Transpose and Inverse of Lorentz Transform Matrix

[devd]

Let $\Lambda$ be a Lorentz transformation. The matrix representing the Lorentz transformation is written as $\Lambda^\mu{}_\nu$, the first index referring to the rows and the second index referring to columns.

The defining relation (necessary and sufficient) for Lorentz transforms is $$g_{\mu\nu}=g_{\alpha\beta}\Lambda^\alpha{}_\mu \Lambda^\beta{}_\nu.$$
In matrix form this reads $g=\Lambda^Tg\Lambda$, where we have used $(\Lambda^T)^\nu{}_\beta=\Lambda^\beta{}_\nu$. From this we can see that $(\Lambda^{-1})^\mu{}_\nu=\Lambda_\nu{}^\mu$.

Up to this point, do I have everything right?

In Wu-Ki Tung's "Group Theory in Physics" (Appendix I.3, equation I.3-1), Tung states that,
$$(\Lambda^T)_\mu{}^\nu=\Lambda^\nu{}_\mu$$. Is this consistent with my definitions/conventions?
For instance, formally taking $\Lambda\rightarrow \Lambda^T,\mu\rightarrow \nu,\nu\rightarrow\mu$, Tung's equation reads,
$$(\Lambda^T)^\mu{}_\nu=\Lambda_\nu{}^\mu.$$
whereas, according to our definition/convention, $(\Lambda^{-1})^\mu{}_\nu=\Lambda_\nu{}^\mu$, and $(\Lambda^T)^\mu{}_\nu=\Lambda^\nu{}_\mu.$ What gives?

 

[vanhees71]

You convention seems to be right to me. Indeed
$$g_{\mu \nu} = g_{\alpha \beta} {\Lambda^{\alpha}}_{\mu} {\Lambda^{\beta}}_{\nu}$$
translates in matrix notation to
$$\hat{g}=\hat{\Lambda}^{T} \hat{g} \hat{\Lambda}.$$
Multiplying this equation from the left with $\hat{g}$ and using $\hat{g}^2=1$ leads to
$$1=\hat{g} \hat{\Lambda}^{T} \hat{g} \hat{\Lambda},$$
which means that $\hat{\Lambda}$ is invertible and that
$$\hat{\Lambda}^{-1}=\hat{g} \hat{\Lambda}^{T} \hat{g}.$$
In index notation that means
$${(\Lambda^{-1})^{\mu}}_{\nu} = g^{\mu \alpha} g_{\nu \beta} {\Lambda^{\beta}}_{\alpha} ={\Lambda_{\nu}}^{\mu}.$$
Of course, you can get this also within the index calculus itself from the very first equation. Contracting it with $g^{\mu \gamma}$ gives
$$\delta_{\nu}^{\gamma} = g_{\alpha \beta} g^{\mu \gamma} {\Lambda^{\alpha}}_{\mu} {\Lambda^{\beta}}_{\nu}.$$
Now contracting this equation with ${(\Lambda^{-1})^{\nu}}_{\delta}$ gives
$${(\Lambda^{-1})^{\gamma}}_{\delta}=g_{\alpha \beta} g^{\mu \gamma} {\Lambda^{\alpha}}_{\mu} \delta_{\delta}^{\beta} ={\Lambda_{\delta}}^{\gamma}. $$

 

[devd]

So, the way Tung defines his transpose is indeed inconsistent with my convention, right?

Also, I was wondering where this freedom to "define" transpose and inverse comes about. Why is there a need for a convention, at all? Given a matrix, its transpose and inverse are uniquely defined. Is it about 2nd-rank tensors not being in exact correspondence with matrices?

 

[vanhees71]

I don't think that there is anything inconsistent except the book you are citing. I don't know it, but it seems to confusing at best, if not just wrong.The author seems to be confused concerning upper and lower indices. Note that ${\Lambda^{\mu}}_{\nu}$ are no tensor components but matrix elements of a basis transformation.

 

[strangerep]

The defining relation (necessary and sufficient) for Lorentz transforms is $$g_{\mu\nu}=g_{\alpha\beta}\Lambda^\alpha{}_\mu \Lambda^\beta{}_\nu.$$
In matrix form this reads $g=\Lambda^Tg\Lambda$, where we have used $(\Lambda^T)^\nu{}_\beta=\Lambda^\beta{}_\nu$. From this we can see that $(\Lambda^{-1})^\mu{}_\nu=\Lambda_\nu{}^\mu$.

