Product of three gamma matrices

[drpepper0708]

I need help proving the identity

[tex]

\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}=\gamma^{\mu}g^{\nu\rho}+\gamma^{\rho}g^{\mu\nu}-\gamma^{\nu}g^{\mu\rho}+i\epsilon^{\sigma\mu\nu\rho}\gamma_{\sigma}\gamma^{5}

[/tex]

 

[arkajad]

Are there any http://en.wikipedia.org/wiki/Gamma_matrices" that you are allowed to use?

Or you want to do it from scratch assuming that nothing at all but definitions can be used?

 

[dextercioby]

Could you, please, use [/tex] ? My advice would be to work on the right hand side and use the Dirac algebra

[tex] \gamma^{(\mu}}\gamma^{\nu)} = 2g^{\mu\nu}\mbox{1}_{4\times4} [/tex]

 

[drpepper0708]

Wiki identities and other identities are allowed, I am trying to start by expressing
[tex]
\gamma^{5}=\frac{i}{24}\epsilon^{\mu\nu\rho\sigma}\gamma_{\mu}\gamma_{v}\gamma_{\rho}\gamma_{\sigma}
[/tex]

from which it follows that

[tex]
\gamma^{5}\gamma^{\sigma}=\frac{i}{24}\epsilon^{\mu\nu\rho\sigma}\gamma_{\mu}\gamma_{v}\gamma_{\rho}\gamma_{\sigma}\gamma^{\sigma}=\frac{i}{6}\epsilon^{\mu\nu\rho\sigma}\gamma_{\mu}\gamma_{v}\gamma_{\rho}

[/tex]

does anyone know of a better way to start

 

[arkajad]

But your identity is in Wiki:

[URL]http://upload.wikimedia.org/math/3/a/b/3ab61b65f059221cc748b304a6a5c7a8.png[/URL]

 

[drpepper0708]

here is what I have so far

[tex]
\gamma^{5}\gamma^{\rho}=\frac{i}{24}\epsilon^{\sigma\mu\nu\rho}\gamma_{\sigma}\gamma_{\mu}\gamma_{\nu}\gamma_{\rho}\gamma^{\rho}=\frac{i}{6}\epsilon^{\sigma\mu\nu\rho}\gamma_{\sigma}\gamma_{\mu}\gamma_{\nu}
[/tex]

[tex]
\gamma^{5}\gamma^{\rho}\gamma^{\nu}=\frac{4i}{6}\epsilon^{\sigma\mu\nu\rho}\gamma_{\sigma}\gamma_{\mu}
[/tex]

[tex]
\gamma^{5}\gamma^{\rho}\gamma^{\nu}\gamma^{\mu}=\frac{8i}{3}\epsilon^{\sigma\mu\nu\rho}\gamma_{\sigma}
[/tex]

then multiplying this result by [tex] \gamma^{5} [/tex] we obtain

[tex]
\gamma^{\rho}\gamma^{\nu}\gamma^{\mu}=-\frac{8i}{3}\epsilon^{\sigma\mu\nu\rho}\gamma_{\sigma}\gamma^{5}
[/tex]

Now,
[tex]
\gamma^{\rho}\gamma^{\nu}\gamma^{\mu}=2\gamma^{\mu}g^{\nu\rho}-2\gamma^{\nu}g^{\mu\rho}+\gamma^{\rho}2g^{\mu\nu}-\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}
[/tex]
so

[tex]
\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}=2\gamma^{\mu}g^{\nu\rho}-2\gamma^{\nu}g^{\mu\rho}+\gamma^{\rho}2g^{\mu\nu}+\frac{8i}{3}\epsilon^{\sigma\mu\nu\rho}\gamma_{\sigma}\gamma^{5}
[/tex]

Now you can see my problem, the factors are wrong. any suggestions?

