# How to express this using the Einstein summation convention

Suppose we are dealing with some mathematical object that has indices; a matrix $M$, say, that has components $M^{\mu\nu}$ for $\mu,\nu=1,2,\dots,n$. Now suppose that I want to express, for instance, the following equations, \begin{eqnarray} M^{11} &=& M^{22} = M^{33} = \dots = M^{nn} = 1 \tag{1} \\ M^{1n} &=& M^{2(n-1)} = M^{3(n-2)} = \dots = M^{n1} = 1 \tag{2} \end{eqnarray} in a compact way. It is tempting to write these equations as \begin{eqnarray} M^{\mu\mu} = 1,\quad M^{\mu(n+1-\mu)} \tag{3} = 1 \end{eqnarray} for all $\mu=1,2,\dots,n$. But if the summation convention understood, then (3) is not something you want to write down; (3) is simply wrong if one sums over $\mu$.

So what is the appropriate way to express equations like (1) and (2), in a context where the Einstein summation convention is understood? Would it be appropriate to use (3), with the explicit remark that the summation convention is not used in that particular equation? (That seems rather artificial.) Or is there some other way?

• You can explicitly state, "no summation", or something similar right next to the equation. – Sean E. Lake Sep 25 '16 at 14:26
• Have you been introduced to the Kronecker delta matrix? – Cosmas Zachos Sep 27 '16 at 1:19
• @cosmos Yes, I thought something like that might be useful, but I don't see how exactly. You do? – Sjorszini Sep 27 '16 at 5:03
• By the way, greek indices ($\mu, \nu$) are usually used for spacetime indices (i.e. starting at 0) while latin indices ($i, j$) are usually used for purely spatial indices (starting at 1) like you are using. – QuantumFool Sep 27 '16 at 6:27
• Did you try looking at the matrices $M^{\mu\nu}\sim \delta ^{\mu\nu}$ or $M^{\mu\nu}\sim \delta ^{\mu(n+1-\nu)}$ ? What does the Einstein summation convention have to do with it? Repeated values for an abstract variable is not the same as repeated abstract variables, as in Es convention. – Cosmas Zachos Sep 27 '16 at 12:42

You could, therefore, write $M^{\mu\mu}$ and state what $\mu$ is.