Self-studying GR. Stuck on Q3.20 in the 3rd edition of Schultz. Orthogonal coordinate transforms in Euclidian space

Question

I am self-studying GR using "A first course in general relativity, 3rd edition". I'm doing my best to be diligent and work though the problems at the end of the chapter. But question 3.20 has me stumped. Abbreviated it asks the following.

Consider 2 coordinate system $O$ and $\bar{O}$. Given vectors transform like so $$ V^{\bar{\alpha}} = \Lambda_{\beta}^{\bar{\alpha}}V^{\beta} $$ and one-forms transform like so $$ P_{\bar{\beta}} = \Lambda_{\bar{\beta}}^{\alpha}P_{\alpha} $$ show that if $\Lambda$ is orthogonal, meaning the transpose of its inverse is itself, that vectors and one-forms transform in the same way. Meaning there is no reason to distinguish between them.

The first eqn implies $e_{\beta}^{\bar{\alpha}} = \Lambda_{\beta}^{\bar{\alpha}}$ for each standard basis vector $\vec{e}_{\beta}$

Example: $\vec{e}_{0}\underset{O}{\rightarrow}(1,0,0,0)$

so $e_{0}^{\bar{\alpha}} = \Lambda_{\beta}^{\bar{\alpha}}e_{0}^{\beta} = \Lambda_{0}^{\bar{\alpha}}$

Similarly for each basis one form $\tilde{\omega}^{\alpha}$ we have $\omega_{\bar{\beta}}^{\alpha} = \Lambda_{\bar{\beta}}^{\alpha}$

I believe the question is asking me to show that $\Lambda_{\beta}^{\bar{\alpha}} = \Lambda_{\bar{\beta}}^{\alpha}$. But here I am stumped. These two transformations are inverses as far as I can tell. But the orthogonality condition stipulates inverse followed by transpose.

The only thing that strikes me as maybe a little odd is the use of $\bar{\alpha}$ as the coordinate index for vectors and $\bar{\beta}$ as the coordinate index for one forms

Answer 1

Because we are talking about transposes, this means that we need to look at these things with some care as to the shape of matrices. In tensors, we tend to manipulate things so that they make sense, first, and then figure out how to express the stuff in matrices correctly. But here, we have to be careful from the start. To that end, it is quite important to write the indices with horizontal placement correct, namely, the left index goes up-down, the right index goes left-right. Other than the Kronecker delta identity matrix $\delta^a_b$ which is obviously not going to be affected by this indexing, we should care that $$\tag1\text{if}\qquad\Lambda^{\bar\alpha}_{\ \ \ \beta}\qquad\text{then}\qquad\left(\Lambda^T\right)_\beta^{\ \ \ \bar\alpha}$$ is how the matrix representation v.s. the index placements are, for a matrix and its transpose. You might also be interested to note that the vertical placement of an index tells us if that index is contravariant or covariant. You have to convince yourself of the correctness of the above considerations before continuing.

---

For 1-forms and vectors to make physical sense, what we really want is that $$\tag2P_\beta V^\beta=P_{\bar\alpha}V^{\bar\alpha}=P_\beta\left(\Lambda^ {-1}\right)^\beta_{\ \ \ \bar\alpha}\,\Lambda^{\bar\alpha}_{\ \ \ \beta}V^\beta\qquad\implies\qquad P_{\bar\beta}=P_\alpha\left(\Lambda^{-1}\right)^\alpha_{\ \ \ \bar\beta}$$ this is the version closest to what you have quoted.

Now, note that we commonly express 1-forms as row matrices and vectors as column matrices, so that the $P_\beta V^\beta$ gets us the single scalar inner product that we expect. Because of that, if we take the transpose of Equation (2) to flip the 1-form row matrix into a column matrix, then we have $$\tag3\left(P^T\right)_{\bar\beta}=\left[\left(\Lambda^{-1}\right)^T\right]_{\bar\beta}^{\ \ \ \alpha}\left(P^T\right)_\alpha$$ and this is now trivially the solution that the question is asking for, when you insert what the question is asking you to assume, and the compare that with the vector transformation law. Note, however, that the up-down index placements are still wrong, showing clearly that this prescribed identification of vectors and 1-forms is still mathematically abusive, that it only looks correct if you ignore contravariance and covariance, concentrating only on the matrix elements.

Answer 2

This is a rare case where the indices might be making it more difficult to see what's going on vs just using matrix notation.

A vector transforms as

$$ V' = \Lambda V $$

whereas a one-form transforms as

$$ \omega' = \omega \Lambda^{-1} $$

However, if $\Lambda$ is orthogonal, then $\Lambda^{-1}=\Lambda^T$, so

$$ \omega' = \omega \Lambda^T $$

Now if we identify the vector version of the one-form as the column vector $\omega^T$ associated with the row vector $\omega$ (this is the key step -- normally you can't do this), then taking the transpose of the above equation yields

$$ (\omega^T)' = \Lambda \omega^T $$

which shows that $\omega^T$ transforms as a vector, meaning our identification of $\omega^T$ as the vector associated with $\omega$ was consistent. (The "magic" is that we were able to convert the one-form to a vector by "just" taking a transpose -- if you were to look at Minkowski-space vectors, you would also need to change the sign of the time component).