Up to this point, do I have everything right?

Well, I'll venture to say that I think your convention is wrong and Tung's is right. One of the reasons for having upper/lower indices on $\Lambda$ here is to be consistent with the summation convention (implicit summation on paired upper+lower indices).

So I think your

we have used $(\Lambda^T)^\nu{}_\beta=\Lambda^\beta{}_\nu$

should be: where we have used $(\Lambda^T)_\mu^{\;~\alpha} = \Lambda^\alpha_{\;~\mu}$

 

[devd]

Well, I'll venture to say that I think your convention is wrong and Tung's is right. One of the reasons for having upper/lower indices on $\Lambda$ here is to be consistent with the summation convention (implicit summation on paired upper+lower indices).

So I think your should be: where we have used $(\Lambda^T)_\mu^{\;~\alpha} = \Lambda^\alpha_{\;~\mu}$

I see what you are saying and that makes sense. But, if $(\Lambda^T)_\mu {}^\nu = \Lambda^\nu{}_\mu$ is true, then, $(\Lambda^T)^\mu {}_\nu = \Lambda_\nu{}^\mu$ is also true, which can be seen by lowering and raising indices by the metric (treating$\Lambda$ as a tensor). But, we know that $(\Lambda^{-1})^\mu {}_\nu=\Lambda_\nu {}^\mu$. This seems to imply that the $\mu \nu$th element of $\Lambda^{-1}$ and $\Lambda^T$ are the same. Which certainly isn't true. How do I reconcile this contradiction?

 

[strangerep]

we know that $(\Lambda^{-1})^\mu {}_\nu=\Lambda_\nu {}^\mu$

That's true for an ordinary orthogonal group. But the Lorentz group is an indefinite orthogonal group, so one must include the metric in the relationship between transpose and inverse. Have a read of that wiki page, under the section titled "Matrix definition".

 

[devd]

That's true for an ordinary orthogonal group. But the Lorentz group is an indefinite orthogonal group, so one must include the metric in the relationship between transpose and inverse. Have a read of that wiki page, under the section titled "Matrix definition".

Yes, I'm aware of that and therein lies the contradiction. Starting from
$$g_{\mu\nu}=g_{\alpha\beta}\Lambda^\alpha{}_\mu\Lambda^\beta{}_\nu\\
\implies\delta^\alpha{}_\nu=g^{\alpha\mu}g_{\mu\nu}=g^{\alpha\mu}g_{\alpha\beta}\Lambda^\alpha{}_\mu\Lambda^\beta{}_\nu\\
\implies \delta^\alpha{}_\nu=\Lambda_\beta{}^\alpha \Lambda^\beta_\nu$$
Now, to write this in Matrix notation, we write $(\Lambda^{-1})^\alpha{}_\beta=\Lambda_\beta{}^\alpha$, and then the previous equation becomes
$$\delta^\alpha{}_\nu=\Lambda_\beta{}^\alpha \Lambda^\beta_\nu\\
\delta^\alpha{}_\nu=(\Lambda^{-1})^\alpha{}_\beta \Lambda^\beta_\nu,$$
which in matrix notation reads, $I=\Lambda^{-1}\Lambda$. Hence, to be consistent with both Einstein summation convention and matrix multiplication, we are forced to write the $\alpha \beta$-th element of $\Lambda^{-1}$ as $\Lambda_\beta{}^\alpha$. Is there an error that I'm inadvertently committing?

If what i wrote is true, i.e., $(\Lambda^{-1})^\alpha{}_\beta=\Lambda_\beta{}^\alpha$, then, if the transpose is defined as in Tung's book, i.e.,$(\Lambda^T)_\mu {}^\nu = \Lambda^\nu{}_\mu$, which is equivalent to $(\Lambda^T)^\mu {}_\nu = \Lambda_\nu{}^\mu$, then this would imply $\Lambda^{-1}=\Lambda^T$ rather than $\Lambda^{-1}=g\Lambda^T g$, as it should be. This is the contradiction, at least apparent, that I'm talking about. What's the way out of this?

By the way, thank you for having the patience for reading my questions and replying!

