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I am following lecture notes on SR. The author writes that the following is equivalent:

$$\Lambda^T\eta\Lambda = \eta \iff \eta_{\mu \nu} {\Lambda^\mu}_\rho{\Lambda^\nu}_\sigma = \eta_{\rho \sigma}. \tag{1}$$ This surprises me, because

$$ {(\Lambda^T)^\mu}_\nu = {\Lambda_\nu}^\mu.\tag{2} $$

And so I expected it to be $$\Lambda^T\eta\Lambda = \eta \iff \eta_{\mu \nu} {\Lambda_\rho}^\mu{\Lambda^\nu}_\sigma = \eta_{\rho \sigma}.\tag{3}$$ Why is this wrong?

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  • $\begingroup$ I think you're put an extra flip in there somewhere. Just think about it like matrices: $ABC = D$ means $A_{ij} B_{jk} C_{k\ell} = D_{i\ell}$. Therefore $A^T BC = D$ means $B_{jk} A_{ji} C_{k\ell} = D_{i\ell}$ where I used the commutativity of multiplication, and replaced $A_{ij}$ with $A_{ji}$. This is exactly the same as equation (1). $\endgroup$ – knzhou Apr 29 '17 at 23:08
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    $\begingroup$ The upstairs/downstairs index placement is a red herring that can only add confusion; it's not a 'real' index because $\Lambda$ is not a tensor. $\endgroup$ – knzhou Apr 29 '17 at 23:09
  • $\begingroup$ @knzhou By your first comment, I believe you confirm that you're puzzled too, correct? Flipping the indecies is exactly what I propose, but that's what the lecture notes do not. $\endgroup$ – Mikkel Rev Apr 29 '17 at 23:21
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  1. OP's three equations should read $$\Lambda^T\eta\Lambda ~=~ \eta \quad\iff\quad (\Lambda^T)_{\rho}{}^{\mu}~\eta_{\mu \nu}~ \Lambda^{\nu}{}_{\sigma} ~=~ \eta_{\rho \sigma} ,\tag{1'}$$ $$ (\Lambda^T)_{\nu}{}^{\mu} ~:=~\Lambda^{\mu}{}_{\nu} ,\tag{2'} $$ $$\Lambda^T\eta\Lambda ~=~ \eta \quad\iff\quad \Lambda^{\mu}{}_{\rho}~\eta_{\mu \nu}~ \Lambda^{\nu}{}_{\sigma} ~=~ \eta_{\rho \sigma} .\tag{3'}$$

  2. In more detail: Let $V$ be $n$-dimensional $\mathbb{R}$-vector space with a basis $(e_{\mu})_{\mu=1, \ldots, n}$. Let $V^{\ast}$ be the dual vector space with the dual basis $(e^{\ast \nu})_{\nu=1, \ldots, n}$. Let $$\Lambda~=~e_{\mu}~ \Lambda^{\mu}{}_{\nu}\otimes e^{\ast \nu}~ \in~V\otimes V^{\ast}~\cong~{\cal L}(V;V)$$ be a linear map from $V$ to $V$. Let us call the positions of the indices on $\Lambda^{\mu}{}_{\nu}$ for the NW-SE convention, cf. a compass rose. Let $$\Lambda^T~=~e^{\ast \nu}~ (\Lambda^T)_{\nu}{}^{\mu}\otimes e_{\mu}~\in~V^{\ast}\otimes V~\cong~{\cal L}(V^{\ast};V^{\ast})$$ be the transposed linear map from $V^{\ast}$ to $V^{\ast}$. Note that $(\Lambda^T)_{\nu}{}^{\mu}$ is written in the SW-NE convention. Let $$\eta~=~e^{\ast \mu}~\eta_{\mu\nu}\odot e^{\ast \nu}~\in~{\rm Sym}^2V^{\ast}~=~V^{\ast}\odot V^{\ast}$$ be an (indefinite) metric, i.e. an invertible element in the symmetrized tensor product. A (pseudo)orthogonal map $\Lambda$ satisfies by definition $$\Lambda^T\eta\Lambda~=~\eta.$$ See also this related Phys.SE post.

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  • $\begingroup$ Wouldn't you have then an ambiguous notation? Wouldn't you use ${\Lambda_\mu}^\nu$ for refer to both $\Lambda^T$ and $\Lambda^{-1}$? $\endgroup$ – falgenint May 8 '17 at 9:41
  • $\begingroup$ @falgenint: Thanks for the feed-back. $\Lambda_{\mu}{}^{\nu}$ does by definition not refer to $(\Lambda^T)_{\mu}{}^{\nu}$ to avoid ambiguity. Rather we use the metric to raise and lower indices, so by definition $\Lambda_{\mu}{}^{\nu}~:=~\eta_{\mu\rho}~\Lambda^{\rho}{}_{\sigma}~(\eta^{-1})^{\sigma\nu}$, which happens to be equal to $((\Lambda^{-1})^T)_{\mu}{}^{\nu}$ if $\Lambda$ is (pseudo)orthogonal. $\endgroup$ – Qmechanic May 8 '17 at 10:05
  • $\begingroup$ Sorry, I made a mistake. I wanted to say: wouldn't you use ${\Lambda^\mu}_\nu$ for refer to both $\Lambda^T$ and $\Lambda^{-1}$? I mean, if (using (2')) this is correct: ${((\Lambda^{-1})^T)_{\mu}}^{\nu}={(\Lambda^{-1})^\nu}_\mu={\Lambda_\mu}^\nu$ it follows ${((\Lambda^{-1})^T)_\mu}^{\nu}={((\Lambda^{T})^{-1})_\mu}^{\nu}$ and from this you can demonstrate that $\Lambda \Lambda^T=I$ with $I$ the identity matrix. Is this correct? $\endgroup$ – falgenint May 8 '17 at 11:06
  • $\begingroup$ The last equation is not correct. When removing indices from a tensor equation, one should re-introduce/put back the metric tensor $\eta$ in pertinent places. $\endgroup$ – Qmechanic May 8 '17 at 11:59
  • $\begingroup$ So I guess I can't say that $(\Lambda^{-1})^T=(\Lambda^T)^{-1}$ although ${((\Lambda^{-1})^T)_\mu}^{\nu}={((\Lambda^{T})^{-1})_\mu}^{\nu}$? $\endgroup$ – falgenint May 8 '17 at 13:57

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