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To phrase the question in a concrete way, I read in a paper:

The super Poincare subalgebra of osp(6,2|4) has bosonic part $so(5,1) \oplus usp(4) \simeq so(5,1) \oplus so(5)$.

It's hard to unpack this sentence without knowing the objects:

  • Can the orthosymplectic group osp(6,2|4) be defined as a group of matrices acting on a super-vector space?
  • Can someone explain this decomposition into its bosonic and fermionic parts?
  • What is the (super) Poincare sub-algebra? Why is usp(4) the same is so(5)?

For a math-physics dictionary: "super" means "Z2-graded" while bosonic means "0-grade" and fermionic means "1-grade".

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1 Answer 1

The way I would understand it that $osp(6,2|4)$ is the group of linear tansformations of a real super vector space with a non-degenerate symmetric inner product. The even (bosonic) vector space has dimension 8 and the inner product is symmetric with signature $(6,2)$ the odd (fermionic) part has dimension 4 and a symplectic form.

The even part is then the product of the groups of these two vector spaces, namely $o(6,2)$ and $sp(4)$. There is an isomophism of rank two Lie algebras $sp(4)\cong so(5)$; to see this note that the spin representation has dimension 4 and has an invariant symplectic form.

I realise you have $so(5,1)$ where I have $so(6,2)$. I don't know what is going on here but $so(6,2)$ is the group of conformal transformations of $R^{5,1}$.

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I imagine that your failure to use the phrase "Poincare subalgebra" cancels your other discrepancy (the last paragraph), but I don't know how. The usual Poincare algebra is the semidirect product of the $so(3,1)$ with $R^{3,1}$. (A super Poincare algebra has that as the bosonic part.) They're probably embedding some generalization of it in a larger group. It seems weird that they could shove the normal subgroup into the fermionic part. Maybe they left out "semisimplification"? –  Ben Wieland May 27 '10 at 22:58

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