For a tensor named T with two indices, there are four possibilities: $T_{ij}$ , $T_i^{\ j}$, $T^i{\ _j}$ and $T^{ij}$. Is there a common convention as to how these tensors would be represented as matrices, i.e. where the entries would go? Is it the left-right order of the indices that determines which matrix entry is meant, or some other convention? What if the the order of the indices in a mixed tensor is not indicated at all (as in $T_i^j$)? Is it true that, for instance, the component with i=2 and j=3 would go on the second row and the third column in all of the above cases? The books will just say "$F_{μν}$ = [some matrix]", and you don't know which is which.

Below is an example that is in itself contradictory. To convey the idea that F is antisymmetric, they use two different conventions in the very same line - here it is the order of the Greek subscripts that determines the order.

$$ F_{\mu \nu} = \left( \begin{array}{cccc} 0 & -E_1 & -E_2 & -E_3 \\ E_1 & 0 & B_3 & -B_2 \\ E_2 & -B_3 & 0 & B_1 \\ E_3 & B_2 & -B_1 & 0 \end{array} \right) = -F_{\nu \mu} $$

  • $\begingroup$ This might be better for math.stackexchange.com $\endgroup$
    – HDE 226868
    Commented Aug 29, 2015 at 17:23
  • $\begingroup$ I thought that indices on top vs indices on bottom had something to do with co- or contravariance, but for what I do (fluid and solid mechanics), we only ever dealt with indices on the bottom so I don't know for sure. $\endgroup$
    – tpg2114
    Commented Aug 29, 2015 at 17:48
  • 1
    $\begingroup$ @AcidJazz This problem only ever comes up in physics, mathematicians tend to be more pedantic about clarity, thankfully. $\endgroup$
    – Pifagor
    Commented Aug 29, 2015 at 17:52
  • $\begingroup$ Well, in that case, and hopefully the author's original setup is maintained throughout the book then, on a practical level, I take the first example he/she gives, identify the element order and trust that the author follows through. But between typos, editors (or lack of them) and the author's inconsistencies, you can be left stranded many times:) Been there..... $\endgroup$
    – user81619
    Commented Aug 29, 2015 at 18:01
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    $\begingroup$ This is a good question that too many authors neglect to clarify. For my own part I always intend for the first index to index rows, and I never stack indices vertically except maybe on $\delta$, but I can't say for sure this is universal. $\endgroup$
    – user10851
    Commented Aug 29, 2015 at 20:51

2 Answers 2


In my experience, reading the indices left to right and top to bottom, the first index is the row and the second is the column.

Your screenshot from Carroll doesn't have to be contradictory (although it's definitely confusing/doesn't make rigorous sense). You can just imagine he omits a little "$_{\mu \nu}$" on the matrix:

$$F_{\mu \nu}=\Bigg( \cdots \Bigg)_{\mu \nu}=-F_{\nu \mu}$$

now it's a true real number equation.


Your example is an outlier, in my experience (personally, I would have written $(F_{\mu\nu})^T$ instead of $F_{\nu\mu}$). Almost always, it's the order of the indices that determines row vs. column. If someone writes $T^i_j$, then while technically there's no way to tell, I would say that it would be far less confusing to make the upper index label the rows and the lower index label the columns. This is because a mixed tensor can be regarded as a linear transformation between vectors:

$$v^i = T^i_j v^j$$

If we want to express this linear transformation as a multiplication of a matrix by a vector, then $j$ should label the columns, since that's the index that is being contracted with the vector $v^j$.

The bottom line, however, is that 99% of the time there will be a left/row index and a right/column index. Representing a tensor as a matrix with any other convention is confusing and should not be done, unless the author has a very strong reason to do so.

  • 2
    $\begingroup$ Good points, thanks. I would still prefer $(F^T)_{μν}$ to your $(F_{μν})^T$, because the transpose of a matrix makes sense, whereas putting the T for transposing outside the brackets makes you think it's all the entries being "transposed" independently. $\endgroup$
    – Pifagor
    Commented Aug 29, 2015 at 21:44

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