I'm trying to show that for an ordinary manifold $X$ and a supermanifold $S$, supermanifold morphisms $\varphi:S\to TX$ are one-to-one to the pairs $(f,F) $ where $f:C^\infty(X)\to C^\infty(S)$ is a super $\mathbb{R}$-algebra homomorphism and $F:C^\infty(X)\to C^\infty(S)$ is an even derivation with respect to $f$, i.e. $F$ is a parity-preserving $\mathbb{R}$-linear map and $F(ab)=F(a)f(b)+f(a)F(b)$ for any $a,b\in C^\infty(X)$.

Intuitively, $f$ is the pullback of smooth maps from the base manifold $X$ to $S$ and $F$ is a global section to the pullback vector bundle $f^*TX$ on $S$. Hence if we let $p:TX\to X$ be the usual projection, then $f= \varphi^*\circ p^* $. The remaining data of $\varphi$ should give $F$, but I cannot see how.

If $S$ is also an ordinary manifold, then $F$ can be defined point-wisely. We can express $\varphi$ point-wisely by $\varphi(p)=(f(p),V_{f(p)})$ for each $p\in S$, where $f(p)\in X$ and $V_{f(p)}\in T_{f(p)}X$. For any $a\in C^\infty(X)$, $F(a)\in C^\infty(S)$ is then the function that sends $p\in S$ to $V_{f(p)}a\in \mathbb{R}$.

However, if $S$ is a supermanifold, then the point-wise definition no longer works. I tried to formulate the point-wise definition in a non-point-wise way but failed.

I guess that $F$ should be a composition $C^\infty(X)\xrightarrow{D}C^\infty(TX)\xrightarrow{\varphi^*}C^\infty(S)$ for some natural derivation $D:C^\infty(X)\to C^\infty(TX)$ with respect to $p^*$, but I cannot see any derivation that arises "naturally".

Thanks in advance for any help.


1 Answer 1


The morphism $$\def\T{{\rm T}} φ:S→\T X$$ can be identified with the homomorphism of algebras $$\def\Ci{{\rm C}^∞} \Ci(\T X)→\Ci(S).$$ The algebra $\Ci(\T X)$ can be identified with the $\Ci$-symmetric algebra $$\def\CiSym{\mathop{\rm\Ci Sym}\nolimits} \CiSym_{\Ci(X)}(\Gamma(\T^*X))$$ of the $\Ci(X)$-module $\Gamma(\T^* X)$ of smooth sections of the cotangent bundle of $X$. Here $\Ci$-symmetric algebras are defined using exactly the same universal property as symmetric algebras, but working in the category of $\Ci$-rings instead of the category of commutative real algebras.

By the universal property of $\Ci$-symmetric algebras, the homomorphism $$\CiSym(\Gamma(\T^*X))→\Ci(S)$$ can be identified with a homomorphism of $\Ci$-algebras (equivalently, commutative real algebras) $$f:\Ci(X)→\Ci(S)$$ together with a morphism of $\Ci(X)$-modules $$Ψ:\Gamma(\T^*X)→\Ci(S).$$ The latter morphism can be identified with a derivation $$F:\Ci(X)→\Ci(S)$$ with respect to $f$: we set $\def\d{{\rm d}} F(g)=Ψ(\d g)$, where $\d g∈Γ(\T^* X)$ is the differential of $g$.

This approach works equally well for ordinary manifolds, supermanifolds, ${\bf Z}$-graded manifolds, derived manifolds, etc.

  • $\begingroup$ Thanks. Is it true that this $\Psi$ is the composition $\Gamma(T^*X)\hookrightarrow C^\infty(TX)\xrightarrow{\varphi^*} C^\infty(S)$, where $\Gamma(T^*X)\hookrightarrow C^\infty(TX)$ sends a global section $s\in \Gamma(T^*X)$, which is a $C^\infty(X)$-linear map $s:\Gamma(TX)\to C^\infty(X)$, to the smooth map on $TX$ that sends $(p,v_p)\in TX$ to $s(V)(p)\in \mathbb{R}$ for an arbitrarily chosen $V\in \Gamma(TX)$ with $V_p=v_p$? To be honest, I know nothing about $C^\infty$-rings, and the introduction on nlab is not quite clear to me. $\endgroup$
    – Shana
    Apr 14, 2022 at 6:02
  • $\begingroup$ Also, if my construction of $\Psi$ is correct, then I get one side of the one-to-one correspondence. But how can I tell the other side, without taking the universal property mentioned above for granted? Or equivalently, how can I show that the universal property is satisfied by $C^\infty(TX)$? I'd appreciate that if you could please give me some detailed materials about this fact. $\endgroup$
    – Shana
    Apr 14, 2022 at 6:30
  • 1
    $\begingroup$ @Shana: Yes, the description of Ψ is correct. Concerning the universal property, it is the fiberwise version of Proposition 1.1 of Models for smooth infinitesimal analysis by Moerdijk and Reyes. The first sections of this book offer a good introduction to C^∞-rings. $\endgroup$ Apr 14, 2022 at 15:44
  • $\begingroup$ Thanks a lot. I'm crystal clear now. $\endgroup$
    – Shana
    Apr 16, 2022 at 11:55

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