The Riemann curvature tensor $R_{ijkl}$ satisfies several algebraic index symmetries:

  1. $R_{ijkl} = -R_{jikl} = -R_{ijlk}$
  2. $R_{ijkl} = R_{klij}$
  3. $R_{i[jkl]} = 0.$

I more or less understand how to interpret the first two identities: the first one says that the Riemann curvature tensor can be interpreted as a linear operator on the space of 2-forms $\Lambda^2(M)$, while the second one says that this linear operator is self-adjoint.

But I don't really know how to think about the third identity, the algebraic Bianchi identity. Does it have any physical intuition?

(This question asks the same thing about the second or "geometric" Bianchi identity.)


1 Answer 1


It is the special case of the first Bianchi identity $$ d{\bf T}^a+ {{\boldsymbol \omega}^a}_b\wedge {\bf T}^b -{{\bf R}^a}_b \wedge {\bf e}^{*a}=0, $$ that follows by taking the exterior derivative of the definition of torsion two-form $$ d{\bf e}^{*a}+ {{\boldsymbol \omega}^a}_b\wedge {\bf e}^{*b}={\bf T}^a, $$ and using ${\bf R}= d{\boldsymbol \omega}+ {\boldsymbol \omega}\wedge {\boldsymbol \omega}$. If the connection is torsion-free ${\bf T}^a=0$ then the first two tems vanish and we have $$ {{\bf R}^a}_b \wedge {\bf e}^{*a}=0. $$ In the coordinate basis the last equation reduces to ${R^\lambda}_{[\mu\nu\sigma]}=0$.

The geometry behind all the Bianchi identies is that $d^2=0$

  • $\begingroup$ Does ${\boldsymbol \omega}$ represent the connection form corresponding to some local frame ${\bf e}^*$? I'm not familiar with those concepts - no need to explain them - but I've seen that notation before. $\endgroup$
    – tparker
    Feb 11, 2023 at 18:32
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    $\begingroup$ Yes. If $X= X^c {\bf e}_c= X^\mu \partial_\mu $ is a vector field and ${\bf e}_a$ a vielbein basis, then $\nabla_X {\bf e}_a = {\bf e}_b {\omega^b}_{ac} X^c= {\bf e}_b {\omega^b}_{a\mu} X^\mu$. Then ${{\boldsymbol \omega}}^a_b ={ \omega^a}_{b \mu}dx^\mu$ I have some online noter at people.physics.illinois.edu/stone/torsion_review.pdf $\endgroup$
    – mike stone
    Feb 11, 2023 at 18:36
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    $\begingroup$ @tparker you may also want to see Ivo Terek’s answer to A doubt on Tensors: Can they be 1-form valued? on MSE. You’ll see that, as mentioned here, the algebraic Bianchi identity is simply a manifestation ‘derivative of zero is zero’ due to lack of torsion, (which can be equivalently described as the covariant exterior derivative of the identity mapping $TM\to TM$). In fact, even if torsion is non-zero, but we merely have that $d_{\nabla}\tau=0$ (i.e the torsion 2-form is covariantly-closed), then we still have the algebraic Bianchi identity. $\endgroup$
    – peek-a-boo
    Feb 11, 2023 at 22:59

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