Poincaré algebra and supersymmetric spaces

Question

If i understand correctly, a supersymmetry algebra should contain as a subalgebra the Poincaré algebra, however for a supersymmetry algebra the corresponding supersymmetric (Minkowski) space has dimension $ 4 + 4\mathcal{N} $ does this mean we should change the $ (3,1) $ D Lorentz group to a $ (4+4\mathcal{N}-1, 1) = (3+4\mathcal{N}, 1) $ dimensional Lorentz group $ \operatorname{SO}^+(3+4\mathcal{N}, 1) $ ? Does this supersymmetric space represent the physical spacetime of our theory, replacing the old 3+1d one?

Answer 1

Introducing extra co-ordinates so that supersymmetry becomes a spacetime symmetry has nothing to do with what you're saying. $so(3 + 4\mathcal{N}, 1)$ is a Lie algebra which means its only bilinear operation obeys the properties of a commutator. It is not a Lie superalgebra which requires anti-commutators as well.

When constructing superspace, the $4\mathcal{N}$ extra co-ordinates that you introduce have to be Grassmann odd. For this reason, there is no such thing as a rotation in a plane spanned by one commuting axis and one anti-commuting axis. The isometries of a superspace like $\mathbb{R}^{3,1 | \mathcal{N}}$ require a new type of structure and this is what a superalgebra is. Analogously to how the non-simple 4d Poincare algebra can be embedded in $so(3, 2)$, the 4d super-Poincare algebra can be embedded in what is called $osp(\mathcal{N} | 4)$.

So it is not correct to treat the extra co-ordinates of superspace as being indistinguishable from space or indistinguishable from time. They are their own thing. Even after recognizing this, there are significant caveats when trying to elevate "higher dimensionality" as a principle of supersymmetric theories. The only superspace formulations which apply uniformly in $\mathcal{N}$ are on-shell. Even if you are only doing on-shell things, 2d theories admit non-linear SUSY algebras where the anti-commutator of two supercharges produces a square of a conserved bosonic charge.

Instead of letting the number of dimensions go from 4 to $4 + 4\mathcal{N}$, there are also claims that we should regard this as 4 to $4 + 2^{8\mathcal{N}}$ because $n$ mutual anti-commutations can be realized with $2^n \times 2^n$ matrices. But this is a naive observation as well because it only holds for one superspace point. With two superspace points, you will need way more than twice this number of co-ordinates as explained in these comments.