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I am confused with morphisms of supermanifolds. Take a simple example $f:R^{0|1}\to R^{0|1}$. By (one of) definition, $f$ is a morphism of superalgebras of functions $C(R^{0|1})\to C(R^{0|1})$. Morphisms of superalgebras preserve the grading, I deduced that $f$ have the form $1\mapsto 1, \theta\mapsto x\theta$, i.e. $Hom(R^{0|1}\to R^{0|1})=R^1$ (as a set?). But I read from a paper that $Hom(R^{0|1},R^{0|1})=R^{1|1}$. What is going on? Thanks in advance!

flag	asked Feb 9 2011 at 20:54 Ma Ming 279●5

Martin O · Accepted Answer · 2011-02-10 08:25:43Z UTC

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You are right that the set of supermanifold morphisms $Hom(\mathbb R^{0|1},\mathbb R^{0|1})$ to itself is $\mathbb R^1$. However, one can define for supermanifolds $X,Y$ with $\dim X=0|d$ a supermanifold $map(X,Y)$ of morphisms from $X$ to $Y$, by $Hom(Z,map(X,Y))=Hom(Z\times X,Y)$ for all supermanifolds $Z$.

And $map(\mathbb R^{0|1},\mathbb R^{0|1})=\mathbb R^{1|1}$.

edited Feb 10 2011 at 8:25

answered Feb 9 2011 at 23:47

Martin O
341●2●4

Very interesting! I reread the paper, it says actrually inter-hom, maybe it your $map$. Should it be $X=R^{0|d}$? – Ma Ming Feb 10 2011 at 1:39

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@Ma: Yes, the inner hom is the same thing as map. Which paper are you reading, by the way? – Dmitri Pavlov Feb 10 2011 at 2:26

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As @Ma Ming said, it suffices that $X$ be $0|\delta$-dimensional for $\operatorname{Maps}(X,Y)$ to be finite-dimensional. However, it's not good enough for $Y$ to be $0|\delta$-dimensional. For example, $\operatorname{Hom}(\mathbb R^{1|1},\mathbb R^{0|1}) = \operatorname{Hom}(\mathbb R[\epsilon], \mathcal C^\infty(\mathbb R)[\epsilon])$, where $\epsilon^2 = 0$ and $\epsilon$ is in odd degree; but such a map is any $\epsilon \mapsto f(x)\epsilon$, so this hom space is infinite-dimensional $\mathcal C^\infty(\mathbb R)$. – Theo Johnson-Freyd Feb 10 2011 at 2:55

Thanks for spotting the typo, which I corrected. – Martin O Feb 10 2011 at 8:30

Thank you all! Could you give a clue how to calculate the $map$, for example, $map(R^{0|d},M)$ for any supermanifold $M$? – Ma Ming Feb 10 2011 at 11:49

Alexander Alldridge · Answer 2 · 2011-08-18 13:46:35Z UTC

@Ma: As an answer to your following question:

Could you give a clue how to calculate the $map$, for example, $map(R^{0∣d},M)$ for any supermanifold $M$?

Take a look at arXiv:math/0307303, where this question is discussed.

For $d=1$, it is well-known (and due to Kontsevich, I think), that $map(R^{0|1},M)$ is the total space of the odd tangent bundle $\Pi TM$ of $M$.

If $U$ is a superdomain of dimension $p|q$, then $map(R^{0|d},U)=U\times R^{pr+qs|ps+qr}$ where $(r+1)|s=2^{d-1}|2^{d-1}$ is the graded dimension of $\bigwedge R^d$. This you can check using the definition of $map$ and the characterisation of morphisms of supermanifolds as given in Leites.

Another good source on this subject (for $d=1$), is the paper "Differential forms and 0-dimensional supersymmetric field theories" by Hohnhold, Kreck, Stolz and Teichner.

Morphisms between supermanifolds R^{0|1}→R^{0|1}

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