# Kronicker Delta, Levi-Civita, Christoffel and tensors

1. Apr 12, 2010

### JustinLevy

Kronicker Delta, Levi-Civita, Christoffel ... and "tensors"

For quick reference in grabbing latex equations:
http://en.wikipedia.org/wiki/Levi-Civita_symbol
http://en.wikipedia.org/wiki/Kronecker_delta
http://en.wikipedia.org/wiki/Christoffel_symbols

Wiki warns that the Christoffel symbols are not actually tensors. But they appears to be defined purely in terms of the metric tensor:
$$\Gamma^i_{k\ell}=\frac{1}{2}g^{im} \left(\frac{\partial g_{mk}}{\partial x^\ell} + \frac{\partial g_{m\ell}}{\partial x^k} - \frac{\partial g_{k\ell}}{\partial x^m} \right) = {1 \over 2} g^{im} (g_{mk,\ell} + g_{m\ell,k} - g_{k\ell,m})$$
so why isn't this a tensor?
And if that is not a tensor, how do we show that the curvature is a tensor if we were forced to start with:
$${R^\rho}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho_{\nu\sigma} - \partial_\nu\Gamma^\rho_{\mu\sigma} + \Gamma^\rho_{\mu\lambda}\Gamma^\lambda_{\nu\sigma} - \Gamma^\rho_{\nu\lambda}\Gamma^\lambda_{\mu\sigma}$$

I'm starting to realize that I've been making a lot of assumptions that basically amount to: if it is in a tensor equation, it must transform like a tensor. So now I'm wondering, are the Kronicker Delta, and Levi-Civita symbols actually tensors ... or should they only be considered a convenient way of writing out constants that appear in front of terms?

For instance, starting with (ct,x) in an inertial frame as my coordinates, the kronicker delta is:
$${\delta^\mu}_\nu = \left(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right)$$
Then any coordinate transformation S, in matrix form would be a similarity transformation $S^{-1} \delta S$, and so should stay the same since delta is just the identity matrix. So can I consider that as a tensor (that just happens to be the same in all coordinate systems)?

In EM when I go to the dual field tensor
$$G^{\mu\nu} = \frac{1}{2} \epsilon^{\mu\nu\alpha\beta}F_{\alpha\beta}$$
how can I know if the dual is really a tensor or not? Are the components of the Levi-Civita symbol the same in all coordinate systems? (if so, is there some quick/obvious way to see this without writing out all the transformations?) ... wikipedia claims they can only scale by an overall constant, but that is not intuitive to me. Even if that is true, it seems that the tensor would then depend on which coordinate system you originally defined it in ... so is this a tensor at all?

Last edited: Apr 12, 2010
2. Apr 12, 2010

### utesfan100

Re: Kronicker Delta, Levi-Civita, Christoffel ... and "tensors"

The real way to see if something is a tensor is to determine how it transforms for some differential in the coordinates. If it is linear it is a tensor.

The Christoffel symbols represent derivatives of the metric and the transformations have a tensor transformation plus a second order term. Thus they are not a tensor.

It is possible to add combination of Christoffel symbols so that the second order terms cancel. The result would then be a tensor, like a curvature tensor.

At least that's what I got from my first read of Reimmian Geometry last week.

3. Apr 12, 2010

### bcrowell

Staff Emeritus
Re: Kronicker Delta, Levi-Civita, Christoffel ... and "tensors"

IMO this is an excellent question, JustinLevy -- thanks for asking it. I only know part of the answer to my own satisfaction.

I can offer two reasons why the Christoffel symbols aren't tensors, one mathematical and one physical.

Reason #1 is that you can't form a tensor by taking a partial derivative, you can only form a tensor by taking a covariant derivative. Here is an explanation of why this is so: http://www.lightandmatter.com/html_books/genrel/ch05/ch05.html#Section5.6 [Broken] (subsection 5.6.2)

Reason #2 is that the Christoffel symbols are the things in GR that play the role of g=9.8 m/s2 from Newtonian mechanics. In GR, there can't be any tensor that plays the role of the gravitational field. To see this, note that the tensor transformation laws will never transform a nonzero tensor to a zero tensor, and yet the equivalence principle guarantees that a change of coordinates can always make g=0 at any desired point in spacetime.

