I have learned that the covariant derive is just the normal derive minus the normal component, 9:53 of this video.

When our space is an intrinsic plane, then the covariant derivative just becomes the normal derivative since in an intrinsic plane since there is no longer a 3rd dimension. The covariant derive of the metric tensor in an intrinsic plane would just be the normal derivative (i.e the rate of change of that metric tensor).

So, since the metric tensor changes across space for a plane that is intrinsically curved, why is the rate of change of that metric tensor (the covariant derivative) zero? An explanation I have heard is that it’s just constrained to be zero by choosing a specific connection. I disagree with this explanation because we get the connection that we are dealing with by just taking the normal derivative of a basis vector, so it is a property of space (refer 17:37 of this video).

I know the proof why it’s zero mathematically, but how do I reason this intuitively the metric tensor is a property that changes from point to point in space?

  • 2
    $\begingroup$ The covariant derivative is chosen to be "metric compatible" because of a degree of freedom when you write down the bare minimum qualities you would like a derivative to have. There is no "physical" why, since we don't directly observe covariant derivatives in nature, like one would observe a curvature. Choosing metric compatibility generates a unique connection via a PDE $\endgroup$ Sep 9, 2020 at 6:00
  • 1
    $\begingroup$ The covariant derivative can be defined intrinsically, but for surfaces in $\mathbb{R}^n$ one can also define it as the projection/restriction of the covariant derivative in $\mathbb{R}^n$ to the surface. These two definitions agree, but it is not so intuitively obvious that they do. $\endgroup$
    – Kajelad
    Sep 9, 2020 at 6:19
  • $\begingroup$ math.stackexchange.com/q/1749725/272127 $\endgroup$
    – C.F.G
    Sep 12, 2020 at 18:26
  • $\begingroup$ I want to see your proof of covariant derivative of metric =0!! $\endgroup$
    – C.F.G
    Sep 12, 2020 at 18:30

1 Answer 1


I think you are confusing several things. In the first video you have linked, the manifold $M$ you are working with is embedded in an ambient flat space; then the covariant derivative of a tensor along some path in $M$ is indeed the usual derivative with respect to the ambient coordinates "minus the normal component". As far as I can understand, what you mean by "intrinsic plane" is NOT embedded isometrically into an ambient flat space, so there is no sense in which there is a "normal derivative"; this whole description is not applicable for manifolds described in intrinsic terms, without an isometric embedding. Thus one uses the machinery of Levi-Civita connections an all that. Now, if you start with a manifold isometrically embedded in a flat space (or, say, use the (difficult) Nash's embedding theorem to embed your $M$ in this way) then you can compute the derivative of metric tensor by differentiating the (flat, constant) ambient metric tensor, and "removing the normal component" - and then restricting to appropriate tensor subbundle corresponding to vectors and covectors tangent to $M$; but of course the ambient metric tensor has zero derivative (being constant). "Removing components" and "restricting" zero still gives zero, so the resulting covariant derivative of the metric tensor on $M$ is also zero, just as the intrinsic computation said.

In some sense, the point is that the ambient metric is constant, and it's the restriction that changes as we move in $M$; and since the covariant differentiation takes derivative first, and restricts afterwards, the result is 0.


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