On the existence of a metric compatible with a given connection

Question: Suppose we are given a torsion-free (i.e. the torsion tensor vanishes) affine connection {\nabla } on a smooth connected manifold {M}. Does there exist a Riemannian metric {g} such that its Levi-Civita connection is {\nabla }? If so, is it unique if we prescribe its value at a point?

Let {(x^i)} be a local coordinates on {M} and let {\nabla _i\partial _j=\Gamma_{ij}^k\partial _k} (using Einstein’s notation). The torsion-free condition is then equivalent to {\Gamma_{ij}^k=\Gamma_{ji}^k}. We want to solve the linear system of partial differential equations (Note that {\Gamma_{ij}^k} are {C^1} as {\nabla } is a connection.)

\displaystyle \begin{array}{rl} \displaystyle  \partial _k g_{ij}=\Gamma_{ki}^pg_{pj}+\Gamma_{kj}^pg_{pi}. \ \ \ \ \ (1)\end{array}

We impose an initial condition at {x_0} by

\displaystyle \begin{array}{rl} \displaystyle g_{ij}(x_0)= h_{ij}\end{array}

for some fixed symmetric {h_{ij}}.

It is clear from the initial condition and from (1) that {g_{ij}} is symmetric if the solution exists. According to the theory of first order partial differential equations system (cf. e.g. Stoker’s Differential Geometry Appendix B), the system (1) is uniquely solvable if and only if the compatibility conditions hold:

\displaystyle \begin{array}{rl} \displaystyle   & \displaystyle \partial _l\Gamma_{ki}^pg_{pj} +\Gamma_{ki}^p \partial _lg_{pj} +\partial _l\Gamma_{kj}^pg_{ip} +\Gamma_{kj}^p \partial _lg_{ip}\\ =& \displaystyle \partial _k\Gamma_{li}^pg_{pj} +\Gamma_{li}^p \partial _kg_{pj} +\partial _k\Gamma_{lj}^pg_{ip} +\Gamma_{lj}^p \partial _kg_{ip}.  \ \ \ \ \ (2)\end{array}

In view of (1), this is equivalent to

\displaystyle \begin{array}{rl} \displaystyle   & \displaystyle \partial _l\Gamma_{ki}^pg_{pj} +\Gamma_{ki}^p (\Gamma_{lp}^m g_{mj}+\Gamma_{lj}^m g_{pm}) +\partial _l\Gamma_{kj}^pg_{ip} +\Gamma_{kj}^p (\Gamma_{li}^m g_{mp}+\Gamma_{lp}^m g_{im})\\ =& \displaystyle \partial _k\Gamma_{li}^pg_{pj} +\Gamma_{li}^p (\Gamma_{kp}^m g_{mj}+\Gamma_{kj}^m g_{pm}) +\partial _k\Gamma_{lj}^pg_{ip} +\Gamma_{lj}^p (\Gamma_{ki}^m g_{mp}+\Gamma_{kp}^m g_{im}).  \ \ \ \ \ (3)\end{array}

To shed some light on the following computation, let us introduce the curvature tensor associated with {\nabla }.

Definition 1 Given a connection {\nabla } and vector fields {X}, {Y}, {Z}, we define the curvature tensor {R(X, Y)Z} by

\displaystyle \begin{array}{rl} \displaystyle  R (X, Y)Z=\nabla _X\nabla _YZ-\nabla _Y\nabla _XZ-\nabla _{[X,Y]}Z. \ \ \ \ \ (4)\end{array}

Here {[\cdot,\cdot]} is the Lie bracket of vector fields. It is readily checked that {R(\cdot, \cdot)\cdot} is a tensor field regardless of whether {\nabla } is torsion-free or not. In local coordinates, we define {R_{ijk}^l} by

\displaystyle \begin{array}{rl} \displaystyle R(\partial _i, \partial _j)\partial _k=R_{ijk}^l\partial _l.\end{array}

It is not hard to see that if {\nabla } is torsion-free and the {\nabla } is compatible with {g} in the sense of (1), then we have the local formula

\displaystyle \begin{array}{rl} \displaystyle R_{ijk}^l = \partial _i \Gamma_{jk}^l - \partial _j \Gamma_{ik }^l + \Gamma _{jk}^m \Gamma_{im}^l- \Gamma _{ik}^m \Gamma _{jm}^l.\end{array}