Regarding the Kronecker $\delta$ and Levi-Civita $\epsilon$ symbols, I am not satisfied with my own understanding and would be thrilled if someone could give a good answer to this one. I was going to ask it myself, as a matter of fact.

It seems clear to me that $\delta$ and $\epsilon$ don't transform like tensors, and yet I guess you can write down certain expressions involving them such that the result *does* transform like a tensor. Therefore it seems that there must be some rule that says what types of expressions involving $\delta$ and $\epsilon$ make tensors and what types don't. I just don't know what that rule could be.

Last edited by a moderator: May 4, 2017
4. Apr 12, 2010

### nicksauce

Re: Kronicker Delta, Levi-Civita, Christoffel ... and "tensors"

It is very clear that the Christoffel symbol is not a tensor. Simple example: In flat space, in cartesian coordinates, all the Christoffel symbols are zero, but in spherical coordinates they are not. Since a tensor that is zero in one coordinate system is zero in all coordinate systems, we cna conclude that the Christoffel symbol is not a tensor.

The Levi-Civita symbol is actually a tensor density. That is $$\sqrt{|\textnormal{det}g|}\epsilon_{\mu_1...\mu_n}$$ transforms like a tensor. This is well explained in chapter 2.8 of Sean Carrol's GR book.

I would say the Kronocker delta is a tensor though. You can define it as $$\delta^{\mu}_{\rho} = g_{\rho_\nu}g^{\nu\mu}$$, then if you write out the transformation of the RHS, you can see that it transforms correctly.

Last edited: Apr 12, 2010
5. Apr 12, 2010

### George Jones

Staff Emeritus
Re: Kronicker Delta, Levi-Civita, Christoffel ... and "tensors"

Yes. Another way to see this is to transform the the left side. Assume that $\delta^{\mu}_{\nu}$ is with respect to one (unprimed, say) coordinate system, apply the appropriate transformation to another (primed, say) coordinate system, and see what happens.

6. Apr 12, 2010

### bcrowell

Staff Emeritus
Re: Kronicker Delta, Levi-Civita, Christoffel ... and "tensors"

Ah, thanks, that helped a lot. For those who don't own a printed copy of Carroll, the discussion of this topic is available in the free online version of the book: http://nedwww.ipac.caltech.edu/level5/March01/Carroll3/Carroll2.html

7. Apr 13, 2010

### George Jones

Staff Emeritus
Re: Kronicker Delta, Levi-Civita, Christoffel ... and "tensors"

Nobody posted it, so I'll post the calculation.

$$\frac{\partial x'^\alpha}{\partial x^\mu} \frac{\partial x^\nu}{\partial x'^\beta} \delta^\mu_\nu = \frac{\partial x'^\alpha}{\partial x^\mu} \frac{\partial x^\mu}{\partial x'^\beta} = \frac{\partial x'^\alpha}{\partial x'^\beta} = \delta'^\alpha_\beta.$$

In the second-to-last equality, the multivariable chain rule is used.

8. Apr 13, 2010

### dx

Re: Kronicker Delta, Levi-Civita, Christoffel ... and "tensors"

The Christoffel symbols are components of a multilinear function, and therefore, in a sense, they are tensors; but they are tensors that are connected with a particular coordinate system, whereas ordinary tensors are defined independently of a coordinate system and therefore their components transform according to the usual trensor transformation law.

This is probably best illustrated by the definition

$$\Gamma^{\mu}_{\alpha \beta} = dx^{\mu} \cdot \nabla_{\beta} \partial_{\alpha}$$

from which one can see how they depend on the coordinate functions xμ.

Last edited: Apr 13, 2010