 

[strangerep]

$$g_{\mu\nu}=g_{\alpha\beta}\Lambda^\alpha{}_\mu\Lambda^\beta{}_\nu\\
\implies\delta^\alpha{}_\nu=g^{\alpha\mu}g_{\mu\nu}=g^{\alpha\mu}g_{\alpha\beta}\Lambda^\alpha{}_\mu\Lambda^\beta{}_\nu\\
\implies \delta^\alpha{}_\nu=\Lambda_\beta{}^\alpha \Lambda^\beta_\nu$$

Your 2nd line is nonsense. On the rhs you have $\alpha$ as both a free index, and as a summation index.

 

[devd]

Your 2nd line is nonsense. On the rhs you have $\alpha$ as both a free index, and as a summation index.

Oops, sorry! I meant,

$$g_{\mu\nu}=g_{\alpha\beta}\Lambda^\alpha{}_\mu\Lambda^\beta{}_\nu\\
\implies\delta^\rho{}_\nu=g^{\rho\mu}g_{\mu\nu}=g^{\rho\mu}g_{\alpha\beta}\Lambda^\alpha{}_\mu\Lambda^\beta{}_\nu\\
\implies \delta^\rho{}_\nu=\Lambda_\beta{}^\rho \Lambda^\beta{}_\nu$$
Now we write $(\Lambda^{-1})^\rho{}_\beta=\Lambda_\beta{}^\rho$. Hence the previous equation becomes,
$$\delta^\alpha{}_\nu=\Lambda_\beta{}^\rho \Lambda^\beta_\nu\\
\delta^\rho{}_\nu=(\Lambda^{-1})^\rho{}_\beta \Lambda^\beta_\nu,$$
which in matrix notation reads, $I=\Lambda^{-1}\Lambda$.

The rest of the argument proceeds the same way. Thanks for being patient!

 

[strangerep]

I just noticed another error in your earlier post...

rather than $\Lambda^{-1}=g\Lambda^T g$, as it should be.

I believe the correct formula is $\Lambda^{-1}=g^{-1}\Lambda^T g ~$.

But I'm getting lost about what your perceived contradiction is (or might have become). You may have to re-write it, just to be clear.

 

[devd]

I just noticed another error in your earlier post...
I believe the correct formula is $\Lambda^{-1}=g^{-1}\Lambda^T g ~$.

But I'm getting lost about what your perceived contradiction is (or might have become). You may have to re-write it, just to be clear.

Indeed, $\Lambda^{-1}=g^{-1}\Lambda^T g ~$. But, for the Minkowski metric tensor, $g^{-1}=g$, since $g^2=1.$

Ok, so my basic confusion, distilled to its essence is thus:

Using the earlier derivation, we see that $(\Lambda^{-1})^\rho{}_\beta=\Lambda_\beta{}^\rho$. Tung's definition of transpose tells us $(\Lambda^T)^\rho{}_\beta=\Lambda_\beta{}^\rho$. If both are to be consistent, it would imply that $\Lambda^{-1}=\Lambda^T$, which is not true for Minkowski space. The actual relation for Minkowski space should be $\Lambda^{-1}=g\Lambda^Tg$. So, is there an inconsistency between Tung's convention/definition and mine, which is generally followed in relativity texts.

Just to get a clarification, how would you write the $\mu\nu$-th element of the $\Lambda^{-1}$ matrix, given the matrix $\Lambda$? As I stated previously, I would write it as $(\Lambda^{-1})^\mu{}_\nu=\Lambda_\nu{}^\mu$

 

[vanhees71]

Well, I'll venture to say that I think your convention is wrong and Tung's is right. One of the reasons for having upper/lower indices on $\Lambda$ here is to be consistent with the summation convention (implicit summation on paired upper+lower indices).

So I think your should be: where we have used $(\Lambda^T)_\mu^{\;~\alpha} = \Lambda^\alpha_{\;~\mu}$

I disagree, since taking the transpose of a matrix means just switching rows and columns, no matter whether the indices are upper or lower indices. Indeed, I think it's easier to stay within the Ricci calculus.

In other words, I think that
$${(\Lambda^T)^{\mu}}_{\nu}={\Lambda^{\nu}}_{\mu}$$
while by definition (although the ${\Lambda^{\mu}}_{\nu}$ are no tensor components)
$${\Lambda_{\mu}}^{\nu}=g_{\alpha \mu} g^{\beta \nu} {\Lambda^{\alpha}}_{\beta}={(\Lambda^{-1})^{\nu}}_{\mu}.$$
At least you come to that conclusion also within the Ricci index calculus (as shown in my previous posting above).

