Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This is my second question on supermanifolds, the previous one is at

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

I've learn the difference between homomorphism and internal-hom of supermanifolds. Also, I know that the homomorphisms are generally easy to calculate, it is defined to be morphisms of superalgebras of functions of supermanifolds, or a bit correctly morphisms of ringed spaces; while the internal-hom is defined indirectly by adjoint functor. I thus want to know how to calculate the internal-hom of supermanifolds, in particularly $Map(\mathbb{R}^{0|n},M)$ for any supermanifold $M$. Thanks for your help!

with regards,

maming

share|improve this question

1 Answer 1

up vote 2 down vote accepted

(Sorry for not answering this on the previous post, you asked this question before. By the way, you didn't say which paper you are reading...)

$Map(\mathbb R^{0|0},M)=M$. Trivial.

$Map(\mathbb R^{0|1},M)=\Pi T M$. ($TM$ is the tangent bundle of $M$, and $\Pi$ reverses grading of a vector bundle. So $\Pi T M$ denotes the total space of the degree-reversed tangent bundle of $M$. If $\dim M=d|\delta$, then $\dim \Pi T M=d+\delta|d+\delta$.)

Since

$Hom(S, Map(\mathbb R^{0|n},M)) = Hom(S\times \mathbb R^{0|n},M)$ $= Hom(S\times \mathbb R^{0|n-1}\times \mathbb R^{0|1},M) = Hom(S\times \mathbb R^{0|n-1},\Pi T M)$,

we obtain inductively $Map(\mathbb R^{0|n},M)=(\Pi T)^n M$. However, the interesting thing about $Map(\mathbb R^{0|n},M)$ is that it has an action by $Diff(\mathbb R^{0|n})$, the supermanifold of invertible maps from $\mathbb R^{0|n}$ to itself. I guess this is not so visible if you write $(\Pi T)^n M$, since this was obtained by destroying the symmetry in the odd coordinates.

For a description which keeps the symmetry: $Map(\mathbb R^{0|2},M)$ is the pullback of $\Pi( T M\oplus T M) \to M$ along $TM\to M$, which I learned from Dan Berwick Evans at Berkeley. I would guess this is as explicit as it gets in general, and that probably more difficult pullbacks squares involving $\Pi T M$ exist for $Map(\mathbb R^{0|n},M)$ with bigger $n$.

One can find a description and discussion of $Map(\mathbb R^{0|n},M)$ in the paper with the nice title "Differential gorms, differential worms", Denis Kochan, Pavol Severa arXiv:math/0307303.

share|improve this answer
    
mm, that's what I am reading. –  Ma Ming Feb 18 '11 at 19:18
    
Updated my answer! –  Martin O Feb 21 '11 at 11:27
    
Is $Diff($\mathbb{R}^{0|n}$)$ simply $Map(\mathbb{R}^{0|n},\mathbb{R}^{0|n})$, only a supersemigroup? ? The action is deduced from 'composition' of $Map$, I suppose it should not hard to see it is associative. What I still can not understand is that why $Maps(R^{0|1},M)=\Pi TM$. Kochan&Severa's approach is use local coordinates, which I do not understand well. Is there a superalgebras-of-functions description of $Map$, like $Hom$ is defined as supermorphisms of superalgebras of functions. –  Ma Ming Feb 21 '11 at 15:19
    
I find another explanation somewhere. Any $f\in Hom(S\times R^{0|1},M)$ or $f:C(M)\to (C(S)\otimes C(\mathbb{R}^{0|1}))$ consists a map of superalgebras $f_0:C(M)\to C(S)$ with an odd derivation w.r.p.t. this map. On the other hand, any map $g:C(\Pi TM)\to C(S)$ or $g:\Omega(M)\to C(S)$ also consists a map $g_0:C(M)\to C(s)$ with an odd derivation. –  Ma Ming Feb 21 '11 at 16:25
    
I think your second explanation is good. Also, it is maybe even better to consider the action of the whole supersemigroup $Map(\mathbb R^{0∣n},\mathbb R^{0∣n})$ instead of the action of the subgroup $Diff(\mathbb R^{0∣n})$ of invertible maps, which can be defined using functor of points again. It is true that associativity follows from associativity of composition of maps between supermanifolds via some abstract nonsense. –  Martin O Feb 22 '11 at 10:19

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.