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Studying special relativity for an exam on an Italian book (Relatività - Barone for those interested), before introducing a geometrical analysis for Lorentz transformations and Minkowski spacetime, the author gives some reminders on the classical laws of physics and their geometry.

He explains all the properties of the Euclidian space $\mathbb E^3$, and what interested me for this question was the definition of distance that was given, mainly

$$\mathbb d\mathbf x^2 = \delta_{ij}\mathbb dx^i\mathbb dx^j$$

which is very clear.

Then the author proceeds to describe the property of the Minkowski spacetime $\mathbb M$, and defines the distance by the metric tensor $g_{\mu\nu}$ as follows

$$\mathbb d s^2=g_{\mu\nu}\mathbb dx^\mu\mathbb dx^\nu$$

for completeness sake the metric tensor is chosen with signature $(+,-,-,-)$, but it won't be necessary.

He then goes by describing the properties of the metric tensor $g_{\mu\nu}$ and here's where my problem lays: he states $$g^{\mu\nu}g_{\nu\rho} = \delta^\mu_\rho $$

Question: Why is now the Kronecker delta shown with one upper index and one lower index? Is there any difference?

I know that clearly the first and the second Kronecker delta have to be the same, but if they are, why showing them differently? Is there any nuances that I should know? Is is just because the first Kronecker delta is in $3$-dimensions and the second in $4$?

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    $\begingroup$ It's not about the dimension but about what type of tensor it is. It seems more a question about Einstein notation than about Kronecker delta. I am not firm on this thus I do not given an answer. Actually that's a bit of the mark it's just about tensors in a way $\endgroup$ – quid Aug 27 '18 at 16:13
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    $\begingroup$ Thank you for the help @quid, really appreciate it! $\endgroup$ – Davide Morgante Aug 27 '18 at 16:26
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I believe the point is that in Einstein's convention, whenever a symbol shows up as both a lower and upper index, there is an implicit summation.

With this convention it is common to use $x^1,\dots,x^n$ as your coordinates so that, for instance,

$$ \delta_{i,j} {\rm d}x^i {\rm d}x^j \text{ means } \sum_{i = 1}^n \sum_{j = 1}^n \delta_{i,j} {\rm d}x^i {\rm d}x^j.$$

If this is the case, that means that your first equation is written incorrectly: $x_i$ instead of $x^i$.

In this convention, the symbols with an upper subscript act on symbols with a lower subscript and vice versa (i.e. they are elements of dual vector spaces). That is, the standard dot product would be written as

$$ x^1y_1 + \dots + x^ny_n = x^i y_i. $$

In particular, upper indices are for "contravariant" vectors and lower indices are for "covariant" vectors (aka covectors) (see e.g. Wikipedia).

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  • $\begingroup$ Yes, thanks for the correction on the first equation. I know that we're using Einstein convention, but it the differences between the two notation for the delta only based on Einstein convention ? $\endgroup$ – Davide Morgante Aug 27 '18 at 16:15
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    $\begingroup$ @DavideMorgante It depends on what kind of tensor it is. See quid's link for instance. $\endgroup$ – Trevor Gunn Aug 27 '18 at 16:21
  • $\begingroup$ I'm not really familiar with tensors. Unfortunately in my physics course the take it for granted and use it without really explaining them first! Thanks for the answers from both you and @quid . I know that I shouldn't ask in comments but just to know: is there any book you can recommend on tensors? Thanks again for the kind help! $\endgroup$ – Davide Morgante Aug 27 '18 at 16:25
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    $\begingroup$ @DavideMorgante For that, you want an introduction to differential geometry. I recommend looking for "intro to differential geometry" or "undergraduate differential geometry" or "differential geometry for physicists." You should be able to find many resources available online (e.g. people.uncw.edu/lugo/COURSES/DiffGeom/dg1.pdf). $\endgroup$ – Trevor Gunn Aug 27 '18 at 19:16

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