 

[drpepper0708]

We have to prove the identity, other identities are allowed, but for obvious reasons not that one

 

[Avodyne]

here is what I have so far

[tex]
\gamma^{5}\gamma^{\rho}=\frac{i}{24}\epsilon^{\sigma\mu\nu\rho}\gamma_{\sigma}\gamma_{\mu}\gamma_{\nu}\gamma_{\rho}\gamma^{\rho}=\frac{i}{6}\epsilon^{\sigma\mu\nu\rho}\gamma_{\sigma}\gamma_{\mu}\gamma_{\nu}
[/tex]

This is wrong. You have illegally used the [itex]\rho[/itex] index twice: once as a summed dummy index, and once as a fixed unsummed index. As a result, your 2nd equality is not valid.

 

[arkajad]

In

[tex]

\gamma^{5}\gamma^{\rho}=\frac{i}{24}\epsilon^{\sig ma\mu\nu\rho}\gamma_{\sigma}\gamma_{\mu}\gamma_{\n u}\gamma_{\rho}\gamma^{\rho}=\frac{i}{6}\epsilon^{ \sigma\mu\nu\rho}\gamma_{\sigma}\gamma_{\mu}\gamma _{\nu}

[/tex]

I think you are making a mistake You are summing over the index rho to get 4, but rho on the LHS is fixed, rho on the rhs is a dummy index. You should not mix them. You should rename the dummy index.

 

[arkajad]

I would suggest that you first check that the identity holds for two of the three indices equal. That is easy. So it remains to check when they are different. For that you can apply epsilon to the lhs and use epsilon-epsilon contraction that will give you a delta. That should simplify the problem. There will be only 4 cases, each easy to check directly.

Perhaps there is a mole elegant way, but this one at least works.

 

[drpepper0708]

I figured it out, here it is

Using

[tex]
\gamma^{\nu}\gamma^{\rho}=2g^{\nu\rho}-\gamma^{\rho}\gamma^{\nu}
[/tex]

we have

[tex]
\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}=\gamma^{\mu}\left(2g^{\nu\rho}-\gamma^{\rho}\gamma^{\nu}\right)
[/tex]

[tex]
=2\gamma^{\mu}g^{\nu\rho}-\gamma^{\mu}\gamma^{\rho}\gamma^{\nu}
[/tex]

next we use

[tex]
\gamma^{\mu}\gamma^{\rho}=2g^{\mu\rho}-\gamma^{\rho}\gamma^{\mu}
[/tex]

and we obtain

[tex]
\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}=2\gamma^{\mu}g^{\nu\rho}-2\gamma^{\nu}g^{\mu\rho}+\gamma^{\rho}\gamma^{\mu}\gamma^{\nu}
[/tex]

now use

[tex]
\gamma^{\mu}\gamma^{\nu}=2g^{\mu\nu}-\gamma^{\nu}\gamma^{\mu}
[/tex]

then we have

[tex]
\gamma^{\rho}\gamma^{\nu}\gamma^{\mu}=2\gamma^{\mu}g^{\nu\rho}-2\gamma^{\nu}g^{\mu\rho}+\gamma^{\rho}2g^{\mu\nu}-\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}
[/tex]

Now we expand out the following

[tex]
6i\epsilon^{\sigma\mu\nu\rho}\gamma_{\sigma}\gamma^{5}=\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}-\gamma^{\mu}\gamma^{\rho}\gamma^{\nu}+\gamma^{\nu}\gamma^{\rho}\gamma^{\mu}-\gamma^{\nu}\gamma^{\mu}\gamma^{\rho}+\gamma^{\rho}\gamma^{\mu}\gamma^{\nu}-\gamma^{\rho}\gamma^{\nu}\gamma^{\mu}

[/tex]

using

[tex]

\gamma^{\rho}\gamma^{\nu}\gamma^{\mu}=2\gamma^{\mu}g^{\nu\rho}-2\gamma^{\nu}g^{\mu\rho}+\gamma^{\rho}2g^{\mu\nu}-\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}

