# Curvature identities

## 1 Introduction

Let $M$$\newcommand{\Z}{\mathbb{Z}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\N}{\mathbb{N}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\F}{\mathbb{F}} \newcommand{\bZ}{\mathbb{Z}} \newcommand{\bR}{\mathbb{R}} \newcommand{\bC}{\mathbb{C}} \newcommand{\bH}{\mathbb{H}} \newcommand{\bQ}{\mathbb{Q}} \newcommand{\bF}{\mathbb{F}} \newcommand{\bN}{\mathbb{N}} \DeclareMathOperator\id{id} % identity map \DeclareMathOperator\Sq{Sq} % Steenrod squares \DeclareMathOperator\Homeo{Homeo} % group of homeomorphisms of a topoloical space \DeclareMathOperator\Diff{Diff} % group of diffeomorphisms of a smooth manifold \DeclareMathOperator\SDiff{SDiff} % diffeomorphism under some constraint \DeclareMathOperator\Hom{Hom} % homomrphism group \DeclareMathOperator\End{End} % endomorphism group \DeclareMathOperator\Aut{Aut} % automorphism group \DeclareMathOperator\Inn{Inn} % inner automorphisms \DeclareMathOperator\Out{Out} % outer automorphism group \DeclareMathOperator\vol{vol} % volume \newcommand{\GL}{\text{GL}} % general linear group \newcommand{\SL}{\text{SL}} % special linear group \newcommand{\SO}{\text{SO}} % special orthogonal group \newcommand{\O}{\text{O}} % orthogonal group \newcommand{\SU}{\text{SU}} % special unitary group \newcommand{\Spin}{\text{Spin}} % Spin group \newcommand{\RP}{\Rr\mathrm P} % real projective space \newcommand{\CP}{\Cc\mathrm P} % complex projective space \newcommand{\HP}{\Hh\mathrm P} % quaternionic projective space \newcommand{\Top}{\mathrm{Top}} % topological category \newcommand{\PL}{\mathrm{PL}} % piecewise linear category \newcommand{\Cat}{\mathrm{Cat}} % any category \newcommand{\KS}{\text{KS}} % Kirby-Siebenmann class \newcommand{\Hud}{\text{Hud}} % Hudson torus \newcommand{\Ker}{\text{Ker}} % Kernel \newcommand{\underbar}{\underline} %Classifying Spaces for Families of Subgroups \newcommand{\textup}{\text} \newcommand{\sp}{^}M$ be a smooth manifold and $E$$E$ a vector bundle over $M$$M$ with covariant derivative $\nabla$$\nabla$. Let $R_{ij} = R(\phi_i,\phi_j) = [\nabla_i,\nabla_j]$$R_{ij} = R(\phi_i,\phi_j) = [\nabla_i,\nabla_j]$ be the curvature tensor with respect to a local parametrization $\phi : \R^n_o \to M$$\phi : \R^n_o \to M$ where $\nabla_i = \nabla_{\phi_i} = \nabla_{\partial_i\phi}$$\nabla_i = \nabla_{\phi_i} = \nabla_{\partial_i\phi}$. By definition,

(1)$R_{ij} = -R_{ji}.$$R_{ij} = -R_{ji}.$

If $E$$E$ carries an inner product $\langle \,,\,\rangle$$\langle \,,\,\rangle$ on each of its fibres and if $\nabla$$\nabla$ is metric, i.e. for any two section $s,\tilde s\in \Gamma E$$s,\tilde s\in \Gamma E$ the inner product satisfies the product rule

$\displaystyle \partial_i\langle s,\tilde s\rangle = \langle \nabla_is,\tilde s\rangle + \langle s,\nabla_i\tilde s\rangle \,,$

then the expression $R_{ijkl} = \langle R_{ij}\phi_k,\phi_l\rangle$$R_{ijkl} = \langle R_{ij}\phi_k,\phi_l\rangle$ is skew symmetric also in the last two indices:

(2)$R_{ijkl} = -R_{ijlk}.$$R_{ijkl} = -R_{ijlk}.$

Now suppose that on the tangent bundle $TM$$TM$ there is another covariant derivative $\nabla$$\nabla$ which is torsion free, $\nabla_i\phi_j = \nabla_j\phi_i$$\nabla_i\phi_j = \nabla_j\phi_i$. Then the tensor derivative $\nabla R$$\nabla R$ of $R$$R$ is defined:

