# Position of an index when raising and lowering indices

I'm reading Carroll's book on GR, page 25, and have a question about raising and lowering indices in 1.72: For the first equation, why do we have (sorry I don't know how to leave spaces for the lower indices) $$T^{\alpha\beta\mu}_{\quad\delta} = \eta^{\mu\gamma}T^{\alpha\beta}_{{\quad}\gamma\delta}$$ rather than $$T^{\mu\alpha\beta}_{\quad\delta} = \eta^{\mu\gamma}T^{\alpha\beta}_{\quad\gamma\delta} \, ?$$ The book said that raising and lowering indices does not change the position of an index relative to other indices. I'm not quite sure what is the relative position of that free index $$\mu$$.

• You can leave spaces for lower indices by using {}. For example, use the code T^a{}_b to get $T^a{}_b$. Jan 17 at 0:44
• Because it could destroy the (anti-)symmetry properties. For example, the Riemann curvature tensor is antisymmetric in its first two and last two indices, so it's important to keep the relative position. Jan 17 at 1:50
• @Níckolas Alves Thanks!!
– IGY
Jan 17 at 10:33

In general (ie, without proving some kind of symmetry property), you cannot permute indices on a single tensor $$$$T_{\mu\nu\rho\sigma} \neq T_{\nu\mu\rho\sigma} \neq T_{\rho\nu\mu\sigma} \neq \cdots$$$$ However, you are perfectly free to commute products of tensors, so long as the indices on each tensor remains the same $$$$\eta_{\mu\nu} T_{\alpha\beta\gamma\delta} = T_{\alpha\beta\gamma\delta} \eta_{\mu\nu}$$$$ In fact this is one of the advantages of index notation, the "non-commutivity" of tensor multiplication is encoded in the index structure, rather than the ordering of the tensor symbols.
Of course you can derive the above facts abstractly from the definition of a tensor. However, I find it easier to justify directly in terms of components. If we think of $$T_{\mu\nu\rho\sigma}$$ as an array of $$4^4=256$$ numbers, there is no reason for $$T_{1000}$$ to be related to $$T_{0100}$$. In general these will simply be different numbers. That rules out the more general assertion $$T_{\mu\nu\rho\sigma}=T_{\nu\mu\rho\sigma}$$. However, $$\eta_{11} T_{0123}=T_{0123}\eta_{11}$$ simply because $$\eta_{11}$$ and $$T_{0123}$$ are ordinary numbers, and multiplication commutes for ordinary numbers. That argument will work for any $$\mu, \nu, \alpha, \beta, \gamma, \delta$$, so $$\eta_{\mu\nu} T_{\alpha\beta\gamma\delta} = T_{\alpha\beta\gamma\delta}\eta_{\mu\nu}$$. With a bit of thought you can also use a similar argument to show that contracting any pair (or pairs) on indices in this expression also allows you to commute $$\eta$$ and $$T$$ (so long as the contracted indices are the same in both expressions and the free indices appear in the same places on $$\eta$$ and $$T$$).