_{av} {c_{ub}}^v
- \frac{1}{2} {\Gamma^u}_{va} {c_{ub}}^v
- \frac{1}{2} {\Gamma^u}_{bv} {c_{ua}}^v
+ \frac{1}{2} {\Gamma^u}_{vb} {c_{ua}}^v
$
...in absense of structure constants...(${c_{ab}}^c = 0, e_a = \partial_a$):
$R_{[ab]} =
\frac{1}{2} \partial_u ( {T^u}_{ba} )
- \partial_{[b} ({\Gamma^u}_{u|a]})
+ \frac{1}{2} {\Gamma^u}_{uv} {T^v}_{ba}
+ \frac{1}{2} {\Gamma^v}_{au} {T^u}_{vb}
- \frac{1}{2} {\Gamma^v}_{bu} {T^u}_{va}
$
I could rewrite the $tr \Gamma$'s as $e(log\sqrt{|g|})$, except I would have to replace the $\Gamma$ with $\hat\Gamma + K$, the contorsion.
Instead I'll just avoid substiting $K_{abc}$ in terms of $T_{abc}$
$R_{[ab]} =
\frac{1}{2} \partial_u ( {T^u}_{ba} )
- \partial_{[b} ({\hat\Gamma^u}_{u|a]} + {K^u}_{u|a]})
+ \frac{1}{2} ({\hat\Gamma^u}_{uv} + {K^u}_{uv}) {T^v}_{ba}
+ \frac{1}{2} {\Gamma^v}_{au} {T^u}_{vb}
- \frac{1}{2} {\Gamma^v}_{bu} {T^u}_{va}
$
$R_{[ab]} =
\frac{1}{2} \partial_u ( {T^u}_{ba} )
+ \frac{1}{2} \partial_v log \sqrt{|g|} {T^v}_{ba}
- \partial_{[b}( {K^u}_{u|a]} )
+ \frac{1}{2} {K^u}_{uv} {T^v}_{ba}
+ \frac{1}{2} {\Gamma^v}_{au} {T^u}_{vb}
- \frac{1}{2} {\Gamma^v}_{bu} {T^u}_{va}
$
...looks very ugly.
...
Shouldn't the first Bianchi identity (worksheet #12) contracted give us something?
${R^a}_{[bcd]} = \nabla_{[b} {T^a}_{cd]} + {T^v}_{bc} {T^a}_{vd}$
I need to separate ${R^a}_{[b|c|d]}$ from ${R^a}_{[bcd]}$
${R^a}_{[bcd]} = \frac{1}{6} (
{R^a}_{bcd}
+ {R^a}_{cdb}
+ {R^a}_{dbc}
- {R^a}_{dcb}
- {R^a}_{cbd}
- {R^a}_{bdc}
)$
...using ${R^a}_{bcd} = {R^a}_{b[cd]}$ (in worksheet #9)
${R^a}_{[bad]} = \frac{1}{3} (
{R^a}_{bad} - {R^a}_{dab} - {R^a}_{abd}
)$
...using $R_{[bd]} = \frac{1}{2} ({R^a}_{bad} - {R^a}_{dab})$
${R^a}_{[bad]} = \frac{2}{3} R_{[bd]} - \frac{1}{3} {R^a}_{abd}$
...using ${R^a}_{abd} = 0$ ...
$R_{[bd]} = \frac{3}{2} {R^a}_{[bad]}$
...using the first Bianchi identity...
$R_{[bd]} = \frac{3}{2} (
\nabla_{[b} {T^a}_{ad]} + {T^v}_{ba} {T^a}_{vd}
)$
How about starting with Riemann, then writing in terms of contorsion?
