I Contravariant derivative of tensor of rank 1

[Apashanka]

If we have two sets of coordinates such that x¹,x²...xⁿ
And y¹,y²,...y^m
And if any yⁱ=f(x¹...,xⁿ)(mutually dependent).
Then dyⁱ=(∂yⁱ/∂x^j)dx^j
Again dyⁱ/dx^k=(∂²yⁱ/∂x^k∂x^j)dx^j+∂yⁱ/∂x^k
Is it the contravariant derivative of a vector??
Or in general dAⁱ/dx^k≠∂Aⁱ/∂x^k

 

[Cryo]

Again dyi/dxk=(∂2yi/∂xk∂xj)dxj+∂yi/∂xk

This makes no sence: $\frac{\partial y^i}{\partial x^k}$ is finite whilst $\frac{\partial^2 y^i}{\partial x^k \partial x^j}dx^j$ is infinitessimal. I don't even know what the left-handside quantity is. Perhaps, what you are trying to say is that the covariant derivative of a vector field $A^i$ is given by:

$\nabla_k Y^i = \frac{\partial Y^i}{\partial x^k} + \Gamma^i_{jk}Y^j$, where $\Gamma^i_{jk}$ is the connection. Note however, that you do not need to define two coordinate systems to define the connection, all you need is a smooth rank (0,2) tensor field (such as metric).

 

[Apashanka]

If we have two sets of coordinates such that x¹,x²...xⁿ
And y¹,y²,...y^m
And if any yⁱ=f(x¹...,xⁿ)(mutually dependent).
Then dyⁱ=(∂yⁱ/∂x^j)dx^j
Again dyⁱ/dx^k=(∂²yⁱ/∂x^k∂x^j)dx^j+∂yⁱ/∂x^k
Is it the contravariant derivative of a vector??
Or in general dAⁱ/dx^k≠∂Aⁱ/∂x^k

In general dAⁱ/dx^k=(∂²Aⁱ/∂x^k∂x^j)A^j+∂Aⁱ/∂x^k

This makes no sence: $\frac{\partial y^i}{\partial x^k}$ is finite whilst $\frac{\partial^2 y^i}{\partial x^k \partial x^j}dx^j$ is infinitessimal. I don't even know what the left-handside quantity is. Perhaps, what you are trying to say is that the covariant derivative of a vector field $A^i$ is given by:

$\nabla_k Y^i = \frac{\partial Y^i}{\partial x^k} + \Gamma^i_{jk}Y^j$, where $\Gamma^i_{jk}$ is the connection. Note however, that you do not need to define two coordinate systems to define the connection, all you need is a smooth rank (0,2) tensor field (such as metric).

and according to this Γⁱ_jk=(∂²Aⁱ/∂x^k∂x^j)
Which turns out to be christoffel symbol??
and covariant derivative is ∇_k=d/dx^k??

 

[Apashanka]

If we have two sets of coordinates such that x¹,x²...xⁿ
And y¹,y²,...y^m
And if any yⁱ=f(x¹...,xⁿ)(mutually dependent).
Then dyⁱ=(∂yⁱ/∂x^j)dx^j
Again dyⁱ/dx^k=(∂²yⁱ/∂x^k∂x^j)dx^j+∂yⁱ/∂x^k
Is it the contravariant derivative of a vector??
Or in general dAⁱ/dx^k≠∂Aⁱ/∂x^k

How is the expression of dyⁱ/dx^k
be related to covariant derivative ??

 

[Cryo]

Which turns out to be christoffel symbol??
and covariant derivative is ∇k=d/dxk??

I am not sure it is so simple. Your logic works very well if we are on a flat manifold, so you can define some sort of Carthesian coordinates, and then relative to those coordinates you can define other coordinate systems. The Christoffel symbols will then arise naturally.

What happens if you are not on a flat manifold ($\mathcal{M}$)? In this case there may be no single coordinate system that covers the manifold fully. That's not a problem. You can cover different patches of the manifold with different coordinate systems. Then at every point ($p\in \mathcal{M}$) on the manifold you can define a different tangent vector space ($T\mathcal{M}|_p$). However, now it is not trivial how you go from one point to another ($T\mathcal{M}|_{p_1}\to T\mathcal{M}|_{p_2}$). Unless you can define this you cannot differentiate vector fields. So you need to connect vector spaces in some way - that's when you need the connection ($\Gamma^\alpha_{\beta\gamma}$). There is more than one way to define a valid connection, if you use metric tensor to define the connection, then you get Christoffel symbols, but there are other ways too.

I think it is treated very well in Lovelock & Rund "Tensors, Differential Forms, and Variational Principles" (Sec 3.3). Another favourite is Bachman "Geometric approach to Differential Forms"

 

[Cryo]

How is the expression of dyi/dxk

I don't know. You need to define it. I assume you don't mean partial derivatives. So then what is this geometrically? Is it a collection of 1D curves ( $\left\{y^{(i)}\right\}$ ) which you then differentiate ...

 

[Apashanka]

I am not sure it is so simple. Your logic works very well if we are on a flat manifold, so you can define some sort of Carthesian coordinates, and then relative to those coordinates you can define other coordinate systems. The Christoffel symbols will then arise naturally.

What happens if you are not on a flat manifold ($\mathcal{M}$)? In this case there may be no single coordinate system that covers the manifold fully. That's not a problem. You can cover different patches of the manifold with different coordinate systems. Then at every point ($p\in \mathcal{M}$) on the manifold you can define a different tangent vector space ($T\mathcal{M}|_p$). However, now it is not trivial how you go from one point to another ($T\mathcal{M}|_{p_1}\to T\mathcal{M}|_{p_2}$). Unless you can define this you cannot differentiate vector fields. So you need to connect vector spaces in some way - that's when you need the connection ($\Gamma^\alpha_{\beta\gamma}$). There is more than one way to define a valid connection, if you use metric tensor to define the connection, then you get Christoffel symbols, but there are other ways too.

I think it is treated very well in Lovelock & Rund "Tensors, Differential Forms, and Variational Principles" (Sec 3.3). Another favourite is Bachman "Geometric approach to Differential Forms"

thank you very much

 

[Apashanka]

I am again raising my question
Representing a point in two set of coordinate system labelled by the coordinates {x_n} and {y_m} mutually dependent
yⁱ=f({x_n})
dyⁱ=(∂yⁱ/∂x^j)dx^j...(1)
From this we define the transformation rule of a contravariant vector from one set of coordinates to another.
e.g Aⁱ(y)=(∂yⁱ/∂x^j)A^j(x)
Now how is the covariant derivative be applied to the above form to get the standard result??
Any proof??

 

[Cryo]

What is your question? How to define a covariant derivative? This is treated in text-books. I gave you a reference. Writing it out here would take too long, and I would probably not be able to do as good of a job as Rund & Lovelock.

Basically, you start with, e.g. a vector field: $A^i$ in coords $\left\{x^i\right\}$ and consider the difference in this vector field as a result of moving along a curve $x^i=x^i\left(s\right)$ by distance $\delta s$:

$dA^i=\partial_j A^i \dot{x}^j \delta s + \Gamma^i_{jk}A^j \dot{x}^k \delta s$

At this point you know nothing about covariant derivatives, connections etc. All you want is for $dA^i$ to be (1) Tensorial quantity (transform the right way), (2) be linear in $A^i$ and $\dot{x}^k \delta s$, as normal derivative would be, (3) be related to partial differentiation. The covariant derivative comes later on...

Also, you don't define the transformation rule. You define the vector space - the transformation rules naturally follow from it.