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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.

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3 Answers 3

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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.

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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.

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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.

$$\bigwedge_{i=1}^{k}\varphi_{i}\mapsto\sum_{\pi}\textrm{sign}\left(\pi\right)\prod_{i=1}^{k}\varphi_{\pi\left(i\right)},$$

or for two $1$-forms

$$(\varphi\wedge\psi)(v,w)=\varphi(v)\psi(w)-\psi(v)\varphi(w).$$

(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

$$(\check{\Theta}[\wedge]\check{\Theta})\left(v,w\right)=2[\check{\Theta}\left(v\right),\check{\Theta}\left(w\right)]$$

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

$$\check{\Theta}[\wedge]\check{\Psi}=\left(-1\right)^{jk+1}\check{\Psi}[\wedge]\check{\Theta},$$

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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