Let $g_{ij}$ be the diagonal Minkowski metric tensor diag$(g) = (1,-1,-1,-1)$, then $g^{ij}$ is defined to be $(g^{-1})^{ij}$, hence $$g_{ik}g^{kj} = g_i^{\ \ j} = \text{diag}(1,1,1,1)=\delta_i^{\ \ j}.$$ There seems to be some freedom of handling indices of mixed tensors. For example, in The Quantum Theory of Fields Vol.1 by Steven Weinberg, p.57 in the line before eq. (2.3.10),
the inverse of $\Lambda^{\mu}{}_{\nu}$ is defined to be $(\Lambda^{-1})^{\nu}{}_{\sigma}$.
But I think ${{(\Lambda^{-1})}_\nu}^\mu $ may be interpreted as an inverse as well depending on how we define contraction and inverse. What's the convention of defining THE inverse of a tensor? General tensors (not restricted to matrices)?
Best Answer
Indices are raised and lowered with the metric tensor. So OP's example
$$\tag{1} (\Lambda^{-1})_{\nu}{}^{\mu}$$
is a shorthand for
$$\tag{2} (\Lambda^{-1})_{\nu}{}^{\mu}~=~g_{\nu\lambda}(\Lambda^{-1})^{\lambda}{}_{\kappa} (g^{-1})^{\kappa\mu}.$$
Or as matrices
$$\tag{3} g\Lambda^{-1}g^{-1}.$$
Equation (3) reduces to the transposed inverse matrix
$$\tag{4} (\Lambda^{-1})^T $$
if $\Lambda$ is a pseudo-orthogonal/Lorentz matrix. Or with indices
$$\tag{5} ((\Lambda^{-1})^T)_{\nu}{}^{\mu} .$$