I am reading Group actions on manifolds by Eckhard Meinrenken (Lecture Notes, University of Toronto, Spring 2003).

In page $45$, definition $5.2$, author introduce the notion of $\mathfrak{g}$-differential graded algebra. Here $\mathfrak{g}$ is a Lie algebra ( I am thinking it is Lie algebra of Lie group $G$ but not sure)

A differential graded algebra is an graded algebra $A = \bigoplus_{I=0}^{\infty}A_i$ with a differential $d$ of degree $1$, such that $d$ is a derivation. It is called a $\mathfrak{g}$-differential algebra if, in addition, there are derivations $L_\xi$ of degree $0$ and $i_\xi$ of degree $-1$ for all $\xi\in \mathfrak{g}$, satisfying the relations of contractions, Lie derivative and differential on a manifold with a $\mathfrak{g}$-action:

  1. $[i_{\xi}, i_{\xi ‘}] = 0$
  2. $[L_\xi,i_{\xi ‘}] = i_{[\xi,\xi’]}$
  3. $[d, L_\xi] = 0$
  4. $[L_\xi,L_{\xi ‘}] = L_{[\xi,\xi ‘]}$
  5. $[d, d] = 0$
  6. $[d,i_\xi] = L_\xi$

I don’t understand the notation of $[d,d]=0$, what Lie algebra structure are we fixing here? The same confusion for all other conditions. Is this the standard way to define a$\mathfrak{g}$-differential graded algebra?


$[x, y]$ here is the super or graded commutator (in the graded algebra of graded endomorphisms of $A$), which you may not be used to; it refers to

$$xy - (-1)^{\deg x \deg y} yx$$

and so if either $x$ or $y$ is even it reduces to the ordinary commutator but if $x, y$ are both odd it's the anticommutator; in particular $[d, d] = 2d^2$ because $d$ is odd.

I've never seen this definition before but the motivation is spelled out pretty clearly:

"satisfying the relations of contractions, Lie derivative and differential on a manifold with a $\mathfrak{g}$-action"

presumably means this definition is intended to abstract the action of $\mathfrak{g}$ on differential forms $\Omega^{\bullet}(M)$ coming from a smooth action of a corresponding Lie group $G$ on $M$.

$L_{(-)}$ and $i_{(-)}$ are presumably also intended to be linear although this isn't stated explicitly.

  • $\begingroup$ I just want to confirm what you mean by action of $\mathfrak{g}$ on $\Omega^{\bullet}(M)$... Given $A\in \mathfrak{g}$, action of $G$ on $M$ gives a vector field $A^*$ on $M$... Given a vector field $X$ on $M$ and a differential form $\omega$ on $M$, I know two ways to associate another differential form.. One way is to consider $L_X\omega$ (degree does not change in this case). another is to consider the $i_X\omega$ (degree reduce by $1$ in this case)... Is this what you are referring to as actions on $\mathfrak{g}$ on $\Omega^{\bullet}(M)$... $\endgroup$ – Praphulla Koushik Sep 16 at 4:35
  • $\begingroup$ Yes, that’s what this definition is intended to abstract. $\endgroup$ – Qiaochu Yuan Sep 16 at 5:21
  • $\begingroup$ Thanks for confirmation... Do you happen to know any references for these... In particular, for "graded algebra of graded endomorphisms of A".. I know the definition but not much properties... $\endgroup$ – Praphulla Koushik Sep 16 at 5:36
  • $\begingroup$ can you suggest some references $\endgroup$ – Praphulla Koushik Sep 22 at 16:33

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.