Transformation of Rank 2 mixed tensor

1. Dec 12, 2007

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?

2. Jan 5, 2008

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.