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Picture below is from pages 50-51 of do Carmo's Riemannian Geometry. I can't understand why the red line means that $\frac{DV}{dt}$ is unique when $V$ is fixed. In my view, there is not any proof to show that the right part of red line is independent to the choice of coordinate.

What I try: The red line can be written as $$ \frac{DV}{dt}=\left( \sum\limits_j \frac{dv^k}{dt} +\sum\limits_{ij}\frac{dx_i}{dt} v^j \Gamma_{ij}^k\right) X_k $$ Then, I have $$ \left( \sum\limits_j \frac{dv^k}{dt} +\sum\limits_{ij}\frac{dx_i}{dt} v^j \Gamma_{ij}^k\right) = f^k(c(t)) $$ namely, the $\left( \sum\limits_j \frac{dv^k}{dt} +\sum\limits_{ij}\frac{dx_i}{dt} v^j \Gamma_{ij}^k\right)$ can be treated as a function of $c(t)$ or $t$. So, I have $$ \frac{DV}{dt}(c(t)) =\sum_k f^k(c(t))X_k(c(t)) \tag{1} $$ But, if in another coordinate $Y_i=\frac{\partial}{\partial y_i}$, similarly, I can get $$ \frac{DV}{dt}(c(t)) =\sum_k \hat f^k(c(t))Y_k(c(t)) \tag{2} $$ how to show that the right parts of (1) and (2) are same ?

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  • $\begingroup$ The sentence immediately below your red line explains it -- if it exists then it must be the RHS in your coordinate patch. Remember uniqueness and existence are different. $\endgroup$ Commented Apr 11, 2021 at 13:00
  • $\begingroup$ Do Carmo asserts that formula (1) is implied by the conditions i), ii), iii). It is unique in the sense that there are no other vector fields that satisfy those three conditions. You appear to be asking about whether or not the right hand side is well-defined, which follows from the chain rule. $\endgroup$
    – treble
    Commented Apr 11, 2021 at 13:20
  • $\begingroup$ @treble Thanks. Could you give the detail process about the well-defined by the chain rule? Maybe, what I real want is it. But I fail to finish the process. $\endgroup$
    – Enhao Lan
    Commented Apr 12, 2021 at 2:35
  • $\begingroup$ @user10354138 what is the mean of RHS ? $\endgroup$
    – Enhao Lan
    Commented Apr 12, 2021 at 2:35

1 Answer 1

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You are correct that the calculation doesn't show that the given expression is independent of the choice of coordinates. However, the argument do Carmo gives doesn't require you to show it and you get it as a byproduct of the proof "for free". Since do Carmo's argument is somewhat subtle and missing in details, let me give you an outline of the argument:

  1. The first thing you need to know is that both $\nabla$ and $\frac{D}{dt}$ are "local operators". This means that $(\nabla_X Y)(p)$ depends only on the values of $Y$ and $X$ in an arbitrary small neighborhood of $p$ and similarly $\frac{DV}{dt}(t_0)$ depends only on the values of $V$ and $c$ near $t_0$. This is a consequence of the linearity, the product rule and the existence of bump functions (see page 50 in Lee's Riemannian Manifolds book). Once you know it, it makes sense to talk about things like $\nabla_{\partial_i} \partial_j$ and $\frac{D \partial_i}{d t}$. The reason is that $\partial_i,\partial_j$ are only local vector fields so a priori you can't plug them into $\nabla$ (or $\frac{D}{dt}$). However, given any point $p \in M$ around which they are defined, you can modify them and extend them using bump functions so that they become global vector fields which remain them same near $p$ and the resulting covariant derivatives do not depend on the specific way you do it. Do Carmo actually mentions it in Remark 2.3 for $\nabla$ (but not for $\frac{D}{dt}$) but his "proof" is wrong because in the proof he uses the expression $\nabla_{\partial_i} \partial_j$ before he actually justifies why it is legitimate.
  2. The next thing is to show uniqueness of $\frac{D}{dt}$. Let $t_0 \in I$ and choose an arbitrary coordinate system $(x^1,\dots,x^n)$ near $c(t_0)$ which gives you a correspodning local frame $(\partial_1, \dots, \partial_n)$. Then using point $(1)$ and performing do Carmo's calculation, you get the formula $$ \frac{DV}{dt}(t_0) = \frac{dv^j}{dt}(t_0) \partial_j|_{c(t_0)} + \frac{d x^i}{d t}(t_0) v^j(t_0) \nabla_{\partial_i}{\partial_j} (c(t_0)) \label{eq:covariant-formula} \tag{1}$$ where I use Einstein's summation to make it shorter. If you would have chosen a different coordinate system you would get a formula in terms of the other coordinate system which a priori could give you a different result but at this point you don't care because this formula is enough to prove uniqueness. If there are two operators $\frac{D_1}{dt}, \frac{D_2}{dt}$ which satisfy the properties of the covariant derivative along $c$, by choosing the same coordinate system around $c(t_0)$ you get the same formula for $\frac{D_1}{dt}(t_0)$ and $\frac{D_2}{dt}(t_0)$ and this holds for all $t_0 \in I$ so $\frac{D_1}{dt} = \frac{D_2}{dt}$.
  3. The next thing to do use equation $\eqref{eq:covariant-formula}$ to define $\frac{D}{dt}$ locally. Choose some $t_0 \in I$, some coordinate system $(x^1,\dots,x^n)$ around $c(t_0)$ and some $\varepsilon > 0$ so that $c \left( (t_0 - \varepsilon, t_0 + \varepsilon) \right)$ lies in that coordinate neighborhood. Define an operator $\frac{D}{dt}$ on vectors fields along $c|_{(t_0 - \varepsilon, t_0 + \varepsilon)}$ using equation $\eqref{eq:covariant-formula}$ and verify that it satisfies all three properties of the covariant derivative. Note that in this definition you are using a specific coordinate system to define $\frac{DV}{dt}$ and this only gives you a "local" covariant derivative. Is this definition depends on the coordinate system? Not really, because of uniqueness. Assume that $\varepsilon$ is small enough so that $c \left( (t_0 - \varepsilon, t_0 + \varepsilon) \right)$ lies in the domain of two coordinate systems $(x^1,\dots,x^n)$ and $(y^1,\dots,y^n)$. A priori you can use the $x^i$'s to define an operator $\frac{D_1}{dt}$ and the $y^i$'s to define an operator $\frac{D_2}{dt}$. However, since they both satisfy the properties of the covariant derivative, the uniqueness shows you that $\frac{D_1}{dt} = \frac{D_2}{dt}$ so in fact the result of the formula is independent of the coordinate system chosen. This can of course be verified directly by a calculation but you can see it is not required once we know uniqueness.
  4. Finally, given a vector field $V$ along $c$ and $t_0 \in I$, choose an arbitrary coordinate system around $c(t_0)$ and using this specific coordinate system to define $\frac{DV}{dt}(t_0)$ by equation $\eqref{eq:covariant-formula}$. This is well-defined by item $(3)$ and since the properties of the covariant derivative are all local, $\frac{D}{dt}$ also satisfies them and you get existence.
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  • $\begingroup$ Thanks very much for your answer. But I still have a problem. In part 2, you get $\frac{D_1}{dt} = \frac{D_2}{dt}$ in the same coordinate. Note that this is the uniqueness in same local coordinate. But in part 3, you define $\frac{D_1}{dt}$ in $(x^1,\dots,x^n)$ , and define $\frac{D_2}{dt}$ in $(y^1,\dots,y^n)$ . I think you can't use the uniqueness of $\frac{D}{dt}$ in same coordinate to get $\frac{D_1}{dt} = \frac{D_2}{dt}$. $\endgroup$
    – Enhao Lan
    Commented Apr 15, 2021 at 2:42
  • $\begingroup$ @lanse7pty: Note that there is no "uniqueness in the same coordinate". The precise statement of "uniqueness" is that given a curve $c$, there exists one and only one operator $\frac{D}{dt} \colon \mathcal{X}(c) \rightarrow \mathcal{X}(c)$ which satisfies properties $(1)-(3)$ of do Carmo's text. There is no mention of a coordinate system. In the proof of uniqueness you choose some coordinate system and obtain a formula for $\frac{D}{dt}(t_0)$ in terms of that coordinate system. In the proof you don't use anything special about $\frac{D}{dt}$ (other than the three properties) or the coordinate $\endgroup$
    – levap
    Commented Apr 15, 2021 at 7:16
  • $\begingroup$ system so you are free to work with any coordinate system. Then given two operators $\frac{D_1}{dt}, \frac{D_2}{dt}$, if you choose the same coordinate system around $c(t_0)$, you get the same formula which shows that $\frac{D_1}{dt}(t_0) = \frac{D_2}{dt}(t_0)$. This works for any $t_0 \in I$ so you get uniqueness. The point is part $(3)$ both $\frac{D_1}{dt}$ and $\frac{D_2}{dt}$ satisfy the three properties of the covariant derivative even though they are defined using different coordinate systems and so by uniqueness they must coincide. $\endgroup$
    – levap
    Commented Apr 15, 2021 at 7:17
  • $\begingroup$ I get you. Really thanks. I am used to deal problem in coordinate, it is not very geometric :-) . Thanks again. $\endgroup$
    – Enhao Lan
    Commented Apr 15, 2021 at 12:24
  • $\begingroup$ Thanks for this detailed answer! I’d like to check if you are suggesting we can infer the formula (1) for the covariant derivative is independent of choice of coordinates WITHOUT knowing the covariant derivative itself is a local operator in Do Carmo’s proof. If so, can we also prove the formula (1) is independent of choice of coordinates by saying the covariant derivative is a local operator before we prove the uniqueness of the covariant derivative (although it seems to be less concise)? $\endgroup$
    – Eric
    Commented Nov 28, 2021 at 11:31

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