# How to caculate the internal hom of supermanifolds?

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

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(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.

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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