# 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.