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Einstein tensor 2-degree as trace-reversal of Ricci curvature:
$ G_{ab} = R_{ab} - \frac{1}{2} R g_{ab}$
Riemann curvature 2-form dual:
$\Omega^{ab} = {R^{ab}}_{cd} e^c \otimes e^d = \frac{1}{2} {R^{ab}}_{cd} e^c \wedge e^d$
$\star \Omega^{ab} = \frac{1}{2} {R^{ab}}_{cd} {\epsilon^{cd}}_{ef} e^e \otimes e^f$
$\star \Omega^{ab} = \frac{1}{4} {R^{ab}}_{cd} {\epsilon^{cd}}_{ef} e^e \wedge e^f$
Einstein tensor 4-degree as double-dual of Riemann curvature tensor, in 4 dimensions:
$ {G^{ab}}_{cd} = {(\star R \ast)^{ab}}_{cd}$
$= \frac{1}{2} {\epsilon^{ab}}_{uv} (\star R)^{uv}$
$= \frac{1}{4} {\epsilon^{ab}}_{uv} {R^{uv}}_{rs} {\epsilon^{rs}}_{cd} $
$= -\frac{1}{4} \delta^{abrs}_{uvcd} {R^{uv}}_{rs}$
Einstein tensors and the basis k-forms, in 4 dimensions:
$\Omega^{ab} \wedge \eta_{ab}$
$= \frac{1}{2} {R^{ab}}_{cd} e^c \wedge e^d \wedge \frac{1}{2} \epsilon_{abuv} e^u \wedge e^v$
$= \frac{1}{4} \epsilon_{uvab} {R^{ab}}_{cd} e^u \wedge e^v \wedge e^c \wedge e^d$
$\star \Omega^{ab} \wedge \eta_{ab}$
$= \frac{1}{4} {R^{ab}}_{cd} {\epsilon^{cd}}_{ef} e^e \wedge e^f \wedge \frac{1}{2} \epsilon_{abuv} e^u \wedge e^v$
$= \frac{1}{8} \epsilon_{uvab} {R^{ab}}_{cd} {\epsilon^{cd}}_{ef} e^u \wedge e^v \wedge e^e \wedge e^f$
$= \frac{1}{2} G_{uvef} e^u \wedge e^v \wedge e^e \wedge e^f$
$= 0$ because of $G_{uvef} = G_{efuv}$
$\frac{1}{2} \eta_{ab} \otimes \star \Omega^{ab}$
$= \frac{1}{2} \epsilon_{abuv} e^u \otimes e_v \otimes \frac{1}{2} {R^{ab}}_{ef} {\epsilon^{ef}}_{cd} e^c \otimes e^d$
$= \frac{1}{4} \epsilon_{uvab} {R^{ab}}_{ef} {\epsilon^{ef}}_{cd} e^u \otimes e_v \otimes e^c \otimes e^d$
$= G_{uvcd} e^u \otimes e_v \otimes e^c \otimes e^d$
$\frac{1}{2} \eta_{abc} \wedge \Omega^{bc}$
$= \frac{1}{2} \epsilon_{abcd} e^d \wedge \frac{1}{2} {R^{bc}}_{uv} e^u \wedge e^v$
$= \frac{1}{4} \epsilon_{adbc} {R^{bc}}_{uv} e^d \wedge e^u \wedge e^v$
...using symmetry of $R_{abcd} = R_{cdab}$...
$= \frac{1}{4} {R_{uv}}^{bc} \epsilon_{bcad} e^u \wedge e^v \wedge e^d$
$= \frac{1}{4} {R_{uv}}^{bc} \epsilon_{bcad} {\epsilon^{uvd}}_e \star e^e$ ... sign?
$= \frac{1}{4} {\epsilon^d}_{euv} {R^{uv}}_{bc} {\epsilon^{bc}}_{ad} \star e^e$
$= {G^d}_{ead} \star e^e$
$= {G^d}_{eda} \star e^e$
$= G_{ea} \star e^e$
... by symmetry?
