Ngô Quốc Anh

November 17, 2009

R-G: Bianchi identities

Filed under: Riemannian geometry — Ngô Quốc Anh @ 0:22

Recall that R_{ikl}^j is defined to be

\displaystyle R_{ikl}^j = \left\langle {R\left( {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right)\frac{\partial }{{\partial {x^i}}},d{x^j}} \right\rangle .

Let

\displaystyle R_{ijkl}=g_{hj} R_{ikl}^h.

The way to understand R_{ijkl} is to look at the following 4-covariant tensor

R(X,Y,Z,T) = g(R(X,Y)Z, T).

As can be seen, the components of R(X,Y,Z,T) are R_{ijkl}.

We first obtain the following result.

Theorem 1. The curvature tensor R_{ijkl}  satisfies the following property

{R_{ijkl}} = - {R_{ijlk}} = - {R_{jikl}} .

Proof.

The proof relies on the definition of the 4-covariant tensor above. To be precise, one has

\displaystyle g\left( {R\left( {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right)\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^j}}}} \right) = g\left( {R_{ikl}^h\frac{\partial }{{\partial {x^h}}},\frac{\partial }{{\partial {x^j}}}} \right) = {g_{hj}}R_{ikl}^h = {R_{ijkl}}

and

\displaystyle g\left( {R\left( {\frac{\partial }{{\partial {x^l}}},\frac{\partial }{{\partial {x^k}}}} \right)\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^j}}}} \right) = {R_{ijlk}}.

Since

\displaystyle R\left( {\frac{\partial }{{\partial {x^l}}},\frac{\partial }{{\partial {x^k}}}} \right)\frac{\partial }{{\partial {x^i}}} = - R\left( {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right)\frac{\partial }{{\partial {x^i}}}

then {R_{ijkl}} = - {R_{ijlk}}. This comes from the definition of curvature tensor and the fact that

\displaystyle\left[ {\frac{\partial }{{\partial {x^m}}},\frac{\partial }{{\partial {x^n}}}} \right] = 0.

Similarly, for the latter case, one can argue as follows

\displaystyle \begin{gathered} g\left( {R\left( {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right)\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^i}}}} \right) \hfill \\ \qquad\qquad= g\left( {{\nabla _{\frac{\partial }{{\partial {x^k}}}}}{\nabla _{\frac{\partial }{{\partial {x^l}}}}}\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^i}}}} \right) - g\left( {{\nabla _{\frac{\partial }{{\partial {x^l}}}}}{\nabla _{\frac{\partial }{{\partial {x^k}}}}}\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^i}}}} \right) - g\left( {{\nabla _{\left[ {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right]}}\frac{\partial }{\partial x^i},\frac{\partial }{{\partial {x^i}}}} \right) \hfill \\ \end{gathered}.

We now use the fact that \nabla is a metric connection. Indeed,

\displaystyle \begin{gathered} \;\;\; g\left( {{\nabla _{\frac{\partial }{{\partial {x^k}}}}}{\nabla _{\frac{\partial }{{\partial {x^l}}}}}\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^i}}}} \right) = \frac{\partial }{{\partial {x^k}}}g\left( {{\nabla _{\frac{\partial }{{\partial {x^l}}}}}\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^i}}}} \right) - g\left( {{\nabla _{\frac{\partial }{{\partial {x^l}}}}}\frac{\partial }{{\partial {x^i}}},{\nabla _{\frac{\partial }{{\partial {x^k}}}}}\frac{\partial }{{\partial {x^i}}}} \right) \hfill \\ - g\left( {{\nabla _{\frac{\partial }{{\partial {x^l}}}}}{\nabla _{\frac{\partial }{{\partial {x^k}}}}}\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^i}}}} \right) = - \frac{\partial }{{\partial {x^l}}}g\left( {{\nabla _{\frac{\partial }{{\partial {x^k}}}}}\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^i}}}} \right) + g\left( {{\nabla _{\frac{\partial }{{\partial {x^k}}}}}\frac{\partial }{{\partial {x^i}}},{\nabla _{\frac{\partial }{{\partial {x^l}}}}}\frac{\partial }{{\partial {x^i}}}} \right) \hfill \\ - g\left( {{\nabla _{\left[ {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right]}}\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^i}}}} \right) = - \left[ {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right]g\left( {\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^i}}}} \right) + g\left( {\frac{\partial }{{\partial {x^i}}},{\nabla _{\left[ {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right]}}\frac{\partial }{{\partial {x^i}}}} \right). \hfill \\\end{gathered}

