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Consider the following equation :

$$\Lambda^{-1}\Lambda^T \Lambda=A$$

Here $\Lambda$ are my lorentz transformations such that $\Lambda^T \eta \Lambda=\eta$. $A$ is some matrix.

I know that in terms of components, $\Lambda \equiv \Lambda ^\mu \space_\nu$.

Similarly, $\Lambda^{-1} \equiv (\Lambda^{-1}) ^\mu \space_\nu = \Lambda_\nu\space^\mu$

However, $\Lambda^{T} \equiv (\Lambda^{T}) _\alpha\space ^\beta = \Lambda^\beta\space_\alpha$

Hence the above equations reads as :

$$ (\Lambda^{-1}) ^\mu \space_\alpha (\Lambda^{T}) _\alpha\space ^\nu\Lambda ^\nu \space_\beta = (A) ^m\space_n$$

$$ \Lambda _\alpha\space ^\mu \Lambda ^\nu \space_\alpha \Lambda ^\nu \space_\beta = (A) ^m\space_n$$

However, the indices are not contracted here so how exactly do we equate both sides ? In essence, how do we exactly find what the indices $m$ and $n$ on the RHS are supposed to be ?

If we had $\Lambda^{T} \equiv (\Lambda^{T}) ^\mu\space_\nu$, then we could contract indices successfully and then $m=\mu$ and $n=\beta$. However, after reading multiple posts on the subject, I know this is incorrect. so, where exactly am I going wrong?

I'm guessing that my conversion from matrix to tensor component notation is correct, if not please tell me how exactly am I going wrong. However I'm still confused, as to how I'll find out the components of the matrix $A$, and how to relate the indices on the RHS to those on the LHS.

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I think you already know that the superscript vs subscript denotes contravariant vs covariant, whereas horizontal placement is whether something is a column or a row. That is $$\bar X ^\mu = \Lambda^\mu_{\ \ \ \nu} X^\nu$$ Now you also know that $$(\Lambda^T)_\mu^{\ \ \ \alpha} \,\eta_{\alpha\beta} \Lambda^\beta_{\ \ \ \nu} = \eta_{\mu\nu}$$ which, with minor alterations, means $$\eta^{\mu\gamma} (\Lambda^T)_\gamma^{\ \ \ \alpha} \,\eta_{\alpha\beta} \Lambda^\beta_{\ \ \ \nu} = \delta^\mu_\nu \qquad \implies \qquad (\Lambda^{-1})^\mu_{\ \ \ \beta } = \eta^{\mu\gamma} (\Lambda^T)_\gamma^{\ \ \ \alpha} \,\eta_{\alpha\beta}$$ Note the horizontal placement of the indices. You made a mistake here.

As for the original matrix multiplication, it is therefore clear that it is in violation of tensor sensibilities. Either you have to insert $\eta$ between each pair of matrices, or the equation is itself physically meaningless.

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  • $\begingroup$ Thanks ! Could you just clarify two things for me ? I understand the relation between $\Lambda^{-1}$ and $\Lambda ^T$. However, I don't see where I made a mistake, as I've related $\Lambda^{-1}$ and $\Lambda^T$ to $\Lambda$. Moreover I see now that the equation doesn't really make sense, purely because it doesn't work out. However, it is unclear as to how you can directly say that the equation is in violation of tensor sensibilities as it is a perfectly valid matrix equation. $\endgroup$ Apr 14, 2023 at 17:12
  • $\begingroup$ No, when you wrote $(\Lambda^{-1})^\mu_{\ \ \ \nu} = \Lambda_\nu^{\ \ \ \mu}$, you were implicitly saying that $\Lambda^{-1} = \Lambda^T$ without realising so. $\endgroup$ Apr 14, 2023 at 17:16
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    $\begingroup$ As for the tensor sensibilities, it is simply a fact that matrices are a mathematical space too wide to express stuff in, and it is very easy to write down "perfectly valid matrix equation(s)" that are physically senseless. That is why we care about tensors. The mathematical machinery of tensors is precisely there to help us write down only physically sensible stuff. Since we can only sensibly contract contravariant indices with covariant indices and vice versa, never contra with contra and co with co, the matrix equation you wrote down is thus unphysical $\endgroup$ Apr 14, 2023 at 17:18
  • $\begingroup$ Okay I see what you're saying. But my idea was that $(\Lambda^{-1})^\mu\space_\nu=(\Lambda^T)^\mu\space_\nu=\Lambda_\nu\space^\mu$ is a valid equation. However this doesn't mean that $\Lambda^T=\Lambda^{-1}$. There is a bit of subtlety here. For example, we can read off $\Lambda^{-1}$ as $(\Lambda^{-1})^\mu\space_\nu$. However, $\Lambda^T \ne (\Lambda^{T})^\mu\space_\nu$. Instead, $\Lambda^T$ should be read as $ (\Lambda^{T})_\mu\space^\nu$ instead, so that things are consistent. $\endgroup$ Apr 14, 2023 at 17:24
  • $\begingroup$ In that sense, I had reached the same relation between the inverse and the transpose as you had. $\endgroup$ Apr 14, 2023 at 17:25

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