$= G_{ab} \star e^b$
so $\star (\eta_{abc} \wedge \Omega^{bc}) = G_{ab} e^b$ ... maybe with a factor or something?
$\eta_{abc} \wedge \star \Omega^{bc}$
$= \epsilon_{abcd} e^d \wedge \frac{1}{4} {R^{bc}}_{ef} {\epsilon^{ef}}_{uv} e^u \wedge e^v$
$= \frac{1}{4} \epsilon_{adbc} {R^{bc}}_{ef} {\epsilon^{ef}}_{uv} e^d \wedge e^u \wedge e^v$
$= G_{aduv} e^d \wedge e^u \wedge e^v$
$= G_{a[duv]} e^d \wedge e^u \wedge e^v$
$= G_{a[[du]v]} e^d \wedge e^u \wedge e^v$
$= 0$ or something by symmetry or something right?
Einstein tensor 4-degree as dual of Riemann curvature 2-form:
$e_a \otimes e_b \otimes {G^{ab}}_{cd} e^c \otimes e^d$
$= \frac{1}{4} e_a \otimes e_b \otimes {\epsilon^{ab}}_{uv} {R^{uv}}_{rs} {\epsilon^{rs}}_{cd} e^c \otimes e^d $
$= \frac{1}{8} e_a \otimes e_b \otimes {\epsilon^{ab}}_{uv} {R^{uv}}_{rs} {\epsilon^{rs}}_{cd} e^c \wedge e^d $
$= \frac{1}{2} e_a \otimes e_b \otimes {\epsilon^{ab}}_{uv} \star \Omega^{uv}$
... in the event you want to abuse the wedge product and apply it to vectors:
$ = \frac{1}{4} e_a \wedge e_b \wedge {\epsilon^{ab}}_{uv} \star \Omega^{uv}$
$ = \frac{1}{16} e_a \wedge e_b \wedge {\epsilon^{ab}}_{uv} {R^{uv}}_{rs} {\epsilon^{rs}}_{cd} e^c \wedge e^d $
... if you want to use the new $\ast$ vector-dual operator:
$= e_a \otimes e_b \otimes \star (\Omega \ast)^{ab}$
$(\star \Omega \ast)^{ua} (e_u)$
$= \frac{1}{2} {\epsilon^{ua}}_{vb} (\star \Omega^{vb}) (e_u)$
$= ({G^{ua}}_{vb} e^v \otimes e^b) (e_u)$
$= {G^{ua}}_{vb} e^b \delta^v_u$
$= {G^{ua}}_{ub} e^b$
$= {G^a}_b e^b$
Einstein tensor 2-degree as contraction of Einstein tensor 4-degree:
${G_b}^a = {G^{ua}}_{ub}$
$= \frac{1}{4} {\epsilon^{ua}}_{cd} {R^{cd}}_{ef} {\epsilon^{ef}}_{ub}$
$= \frac{1}{4} \epsilon^{uacd} {R_{cd}}^{ef} \epsilon_{efub}$
$= \frac{1}{4} \epsilon^{acdu} \epsilon_{efbu} {R_{cd}}^{ef}$
$= -\frac{1}{4} \delta^{acdu}_{efbu} {R_{cd}}^{ef}$
$= -\frac{1}{4} \delta^{acd}_{efb} {R_{cd}}^{ef}$
$= -\frac{6!}{4} \delta^a_{[e} \delta^c_f \delta^d_{b]} {R_{cd}}^{ef}$
$= -\frac{1}{4} (
\delta^a_e \delta^c_f \delta^d_b
+ \delta^a_b \delta^c_e \delta^d_f
+ \delta^a_f \delta^c_b \delta^d_e
- \delta^a_b \delta^c_f \delta^d_e
- \delta^a_f \delta^c_e \delta^d_b
- \delta^a_e \delta^c_b \delta^d_f
) {R_{cd}}^{ef}$
$= \frac{1}{4} (
- {R_{cb}}^{ac}
- \delta^a_b {R_{cd}}^{cd}
- {R_{bd}}^{da}
+ \delta^a_b {R_{cd}}^{dc}
+ {R_{cb}}^{ca}
+ {R_{bd}}^{ad}
)$
$= {R_{cb}}^{ca} - \frac{1}{2} \delta^a_b {R_{cd}}^{cd}$
$= {R_b}^a - \frac{1}{2} \delta^a_b R$
It looks like the symmetry of $G_{ab}$ can be inferred from the two $\epsilon^{acdu} \epsilon_{efbu}$'s , regardless of any symmetry of $R_{abcd}$, since those turn into the $\delta^{acd}_{efb}$, which can be rearranged.
