2

A connection $\omega$ on a Principle G bundle is a Lie algebra valued one form, defined as follows:

  1. $(R_g)^*\omega=ad_{g^{-1}}(\omega)$
  2. $ \omega(A^\#)=A$
    where $R_g$ is the right translation map for $g \in G$, $ad$ is the adjoint map, $A^\#$ is the fundamental vector field generated by the element $A$ of Lie algebra.

What does the first condition (right equivariance) mean? I understand that the LHS is the pullback of the one form. But I do not understand the RHS. I cannot intuitively comprehend what the adjoint map does with the one form.

When we speak about left invariant vector field, I can intuitively think that, a vector in the vector field at a point $p$, when pushed forward by a left translation map $L_g$, should give a vector in the same vector field at the point $g.p$, and not any other vector in $g.p$

Likewise, can anyone explain condition number 1.

amWhy
  • 210,739
rishitharun
  • 57
  • $ad_{g^{-1}} \omega$ is to be understood as follows. While $\omega$ provides a linear map from $T_p(P)$ to $\mathfrak{g}$, you then compose that map with $Ad_{g^{-1}}$, which is a linear map from $\mathfrak{g}$ to itself preserving the Lie algebra structure. So, in other words, it is a composition of maps. – Malkoun Aug 01 '24 at 19:45
  • Thanks! Can you help me comprehend the first condition as a whole, intuitively? – rishitharun Aug 02 '24 at 04:01
  • The way I think of it is as follows. Suppose the base manifold is a single point. Given $g_1 \in G$, what is $R_{g_1}^*( g^{-1} dg )$? – Malkoun Aug 02 '24 at 07:28

1 Answers1

4

In my personal opinion, it is easiest to think about this in terms of vector bundles $\pi:E\to M$. That is to say, to view $G\subseteq\text{GL}(r,\mathbb{R})$ for some $r> 0$, such that the associated bundle is isomorphic to $E$. Then the adjoint representation becomes conjugation, i.e. $\text{Ad}_g(\omega)(X)=g\cdot\omega(X)\cdot g^{-1}$.

In this case, the condition $(R_g)^*\omega=\text{Ad}_{g^{-1}}\omega$ just says: suppose you have a frame at $x\in M$, i.e a choice of basis in $E_x$. Suppose you make a change of basis with $g\in\text{GL}(r,\mathbb{R})$. Then the connection $1$-form respects this change of basis, in the sense that there is a commutative diagram $$\require{AMScd} \begin{CD} E_x @>{\omega(dR_gX)}>> E_x\\ @VVgV @VVgV \\ E_x @>{\omega(X)}>> E_x \end{CD}$$ where we view $\mathfrak{gl}(r,\mathbb{R})$ as the endomorphisms of $\mathbb{R}^r$.

For an abstract Lie group the idea is much the same, but I think it is easier to intuit the concept for subgroups of the linear group.

Quaere Verum
  • 4,394
  • So that is to say, the Right invariance condition $(R_g)^*\omega=\text{Ad}_{g^{-1}}\omega$ is precisely the $covariance$ condition of a connection, right? – Z. Liu May 13 '25 at 23:15
  • No, I do not think that is right. But maybe you can elaborate a bit because it's not clear to me what you precisely mean. – Quaere Verum May 14 '25 at 05:36