Let $G$ be a connected, reductive Lie group, and $W\mathfrak g = (S[\mathfrak g^\vee] \otimes \Lambda[\mathfrak g^\vee],\delta)$ the associated Weil algebra. This is a CDGA equipped with an action of $\mathfrak g$ by interior multiplication $\iota$ and a resulting action $\mathcal L$ of $\mathfrak g$ by Lie derivatives, $\mathcal L$-equivariantly DGA-isomorphic to the standard Koszul complex $(S[\mathfrak g^\vee] \otimes \Lambda[\mathfrak g^\vee],k)$. In particular, both it and the subalgebra $(W\mathfrak g)^G$ annihilated by the $\mathcal L$-action are contractible.

The basic subalgebra of $W\mathfrak g$, that annihilated by both the $\mathcal L$- and $\iota$-actions, is easily seen to be just the symmetric invariants $S[\mathfrak g^\vee]^G$, on which the Weil differential $\delta$ vanishes identically. By acyclicity of the big algebra, each symmetric invariant $x$ is $\delta y$ for some $y \in (W\mathfrak g)^G$, and we can consider its projection $z$ to the subalgebra $\Lambda[\mathfrak g^\vee]^G$, sending the symmetric generators to zero. The resulting element $z$ is independent of the choice of $y$, and the $x \mapsto z$ corresponds to the cohomology suspension $H^*BG \to H^* G$ in the universal bundle $G \to EG \to BG$ and induces a linear bijection between generators of $\Lambda[\mathfrak g^\vee]^G$ and $S[\mathfrak g^\vee]^G$.

There are natural elements of $S^2[\mathfrak g^\vee]^G$ and $\Lambda^3[\mathfrak g^\vee]^G$ which pull back along homomorphisms of Lie groups and are nonzero when $G$ is nonabelian, namely the Killing form $B$ and the Cartan form $(X,Y,Z) \mapsto B\big([X,Y],Z\big)$. From naturality and lack of other options (also, through hearsay), the Cartan form must be the suspension of the Killing form, but I've never seen anyone show this. But it must be possible to do very explicitly; very explicit formulae for the Weil differential and the two forms in question are possible.

Pick corresponding $B$-orthonormal bases $\theta^i$ and $u^i$ of $\Lambda^1[\mathfrak g^\vee]$ and $S^1[\mathfrak g^\vee]^G$. Let $\iota_i$ and $\mathcal L_i$ be the action of the dual basis of $\mathcal g$ and $c^i_{jk}$ the structure contents. If $x$ is the Killing form and $z$ is the Cartan form, find a simple expression in these symbols for $y$.


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This is the standard Chern-Simons calculation. Let me use gauge-theoretical notation and write $A$ and $F$ for the odd and even, respectively, generators of the Weyl algebra. If we choose canonical dual bases $\theta^a$ for $\mathfrak{g}^*$ and $X_a$ for $\mathfrak{g}$, then $$ A = A^a X_a \qquad\text{and}\qquad F = F^a X_a$$ and the differential $d$ (sorry to change notation) in the Weyl algebra can be read from the usual $$ F = dA + \tfrac12 [A,A] $$ or, in components, $$ F^a = dA^a + \tfrac12 C^a{}_{bc} A^b A^c $$ with $C^a{}_{bc}$ the structure constants in that basis: $[X_a,X_b] = C^c{}_{ab} X_c$.

Now let $B_{ab} = B(X_a, X_b)$ be the components of the Killing form. Your element $x = B_{ab} F^a F^b$ indeed satisfies $x = d y$, where $y$ is the Chern-Simons form $$ y = B_{ab} A^a dA^b + \tfrac13 C_{abc} A^a A^b A^c $$ where $C_{abc} = C^d{}_{bc} B_{ad}$.

Finally, re-express $dA$ in the Chern--Simons form in terms of $F$: $$y = B_{ab}A^a F^b - \tfrac16 C_{abc} A^a A^b A^c$$ and projecting to $\Lambda \mathfrak{g}^*$ (setting $F=0$) we arrive at $$z = - \tfrac16 C_{abc} A^a A^b A^c$$ which seems to be the negative of the Cartan form. (I can't discard having made a sign error or perhaps chosen different conventions.)

  • $\begingroup$ I've always heard that pair of names but never knew that was what their paper said! I calculated $dy$ and got the same signs. One reason for the sign mismatch might be either carelessness on the part of my source or that the "correct" choice is to replace $B$ with $-B$ so that the inner product is positive-semidefinite. Thanks! $\endgroup$ – jdc Jun 11 '17 at 22:54
  • $\begingroup$ @jdc Alternatively, you could presumably redefine the differential in the Weyl complex by $F = dA - \tfrac12 [A,A]$ and then keeping the same definition of $x$, you get a different $y$ and hence a different $z$ which is now precisely the Cartan $3$-form. $\endgroup$ – José Figueroa-O'Farrill Jun 12 '17 at 6:29
  • $\begingroup$ You're right! I'd forgotten the sign! But pretty much the entire construction of the Weil algebra is forced on you. Picking a connection, you have a map $\chi$ from the exterior factor on the $\theta^i$ into the 1-forms $\omega^i$ on your principal bundle, and you have derivations on either which don't match. The curvature is the difference $\Omega^i = (d\chi - \chi d)\theta^i$, so the only way to fix the map failing to be a DGA map is to say $\delta \theta^i = d \theta^i + u^i$ where $u^i$ is some new element. and $\chi(u^i) = \Omega^i$. Expanding out in coordinates gives the "minus" version. $\endgroup$ – jdc Jun 12 '17 at 9:15

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