4
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $$R_{212}^{1} = \frac{1}{4},$$ $$R_{323}^{2} = \frac{1}{4},$$ $$R_{323}^{1} = -\frac{3}{4}.$$

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

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.