In my textbook, there is a proof that the dot product of 2 four-vectors is invariant under a Lorentz transformation. While I understood most of the derivation (I am a beginner and we haven't done any math regarding this notation), there is one step which I do not understand:

$$ (\Lambda^μ_α)(\Lambda_{μβ})x^αy^β =(\Lambda^μ_α)(\Lambda^β_μ)x^αy_β.$$

As you can see the $\beta$ subscript on the $\Lambda$ becomes a superscript, while the $\beta$ superscript on $y$ vector becomes a subscript. I know for a fact that you can change the index position on the $\Lambda$ but the index also changes, meaning we use a new character when the index goes up or down after we multiply $\Lambda$ with the metric tensor. So how does the book do it here?

  • $\begingroup$ You can raise and lower indices with the Minkowski metric. $\endgroup$ Commented Jan 3, 2021 at 19:31
  • $\begingroup$ yes but you also have to change the index greek letter when you do that. Meaning if you change a superscript to a subscript you also have to change the letter i.e from beta to sigma. $\endgroup$
    – imbAF
    Commented Jan 3, 2021 at 19:32
  • $\begingroup$ They are dummy indices as you sum over them. I.e you can call them whatever you want to. Hence $u^\mu v_\mu = u_\sigma v^\sigma $. $\endgroup$ Commented Jan 3, 2021 at 19:36
  • $\begingroup$ honestly i do not understand it.Once that β goes up it changes lets say to a σ. So the RHS of the equation is the same as above but with the difference that the β superscript on the 2nd Λ is a σ. And you say i can simply just change it to β ? What about the β on the y vector. that also goes down but it stays as β. $\endgroup$
    – imbAF
    Commented Jan 3, 2021 at 19:41
  • $\begingroup$ It is true that $x^\mu y_\mu=x_\mu y^\mu$. In the summation convention, repeated indices are summed over implicitly (i.e. we remove the summation symbol and simply define a repeated index, one up one down, to mean that index is summed over), you cannot arbitrarily raise and lower indices unless they are summed over, in which case you can move one up as long as you move the other down. Indices that are repeated are called "dummy indices", indices that are not repeated at called "free indices". $\endgroup$
    – Charlie
    Commented Jan 3, 2021 at 20:24

2 Answers 2


If you have a quantity with a down index, say $A_\mu$, you can raise the index using the metric (in this case the Minkowski metric $\eta$) as $A_\mu=A^\nu\eta_{\nu\mu}$. In an analogous way, we can lower the index of $B^\mu$ as $B^\mu=B_\nu\eta^{\nu\mu}$.

Now suppose we have a term as in your example where there is a quantity with a down-index next to one with an up-index, and they are contracted (i.e. summed over / named the same) as in $A_\mu B^\mu$. We can raise the index of $A_\mu$ and lower the one of $B^\mu$. It's always good to introduce a new letter when raising or lowering an index if you are not sure. So we would do $A_\mu=A^\nu\eta_{\nu\mu}$ just as before, and $B^\mu=B_\sigma\eta^{\sigma\mu}$ (where I have introduced a new letter $\sigma$ instead of $\nu$). With this we have

\begin{align} A_\mu B^\mu&=A^\nu\eta_{\nu\mu}B_\sigma\eta^{\sigma\mu}\\ &=A^\nu B_\sigma\eta_{\nu\mu}\eta^{\sigma\mu}. \end{align}

If we think of the metric $\eta_{\mu\nu}$ as a matrix, notice that the product $\eta_{\nu\mu}\eta^{\sigma\mu}$ is the way to write the matrix multiplication of a matrix times its inverse in the Einstein convention. A matrix times its inverse gives the identity matrix, whose components are given by a Kronecker delta $\delta_\nu^\sigma$. So if we write $\eta_{\nu\mu}\eta^{\sigma\mu}=\delta_\nu^\sigma$, we are left with

$$A_\mu B^\mu=A^\nu B_\sigma\delta_\nu^\sigma.$$

This last expression is summed over $\nu$ and over $\sigma$, but only the terms with $\sigma=\nu$ survive the sum because of the Kronecker delta: $A^\nu B_\sigma\delta_\nu^\sigma=A^\nu B_\nu$. Renaming now the index $\nu$ as $\mu$ (we can do that, it's a dummy index) we end up with

$$A_\mu B^\mu=A^\mu B_\mu.$$

So with a little more practice you'll be able to identify the change $A_\mu B^\mu\to A^\mu B_\mu$ as something trivial.

  • $\begingroup$ Thanks, now i get it. I was trying something similar,but because this is like day 2 of having to deal with this type of notation + no mathematical background, it was hard to understand! Thanks $\endgroup$
    – imbAF
    Commented Jan 3, 2021 at 21:40

If I understand OP's question correctly, he does not understand why $\sum_\beta \Lambda_{\mu\beta} y^\beta = \sum_\beta\Lambda_{\mu}^{\;\;\beta} y_\beta$. This is how it works in detail $$ \sum_\beta \Lambda_{\mu\beta} y^\beta =\sum_{\beta,\sigma} \Lambda_\mu^{\;\;\sigma} \eta_{\sigma\beta} y^\beta = \sum_\sigma \Lambda_\mu^{\;\;\sigma} y_\sigma = \sum_\beta \Lambda_\mu^{\;\;\beta} y_\beta $$ I have written out the sum symbol explicitly, but in these expression repeated indices are understood to be summed over.

Note also that I write $\Lambda_\mu^{\;\;\sigma}$ and not $\Lambda_\mu^{\sigma}$ as it is important to know which of the two indices is raised.

  • $\begingroup$ Thanks a lot, now i get it. So you can change the index once you see that they sum up? $\endgroup$
    – imbAF
    Commented Jan 3, 2021 at 21:39
  • $\begingroup$ Pls accept the answer if you can. Thanks. $\endgroup$ Commented Jan 3, 2021 at 23:00

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