# How does a tensor from cotangent and tangent spaces transform?

In Sean Carroll's Spacetime and Geometry An Introduction to General Relativity Chapter 2, there is an example of tensor transformation from $$x,y$$ coordinates to primed ones using $$(x',y') = (\frac{2x}{y},\frac{y}{2}).\tag{1}$$ The given tensor is $$S_{\mu\nu} = \left(\begin{matrix} 1 & 0 \\ 0 & x^2 \\ \end{matrix}\right)\tag{2}$$ Which is a (0,2) tensor on a 2D manifold.

I want to calculate $$S_{\mu'\nu'}$$ which from what I understand is

$$S_{\mu'\nu'} = \frac{\partial{x^\mu}}{\partial{x^{\mu'}}} \frac{\partial{x^\nu}}{\partial{x^{\nu'}}} S_{\mu\nu} \tag{3}$$ in the given coordinates. (I mean the indices match right?)

Can someone show (refer) a detailed derivation of this? I essentially want to know what the form of the matrices of the transformation are. This will give a lot of clarification to the framework to me (and hopefully others).

Edit I: I have calculated all partials (correctly), the problem I have is in intuition, as to what the exact form of the matrices are, I have tried a LOT of combinations but I just can't get it right.

Edit II(answer): As a clarification of the notation used and the operations I'll post a detailed answer.

Given a tensor $$S_{\mu\nu}$$ in a 2 dimensional manifold the transformation to $$S_{\mu'\nu'}$$ follows from the equation above. The equation is perfectly fine when we consider the indices and the elements of each matrix. In matrix form though the equation is a little bit different.

Assume $$x^1,x^2 = x,y$$ and $$x^{1'},x^{2'} = x',y'$$

$$S_{\mu'\nu'} = \left(\begin{matrix} \frac{\partial{x^1}}{\partial{x^{1'}}} & \frac{\partial{x^2}}{\partial{x^{1'}}}\\ \frac{\partial{x^1}}{\partial{x^{2'}}} & \frac{\partial{x^2}}{\partial{x^{2'}}}\\ \end{matrix}\right) \left(\begin{matrix} 1 & 0\\ 0 & x^2\\ \end{matrix}\right) \left(\begin{matrix} \frac{\partial{x^1}}{\partial{x^{1'}}} & \frac{\partial{x^1}}{\partial{x^{2'}}}\\ \frac{\partial{x^2}}{\partial{x^{1'}}} & \frac{\partial{x^2}}{\partial{x^{2'}}}\\ \end{matrix}\right)$$ Where the far right matrix is the transpose of the far left. In this example $$x' = \frac{2x}{y}$$ and $$y' = \frac{y}{2}$$ which gives $$x = x'y'$$ and $$y = 2y'$$. Substitution of those gives:

$$S_{\mu'\nu'} = \left(\begin{matrix} y' & 0\\ x' & 2\\ \end{matrix}\right) \left(\begin{matrix} 1 & 0\\ 0 & x^2\\ \end{matrix}\right) \left(\begin{matrix} y' & x' \\ 0 & 2 \\ \end{matrix}\right)$$ Which eventually gives $$S_{\mu'\nu'} = \left(\begin{matrix} (y')^2 & y'x'\\ x'y' & (x')^2 + 4(x'y')^2\\ \end{matrix}\right)$$

• Have you computed the partial derivatives? – G. Smith Mar 15 '19 at 16:33

You have $$y = 2 y'$$ and $$x = x' y'$$ so
$$\frac{\partial x^0}{\partial x^{0'}} = y' \\ \frac{\partial x^0}{\partial x^{1'}} = 0 \\ \mbox{etc.}$$
After that it goes $$S_{\mu'\nu'} = \sum_{\mu=0}^1 \sum_{\nu=0}^1 K^{\;\mu}_{\mu'} K^{\;\nu}_{\nu'} S_{\mu\nu}$$ where $$K^{\;\mu}_{\mu'} = \frac{\partial x^\mu}{\partial x^{\mu'}}$$ This can also be seen as the matrix product $$S' = K S K^T$$ (to check your understanding of this point, note which row/column index is being summed over in the standard rules of matrix multiplication). But you don't have to use matrices if you don't want to.