Explicit formula for the principal connection 1-form induced by a Cartan connection

Question

Let $P \subseteq G$ be a closed Lie subgroup. Suppose that a principal $P$-bundle $\mathcal{P} \to M$ is equipped with a Cartan connection $\omega: T\mathcal{P} \to \mathfrak{g}$. Then the extended principal $G$-bundle $\mathcal{P} \times_P G \to M$ admits a principal connection $\overline{\omega}$. However, I cannot find a nice explicit formula given in Sharpe's or Cap & Slovak's book for this principal connection $1$-form.

Note that the tangent bundle of the extended principal $G$-bundle is $$T(\mathcal{P} \times_P G) \cong T\mathcal{P} \times_{TP} TG,$$ where $TP$ is the tangent group of $P$. For those unaware of this term, note that the tangent bundle of a Lie group is again a Lie group when we apply the tangent functor to the multiplication and inversion operations (since the tangent functor preserves limits).

Thus, for some tangent vector $[\gamma, \tau] \in T\mathcal{P} \times_{TP} TG$, I would naively guess that the formula should be something like $$\overline{\omega} [\gamma, \tau] = \omega(\gamma) + \omega_G(\tau) $$ where $\omega_G$ is the Maurer-Cartan form for $G$.

However, while I have a formula for how the Maurer-Cartan form behaves with respect to the action of the tangent group $TP$, I don't see how to get a similar formula for the Cartan connection with respect to the action of $TP$.

Answer 1

The answer to your question is contained in the proof of Theorem 1.5.6 of Cap & Slovak.

That proof deals with a more general situation of $i: P\to K$ (your $P$ is their $H$), but the inclusion $i: P\to G$ naturally satisfies the conditions.

It seems that what you proposed is what they do at the trivial section $$ j : {\mathcal P}\to {\mathcal P}\times_P G;\ u\mapsto [(u, e)]. $$ Then they shift this vertically to define the connection at a general point $j(u)\cdot g=[(u, g)]$ by equivariancy.

Putting these two steps in the proof together, I propose the following fix to your guess. You need to be very careful about the base point of a tangent vector. Suppose now $[(u, g)]\in {\mathcal P}\times_P G$ and $[(\gamma, \tau)]\in T_{[(u, g)]} ({\mathcal P}\times_P G)$. Then $$ \bar\omega([\gamma, \tau]) = \operatorname{Ad}(g^{-1})\omega(\gamma) + \omega_G(\tau). $$

We can directly check that this $\bar \omega$ is well-defined on ${\mathcal P}\times_P G$. For by definition, $[(u, g)]=[(uh, h^{-1}g)]$ for $h\in P$, and $[(\gamma, \tau)]=[(Tr^h(\gamma), Tl^{h^{-1}}(\tau))]$, where $T$ stands for the tangent map, and $r$ and $l$ are right and left multiplications. Now \begin{align} \bar\omega([(Tr^h(\gamma), Tl^{h^{-1}}(\tau))]) &=\operatorname{Ad}((h^{-1}g)^{-1})\omega(Tr^h( \gamma)) + \omega_G(Tl^{h^{-1}}( \tau))\\ &= \operatorname{Ad}(g^{-1}h)(\operatorname{Ad}(h^{-1})\omega(\gamma)) + \omega_G(\tau)\\ &= \operatorname{Ad}(g^{-1})\omega(\gamma) + \omega_G(\tau)\\ &=\bar\omega([\gamma, \tau]), \end{align} where the second line uses the equivariancy of the Cartan connection $\omega$, and the invariance of the Maurer-Cartan form.

In response to the comments As pointed by the OP, $T({\mathcal P}\times_P G)=T{\mathcal P}\times_{TP} TG$, where $TP$ is the tangent group of $P$. So there is another type of equivalence under which I need to show that $\bar\omega$ is invariant.

Let me indicate what this is. Let $A\in {\mathfrak p}$, the Lie algebra of $P$. This type of equivalence is at a fixed point $[(u, g)]$, we have $$ [(\gamma, \tau)]= [(\gamma + \zeta_A(u), -Tr^g A+\tau)], $$ where $\zeta_A(u)=\frac{d}{dt}\big|_{t=0} u\cdot \exp(tA)$ is the fundamental vector field of $A$ at $u$ by the right action of $P$ on ${\mathcal P}$. Similarly, $-Tr^g A = \frac{d}{dt}\big|_{t=0}\exp(-tA)g$ is induced from the left multiplication of the inverse on $g$. (Remember $Tr^g = (R_g)_*$ is the tangent map.) We compute again \begin{align} \bar \omega([(\gamma + \zeta_A(u), -Tr^g A+\tau)]) &= Ad(g^{-1})\omega(\gamma + \zeta_A(u)) + \omega_G(-Tr^g A+\tau)\\ &=Ad(g^{-1})\omega(\gamma) + Ad(g^{-1})\omega(\zeta_A(u)) - \omega_G((R_g)_*A) + \omega_G(\tau)\\ &=Ad(g^{-1})\omega(\gamma) + Ad(g^{-1})A - Ad(g^{-1})A + \omega_G(\tau)\\ &=Ad(g^{-1})\omega(\gamma)+\omega_G(\tau) = \bar\omega([(\gamma,\tau)]), \end{align} where in line 3, we again use the properties of the connection $\omega$ on fundamental vector fields and the Maurer-Cartan form on right translations now.