 

[strangerep]

After sleeping on it, and doing some Googling, I am persuaded.

I.e., I now agree that $$ (\Lambda^T)^\mu_{~\;\nu} ~=~ \Lambda^\nu_{~\;\mu}$$is correct.

In my Googling, I noticed that quite a few authors (not just Tung) also make the same mistake. The Wiki entry for Ricci calculus doesn't even mention the subtleties involved in taking a transpose. Wald's GR book doesn't mention it, MTW only talk about transpose as interchanging index positions (without mentioning raising/lowering), and Weinberg's GR book confuses things more by his eq(3.6.2): $$D_{\alpha\mu} ~\equiv~ \frac{\partial \xi^\alpha}{\partial x^\mu}$$ which has the $\alpha$ downstairs on the left, but upstairs on the right. Then he writes $$D^T_{\mu\alpha} ~\equiv~ D_{\alpha\mu}$$which is$$\frac{\partial \xi^\mu}{\partial x^\alpha} ~,$$so he gets it right in the end, but only after a confusing detour.

The source of this error seems to lie in not properly understanding the abstract meaning of the transpose operation. If $V,W$ are vector spaces, and $L$ is a linear map $L: V \to W$, then the the transposed map is $L^T : W^* \to V^*$, where the asterisk denotes the dual space. (In Ricci calculus, taking the dual corresponds to raising/lowering indices.)

 

[vanhees71]

The problem is that taking the transpose of a matrix is not reallymatching the difference between co- and contravariant components, which is so elegantly encoded in the Ricci-index calculus notation. For me the greatest obstacle of coordinate-free notations is to keep in mind which kind of objects one is dealing with, because it's difficult to find different symbols for the different kinds of objects, e.g., to distinguish a one-form clearly from a vector etc.

 

[strangerep]

The problem is that taking the transpose of a matrix is not really matching the difference between co- and contravariant components, [...]

I'm not 100% sure what you mean by this sentence. If you mean that taking the transpose of a matrix does not explicitly display the swap between primal and dual spaces, then yes, I agree.

 

[vanhees71]

Then we agree.

 

[DrGreg]

The source of this error seems to lie in not properly understanding the abstract meaning of the transpose operation. If $V,W$ are vector spaces, and $L$ is a linear map $L: V \to W$, then the the transposed map is $L^T : W^* \to V^*$, where the asterisk denotes the dual space.

Would it cause less confusion to call it an "adjoint" $L^*$ instead of a "transpose" $L^T$?

 

[strangerep]

Would it cause less confusion to call it an "adjoint" $L^*$ instead of a "transpose" $L^T$?

According to Wikipedia's entry on linear maps, there's a notable distinction:

The definition of the transpose may be seen to be independent of any bilinear form on the vector spaces, unlike the adjoint.

 

[Orodruin]

For me the greatest obstacle of coordinate-free notations

I assume you mean index-free? Coordinate free notations should not include transformation coefficients at all as they are related to coordinate transformations.

 

[vanhees71]

What I mean is to work with abstract tensors, i.e., without introducing a concrete basis and not distinguishing the different objects by notation.

 

[cianfa72]

You convention seems to be right to me. Indeed
$$g_{\mu \nu} = g_{\alpha \beta} {\Lambda^{\alpha}}_{\mu} {\Lambda^{\beta}}_{\nu}$$
translates in matrix notation to
$$\hat{g}=\hat{\Lambda}^{T} \hat{g} \hat{\Lambda}.$$
Multiplying this equation from the left with $\hat{g}$ and using $\hat{g}^2=1$ leads to
$$1=\hat{g} \hat{\Lambda}^{T} \hat{g} \hat{\Lambda},$$
which means that $\hat{\Lambda}$ is invertible and that
$$\hat{\Lambda}^{-1}=\hat{g} \hat{\Lambda}^{T} \hat{g}.$$
In index notation that means
$${(\Lambda^{-1})^{\mu}}_{\nu} = g^{\mu \alpha} g_{\nu \beta} {\Lambda^{\beta}}_{\alpha} ={\Lambda_{\nu}}^{\mu}.$$

Sorry, re-reading this old post, I believe last equation makes sense just because the tensor $g_{\mu \nu}$ is symmetric.

