I Deriving tensor transformation laws

Summary
I'm attempting to understand how the definition of a tensor in terms of how it transforms arises from demanding invariance of the tensor when we define it as a multilinear map
Hi, I'm worried I've got a grave misunderstanding. Also, throughout this post, a prime mark (') will indicate the transformed versions of my tensor, coordinates, etc.

I'm going to define a tensor.

$$T^\mu_\nu \partial_\mu \otimes dx^\nu$$

Now I'd like to investigate how the tensor transforms under an arbitrary coordinate transformation where I require that the tensor is invariant under the coordinate transformation. In order to understand how partial-mu transforms, I'm going to apply the chain rule for partial derivatives.

$$\begin{equation} \partial'_\alpha = \frac{\partial}{\partial x'^\alpha} = \frac{\partial x^\mu}{\partial x'^\alpha}\partial_\mu \end{equation}$$

Similarly, I can write down how the basis one-forms transform.

$$\begin{equation} dx'^\alpha = \frac{\partial x'^\alpha}{\partial x^\mu} dx^\mu \end{equation}$$

Therefore, the basis of my tensor transforms like this:

$$\begin{equation} T'^\alpha_\beta \partial'_\alpha \otimes dx'^\beta = T^\mu_\nu \frac{\partial x^\mu}{\partial x'^\alpha} \partial_\mu \otimes \frac{\partial x'^\alpha}{\partial x^\mu} dx^\nu \end{equation}$$

Which mirrors my more familiar definition of a tensor in terms of how it's components transform:

$$\begin{equation} T'^\alpha_\beta = \frac{\partial x'^\alpha}{\partial x^\mu}\frac{\partial x^\nu}{\partial x'^\beta} T^\mu_\nu \end{equation}$$

Is it correct to write down the transformation of the tensor in the way I did in equation (3)? Is this equivalent to equation (4)?

Please help me if I've had a misunderstanding. Thank you very much for any help.
 

Orodruin

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Summary: I'm attempting to understand how the definition of a tensor in terms of how it transforms arises from demanding invariance of the tensor when we define it as a multilinear map

Is it correct to write down the transformation of the tensor in the way I did in equation (3)? Is this equivalent to equation (4)?
Yes. Yes (the base tensors ##\partial_\mu \otimes dx^\nu## form a complete and linearly independent basis).

Edit: Also note that the tensor product is linear and that the partial derivatives are just numbers that you can move out of it.
 
Yes. Yes (the base tensors ##\partial_\mu \otimes dx^\nu## form a complete and linearly independent basis).

Edit: Also note that the tensor product is linear and that the partial derivatives are just numbers that you can move out of it.
Thanks so much for the quick reply!
 
Therefore, the basis of my tensor transforms like this:

$$T'^\alpha_\beta \partial'_\alpha \otimes dx'^\beta = T^\mu_\nu \frac{\partial x^\mu}{\partial x'^\alpha} \partial_\mu \otimes \frac{\partial x'^\alpha}{\partial x^\mu} dx^\nu$$...
Is it correct to write down the transformation of the tensor in the way I did in equation (3)? Is this equivalent to equation (4)?
Note that the RHS of your equation 3 is not a valid expression as you have four ##\mu## indices. You need to change either the inner two ##\mu##’s or the outer two for this to make sense. But even then, if you’re wanting to use equations 1 and 2 and the fact that a tensor remains the same despite a change in basis in order to derive eq. 4, then this might be a more intuitive approach:
$$
\begin{equation*}
\begin{split}
\mathbf T = T^{\bar \alpha}_{\bar \beta} ~ \partial_{\bar \alpha} \otimes dx^{\bar \beta} & = T^{\bar \alpha}_{\bar \beta} (\frac{\partial x^\mu}{\partial \bar x^{\bar \alpha}} \partial_\mu ) \otimes (\frac{\partial \bar x^{\bar \beta}}{\partial x^\nu} dx^\nu) \\
& = (T^{\bar \alpha}_{\bar \beta} \frac{\partial x^\mu}{\partial \bar x^{\bar \alpha}} \frac{\partial \bar x^{\bar \beta}}{\partial x^\nu}) \partial_\mu \otimes dx^\nu
\end{split} \tag{5}
\end{equation*}
$$And because ##\mathbf T## is unchanged by a change in basis, we can say that
$$T^{\bar \alpha}_{\bar \beta} ~ \partial_{\bar \alpha} \otimes dx^{\bar \beta} = T^\mu_\nu ~ \partial_\mu \otimes dx^\nu$$
And therefore
$$T^\mu_\nu = T^{\bar \alpha}_{\bar \beta} \frac{\partial x^\mu}{\partial \bar x^{\bar \alpha}} \frac{\partial \bar x^{\bar \beta}}{\partial x^\nu}$$
We can rearrange like we did in the 2nd line of eq. 5 precisely because of the reason @Orodruin gave in the edit of his post.
 

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