Bianchi differential identity

The Riemann tensor satisfies the following identity:

(1)\qquad \nabla _{i}R_{\;rjk}^{s}+\nabla _{j}R_{\;rki}^{s}+\nabla _{k}R_{\;rij}^{s}=0

{\ displaystyle (1) \ qquad \ nabla _ {i} R _ {\; rjk} ^ {s} + \ nabla _ {j} R _ {\; rki} ^ {s} + \ nabla _ {k} R_ { \; rij} ^ {s} = 0}

(1) \ qquad \ nabla _ {i} R _ {{\; rjk}} ^ {s} + \ nabla _ {j} R _ {{\; rki}} ^ {s} + \ nabla _ {k} R_ {{\; rij}} ^ {s} = 0

which is called the Bianchi differential identity or the second Bianchi identity in differential geometry .

Proof using a special coordinate system

On the manifold , we choose a single arbitrary point $P$ ${\ displaystyle P}$ $P$ and prove equality (1) at this point. Since the point $P$ ${\ displaystyle P}$ $P$ arbitrary, then the validity of identity (1) on the whole manifold will follow.

At the point $P$ ${\ displaystyle P}$ $P$ we can choose such a special coordinate system that all Christoffel symbols (but not their derivatives) turn to zero at the point $P$ ${\ displaystyle P}$ $P$ . Then for the covariant derivatives at the point $P$ ${\ displaystyle P}$ $P$ we have:

(2)\qquad \nabla _{i}R_{\;rjk}^{s}=\partial _{i}R_{\;rjk}^{s}

{\ displaystyle (2) \ qquad \ nabla _ {i} R _ {\; rjk} ^ {s} = \ partial _ {i} R _ {\; rjk} ^ {s}}

(2)\qquad \nabla _{i}R_{{\;rjk}}^{s}=\partial _{i}R_{{\;rjk}}^{s}

Insofar as

(3)\qquad R_{\;rjk}^{s}=\partial _{j}\Gamma _{kr}^{s}-\partial _{k}\Gamma _{jr}^{s}+\Gamma _{jp}^{s}\Gamma _{kr}^{p}-\Gamma _{kp}^{s}\Gamma _{jr}^{p}

{\ displaystyle (3) \ qquad R _ {\; rjk} ^ {s} = \ partial _ {j} \ Gamma _ {kr} ^ {s} - \ partial _ {k} \ Gamma _ {jr} ^ { s} + \ Gamma _ {jp} ^ {s} \ Gamma _ {kr} ^ {p} - \ Gamma _ {kp} ^ {s} \ Gamma _ {jr} ^ {p}}

(3)\qquad R_{{\;rjk}}^{s}=\partial _{j}\Gamma _{{kr}}^{s}-\partial _{k}\Gamma _{{jr}}^{s}+\Gamma _{{jp}}^{s}\Gamma _{{kr}}^{p}-\Gamma _{{kp}}^{s}\Gamma _{{jr}}^{p}

then at the point $P$ ${\ displaystyle P}$ $P$ we have:

(4)\qquad \nabla _{i}R_{\;rjk}^{s}=\partial _{i}\partial _{j}\Gamma _{kr}^{s}-\partial _{i}\partial _{k}\Gamma _{jr}^{s}

{\ displaystyle (4) \ qquad \ nabla _ {i} R _ {\; rjk} ^ {s} = \ partial _ {i} \ partial _ {j} \ Gamma _ {kr} ^ {s} - \ partial _ {i} \ partial _ {k} \ Gamma _ {jr} ^ {s}}

(4)\qquad \nabla _{i}R_{{\;rjk}}^{s}=\partial _{i}\partial _{j}\Gamma _{{kr}}^{s}-\partial _{i}\partial _{k}\Gamma _{{jr}}^{s}

Rearranging indices in (4) $i j k$ ${\ displaystyle ijk}$ $ijk$ we get two more equalities:

(5)\qquad \nabla _{j}R_{\;rki}^{s}=\partial _{j}\partial _{k}\Gamma _{ir}^{s}-\partial _{j}\partial _{i}\Gamma _{kr}^{s}

{\ displaystyle (5) \ qquad \ nabla _ {j} R _ {\; rki} ^ {s} = \ partial _ {j} \ partial _ {k} \ Gamma _ {ir} ^ {s} - \ partial _ {j} \ partial _ {i} \ Gamma _ {kr} ^ {s}}

(5)\qquad \nabla _{j}R_{{\;rki}}^{s}=\partial _{j}\partial _{k}\Gamma _{{ir}}^{s}-\partial _{j}\partial _{i}\Gamma _{{kr}}^{s}

(6)\qquad \nabla _{k}R_{\;rij}^{s}=\partial _{k}\partial _{i}\Gamma _{jr}^{s}-\partial _{k}\partial _{j}\Gamma _{ir}^{s}

{\ displaystyle (6) \ qquad \ nabla _ {k} R _ {\; rij} ^ {s} = \ partial _ {k} \ partial _ {i} \ Gamma _ {jr} ^ {s} - \ partial _ {k} \ partial _ {j} \ Gamma _ {ir} ^ {s}}

(6)\qquad \nabla _{k}R_{{\;rij}}^{s}=\partial _{k}\partial _{i}\Gamma _{{jr}}^{s}-\partial _{k}\partial _{j}\Gamma _{{ir}}^{s}

It is easy to see that with the addition of equalities (4), (5) and (6) on the left side of the equation will be the expression (1), and on the right, taking into account the commutativity of partial derivatives , all terms are mutually annihilated and we get zero.

Bianchi differential identity

Proof using a special coordinate system

See also

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