18.2 Covariant derivative of tensor fields

中文

For every vector field $X$ we define a map $D_X:{\mathcal{T}}^{(r,s)}(M)\to {\mathcal{T}}^{(r,s)}(M)$ that extends the covariant derivative to general tensor fields by requiring:

(Cov1) For scalar fields, $f\in \mathcal{F}(M)=T^{(0,0)}(M),\mathrm{we}\mathrm{set}{D}_{X}f=Xf$

(Cov2) For 1-forms $\omega \in {\mathcal{T}}^{(0,1)}(M)$ assume a Leibnitz rule for , ,

$$D_X⟨\omega ,Y⟩=⟨D_X\omega ,Y⟩+⟨\omega ,D_{X}Y⟩.$$

(Cov3) $D_X(T+S)=D_{X}T+D_{X}S$ for any pair of tensor fields $T$, $S\in {\mathcal{T}}^{(r,s)}(M)$

(Cov4) The Leibnitz rule holds with respect to tensor products

$$D_X(T\otimes S)=(D_{X}T)\otimes S+T\otimes D_{X}S.$$

These requirements define a unique tensor field $D_{X}T$ for any smooth tensor field $T$. Firstly, let $\omega =w_i\mathrm{d}{x}^i$ be any 1-form defined on a coordinate chart $(U;x^i)$ covering $p.$ Setting $Y={\partial }_{x^i}$ , condition (Cov1) gives $D_X⟨\omega ,{\partial }_{x^i}⟩=D_X(w_i)=X^k{w}_{i,k}$ , while (Cov2) implies

$$D_X⟨\omega ,{\partial }_{x^i}⟩=⟨D_X\omega ,{\partial }_{x^i}⟩+⟨\omega ,X^k{D}_k{\partial }_{x^i}⟩.$$

Hence, using Eq. (18.5) and $(D_X\omega {)}_i=⟨D_X\omega ,{\partial }_{x^i}⟩$ , we find

$${\left(D_X\omega \right)}_i=w_{i;k}{X}^k\text{\hspace{1em}(18.12)}$$

where

$$w_{i;k}=w_{i,k}-{\Gamma }_{ik}^j{w}_j.\text{\hspace{1em}(18.13)}$$

Exercise

Verify that Eq. (18.13) implies the coordinate expression for condition (Cov2),

$${\left(w_i{Y}^i\right)}_{,k}{X}^k=w_{i;k}{X}^k{Y}^i+w_i{Y}_{;k}^i{X}^k.$$

For a general tensor $T_{:}$ , expand in terms of the basis consisting of tensor products of the ${\partial }_{x^i}$ and $\mathrm{d}{x}^j$ and use (Cov3) and (Cov4). For example, if

$$T=T_{kl\dots }^{ij\dots }{\partial }_{x^i}\otimes {\partial }_{x^j}\otimes \cdots \otimes \mathrm{d}{x}^k\otimes \mathrm{d}{x}^l\otimes \dots$$

a straightforward calculation results in

$$D_{X}T=T_{kl\dots ;p}^{ij\dots }{X}^p{\partial }_{x^i}\otimes {\partial }_{x^j}\otimes \cdots \otimes \mathrm{d}{x}^k\otimes \mathrm{d}{x}^l\otimes \dots$$

where

$$\begin{array}{c}{T}_{kl\dots ;p}^{ij\dots }=T_{kl\dots ,p}^{ij\dots }+{\Gamma }_{ap}^i{T}_{kl\dots }^{aj\dots }+{\Gamma }_{ap}^j{T}_{kl\dots }^{ia\dots }+\dots \\ -{\Gamma }_{kp}^a{T}_{al\dots }^{ij\dots }-{\Gamma }_{lp}^a{T}_{ka\dots }^{ij\dots }-\dots \end{array}\text{\hspace{1em}(18.14)}$$

This demonstrates that (Cov1)–(Cov4) can be used to compute the components of $D_{X}T$ at any point $p\in M$ with respect to a coordinate chart $(U;x^i)$ covering $p,$ , and thus uniquely define the tensor field $D_{X}T$ throughout $M$.

For every tensor field $T$ of type $(r,s)$ its covariant derivative $DT$ is the tensor field of type $(r,s+1)$ defined by

$$DT({\omega }^1,{\omega }^2,\dots ,{\omega }^r,Y_1,Y_2,\dots ,Y_s,X)=D_{X}T({\omega }^1,{\omega }^2,\dots ,{\omega }^r,Y_1,Y_2,\dots ,Y_s).$$

Exercise

Show that, with respect to any local coordinates, the tensor field $DT$ has component $T^{ij\dots }{}_{kl\dots ;p}$ defined by Eq. (18.14).

Exercise

Show that

$${\delta }_{j;k}^i=0.\text{\hspace{1em}(18.15)}$$

The covariant derivative commutes with all contractions on a tensor field

$$D\left(C_l^{k}T\right)=C_l^{k}DT.\text{\hspace{1em}(18.16)}$$

This relation is most easily shown in local coordinates, where it reads

$$(T_{j_1\dots a\dots j_s}^{i_1\dots a\dots i_r}{)}_{;k}=T_{j_1\dots a\dots j_s;k}^{i_1\dots a\dots i_r},\text{\hspace{1em}(18.17)}$$

the upper index $a$ being in the $k$th position, the lower in the $l$th. If we expand the right-hand side according to Eq. (18.14) the terms corresponding to these indices are

$${\Gamma }_{bk}^a{T}_{j_1\dots a\dots j_s}^{i_1\dots b\dots i_r}-{\Gamma }_{ak}^b{T}_{j_1\dots b\dots j_s}^{i_1\dots a\dots i_r}=0$$

and what remains reduces to the expression formed by expanding the left-hand side of Eq. (18.17).

A useful corollary of this property is the following relation:

$$\begin{array}{l}X(T({\omega }^1,\dots ,{\omega }^r,Y_1,\dots ,Y_s))=D_{X}T({\omega }^1,\dots ,{\omega }^r,Y_1,\dots ,Y_s)\\ +T(D_X{\omega }^1,\dots ,{\omega }^r,Y_1,\dots ,Y_s)+\cdots +T({\omega }^1,\dots ,D_X{\omega }^r,Y_1,\dots ,Y_s)\\ +T({\omega }^1,\dots ,{\omega }^r,D_X{Y}_1,\dots ,Y_s)+\cdots +T({\omega }^1,\dots ,{\omega }^r,Y_1,\dots ,D_X{Y}_s).\end{array}\text{\hspace{1em}(18.18)}$$

Problems

Problem 18.3

Show directly from (Cov1)–(Cov4) that $D_{fX}T=f{D}_{X}T$ for all vector fields $X$ tensor fields $T$ and scalar functions $f:M\to \mathbb{R}$

Problem 18.4

Verify from the coordinate transformation rule Eq. (18.11) for ${\Gamma }_{jk}^i$ that the components of the covariant derivative of an arbitrary tensor field, defined in Eq. (18.14), transform as components of a tensor field.

Problem 18.5

Show that the identity Eq. (18.18) follows from Eq. (18.16).