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.

  • $\begingroup$ 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. $\endgroup$ Mar 1, 2022 at 16:35
  • $\begingroup$ @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? $\endgroup$
    – Shana
    Mar 2, 2022 at 3:15
  • $\begingroup$ @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. $\endgroup$
    – Shana
    Mar 2, 2022 at 3:22

1 Answer 1


My advisor told me the answer; I was just one step away from it.

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.


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