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Let $(M,g)$ be a Riemannian manifold and $\nabla$ be the Levi-Civita connection. I understand that the total covariant derivative $\nabla g$ of the metric tensor is zero. However, In general relativity's book, it is saying that the covariant derivative $\nabla_{\partial_k}g_{ij}$ is zero. It seem to me that $\nabla_{\partial_k}g_{ij}$ is just the function $\partial_{k}g_{ij}$ since $g_{ij}$ is a smooth function defined on the manifold. Does it mean that partial derivatives of metric components are zero or is there something wrong with my understanding? Many thanks to every answer!

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    $\begingroup$ $g_{ij}$ are the components of the metric tensor in some coordinate system. So $g_{ij}$ is (not just) a smooth function. Similarly, $\nabla_{\partial_k} g_{ij}$ is the expression for the covariant derivative of the covariant derivative of the metric tensor in a coordinate system. Since the metric tensor is known to be parallel wrt the Levi Civita Connection, this is $=0$. And this is not just true in general relativity, but in Riemannian Geometry as well. $\endgroup$
    – Thomas
    Dec 11, 2022 at 21:08
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    $\begingroup$ No because $\partial_kg_{ij}$ is in general nonzero. $\endgroup$
    – J.G.
    Dec 11, 2022 at 21:15
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    $\begingroup$ No. $\partial_k g_{ij}$ is the partial derivative of the function $g_{ij}$ in the local coordinate system. You may say that in the origin of a normal coordinate system, but not in general. $\endgroup$
    – Thomas
    Dec 11, 2022 at 21:15
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    $\begingroup$ Notation is a funny thing, but here's a little fact which you should remember. People are lazy, and they omit brackets often, so it is declared by convention that you should always perform covariant derivatives first, then extract components. Quiz: how should you interpret $\nabla_i\nabla_jf$? $\endgroup$
    – peek-a-boo
    Dec 11, 2022 at 22:15
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    $\begingroup$ A Riemannian metric defines an inner product on each tangent space $T_pM$ that depends smoothly on $p$. That $\nabla g = 0$ means simply that if you parallel translate two tangent vectors along a curve, then their inner product remains constant. $\endgroup$
    – Deane
    Dec 11, 2022 at 22:35

3 Answers 3

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The part you're missing is that although you originally define the Levi-Civita connection on vector fields, you naturally extend it to general tensors essentially via the product rule. Namely let $X,Y$ be vector fields, then a general $(0,2)$-tensor satisfies the following product rule $$\partial_Z(g(X,Y))=\nabla_Z (g(X,Y))=(\nabla_Z g)(X,Y)+g(\nabla_Z X,Y)+g(X,\nabla_ZY);$$ in other words, $\nabla_Z g$ is defined as the $(0,2)$-tensor to satisfy the above formula. The requirement of metric-compatibility of the Levi-Civita connection is exactly $\nabla_Z g\equiv 0$ for all vector fields $Z$. If you want to work with local coordinates, metric-compatibility tells us for $X=\partial_i,Y=\partial_j,Z=\partial_k$, $$\partial_k g_{ij}=g(\Gamma^l_{ki}\partial_l,\partial_j)+g(\partial_i,\Gamma^l_{kj}\partial_l)=\Gamma^l_{ki}g_{lj}+g_{il}\Gamma^l_{kj}.$$

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  • $\begingroup$ Does it mean that the notation $\nabla_{\partial_k}g_{ij}$ means the $ij$-th component of the (0,2)-tensor $\nabla_{\partial_k}g$? $\endgroup$
    – 5Somebody
    Dec 11, 2022 at 21:48
  • $\begingroup$ @5Somebody Usually we would write $\nabla_kg_{ij}$ to mean $(\nabla g)_{ij~k}$. Note that $\nabla g$ is a (0,3), not a (0,2), tensor. $\endgroup$
    – K.defaoite
    Dec 11, 2022 at 21:50
  • $\begingroup$ @5Somebody As you know, notation in DG is a nightmare. If the author claims that $\nabla_k g_{ij}\equiv 0$ for all $k$, then yes, I would interpret his writing as what you've written. I think the closest I've seen to this notation is $g_{ij;k}$ where the $;$ stands for covariant derivative. $\endgroup$
    – Mr. Brown
    Dec 11, 2022 at 21:52
  • $\begingroup$ @ZackFox Best to stay coordinate-free as much as possible and just say $\nabla g=0$. $\endgroup$
    – K.defaoite
    Dec 11, 2022 at 21:55
  • $\begingroup$ Different strokes for different folks. It's probably nice to see every approach at least once, since everyone has their own favorite method. $\endgroup$
    – Mr. Brown
    Dec 11, 2022 at 21:58
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As I mentioned in my comments, usually it is a good idea to stay away from indices and coordinates as much as possible. So I will try to do that.


