I Kronecker Delta: Order of Indices Explained

[Jim Lai]

Hi everyone,

I am a new member and would like to ask a naive simple (my guess) question.

I am reading Weinberg’s Gravitation and Cosmology. On page 59, Eq. 2.12.10 therein reads
$$
\begin{aligned}
\left[\sigma_{\alpha \beta}\right]_{\gamma \delta}{}^{\varepsilon \zeta}
&=\eta_{\alpha \gamma} \delta_{\beta}{}^{\varepsilon} \delta^{\zeta}{}_{\delta}
-\eta_{\beta \gamma} \delta_{\alpha}{}^{\varepsilon} \delta^{\zeta}{}_{\delta}\\
&+\eta_{\alpha \delta} \delta_{\beta}{}^{\zeta} \delta^{\varepsilon}{}_{\gamma}
-\eta_{\beta \delta} \delta_{\alpha}{}^{\zeta} \delta^{\varepsilon}{}_{\gamma}
\end{aligned}
$$

I wonder if \delta_{\beta}{}^{\varepsilon} is equal to \delta^{\varepsilon}{}_{\beta}. Would anyone enlighten me?

Regards,
Jim Lai

 

[Orodruin]

Yes, those are equal. Both are equal to one if $\varepsilon = \beta$ and zero otherwise. Generally there is therefore no need to worry about the horizontal positioning of the indices on the $\delta$.

 

[vanhees71]

That's because the Kronecker symbol can be read as the mixed components of the "metric" (it's rather a pseudometric) tensor,
$${\delta_{\alpha}}^{\beta}=g_{\alpha \gamma} g^{\gamma \beta} = g_{\gamma \alpha} g^{\gamma \beta} = {\delta^{\beta}}_{\alpha}.$$

 

[Orodruin]

That's because the Kronecker symbol can be read as the mixed components of the "metric" (it's rather a pseudometric) tensor,
$${\delta_{\alpha}}^{\beta}=g_{\alpha \gamma} g^{\gamma \beta} = g_{\gamma \alpha} g^{\gamma \beta} = {\delta^{\beta}}_{\alpha}.$$

I thought about writing that, but then I thought ”… of any non-degenerate (0,2) tensor and its inverse really … or wait, it is just the identity map on the tangent space vs the identity map on the cotangent space …” and then I realized I knew too much and left it there

 

[pervect]

The Kronecker delta is a symmetric tensor, so interchanging the order of the indices doesn't matter. One often sees people omitting the order of indices for symmetric tensors in general, i.e. they're write $\sigma^a_b$. Since for symmetric tensors $\sigma^a{}_b = \sigma^b{}_a$, the order doesn't matter. In general though, the order of indices does matter, and specifying the order of indices is required. For an anti-symmetric tensor $\sigma^a{}_b$ = -$\sigma^b{}_a$, for example.

 

[Orodruin]

The Kronecker delta is a symmetric tensor, so interchanging the order of the indices doesn't matter. One often sees people omitting the order of indices for symmetric tensors in general, i.e. they're write $\sigma^a_b$.

This is incorrect. The Kronecker delta is a (1,1) tensor and if therefore really does not make much sense to discuss symmetries as it only has one index of each type. In a Euclidean space in Cartesian coordinates, it is common to write the metric tensor as $\delta_{ij}$ with both indices down, but this is particular for that coordinate system.

Since for symmetric tensors $\sigma^a{}_b = \sigma^b{}_a$, the order doesn't matter. In general though, the order of indices does matter, and specifying the order of indices is required. For an anti-symmetric tensor $\sigma^a{}_b$ = -$\sigma^b{}_a$, for example.

None of those relations are viable as they mix covariant and contravariant indices (or, in abstract index notation, mixes vector and dual vector arguments).