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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)?

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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

$$\tag{5} ((\Lambda^{-1})^T)_{\nu}{}^{\mu} .$$

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    $\begingroup$ Thanks. I guess I was not clear in my post about what I want. Your interpretation is in terms of matrix multiplication. Is there a generalisation to higher rank tensors? $\endgroup$ – DarKnightS Jan 8 '15 at 15:27
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In the case of $\Lambda_\mu^\nu$ we are talking about an orthogonal tensor (remember in 4d the Lorentz group is SO(1, 3) ), and for an orthogonal matrix we have the definition $$O^T O = O O^T = I $$ Thus we can simply transpose these Lorentz matrices (swapping the indices) to obtain their inverse. This isn't in general true for the inverse of an arbitrary tensor!

In general the inverse of $T^\mu_\nu$ is some other tensor $T^{-1}$ which also has two indices (note the name of the indices is arbitrary, so we might call it $(T^{-1})^\lambda_\sigma$ or whatever we like, as long as it has two indices. However, using the notation $T^{-1}$ makes clear we are talking about the inverse of $T$ here.

In the case of $\delta^i_j$ we have a symmetric tensor, so it doesn't matter when we transpose it (swap the indices). This is true for any symmetric tensor: $$S = S^T \;\;\;\;\;\;\;\;\;\; S^i_j = S^j_i$$

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  • $\begingroup$ Yes. I understand this. I wonder why Weinberg and some others want to emphasize the order of indices. $\endgroup$ – DarKnightS Jan 8 '15 at 14:34
  • $\begingroup$ The order is indeed important. Changing it is like conjugating with the metric, which for Lorentz transforms equals transposition $\endgroup$ – Phoenix87 Jan 8 '15 at 14:55
  • $\begingroup$ Yes, it is important in the sense that $(v_i)^j \in V^*\otimes V$ or$V \otimes V^*$ (depending on how we define tensor product) , but we can also "consolidate" it to be $(v^j)_i$ (in the language of Jefferey Lee in Manifolds and Differential Geometry) by moving say $V$ to the right, then the order becomes "unimportant". $\endgroup$ – DarKnightS Jan 8 '15 at 15:23
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When you have a rank (1,1) tensor (one raised and one lower index and no other indices) you can treat the nxn coordinates in a basis as a matrix, and then as a matrix you can compute the inverse, then interpret those numbers in the matrix as the coordinates in the same basis for a (possibly different) rank (1,1) tensor.

That's literally all that is going on. There is no general tensor inverse operation and not even all rank (1,1) tensors when treated like matrices have inverses.

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