(TODO put this in the Riemann curvature worksheet)
$R_{ab} = {R^u}_{aub}$
$=
e_u ({\Gamma^u}_{ba})
- e_b ({\Gamma^u}_{ua})
+ {\Gamma^u}_{uv} {\Gamma^v}_{ba}
- {\Gamma^u}_{bv} {\Gamma^v}_{ua}
- {\Gamma^u}_{va} {c_{ub}}^v$
$=
e_u ({K^u}_{ba} + {\hat\Gamma^u}_{ba})
- e_b ({K^u}_{ua} + {\hat\Gamma^u}_{ua})
+ ({K^u}_{uv} + {\hat\Gamma^u}_{uv}) ({K^v}_{ba} + {\hat\Gamma^v}_{ba})
- ({K^u}_{bv} + {\hat\Gamma^u}_{bv}) ({K^v}_{ua} + {\hat\Gamma^v}_{ua})
- ({K^u}_{va} + {\hat\Gamma^u}_{va}) {c_{ub}}^v$
$=
e_u ({\hat\Gamma^u}_{ba})
- e_b ({\hat\Gamma^u}_{ua})
+ {\hat\Gamma^u}_{uv} {\hat\Gamma^v}_{ba}
- {\hat\Gamma^u}_{bv} {\hat\Gamma^v}_{ua}
- {\hat\Gamma^u}_{va} {c_{ub}}^v$
$
+ e_u ({K^u}_{ba})
- e_b ({K^u}_{ua})
+ {K^u}_{uv} {K^v}_{ba}
- {K^u}_{bv} {K^v}_{ua}
- {K^u}_{va} {c_{ub}}^v $
$
+ {K^u}_{uv} {\hat\Gamma^v}_{ba}
+ {\hat\Gamma^u}_{uv} {K^v}_{ba}
- {K^u}_{bv} {\hat\Gamma^v}_{ua}
- {\hat\Gamma^u}_{bv} {K^v}_{ua}
$
$= \hat{R}_{ab}
+ e_u ({K^u}_{ba})
- e_b ({K^u}_{ua})
+ {K^u}_{uv} {K^v}_{ba}
- {K^u}_{bv} {K^v}_{ua}
- {K^u}_{va} {c_{ub}}^v
+ {K^u}_{uv} {\hat\Gamma^v}_{ba}
+ {\hat\Gamma^u}_{uv} {K^v}_{ba}
- {K^u}_{bv} {\hat\Gamma^v}_{ua}
- {\hat\Gamma^u}_{bv} {K^v}_{ua}
$
...where $\hat{R}_{ab} = {\hat{R}^u}_{aub} =
e_u ({\hat\Gamma^u}_{ba})
- e_b ({\hat\Gamma^u}_{ua})
+ {\hat\Gamma^u}_{uv} {\hat\Gamma^v}_{ba}
- {\hat\Gamma^u}_{bv} {\hat\Gamma^v}_{ua}
- {\hat\Gamma^u}_{va} {c_{ub}}^v$ is the curvature of the Levi-Civita connection.
TODO seems I've seen a few sources that say the difference between the curvature and Levi-Civita connection curvature should only be in terms of contorsion, so the (Levi-Civita * contorsion) components should cancel maybe?
I could isolate the $\nabla T$ term, starting with ${c_{ab}}^c = 0$:
$R_{[ab]} =
\frac{1}{2} (
\partial_u {T^u}_{ba}
+ {\Gamma^u}_{uv} {T^v}_{ba}
- {\Gamma^v}_{ub} {T^u}_{va}
- {\Gamma^v}_{ua} {T^u}_{bv}
- 2 {\Gamma^v}_{[au]} {T^u}_{bv}
- 2 {\Gamma^v}_{[bu]} {T^u}_{va}
)
- \partial_{[b} ({\Gamma^u}_{u|a]})
$
$R_{[ab]} =
\frac{1}{2} \nabla_u {T^u}_{ba}
- \partial_{[b} ({\Gamma^u}_{u|a]})
$
$R_{[ab]} =
\frac{1}{2} \nabla_u {T^u}_{ba}
+ \frac{1}{2} (
- {\Gamma^v}_{au} {T^u}_{bv}
+ {\Gamma^v}_{ua} {T^u}_{bv}
- {\Gamma^v}_{bu} {T^u}_{va}
+ {\Gamma^v}_{ub} {T^u}_{va}
- \partial_b {T^u}_{ua}
+ \partial_a {T^u}_{ub}
+ \partial_a {\Gamma^u}_{bu}
- \partial_b {\Gamma^u}_{au}
)
$
...get rid of the ${\Gamma^a}_{bc}$'s ...
