I am learning about vector-valued differential forms, including forms taking values in a Lie algebra. On Wikipedia there is some explanation about these Lie algebra-valued forms, including the definition of the operation $[-\wedge -]$ and the claim that "with this operation the set of all Lie algebra-valued forms on a manifold M becomes a graded Lie superalgebra". The explanation on Wikipedia is a little short, so I'm looking for more information about Lie algebra-valued forms. Unfortunately, the Wikipedia page does not cite any sources, and a Google search does not give very helpful results.

Where can I learn about Lie algebra valued differential forms?

In particular, I'm looking for a proof that $[-\wedge -]$ turns the set of Lie algebra-valued forms into a graded Lie superalgebra. I would also appreciate some information about how the exterior derivative $d$ and the operation $[-\wedge -]$ interact.


3 Answers 3


A $\frak g$-valued differential form is , as far as I know, just a section $\alpha$ of the tensor product of the exterior power of the cotangent bundle $\Lambda^{\bullet}T^*M$ of some manifold $M$ with the trivial vector bundle $M\times\frak{g}$. As such, locally over some chart domain $U$, $\alpha$ can be cast in the follwing form $$\alpha\equiv\alpha_1\otimes x_1+\cdots+\alpha_n\otimes x_n$$ where $\alpha_1,\dots,\alpha_n$ are local differential forms on $M$ defined over the chart domain $U$, and $x_1,\dots,x_n$ is a basis of $\frak g$. The differential is then calculated by ignoring the Lie algebra terms: $$d\alpha\equiv (d\alpha_1)\otimes x_1+\cdots+(d\alpha_n)\otimes x_n$$ Similarly, the product is defined by treating the differential forms and the Lie algebra elements as separate entities: $$[\alpha\wedge\beta]=\sum_{1\leq i,j\leq n}\alpha_i\wedge\beta_j\otimes[x_i,x_j]$$ For instance, for a pure form $\alpha$ of degree $p$, what you know about the exterior differential immediately implies that $$d[\alpha\wedge\beta]=[(d\alpha)\wedge\beta]+(-1)^p[\alpha\wedge(d\beta)]$$ Also, if $\alpha$ has degree $p$, and $\beta$ has degree $q$, then $$[\beta\wedge\alpha]=(-1)^{pq+1}[\alpha\wedge\beta]$$

I think the algebraic questions that arise are easy enough that I'm sure you can find all the relations you want on your own. However, you can always take a look at Peter W. Michor's Topics in Differential Geometry, in particular his chapter IV, §19, or Morita's Geometry of Characteristic Classes.


I studied it from Differential Geometry Connections,Curvature and Characteristic Classes by Loring W.Tu in the Chapter 4, Section 21(specially subsection 21.5). The whole section 21 (Vector-valued forms) is very well treated. As a Special case Lie-Algebra valued differential forms are discussed. Also how a vector valued differential form differ from a real valued form is discussed in 21.10. Say for example on the Lie Algebra of $Gl(n,\mathbb{R})$, $\alpha\wedge\alpha$ which is given by Matrix Multiplication is not always zero when the deree of $\alpha $ is odd which is contradictory to the usual notion of Real Valued differential form.


Although this is an old post, it comes up prominently in searches and so another viewpoint might be helpful. I’ll view the finite-dimensional real Lie algebra $\mathfrak{g}$ as real matrices under the Lie commutator (such a faithful rep always exists). The reference the treatment is taken from is available online here.

At a single point $p$ on a manifold $M$, a $\mathfrak{g}$-valued differential $1$-form can be viewed as a linear mapping

$$\check{\Theta}:T_{p}M\to \mathfrak{g}.$$

These $1$-forms comprise a real vector space with basis the tensor product of the bases of $T_{p}^{*}M$ and $\mathfrak{g}$ as mentioned in the previous answer. Therefore we can take the exterior product of any number of them to obtain $\mathfrak{g}$-valued $k$-forms.

With real-valued $1$-forms $\varphi$, we usually turn $k$-forms into alternating multilinear mappings via an isomorphism which multiplies the values of the forms, e.g.


or for two $1$-forms


(Note that this isomorphism is not unique, and others are in use). With $\mathfrak{g}$-valued forms we can do the same, making sure to keep the order of the factors consistent to ensure anti-symmetry. So for example the exterior product of two $\mathfrak{g}$-valued $1$-forms is

\begin{aligned}(\check{\Theta}[\wedge]\check{\Psi})\left(v,w\right) & =[\check{\Theta}\left(v\right),\check{\Psi}\left(w\right)]-[\check{\Theta}\left(w\right),\check{\Psi}\left(v\right)]\\ & =\check{\Theta}\left(v\right)\check{\Psi}\left(w\right)-\check{\Psi}\left(w\right)\check{\Theta}\left(v\right)-\check{\Theta}\left(w\right)\check{\Psi}\left(v\right)+\check{\Psi}\left(v\right)\check{\Theta}\left(w\right)\\ & =(\check{\Theta}\wedge\check{\Psi})\left(v,w\right)+(\check{\Psi}\wedge\check{\Theta})\left(v,w\right), \end{aligned}

where we denote the exterior product using the Lie commutator by $\check{\Theta}[\wedge]\check{\Psi}$ to avoid ambiguity (note that $[\check{\Theta},\check{\Psi}](v,w)$ and $[\check{\Theta}(v),\check{\Psi}(w)]$ give different results that are in general not even anti-symmetric). We can therefore view $\mathfrak{g}$-valued $k$-forms as alternating multilinear mappings from $k$ vectors to $\mathfrak{g}$.

The above also shows that


does not in general vanish; instead, for $\mathfrak{g}$-valued $j$- and $k$-forms $\check{\Theta}$ and $\check{\Psi}$ we have the graded commutativity rule


with an accompanying graded Jacobi identity, forming a $\mathbb{N}$-graded Lie algebra which decomposes into a direct sum of each $\Lambda^{k}$. Alternatively, we can decompose into even and odd grade forms, comprising a $\mathbb{Z}_{2}$ gradation, i.e. a Lie superalgebra (since the sign of $(-1)^{jk+1}$ is determined by whether $j$ and $k$ are odd or even).

The exterior derivative can be viewed as operating on the form component of $\mathfrak{g}$-valued forms, as mentioned in the previous answer; this can be also seen by examining the definition of the exterior form in light of the basis for Lie algebra valued forms given above.


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