Why are the equations for dt/du and Dt[a]/Du equal in the geodesic equation?

[PAllen]

Okay, but in the context of talking about a smooth manifold with connection (which is the case for GR), the notation $\frac{d\mathbf{V(s)}}{ds}$ is unambiguous.

Unambiguous, yes , if defined by the author. Confusing and nonstandard, also yes. Also, my argument was in response to an argument that connections were only relevant in a coordinate basis - which is not correct.

 

[vanhees71]

The entire vector analysis on differentiable manifolds do not need additional structures a priori. As long as you restrict yourself to alternating differential forms you don't need covariant derivatives.

And now, with more thought, I question your whole justification. Independent of any coordinate basis, there is no notion of vector derivative without a connection. Otherwise there is no way to compare or subract vectors at different points, even along a given curve, because the vectors are in different tangent spaces - and the connection is what maps between tangent spaces of nearby points. Given this need for connection to even define a vector derivative, it is confusing to use an ordinary derivative symbol for this.

[edit: I see I cross posted with @martinbn ]

Why then don't you need a connection when dealing with plain differentiable manifolds without any additional structures? You can go pretty far, introducing tangent and cotangent spaces and introducing differential forms. For this I don't need a connection, let alone a fundamental form (i.e., metric of pseudometric).

Also see the remark on Wikipedia:

https://en.wikipedia.org/wiki/Covariant_derivative#Remarks

 

[martinbn]

If you want to differentiate objects other than functions in the direction of a vector you need a connection.

 

[vanhees71]

Then, why don't I need a connection to define (alternating) differential forms?

 

[martinbn]

Then, why don't I need a connection to define (alternating) differential forms?

Why would you?

 

[PeterDonis]

I was thinking of the case of a scalar-valued function $f(\lambda)$ defined on a curve $\gamma^i(\lambda)$. Calculating its derivative involves $f(\lambda+\epsilon) - f(\lambda)$.

But the curve plays no role in this whatsoever; $\lambda$ is just the independent variable for the function $f(\lambda)$ and the derivative makes no use of the curve $\gamma^i(\lambda)$ at all. So you're not subtracting scalars at "different points" in some manifold, or in different tangent spaces; you're just subtracting values of a function at different values of its independent variable. Or, to put it another way, there is no need to define a "connection" (along a curve or by any other means) in order to take the ordinary derivative of a function, which is all you are doing here. But that's not the same thing as trying to compare scalars at different events in spacetime.

 

[PAllen]

But the curve plays no role in this whatsoever; $\lambda$ is just the independent variable for the function $f(\lambda)$ and the derivative makes no use of the curve $\gamma^i(\lambda)$ at all. So you're not subtracting scalars at "different points" in some manifold, or in different tangent spaces; you're just subtracting values of a function at different values of its independent variable. Or, to put it another way, there is no need to define a "connection" (along a curve or by any other means) in order to take the ordinary derivative of a function, which is all you are doing here. But that's not the same thing as trying to compare scalars at different events in spacetime.

Suppose the scalar is function on the manifold. Then you can take its derivative along along a curve, and this does depend on the curve, and it subtracts scalar values at different points of the manifold (along the curve). However, none of this need a connection for the reason @martinbn pointed out several posts ago: the scalar is a mapping from the manifold to reals; there is no notion of a different set of reals for each manifold point.

 

[strangerep]

[...] So you're not subtracting scalars at "different points" in some manifold, or in different tangent spaces; you're just subtracting values of a function at different values of its independent variable. [...]

My response would have been similar to PAllen's in post #37.

there is no notion of a different set of reals for each manifold point.

But isn't that merely because we have tacitly identified the different copies of $\mathbb R$ at each manifold point, and thus treat them all as one and the same?

BTW, @Apashanka Das : I'm assuming that your original question has been answered sufficiently, (hence a bunch of SAs debating like this is not an impolite hijack of your thread). If your original question remains insufficiently answered, please say so.

