Connections on frame bundles with filtrations

Question

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?

Answer 1

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.