We know that, given a Lie group, we can build its Maurer-Cartan form $\omega_G=P^{-1}dP$ (I don't explain now the meaning of these symbols, which I think to be generally known). This is a left-invariant form and satisfies the Maurer-Cartan equation $d\omega_G+\omega_G\wedge\omega_G=0$.

My question is: how can we relate this form with the so-called "curvature forms" on the left-invariant metric (I use "the" since they're all equivalent)? I'll explain this question by steps:

1) By "connection form" I mean a matrix-valued 1-form on a manifold which satisfies $\nabla e_i=\omega^j_i\otimes e_j$ given a frame $\{e_j\}$. Is there a clear way to obtain the connection form from the M-C form?

2) Given the connection form $\omega$, we define the curvature form $\Theta=d\omega+\omega\wedge\omega$ (i.e. $\Theta^j_i=d\omega_i^j+\omega^j_k\wedge\omega^k_i$). It turns out that $\Theta$ is of the form $\frac{1}{2}R^i_{jkt}\theta^k\wedge\theta^t$, where $\{\theta^j\}$ is the dual coframe of $\{e_j\}$ and $R^i_{jkt}=R_{ijkt}$ is the $(4,0)$ curvature tensor $$R(X,Y,Z,W)=g(\tilde R(Z,W)Y,X)$$ where $\tilde R$ is the well-known curvature $(3,1)$ tensor $\tilde R(Z,W)Y=\nabla_Z\nabla_W Y-\nabla_Z\nabla_W Y-\nabla_{[Z,W]}Y$.

3) Now, if a relation between the M-C form and the connection form $\omega$ exists, it must not be that they simply are the same thing, since in that case the Maurer-Cartan equation would say that $\Theta$ is $0$, and therefore so is the curvature tensor, which isn't true for all Lie groups. But since the expression of $\Theta$ and the Maurer-Cartan equation are so similar I suspect that a (less trivial) relation must exist.


2 Answers 2


Let me see if I remember correctly. To be the Levi Civita connection, the connection forms must satisfy the following equations:

TorsionLess $$\mbox{d}\theta^{i}={\sum}\theta^{j}\wedge\theta_{j}^{i}$$ Compatibility with the metric $$\mbox{d}g_{ij}={\sum}\left(\theta_{i}^{k}g_{kj}+\theta_{j}^{k}g_{ki}\right). $$ I think Cartan structure equations gives you the first condition and not the second. I'll explain everything with a concrete example you can then generalize easily. Let us consider the Heisenberg group. We have that the Maurer Cartan form is given by $$A^{-1}\mbox{d}A=\left(\begin{array}{ccc} 0 & 1 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{array}\right)\mbox{d}x+\left(\begin{array}{ccc} 0 & 0 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{array}\right)\left(\mbox{d}y-x\mbox{d}z\right)+\left(\begin{array}{ccc} 0 & 0 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{array}\right)\mbox{d}z,$$ so we have that the invariant forms are $$\theta^{1}=\mbox{d}x, \theta^{2}=\mbox{d}y-x\mbox{d}z, \theta^{3}=\mbox{d}z. $$ If we apply the external derivative we obtain $$\mbox{d}\theta^{1} =0,$$ $$\mbox{d}\theta^{2} =-\mbox{d}x\wedge\mbox{d}z=-\theta^{1}\wedge\theta^{2},$$ $$\mbox{d}\theta^{3} =0.$$ The dual vectors of theinvariant differential form $\theta^{1}$, $\theta^{2}$ and $\theta^{3}$ are the vectors $$\tilde{E}_{1} = \frac{\partial}{\partial x},$$ $$\tilde{E}_{2} = \frac{\partial}{\partial y},$$ $$\tilde{E}_{3} = x\frac{\partial}{\partial y}+\frac{\partial}{\partial z}. $$ So in the moving frame $ \left\{ \tilde{E}_{1},\,\tilde{E}_{2},\,\tilde{E}_{3}\right\} $ we have that $\left(g_{A}\right)_{ij}$ is the identity and then the equations for the compatibility with the metric (if we want to find the Levi Civita Connection) are $$\omega_{j}^{i}+\omega_{i}^{j}=0.$$

