Let $Q$ be a principal $H$-bundle for some (Lie) group $H$, and $\omega_Q\in \Omega^1(Q,\mathfrak{h})$ a connection 1-form on $Q$. Given an $H$-representation it is straightforward to construct a covariant derivative on the associated vector bundle.
I'm interested in the case where you have a group homomorphism $\alpha: H\to H'$ into some other group and want to construct a connection 1-form on the associated principal $H'$-bundle $Q\times_{\alpha}H'$ (where $H$ acts on $H'$ via left multiplication by $\alpha(H)$).
Is there some canonical way that $\omega_Q$ induces a connection form on this bundle? The only construction I could think of is the following:
Let $p:Q\times H\to H $ be projection onto the second factor, and $\omega'\in \Omega^1(H',\mathfrak{h'})$ the Maurer-Cartan form on $H'$. I have computed that the 1-form $\omega:= p^*\omega' \in \Omega^1(Q\times H', \mathfrak{h'})$ descends to a form on $Q\times_{\alpha}H'$ and satisfies the appropriate equivariance property $R_{h'}^*\omega= \text{Ad}_{(h')^{-1}}\omega $ (and the verticality condition) and thus should be a connection 1-form on $Q\times_{\alpha} H'$.
However, $\omega$ doesn't "know" anything about the original connection $\omega_Q$ that we started with. Is there some way to put a connection on $Q\times_{\alpha}H'$ that is in some way constructed from $\omega_Q$?
Thanks!
Best Answer
I don't think this is true: firstly as Ted points out, the Lie group homomorphism $\alpha:H\to H'$ induces a Lie algebra homomorphism $\dot{\alpha} = T_e\alpha:\mathfrak{h}\to\mathfrak{h}'$. For $(q,h')\in Q\times H'$ and $\xi\in\mathfrak{h}$, the vector along the quotient direction $$ (q\cdot \xi, -\dot\alpha(\xi)\cdot h')\in T_{(q,h')}(Q\times H') $$ when contracted with $\omega = p^*\omega'$ takes the value $-\dot\alpha(\xi)$, which is non-zero for $\xi\notin \ker \dot\alpha$. However this vector descends to the zero vector on $Q\times_\alpha H'$.
I haven't worked through the details, but I think the correct form to consider would be $$ \omega = \dot\alpha(p_1^*\omega_Q) + p_2^*\omega', $$ where $p_1:Q\times H'\to Q$ and $p_2:Q\times H'\to H'$ denote the respective projections. It doesn't suffer from the above problem: applying it to the quotient direction $(q\cdot \xi, -\dot\alpha(\xi)\cdot h')$ yields 0, and so it should descend to a 1-form on $Q\times_\alpha H'$, which will be a connection.
Edit: I went back to work out the details of this, and it seems I was incorrect in my guess. The correct form should be $$ \omega_{(q,h')} :=\mathrm{Ad}_{(h')^{-1}}\left(\dot{\alpha}(p_1^*\omega_Q )_{(q,h')}\right) + (p_2^*\omega_L')_{(q,h')}=\mathrm{Ad}_{(h')^{-1}}\left(\dot{\alpha}(p_1^*\omega_Q )_{(q,h')} + (p_2^*\omega_R')_{(q,h')}\right) $$ Here $\omega_L' (\omega_R')$ is the left-(right-)invariant Maurer-Cartan form on $H'$. Letting $\pi_\alpha:Q\times H' \to Q\times_\alpha H'$ denote the quotient map, $\omega$ vanishes on vectors of the form $(q\cdot\xi, -\dot\alpha(\xi)\cdot h')$ (i.e. vectors along the fibres of $\pi_\alpha$), and is invariant under the quotient action. So it reduces to a form $\omega_\alpha$ on $Q\times_\alpha H'$ $$ \omega = \pi_\alpha^*\omega_\alpha. $$ The obvious right actions $R_\cdot$ and $\check{R}_\cdot$ of $H'$ on $Q\times H'$ and $Q\times_\alpha H'$ commute with $\pi_\alpha$, $$ \pi_\alpha\circ R_{h'} = \check{R}_{h'}\circ \pi_\alpha, $$ and it is straightforward to verify that $R_{k'}^*\omega = \mathrm{Ad}_{(k')^{-1}}\omega$ for all $k'\in H'$. It follows that $$ R_{k'}^*\pi_\alpha^*\omega_\alpha = \mathrm{Ad}_{(k')^{-1}}\pi_\alpha^*\omega_\alpha \implies \pi_\alpha^*\check{R^*}_{k'}\omega_\alpha = \pi_\alpha^*(\mathrm{Ad}_{(k')^{-1}}\omega_\alpha). $$ Since $\pi_\alpha$ is a submersion, this implies $\check{R^*}_{k'}\omega_\alpha = \mathrm{Ad}_{(k')^{-1}}\omega_\alpha$. Also $\omega_{(q,h)}((q,h)\cdot \xi) =\omega_{(q,h)}(0_q,h\cdot \xi) = \xi$ and the fact that $\pi_\alpha$ commutes with the right action implies the equivalent result for $\omega_\alpha$. So $\omega_\alpha$ is a connection form on $Q\times_\alpha H'$.