$$g^{\mu \alpha} g_{\nu \beta} {\Lambda^{\beta}}_{\alpha} = (g^{\mu \alpha} {\Lambda^{\beta}}_{\alpha}) g_{\nu \beta}$$
and using the "mapping" into matrix notation $g^{\mu \alpha} \to \hat{g}, {\Lambda^{\beta}}_{\alpha} \to \hat{\Lambda}$

the bracket $(g^{\mu \alpha} {\Lambda^{\beta}}_{\alpha})$ translates in matrix notation to $\hat{g}\hat{\Lambda}^{T} \to (\hat{g} \hat{\Lambda}^{T})^{\mu \beta}$

then, using $g_{\nu \beta} = g_{\beta \nu}$ we get
$$(\hat{g}\hat{\Lambda}^{T})^{\mu \beta} g_{\nu \beta} = (\hat{g}\hat{\Lambda}^{T})^{\mu \beta} g_{\beta \nu}$$
or in matrix notation
$$\hat{g} \hat{\Lambda}^{T} \hat{g}$$

Does it make sense ? thanks

 

[vanhees71]

Looks good to me.

 

[cianfa72]

Thanks. Btw, given a matrix $A$ (for instance the Lorentz matrix $\Lambda$) the 'mapping' into the index form $A^{\mu}{}_{\nu}$ (where $\mu$ is the matrix row number and $\nu$ the matrix column number) is just a convention ?

 

[vanhees71]

In this context, the matrix notation is just a short-hand notation for the operations you do with the Ricci calculus. The advantage is that it is less cumbersome to write and you have simple rules of calculation often making a calculation shorter than taking care of all the indices. A great disadvantage however is that the information concerning the transformation properties of the objects you deal with, which is just "encoded" in the placement of an index as an "upper" or a "lower" index is lost.

 

[cianfa72]

So basically what you are saying is that - starting from Ricci calculus form - in order to simplify calculus we can choose to translate each of the involved tensor objects into a corresponding matrix 'keeping in mind' which kind of tensor object each of them is actually representing (as you pointed out that information is not 'encoded' in the matrix itself).

Thus, for instance, Lorentz transformation as linear transformation is actually a (1,1) tensor hence we need to 'keep in mind' that row and column indices $\mu , \nu$ of the associated $\Lambda$ matrix are actually of type $^{\mu}{}_{\nu}$ (one covariant & one contravariant slot).

 

[cianfa72]

Thus, for instance, Lorentz transformation as linear transformation is actually a (1,1) tensor hence we need to 'keep in mind' that row and column indices of the associated matrix are actually of type (one covariant & one contravariant slot).

Btw, would it have been possible to choose the other possible notation $_{\mu}{}^{\nu}$ i.e. $$\Lambda _{\mu}{}^{\nu}$$?

 

[vanhees71]

It is important to keep in mind that the matrix elements of a Lorentz transformation, ${\Lambda^{\mu}}_{\nu}$, are not tensor components.

Nevertheless one defines
$${\Lambda_{\mu}}^{\nu}=\eta_{\mu \rho} \eta^{\nu \sigma} {\Lambda^{\rho}}_{\sigma}.$$
From the Lorentz-transformation property then it follows that
$${\Lambda_{\mu}}^{\nu} {\Lambda^{\mu}}_{\alpha}=\eta_{\mu \rho} \eta^{\nu \sigma} {\Lambda^{\rho}}_{\sigma} {\Lambda^{\mu}}_{\alpha}=\eta_{\sigma \alpha} \eta^{\nu \sigma}=\delta_{\alpha}^{\nu},$$
which again leads back to the properties discussed in the beginning of this thread, i.e.,
$${(\Lambda^{-1})^{\mu}}_{\nu} = {\Lambda_{\nu}}^{\mu}.$$

 

[cianfa72]

It is important to keep in mind that the matrix elements of a Lorentz transformation, ${\Lambda^{\mu}}_{\nu}$, are not tensor components.

Why not ? A Lorentz transformation is a linear transformation hence is a (1,1) tensor. The $\mu, \nu$ entries in the associated $\Lambda$ matrix should be the components of such (1,1) tensor in a fixed basis, I believe...

