# Hermitian conjugation and representations of the Lorentzian Clifford algebras

The Clifford algebra $\mathcal{C}\ell _{1,2d-1}$ is central and simple (L), and hence has a unique faithful, irreducible representation (over $\mathbb{R}$) (A). Denote this representation by $\gamma :\mathcal{C}\ell _{1,2d-1}\rightarrow \mathfrak{gl}(V)$, for some complex vector space $V$, let $e_0,\ldots ,e_{2d-1}$ be an orthonormal basis for $\mathcal{C}\ell _{1,2d-1}$, and define $\gamma ^\mu :=\gamma (e_\mu )$.

I have been told that we may always choose the $\gamma$ matrices so that $(\gamma ^\mu )^*=\gamma _\mu=\begin{cases}(-\gamma ^0,\vec{\gamma}) & \text{for signature }(-,+,\cdots ,+) \\ (\gamma ^0,-\vec{\gamma}) & \text{for signature }(+,-,\cdots ,-)\end{cases},$ where $^*$ denotes the Hermitian conjugate.

($\vec{\gamma}=(\gamma ^1,\ldots ,\gamma ^{2d-1})$ is the spatial part of $\gamma ^\mu$.)

How does one show that there is always a representation in the equivalence class of the given $\gamma$ that satisfies this property? (To be clear, I was only told this to be true in the case relevant for physics, i.e. $d=2$, but presumably it is true in general).

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I'm having a hard time parsing the notation. Since $\gamma$ is a representation of the Clifford algebra, it looks like $(\gamma^\mu)^\ast$ is a Hermitian conjugate (maybe). The braces with two ordered pairs kind of looks like a matrix, but I have no idea how to interpret an ordered pair consisting of a matrix and the reresentation $\gamma$. Can you help us read this a little? –  rschwieb Feb 4 '13 at 1:26
@rschwieb I edited the post in an attempt to make my notation more lucid. Please let me know if something is still unclear. –  Jonathan Gleason Feb 4 '13 at 1:49
OK, that helps a bit, although I still have a question. I can see now that the ordered pair of gammas consists of the coordinate that's of different signature from the rest, which you are lumping together into a single vector. Does that sound right? –  rschwieb Feb 4 '13 at 14:02
@rschwieb If I understand you correctly, then yes, that is right. I edited the post again to further clarify. Let me know if something is still unclear. –  Jonathan Gleason Feb 4 '13 at 14:34
Thanks! I think it's clear enough to think about now :) –  rschwieb Feb 4 '13 at 14:40