...try again for arbitrary dimension...
${G_b}^a = {G^{ua}}_{ub}$
$= -\frac{1}{4} \delta^{uars}_{ubpq} {R^{pq}}_{rs}$
TODO redo the substitution I just did but for arbitrary dimensions.
Using $d^2 e_b = e_a \wedge {\Omega^a}_b$...
...so $g^{bc} d^2 e_c = e_a \wedge \Omega^{ab}$...
...so $e^a ( g^{bc} d^2 e_c ) = \Omega^{ab}$ ... (right? can I do this? probably not with that Hodge dual outside the expression ...)
${G^a}_b e^b = \frac{1}{2} {\epsilon^{ua}}_{vb} (\star e^v (g^{bc} d^2 e_c)) (e_u)$
...maybe I can define it this way ...
$(\star d^2 e_v) ( e_a \otimes e_b ) = \frac{1}{2} e_u \wedge {R^u}_{vcd} {\epsilon^{cd}}_{ab}$
...but somehow I need to apply a Levi-Civita permutation tensor between those u-indexes...
Also ...
$(d^2 e_v) (e_a \otimes e_b) = e_u \wedge {R^u}_{vab}$
$(d^2 e_v) (e_a) = e_u \wedge {R^u}_{vab} \wedge e^b$
$-g^{va} (d^2 e_v) (e_a)
= -g^{va} e_u \wedge {R^u}_{vab} \wedge e^b
= g^{va} e_u \wedge {{R_v}^u}_{ab} \wedge e^b
= e_u \wedge {R^u}_b \wedge e^b
= e_u \wedge R^u
$
Jury is still out on interior product of one-forms to vectors-wedge-one-forms:
$(e_b \wedge e^c) \lrcorner e^a = e^a ( e_b \wedge e^c ) = \delta^a_b e^c$ ... right?
But if I did want to use it ...
$e^a(d^2 e_b) = e^a(e_c \wedge {\Omega^c}_b) = {\Omega^a}_b$
$g^{cb} e^a(d^2 e_c) = \Omega^{ab}$
$(e_u \otimes e_v) \lrcorner g^{cb} e^a (d^2 e_c) = g^{cb} e^a(d^2 e_c) (e_u \otimes e_v) = {R^{ab}}_{uv}$
$(e_u \otimes e_v) \lrcorner g^{cb} (\star e^a (d^2 e_c)) = g^{cb} (\star e^a(d^2 e_c)) (e_u \otimes e_v) = \frac{1}{2} {R^{ab}}_{cd} {\epsilon^{cd}}_{uv}$
$\frac{1}{2} {\epsilon^{pq}}_{ab} g^{cb} (\star e^a(d^2 e_c)) (e_u \otimes e_v)
= \frac{1}{4} {\epsilon^{pq}}_{ab} {R^{ab}}_{cd} {\epsilon^{cd}}_{uv}
= {G^{pq}}_{uv}$
So...
