General variance and contravariance

Suppose that a group $G$ acts transitively on a space $X$ . A transformation $T \in G$ can be thought of indistinctly as either "moving" the objects in $X$ or as a change of "point of view" in the following way:

Suppose we have a comfortable, mathematically speaking, space with a distinguished point, $(S, s_{0})$ . We are going to think of $(Z, 0)$ , for greater ease, and a simplified physical world $X$ (the red drawings in the picture). Suppose also a bijection $b_{1} : S \to X$ (in the line of reasoning of the notes homogeneous space#Intuitive approach and basis and change of basis). It is better to think in terms of $b_{1}^{- 1}$ . That is, for every $x \in X$ we have a kind of "coordinates" $s$ for $x \in X$ , given by $s = b_{1}^{- 1} (x) \in S$ . Think, for example, of a vector space $V$ and the isomorphism $b_{1} : R^{2} \to V$ (fixing a basis in $V$ ).
We can think of this as our initial point of view, and we can imagine as if we were physically located at $x_{0} = b_{1} (0) \in X$ , provided with some devices to take measurements that have allowed us to create $b_{1}$ .
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An element of the group $T \in G$ is a transformation, a movement, of $X$ into itself (think of a kind of earthquake in this simplified world). The application of this transformation yields the yellow drawings in the picture:
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We have a description of the "new world" $T (X)$ given by $b_{2}^{- 1} = b_{1}^{- 1} \circ T$ . And we can think that to translate from the description of "the red world" to the description of the "yellow world" we only have to add 6 units. In other words, the transformation $T$ induces a map in $Z$ :

\tilde{T} = b_{2}^{- 1} \circ b_{1} = b_{1}^{- 1} \circ T \circ b_{1} \equiv + 6

But there is another interpretation for this fact, which ends up with the same result. What if the transformation (the earthquake) affected only me and my measurement system just in the opposite way? That is, we have the map $T^{- 1}$ applied to me.
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The point of view 2 is how the red drawings is seen from the purple person.
and is given by the same $b_{2}$ .

It is impossible to distinguish whether the new description $b_{2}^{- 1}$ is caused by the world moving ahead, or by myself moving in the opposite direction.
Since the change of the "components" (+6) is opposite to the "transformation of myself" (-6), it is said that the components are contravariant.

Very important example: covariance and contravariance in linear algebra.

Behavior of functions

Suppose a function $h : X \to R$ (in the pictures above think of the "height" of every object). The description given by $b_{1}$ let us create a "more comfortable" function $F_{1} : Z \to R$ given by $F_{1} = h \circ b_{1}$ , encoding the same data as in the diagram

\begin{array}{ccc} X & \overset{h}{⟶} & R \\ b_{1} ↑ \\ Z \end{array}

Consider a transformation $T$ of $X$ (think of a shift of +6, por example, in the picture above). What is the new comfortable description $F_{2}$ of $h$ with the new description $b_{2}$ of $X$ ? Since $b_{2}$ is the red drawings as seen by the purple person, it correspond to shift the graph by 6 units to the left:
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Analytically, the new description of $X$ is $b_{2} = T \circ b_{1}$ ,

\begin{array}{ccc} R & R \\ h ↑ & h ↑ \\ X & \overset{T}{⟶} & X \\ b_{1} ↑ \\ Z \end{array}

so the new comfortable function is

F_{2} = h \circ b_{2} = h \circ T \circ b_{1} = h \circ b_{1} \circ b_{1}^{- 1} \circ T \circ b_{1} =

and

= F_{1} \circ {\tilde{T}}_{1}

So if, for example, $F_{1} (x) = x^{2} + 1$ then $F_{2} (x) = (x + 6)^{2} + 1$ . Observe then that this corresponds to the elemental fact in high school mathematics that to shift the graph of a function f(x) by $a$ units you have to substitute $f$ by $f (x - a)$ .

On the other hand, we may be interested in the implicit description of subsets of $X$ . How does the zero set of functions transform under $T$ ?
Consider $C = {h (x) = 0}$ , the zero set of $h$ . The transformation $T$ moves the set $C$ to the set $T (C)$ . Since $y \in T (C)$ if and only if there exists $y^{'} \in C$ such that $y = T (y^{'})$ , with $h (y^{'}) = 0$ . Then, $h (T^{- 1} (y)) = 0$ , and therefore

T (C) = {x \in X : (h \circ T^{- 1}) (x) = 0}

i.e. the transformed function $h \circ T^{- 1}$ is just the function such that its zero set is the transformed of the zero set of $h$ . If we had started with $T^{- 1}$ instead of $T$ we have arrived at the conclusion that $h \circ T$ is the function whose zeroes are the translated by $T^{- 1}$ , and then the function $F_{2} = h \circ b_{2} = (h \circ T) \circ b_{1}$ have an interpretation different than above: is the description, in the first point of view, of the implicit function of the translation of $C$ by $T^{- 1}$ .
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Particular cases: