# Connections on frame bundles with filtrations

I have a filtered vector bundle $$\underline{\omega} \subset \mathcal{H}$$ over a space $$X$$. The subbundle $$\underline{\omega}$$ is rank $$1$$, and the full bundle $$\mathcal{H}$$ is rank $$2$$. I also have a connection on the full vector bundle $$\nabla \colon \mathcal{H} \to \mathcal{H} \otimes \Omega^1_X$$, which does not restrict to a connection on the subbundle $$\underline{\omega}$$. (i.e., even if $$s$$ is a section of $$\underline{\omega}$$, we expect that $$\nabla(s) \not\in \underline{\omega} \otimes \Omega^1_X$$.)

I know that we can use the Leibniz rule and $$\nabla$$ to induce connections on each of $$\operatorname{Sym}^k\mathcal{H}$$, $$\operatorname{Sym}^k\mathcal{H}^*$$, and $$\left(\bigwedge^2 \mathcal{H} \right)^{\otimes k}$$ for any $$k$$, which taken together give a connection on the frame bundle $$\operatorname{Frames}(\mathcal{H})$$, a $$\operatorname{GL}_2$$-torsor. It does not give a connection on $$\operatorname{Frames}(\underline{\omega})$$, a $$\mathbb{G}_m$$-torsor, since it doesn't even give a connection on $$\underline{\omega}$$.

However, there is one more torsor around: the "filtered frame bundle" $$N$$, a torsor for the upper triangular matrices $$B$$, consisting locally of bases $$(\omega_1,\omega_2)$$ for $$\mathcal{H}$$ with the condition that $$\omega_1 \in \underline{\omega}$$. Can $$\nabla$$ determine a connection on $$N$$? Or does a connection on the "filtered frame bundle" $$N$$ determine a connection on $$\operatorname{Frames}(\underline{\omega})$$?

One extra buzzword that I know is relevant to the situation is Griffiths transversality, which states in this case that the induced operator $$\underline{\omega}^{k-r} \otimes \operatorname{Sym}^r\mathcal{H} \hookrightarrow \operatorname{Sym}^k\mathcal{H} \xrightarrow{\nabla} \operatorname{Sym}^k\mathcal{H} \otimes \Omega^1_X$$ actually lands in $$\underline{\omega}^{k-r-1} \otimes \operatorname{Sym}^{r+1}\mathcal{H} \otimes \Omega^1_X \subset \operatorname{Sym}^k \mathcal{H} \otimes \Omega^1_X$$. Does this relate in any way to the answer to the question above?

EDIT: Maybe another question, trying to get at the same information. What's the difference between a connection on $$\mathcal{H}$$ that restricts to a connection on the subspace $$\underline{\omega}$$, versus one that simply satisfies Griffiths transversality? If the second does not induce a connection on the $$B$$-torsor $$N$$, what kind of differential operator does it induce?

I think I've figured this out. The way that I tend to manipulate connections on $$G$$-torsors $$p \colon E \to X$$ is by thinking about the equivalent data of a $$G$$-equivariant splitting of the exact sequence $$0 \to p^*\Omega^1_X \to \Omega^1_E \to \Omega^1_G \to 0.$$ If $$E^\prime \subset E$$ is a $$G^\prime$$-torsor for $$G^\prime \subset G$$, there's a way to "restrict the connection" in certain cases (especially when $$G^\prime$$ is an open subgroup of $$G$$). When I see this done, the new splitting is said to be $$G^\prime$$-equivariant because $$G^\prime \subset G$$. But the $$GL_2$$-equivariance should make the operator descend, e.g., to a connection $$\operatorname{Sym}^k\mathcal{H} \to \operatorname{Sym}^k\mathcal{H} \otimes \Omega^1_X$$, while the $$B$$-equivariance should give connections $$\underline{\omega}^r \otimes \operatorname{Sym}^k\mathcal{H} \to \underline{\omega}^r \otimes \operatorname{Sym}^k\mathcal{H} \otimes \Omega^1_X$$, which it does not e.g. for $$k=0$$ by assumption. So even if we can still get a splitting of that exact sequence, it shouldn't be $$B$$-equivariant.
My vague, naive reasoning for this is that $$GL_2$$ treats all vectors in the representation $$\operatorname{Sym}^k\mathcal{H}$$ the same, while $$B$$ does not. So when you move to a different element of $$\operatorname{Sym}^k\mathcal{H}$$, $$GL_2$$ has no problem compensating, but $$B$$ can't. I would love to hear if this reasoning is correct, and it would be useful to hear if all of this is wrong.