[/tex]

this becomes,

[tex]
=2\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}-\gamma^{\mu}\gamma^{\rho}\gamma^{\nu}+\gamma^{\nu}\gamma^{\rho}\gamma^{\mu}-\gamma^{\nu}\gamma^{\mu}\gamma^{\rho}+\gamma^{\rho}\gamma^{\mu}\gamma^{\nu}-2\gamma^{\mu}g^{\nu\rho}+2\gamma^{\nu}g^{\mu\rho}-2\gamma^{\rho}g^{\mu\nu}

[/tex]

moreover, using

[tex]
\gamma^{\mu}\gamma^{\rho}\gamma^{\nu}=2\gamma^{\nu}g^{\rho\mu}-2\gamma^{\rho}g^{\nu\mu}+2\gamma^{\mu}g^{\nu\rho}-\gamma^{\nu}\gamma^{\rho}\gamma^{\mu}
[/tex]

we obtain,

[tex]

6i\epsilon^{\sigma\mu\nu\rho}\gamma_{\sigma}\gamma^{5}=2\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}+2\gamma^{\nu}\gamma^{\rho}\gamma^{\mu}-\gamma^{\nu}\gamma^{\mu}\gamma^{\rho}+\gamma^{\rho}\gamma^{\mu}\gamma^{\nu}-4\gamma^{\mu}g^{\nu\rho}

[/tex]

once more into the breech! use,

[tex]
\gamma^{\nu}\gamma^{\mu}\gamma^{\rho}=2\gamma^{\rho}g^{\mu\nu}-2\gamma^{\mu}g^{\rho\nu}+2\gamma^{\nu}g^{\rho\mu}-\gamma^{\rho}\gamma^{\mu}\gamma^{\nu}
[/tex]

then the foregoing becomes

[tex]
3i\epsilon^{\sigma\mu\nu\rho}\gamma_{\sigma}\gamma^{5}=\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}+\gamma^{\nu}\gamma^{\rho}\gamma^{\mu}+\gamma^{\rho}\gamma^{\mu}\gamma^{\nu}-\gamma^{\rho}g^{\mu\nu}-\gamma^{\nu}g^{\rho\mu}-\gamma^{\mu}g^{\nu\rho}
[/tex]

Now, we use

[tex]
\gamma^{\mu}\gamma^{\nu}=2g^{\mu\nu}-\gamma^{\nu}\gamma^{\mu}
[/tex]

on the third term to get

[tex]
3i\epsilon^{\sigma\mu\nu\rho}\gamma_{\sigma}\gamma^{5}=\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}+\gamma^{\nu}\gamma^{\rho}\gamma^{\mu}+\gamma^{\rho}g^{\mu\nu}-\gamma^{\rho}\gamma^{\nu}\gamma^{\mu}-\gamma^{\nu}g^{\rho\mu}-\gamma^{\mu}g^{\nu\rho}
[/tex]

[tex]
=2\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}+\gamma^{\nu}\gamma^{\rho}\gamma^{\mu}-\gamma^{\rho}g^{\mu\nu}-3\gamma^{\mu}g^{\nu\rho}+\gamma^{\nu}g^{\mu\rho}
[/tex]

then, using

[tex]
\gamma^{\nu}\gamma^{\rho}=2g^{\nu\rho}-\gamma^{\rho}\gamma^{\nu}
[/tex]

we finally get

[tex]
3i\epsilon^{\sigma\mu\nu\rho}\gamma_{\sigma}\gamma^{5}=3\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}-3\gamma^{\mu}g^{\nu\rho}+3\gamma^{\nu}g^{\mu\rho}-3\gamma^{\rho}g^{\mu\nu}
[/tex]

or

[tex]
\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}=\gamma^{\mu}g^{\nu\rho}+\gamma^{\rho}g^{\mu\nu}-\gamma^{\nu}g^{\mu\rho}+i\epsilon^{\sigma\mu\nu\rho}\gamma_{\sigma}\gamma^{5}
[/tex]

QED

 

[drpepper0708]

Thank you everyone for your help in pointing out my mistake, it helped me narrow in on the solution.

 

[arkajad]

And how did you get this?

[tex]

6i\epsilon^{\sigma\mu\nu\rho}\gamma_{\sigma}\gamma ^{5}=\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}-\gamma^{\mu}\gamma^{\rho}\gamma^{\nu}+\gamma^{\nu} \gamma^{\rho}\gamma^{\mu}-\gamma^{\nu}\gamma^{\mu}\gamma^{\rho}+\gamma^{\rho }\gamma^{\mu}\gamma^{\nu}-\gamma^{\rho}\gamma^{\nu}\gamma^{\mu}

[/tex]

What kind of expansion?

 

[Jakob_L]

But your identity is in Wiki:

[PLAIN]http://upload.wikimedia.org/math/3/a/b/3ab61b65f059221cc748b304a6a5c7a8.png[/QUOTE]
So continuing the discussion after 30 months :-)

Thank you for outlining the proof, drpepper0708.
Since my question below involves the detailes of conventions, etc, let me summarize all the ingredients of the proof.
If one starts from the identity
\begin{equation}
i\epsilon^{\sigma\mu\nu\rho}\gamma_\sigma \gamma_5 = \gamma^{\mu\nu\rho} ,\;\;\;\;\;\;\;\;\;(\ast)
\end{equation}
expanding the right hand side and using only the Clifford algebra,
\begin{equation}
\{\gamma_\mu ,\gamma_\nu \} = 2g_{\mu\nu},
\end{equation}
one finds the identity
\begin{equation}
\gamma^\mu\gamma^\nu \gamma^\rho = g^{\mu\nu}\gamma^\rho + g^{\nu\rho}\gamma^\mu-g^{\mu\rho}\gamma^\nu + i\epsilon^{\sigma \mu\nu\rho}\gamma_\sigma \gamma_5 .\;\;\;\;\;\;\;\;(\ast \ast)
\end{equation}
Now, the identity (*) can be proven be expanding the left hand side using
\begin{equation}
\gamma_5 = \frac i{4!}\epsilon_{\mu\nu\rho\sigma}\gamma^\mu\gamma^\nu \gamma^\rho \gamma^\sigma ,
\end{equation}
and then using the Clifford algebra as well as
\begin{equation}
\{\gamma_5 , \gamma_\mu\} =0
\end{equation}
and
\begin{equation}
\epsilon^{\alpha\mu\nu\rho}\epsilon_{\alpha\sigma\lambda\kappa} = -3! \delta^{[\mu}_{\sigma}\delta^\nu_\lambda\delta^{\rho]}_\kappa .
\end{equation}
The last equation is valid on four-dimensional manifolds with Lorentzian signature. It can be found in more generality in e.g. Carroll.

Using only these ingredients, I find that the proof goes through.

Now to the question!

As noted by arkajad, the identity (**) is also on wikipedia
http://en.wikipedia.org/wiki/Gamma_matrices,
but the last term differs by a sign.
I can't seem to figure out how this sign comes in. I assume it is a matter of conventions. However, both wikipedia and the proof by drpepper0708 use the same conventions,
\begin{equation}
\{\gamma_\mu ,\gamma_\nu \} = 2g_{\mu\nu},
\end{equation}
and
\begin{equation}
\gamma_5 = \frac i{4!}\epsilon_{\mu\nu\rho\sigma}\gamma^\mu\gamma^\nu \gamma^\rho \gamma^\sigma .
\end{equation}

Am I missing an obvious mistake?
I very much invite anyone to redo these computations and find a possible reason for the differing sign.

Cheers!

 

[arkajad]

What if there is a sign mistake in the formula of drpepper0708 that I have asked about in my post of Oct27,10?

 

[Bill_K]

Since you know the answer, and it is just now a matter of sign, why don't you try by hand one of the components and see whether it comes out plus or minus? This will take you a grand total of three matrix multiplications!

