# Indicating that indices are equal in Einstein notation

tl;dr: I have an expression like this: (dramatization) $$R_{\mu\nu} = \begin{pmatrix} B^{00}C_{00} & 0 & 0 & 0 \\ 0 & B^{11}C_{10} & 0 & 0 \\ 0 & 0 & B^{22}C_{20} & 0 \\ 0 & 0 & 0 & B^{33}C_{30} \end{pmatrix},$$ and I want to express it compactly within the confines of (implied summation) tensor notation. Preferrably in a way which can be safely substituted for $R_{\mu\nu}$ inside larger expressions (modulo index renaming). Halp.

The following attempt at an expression is dangerous:

$$R_{\mu\nu} = B^{\alpha\alpha}C_{\alpha0} \delta^\alpha_\mu\delta^\alpha_\nu$$

This breaks the general rule of thumb that an implied sum must always be between exactly two copies of the same free index (preferably one upper, one lower); here we have five! More importantly, by breaking this rule, we have thrown away our ability to look at arbitrary subexpressions and evaluate them out of context.

For instance, in any equation which follows the "two copy rule," it would be perfectly valid to substitute $\delta^\alpha_\mu\delta^\alpha_\nu=\delta^\mu_\nu$ regardless of context, because we know the full scope of $\alpha$. This is not the case in the above expression. Simply put, breaking this rule leads to madness, and I'm not one to enjoy setting myself up for error!

So what alternatives do I have?

I can write out the elements explicitly... \begin{align*} R_{\mu\nu} & = 0 \qquad (\mu \ne \nu) \\ R_{00} & = B^{00}C_{00} \qquad\qquad\qquad R_{11} = B^{11}C_{10} \\ R_{22} & = B^{22}C_{20} \qquad\qquad\qquad R_{33} = B^{33}C_{30} \end{align*} But I can't substitute this into other expressions, and it is hardly compact!

I can expand the "dangerous expression" above into something safe... $$R_{\mu\nu} = B^{00}C_{00}\delta^0_\mu\delta^0_\nu +B^{11}C_{10}\delta^1_\mu\delta^1_\nu +B^{22}C_{20}\delta^2_\mu\delta^2_\nu +B^{33}C_{30}\delta^3_\mu\delta^3_\nu$$ which can be substituted for $R\mu\nu$ in other expressions, but geeze... what a mouthful!

What if I write an explicit sum to make it stand out?

$$R_{\mu\nu} = \sum_\alpha\left(B^{\alpha\alpha}C_{\alpha0} \delta^\alpha_\mu\delta^\alpha_\nu\right)$$

If I did this, would it be reasonable? I mean, like, does anybody ever do this?...

So I'm out of ideas. What do I do?

• Are you totally sure that expression is actually a tensor? Einstein notation is built to make it hard to make non-tensors. (Tensors are not just 2d arrays of numbers...) Apr 28 '16 at 22:45
• @knzhou Good catch. The example above is just a strawman, however I can confirm that the motivating example was not a tensor; it was a Kristoffel symbol. Nonetheless I feel that non-tensorlike forms do have a tendency to pop up after substituting specific details to a problem and performing some simplifications. (E.g. ever see someone apply the fact that $\eta_{\mu\nu}=\eta^{\mu\nu}$ in the Minkowski metric?) Apr 29 '16 at 20:40
• Yeah, that is the very first place this kind of issue pops up. I've seen a few sources try to handle it, and they all do it by abandoning indices and using matrix notation, which works fine for tensors up to rank 2 (they write $\eta = \Lambda^T \eta \Lambda$). But this probably isn't what you're looking for. Apr 29 '16 at 20:52
• @knzhou Your example $\eta = \Lambda^T \eta \Lambda$ doesn't actually require the (semi-)abuse of notation $\eta_{\mu \nu} = \eta^{\mu \nu}$. If you restore the indices, they're actually lowered for both $\eta$'s. Aug 4 '17 at 7:58

Your problem is that the thing you want to express is only true in the particular basis you're working in. But equations in the summation convention are true in every basis.

Your expression involving an explicit summation is fine and is probably what any reasonable mathematician would do.

But if you want to show unreasonable devotion to the summation convention then define a new tensor such that

$$\delta_{\alpha\beta\gamma\delta}^{\phantom{\alpha\beta\gamma\delta}\varepsilon}=\begin{cases} 1&\text{if}\;\alpha=\beta=\gamma=\delta=\varepsilon\\ 0&\text{otherwise}\\ \end{cases}$$

in the particular basis that you are working in.

Then write

$$R_{\mu\nu}=\delta_{\mu\nu\gamma\delta}^{\phantom{\mu\nu\gamma\delta}\varepsilon}B^{\gamma\delta}C_{\varepsilon 0}$$

(The tensor $\delta$ can be seen as a way of encoding your basis as a tensor so that it can be used in expressions. See also my answer here.)

• Good point. I've been following this dogmatic approach that everything needs to use the summation convention... but in this case I'm trying to express something that is not manifestly covariant; Trying to shoehorn the notation in rather corrupts its purpose! (then again, I may still want to take the course grader's dogmatism into account...) May 7 '16 at 2:42
• Also the ad-hoc definition of $\delta_{\mu\nu\gamma\delta}{}^\epsilon$ stands as a rather beautiful testament to how simple it is finish tensor expressions by just looking at the unmatched indices. May 7 '16 at 2:55
• While I don't think the convention here is totally established, I believe that the Einstein summation convention is usually only used (in the context of relativity) to represent actual tensor contractions. If you want to sum over an index which happens to be repeated twice in an expression that only holds in a particular frame (which is therefore not a tensor contraction), it's better to write the sum out explicitly. Aug 4 '17 at 6:01
• @tparker I think this is an actual tensor contraction, and holds in every frame. It's just that I've given the definition of $\delta_{\alpha\beta\gamma\delta}^{\phantom{\alpha\beta\gamma\delta}\varepsilon}$ in one particular frame. In other frames it will have different entries, given by using the tensor transformation rule to transform from the basis where I gave the definition. Aug 4 '17 at 7:17
• Whether that's true depends on how the contracted quantity $R_{\mu\nu}$ transforms, which the OP didn't specify beyond saying that it isn't a tensor (which is the origin of the whole subtlety). If it transforms as part of some rank-three tensor, i.e. $R_{\mu \nu} = C_{\mu \nu 0}$, then you are correct that your $\delta$ symbol must transform as a rank-five tensor. If $R_{\mu\nu}$ doesn't transform that way, then $\delta$ isn't a rank-five tensor. Aug 4 '17 at 8:03