I know that given a manifold $M$ and its corresponding tangent bundle $TM$ we can call $\Pi TM$ the space of forms parametrized (via charts) by $\{x_i\}_{i=1,\dotsc,n}$ and its corresponding cotangent basis $\{dx_i\}_{i=1,\dotsc,n}$. The notation is due to the fact that this space is a $n|n$-supermanifold, since the coordinates of cotangent bundle is anti-commutative.

My question is: Given a Lie algebra $\mathfrak{g}$, what would be $\Pi\mathfrak{g}$? I currently see this notation and don't know what it means.

  • 3
    $\begingroup$ Where do you see this notation? $\endgroup$
    – LSpice
    Aug 9, 2020 at 1:19
  • 2
    $\begingroup$ Suppose $\mathfrak{g}$ is identified with the tangent space of a Lie group $G$ at identity. In this case, $\Pi \mathfrak{g}$ is the fiber of $\Pi TG$ over the identity element. $\endgroup$
    – S. Carnahan
    Aug 9, 2020 at 1:54
  • $\begingroup$ @OP prob worth noting that the shift of \mathfrak{g} is no longer super lie alg in int way as brackets are forced to vanish by the parity condition. Nonetheless it's endowed with some extra structure, there is an odd square zero vector field on its symmetric algebra $\endgroup$
    – EBz
    Nov 20, 2020 at 11:28

1 Answer 1


Since a Lie algebra is a vector space, $\Pi\mathfrak{g}$ is just the vector space $\mathfrak{g}$ with a $\mathbb{Z}_2$-grading where every element is odd. In particular, if $(t_a)$ is a basis of $\mathfrak{g}$ and $(c^a)$ is the dual basis, then we can take $c^a$ as coordinates on $\Pi\mathfrak{g}$. Then functions on $\Pi\mathfrak{g}$ are polynomials in the $c$ variables where we identify $c^ac^b$ with $-c^bc^a$. A good resource for these matters is D. Fiorenza's paper An introduction to the Batalin-Vilkovisky formalism (arXiv link).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.