# Kronecker symbol

1. Jul 20, 2011

### 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?

Last edited: Jul 20, 2011
2. Jul 20, 2011

### dextercioby

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

3. Jul 20, 2011

### 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.

4. Jul 20, 2011

### Fredrik

Staff Emeritus
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 $\delta'_{\alpha\beta}= \delta_{\alpha\beta}$.

5. Jul 20, 2011

### Rasalhague

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

- 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 $\delta^\alpha_\beta$.

6. Jul 20, 2011

### Fredrik

Staff Emeritus
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.

Last edited: Jul 20, 2011
7. Jul 20, 2011

### Rasalhague

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

8. Jul 21, 2011

### 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.

9. Jul 21, 2011

### intervoxel

Thanks to everyone for clarifying my query!

10. Jul 21, 2011

### Rasalhague

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