Curvature identities
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1 Introduction
Let be a smooth manifold and
a vector bundle over
with
covariant derivative
. Let
be the curvature tensor with respect to a local parametrization
where
.
By definition,
![R_{ij} = -R_{ji}.](/images/math/4/1/5/41508dfe6a906e60d286c177d204c725.png)
If carries an inner product
on each of its fibres and if
is metric, i.e. for any two section
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 \,,](/images/math/7/7/3/773fdbdd20e5941fb164b320dbca60dd.png)
then the expression is skew symmetric also in
the last two indices:
![R_{ijkl} = -R_{ijlk}.](/images/math/0/0/0/0005528a3e11fc14434879a1a9ccd0ac.png)
Now suppose that on the tangent bundle there is another covariant derivative
which is torsion free,
. Then
the tensor derivative
of
is defined:
![\displaystyle (\nabla_iR)_{jk} = \nabla_i(R_{jk}) - R(\nabla_i\phi_j,\phi_k) - R(\phi_j,\nabla_i\phi_k).](/images/math/5/1/1/511e541861227b920aa452befe1d5802.png)
Abbreviating (symmetric in the first two indices
and
),
we have
, using the antisymmetry (1)
and the torsion freeness. Now the cyclic sum of the last two
![\displaystyle -(ij,k)+(ik,j)-(jk,i)+(ji,k) - (ki,j)+(kj,i) = 0.](/images/math/8/e/e/8ee3c29450e44341b1e36f4135ac1700.png)
Tex syntax erroris
![\displaystyle \nabla_i(R_{jk}) = [\nabla_i,R_{jk}] = [\nabla_i,[\nabla_j,\nabla_k]].](/images/math/e/4/d/e4dc3588187f55f39b0008677a43af2c.png)
Thus we obtain an identity for , sometimes called the Second Bianchi Identity
![(\nabla_iR)_{jk}+(\nabla_jR)_{ki}+(\nabla_kR)_{ij} = 0.](/images/math/a/4/5/a457c420bf5c14dbc775423c014f556e.png)
When is a torsion free covariant derivative on the tangent bundle
, there is
in addition the {\it First Bianchi Identity}: Putting
, we have
![R_{ijk} + R_{jki} + R_{kij} = 0 .](/images/math/4/6/5/465d02b15de22ff717ca90bebb847842.png)
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](/images/math/b/3/7/b378fefb24f33c6dfd645b26298bf3f6.png)
An algebraic consequence is the block symmetry:
![R_{ijkl} = R_{klij}](/images/math/6/9/b/69b56262d87204c4bdd3834d8f9dfc7e.png)
In fact, putting 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)](/images/math/5/0/d/50dee8473db55d9d4ff555a718436203.png)
Equations (1), (2), (4), (5) are the algebraic identities for the
curvature tensor of the Levi-Civita derivative, the {\it Riemannian curvature tensor}.
By (1), (2), (5), a Riemannian curvature tensor
can be viewed as a section of , a symmetric bilinear form on
.
The antisymmetric 4-forms form another subspace
, and the
additional identity (4) characterizes precisely
the orthogonal complement of
in
.