# Dimension of the gamma matrices in general dimension

#### ismaili

Dear guys,

I read a derivation of the dimension of gamma matrices in a $$d$$ dimension space, which I don't quite understand.

First of all, in $$d$$ dimension, where $$d$$ is even.
One assumes the dimension of gamma matrices which satisfy
$$\{ \gamma^\mu , \gamma^\nu \} = 2\eta^{\mu\nu} \quad\quad\cdots(*)$$
is $$m$$.
A general a m by m matrix with complex arguments should have $$2m^2$$ independent components.
Now, eq(*) gives $$m^2$$ constraints. (<= I don't quite understand this.)

So, the independent components of a single gamma matrix should be $$m^2$$.

On the other hand, one finds that the anti-symmetrization of gamma matrices can produce space-time tensors under Lorentz transformation, i.e. for example,
$$\bar{\psi}\gamma^{\mu\nu}\psi \rightarrow \Lambda^\mu{}_\rho\Lambda^{\nu}{}_\sigma\bar{\psi}\gamma^{\rho\sigma}\psi$$
where $$\gamma^{\mu\nu} = \gamma^{[\mu}\gamma^{\nu]}$$

Now, the various antisymmetric tensors decompose the Lorentz group into different pieces which do not mix. We now calculate the independent components of each anti-symmetric tensor, and add it up:
$$C^d_0 + C^d_1 +C^d_2 + \cdots + C^d_d = 2^d$$

Now we match the two independent components we calculated (Why?! why they should match?!)
$$m^2 = 2^d$$
This concludes that $$m = 2^{d/2}$$.

Now, for d = 2k+1 being odd, one can easily add $$\gamma^{2k} \sim \gamma^0\gamma^1\cdots\gamma^{2k-1}$$, together with the original $$\gamma^0,\gamma^1,\cdots,\gamma^{2k-1}$$ to form gamma matrices in d = 2k+1.

Since the anti-symmetric tensors has a linear relation,
$$\gamma^{\mu_0\mu_1\cdots\mu_r} = \epsilon^{\mu_0\mu_1\cdots\mu_{2k}}\gamma_{\mu_{r+1}\cdots\mu_{2k}}$$.
So there are actually $$2^{d}/2$$ independent components for odd $$d$$.
Hence, the dimension of gamma matrices in odd spacetime dimension should be $$2^{\frac{d-1}{2}}$$.
My question is that, isn't the linear relation between anti-symmetric tensors also hold in d = even spacetime dimension?

Anyone help me go through the puzzles? thanks so much!

ismaili

----

Oh, by the way, I found that from the Dirac representation method which I described in another nearby thread titled "spinors in various dimensions", one can easily realize the dimension of gamma matrices in even dimension d should be $$2^{d/2}$$.

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#### fermi

I think you are trying to do this the hard way: by brute force. There are several really cute and powerful theorems that help a lot on the way. I really recommend that you take a look at Appendix-L, in the book "Group Theory in Physics, Volume-II, by J.F. Cornwell", published by Academic Press in 1984. It may or may not be out of print these days, but I am sure you can find it in a library if it is out of print.

In this appendix which is only 25 pages or so, Cornwell does better than most books do in a 100 pages on the application of Clifford Algebras to Physics. The theorems are provided, many are actually proven on the way, including the ones you need. Where needed, he also provides good physical insight. I hope it will be helpful to you.

#### dextercioby

Homework Helper
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#### fermi

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#### ismaili

Dear guys,

I read a derivation of the dimension of gamma matrices in a $$d$$ dimension space, which I don't quite understand.

First of all, in $$d$$ dimension, where $$d$$ is even.
One assumes the dimension of gamma matrices which satisfy
$$\{ \gamma^\mu , \gamma^\nu \} = 2\eta^{\mu\nu} \quad\quad\cdots(*)$$
is $$m$$.
A general a m by m matrix with complex arguments should have $$2m^2$$ independent components.
Now, eq(*) gives $$m^2$$ constraints. (<= I don't quite understand this.)

So, the independent components of a single gamma matrix should be $$m^2$$.

On the other hand, one finds that the anti-symmetrization of gamma matrices can produce space-time tensors under Lorentz transformation, i.e. for example,
$$\bar{\psi}\gamma^{\mu\nu}\psi \rightarrow \Lambda^\mu{}_\rho\Lambda^{\nu}{}_\sigma\bar{\psi}\gamma^{\rho\sigma}\psi$$
where $$\gamma^{\mu\nu} = \gamma^{[\mu}\gamma^{\nu]}$$

Now, the various antisymmetric tensors decompose the Lorentz group into different pieces which do not mix. We now calculate the independent components of each anti-symmetric tensor, and add it up:
$$C^d_0 + C^d_1 +C^d_2 + \cdots + C^d_d = 2^d$$

Now we match the two independent components we calculated (Why?! why they should match?!)
$$m^2 = 2^d$$
This concludes that $$m = 2^{d/2}$$.

Now, for d = 2k+1 being odd, one can easily add $$\gamma^{2k} \sim \gamma^0\gamma^1\cdots\gamma^{2k-1}$$, together with the original $$\gamma^0,\gamma^1,\cdots,\gamma^{2k-1}$$ to form gamma matrices in d = 2k+1.

Since the anti-symmetric tensors has a linear relation,
$$\gamma^{\mu_0\mu_1\cdots\mu_r} = \epsilon^{\mu_0\mu_1\cdots\mu_{2k}}\gamma_{\mu_{r+1}\cdots\mu_{2k}}$$.
So there are actually $$2^{d}/2$$ independent components for odd $$d$$.
Hence, the dimension of gamma matrices in odd spacetime dimension should be $$2^{\frac{d-1}{2}}$$.
My question is that, isn't the linear relation between anti-symmetric tensors also hold in d = even spacetime dimension?

Anyone help me go through the puzzles? thanks so much!

ismaili

----

Oh, by the way, I found that from the Dirac representation method which I described in another nearby thread titled "spinors in various dimensions", one can easily realize the dimension of gamma matrices in even dimension d should be $$2^{d/2}$$.
I reconsidered this recently.
I think for my first question, the reason that reduces the independent components $$2m^2$$ into $$m^2$$ is the (anti-)hermitian properties of the gamma matrices, not the Clifford algebra. Then, the antisymmetrized products of gamma matrices form a basis for the algebra, hence, by matching independent components $$m^2$$ and the number of basis $$2^d$$, we see that $$m = 2^{d/2}$$ for even dimension $$d$$.

The solution to the second equation is due to the role of $$\gamma_5$$.
That's why only in odd dimension, those basis are related by Levi-Civita tensor,
and the number of basis is reduced to $$2^{(d-1)/2}$$ for odd $$d$$.

#### ismaili

I reconsidered this recently.
I think for my first question, the reason that reduces the independent components $$2m^2$$ into $$m^2$$ is the (anti-)hermitian properties of the gamma matrices, not the Clifford algebra. Then, the antisymmetrized products of gamma matrices form a basis for the algebra, hence, by matching independent components $$m^2$$ and the number of basis $$2^d$$, we see that $$m = 2^{d/2}$$ for even dimension $$d$$.

The solution to the second equation is due to the role of $$\gamma_5$$.
That's why only in odd dimension, those basis are related by Levi-Civita tensor,
and the number of basis is reduced to $$2^{(d-1)/2}$$ for odd $$d$$.
I found this reference which deals with Clifford algebra in a mathematical rigorous way.
http://arxiv.org/abs/hep-th/9811101
But at least those $$B,C$$ matrices are not suddenly popped.

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