QFT: Why Kronecker Symbol \delta_j^i is Basis Invariant?

[intervoxel]

I'm preparing myself for a QFT course and I have the following question about the Kronecker symbol:

Why $\delta_j^i$ is invariant to a change of basis and $\delta_{ij}$ is not?

 

[dextercioby]

?? $\delta_{i}^{j}$ is nothing but the unit matrix. It's supposed to be invariant, now matter how write the indices...

 

[Rasalhague]

$\delta^\alpha_{\beta}$ is the identity tensor; its indices with respect to any basis are the identity matrix. If $\Lambda^{\alpha}_{\enspace\beta}$ is a Lorentz transformation, its inverse is represented in Einstein's summation convention by lowering one index and raising the other. (This convention for denoting the inverse applies only to Lorentz tarnsformations, not tensors in general.) With these notational rules

$$\delta^\alpha_\beta=\Lambda^\alpha_{\enspace\gamma}\,\Lambda_\beta^{\enspace\gamma}=\eta^{\nu\gamma}\eta_{\mu\beta}\,\Lambda^\alpha_{\enspace\gamma}\,\Lambda^{\mu}_{\enspace\nu}$$

where $\eta_{\rho\sigma}$ is the metric tensor for Minkowski space, and has the same components as its inverse, $\eta^{\rho\sigma}$, namely Diag(-1,1,1,1) or Diag(1,-1,-1,-1), depending on which sign convention is used. The equation just says that the transformation composed with its inverse is (by definition) the identity tensor.

But

$$\delta'_{\alpha\beta}=\Lambda_{\alpha}^{\enspace \gamma}\,\delta_{\gamma\delta}\,\Lambda_{\beta}^{ \enspace \delta} \neq \delta_{\alpha\beta}$$

is rather the composition of the two matrices representing the Lorentz transformation. $\delta_{\gamma\delta}$ is the identity matrix, but $\delta'_{\alpha\beta}$ will not be, except in the trivial case where $\Lambda_{\alpha}^{\enspace \gamma}$ is the identity matrix too.

 

[Fredrik]

$\delta_{\gamma\delta}$ is the identity matrix, but $\delta'_{\alpha\beta}$ will not be, except in the trivial case where $\Lambda_{\alpha}^{\enspace \gamma}$ is the identity matrix too.

If $\delta_{\alpha\beta}$ is defined by $\eta_{\alpha\gamma}\delta^\gamma_\beta$, then $\delta_{\alpha\beta}=\eta_{\alpha\beta}$. The right-hand side of $\delta'_{\alpha\beta}=\Lambda_{\alpha}^{\enspace \gamma}\,\delta_{\gamma\delta}\,\Lambda_{\beta}^{ \enspace \delta}$ is the $\alpha\beta$ component of the matrix $(\Lambda^{-1})^T\eta\Lambda^{-1}$. This is equal to $\eta$ since the inverse of a Lorentz transformation is a Lorentz transformation. So $\delta'_{\alpha\beta}= \delta_{\alpha\beta}$.

 

[Rasalhague]

Ah yes, that seems like a more consistent definition. Is it more usual than the other convention?

However, a second rank covariant tensor whose components in the $x^i$-frame are the Kronecker deltas (in this case denoted by $\delta_{ij}$) has different components in other frames and is accordingly of no special interest.

- D.F. Lawden: An Introduction to Tensor Calculus, Relativity and Cosmology, 3rd ed., p. 92.

And I think I remember Leonard Susskind talking in one of his online GR lectures about the deltas with both indices down, or both up, as "not tensors" in constrast to $\delta^\alpha_\beta$.

 

[Fredrik]

If the other convention is to have the components of $\delta_{\alpha\beta}$ be the components of the identity matrix, the components would be preserved by those coordinate transformations that correspond to orthogonal transformations (rotations) of the tangent space.

Sounds like the quote is about arbitrary coordinate transformations, not specifically about Lorentz transformations or rotations.

 

[Rasalhague]

Sounds like the quote is about arbitrary coordinate transformations, not specifically about Lorentz transformations or rotations.

You're right, it is about general coordinate transformations.

 

[dextercioby]

The <delta with 2 indices down or up> is nothing but the metric tensor or its inverse, it's no longer the delta Kronecker which always has one index up and one index down and always denotes the unit matrix in 2,3,4,... dimensions, no matter if the underlying manifold is flat or curved.

 

[intervoxel]

Thanks to everyone for clarifying my query!

 

[Rasalhague]

Thank you, intervoxel, for asking - I learned something too : )