most recent 30 from http://mathoverflow.net 2013-05-21T14:46:13Z http://mathoverflow.net/feeds/question/17811 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/17811/morphisms-of-supermanifolds Vamsi 2010-03-11T03:19:41Z 2010-03-11T05:44:27Z

<p>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?</p>

http://mathoverflow.net/questions/17811/morphisms-of-supermanifolds/17812#17812 Mariano Suárez-Alvarez 2010-03-11T03:36:31Z 2010-03-11T04:14:24Z

<p>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.</p>

http://mathoverflow.net/questions/17811/morphisms-of-supermanifolds/17823#17823 S. Carnahan 2010-03-11T04:40:46Z 2010-03-11T05:44:27Z

<p>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.</p>

http://mathoverflow.net/questions/17811/morphisms-of-supermanifolds/17824#17824 Chris Schommer-Pries 2010-03-11T04:44:34Z 2010-03-11T04:44:34Z

<p>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: $$ x \mapsto x' + a'b'$$ $$ a \mapsto a'$$ $$ b \mapsto b'$$</p> <p>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 $\mathbb{R}$.</p>