$R_{[ab]} =
\frac{1}{2} \nabla_u {T^u}_{ba}
+ \frac{1}{2} (
- {\Gamma^v}_{au} {T^u}_{bv}
+ {\Gamma^v}_{ua} {T^u}_{bv}
- {\Gamma^v}_{bu} {T^u}_{va}
+ {\Gamma^v}_{ub} {T^u}_{va}
- \partial_b {T^u}_{ua}
+ \partial_a {T^u}_{ub}
+ \partial_a ({\hat\Gamma^u}_{bu} + {K^u}_{bu})
- \partial_b ({\hat\Gamma^u}_{au} + {K^u}_{au})
)
$
...substitute ${\Gamma^u}_{au} = e_a(log\sqrt{|g|})$...
$R_{[ab]} =
\frac{1}{2} \nabla_u {T^u}_{ba}
+ \frac{1}{2} (
- {\Gamma^v}_{au} {T^u}_{bv}
+ {\Gamma^v}_{ua} {T^u}_{bv}
- {\Gamma^v}_{bu} {T^u}_{va}
+ {\Gamma^v}_{ub} {T^u}_{va}
- \partial_b {T^u}_{ua}
+ \partial_a {T^u}_{ub}
+ \partial_a \partial_b log\sqrt{|g|}
- \partial_b \partial_a log\sqrt{|g|}
+ \partial_a {K^u}_{bu}
- \partial_b {K^u}_{au}
)
$
...substitute contorsion for torsion (or should we be substituting torsion for contorsion?)...
$R_{[ab]} =
\frac{1}{2} \nabla_u {T^u}_{ba}
+ \frac{1}{2} (
- {\Gamma^v}_{au} {T^u}_{bv}
+ {\Gamma^v}_{ua} {T^u}_{bv}
- {\Gamma^v}_{bu} {T^u}_{va}
+ {\Gamma^v}_{ub} {T^u}_{va}
- \partial_b {T^u}_{ua}
+ \partial_a {T^u}_{ub}
+ \frac{1}{2} \partial_a ({{T_u}^u}_b + {{T_b}^u}_u + {T^u}_{bu})
- \frac{1}{2} \partial_b ({{T_u}^u}_a + {{T_a}^u}_u + {T^u}_{au})
)
$
... distribute partials, cancel $T_{abc} g^{bc}$ since T is antisymmetric and g is symmetric..
$R_{[ab]} =
\frac{1}{2} \nabla_u {T^u}_{ba}
+ \frac{1}{2} (
- {\Gamma^v}_{au} {T^u}_{bv}
+ {\Gamma^v}_{ua} {T^u}_{bv}
- {\Gamma^v}_{bu} {T^u}_{va}
+ {\Gamma^v}_{ub} {T^u}_{va}
+ \partial_a {T^u}_{ub}
- \partial_b {T^u}_{ua}
)
$
...combine covariant derivatives...
$R_{[ab]} = \frac{1}{2} (
\nabla_u {T^u}_{ba}
+ \nabla_a {T^u}_{ub}
+ \nabla_b {T^u}_{au}
+ {T^u}_{vu} {T^v}_{ba}
)$
$R_{[ab]} = \frac{3}{2} \nabla_{[u} {T^u}_{ba]} + \frac{1}{2} {T^u}_{vu} {T^v}_{ba}$
$R_{[ab]} = \frac{3}{2} (\nabla_{[u} {T^u}_{ba]} + {T^u}_{v[u} {T^v}_{ba]})$
Comparing this with the Bianchi identity, I wonder if this means that ${T^u}_{vu} {T^v}_{ba} = 3 {T^v}_{au} {T^u}_{vb}$?
Also, what is $R_{[ab]}$ in terms of ${K^a}_{bc}$?
In fact, rule of thumb, how do you turn a two-index antisymmetry into a 3-index antisymmetry?
For any aribtrary rank-3 tensor (put this back in worksheet #1):
$x_{[abc]} = \frac{1}{6} (x_{abc} + x_{bca} + x_{cab} - x_{cba} - x_{bac} - x_{acb})$
$ = \frac{1}{3} (x_{a[bc]} + x_{b[ca]} + x_{c[ab]})$
For a tensor which is antisymmetric in 2nd and 3rd indexes (similarly works with any other antisymmetric pair):
$x_{[abc]} = \frac{1}{6} (x_{abc} + x_{bca} + x_{cab} - x_{cba} - x_{bac} - x_{acb})$
$x_{[abc]} = \frac{1}{6} (x_{abc} + x_{bca} + x_{cab} + x_{cab} + x_{bca} + x_{abc})$
$x_{[abc]} = \frac{1}{3} (x_{abc} + x_{bca} + x_{cab})$
(The same is used with the Jacobi identity applied to the Faraday tensor, which is (typically) antisymmetric)
TODO Is the Faraday tensor antisymmetric in torsion or anholonomic geometry?
$F_{ab} = 2 \nabla_{[b} A_{a]}$ means it is antisymmetric by definition. So does $F = dA$.
$F_{ab} = 2 \nabla_{[b} A_{a]}$ means it is antisymmetric by definition. So does $F = dA$.
However, next question, is the Faraday tensor invariant of connection in torsion or anholonomic geometry? (In a coordinate basis it is connection invariant.)
$F_{ab} = 2 \nabla_{[b} A_{a]} = 2 ( e_{[b} ( A_{a]} ) - {\Gamma^u}_{[ba]} A_u ) = 2 e_{[b} ( A_{a]} ) + ( {T^u}_{ab} + {c_{ab}}^u ) A_u $
Hmm, does the presence of torsion (which comes from the asymmetry of the connection) imply that the differential form definition of the Faraday tensor for curved space should instead be...
$F = D A = d A + \omega \wedge A$
(put Faraday and Maxwell stuff in its own worksheet ... it might be signature-specific, mind you)
Ricci antisymmetry, with torsion and commutation:
$R_{[ab]} =
e_u ({\Gamma^u}_{[ba]})
- e_{[b} ({\Gamma^u}_{u|a]})
+ {\Gamma^u}_{uv} {\Gamma^v}_{[ba]}
- {\Gamma^u}_{[b|v} {\Gamma^v}_{u|a]}
- {\Gamma^u}_{v[a} {c_{u|b]}}^v$
$R_{[ab]} = \frac{1}{2} (
e_u( {T^u}_{ba} + {c_{ba}}^u )
- 2 e_{[b} ({\Gamma^u}_{u|a]})
+ {\Gamma^u}_{uv} ({\Gamma^v}_{ba} - {\Gamma^v}_{ab})
- {\Gamma^u}_{bv} {\Gamma^v}_{ua}
+ {\Gamma^u}_{av} {\Gamma^v}_{ub}
- {\Gamma^u}_{va} {c_{ub}}^v
+ {\Gamma^u}_{vb} {c_{ua}}^v
)$
$R_{[ab]} = \frac{1}{2} (
e_u( {T^u}_{ba} + {c_{ba}}^u )
- 2 e_{[b} (e_{|a]} ( log \sqrt{|g|} ) + {c_{u|a}}^u + {T^u}_{u|a]} )
+ {\Gamma^u}_{uv} ({T^v}_{ba} + {c_{ba}}^v)
- {\Gamma^u}_{bv} ({\Gamma^v}_{au} + {T^v}_{ua} + {c_{ua}}^v)
+ {\Gamma^u}_{av} ({\Gamma^v}_{bu} + {T^v}_{ub} + {c_{ub}}^v)
- {c_{ub}}^v ({\Gamma^u}_{av} + {T^u}_{va} + {c_{va}}^u)
+ {c_{ua}}^v ({\Gamma^u}_{bv} + {T^u}_{vb} + {c_{vb}}^u)
)$
$R_{[ab]} = \frac{1}{2} (
e_u ( {T^u}_{ba} )
+ e_a ( {T^u}_{ub} )
+ e_b ( {T^u}_{au} )
+ e_u ( {c_{ba}}^u )
+ e_b ( {c_{au}}^u )
+ e_a ( {c_{ub}}^u )
+ {\Gamma^u}_{uv} {T^v}_{ba}
+ {\Gamma^u}_{bv} {T^v}_{au}
+ {\Gamma^u}_{av} {T^v}_{ub}
+ {c_{ba}}^v {c_{uv}}^u
+ {c_{ba}}^v {T^u}_{uv}
+ {c_{au}}^v {T^u}_{bv}
+ {c_{ub}}^v {T^u}_{av}
)$
...isolate covariant derivatives?