We subtract RHS from LHS of (3) and get

\displaystyle \begin{array}{rl} \displaystyle   & \displaystyle \partial _l\Gamma_{ki}^pg_{pj} +\Gamma_{ki}^p (\Gamma_{lp}^m g_{mj}+\Gamma_{lj}^m g_{pm}) +\partial _l\Gamma_{kj}^pg_{ip} +\Gamma_{kj}^p (\Gamma_{li}^m g_{mp}+\Gamma_{lp}^m g_{im})\\ & \displaystyle -\left(\partial _k\Gamma_{li}^pg_{pj} +\Gamma_{li}^p (\Gamma_{kp}^m g_{mj}+\Gamma_{kj}^m g_{pm}) +\partial _k\Gamma_{lj}^pg_{ip} +\Gamma_{lj}^p (\Gamma_{ki}^m g_{mp}+\Gamma_{kp}^m g_{im})\right)\\ =& \displaystyle  \left(\partial _l \Gamma_{ki}^m-\partial _k \Gamma_{li}^m+\Gamma_{ki}^p\Gamma_{lp}^m-\Gamma_{li}^p\Gamma_{kp}^m\right)g_{mj} +\left(\partial _l \Gamma_{kj}^m-\partial _k \Gamma_{lj}^m+\Gamma_{kj}^p\Gamma_{lp}^m-\Gamma_{lj}^p\Gamma_{kp}^m\right)g_{mi}\\ & \displaystyle +\Gamma_{ki}^p\Gamma_{lj}^m g_{pm}+\Gamma_{kj}^p\Gamma_{li}^m g_{mp}-\Gamma_{li}^p \Gamma_{kj}^m g_{pm}-\Gamma_{lj}^p\Gamma_{ki}^m g_{mp}\\ =& \displaystyle  \left(\partial _l \Gamma_{ki}^m-\partial _k \Gamma_{li}^m+\Gamma_{ki}^p\Gamma_{lp}^m-\Gamma_{li}^p\Gamma_{kp}^m\right)g_{mj} +\left(\partial _l \Gamma_{kj}^m-\partial _k \Gamma_{lj}^m+\Gamma_{kj}^p\Gamma_{lp}^m-\Gamma_{lj}^p\Gamma_{kp}^m\right)g_{mi}\\ =& \displaystyle R_{lki}^m g_{mj}+R_{lkj}^m g_{mi}.  \ \ \ \ \ (5)\end{array}

Let us show that (1) implies {R_{lki}^m g_{mj}+R_{lkj}^m g_{mi}=0}, so by (5), we conclude that the compatibility condition (2) holds. To see this, using the compatibility of metric (1) (We do not sum over {k} in the following):

\displaystyle \begin{array}{rl} \displaystyle   \frac{1}{2}\partial _i\partial _j g_{kk} =\partial _i\left(\Gamma_{jk}^l g_{lk}\right) =\partial _i\Gamma_{jk}^l g_{lk}+\Gamma_{jk}^l\Gamma_{il}^m g_{mk} +\Gamma_{jk}^l\Gamma_{ik}^m g_{ml} \end{array}

and similarly

\displaystyle \begin{array}{rl} \displaystyle \frac{1}{2}\partial _j \partial _i g_{kk} =\partial _j\Gamma_{ik}^l g_{lk}+\Gamma_{ik}^l\Gamma_{jl}^m g_{mk} +\Gamma_{ik}^l\Gamma_{jk}^m g_{ml}. \end{array}

Subtracting the two equations,

\displaystyle \begin{array}{rl} \displaystyle   0 =\partial _i\Gamma_{jk}^l g_{lk}-\partial _j\Gamma_{ik}^l g_{lk}+\Gamma_{jk}^l\Gamma_{il}^m g_{mk} -\Gamma_{ik}^l\Gamma_{jl}^m g_{mk} = R_{ijk}^lg_{lk}. \end{array}

This implies that {g(R(X, Y)Z, Z)=0} if {g} is defined by {g:=g_{ij}dx^idx^j}. This is easily seen to be equivalent to {R_{lki}^m g_{mj}+R_{lkj}^m g_{mi}=0}. We conclude that (2) holds, and therefore we have proved the existence and uniqueness of the solution to the system

\displaystyle \begin{array}{rl} \displaystyle   \begin{cases} g_{ij,k}=\Gamma_{ki}^pg_{pj}+\Gamma_{kj}^pg_{ip}\\ g_{ij}(x_0)=h_{ij}. \end{cases} \end{array}

Gometrically, if we prescribe an inner product on {T_{p_0}M} for some {p_0\in M}, then there is a unique Riemannian metric which is compatible with the torsion-free connection {\nabla }. So we have proved

Theorem 2 Suppose {\nabla } is a torsion-free affine connection on a smooth connected manifold {M}. Let {g_0} be an inner product on {T_{p_0}M}, where {p_0\in M}. Then there exists a unique Riemannian metric {g} such that {\nabla } is the Levi-Civita connection of {g} and {g(p_0)=g_0}.

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2 Responses to On the existence of a metric compatible with a given connection

  1. tong cheung yu says:

    hallmarkXmas

    ________________________________
    寄件者: Mathematics@CUHK
    寄件日期: 2016年12月5日 5:53
    收件者: yiuyu4@hotmail.com
    主旨: [New post] On the existence of a metric compatible with a given connection

    KKK posted: ” Question: Suppose we are given a torsion-free (i.e. the torsion tensor vanishes) affine connection {\nabla } on a smooth connected manifold {M}. Does there exist a Riemannian metric {g} su”

  2. tong cheung yu says:

    hallmarkXmas

    ________________________________
    寄件者: Mathematics@CUHK
    寄件日期: 2017年5月20日 15:24
    收件者: yiuyu4@hotmail.com
    主旨: [New post] The Cauchy-Schwarz inequality and the Lagrange identity

    KKK posted: ” The classical Lagrange identity is the following: $latex \displaystyle \begin{array}{rl} \displaystyle \left(\sum_{i=1}^n a_ib_i\right)^2+\sum _{1\le i<j\le n} \left(a_ib_j-a_jb_i\right)^2 =\left(\sum_{i=1}^{n}a_i^2\right)\left(\sum_{j=1}^{n}b_j^2\rig"

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