See also here - https://www.physicsforums.com/threa...on-matrix-and-its-inverse.830122/post-5217794

 

[vanhees71]

A tensor is not a linear transformation but a multi-linear map $T:V^m \times V^{*n} \rightarrow \mathbb{R}$. The corresponding tensor components are given by putting the corresponding basis vectors and their dual-basis vectors into the "slots":
$${T_{\mu_1\ldots \mu_m}}^{\nu_1\ldots \nu_n}=T(e_{\mu_1},\ldots,e_{\mu_m},e^{\nu_1},\ldots,e^{\nu_n}).$$

 

[cianfa72]

A tensor is not a linear transformation but a multi-linear map $T:V^m \times V^{*n} \rightarrow \mathbb{R}$. The corresponding tensor components are given by putting the corresponding basis vectors and their dual-basis vectors into the "slots":
$${T_{\mu_1\ldots \mu_m}}^{\nu_1\ldots \nu_n}=T(e_{\mu_1},\ldots,e_{\mu_m},e^{\nu_1},\ldots,e^{\nu_n}).$$

Consider the multi-linear map $T:V^{*} \times V \rightarrow \mathbb{R}$, $T \in V \otimes V^{*}$

$$T^{\mu}{}_{\nu} \text{ } e_{\mu} \otimes e^{\nu}$$
If we contract it with a vector $v = v^{\alpha} e_{\alpha}$ (basically filling in its correspondent slot) we get the vector $T^{\mu}{}_{\nu}v^{\nu} e_{\mu}$ that is any linear transformation of the vector space $V$ in itself (an endomorphism) is actually a (1,1) tensor, dont'you ?

 

[vanhees71]

The Lorentz-transformation matrix is just the transformation from one pseudo-Cartesian (in the following called "Lorentzian") basis (and its corresponding dual basis) to another. It has thus no basis-independent meaning and hence its matrix elements are no tensor components.

This reminds me that I should also write the general mathematical part in the Appendix about tensor algebra and calculus in Minkowski space...

 

[cianfa72]

The Lorentz-transformation matrix is just the transformation from one pseudo-Cartesian (in the following called "Lorentzian") basis (and its corresponding dual basis) to another. It has thus no basis-independent meaning and hence its matrix elements are no tensor components.

Sorry (I'm not an expert )...even if we think of it from an active point of view (as a mapping of a vector space in itself) ?

Just to be clear: I'm not saying Lorentz-transformation matrix has a physical basis-indipendent meaning, though..

 

[ergospherical]

Just to be clear: I'm not saying Lorentz-transformation matrix has a physical basis-indipendent meaning, though..

That's a shame, because you'd be correct.
You can of course view it either actively or passively.

The active point of view: A Lorentz transformation $L : \mathbb{R}^4 \rightarrow \mathbb{R}^4$ is an isometry (read: preserves the scalar square of vectors) of spacetime and the Lorentz matrix $\Lambda = [{\Lambda^{\mu}}_{\nu}]$ of this transformation in some orthonormal basis $\{ \hat{\mathbf{e}}_{\mu} \}$ is the matrix that satisfies $L(\hat{\mathbf{e}}_{\nu}) = {\Lambda^{\mu}}_{\nu} \hat{\mathbf{e}}_{\mu}$.

[The set of all these transformations $L$ (along with a binary operation $\circ$) forms the Lorentz group $O(3,1)$]

 

[hartmutneff]

In my opinion, the clean way of doing this is to take the transpose of metric type matrices only. So of a matrix with either to upper or two lower indices.

For the coefficients of tensors, there is no rule of where the upper or the lower indices have to go, with one exception. If you want to use the convention of how to multiply matrices and vectors, so the row times column rule. There is no need to follow this convention, as you can always write down the full tensor equations with all their indices. But if you want to follow the convention, then you will have to stick to some rules to be consistent.
The basic rule for this convention is, that upper indices run along columns and lower indices along rows. That immediately poses a problem with a metric, with two lower indices. A metric doesn't have an upper index that runs along columns. Hence a metric can't really be written as a matrix.

When it comes to the orthogonality equation of the Lorentz transformation, people often fix this by swapping the indices of the Lorentz matrix horizontally and denote this by the transpose. I think the consistent way of doing this is to raise the first index of the metric instead. That is twisting the notation as well, but less so than swapping indices horizontally. With one upper index, the metric and the whole equation can be written in matrix form. Then one can swap the index of the Lorentz transformation between upper left and lower right corner to get the transpose.

So again, I think the confusion with this equations is due to the fact that for this kind of tensors, namely vectors, dual vectors and (1,1) tensors representing linear transformations, one would like to use the matrix vector multiplication convention.