Parity reversed tangent bundle as a supermanifold

I encountered an example in a paper telling that $$\underline{SM}(\mathbb{R}^{0|1},X)\cong \pi TX$$, where $$X$$ is some fixed ordinary Riemannian manifold, $$\pi TX$$ is the supermanifold with base manifold $$X$$ and structural sheaf $$\pi(\wedge^*(TX)^\vee)$$ according to my advisor, seen as a functor $$\pi TX:=SM(-,\pi TX): SM^{\mathrm{op}} \to Set$$, $$S\mapsto SM(S,\pi TX)$$ where $$SM$$ stands for the category of supermanifolds. $$\underline{SM}(\mathbb{R}^{0|1},X)$$ is the functor which acts on objects by $$\underline{SM}(\mathbb{R}^{0|1},X)(S)= SM(S\times \mathbb{R}^{0|1},X)$$, with the obvious action on morphisms. "$$\cong$$" means that there is a natural isomorphism between the two functors.

Now, by the theorem that supermanifolds are "affine" we have $$\underline{SM}(\mathbb{R}^{0|1},X)(S) \cong SAlg_{\mathbb{R}}(C^\infty(X), C^\infty(S)\otimes C^\infty(\mathbb{R}^{0|1}))$$

where $$SAlg_{\mathbb{R}}$$ is the category of super $$\mathbb{R}$$-algebra in which the morphisms are parity-preserving $$\mathbb{R}$$-algebra homomorphisms. Let $$\theta$$ be the odd coordinate of $$\mathbb{R}^{0|1}$$ and we see that an element in $$\underline{SM}(\mathbb{R}^{0|1},X)(S)$$ is identified with a super $$\mathbb{R}$$-algebra morphism $$\Phi^*:C^\infty(X)\to C^\infty(S)\otimes C^\infty(\mathbb{R}^{0|1})\cong C^\infty(S) \oplus C^\infty(S)\cdot \theta$$.

Decompose $$\Phi^*$$ using the direct sum and we can write $$\Phi^*=f+\phi\theta$$ where $$f:C^\infty(X)\to C^\infty(S)$$ is a super $$\mathbb{R}$$-algebra morphism and $$\phi:C^\infty(X)\to C^\infty(S)$$ is a parity-reversing map such that

$$\phi(ab)=\phi(a)f(b)+f(a)\phi(b)= \phi(a)f(b)+(-1)^{p(a)}f(a)\phi(b)$$ for any $$a,b\in C^\infty(X)$$, where $$p(a)$$ stands for the parity of $$a$$ which is always zero.

The paper refers to Deligne & Morgan's Notes on Supersymmetry (following Joseph Bernstein) and says that this is the standard description of $$\pi TX$$ in terms of its $$S$$-points so we get $$\underline{SM}(\mathbb{R}^{0|1},X)\cong \pi TX$$, but Deligne & Morgan's description is not clear to me at all. My advisor told me that $$\pi TX$$ is supposed to be as above and left the rest to me as an exercise.

I have no difficulty understanding $$\pi TX$$ for sure, but I don't have a clue how a morphism $$\varphi: S\to \pi TX$$, which can be identified with a super $$\mathbb{R}$$-algebra morphism $$\varphi^*:\pi(\wedge^*(TX)^\vee) \to C^\infty(S)$$, could be translated to some $$\Phi^*:C^\infty(X)\to C^\infty(S)\oplus C^\infty(S)\cdot \theta$$ satisfying the axioms above.

For the simplest case, say $$X=\mathbb{R}^1$$, we have $$\pi(\wedge^*(TX)^\vee) =C^\infty(\mathbb{R})\oplus C^\infty(\mathbb{R})\cdot dx$$ where $$dx$$ is considered to be odd. From this $$\varphi^*$$ gives a parity-preserving $$\mathbb{R}$$-algebra $$\bar f:C^\infty(X)\to C^\infty(S)$$ which I guess is the candidate of $$f$$ (or not?). Put $$s:=\varphi^*(dx)$$ and we see that $$\varphi^*$$ is determined by $$\bar f$$ and $$s$$, as

$$\varphi^* (a+bdx)=\bar f(a)+\bar f(b)s.$$

Unless $$\phi=\bar f\cdot s$$, which doesn't make any sense, I cannot see where $$\phi$$ could possibly come from. I think I do need some help about this. Thanks in advance.

