supergroup in nLab

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$\array{ && Grp \\ & {}^{(G,\cdot)}\nearrow & \downarrow \\ SDiff^{op} &\stackrel{Y(G)}{\to}& Set } \,.$

This gives for each supermanifold $S$ an ordinary group $(G(S), \cdot)$ , so in particular a product operation

$\cdot_S : G(S) \times G(S) \to G(S) \,.$

Moreover, since morphisms in $Grp$ are group homomorphisms, it follows that for every morphism $f : S \to T$ of supermanifolds we get a commuting diagram

$\array{ G(S) \times G(S) &\stackrel{\cdot_S}{\to}& G(S) \\ \uparrow^{G(f)\times G(f)} && \uparrow^{G(f)} \\ G(T) \times G(T) &\stackrel{\cdot_T}{\to}& G(T) }$

of representable presheaves. By the Yoneda lemma, this uniquely comes from a morphism $\cdot : G \times G \to G$ , which is the product of the group structure on the object $G$ that we are after.

etc.

This way of thinking about supergroups is often explicit in some parts of the literature on supergeometry: some authors define a supergroup or super Lie algebra as a rule that assigns to every Grassmann algebra $A$ over an ordinary vector space an ordinary group $G(A)$ or Lie algebra and to a morphism of Grassmann algebras $A \to B$ covariantly a morphism of groups $G(A) \to G(B)$ . But the Grassmann algebra on an $n$ -dimensional vector space is naturally isomorphic to the function ring on the supermanifold $\mathbb{R}^{0|n }$ . So the definition of supergroups in terms of Grassmann algebras is secretly the same as the above definition in terms of the Yoneda embedding.

The additive group structure on $\mathbb{R}^{1|1}$ is given on generalized elements in (i.e. in the logic internal to) the topos of sheaves on the category SCartSp of cartesian superspaces by

$(t_1, \theta_1), (t_2, \theta_2) \mapsto (t_1 + t_2 + \theta_1 \theta_2, \theta_1 + \theta_2) \,.$

and we also have the theorem, discussed at supermanifolds, that maps from some $S \in SDiff$ into $\mathbb{R}^{p|q}$ is given by a tuple of $p$ even section $t_i$ and $q$ odd sections $\theta_j$ . The above notation specifies the map of supermanifolds by displaying what map of sets of maps from some test object $S$ it corresponds to under the Yoneda embedding.

Now, for each $S \in$ SDiff there is a group structure on the hom-set $SDiff(S, \mathbb{R}^{1|1}) \simeq C^\infty(S)^{ev} \times C^\infty(S)^{odd}$ given by precisely the above formula for this given $S$

$\mathbb{R}^{1|1}(S) \times \mathbb{R}^{1|1}(S) \to \mathbb{R}^{1|1}(S)$

$(t_1, \theta_1), (t_2, \theta_2) \mapsto (t_1 + t_2 + \theta_1 \theta_2, \theta_1 + \theta_2) \,.$

where $(t_i, \theta_i) \in C^\infty(S)^{ev} \times C^\infty(S)^{odd}$ etc and where the addition and product on the right takes place in the function super algebra $C^\infty(S)$ .

Since the formula looks the same for all $S$ , one often just writes it without mentioning $S$ as above.

The super-translation group is the $(1|1)$ -dimensional case of the super Euclidean group.

Super-Minkowski Lie group

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:

(1) $\mathbb{R}^{1,d\vert\mathbf{N}} \;\simeq\; Iso\big( \mathbb{R}^{1,d\vert\mathbf{N}} \big) \big/ Spin(1,d) \,.$

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

$\mathbb{R}^{1,d\vert\mathbf{N}} \;\coloneqq\; \mathbb{R}^{1+d} \times \mathbb{R}^N_{odd}$

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

(2) $\begin{array}{ccl} \mathrm{d}\, e^a &\coloneqq& \big( \overline{\psi} \,\Gamma^a\, \psi \big) \\ \mathrm{d}\, \psi &=& 0 \end{array}$

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

$(e,\psi) \;\colon\; T\mathbb{R}^{1,d\vert\mathbf{N}} \xrightarrow{\;} \mathbb{R}^{1,10\vert\mathbf{N}}$

(where on the left we have the tangent bundle and on the right its typical fiber super vector space) given by

(3) $\begin{array}{ccl} e^a &\coloneqq& \mathrm{d}x^a + \big(\overline{\theta} \,\Gamma^a \mathrm{d}\theta\big) \\ \psi &\coloneqq& \mathrm{d}\theta \end{array}$

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

$T_{x}\mathbb{R}^{1,d\vert\mathbf{N}} \xrightarrow{\;\sim\;} \mathbb{R}^{1,d\vert\mathbf{N}}$

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

$\begin{array}{ccl} D_a &\coloneqq& \partial_{x^a} \\ D_\alpha &\coloneqq& \partial_{\theta^\alpha} + \overline{\theta}\Gamma^a \partial_{v^a} \end{array} \;\;\;\;\;\;\; \text{such that} \;\;\;\; \begin{array}{ll} e^a(D_b) \,=\, \delta^a_b \,, & e^a(D_\alpha) \,=\, 0 \\ \psi^\alpha(D_a) \,=\, 0 \,, & \psi^\alpha(D_\beta) \,=\, \delta^\alpha_\beta \end{array}$

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:

(4)

(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):

$\begin{array}{ccl} \mathrm{act}^\ast e^a &=& \mathrm{act}^\ast \Big( \mathrm{d}x^a + \big(\overline{\theta} \,\Gamma^a\, \mathrm{d}\theta\big) \Big) \\ &=& \mathrm{d}\,\mathrm{act}^\ast x^a + \big(\overline{\mathrm{act}^\ast\theta} \,\Gamma^a\, \mathrm{d}\,\mathrm{act}^\ast\theta\big) \\ &=& \mathrm{d} \Big( x^a_{'} + x^a - \big( \overline{\theta}_{'} \,\Gamma^a\, \theta \big) \Big) + \big( (\overline{\theta}_{'} + \overline{\theta}) \,\Gamma^a\, \mathrm{d}( \theta_{'} + \theta ) \big) \\ &=& \mathrm{d}x^a - \big( \overline{\theta}_{'} \,\Gamma^a\, \mathrm{d}\theta \big) \,+\, \big( \overline{\theta}_{'} \,\Gamma^a\, \mathrm{d}\theta \big) \,+\, \big( \overline{\theta} \,\Gamma^a\, \mathrm{d}\theta \big) \\ &=& \mathrm{d}x^a \,+\, \big( \overline{\theta} \,\Gamma^a\, \mathrm{d}\theta \big) \\ &=& e^a \mathrlap{\,,} \end{array} \;\;\;\;\; \begin{array}{ccl} \mathrm{prd}^\ast \psi &=& \mathrm{prd}^\ast \mathrm{d}\theta \\ &=& \mathrm{d} \, \mathrm{prd}^\ast \theta \\ &=& \mathrm{d}\big( \theta_{'} + \theta \big) \\ &=& \mathrm{d}\theta \\ &=& \psi \mathrlap{\,.} \\ {} \end{array}$

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:

neutral element:

inverse elements:

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:

$\Box$

There is a finite analog for super-groups that does not quite fit in the framework presented here: