I am confused regarding supermanifolds. Suppose I consider R^(1,2) (1 "bosonic", 2 "fermionic"), This map (x,a,b) -> (x+ab, a,b) (a,b are fermionic) is supposed to be a morphism of this supermanifold. But I thought a morphism should be a continuous map from R->R together with a sheaf map of the sheaf of supercommutative algebra of smooth functions. How is this (x-> x+ab) giving me a continuous map from R->R?
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1$\begingroup$ I think the answer may just be that (x,a,b) -> (x+ab, a,b) isn't supposed to give you a map of supermanifolds. I'm not an expert in this stuff (which is why I didn't write an answer), but I don't think you should think of ab as a real number. $\endgroup$– Ben Webster ♦Commented Mar 11, 2010 at 4:21
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1$\begingroup$ A good reference is the big "Quantum Fields and Strings" book. Volume 1 contains an article "Notes on Supersymmetry" by Deligne and Morgan which discusses in particular the reduction stuff that Scott and Chris mention below. $\endgroup$– Kevin H. LinCommented Mar 11, 2010 at 5:55
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3 Answers
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The ring of functions on your supermanifold is $C^\infty(\mathbb{R}) \otimes \mathbb{C}[a,b]$, where $a$ and $b$ are odd. The even part is then $C^\infty(\mathbb{R}) \oplus C^\infty(\mathbb{R})ab$, where $(ab)^2 = 0$, so there is an even nilpotent direction. You might want to view it as a thickening in a perpendicular direction. The map in question is the identity on the reduced quotient $C^\infty(\mathbb{R})$, and this yields the identity map of manifolds (which is continuous). The $(x \mapsto x+ab)$ can be viewed as an infinitesimal shearing on the even part.
answered Mar 11, 2010 at 4:40
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2$\begingroup$ Careful! It is dangerous to think of $C^\infty(\mathbb{R})$ as sitting inside the functions on $\mathbb{R}^{1|2}$ because such an inclusion is not natural/functorial/canonical. If you act by an automorphism of $\mathbb{R}^{1|2}$, this inclusion of algebras changes. You are not really "restricting" to $C^\infty(\mathbb{R}$. Rather it is better to quotient by nilpotents, which is a functorial construction. You then obtain the map on $C^\infty(\mathbb{R})$ as the induced map. $\endgroup$ Commented Mar 11, 2010 at 5:25
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$\begingroup$ Oh, yeah. Reduction is a quotient. Editing... $\endgroup$– S. Carnahan ♦Commented Mar 11, 2010 at 5:43
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Unlike many schemes, but similar to ordinary manifolds, a map of super-manifolds $$(X, \mathcal{O}_X) \to (Y, \mathcal{O}_Y)$$ determines and is completely determined by the map of superalgebras obtained by looking at global sections: $$\mathcal{O}_Y(Y) \to \mathcal{O}_X(X)$$ In the example at hand this is the graded ring map: $$ x \mapsto x' + a'b'$$ $$ a \mapsto a'$$ $$ b \mapsto b'$$
This map induces a map of rings after we mod out by nilpotents: $$C^\infty(Y) = \mathcal{O}_Y/Nil \to \mathcal{O}_X / Nil = C^\infty(X)$$ This map in turn induces a smooth map $X \to Y$ (in fact it is equivalent to such a map). In this case, after modding out by nilpotents we get the map $x \mapsto x'$, i.e. the identity on the underlying manifold $\mathbb{R}$.
answered Mar 11, 2010 at 4:44
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1$\begingroup$ Ordinary smooth manifolds, right? I am pretty sure that this is not true for analytic or complex manifolds because this approach of determining the manifold by its global algebra of smooth functions is only valid by using partitions of unity. $\endgroup$ Commented Mar 11, 2010 at 4:51
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1$\begingroup$ For supermanifolds of a more exotic flavor the same essential argument works, but you need to work with maps of sheaves rather then algebras. You can still mod out by nilpotents and obtain a map of underlying ordinary manifolds. I'm pretty sure the question was about the generic smooth setting, where the above answer applies directly. $\endgroup$ Commented Mar 11, 2010 at 5:20
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A morphism of supermanifolds is a continuous map and a map of sheaves of superfunctions in the opposite direction. What you've given is the second part of the datum. In your example, I guess the continuous map $\mathbb R\to\mathbb R$ between the underlaying manifolds is just the identity.
answered Mar 11, 2010 at 3:36