Just pick a representation of the gamma matrices, set μ = 1, ν = 2, λ = 3 and see what you get.

 

[arkajad]

And that is exactly what I did.

 

[Bill_K]

And so what is the remaining question? BTW, what sign convention are you using for ε^μνστ?

 

[arkajad]

I think there is no convention needed. The natural order is 0,1,2,3. My remaining question is where drpepper0708 got his "expansion" from. But after 30 months I am not expecting an answer.

 

[Bill_K]

I think there is no convention needed. The natural order is 0,1,2,3.

Any of ε₀₁₂₃, ε⁰¹²³, ε₁₂₃₄ or ε¹²³⁴ could be +1, depending on your convention.

 

[arkajad]

Then if someone writes a formula that has several possible meanings, one should explain the meaning of the symbols used. I do not see many indices "4" in modern space-time textbooks. As for the upper and lower indices - one should remember that epsilon is a tensor density (or pseudo-tensor), not a tensor. Therefore one should not rise and lower indices the same way as you do it for tensors. Unfortunately some authors call it a "tensor". Surely it adds to confusion.

 

[Bill_K]

As for the upper and lower indices - one should remember that epsilon is a tensor density (or pseudo-tensor), not a tensor. Therefore one should not rise and lower indices the same way as you do it for tensors.

Of course you do! In Minkowski coordinates there's no difference between tensors and tensor densities anyway. And in either case, indices are raised and lowered exactly the same way - using the metric.

 

[arkajad]

Of course you do! In Minkowski coordinates there's no difference between tensors and tensor densities anyway.

There is a difference if you allow, for instance, for space inversion.

 

[arkajad]

But you asked what is my convention. My convention is:

[tex]\varepsilon^{0123}=\varepsilon_{0123}=1[/tex]

Unusual? Perhaps. But there are reasons for it.

 

[Bill_K]

Sorry, that's inconsistent. If you raise all four indices there must be a sign change.

Here's one set of conventions, copied from the nearest GR book. Let [μνστ] denote a purely numerical quantity, the permutation symbol, always equal to ±1, 0. We define a tensor (not a density) ε_μνστ as follows. In Minkowski coordinates, its covariant components are ε_μνστ = [μνστ]. Consequently its contravariant components are ε^μνστ = - [μνστ]. (Note the minus sign! All four indices must be raised using η^μν = Diag(-1, 1, 1, 1), and one of those is negative.) Again, ε_μνστ is a genuine, honest-to-gosh tensor.

Transforming to a general coordinate system, one can show that ε_μνστ = (-g)^1/2 [μνστ]. The (-g)^1/2 is the fault of the permutation symbol, not the fault of ε_μνστ. It shows that [μνστ] can be regarded as a covariant tensor density of weight -1.

One can also show that ε^μνστ = - (-g)^-1/2 [μνστ]. This shows that [μνστ] can also be regarded as a contravariant tensor density of weight +1.

 

[arkajad]

Sorry, that's inconsistent.

This is consistent if you consider epsilon as a pseudotensor, not a tensor, and if you consider one as a dual object for the other. Of course the rule of lowering and rising the indices is not the same for tensors and pesudotensors. But why should it be the same? These are two different kinds of geometric objects.

But I have no objections for other conventions. Each has its advantages and disadvantages.

My reference book: "Differential geometry and Lie groups for physicists", Marian Fecko, Cambridge 2006, p. 107

 

[Bill_K]

Sorry I was not able to help you. At least now I understand why you're having trouble with the sign.

 

[arkajad]

It was not me who asked for help. It was first drepper0708 and then Jacob_L. But I think that you helped a lot with your explanation of different convention used in physics book that are not always in agreement with the mathematical nature of the objects. Physicists often simply play with index rules, mathematicians pay usually more attention to to the true nature of these objects. But even mathematicians may use different conventions for essentially the same entities. I think that it is good to be aware of these differences.