0

I am following Rogers's Supermanifolds: Theory and Applications and I might be getting something wrong, because I reach a definition that, as I understand it, doesn't imply what the author states.

Letting $\mathbb{R}_{\infty}$ be a real Grassmann algebra generated by anticommuting $\{\xi_j\}_{j\in\mathbb{N}}$, with even and odd parts $\mathbb{R}_{\infty,0}$ and $\mathbb{R}_{\infty,1}$ respectively, we call superspace the product $$\mathbb{R}^{m,n}_{\infty}:= (\mathbb{R}_{\infty,0})^{\times m}\times(\mathbb{R}_{\infty,1})^{\times n}$$ and we specify a point in it as $({\bf x};{\bf \xi})$ for ${\bf x}\in (\mathbb{R}_{\infty,0})^{\times m}$ and ${\bf \xi}\in (\mathbb{R}_{\infty,1})^{\times n}$.

Meanwhile we note $G^\infty(U)$ the $\mathbb{R}_\infty$-module of maps $U\to\mathbb{R}_\infty$ for any open $U$ in a DeWitt supermanifold. Furthermore, the author defines the superderivative $\mathcal{D}$ of functions from $\mathbb{R}^{1,1}_\infty$ as acting in the following way: $$\mathcal{D}f(t;\tau) = \frac{\partial^E}{\partial t} f(t;\tau) + \tau\frac{\partial^O}{\partial \tau} f(t;\tau)$$ where the $E$ (respectively $O$) index of a derivative signals it is even (odd).

Now the author affirms that $\mathcal{D}^2 = \frac{\partial^E}{\partial t}=:\partial_t^E$, but I cannot see that. In my development $$\begin{align} \mathcal{D}^2 &= (\partial_t^E)^2 + \tau \partial_t^E\partial_\tau^O + \tau \partial_\tau^O\partial_t^E + \tau \partial_\tau^O - \tau^2(\partial_\tau^O)^2 \\ & = (\partial_t^E)^2 + 2 \tau \partial_t^E\partial_\tau^O + \tau \partial_\tau^O \end{align}$$ because, as far as I understand, $\tau$ being odd implies $\tau^2 = 0$, and even derivatives commute with odd derivatives as well as with odd scalars. This result is visibly not equal to $\partial_t^E$. What is wrong?

Albert
  • 747
  • The relevant passage is found in section 6.3 of the first edition of the book. – Albert Mar 21 '24 at 18:10
  • If we consider the classical function f(t) would D just be d/dt? – Liding Yao Mar 22 '24 at 15:48
  • Then why $D^2=d/dt$ not obviously wrong? Since in this case $D^2=d^2/dt^2$. – Liding Yao Mar 23 '24 at 18:39
  • ‍♂️ That this might be some sort of typo is not impossible, but I prefer letting that be the last possible explanation. Another problem is I have not found an errata of the book. – Albert Mar 24 '24 at 11:10
  • Also, I was probably wrong when I said $D$ reduces to $\partial_t$ on classical functions, first and foremost because classical functions are absent on the space on which $D$ acts; the closest thing to a classical function in $G^\infty$ is its Grassmanian continuation. Also note that $t$ is not a real number (nor in any $\mathbb{R}^n$), so you shouldn't think that a $G^\infty$‐function of $t$ that is independent of $\tau$ is a classical function. – Albert Mar 24 '24 at 11:38
  • In case you were curious, @LidingYao, you can check below to see what the problem was. – Albert Mar 28 '24 at 12:18
  • I am actually more interested in what is the way to settle down the non-uniqueness of the odd derivative on supermanifold. I knew a bit for supermanifolds from my friend. – Liding Yao Mar 29 '24 at 00:44
  • @LidingYao , both the odd and the even derivative can be univocally defined by defining a superanalytic function $f$ as a decomposition over the Grassmann continuation $\hat{f}\alpha$ of a set of (not super-)analytic functions $f\alpha: \mathbb{R}^m\to\mathbb{R}\infty$, for some multi-index $\alpha$, such that $$f(x;\xi) = \sum\alpha \hat{f}\alpha(\epsilon{m,n}(x)) \xi^\alpha$$ for the body function $\epsilon_{m,n}: \mathbb{R}^{m,n}\infty\to\mathbb{R}^m$, and then defining the derivatives as acting in an explicit manner on each $ \hat{f}\alpha(\epsilon_{m,n}(x)) \xi^\alpha$. – Albert Apr 03 '24 at 10:07
  • If I remember correct the odd derivative has an issue when you consider super-diffeomorphism. That is, it's not preserved under different coordinate chart. – Liding Yao Apr 03 '24 at 14:54

1 Answers1

1

After further reading and waking up this morning with an epiphany, I have arrived to the conclusion that the problem is double: in one hand I was misunderstanding some notions, and in the other the author committed a typo. Here I present the solution.

First and foremost, I got wrong what is meant by $\mathcal{D}^2$. The space of derivations $D(\mathbb{R}^{1,1})$ to which $\mathcal{D}^2$ belongs is not a super algebra over $\mathbb{R}^{1,1}$ but a super Lie module over $\mathbb{R}^{1,1}$, so by $\mathcal{D}^2$ one should understand $[\mathcal{D},\mathcal{D}]$, for the corresponding super Lie bracket $[\cdot,\cdot]$. The latter, for any two derivations $\delta_1, \delta_2$ and with $|\cdot|$ being the $\mathbb{Z}_2$-graded degree, is specifically defined as $$[\delta_1, \delta_2] :=\delta_1 \delta_2 - (-1)^{|\delta_1||\delta_2|}\delta_2\delta_1.\tag{1}$$ Secondly, the author seems to have committed a typo, as confirmed by the expressions used later in the same book, and the proper definition of $\mathcal{D}$ is $$\mathcal{D}:= \partial^O_\tau + \tau\partial_t^E.\tag{2}$$ Consequently, dropping the indices $E$ and $O$ for notational convenience and considering that $|\partial_\tau|=|\tau|=1$ while $|\partial_t|=0$, we combine $(1)$ and $(2)$ to obtain $$\begin{split} \mathcal{D}^2&=[\partial_\tau,\partial_\tau]+ [\partial_\tau, \tau\partial_t] + [\tau\partial_t, \partial_\tau] + \underbrace{[\partial_t, \partial_t]}_{=0}\\ &=(2\underbrace{\partial_\tau\partial_\tau}_{=0})+(\partial_t-\tau\partial_t\partial_\tau +\tau\partial_t\partial_\tau)+(\tau\partial_t\partial_\tau+\partial_t-\tau\partial_t\partial_\tau)\\ &=2\partial_t \end{split}$$ In the context of supergeometry it is not uncommon to say that the superderivative $(2)$—modulo a factor $\frac{1}{\sqrt{2}}$—is the square root of the classical derivative $\partial_t$ precisely because $(\frac{1}{\sqrt{2}}\mathcal{D})^2=\partial_t$.

Albert
  • 747