# How to differentiate natural transformations?

Let $$G_{\bullet} = (G_1 \rightrightarrows G_0)$$ and $$H_{\bullet} = (H_1 \rightrightarrows H_0)$$ be Lie groupoids and $$\varphi_\bullet, \psi_\bullet : G_\bullet \to H_\bullet$$ be Lie groupoid morphisms, i.e. smooth functors. Suppose that $$\eta : \varphi_\bullet \Rightarrow \psi_\bullet$$ is a natural transformation, i.e. a smooth map $$\eta : G_0 \to H_1$$ such that $$\eta(t(g))\varphi(g) = \psi(g) \eta(s(g))$$ for all $$g \in G_1$$.

Thinking of $$G_\bullet$$ and $$H_\bullet$$ as representing stacks, this says that $$\varphi_\bullet$$ and $$\psi_\bullet$$ represent the same map $$[G_0/G_1] \to [H_0/H_1]$$. Hence (if my intuition is correct), there should be some kind of chain homotopy between the corresponding maps on the tangent complexes: $$\begin{array}{ccccccc} 0 & \rightarrow & \operatorname{Lie}(G_\bullet) & \rightarrow & TG_0 & \rightarrow & 0\\ & &\varphi_1\downdownarrows\psi_1& \swarrow &\varphi_0\downdownarrows\psi_0 \\ 0 & \rightarrow & \operatorname{Lie}(H_\bullet) & \rightarrow & TH_0 & \rightarrow & 0. \end{array}$$ Is that correct, and if yes, how do we get the map $$\swarrow$$? One issue is that I'm not even sure what a chain homotopy would be in that context since these complexes of vector bundles do not live on the same manifolds. Is there a way to get a sort of chain homotopy capturing this idea? Note that differentiating $$\eta$$ doesn't automatically give us a map $$TG_0 \to \operatorname{Lie}(H_\bullet)$$.

Let's take a look at Lie groups first. We have morphisms $$\varphi,\psi:G\to H$$ and a natural transformation $$\varphi\Rightarrow\psi,$$ which is just given by some $$h\in H\,,$$ so that $$\psi=\text{Ad}_{h}\varphi\,.$$ Then differentiating we get morphisms $$d\varphi,d\psi:\mathfrak{g}\to\mathfrak{h}\,,$$ and they satisfy $$d\psi=\text{Ad}_h\,d\varphi,$$ so it would make sense to say that this is how we "differentiate" the natural transfomation in this case.
Now going back to groupoids, we could try to do the same thing but note that the adjoint action doesn't exist (instead there is something called a representation up to homotopy which replaces the adjoint action, but it's not a strict action, e.g. see Abad and Crainic - Representations up to homotopy and Bott's spectral sequence for Lie groupoids). However, while $$H$$ doesn't act on $$\mathfrak{h}\,,$$ it is still true that $$TH$$ acts on $$\mathfrak{h}\,,$$ and this is really what we need. The action of $$TH$$ on $$\mathfrak{h}$$ is induced by the action of $$TH$$ on itself (Lie groupoids all act on themselves by translation, e.g. page 94 here: Lackman - The van Est Map on Geometric Stacks ). If I haven't made a mistake, then for $$X\in\mathfrak{g}$$ we have that $$d\psi(X)=d\eta(\alpha(X))\cdot d\varphi(X)\,.$$
This is obtained by differentiating the formula $$\psi(g)=\eta(t(g))\varphi(g)\eta(s(g))^{-1}$$ along a source fiber of $$G\,.$$
Here $$d\eta:TG^0\to TH^{(1)}\,.$$ This generalizes the construction in the first paragraph, but of course as we see this natural transformation (even in the case of groups) uses the groupoid $$H$$ and not just $$\mathfrak{h}\,.$$