4

[Throughout we're considering the intrinsic version of the covariant derivative. The extrinsic version isn't of any concern.]

I'm having trouble reconciling different versions of the properties to be satisfied by the covariant derivative. Essentially $\nabla$ sends $(p,q)$-tensors to $(p,q+1)$-tensors. I'll write down the required properties for $\nabla$ from the two sources.

This lecture (relevant timestamp linked)

If $X$ is a vector field,

  1. $\nabla_Xf=Xf$, for a scalar field $f$
  2. $\nabla_X(T+S)=\nabla_XT+\nabla_XS$
  3. $\nabla_X(T(\omega,Y))=(\nabla_XT)(\omega,Y)+T(\nabla_X\omega,Y)+T(\omega,\nabla_XY)$
  4. $\nabla_{fX+Z}\ T=f\nabla_XT+\nabla_ZT$

Core principles of special and general relativity (Luscombe):

  1. $\nabla_if=\partial_if$
  2. $\nabla(aT+bS)=a\nabla T+b\nabla S$ for real $a,b$
  3. $\nabla(S\otimes T)=(\nabla S)\otimes T+S\otimes (\nabla T)$
  4. $\nabla$ commutes with contractions, $\nabla_i(T^j_{\ \ jk})=(\nabla T)^j_{\ \ ijk}$

At least the second property is consistent. The first property from the book is a more restrictive version of the first property from the lecture. In fact, $\nabla_i$ means $\nabla_{\partial_i}$ and $\partial_i$ isn't even a vector field!

As for the last two properties from the two sources, I have no idea on how to relate them. Are these requirements incomplete for either of the sources?

If not, how can these two sets of requirements be shown to be equivalent?

Shirish Kulhari
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  • On a general manifold, $\partial_i$ does represent a vector field - the $i$th coordinate vector field, namely $$ \partial_i f|_p = \frac{\partial (f\circ \phi^{-1})}{\partial x_i}(\phi(p)), $$ where $\phi : U \subset M \to \mathbb{R^n}$ is a chart. – Chris Jul 31 '20 at 15:17
  • @Chris: True, but it's only defined as a vector field in local coordinates as far as I know. It's not like the standard vector field that specifies a unique vector at each point. e.g. see the first comment to my question here: https://math.stackexchange.com/questions/3775553/if-x-is-a-vector-field-and-f-in-c-inftym-then-is-xf-in-c-inftym – Shirish Kulhari Jul 31 '20 at 15:20
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    True, we start by defining it in local coordinates. But if we define a vector field in local coordinates, that object's still a vector field - it just will not have the same expression in different coordinates. E.g. $\partial_i$ in one coordinate patch $(\phi, U)$ will not in general be $\partial_i$ in a different coordinate patch $(\tilde{\phi}, \tilde{U})$ (with $U \cap \tilde{U} \neq \emptyset)$. Think of, for example, how vector fields on $\mathbb{R}^2$ change form (but represent the same vector field) going from Cartesian to polar coordinates. – Chris Jul 31 '20 at 15:57
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    Although note that a priori if we define a vector field in local coordinates, it's only defined on the coordinate patch $U \subset M$. – Chris Jul 31 '20 at 15:59
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    @Chris: Thanks! That gives me a fresh perspective on local vector fields. – Shirish Kulhari Jul 31 '20 at 16:57

1 Answers1

2

The key point is that the covariant derivative is a local operator: If two tensors agree on an open set $U$, then their covariant derivatives agree on that open set. Moreover, if $X(p) = Y(p)$, then $(\nabla_X T)(p) = (\nabla_Y T)(p)$. So, given a vector field $X$ defined on an open set (in your case, $\partial_i$ on some coordinate chart $U$), using a bump function we can extend it to a globally defined vector field $\tilde X$ that agrees with $X$ on a pre-determined non-empty subset $V\subset U$. That means that your "local" definition of $\nabla$ on $V$ agrees with the "global" definition.

Ted Shifrin
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  • I have a follow-up. I may have misunderstood some point so please correct me if I'm wrong. If we have the "local" vector field $\partial_i$ defined in some $U$, and if $p\in U$, the vector itself specified by $\partial_i(p)$ will change depending on what coordinate system we pick. So if we try to extend this to some global field $\bar X$, since $\bar X$ is global, $\bar X(p)$ is a unique vector in $T_pM$. How can it agree with $\partial_i$ at $p$, given that $\partial_i(p)$ depends on choice of coordinates? – Shirish Kulhari Jul 31 '20 at 17:18
  • Secondly, I'm looking for some details on how the 3rd and 4th properties from both the sources can be reconciled with each other, i.e., a demonstration that both sets of properties as given by the two sources are equivalent. Have edited the question to emphasize that. – Shirish Kulhari Jul 31 '20 at 17:26
  • I'm concerned only with your one coordinate chart $U$, as I said. As I said, you can turn that locally defined vector field into a global one by using usual bump function techniques. I'm saying nothing about the transformation laws for the connection if I have two different coordinate charts on the same neighborhood. That hasn't been discussed in what you posted. The usual Christoffel symbol presentation of the connection has transformation rules when you change coordinate systems. – Ted Shifrin Jul 31 '20 at 17:27
  • The lists of properties do not align. 2 and 4 in the first list turn into 2 in the second list. 3 in the first list is a specific, not general, version of 3 and 4. The first list needs to be fixed up a bit. – Ted Shifrin Jul 31 '20 at 17:30
  • Could you elaborate on how 2 and 4 in the first list turn into 2 in the second list? I mean, 4 in the first list indicates $C^{\infty}(M)$-linearity. I can understand that 2 in the first list corresponds to 2 in the second list. But 2 in the second list doesn't indicate $C^{\infty}(M)$-linearity in the subscript vector field. Also, I'm a bit lost on what the 4th point in the second list means. Sorry if I sound dense - it's my first time studying the subject and this topic in particular. – Shirish Kulhari Jul 31 '20 at 17:41
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    Yes, you're right. There's sloppiness in both lists. I suggest you read a reliable text on differential geometry :) – Ted Shifrin Jul 31 '20 at 17:54
  • Let us continue this discussion in chat. – Shirish Kulhari Jul 31 '20 at 17:57