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I'm reading Volume 1 of Quantum Fields and Strings: A Course for Mathematicians.

Let $V$ be a Minkowski vector space of signature $(1,d-1)$, and let $\mathrm{Spin}(V) \curvearrowright S$ be an irreducible real spin representation.
Fix the gamma map $\Gamma: S^* \otimes S^* \to V$.

The super Lie algebra $\mathcal{L}$ is defined as $V \oplus \Pi S^*$ as a vector space, where $V$ is central, and

$$ [s,t] = -2 \Gamma(s,t) $$

for $s,t \in \Pi S^*$. Here $\Pi$ denotes odd parity.

Let $\exp(\mathcal{L})$ be the Lie supergroup with Lie algebra $\mathcal{L}$.
Fix bases $\{e_\mu\} \subset V$ and $\{f_a\} \subset \Pi S^*$.

I want to show that

$$ D_a = \partial_a - \theta^b \Gamma_{ab}^\mu \partial_\mu $$

is the left-invariant vector field on $\exp(\mathcal{L})$ corresponding to $f_a \in \mathcal{L}$, but I have encountered a sign problem.

From the BCH formula, I think the group law is

$$ (x^\mu, \theta^a) = (x_1^\mu + x_2^\mu - \Gamma_{bc}^\mu\theta^b_1 \theta^c_2, \theta^a_1 + \theta^a_2) $$

Let $m: \exp(\mathcal{L}) \times \exp(\mathcal{L}) \to \exp(\mathcal{L})$ be the group multiplication. The left-invariance condition means

$$ \begin{aligned} (0, D_a)m^* x^\mu &= m^*(D_a x^\mu), \\ (0, D_a)m^* \theta^b &= m^*(D_a \theta^b). \end{aligned} $$

The second equation poses no problem. However,

$$ \begin{aligned} (0, D_a)m^* x^\mu &= (\partial_{a, 2} - \theta^b_2 \Gamma_{ab}^\nu \partial_{\nu, 2})(x_1^\mu + x_2^\mu - \Gamma_{bc}^\mu\theta^b_1 \theta^c_2) \\ &= -\Gamma_{bc}^\mu\partial_{a, 2}(\theta^b_1 \theta^c_2) - \theta^b_2 \Gamma_{ab}^\mu \\ &= \Gamma_{bc}^\mu\theta^b_1\partial_{a, 2}(\theta^c_2) - \theta^b_2 \Gamma_{ab}^\mu \\ &= \Gamma_{ba}^\mu\theta^b_1 - \theta^b_2 \Gamma_{ab}^\mu \\ &= \theta^b_1\Gamma_{ab}^\mu - \theta^b_2 \Gamma_{ab}^\mu, \\ m^*(D_a x^\mu) &= m^*(-\theta^b \Gamma_{ab}^\mu) = -(\theta_1^b + \theta_2^b) \Gamma_{ab}^\mu, \end{aligned} $$

which do not coincide.

hipokaba
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1 Answers1

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I’ve found the answer myself.

The mistake was in the group law. The correct group law is: $$ \begin{aligned} (x^\mu, \theta^a) = (x_1^\mu + x_2^\mu + \Gamma_{bc}^\mu \theta_1^b \theta_2^c,\; \theta_1^a + \theta_2^a) \end{aligned} $$

PBW can be applied to $S$-points, not to coordinates.
The $S$-point map of $G \times G \to G$ is $$ \begin{aligned} &C^\infty(S, G \times G) = C^\infty(S, G) \times C^\infty(S, G) = (\mathcal{O}(S) \otimes \mathcal{L})_0 \times (\mathcal{O}(S) \otimes \mathcal{L})_0 \ni \\ &\quad (s^\mu \otimes e_\mu + \sigma^a \otimes f_a,\; t^\mu \otimes e_\mu + \tau^a \otimes f_a) \\ &\quad \mapsto (s^\mu + t^\mu) \otimes e_\mu + (\sigma^a + \tau^a) \otimes f_a + \tfrac{1}{2}[\sigma^a \otimes f_a,\; \tau^b \otimes f_b] \\ &\qquad = (s^\mu + t^\mu + \Gamma_{ab}^\mu \sigma^a \tau^b) \otimes e_\mu + (\sigma^a + \tau^a) \otimes f_a \\ &\in (\mathcal{O}(S) \otimes \mathcal{L})_0 = C^\infty(S, G) \end{aligned} $$

Now, set $S = G \times G$ and consider $\mathrm{id}_{G \times G}$.
Then we obtain the coordinate expression of the group law: $$ \begin{aligned} \mathrm{id}_{G \times G} &= (x_1^\mu \otimes e_\mu + \theta_1^a \otimes f_a,\; x_2^\mu \otimes e_\mu + \theta_2^a \otimes f_a) \\ &\mapsto (x_1^\mu + x_2^\mu + \Gamma_{ab}^\mu \theta_1^a \theta_2^b) \otimes e_\mu + (\theta_1^a + \theta_2^a) \otimes f_a \\ &= (x_1^\mu + x_2^\mu + \Gamma_{ab}^\mu \theta_1^a \theta_2^b,\; \theta_1^a + \theta_2^a) \end{aligned} $$

hipokaba
  • 376