So I found this a bit confusing, and I am wondering if in general is possible to apply the Kronecker Delta only on a “subset” of indices.

So, if we sum over all possible indices we can say for example ($E$ is a generic 4-vector): $$\delta^{t}_{\hspace{0.2 cm}\mu}E^{\mu}=E^{t}$$

But if we restrict the sum only on the spatial indices it doesn’t hold anymore:

$$\delta^{t}_{\hspace{0.2 cm}i}E^{i}\neq E^{t}$$

$$\delta^{t}_{\hspace{0.2 cm}i}E^{i} =0$$

But if we do the opposite way, i.e. restricting only over the spatial indices, the expression is again true:

$$\delta^{x}_{\hspace{0.2 cm}i}E^{i}=E^{x}$$

But my question now is: Can the last expression be used? Or we are forced to sum over all possible indices?

Generalizing, if we have a generic manifold, and {${a,b,c,d,\dots}$}, {${\alpha, \beta,\dots}$} are 2 distinct subsets of the “whole set” {${A,B,C,\dots}$}, then can we say that:

$$\delta^{c}_{\hspace{0.2 cm}A}E^{A}=E^{c}$$

$$\delta^{c}_{\hspace{0.2 cm}\beta}E^{\beta}\neq E^{c}$$

$$\delta^{c}_{\hspace{0.2 cm}\beta}E^{\beta} =0$$

$$\delta^{c}_{\hspace{0.2 cm}d}E^{d}=E^{c}$$

The reason is just that we can use the delta only if the indice over which we are summing can assume that value (e.g. $x$ or $t$), right? Are there others explainations?


1 Answer 1


If you think of indices as something to be summed-over, and you use the summation convention, then it seems you must sum over the entire range of the index.

If you wish to use a subset, then write out the summation symbol and the restricted range explicitly.

if you think of the indices abstractly ["slots", if you will], as in the https://en.wikipedia.org/wiki/Abstract_index_notation , then there is no summation---it's just "index-substitution". (But I suspect that this is not what you are thinking of.)

From your example $$\delta^{x}_{\hspace{0.2 cm}i}E^{i}=E^{x},$$ it seems this is not a vector, but just the "$x$-component of $\vec E$".
You may just be looking for a dot product with a unit basis vector along the $x$-direction: $$g_{ij}\hat x^i E^{j}=E^{x}.$$

To make this more general, you may want to use two different type of indices: one to identify the object as a vector, and another to identify which vector.

It may help to be explicit about what you have to start with (definitions and conventions) and what you are trying to express.

  • $\begingroup$ I have edited the last part of the post, now it should be more clear. Anyway I was trying studying a problem in generic d dimension, with 4 “usual” dimensions and d-4 compactified dimension. So I wanted to use the Greek letters for the 4 usual dimensions, Lowercase Latin letters for the compactified dimension and Uppercase Latin letters in general. So for example I can’t write: “T^{A}S_{A}=T^{\mu}S_{\mu}+ T^{a}S_{a}” because it is quite easy if I distinct the two type of dimensions. $\endgroup$ May 27, 2021 at 23:40
  • $\begingroup$ The only problem was indeed with the Kroneker Delta because it would be easy to do the error “\delta^{a}_{\mu}T^{a}=T^{\mu}” that is not true since the “\delta^{a}_{\mu}=0” since in the subset of the lowercase Latin Letters there are no Greeks letters. Yes in the first example E^{x} was the x component of the 4-vector x, and similarly for t. $\endgroup$ May 27, 2021 at 23:43
  • 1
    $\begingroup$ Use can use $\TeX$ in the comments.... use the dollar-signs as usual. $\endgroup$
    – robphy
    May 27, 2021 at 23:45
  • $\begingroup$ You probably should look into "projection tensors" math.stackexchange.com/questions/1908910/… , en.wikipedia.org/wiki/Projection_(linear_algebra) , people.vcu.edu/~rgowdy/phys691/pdf/pt.pdf $\endgroup$
    – robphy
    May 27, 2021 at 23:47

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