Thus

\displaystyle \begin{gathered} g\left( {R\left( {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right)\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^i}}}} \right) \hfill \\ \qquad\qquad= \frac{\partial }{{\partial {x^k}}}g\left( {{\nabla _{\frac{\partial }{{\partial {x^l}}}}}\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^i}}}} \right) - \frac{\partial }{{\partial {x^l}}}g\left( {{\nabla _{\frac{\partial }{{\partial {x^k}}}}}\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^i}}}} \right) - \frac{1}{2}\left[ {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right]g\left( {\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^i}}}} \right) = 0. \hfill \\\end{gathered}

Hence

\displaystyle g\left( {R\left( {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right)\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^i}}}} \right) = 0.

The above identity also holds if we replace \frac{\partial}{\partial x^i} by a vector field X. Thus

\displaystyle g\left( {R\left( {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right)\left( {\frac{\partial }{{\partial {x^i}}} + \frac{\partial }{{\partial {x^j}}}} \right),\frac{\partial }{{\partial {x^i}}} + \frac{\partial }{{\partial {x^j}}}} \right) = 0

which implies

\displaystyle g\left( {R\left( {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right)\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^j}}}} \right) = - g\left( {R\left( {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right)\frac{\partial }{{\partial {x^j}}},\frac{\partial }{{\partial {x^i}}}} \right).

Therefore, {R_{ijkl}} = - {R_{jikl}} .

Corollary 1. R(X,Y)Z=-R(Y,Z)Z and \left\langle {R\left( {X,Y} \right)Z,W} \right\rangle = - \left\langle {R\left( {X,Y} \right)W,Z} \right\rangle.

Theorem 2 (the first Bianchi identity). The curvature tensor R_{ijkl}  satisfies the following property

{R_{ijlk}} + {R_{iklj}} + {R_{iljk}} = 0.

Proof. Since

\displaystyle R\left( {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right)\frac{\partial }{{\partial {x^i}}} = {\nabla _{\frac{\partial }{{\partial {x^k}}}}}{\nabla _{\frac{\partial }{{\partial {x^l}}}}}\frac{\partial }{{\partial {x^i}}} - {\nabla _{\frac{\partial }{{\partial {x^l}}}}}{\nabla _{\frac{\partial }{{\partial {x^k}}}}}\frac{\partial }{{\partial {x^i}}} - \underbrace {{\nabla _{\left[ {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right]}}\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^i}}}}_0

then

\displaystyle R\left( {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right)\frac{\partial }{{\partial {x^i}}} = {\nabla _{\frac{\partial }{{\partial {x^k}}}}}{\nabla _{\frac{\partial }{{\partial {x^l}}}}}\frac{\partial }{{\partial {x^i}}} - {\nabla _{\frac{\partial }{{\partial {x^l}}}}}{\nabla _{\frac{\partial }{{\partial {x^k}}}}}\frac{\partial }{{\partial {x^i}}}.

