# How can we understand a group-scheme action as a natural transformation

I am trying to understand the paragraph from Wikipedia on group-scheme action (to make this paragraph complete or "self-contained", I put the definition and basic settings at the end of the post):

One can also consider (at least some special case of) an action of a group functor: viewing $G$ as a functor, an action is given as a natural transformation satisfying the conditions analogous to the above.

In details, given a group-scheme action $\sigma$, for each morphism $T \to S$, $\sigma$ determines a group action $$G(T) \times X(T) \to X(T);$$ i.e., the group $G(T)$ acts on the set of $T$-points $X(T)$.

Conversely, if for each $T \to S$, there is a group action $$\sigma_T: G(T) \times X(T) \to X(T)$$ and if those actions are compatible; i.e., they form a natural transformation, then, by the Yoneda lemma, they determine a group-scheme action $$\sigma: G \times_S X \to X.$$

I understand the direction from group-scheme action $\sigma$ to natural transformation. But I am confused about how to determine $\sigma$ by given natural transformation. Here is my unsuccessful attempt:

We know $\sigma$ is the transformation between $G(-)\times X(-)$ to $X(-)$, where $$G(-)=h_G=\mathrm{Hom}(-,G).$$

However, the Yoneda Lemma says $$F(A)\cong \mathrm{Hom}(h_A,F)$$ where $F\in \mathrm{Fun}(\mathcal{C}^{\mathrm{op}},\mathbf{Set})$.

Take $F=h_G\times h_X$ and $A=X$, we have $$\mathrm{Hom}(X,G)\times \mathrm{Hom}(X,X)\cong \mathrm{Hom}(h_X,h_G\times h_X).$$ If the "arrows" are reversing, that is, $$\sigma\in \mathrm{Hom}(h_X,h_G\times h_X)$$ (in fact, it is in $\mathrm{Hom}(h_G\times h_X,h_X)$), so we can find a unique corresponding element in $$\hat{\sigma}\in\mathrm{Hom}(X,G)\times \mathrm{Hom}(X,X).$$

By the universal property of fiber product, we have $$\mathrm{Hom}(X,G)\times \mathrm{Hom}(X,X)\cong \mathrm{Hom}({X,G\times_S X})$$ If we treat the arrow as its reverse again, we have $\sigma$ corresponds to $$G\times_S X\to X$$ (in fact, it should be $X \to G\times_S X$)

Obviously, it is absurd to reverse the arrows. So probably I was missing something small but important. Or my try is totally wrong and meaningless. Any hint and answer are welcome! Thanks!

The definition of group-scheme action is as follows:

given a group $S$-scheme $G$, a left action of $G$ on an $S$-scheme $X$ is an $S$-morphism $$\sigma: G \times_S X \to X$$ such that

• (associativity) $\sigma \circ (1_G \times \sigma) = \sigma \circ (m \times 1_X)$, where $m: G \times_S G \to G$ is the group law,
• (unitality) $\sigma \circ (e \times 1_X) = 1_X$, where $e: S \to G$ is the identity section of $G$.

The Yoneda lemma, when applied to the case where $F$ is itself a representable functor $h_B$, gives you an isomorphism $\mathrm{Hom}(A,B)=h_B(A)\cong \mathrm{Hom}(h_A,h_B)$ (note that the two $\mathrm{Hom}$ don't apply to the same category); this means that the Yoneda embedding $$\mathcal{C}\to \mathrm{Fun}(\mathcal{C}^{\mathrm{op}},\mathbf{Set}):A\to h_A=\mathrm{Hom}(-,A)$$ is fully faithful, and sometimes this fact is also called the Yoneda lemma.
In your case, you have a natural transformation $\sigma :h_G\times h_X\Rightarrow h_X$, and as you note, the universal property of the fiber product implies that $h_G\times h_X\cong h_{G\times_S X}$; so you can also see $\sigma$ as a natural tansformation $h_{G\times_S X}\Rightarrow h_X$, and as explained above, this must be induced by a map $G\times_S X\to X$.
To see that this map satisfies the associativity and unitality axiom, you apply the Yoneda lemma again, more precisely, the fact that the map inducing a specific transformation must be unique : indeed, this implies that in order to check for example that $\sigma \circ (1_G \times \sigma) = \sigma \circ (m \times 1_X)$, it is enough to check that they induce the same natural transformation $h_G\times h_G\times h_X\Rightarrow h_X$; and this will just follow from the fact that every $$\sigma_T:G(T)\times X(T)\to X(T)$$ is itself a group action.