• I give a hint; maybe someone else will spell it out. The component $f$ of your $\Phi^*$ is (dual to) the composition $S \overset\varphi\to \pi T X \to X$. The remaining data is basically a vector field on $S$; more precisely, it is a field on $S$ valued the pulled-back tangent bundle $f^* \pi T_X$. Now recall that vector fields are the same as derivations. In this case, you find that sections of $f^* \pi T_X$ are precisely (odd) "derivations along $f$", i.e. functions $\phi$ satisfying the Leibniz-like relation that you wrote down. Mar 1 at 16:35
• @TheoJohnson-Freyd Thanks for the hint; so my naive observation for candidate of $f$ is correct. But how does the tangent bundle $\pi TX$ pullback to a bundle on $S$? It seems that it should be compatible with post-composition a derivation $D:C^\infty(X)\to C^\infty(X)$ by $f$, getting $f\circ D: C^\infty(X)\to C^\infty(S)$, but for an arbitrary open subset $U\subset S$, what is $(f^*\pi TX)(U)$ supposed to be? To have a locally free sheaf we can't take simply the composition $C^\infty(X)\to C^\infty(S)\to C^\infty(U)$; if $X$ should be replaced by some open subset of $X$, what is our choice? Mar 2 at 3:15
• @TheoJohnson-Freyd Moreover, even with $f^*\pi TX$ defined as a sheaf, I can't see how a section might be related to $\varphi:S\to \pi TX$, with $\pi TX$ the supermanifold. I read similar things in Deligne & Morgan before, but nothing is clear to me. Mar 2 at 3:22

A morphism $$\varphi:S\to \pi TX$$ is determined by the natural transformation between the structural sheaves, which is locally determined, so it suffices to assume that $$X$$ has coordinates $$x_1,\cdots,x_n$$. Now $$C^\infty(\pi TX)=\pi(\wedge^*(TX)^\vee)$$ is finitely generated by $$d x_1,\cdots,dx_n$$ over $$C^\infty(X)$$, the composition $$C^\infty(X)\hookrightarrow \pi(\wedge^*(TX)^\vee)\to C^\infty(S)$$ gives our $$f$$ and the remaining data of $$\varphi^*:\pi(\wedge^*(TX)^\vee)\to C^\infty(S)$$ is given by the images of $$d x_1,\cdots,dx_n$$.
Let $$s^i:= \varphi^*(dx_i)$$, then $$\varphi^*$$ on $$dx_i$$'s can be expressed by (with abuse of notation) $$\varphi^*=\sum_i \partial_{x_i}\otimes_f s^i$$ in light of $$\left(\sum_i\partial_{x_i}\otimes _f s^i\right)(dx_j)= \sum_i dx_j(\partial_{x_i})\otimes_f s^i =s^j$$.
Now, $$\sum_i \partial_{x_i}\otimes_f s^i$$ gives an element in $$\text{Der}(C^\infty(X),C^\infty(X))\otimes_f C^\infty(S)$$, which is by definition (see Wikipedia) a global section of the pullback bundle $$f^*\pi TX$$ on $$S$$. Finally, as $$\text{Der}(C^\infty(X),C^\infty(X))\otimes_f C^\infty(S)\cong \text{Der}_f(C^\infty(M),C^\infty(N))$$, where $$\text{Der}_f$$ denotes derivations with respect to $$f$$, via $$W\otimes s\mapsto s(f^*\circ W)$$ with inverse $$V\mapsto \sum_i \partial_{x_i}\otimes _f V(x_i)$$, the above relates $$\varphi$$ with $$(f,\phi)$$ bijectively.