Similarly,

\displaystyle \begin{gathered} R\left( {\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^k}}}} \right)\frac{\partial }{{\partial {x^l}}} = {\nabla _{\frac{\partial }{{\partial {x^i}}}}}{\nabla _{\frac{\partial }{{\partial {x^k}}}}}\frac{\partial }{{\partial {x^l}}} - {\nabla _{\frac{\partial }{{\partial {x^k}}}}}{\nabla _{\frac{\partial }{{\partial {x^i}}}}}\frac{\partial }{{\partial {x^l}}}, \hfill \\ R\left( {\frac{\partial }{{\partial {x^l}}},\frac{\partial }{{\partial {x^i}}}} \right)\frac{\partial }{{\partial {x^k}}} = {\nabla _{\frac{\partial }{{\partial {x^l}}}}}{\nabla _{\frac{\partial }{{\partial {x^i}}}}}\frac{\partial }{{\partial {x^k}}} - {\nabla _{\frac{\partial }{{\partial {x^i}}}}}{\nabla _{\frac{\partial }{{\partial {x^l}}}}}\frac{\partial }{{\partial {x^k}}}. \hfill \\ \end{gathered}.

Since \nabla is torsion free, one gets

\displaystyle {\nabla _{\frac{\partial }{{\partial {x^l}}}}}\frac{\partial }{{\partial {x^i}}} = {\nabla _{\frac{\partial }{{\partial {x^i}}}}}\frac{\partial }{{\partial {x^l}}}, \quad {\nabla _{\frac{\partial }{{\partial {x^k}}}}}\frac{\partial }{{\partial {x^i}}} = {\nabla _{\frac{\partial }{{\partial {x^i}}}}}\frac{\partial }{{\partial {x^k}}}, \quad {\nabla _{\frac{\partial }{{\partial {x^k}}}}}\frac{\partial }{{\partial {x^l}}} = {\nabla _{\frac{\partial }{{\partial {x^l}}}}}\frac{\partial }{{\partial {x^k}}}.

As a consequence,

\displaystyle R\left( {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right)\frac{\partial }{{\partial {x^i}}} + R\left( {\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^k}}}} \right)\frac{\partial }{{\partial {x^l}}} + R\left( {\frac{\partial }{{\partial {x^l}}},\frac{\partial }{{\partial {x^i}}}} \right)\frac{\partial }{{\partial {x^k}}} = 0.

Now

\displaystyle g\left( {R\left( {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right)\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^j}}}} \right) + g\left( {R\left( {\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^k}}}} \right)\frac{\partial }{{\partial {x^l}}},\frac{\partial }{{\partial {x^j}}}} \right) + g\left( {R\left( {\frac{\partial }{{\partial {x^l}}},\frac{\partial }{{\partial {x^i}}}} \right)\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^j}}}} \right) = 0

which implies

\displaystyle {R_{ijkl}} + {R_{ljik}} + {R_{kjli}} = 0.

If we change i by j, j by i we then obtain

\displaystyle \underbrace {{R_{ijkl}}}_{{R_{jikl}}} + \underbrace {{R_{ljik}}}_{{R_{lijk}}} + \underbrace {{R_{kjli}}}_{{R_{kilj}}} = 0

which implies, by using Theorem 1,

\displaystyle - {R_{ijkl}} - {R_{iljk}} - {R_{iklj}} = 0.

Corollary 2. R\left( {X,Y} \right)Z + R\left( {Z,X} \right)Y + R\left( {Y,Z} \right)X = 0.

Corollary 3. Followed from the proof of Theorem 2, by pairing with dx^m to the both sides one has

\displaystyle R_{ikl}^m + R_{lki}^m + R_{kil}^m = 0.

Theorem 3 . The curvature tensor R_{ijkl}  satisfies the following property

R_{ijkl}=R_{klij}.

Proof. By the first Bianchi indentity,

\displaystyle \begin{gathered} {R_{ijkl}} + {R_{iljk}} + {R_{iklj}} = 0, \hfill \\ {R_{jikl}} + {R_{jlik}} + {R_{jkli}} = 0, \hfill \\ \end{gathered}

which implies

\displaystyle 2{R_{ijkl}} + {R_{iljk}} - {R_{jlik}} + {R_{iklj}} - {R_{jkli}} = 0.

Thus

\displaystyle 2{R_{ijkl}} + {R_{iljk}} + {R_{ikjl}} + {R_{iklj}} + {R_{lijk}} = 0.

Similarly, by changing i \to k, j \to l, k \to i and l \to j one gets

\displaystyle 2{R_{klij}} + {R_{kjli}} + {R_{kilj}} + {R_{kijl}} + {R_{jkli}} = 0.

