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I have some open-ended questions on the use of staggered indices in writing Lorentz transformations and their inverses and transposes.

What are the respective meanings of $\Lambda^\mu{}_\nu$ as compared to $\Lambda_\mu{}^\nu$? How does one use this staggered index notation to denote transpose or inverse?

If I want to take any of these objects and explicitly write them out as matrices, then is there a rule for knowing which index labels row and which labels column for a matrix element? Is the rule: "(left index, right index) = (row, column)" or is it "(upper index, lower index) = (row, column)" or is there a different rule for $\Lambda^\mu{}_\nu$ as compared to $\Lambda_\mu{}^\nu$?

Are there different conventions for any of this used by different authors?

As a concrete example of my confusion, let me try to show two definitions of a Lorentz transformation are equivalent.

Definition-1 (typical QFT book): $\Lambda^\mu{}_\alpha \Lambda^\nu{}_\beta \eta^{\alpha\beta} = \eta^{\mu\nu}$

Definition-2 ($\Lambda$ matrix must preserve pseudo inner product given by $\eta$ matrix): $(\Lambda x)^T \eta (\Lambda y) = x^T \eta y$, for all $x,y \in \mathbb{R}^4$. This implies, in terms of matrix components (and now I'll switch to linear algebra notation, away from physics-tensor notation): $\sum_{j,k}(\Lambda^T)_{ij} \eta_{jk} \Lambda_{kl} = \eta_{il}$. This last equation is my "Definition-2" of a Lorentz transformation, $\Lambda$, and I can't get it to look like "Definition-1", i.e., I can't manipulate-away the slight difference in the ordering of the indices.

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By convention, vectors are written as column vectors, whereas dual vectors are written as row vectors. This means that in principle, upper indices should index columns and lower indices should index rows. However, in practice, we normally translate rank-2 tensors to matrices by order of the indices, the first one indexing rows, the second one indexing columns.

The only way I can think of to make this translation from tensors to matrices structurally well-defined (which I've never seen done in the literature), is to force all rank-2 tensors into the form $\cdot\;^\mu{}_\nu$, which can be achieved by contraction with appropriate 'Kronecker tensors', by which I mean rank-2 tensors whose components are 1 if the indices agree and 0 otherwise.

Let's call these tensors $\overline\delta^{\mu\nu}$ and $\underline\delta_{\mu\nu}$.

Then, the matrix product given in your question $$ x^T\cdot\eta\cdot y $$ would translate to $$ \left(x^\mu\underline\delta_{\mu\nu}\right)\cdot\left(\overline\delta^{\nu\alpha}\,\eta_{\alpha\beta}\right)\cdot\left(y^\beta\right) $$ The first term has a single free lower index (aka a row vector), the second term a free upper and lower index (aka a matrix) and the third one a free upper index (aka a column vector).

As all Kronecker tensors can be removed through index adjustement, this is equivalent to the far simpler expression $$ x^\mu\,\eta_{\mu\beta}\,y^\beta $$

As you can see, while there is no special symbol for transposition in index notation - it is normally implied by which index is summed over - it could be made explicit by using the 'Kronecker tensors' - but all you'd gain is adding unnecessary complexity.

Now after this round of useless musings, let's get back to something that actually is important when reading literature:

Indices are lowered and raised by contraction with the metric tensor and its inverse. So for example given a tensor $A^\mu{}_\nu$, then $$ A_\mu{}^\nu \equiv A^\alpha{}_\beta\; \eta_{\alpha\mu}\; (\eta^{-1})^{\beta\nu} $$

For the metric tensor itself, we have $$ (\eta^{-1})^{\mu\nu} = \eta^{\mu\nu} $$ proven over here and for Lorentz transformations $$ (\Lambda^{-1})^\tau{}_\mu = \Lambda_\mu{}^\tau $$ proven over here.

This is a special property of these specific tensors and does not hold for arbitrary ones.

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  • $\begingroup$ Thanks - great answer! Also, for a totally-lowered or totally-raised rank-2 tensor, how do we identify the matrix elements? Is it always (left, right) = (row, column)? $\endgroup$ – Nahsik May 15 '16 at 1:53
  • $\begingroup$ @Nahsik: I believe so; if both indices are of the same type, there is no intrinsic mapping to rows and columns, so we go by the convention for matrix indices; that's also my guess why we normally write linear maps as $A^\mu{}_\nu$ instead of the equally valid choice $A_\nu{}^\mu$ - only with the former, the order of indices is row, column $\endgroup$ – Christoph May 15 '16 at 3:56
  • $\begingroup$ I'm unsure if your original answer's remark about rows/columns is correct. You say upper index labels the column. @joshphysics says left-most index labels the row. That is, you say the upper/lower distinction is what matters to identify a matrix element, but he says the left/right distinction is what matters. He says that in this answer: physics.stackexchange.com/a/118580/116779 $\endgroup$ – Nahsik May 15 '16 at 4:05
  • $\begingroup$ To clarify, under Joshphysics's interpretation of notation, we would be wise to avoid using the notation of $\Lambda_\mu{}^\tau$, since it is the same as $(\Lambda^{-1})^\tau{}_\mu$, which reveals the true left-most index is actually $\tau$. $\endgroup$ – Nahsik May 15 '16 at 4:32
  • $\begingroup$ And to elaborate as to why I'm inclined to think Joshphysics is correct in saying that the left-most indicates the row, consider the product $B^\mu{}_\nu v^\nu$. Here, $v^\nu$ is a vector (i.e., a column vector), so its one index labels rows (i.e., components of a column vector). The matrix index contracted with this index must of course be the matrix's column index. So $\nu$ (the right-most index) labels the matrix's columns, and $\mu$ is then left-most index labeling the rows, consistent with Joshphysics. $\endgroup$ – Nahsik May 15 '16 at 4:39
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Not correct. $(\Lambda^{T})^\mu{}_\tau = \Lambda_\tau{}^\mu $ The source you used didn't take into account the fact that index notation doesn't distinguish matrices from their transposes, hence the error. The correct inverse is : $(\Lambda^{-1})^\mu{}_\tau = \Lambda^{\tau}{}_{\mu}$ This notation can be found in Schutz, for example, and is consistent with the Kronecker tensors, as well as the primed/unprimed convention. These rules apply only to LT matrices - an arbritrary (non-LT) matrix would still use the standard transposition rule.

