Curvature identities

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1 Introduction

Let $M$${{Authors|Jost Eschenburg}}{{Stub}} == Introduction == <wikitex>; Let $M$ be a smooth manifold and $E$ a vector bundle over $M$ with [[Covariant derivative|covariant derivative]] $\nabla$. Let $R_{ij} = R(\phi_i,\phi_j) = [\nabla_i,\nabla_j]$ be the [[Curvature tensor and second derivative|curvature tensor]] with respect to a local parametrization $\phi : \R^n_o \to M$ where $\nabla_i = \nabla_{\phi_i} = \nabla_{\partial_i\phi}$. By definition, \begin{equation} \label{Rij} R_{ij} = -R_{ji}. \end{equation} If $E$ carries an inner product $\langle \,,\,\rangle $ on each of its fibres and if $\nabla$ is ''metric'', i.e. for any two section $s,\tilde s\in \Gamma E$ the inner product satisfies the product rule $$ \partial_i\langle s,\tilde s\rangle = \langle \nabla_is,\tilde s\rangle + \langle s,\nabla_i\tilde s\rangle \,, $$ then the expression $R_{ijkl} = \langle R_{ij}\phi_k,\phi_l\rangle $ is skew symmetric also in the last two indices: \begin{equation} \label{Rkl} R_{ijkl} = -R_{ijlk}. \end{equation} Now suppose that on the tangent bundle $TM$ there is another covariant derivative $\nabla$ which is [[Torsion tensor|''torsion free'']], $\nabla_i\phi_j = \nabla_j\phi_i$. Then the [[Tensor derivative|tensor derivative]] $\nabla R$ of $R$ is defined: $$ (\nabla_iR)_{jk} = \nabla_i(R_{jk}) - R(\nabla_i\phi_j,\phi_k) - R(\phi_j,\nabla_i\phi_k). $$ Abbreviating $R(\nabla_i\phi_j,\phi_k) = (ij,k)$ (symmetric in the first two indices $i$ and $j$), we have $(\nabla_iR)_{jk} = \nabla_iR_{jk} - (ij,k) + (ik,j)$, using the antisymmetry (1) and the torsion freeness. Now the cyclic sum of the last two terms vanishes: $$-(ij,k)+(ik,j)-(jk,i)+(ji,k) - (ki,j)+(kj,i) = 0.$$ But the cyclic sum of the first term vanishes also by Jacobi identity, since the tensor derivative of $R_{ij} \in \Gamma{\rm End}(TM)$ is $$ \nabla_i(R_{jk}) = [\nabla_i,R_{jk}] = [\nabla_i,[\nabla_j,\nabla_k]]. $$ Thus we obtain an identity for $\nabla R$, sometimes called the ''Second Bianchi Identity'' \begin{equation} \label{BII} (\nabla_iR)_{jk}+(\nabla_jR)_{ki}+(\nabla_kR)_{ij} = 0. \end{equation} When $\nabla$ is a torsion free covariant derivative on the tangent bundle $TM$, there is in addition the ''First Bianchi Identity'': Putting $R_{ijk} = R_{ij}\phi_k$, we have \begin{equation} \label{BI} R_{ijk} + R_{jki} + R_{kij} = 0 . \end{equation} In fact, $$ \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 $$ An algebraic consequence is the block symmetry: \begin{equation} \label{block} R_{ijkl} = R_{klij} \end{equation} In fact, putting $ij|kl := R_{ijkl}$ and applying (\ref{BI}), (\ref{Rij}), (\ref{Rkl}) we obtain (expressions with equal parentheses cancel each other): $$ 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) $$ Equations (\ref{Rij}), (\ref{Rkl}), (\ref{BI}), (\ref{block}) are the algebraic identities for the curvature tensor of the [[Levi-Civita connection|Levi-Civita derivative]], the ''Riemannian curvature tensor''. By (\ref{Rij}), (\ref{Rkl}), (\ref{block}), a Riemannian curvature tensor can be viewed as a section of $S^2\Lambda^2TM$, a symmetric bilinear form on $\Lambda^2TM$. The antisymmetric 4-forms form another subspace $\Lambda^4TM\subset S^2\Lambda^2TM$, and the additional identity (\ref{BI}) characterizes precisely the orthogonal complement of $\Lambda^4(TM)$ in $S^2\Lambda^2(TM)$. </wikitex> == References == {{#RefList:}} [[Category:Definitions]] [[Category:Connections and curvature]]M$ be a smooth manifold and $E$$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

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

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:

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

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:

$\displaystyle -(ij,k)+(ik,j)-(jk,i)+(ji,k) - (ki,j)+(kj,i) = 0.$ $$\displaystyle -(ij,k)+(ik,j)-(jk,i)+(ji,k) - (ki,j)+(kj,i) = 0.$$

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

$\displaystyle \nabla_i(R_{jk}) = [\nabla_i,R_{jk}] = [\nabla_i,[\nabla_j,\nabla_k]].$ $$\displaystyle \nabla_i(R_{jk}) = [\nabla_i,R_{jk}] = [\nabla_i,[\nabla_j,\nabla_k]].$$

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,

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

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):

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

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)$.