special relativity - Why is not ${(Lambda^T)^mu}_nu = {Lambda_nu}^mu$?

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?

Answer

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.