# Do I sum over these indices?

Question 2.11 from A First Course In General Relativity, 3rd Edition by Bernard Schutz, asks the reader to verify the following equation:

$$\Lambda^\nu_\beta(v) \Lambda^\beta_\alpha(-v) = \delta^\nu_\alpha$$

After reintroducing the sums and iterating over $$\beta$$ we have:

\begin{align} & \sum\ \Lambda^\nu_\beta(v) \Lambda^\beta_\alpha(-v) \\[8pt] = {} & \Lambda^\nu_0(v) \Lambda^0_\alpha(-v) + \Lambda^\nu_1(v) \Lambda^1_\alpha(-v) + \Lambda^\nu_2(v) \Lambda^2_\alpha(-v) + \Lambda^\nu_3(v) \Lambda^3_\alpha(-v) \end{align}

I am only somewhat familiar with the Einstein Summation Convention at this point, so I am a little confused. Am I supposed to now sum each term over $$\nu$$ and $$\alpha$$? Since $$\delta^\nu_\alpha$$ is the Kronecker Delta clearly there is going to be a lot of cancellation, but I am not entirely sure how to move forward, and how do I tell the next step from the notation.

For reference I am teaching my self GR from this book, so I do not have reliable access to anyone fluent in GR. For this reason I want to be sure my understanding is sound before I move on.

## 2 Answers

Do not sum over $$\alpha$$ and $$\nu$$. To clarify, here is the Einstein summation convention in its most accurate form:

If a covariant index is followed by the same contravariant index and vice versa, we must sum over that particular index. It is needed to make sure that after the summation, the indices of the LHS must coincide with the indices of the RHS. Any violation of this means that there was some mistake in the calculations.

In short, you must not sum over indices other than $$\beta$$, because they are separated by an "=" sign and they don't follow the rule mentioned above.

The equations would go like this: $$\Lambda^0_\beta\Lambda^\beta_1 = \delta^0_1$$ where you sum over $$\beta$$, this was one of the components. In this way, you verify all the components one by one, inserting values to $$\alpha$$ and $$\nu$$ one by one.

• Okay, so I don't sum over them like I would a repeated index. However, I am still unsure how to proceed from here. Another comment says that this should work like matrix multiplication, but that would mean i'm adding terms for corresponding alphas and nus Commented May 25 at 1:39
• I have edited the asnwer, I think this must clarify. Commented May 25 at 1:45
• The singular of "indices" is "index," not "indice."
– hft
Commented May 25 at 1:46
• Okay, my bad, sorry Commented May 25 at 1:47
• So you must evaluate the free index before you sum over the dummy index is that right? Commented May 25 at 1:55

Do not sum over $$\nu,\alpha$$. You would only sum over them if they were repeated. The next step would be to plug in the specific formulas for each matrix element and see the cancellation explicitly. It's not that bad if you just write out the explicit matrices for each boost and do that matrix multiplication.

• So I should iterate through nu and alpha at the same time like 0,0 1,1 ... etc? But if it's like matrix multiplication wouldn't I still have to add each of those terms? Commented May 25 at 0:53
• @RudyJD - You might have gotten this in the meantime, but I think what confused you is that you are (or were) thinking about this as a single expression that has 4 terms, and that's it. But this sum is a shorthand for a whole list of separate sums of the same form, one for each combination of $\nu,\alpha$. Each entire sum is a distinct matrix element $m^\nu_\alpha$, a row-column dot product, if you will. The $\delta^\nu_\alpha$ means that each sum (when evaluated with actual values) will either equal zero or one depending on the $\nu,\alpha$ combination. I.e., they give the identity matrix. Commented May 25 at 13:08
• @FilipMilovanović yes this was precisely my confusion. I didn't realize that you have to expand it into separate expressions, then sum over beta. Thank you for confirming this as well, the responses have been very helpful in correcting this misunderstanding. Commented May 25 at 15:24