Then from the structure equations we have $$\mbox{d}\theta^{1}= 0 =\left(\mbox{d}y -x\mbox{d}z\right)\wedge\omega_{2}^{1}+\mbox{d}z\wedge\omega_{3}^{1},$$ $$\mbox{d}\theta^{2}= -\mbox{d}x\wedge\mbox{d}z =-\mbox{d}x\wedge\omega_{2}^{1}+\mbox{d}z\wedge\omega_{3}^{2},$$ $$\mbox{d}\theta^{3}= 0 =-\mbox{d}x\wedge\omega_{3}^{1}-\left(\mbox{d}y-x\mbox{d}z\right)\wedge\omega_{3}^{2}.$$ If we solve we then find the connection forms $$\omega_{2}^{1} =-\omega_{1}^{2} =\frac{1}{2}\mbox{d}z=\frac{1}{2}\theta^{3},$$ $$\omega_{3}^{2} =-\omega_{2}^{3} =\frac{1}{2}\mbox{d}x=\frac{1}{2}\theta^{1},$$ $$\omega_{3}^{1} =-\omega_{1}^{3} =\frac{1}{2}\left(\mbox{d}y-x\mbox{d}z\right)=\frac{1}{2}\theta^{2}.$$

And then you have the curvature forms $$\Omega_{2}^{1}= \theta_{3}^{1}\wedge\theta_{2}^{3} =\frac{1}{4}\theta^{1}\wedge\theta^{2},$$ $$\Omega_{3}^{2}= \theta_{1}^{2}\wedge\theta_{3}^{1} =\frac{1}{4}\theta^{2}\wedge\theta^{3},$$ $$\Omega_{3}^{1}= -\frac{1}{2}\theta^{1}\wedge\theta^{3}+\theta_{2}^{1}\wedge\theta_{3}^{2} =-\frac{3}{4}\theta^{2}\wedge\theta^{3}.$$ And at the end the Riemann coefficients $$R_{212}^{1} = \frac{1}{4},$$ $$R_{323}^{2} = \frac{1}{4},$$ $$R_{323}^{1} = -\frac{3}{4}.$$


A) We consider the trivial principal fiber bundle $P(M,G)$, with $P=M\times G$. Then a regular flat connection is defined on $P$ by considering as horizontal subspace on each $u=(x,a) \in P$ the tangent space of $M\times \{a\}$. We consider now the Maurer-Cartan form $ω_{G}$ οn $G$ and the projection map $p:M\times G \rightarrow G$ which induces the form $ω=p^{*}ω_{G}$ on $P$. This is the connection form for the regular flat connection of $P=M\times G$. It can be proved the curvature is zero: $dω=d(p^{*}ω_{G})=p^{*}(dω_{G})=p^{*}(-[ω_{G},ω_{G}])=-[p^{*}ω_{G},p^{*}ω_{G}]=-[ω,ω]$.

Now, set on the above $M=G$, so we have the trivial primary bundle $p:G\times G\rightarrow G$ with connection form to be the pullback of the Maurer-Cartan on $G$. Thus, for this connection the curvature is zero.

B) Let's now consider the same question by using another more general bundle: the primary fiber bundle $L$ of linear frames on $G$. A connection on that bundle is by definition a linear connection on the corresponding relative budle $E$ of $G$. We remark that the connection notation $∇(X,Y)$ refers to the relative bundle $E$ and it is the "infinitesimal analogue" of the connection structure on $E$, the latter induced by the connection structure on the principal bundle $L$. By Helgason's book, there is a 1-1 correspondence between the set of the left invariant connections on a Lie group $G$ and the set of the bilinear functions $α:g \times g \rightarrow g$, where $g$ is the Lie algebra, with $α(X,Y)=∇(X',Y')$, where $X',Y'$ are the left invariant fields corresponding to $X,Y$. We choose $α$ to be identically zero. So, $α(X,Y)=∇(X',Y')=0$. Then we have a corresponding linear connection on $G$. For this connection we have that the Christoffel symbols $Γ^{k}_{ij}=0$, since it is identically a zero connection, so for the connection forms we have $ω_{j}^{i}=\sum Γ^{i}_{kj}ω^{k}=0$, where $ω^{k}$ give the dual basis locally. So in the general formula $dω_{i}=-\sum ω_{k}^{i} \wedge ω^{k}+\frac{1}{2}\sum T_{jk}^{i}ω^{j}\wedge ω^{k}$ the first sum in the right side is zero and we get $dω^{i}=\frac{1}{2}\sum T_{jk}^{i}ω^{j}\wedge ω^{k}$. Here $T$ is the torsion tensor. But it can be proved $T_{jk}^{i}=Γ_{jk}^{i}-Γ_{κξ}^{i}-c_{jk}^{i}=-c_{jk}^{i}$, where $c$ are the structural constants of the group $G$. So we get the formula $dω^{i}=-\frac{1}{2}\sum c_{jk}^{i}ω^{j}\wedge ω^{k}$ which is the Maurer-Cartan equation for that connection and it s not related to the usual Maurer-Cartan equation for $ω_{G}$, see (A). The curvature of that connection is also zero, not because of the last formula, but because the connection form is identically zero.


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