Hence R_{ijkl}=R_{klij} by using Theorem 1.

Theorem 4 (the second Bianchi identity). The curvature tensor R_{ijkl}  satisfies the following property

{R_{ijkl,h}} + {R_{ijlh,k}} + {R_{ijhk,l}} = 0.

Proof. One can use the normal coordinates in order to simplify the calculation. Indeed, normal coordinates tell us at a given point that

g_{ij}=\delta_{ij} and g_{ij,k}=\Gamma_{ij}^k=0

for all i, j, k. Thus,

\displaystyle\begin{gathered} R_{ikl}^h\frac{\partial }{{\partial {x^h}}} \;= R\left( {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right)\frac{\partial }{{\partial {x^i}}} = {\nabla _{\frac{\partial }{{\partial {x^k}}}}}{\nabla _{\frac{\partial }{{\partial {x^l}}}}}\frac{\partial }{{\partial {x^i}}} - {\nabla _{\frac{\partial }{{\partial {x^l}}}}}{\nabla _{\frac{\partial }{{\partial {x^k}}}}}\frac{\partial }{{\partial {x^i}}} \hfill \\ \qquad\qquad = {\nabla _{\frac{\partial }{{\partial {x^k}}}}}\left( {\Gamma _{li}^m\frac{\partial }{{\partial {x^m}}}} \right) - {\nabla _{\frac{\partial }{{\partial {x^l}}}}}\left( {\Gamma _{ki}^n\frac{\partial }{{\partial {x^n}}}} \right) = \frac{{\partial \Gamma _{li}^m}}{{\partial {x^k}}}\frac{\partial }{{\partial {x^m}}} - \frac{{\partial \Gamma _{ki}^n}}{{\partial {x^l}}}\frac{\partial }{{\partial {x^n}}} \hfill \\\qquad\qquad = \left( {\frac{{\partial \Gamma _{li}^m}}{{\partial {x^k}}} - \frac{{\partial \Gamma _{ki}^m}}{{\partial {x^l}}}} \right)\frac{\partial }{{\partial {x^m}}} \hfill \\ \end{gathered}

which implies

\displaystyle \begin{gathered}R_{ikl}^h \;= \frac{{\partial \Gamma _{li}^h}}{{\partial {x^k}}} - \frac{{\partial \Gamma _{ki}^h}}{{\partial {x^l}}} \hfill \\ \qquad= \frac{\partial }{{\partial {x^k}}}\left( {\frac{1}{2}{g^{hm}}\left( {{g_{ml,i}} + {g_{mi,l}} - {g_{li,m}}} \right)} \right) - \frac{\partial }{{\partial {x^l}}}\left( {\frac{1}{2}{g^{hn}}\left( {{g_{nk,i}} + {g_{ni,k}} - {g_{ki,n}}} \right)} \right) \hfill \\ \qquad= \frac{1}{2}{g^{hm}}\left( {{g_{ml,ik}} + {g_{mi,lk}} - {g_{mk,il}} - {g_{mi,kl}} - {g_{li,mk}} + {g_{ki,ml}}} \right) \hfill \\ \qquad = \frac{1}{2}{g^{hm}}\left( {{g_{ml,ik}} - {g_{mk,il}} - {g_{li,mk}} + {g_{ki,ml}}} \right). \hfill \\ \end{gathered}

Hence

\displaystyle {R_{ijkl}} = {g_{jh}}R_{ikl}^h = \frac{1}{2}\left( {{g_{jl,ik}} - {g_{jk,il}} - {g_{li,jk}} + {g_{ki,jl}}} \right)

which implies

\displaystyle{R_{ijkl,h}} = \frac{1}{2}\left( {{g_{jl,ikh}} - {g_{jk,ilh}} - {g_{li,jkh}} + {g_{ki,jlh}}} \right).

Similarly, we can write down R_{ijlh,k} and R_{ijhk,l}. Summing up we get the desired result.

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