 

[PeterDonis]

Suppose the scalar is function on the manifold. Then you can take its derivative along along a curve, and this does depend on the curve, and it subtracts scalar values at different points of the manifold (along the curve).

But if you say that the derivative involves $f(\lambda + \epsilon) - f(\lambda)$ (with all the appropriate language about limits, etc.), you aren't doing that. You're just taking the ordinary derivative of a function with one variable.

To do what you're talking about, you have to start with the parameterization of the curve, and use it in defining the derivative. And if you do that, since, as you say, the result will depend on the curve you choose, it is the curve that is providing the connection between scalars at different points on the manifold.

 

[PAllen]

To do what you're talking about, you have to start with the parameterization of the curve, and use it in defining the derivative. And if you do that, since, as you say, the result will depend on the curve you choose, it is the curve that is providing the connection between scalars at different points on the manifold.

This is what I am saying. The point is exactly that different curves through a point lead to different scalar derivatives at that point, and in all cases, the limit defining the derivative involves subtracting scalar function values at different points in the manifold. Again, this is needs no extra structure beyond a differential manifold because the scalar is a mapping between the manifold and R.

 

[PAllen]

But isn't that merely because we have tacitly identified the different copies of $\mathbb R$ at each manifold point, and thus treat them all as one and the same?

BTW, @Apashanka Das : I'm assuming that your original question has been answered sufficiently, (hence a bunch of SAs debating like this is not an impolite hijack of your thread). If your original question remains insufficiently answered, please say so.

No , it is because a scalar is defined as mapping between the manifold and R.

You could, for example, define a mapping between the manifold and some fixed vector space isomorphic to Rⁿ. Then, derivatives of this vector valued function along curves, or in different directions on the manifold (e.g. divergences) would not need a connection. But this vector function would be invariant, not contravariant; it would be totally unaffected by coordinate transforms on the manifold. I am not aware of any physical theory making use of such a construct, but it is perfectly well defined mathematically.

However, in GR, we normally want to say that a vector field, or vectors along a curve, are vectors from the tangent space at each point. Since this is a different vector space at each point, this is what makes it impossible to subtract vectors for different points without a connection, and thus a connection is required for derivative of such a vector function.

 

[vanhees71]

$\newcommand{\dd}{\mathrm{d}}$ $\newcommand{\vv}[2]{\begin{pmatrix} #1 \\ #2 \end{pmatrix}}$ $\newcommand{\vvv}[3]{\begin{pmatrix} #1 \\ #2 \\3 \end{pmatrix}}$ $\newcommand{\bvec}[1]{\boldsymbol{#1}}$Well after reading to last few postings, I think, @martinbn is right after all. The point indeed is that we don't just consider some vector-valued function of a scalar parmater $\lambda$ as I did in one of my previous postings but a vector field $\bvec{V}:M \rightarrow \mathbb{R}$ (or $\mathbb{C}$ if needed), and we want a "canonical" (i.e., basis independent) derivative of $\bvec{V}$ along a curve or in direction of the tangent vectors along the curve. For this you have various options like the Lie derivative, for which you don't need an affine connection but also the covariant derivative, where you need some rule to infinitesimally move the vector from one tangent space at point $P$ to some point $P'$ along the given direction, and how you do this defines the connection.

In my derivation for the derivative of the vector field along a curve that's hidden in the fact that I've not only used the dependence of $\bvec{V}$ on the parameter $\lambda$, which parametrizes the curve, but the dependence of the complete basis $\bvec{b}_{\mu}(\lambda)$ along the curve, and I can choose this dependence quite arbitrarily, defining different affine connections along the curve.

Of course, as I said before, you can do a lot of vector analysis on a plain differentiable manifold without additional structures like an affine connection (making the manifold to a affine manifold) or fundamental bilinear form (making the manifold to a Riemann or pseudo-Riemann space when choosing the torsion free affine connection compatible with this (pseudo-)metric). Then you are however more or less restricted to alternating forms and their derivatives.