$R_{[ab]} = \frac{1}{2} (
\nabla_u ( {T^u}_{ba} + {c_{ba}}^u )
+ \nabla_a ( {T^u}_{ub} + {c_{ub}}^u )
+ \nabla_b ( {T^u}_{au} + {c_{au}}^u )
+ {T^u}_{vu} {T^v}_{ba}
+ {T^u}_{va} {T^v}_{ub}
+ {T^u}_{vb} {T^v}_{au}
+ {c_{ab}}^v {c_{uv}}^u
- {c_{au}}^v {\Gamma^u}_{bv}
- {c_{ub}}^v {\Gamma^u}_{av}
- {c_{ba}}^v {\Gamma^u}_{vu}
+ {c_{ab}}^v {T^u}_{uv}
+ {c_{au}}^v {T^u}_{bv}
+ {c_{ub}}^v {T^u}_{av}
+ {c_{uv}}^u {T^v}_{ab}
)$
...making great use of ${\Gamma^a}_{bc} = {\Gamma^a}_{cb} + {T^a}_{bc} + {c_{bc}}^a$...
$R_{[ab]} = \frac{1}{2} (
3 \nabla_{[u} ( {T^u}_{ba]} + {c_{ba]}}^u )
+ 3 ({T^u}_{v[u} + {c_{v[u}}^u) ({T^v}_{ba]} + {c_{ba]}}^v)
- {\Gamma^u}_{vb} {c_{au}}^v
- {\Gamma^u}_{va} {c_{ub}}^v
+ {\Gamma^u}_{vu} {c_{ba}}^v
+ {c_{uv}}^u ({T^v}_{ab} + {c_{ba}}^v)
- {T^v}_{ba} {c_{vu}}^u
- {T^v}_{uv} {c_{ba}}^u
)$
... Is the following denotation correct? Does $x_{[a} (y_{bc]} + z_{bc]})$ equal to $x_{[a} (w_{bc]} - w_{cb]})$
for $w_{bc} - w_{cb} = y_{bc} + z_{bc}$? Do the indexes expand correctly?
$R_{[ab]} = \frac{3}{2} (
\nabla_{[u} ( {\Gamma^u}_{ba]} - {\Gamma^u}_{ab]} )
+ ( {\Gamma^u}_{v[u} - {\Gamma^u}_{[u|v} ) ( {\Gamma^v}_{ba]} - {\Gamma^v}_{ab]} )
)
+ \frac{1}{2} (
- {\Gamma^u}_{vb} {c_{au}}^v
- {\Gamma^u}_{va} {c_{ub}}^v
+ {\Gamma^u}_{vu} {c_{ba}}^v
+ {c_{uv}}^u ({T^v}_{ab} + {c_{ba}}^v)
- {T^v}_{ba} {c_{vu}}^u
- {T^v}_{uv} {c_{ba}}^u
)$
How to simplify the rest further ...
This reminds me, what is $tr [\nabla, \nabla] (z)$ for some arbitrary degree z?
Ricci curvature one-form as partial-application:
${\Omega^c}_a (e_c)$
$= (\frac{1}{2} {R^c}_{aub} e^u \wedge e^b) (e_c)$
$= ({R^c}_{aub} e^u \otimes e^b) (e_c)$
... and here we use our partial-application rule ...
$= {R^c}_{aub} e^b \delta^u_c$
$= {R^u}_{aub} e^b$
$= R_{ab} e^b$
Ricci curvature one-form as interior product:
$e_c \lrcorner {\Omega^c}_a$
$e_c \lrcorner {R^c}_{auv} e^u \otimes e^v$
${R^c}_{auv} \delta_c^u e^v$
${R^u}_{auv} e^v$
$R_{av} e^v$
Let $R_a = R_{ab} e^b$
So $R_a = {\Omega^b}_a (e_b) = e_b \lrcorner {\Omega^b}_a$
Notice that even though $R_a$ is indexed covariant, it is still not considered a 1-form. This is only for summation convention.
So maybe it is more convenient to denote it in contravariant form:
Let $R^a
= g^{ab} R_b
= g^{ab} {\Omega^c}_b (e_c)
= \Omega^{ca} (e_c)
= g^{ab} R_{bc} e^c
= {R^a}_b e^b$
In fact, maybe it would be better to define the curvature 2-form as both contravariant indexed?
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