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Let's say I have a supermanifold $\mathbb R^{1|2}$ with the standard volume form $dx d \psi_1 d \psi_2$, regarded as a section of $\operatorname{Ber} \Omega^1$, in the sense of Deligne-Morgan-Bernstein. Let $$X = \psi_2 \frac{d}{dx} + f(x) \frac{d}{d \psi_1}$$ be an odd vector field, here $f : \mathbb R \to \mathbb R$ be a smooth function (purely even). I want to show that $\mathcal{L}_{X} (dx d\psi_1 d\psi_2) = 0$.

I am a bit unsure which transformation rules I am allowed to use. Naively, using the Leibniz rule and the fact that Lie derivative commutes with exterior differential, I will get $d\psi_2 d\psi_1 d \psi_2$, which is not $0$. I see at least three potential issues why it doesn't work:

  1. The object $dx d\psi_1 d\psi_2$ regarded as a section of $\operatorname{Ber} \Omega^1$ is not the same as literally the wedge product of $dx, d\psi_1$ and $d\psi_2$ so Leibniz rule is unjustified
  2. There is no theorem that $[\mathcal L_{X}, d] = 0$ in the aforementioned notes, although I think it should be true in some sense
  3. I can "ignore" mixed terms and forcefully set them $=0$ for some reason, but I cannot justify that rigorously.

So my question is: exactly where does the naive computation go wrong, and how to actually show that $\mathcal{L}_{X} (dx d\psi_1 d\psi_2) = 0$? Definition 3.7.1 seems too implicit, even in this simple case.

Myungheon Lee
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Mykola Pochekai
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1 Answers1

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I assume that you are workinig with the definition \begin{align} L_X(\mu) = \left.\frac{d}{dt}\right\vert_{t=0}(\phi_t^X)^*\mu, \end{align} where $\phi_t^X\in\mathrm{Diff}(M)$ is the flow generated by $X\in\mathcal T_M$, and $\mu\in\Gamma(M,E)$ is a section of a vector bundle $E\to M$ canonically associated with $M$ (e.g., the Berezinian bundle). Note here that $t$ should really be thought of as some parameter from the base ring $\mathcal O _S$ via the functor-of-points picture (i.e., we parametrise $M$ via a smooth submersion of supermanifolds $M\to S$).

The following two lemmas are useful for the problem at hand.

Lemma 1: Let $U\subset M$ such that there exists a coordinate chart $(x^\bullet):U\to U^{m|n}\subset\mathbb R^{m|n}$, and let $[-] : \left((\Omega^1)^{m|n}\right)^\times \to \mathcal B er(\Omega^1)$ be the Berezinian of a local basis of $\Omega^1$. It holds $$L_{\partial/\partial x^i} [\mathrm{d} x^1 \cdots \mathrm{d} x^{m+n}] = 0.$$ Proof: Note that by definition of the Berezinian sheaf, $$ \phi^* [\mathrm d x^\bullet] = ber(D\phi)[\mathrm d x^\bullet]$$ for all diffeomorphisms $\phi\in\mathrm{Diff}(M)$. Now, to compute the Lie derivative, we need to compute the flow of the coordinate vector fields $$ x^j \circ \phi^{\partial / \partial x^i}_t = x^j + \delta^{ij} t, $$ or, in other words, $$\phi^{\partial/\partial x^i}_t \begin{pmatrix}x^1\\ \vdots \\ x^{i-1} \\ x^i\\ x^{i+1}\\ \vdots \end{pmatrix} = \begin{pmatrix}x^1\\ \vdots \\ x^{i-1} \\ x^i + t\\ x^{i+1}\\ \vdots \end{pmatrix}.$$ Here, $t$ should really be thought of as a parameter from the base via the funtor-of-points picture: In particular, whenever $x^i$ is an odd coordinate function, $t$ should also be an odd parameter. In either case, for all $t\in\mathcal O _S$, the Jacobian is given by $D\phi^{\partial/\partial x^i}_t = \mathrm{id}$, such that \begin{align} L_{\partial/\partial x^i} [\mathrm d x^\bullet] &= \left.\frac{d}{dt}\right\vert_{t=0} \left(\phi^{\partial/\partial x^i}_t\right)^*[\mathrm d x^\bullet]\\ &= \left.\frac{d}{dt}\right\vert_{t=0} ber(D\phi^{\partial/\partial x^i}_t ) [\mathrm d x^\bullet]\\ &= \left.\frac{d}{dt}\right\vert_{t=0} ber(\mathrm{id}) [\mathrm d x^\bullet]\\ &=0, \end{align} which proves the claim.

Lemma 2: Let $E\to M$ be a vector bundle canonically associated with $TM$, and denote $\mathcal E = \Gamma(M,E)$. For all $\mu\in\mathcal E$, $X\in\mathcal T_M$ and $f\in\mathcal O_M$, it holds $$ L_{fX}(\mu) = (-1)^{p(X)p(f)}L_X(f\mu) = (-1)^{p(X)p(f)}X(f)\mu + fL_X(\mu).$$ Proof: This is just a reincarnation of the Leibniz rule on $\mathbb R$ (or more precisely, the base ring $\mathcal O _S$), together with $L_Xf = X(f)$. Detailed proofs can be found in Deligne & Morgan's Notes on supersymmetry.

The solution to the problem at hand now simply follows from $$\frac{\partial \psi^2}{\partial x} = \frac{\partial f}{\partial \psi^1}=0$$ together with the two lemmas above.

Johannes Moerland
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