# Clifford algebra in higher dimensions

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1. Jul 26, 2017

### Ken Gallock

1. The problem statement, all variables and given/known data
Consider gamma matrices $\gamma^0, \gamma^1, \gamma^2, \gamma^3$ in 4-dimension. These gamma matrices satisfy the anti-commutation relation
$$\{ \gamma^\mu , \gamma^\nu \}=2\eta^{\mu \nu}.~~~(\eta^{\mu\nu}=diag(+1,-1,-1,-1))$$
Define $\Gamma^{0\pm}, \Gamma^{1\pm}$ as follows:
\begin{align}
\Gamma^{0\pm}:=\dfrac12 (\gamma^0 \pm \gamma^1),\\
\Gamma^{1\pm}:=\dfrac12 (i\gamma^2 \pm \gamma^3).
\end{align}

They satisfy following anti-commutation relations.
$$\{ \Gamma^{a+}, \Gamma^{b-} \}=\delta^{ab}, \{ \Gamma^{a+}, \Gamma^{b+} \} = \{ \Gamma^{a-}, \Gamma^{b-} \}=0. ~~(a,b=0,1)$$

Now, we can construct other Clifford algebra in higher dimensions.

Question: Construct Clifford algebra for 1+10 dimension and 2+10 dimension as we did in above.

2. Relevant equations

3. The attempt at a solution
The reason this problem is difficult for me is that 1+10 dimension is 11-dimension space-time and they are odd number dimension. If it were even number dimension, then it is easy to construct Clifford algebra; we only have to add new terms such like
$$\Gamma^{2\pm}=\dfrac12 (i\gamma^3 \pm \gamma^4).$$
But in 11-dimension, we can't do this. I tried to think in 3-dimension, but I couldn't figure out how to construct Clifford algebra.

Thanks.

2. Jul 26, 2017

### Staff: Mentor

I don't really understand the problem. Firstly, the pairing which leads to the $\Gamma^a$ is simply a basis transformation of the $\gamma^a$, so we don't have do bother them. Now $\eta$ is all that counts for the definition, since this is the bilinear form. So why isn't $\eta = \operatorname{diag}(+1,-1,\ldots ,-1)$ the solution with as many $-1$ as desired?

The grading can be done for all Clifford algebras. All you need is an involution. The difference between the given $(1,3)$ and the requested $(2,10)$ example is bigger than between $(1,3)$ and $(1,10)$.

3. Jul 26, 2017

### Ken Gallock

I think I should have noted the motivation of this problem. We want to think about a generalization of 4D Clifford algebra. For example, in $1+3$ dimension, we can set $\Gamma$s as
\begin{align}
\Gamma^{0\pm}:=\dfrac12 (\gamma^0 \pm \gamma^1),\\
\Gamma^{1\pm}:=\dfrac12 (i\gamma^2 \pm \gamma^3).
\end{align}

The anti-commutation relations are
$$\{ \Gamma^{a+}, \Gamma^{b-} \}=\delta^{ab}, \{ \Gamma^{a+}, \Gamma^{b+} \} = \{ \Gamma^{a-}, \Gamma^{b-} \}=0. ~~(a,b=0,1)$$

These $\Gamma^{0+}$ and $\Gamma^{1+}$ are kind of raising operators, and $\Gamma^{0-}$ and $\Gamma^{1-}$ are lowering operators. Since the anti-commutation relations are similar to that of spinor fields', we could define a vacuum state $\Omega$ and set
$$\Gamma^{a-}\Omega = 0.$$

Then, states like $\Gamma^{a+}\Omega$ can be thought as a one-particle state.

If we construct Clifford algebra by using these $\Gamma$s, we can think of Clifford algebra in various dimensions such like 1+10 or 2+10.
This is used in string theory.

4. Jul 26, 2017

### Staff: Mentor

I know. But it doesn't change my answer. You can extend $v \longmapsto -v$ to an involution $\varphi$ of $\mathcal{Cl}(V,\eta)$ and the grading is then $\mathcal{Cl}^{(i)}(V,\eta)= \operatorname{im}(\operatorname{id} \pm \varphi)$. With that you can think about an appropriate basis.

5. Jul 29, 2017

### Ken Gallock

Thanks.
I will do my best.