$\displaystyle (\nabla_iR)_{jk} = \nabla_i(R_{jk}) - R(\nabla_i\phi_j,\phi_k) - R(\phi_j,\nabla_i\phi_k).$

Abbreviating $R(\nabla_i\phi_j,\phi_k) = (ij,k)$$R(\nabla_i\phi_j,\phi_k) = (ij,k)$ (symmetric in the first two indices $i$$i$ and $j$$j$), we have $(\nabla_iR)_{jk} = \nabla_iR_{jk} - (ij,k) + (ik,j)$$(\nabla_iR)_{jk} = \nabla_iR_{jk} - (ij,k) + (ik,j)$, using the antisymmetry (1) and the torsion freeness. Now the cyclic sum of the last two

terms vanishes:
$\displaystyle -(ij,k)+(ik,j)-(jk,i)+(ji,k) - (ki,j)+(kj,i) = 0.$
But the cyclic sum of the first term vanishes also by Jacobi identity, since the tensor derivative of
Tex syntax error
$R_{ij} \in \Gamma{\rm End}(TM)$ is
$\displaystyle \nabla_i(R_{jk}) = [\nabla_i,R_{jk}] = [\nabla_i,[\nabla_j,\nabla_k]].$

Thus we obtain an identity for $\nabla R$$\nabla R$, sometimes called the Second Bianchi Identity

(3)$(\nabla_iR)_{jk}+(\nabla_jR)_{ki}+(\nabla_kR)_{ij} = 0.$$(\nabla_iR)_{jk}+(\nabla_jR)_{ki}+(\nabla_kR)_{ij} = 0.$

When $\nabla$$\nabla$ is a torsion free covariant derivative on the tangent bundle $TM$$TM$, there is in addition the First Bianchi Identity: Putting $R_{ijk} = R_{ij}\phi_k$$R_{ijk} = R_{ij}\phi_k$, we have

(4)$R_{ijk} + R_{jki} + R_{kij} = 0 .$$R_{ijk} + R_{jki} + R_{kij} = 0 .$

In fact,

$\displaystyle \left\{\begin{matrix} & R_{ijk} \cr + & R_{jki} \cr + & R_{kij}\end{matrix}\right\}\ \ =\ \left\{\begin{matrix} &\nabla_i\nabla_j\phi_k &-& \underline{\nabla_j\nabla_i\phi_k} \cr + &\underline{\nabla_j\nabla_k\phi_i} &-& \underline{\underline{\nabla_k\nabla_j\phi_i}} \cr + &\underline{\underline{\nabla_k\nabla_i\phi_j}} &-& \nabla_i\nabla_k\phi_j \end{matrix}\right\} \ \ =\ \ 0$

An algebraic consequence is the block symmetry:

(5)$R_{ijkl} = R_{klij}$$R_{ijkl} = R_{klij}$

In fact, putting $ij|kl := R_{ijkl}$$ij|kl := R_{ijkl}$ and applying (4), (1), (2) we obtain (expressions with equal parentheses cancel each other):

$\displaystyle 0 = \left\{\begin{matrix} \ \ \,{ij|kl} \cr +\, (jk|il) \cr +\, [ki|jl] \end{matrix}\right\} + \left\{\begin{matrix} \ \ \,{ji|lk} \cr +\, \langle il|jk\rangle \cr + \{lj|ik\} \end{matrix} \right\}- \left\{\begin{matrix} \ \ \,{kl|ij} \cr +\, \langle li|kj\rangle \cr +\, [ik|lj] \end{matrix} \right\}- \left\{\begin{matrix} \ \ \,{lk|ji} \cr +\, (kj|li) \cr + \{jl|ki\} \end{matrix}\right\} = 2(ij|kl - kl|ij)$

Equations (1), (2), (4), (5) are the algebraic identities for the curvature tensor of the Levi-Civita derivative, the Riemannian curvature tensor. By (1), (2), (5), a Riemannian curvature tensor can be viewed as a section of $S^2\Lambda^2TM$$S^2\Lambda^2TM$, a symmetric bilinear form on $\Lambda^2TM$$\Lambda^2TM$. The antisymmetric 4-forms form another subspace $\Lambda^4TM\subset S^2\Lambda^2TM$$\Lambda^4TM\subset S^2\Lambda^2TM$, and the additional identity (4) characterizes precisely the orthogonal complement of $\Lambda^4(TM)$$\Lambda^4(TM)$ in $S^2\Lambda^2(TM)$$S^2\Lambda^2(TM)$.