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I'm working with Clifford algebras, of which the first few are $C_0 = \mathbb{R}$, $C_1 = \mathbb{C}$, $C_2 = \mathbb{H}$, $C_3= \mathbb{H}^2$, $C_4 = M_{2,2}(\mathbb{H})$, $C_5= M_{4,4}(\mathbb{C})$, $C_6=M_{8,8}(\mathbb{R})$, $C_7 = M_{8,8}(\mathbb{R})^2$, $C_8 = M_{16,16}(\mathbb{R})$, etc. (The operations on the product of two algebras are componentwise.)

The notes I'm working with claim without proof that the minimum dimensions of representations are 1, 2, 4, 4, 8, 8, 8, 8, 16. (Sorry I probably could've formatted this better.) I assume that they mean faithful representations, and that they're representations as $\mathbb{R}$-algebras (and not just as rings), i.e. they're $\mathbb{R}$-linear maps $C_k \rightarrow End(V)$ (or equivalently maps $C_k \otimes_\mathbb{R} V \rightarrow V$, I think). I have no idea why these numbers are right, though. The first few I can see by ad hoc arguments, but that's not really that satisfying. I know no representation theory beyond a few definitions, so this might be really obvious and I just don't know the right theorems/arguments?

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  • $\begingroup$ Based on the list, I'd say the representations are not required to be faithful, but they are required to be unital. $\endgroup$
    – S. Carnahan
    Sep 21, 2010 at 3:25
  • $\begingroup$ Hmm. Yeah, you're obviously right. Faithful means that the map $C_k \rightarrow End(V)$ is a linear injection. I guess I meant nontrivial. $\endgroup$ Sep 21, 2010 at 4:46

1 Answer 1

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Yes. These can be proven by some very simple principles.

  • The smallest representation of a division ring is the ring itself (this covers the first 3), since it has no left or right ideals.
  • When you take $n\times n$ matrices in a division ring $D$, the smallest irrep is the obvious one $D^n$. This is just doing a computation of all the left ideals in matrices.
  • And, of course, if you take the sum of two rings, the dimension of the smallest representation is just minimum of the two summands.
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  • $\begingroup$ This is exactly what I was looking for. Thanks a bunch. $\endgroup$ Sep 21, 2010 at 4:46

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