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I was thinking whether the identity matrix could be considered as a rank 2 tensor or not, but then I found out that there is a tensor called identity tensor that has the same function.

So what do you think?

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A tensor is an object satisfying a certain transformation law relating its components in different coordinate systems. Any tensor which has the same components as the identity matrix in one Cartesian coordinate system also achieves that in all others obtainable by an orthogonal transformation. But more general transformations can have different effects. For example, a Euclidean plane satisfies$$\mathrm{d}s^2=\mathrm{d}x^2+\mathrm{d}y^2=\mathrm{d}r^2+r^2\mathrm{d}\theta^2=g_{ab}\mathrm{d}x^a\mathrm{d}x^b.$$In Cartesian coordinates, the metric tensor's components are those of the identity matrix, viz. $g_{ab}=\delta_{ab}$. In polar coordinates, the matrix we get is $\operatorname{diag}(1,\,r^2)$ instead.

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  • $\begingroup$ Thank you. So the identity matrix can't be considered as a rank 2 identity tensor. But the rank 2 identity tensor has the same components of identity matrix. Right? $\endgroup$
    – user833434
    Oct 9, 2020 at 9:46
  • $\begingroup$ @Unknown It's precisely because the components vary under general coordinate transformations that we can't talk of an "identity tensor", unless it's a mixed tensor viz. $g_{ab}g^{bc}=\delta_a^c$, but even then we'd usually call it the Kronecker delta. The correct name for the tensor denoted $g_{ab}$ is the (index-lowering) metric tensor. In a non-Euclidean manifold, its components won't be those of an identity matrix even in Cartesian coordinates, because its eigenvalues will be of mixed sign. $\endgroup$
    – J.G.
    Oct 9, 2020 at 10:41

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