Linear Ehresmann connection on a vector bundle

This article says that for an Ehresmann connection $$H \subseteq TE$$ of a vector bundle $$E \to M$$ to be linear (and so to define a connection in the usual sense as a linear covariant derivative operator) all you need is

$$D(S_\lambda)_e(H_e) = H_{\lambda e}$$ for $$\lambda \in \mathbb{R}$$ and $$e \in E$$, where $$S_\lambda : E \to E$$ is the multiplication by $$\lambda$$ map. But to me it seems like we also need $$D\sigma_{(e,e')} (H_e \oplus H_e') = H_{e + e'}$$ where $$\sigma : E \oplus E \to E$$ is the addition map and $$e,e'$$ lie on the same fiber. Can this somehow be deduced from the previous property, or is the claim in the article mistaken?

• It sure looks like you're correct. We certainly need the sum of parallel sections to be parallel. Mar 9 '21 at 17:35
• Actually it seems that the second property can indeed be deduced from the first, by the argument in this answer to what was basically the same question here math.stackexchange.com/a/2659492/9921 Mar 10 '21 at 3:24
• More specifically, the first property is enough to show (via the usual Leibniz rule trick) that the covariant derivative of a section at a point only depends on the 1-jet of that section, and that this map $(J^1 E)_p \to E_p$ given by covariant derivative along a fixed $X$ is homogeneous; by the argument in that answer it is linear, which is equivalent to the second property. Mar 10 '21 at 3:27