We spell out the super-translation super Lie group-structure on the supermanifold $\mathbb{R}^{1,d\vert\mathbf{N}}$ underlying super Minkowski spacetime, hence equivalently of the quotient super Lie group of the super Poincaré group (the “supersymmetry” group) by its Lorentzian spin-subgroup:
Here
$d \in \mathbb{N}$ is the spatial dimension (a natural number),
$\mathbf{N} \in Rep_{\mathbb{R}}\big(Spin(1,d)\big)$ is a real spin representation equipped with a linear map
which is symmetric and $Spin(1,d)$-equivariant.
First, the super-Minkowski super Lie algebra structure on the super vector space
is defined, dually, by the Chevalley-Eilenberg dgc-superalgebra with generators of $\mathbb{Z} \times \mathbb{Z}/2$ bidegree
generator | bidegree |
---|---|
$e^a$ | $(1,evn)$ |
$\psi^\alpha$ | $(1,odd)$ |
for $a \in \{0,1, \cdots, d\}$ indexing a linear basis of $\mathbb{R}^D$ and $\alpha \in \{1,\cdots, N\}$ indexing a linear basis of $\mathbf{N}$ by the differential equations
The first differential is the linear dual of the archetypical super Lie bracket in the supersymmetry super Lie algebra which takes two odd elements to a spatial translation. The second differential is the linear dual of the fact that in the absence of rotational generators, no Lie bracket in the supersymmetry alegbra results in a non-vanishing odd element.
Next we regard $\mathbb{R}^{1,10\vert\mathbf{N}}$ not just as a super vector space but as a Cartesian supermanifold. As such it has canonical coordinate functions
generator | bidegree |
---|---|
$x^a$ | $(0,evn)$ |
$\theta^\alpha$ | $(0,odd)$ |
On this supermanifold, consider the super coframe field
(where on the left we have the tangent bundle and on the right its typical fiber super vector space) given by
It is clear that this is a coframe field in that for all $x \in \mathbb{R}^{1,d\vert\mathbf{N}}$ it restricts to an isomorphism
and the peculiar second summand in the first line is chosen such that its de Rham differential has the same form as the differential in the Chevalley-Eilenberg algebra (2).
(Incidentally, a frame field linear dual to the coframe field (3) is
which are the operators often stated right away in introductory texts on supersymmetry.)
This fact, that the Maurer-Cartan equations of a coframe field (3) coincide with the defining equations (2) of the Chevalley-Eilenberg algebra of a Lie algebra of course characterizes the left invariant 1-forms on a Lie group, and hence what remains to be done now is to construct a super Lie group-structure on the supermanifold $\mathbb{R}^{1,d\vert\mathbf{N}}$ with respect to which the coframe (3) is left invariant 1-form.
Recalling (from here) that a morphism of supermanifolds is dually given by a reverse algebra homomorphism between their function algebras, which in the present case are freely generated by the above coordinate functions, we denote the canonical coordinates on the Cartesian product $\mathbb{R}^{1,d\vert\mathbf{N}} \times \mathbb{R}^{1,d\vert\mathbf{N}}$ by $(x^a_{'}, \theta^\alpha_{'})$ for the first factor and $(x^a, \theta^\alpha)$ for the second, and declare a group product operation as follows:
(cf. CAIP99, (2.1) & (2.6))
Here the choice of notation for the coordinates on the left is adapted to thinking of this group operation equivalently as the left multiplication action of the group on itself, which makes the following computation nicely transparent.
Indeed, the induced left action of the super-group on its odd tangent bundle
is dually given by
and left-invariance of the coframe (2) means that it is fixed by this operation (so the differential $\mathrm{d}$ in the following computation is just that of the second factor, hence acting on unprimed coordinates only):
This shows that if (4) is the group product of a group object in SuperManifolds then the corresponding super Lie algebra is the super-Minkowski super translation Lie algebra and hence that this group object is the desired super-Minkowski super Lie group.
So, defining the remaining group object-operations as follows:
we conclude by checking the group object-axioms:
For associativity we need to check that the following diagram commutes:
and indeed it does — the term $\big(\overline{\theta} \Gamma^a \theta\big)$ vanishes because the $\theta^\alpha$ anti-commute among themselves, while the pairing (1) is symmetric:
For unitality we need to check that the following diagram commutes:
and indeed it does:
And finally, for invertibility we need to check that the following diagram commutes:
and indeed it does:
$\Box$
Last revised on September 3, 2024 at 15:00:26. See the history of this page for a list of all contributions to it.