Staggered Indices ($\Lambda^\mu{}_\nu$ vs. $\Lambda_\mu{}^\nu$) on Lorentz Transformations

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $(1)$.