The dimension of the Clifford algebra $C_p$ generated by a vector space $V^p$ is given by $2^p$, where $p$ is the dimension of the vector space (T. Frankel, the geometry of physics). Based on the top-rated answer to this post Dimension of Dirac $\gamma$ matrices, it seems that the vector space that generates this Clifford algebra has dimension $p=[d/2]$, where $d$ is the general spacetime dimension and $[\cdot]$ takes the integer part. Also it is shown that the irreducible representation of the $C_p$ has dimension $2^p=2^{[d/2]}$.

First of all, it seems that the dimension of the Clifford algebra $C_p$ in Frankel's book is the same with the dimension of its irreducible representation. This makes sense to me, but I would like to confirm whether this is true or not.

Secondly, as mentioned above, there is a mysterious relation between spacetime dimension $d$ and the dimension of the vector space $p$ given by $p=[d/2]$. The reason that I ask about this is we need proper $\gamma$ matrices to write down a theory for spinor fields and the aforementioned relation is crucial for doing this when the theory lives in spacetime with generic dimension $d$. I would like know why the spacetime dimension is related to the dimension of the vector space used to generate the Clifford algebra in this way.

Thirdly, I will take $d=1+1$ for this part, for simplicity. In this simple case, the irreducible rep has dimension $2^1=2$, i.e. we can just use Pauli matrices here. However, nothing can stop us from using the $4\times 4$, or any integer multiple of 2, reducible rep, i.e. we will have 4-component spinors. Then my questions is based on the underlying physics, how to decide which representation to use? People may think that different rep will give the same physics, but here is a paper https://journals.aps.org/prd/abstract/10.1103/PhysRevD.33.3704 that actually treated them differently.


1 Answer 1


Wikipedia reviews and proves the fundamental fact of the irreps of the Clifford in d dimensions, mysterious or not (!?). It actually constructs them in a given chiral basis. They are, as you indicate, $ 2^{[d/2]}\times 2^{[d/2]}$ independent matrices. One should not waste time on Frankel's book.

Now for your problem. In d=1+1 the chiral Γ is non trivial, but the reference you are citing is working in d=1+2, a huge difference, and in odd dimensions the chiral matrix is trivial in the irrep, so proportional to the identity. In your case, the product of the 3 Pauli matrices (your gamma matrices) of course collapses to the identity.

  • "People may think ..." is spectacularly misguided: different reps correspond to different wealth of degrees of freedom, hence very different physics. This is what model-building is all about: picking and choosing a suitable rep exemplifying the features of interest to one.

In your cited ref, the authors note the $2\times 2 $ irrep chiral matrix is trivial, and cannot tell them much about chiral symmetry breaking! So they move up to the reducible ${\mathbf 2}\oplus{\mathbf 2}$ spinors acted upon by $4\times 4$ gammas, where they contrive two chiral matrices, (2.5), connecting the sub representations to yield a rich chiral structure they detail.

Manifestly different physics, so what is the problem?

  • $\begingroup$ Thank you for answer. From the wiki page, it doesn't seem to me to be explicit about why the dimension of the matrices should be $2^{[d/2]}$, which is my main confusion here. Btw, which book would you recommend if you think Frankel is not good? Thanks again. $\endgroup$
    – M. Zeng
    Commented May 30, 2018 at 17:27
  • 1
    $\begingroup$ The book by de Wit & Smith, appendix E does the counting and the proofs quite well. $\endgroup$ Commented May 30, 2018 at 19:02
  • 1
    $\begingroup$ The book "Geometry, Topology and Physics" by M Nakahara is also superior to Frankel's. You probably missed the counting arguments on the WP article linked. $\endgroup$ Commented May 30, 2018 at 19:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.