Transformation of Rank 2 mixed tensor

[cathalcummins]

Thanks for the help on the other questions.

I am having trouble with another derivation. Unlike the others, it's not abstract whatsoever.

Okay I wish to find the transformation Law for the components of a rank 2 tensor.

Easy, I know: T: V^* \times V \mapsto \mathbb{R}

So

T = T^i_{\phantom{i} j} e_i \otimes e^j

I wish to find

T^i'_{\phantom{i'} j'}

Where the following hold:

e_{i'}=a^{J}_{\phantom{J} i'} e_J

and

e^{i'}=b^{i'}_{\phantom{i'} J} e^J

where, the coefficients are just real numbers. Now T^i_{\phantom{i} j}=T(e^i, e_j) so that:

T^{i'}_{\phantom{i'} j'}=T(e^{i'}, e_{j'})

=T(b^{i'}_{\phantom{i'} J} e^J, a^{L}_{\phantom{L} j'} e_L)

By linearity of \otimes we have:

=b^{i'}_{\phantom{i'} J} a^{L}_{\phantom{L} j'} T( e^J, e_L)

=b^{i'}_{\phantom{i'} J} a^{L}_{\phantom{L} j'} T^{J}_{\phantom{J} L}

Now my lecturer done a funny thing and said,

"it may be shown that b=a"

which confuses the hell outta me because,

e_{i'}=a^{J}_{\phantom{J} i'} e_J

e^{i'}=b^{i'}_{\phantom{i'} J} e^J

abide by the duality relation so that:

e^{i'} e_{j'}=b^{i'}_{\phantom{i'} J} e^J a^{K}_{\phantom{K} j'} e_K=\delta^{i'}_{j'}

So do the original bases obey their own set of duality relations so that:

e^{i'} e_{j'}=b^{i'}_{\phantom{i'} J} a^{K}_{\phantom{K} j'} \delta^J_K=\delta^{i'}_{j'}

So

e^{i'} e_{j'}=b^{i'}_{\phantom{i'} K} a^{K}_{\phantom{K} j'} =\delta^{i'}_{j'}

Is this not the definition of a being the inverse of b. Of course, I know a priori I am wrong as this would give T^{i'}_{\phantom{i'} j'}=T^{i}_{\phantom{i} j} regardless of transformation.

In my definition of a and b, the superindex refers to row and the lower index refers to column.

My gut feeling is that I am using the matrix notation all wrong.

Any takers?

 

[dextercioby]

My take is this one:

T=T_{i}{}^{j}e^{i}\otimes e_{j}=T_{i'}{}^{j'} e^{i'}\otimes e_{j'}

Assume

e^{i'}= A^{i'}{}_{i} e^{i} and

e_{j'}= B_{j'}{}^{j} e_{j} .

You know that e_{j}(e^{i})=\delta_{j}^{i}.

Now, if you want to have e_{j'}(e^{i'})=\delta_{j'}^{i'}, that is keep the orthonormality relation, they you must necessarily have

A^{-1}=B^{T}

which means that the components of T are invariant under a transformation, bacause the basis in the tensor product space is invariant.