Let $ \mathcal C $ be the set of multivectors which commute with every vector in $ \mathbb G^n $. Show that:
1) When n is even, then $ \mathcal C $ is the set of all scalars $ a $.
2) When n is odd, then $ \mathcal C $ is the set of all scalars plus pseudoscalars $ a + bI $.
Not able to make much headway at all, so any help would be appreciated.
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user997712
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Are you specifically looking for objects which commute with only vectors, or with arbitrary multivectors? – Muphrid Jul 01 '13 at 20:25
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with only vectors – user997712 Jul 01 '13 at 20:38
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Hm, proving that the peudoscalar commutes in odd dimensions and not in even dimensions is easy (see your previous question on $k$-vectors commuting with the pseudoscalar). Proving that there are no other objects that commute with general vectors is what's got me stumped. – Muphrid Jul 01 '13 at 21:05
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I came up with this proof which is inelegant but hopefully correct:
I consider a general k-vector $ e_{i1}e_{i2}...e_{ik} $. If i can prove that there is a vector in $ G^n $ that it does not commute with, we can rule out all multi-vectors containing k-grade elements from $ C $.
Case 1: n is even
Consider a vector $ e_{i1} $. since n is even, it anti-commutes with the k-grade vector.
$$ e_{i1}e_{i1}e_{i2}...e_{ik} = e_{i2}...e_{ik} = -e_{i1}e_{i2}...e_{ik}e_{i1} $$ because the number of anticommuting "swaps" required is odd. Therefore there are no k-vectors which commute with arbitrary vectors if $ k > 0 $. Which leaves only the scalars in $ C $.
Case 2: n is odd Consider the case where $ 0 < k < n $. For any k-vector $ e_{i1}e_{i2}...e_{ik} $ we can find a vector $ e_{i(k+1)} $ which is not already included in the k-vector. This vector anticommutes with the k-vector being considered: $$ e_{i1}e_{i2}...e_{ik}e_{i(k+1)} = - e_{i(k+1)}e_{i1}e_{i2}...e_{ik} $$ again because the number of "swaps" is odd. However in the cases of $ k = 0 $ and $ k = n $ the above doesn't hold and all vectors commute. This means that $ C $ consists of all $ a + bI $.
user997712
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