What does it mean for right equivariant maps to be left translations?

Question

My book is Connections, Curvature, and Characteristic Classes by Loring W. Tu (I'll call this Volume 3), a sequel to both Differential Forms in Algebraic Topology by Loring W. Tu and Raoul Bott (Volume 2) and An Introduction to Manifolds by Loring W. Tu (Volume 1).

I refer to Section 27.1 (part 1), Section 27.1 (part 2), Section 27.1 (part 3) and Section 27.1 (part 4).

Question: About Lemma 27.7, what exactly is going on?

Based on my understanding, given next, I think $f$ should be assumed to have the action on $G$ in the range as $G$'s law of composition. There might be some understanding, for those already familiar with this topic, of Lemma 27.7 based on its use later in Section 27.1. But, as a beginner, I'm not sure that I dare to read the rest of Section 27.1 without first understanding Lemma 27.7. (Anyway, the rest of Section 27.1 doesn't seem to be used in the rest of this book except in Example 27.10 and Problem 31.1.)

This is what I understand:

1. Let $G$ be any group, not necessarily a Lie group or a topological group, with law of composition $\text{Law}: G \times G \to G$, $\text{Law}(g,h) \in G$ for all $g,h \in G$. Let $1_G$ be the identity of $G$. Let $N$ and $M$ be any sets, not necessarily manifolds or topological spaces. Let $\zeta: N \times G \to N$, $\sigma: M \times G \to M$ be right actions by $G$ to the sets, respectively, $N$ and $M$. Let $f: N \to M$ be a right $G$-equivariant map. Let $x \in N$, and let $g \in G$. Then $f(\zeta(x,g)) = \sigma(f(x),g)$.

2. In (1), choose $N = M = G$ and then choose $x = 1_G$. Then $f(g) =$ $ f(\zeta(1_G, g))$ $ = \sigma(f(1_G),g)$.

   • 2.1. Edit: I made a mistake. I assumed that we have $\zeta(1_G, g) = g$ for any action $\zeta$. Rather, we have, similar to (4) below, that $\zeta = \text{Law}$.

3. "Left translation" is any map $l_h: G \to G$, where $l_h(g) = \text{Law}(h,g)$, with $h \in G$.

4. Somehow, $\sigma = \text{Law}$.

5. By (2) and (4), $f(g) = \text{Law}(f(1_G),g)$

6. By (3) and (5), $f = l_h$, with $h=f(1_G)$.

Answer 1

Your understanding is perfectly correct.

To say that a map is $G$-equivariant requires an action of $G$ on both the domain and target of the map. When both the domain and target are the group $G$ itself, a 'canonical' action is defined by the group law. That is, in a setting such as Lemma 27.7. in the book, the action is not usually explicitly specified, as there is only one natural interpretation of the statement.