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  • $\begingroup$ I think that is because he uses $\Lambda ^T \eta \Lambda = \eta$ to get the given index notation... I saw this kind of calculation in some article, but I forget what it is. $\endgroup$ – ChoMedit Sep 6 at 1:14
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The way I think about it is by trying to contract always the nearest index. So $\omega_\mu$ transforms as $$ \omega_\mu \;\to\;\omega_\mu^{\,\prime} = \Lambda_\mu^{\phantom{\mu}\nu}\, \omega_\nu\,, $$ and $v^\mu$ transforms as $$ v^\mu \;\to\;{v^\mu}' = \Lambda^\mu_{\phantom{\mu}\nu}\, v^\nu\,\,. $$ The rule for raising/lowering must leave intact the order of the indices (i.e. move them only vertically), so $$ g^{\mu \sigma} g_{\nu\lambda}\,\Lambda_\sigma^{\phantom{\sigma}\lambda} = \Lambda^\mu_{\phantom{\mu}\nu}\,,\qquad g_{\mu \sigma} g^{\nu\lambda}\,\Lambda^\sigma_{\phantom{\sigma}\lambda} = \Lambda_\mu^{\phantom{\mu}\nu}\,. $$ Let me denote $g$ the matrix $(g^{\mu\nu})_{\substack{\mu&\to\mathrm{row}\\\nu&\to\mathrm{col}\\}}$, by $\Lambda$ the matrix $(\Lambda^\mu_{\phantom{\mu}\nu})_{\substack{\mu&\to\mathrm{row}\\\nu&\to\mathrm{col}\\}}$ and by $\tilde\Lambda$ the matrix $(\Lambda_\mu^{\phantom{\mu}\nu})_{\substack{\mu&\to\mathrm{row}\\\nu&\to\mathrm{col}\\}}$. Note that, due to $g^{\mu\nu}g_{\nu\rho} = \delta^\mu_{\phantom{\mu}\rho}$, one has also $g^{-1} =(g_{\mu\nu})_{\substack{\mu&\to\mathrm{row}\\\nu&\to\mathrm{col}\\}}$.

The above equations read in this notation $$ g\,\tilde{\Lambda}\,(g^{\mathsf{T}})^{-1} = \Lambda \,, \qquad g^{-1}\,\Lambda\,g^{\mathsf{T}} = \tilde{\Lambda}\,. $$ By doing obvious manipulations and using $g = g^{\mathsf{T}}$ we get, for instance $$ \Lambda^{-1} \,g\,\tilde{\Lambda} = g\,. $$ A priori $\Lambda$ and $\tilde{\Lambda}$ are independent matrices, but the equation above, in light of the definition of Lorentz transformations, suggests to define $$ \Lambda^{-1} =\tilde{\Lambda}^{\mathsf{T}}\,.\tag{1} $$ The scalar products work out as well. Indeed $$ \omega \cdot v \equiv \omega_\mu v^\mu \;\to\;\omega_\mu^{\,\prime} {v^\mu}' = \omega_\nu \,\Lambda_\mu^{\phantom{\mu}\nu}\,\Lambda^\mu_{\phantom{\mu}\rho}\,v^\rho = \omega \,\cdot\,\tilde{\Lambda}^{\mathsf{T}}\,\Lambda\,\cdot v = \omega \cdot v\,. $$ And clearly the same can be shown by manipulating the indices only $$ \Lambda_\mu^{\phantom{\mu}\nu}\,\Lambda^\mu_{\phantom{\mu}\rho} = g^{\mu\sigma}g_{\rho\lambda}\,\Lambda_\mu^{\phantom{\mu}\nu}\,\Lambda^\lambda_{\phantom{\lambda}\sigma} = g_{\rho\lambda} g^{\nu\lambda} = \delta_\rho^{\phantom{\rho}\nu}\,. $$ In my opinion the clearest approach to explain them would be to say that the up-down matrix and the down-up matrix are a priori independent matrices, and then introduce the constraint $(1)$.

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  • $\begingroup$ In your 4th displayed equation (it's actually a pair of equations), I believe the "T" superscript is on the wrong g in each equation of that pair. Also, in your final Kroncker delta, I think we're morally obligated to give full respect to index-placement (at least in the present discussion), thus the rho and nu should be staggered, with the rho coming first. $\endgroup$ – Nahsik Jul 10 at 0:39
  • $\begingroup$ Thanks for the comment. I think the placement of $\mathsf{T}$ was correct but to be fair I didn't define the matrix associated to $g$ with lower indices. I think now it's clearer. I also shifted the indices on the $\delta$ even though in my opinion it would be better to leave them as they were, to emphasize $\delta^\mu_{\;\rho} = \delta^{\;\mu}_\rho$. $\endgroup$ – MannyC Jul 10 at 0:59
  • $\begingroup$ Small correction: most of the time you correctly identify $\Lambda$ as a matrix, but at one point you call it a tensor, when it clearly isn't. $\endgroup$ – knzhou Sep 6 at 0:38

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