# Spacing in indices, and relation to matrices, in special relativity notation

I have some general confusion regarding notation on tensors in special relativity, and how indices correspond to the matrix representation of second-rank tensors.

When one has a second-rank tensor $$T_\mu^\nu$$, is it necessary to stagger the indices, e.g. $${T_\mu}^\nu$$ or $${T^\nu}_\mu$$?

I also read elsewhere that it's common convention to take the first index as the row of the matrix representation of a second-rank tensor and the second index as a column. But when one has a mixed tensor like $$T_\mu^\nu$$, there seems to be no way to tell. It only seems to become a little more clear in products like $$T_\mu^\nu p_\nu$$, because you know which one has to be which in order for the product to make any sense from a linear algebra perspective ($$\nu$$ must be the column index).

So I guess my questions are

1. Are the above generally correct statements?
2. Is it convention, or necessary, or both/neither, to stagger indices (e.g. $${T_\mu}^\nu$$), or is no spacing (e.g. $$T_\mu^\nu$$) valid?
• You always should stagger the indices. The common exceptions are the Dirac delta $\delta^\mu_\nu$ which is symmetric anyway, so switching the indices doesn't matter, and the Lorentz transformation $\Lambda^\mu_{\ \ \nu}$, which aren't tensors anyway. – knzhou Sep 6 '19 at 0:29
• Then for a mixed rank $2$ tensor the index order just tells you whether the one-form is the first or the second argument. The one-form is the first argument if the upper index is first. – knzhou Sep 6 '19 at 0:46
• It's fine to write things out in components to see them clearer, but you can do this any way you want. You can take the first index to be the row, or the second index to be the row, doesn't matter. The only thing that matters is that the eventual matrix multiplication you write down matches the tensor contraction you actually want to compute. – knzhou Sep 6 '19 at 0:48
• @flevinBombastus Suppose $F_{ab}=A_a B_b$ with $A\neq B$. What is $F_a^b$? Is is $A_a B^b$ or $A^b B_a$? – Chiral Anomaly Sep 6 '19 at 0:57
• Well, suppose you raise the index on $F^\mu_\nu$. Do you get $F^{\mu\nu}$ or $F^{\nu\mu}$? You would keep track of index order for non-mixed rank $2$ tensors, so to avoid losing information you need to do the same when they're mixed. – knzhou Sep 6 '19 at 0:59

The only time you don't need to stagger a tensor with two indices is when the unstaggered version is unambiguous, i.e. iff the tensor would be symmetric if its indices weren't mixed.

For example, the metric tensor (whose mixed version is the Kronecker delta) doesn't need staggering, and in general relativity the Ricci and stress tensors don't need it either.

On the other hand, the electromagnetic $$F_{\mu\nu}$$ is antisymmetric, so if it becomes mixed it must be staggered! (But it's probably best to write expressions so it isn't mixed.)

The rules for tensors with more indices are left as an exercise.

• I prefer all the indices below: For me it is more clear and simple. – Sebastiano Sep 6 '19 at 15:46
• @Sebastiano With all lower indices, you can’t have contractions. So, for example, you can’t write Maxwell’s equations or the Lorentz force law. – G. Smith Sep 6 '19 at 16:57
• @G.Smith You kinda can, but it starts looking silly, e.g. $x_iy_i=x_iy_j\eta_{ij}$. – J.G. Sep 6 '19 at 17:00
• @J.G.That is often done for spatial indices, but never for spacetime indices. – G. Smith Sep 6 '19 at 17:04
• @G.Smith It shouldn't ever be done for spacetime indices, but I've definitely seen it done. – J.G. Sep 6 '19 at 17:09