$\frac{1}{2} {\epsilon^{ua}}_{ef} g^{cf} (\star e^e(d^2 e_c)) (e_u \otimes e_b)
= \frac{1}{4} {\epsilon^{ua}}_{ef} {R^{ef}}_{cd} {\epsilon^{cd}}_{ub} = {G^{ua}}_{ub} = {G^a}_b$
$\frac{1}{2} {\epsilon^{ua}}_{ef} g^{cf} (\star e^e(d^2 e_c)) (e_u)
= {G^a}_b e^b$
$-\frac{1}{2} {\epsilon^{uac}}_e (\star e^e(d^2 e_c)) (e_u) = {G^a}_b e^b$
$-\frac{1}{2} {\epsilon^{uac}}_e e_u \lrcorner (\star e^e(d^2 e_c)) = {G^a}_b e^b$
$g^{cf} \frac{1}{2} {\epsilon_{afe}}^u e_u \lrcorner (\star e^e(d^2 e_c)) = G_{ab} e^b$
$g^{cf} \frac{1}{2} (\eta_{afe})^\sharp \lrcorner (\star e^e(d^2 e_c)) = G_{ab} e^b$
Einstein tensor from exterior algebra:
$G_a = G_{ab} e^b$
$= R_{ab} e^b - \frac{1}{2} R g_{ab} e^b$
$= R_a - \frac{1}{2} e_a R$
$= R_a - \frac{1}{2} e_a (R_u \cdot e^u)$
Using inner product $\cdot$ (should I use $\langle, \rangle$ to denote inner product?):
$= R_u \cdot (\delta^u_a - \frac{1}{2} e^u \otimes e_a)$
Using Hodge dual:
$= R_a - \frac{1}{2} \sigma \star (R_u \wedge \star e^u) e_a$
... and that last $e_a$ outside the Hodge dual star is what keeps me from factoring any further ...
Now what if we're not in just 4 dimensions?
Einstein tensor $2(n-2)$-degree as double-dual of Riemann curvature tensor in n dimensions:
${G^{a_1 ... a_{n-2}}}_{b_1 ... b_{n-2}} = {(\star R \ast)^{a_1 ... a_{n-2}}}_{b_1 ... b_{n-2}}$
$ = \frac{1}{4} {\epsilon^{a_1 ... a_{n-2}}}_{c d} {R^{cd}}_{ef} {\epsilon^{ef}}_{b_1 ... b_{n-2}}$
... by raising/lowering...
$ = \frac{1}{4} \epsilon^{a_1 ... a_{n-2} c d} {R_{cd}}^{ef} \epsilon_{ef b_1 ... b_{n-2}}$
... substitute the generalized Kronecher delta definition ...
$ = \frac{1}{4} \sigma \delta^{a_1 ... a_{n-2} c d}_{e f b_1 ... b_{n-2}} {R_{cd}}^{ef}$
... cycle the sum indexes of the Levi-Civita to the end. $2(n-2)$ sign flips.
$ = \frac{1}{4} \sigma \delta^{a_1 ... a_{n-2} c d}_{b_1 ... b_{n-2} e f} {R_{cd}}^{ef}$
Einstein tensor 2-degree as contraction of Einstein tensor $2(n-2)$-degree:
${G^a}_b = {G^{u_1 ... u_{n-3} a}}_{u_1 ... u_{n-3} b}$
...substitute definition:
$= \frac{1}{4} \sigma \delta^{u_1 ... u_{n-3} a c d}_{u_1 ... u_{n-3} b e f} {R_{cd}}^{ef}$
...substitute definition of kronecher delta constractions:
$= \frac{1}{4} (n-3)! \sigma \delta^{a c d}_{b e f} {R_{cd}}^{ef}$
...substitute definition of kronecher delta:
$= \frac{1}{4} 3! (n-3)! \sigma \delta^a_{[b} \delta^c_e \delta^d_{f]} {R_{cd}}^{ef}$
...substituting the progress in the n=4 definition:
$= -\sigma (n-3)! ({R^a}_b - \frac{1}{2} \delta^a_b R)$
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