Start with a pseudo-Riemannian manifold $(\mathcal M,g)$. It was proved by Lev-Civita that there is a unique torsion-free affine connection which we denote $\nabla$ that preserves the metric, i.e $\nabla g=0$.

Why do we like this connection? Well, it makes computations much easier. For one, it is torsion free, which means that we don't need to worry about the order of the lower indices of our connection coefficients. This is already convenient. And, the other key point is that if we let $v,\omega$ be a vector and covector in $\mathrm T\mathcal M,\mathrm T^*\mathcal M$ respectively, if we define the sharp and flat of $\omega,v$ as $$\omega_{\sharp}=g^{-1}(\omega,\cdot) \\ v^\flat=g(v,\cdot)$$ The metric compatibility means that sharping and flatting commutes with covariant differentiation, that is $$\nabla^*(\omega_{\sharp})=(\nabla^*\omega)_{\sharp}\\\nabla(v^\flat)=(\nabla v)^{\flat} $$ I am using the notation $\nabla^*=\nabla_\sharp$ to denote basically the dual of the covariant derivative, and we have extended the notion of the musical isomorphism to tensor products in the obvious way.

More concretely, in terms of coordinates, this basically means that you can do things like take the divergence whichever way you please: $$\operatorname{div}u=(\nabla u)^k{}_k=\nabla_ku^k\\=g_{ik}\nabla^i(g^{jk}u_j) =g_{ik}\big(g^{jk}\nabla ^i u_j+u_j\underbrace{\nabla^ig^{jk}}_{=0}\big) \\=g^{jk}g_{ik}\nabla^iu_j \\ =\nabla^ku_k$$ So $\nabla \cdot u=\nabla^*\cdot u^{\flat}$.

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This seems like a rightful confusion about notation. As Mr. Brown points out in his answer, for any smooth vector field $X\in\mathfrak X(M)$, a connection induces a map \begin{align} \nabla_X : \mathcal T^k_l(M)\to \mathcal T^k_l(M), \qquad T \mapsto \nabla_X T, \end{align} where $\mathcal T^k_l(M)$ denotes the space of tensor fields of type $(k,l)$.

As you correctly say, one of the defining properties of the Levi-Civita connection is that the covariant derivative $\nabla g$ of the metric tensor vanishes, which comes down to the property that $\nabla_X g =0$ for all $X\in\mathfrak X(M)$. Since two of the defining properties of a connection are that $\nabla_{fX} = f\nabla_X$ and $\nabla_{X+Y} = \nabla_X+\nabla_Y$ for any smooth function $f$ and smooth vector fields $X,Y$, and since, locally, we may write $X = X^i\partial_i$, this is equivalent to the property that $X^i\nabla_{\partial_i}g=0$ for any $X$, and hence equivalent to $$\nabla_{\partial_i}g=0, \qquad i=1,\dots,\dim M.$$

Now, since $g=g_{ij}dx^i\otimes dx^j$ is a $(0,2)$ tensor field, $\nabla_{\partial_i}g$ is again a $(0,2)$ tensor field, so it can be expressed as

$$\nabla_{\partial_i}g = T_{ijk} dx^j\otimes dx^k$$

for a set of components functions $T_{ijk}$. It's common practice to denote the components by $T_{ijk}=\nabla_ig_{jk}$, or in other words, one defines

$$\nabla_ig_{jk} \equiv (\nabla_{\partial_i}g)_{jk},\qquad \text{i.e.,}\qquad \nabla_{\partial_i}g \equiv \nabla_ig_{jk}\, dx^j\otimes dx^k,$$

and with this definition it immediately follows, of course, that

$$\nabla g=0 \qquad\text{if and only if}\qquad \nabla_ig_{jk}=0 \qquad \text{for all $i,j,k$}.$$

You rightfully observe that this standard notation can be quite confusing: $\nabla_ig_{jk}$ could also be interpreted as $\nabla_ig_{jk} = \partial_ig_{jk}$, since $\nabla_i = \partial_i$ with respect to its action on smooth functions. But, whenever the latter is meant most authors would just write $\partial_ig_{jk}$ rather than $\nabla_ig_{jk}$ precisely in order to avoid this confusion.

I would say that your book is apparently using the convention that $\nabla_{\partial_k}g_{ij} = \nabla_{k}g_{ij}$, which is less standard, and makes things even more confusing in my personal opinion.

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