Meaning of a subscript asterisk?

Question

For example, I have a definition of an inner automorphism:

$k_*: G \rightarrow G$

Taken from Mathematics Form and Function by Saunders Mac Lane, p.137.

Answer 1

I guess the comments that it is just notation to mean the automorphism $g \mapsto kgk^{-1}$ could be right, but seeing as the author is Saunders Mac Lane, who was a hugely influential figure in the creation of category theory it seems plausible that there is more to this than a random decoration.

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In general a lower asterisk denotes a pushforward and an upper asterisk a pullback, that is, in general terms if $g \colon X \to Y$ is a map between mathematical objects and $H(X)$ and $H(Y)$ are some other objects associated to $X$ and $Y$ in a way that $g$ induces a map from $H(X) \to H(Y)$ then $g_*$ is used to denote this induced map.

For example if $X, Y$ are just sets and $T$ is another set then functions $T \to X$ and $T\to Y$ are sets again. So in the notation from before we can define $$H(?) = \text{functions from } T \text{ to }?$$ Now given $g\colon X \to Y$ and $\phi\colon T \to X$ (i.e. $\phi \in H(X)$) we can compose $g$ and $T$ to get a new function $(g \circ \phi)\colon T \to Y$. That is $g$ gives us a way of turning functions from $T$ to $X$ into functions $T$ to $Y$. So we have a function $H(X)\to H(Y)$, which as it goes the same way (from the $X$ side to the $Y$ side) as our original $g$ makes it a pushforward, as we push things about $X$ forward in the direction of $g$ to become things about $Y$. Then the convention is that this pushforward map $H(X) \to H(Y)$ is denoted $g_*$ as we get it from $g$ and it pushes forward.

The reversed version of this is a pullback, for example in the same set-up if we instead defined $H(?) = \text{functions }?\to T$ (so the map is going to $T$ this time) then we would be able to take an element $\phi\colon Y\to T$ and compose with $g\colon X\to Y$ to get a function $(\phi \circ g)\colon X \to T$ so we would have a way of taking elements of $H(Y)$ to $H(X)$ instead, so we are pulling back via $g$ this time.

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Back in the context of your book, how is $k_*$ a pushforward?

Well Mac Lane gives us a bit more about his thought process, on the same page 137 he says:

For a transformation group G, conjuga­tion by k has "geometric" meaning; for instance (§III.9), it carries the subgroup fixing one point x isomorphically onto the subgroup fixing the point k(x ); cf. (III.9.3) and (III.9. 4).

In this geometric picture in the earlier chapter we have a group $G$ acting on a set $X$, so that each element $k\in G$ sends each element $x\in X$ to some other element $k(x)$. So we can think of elements of $G$ as maps $X\to X$. The interesting thing to consider there is the set that Mac Lane denotes $F_x$ the set of all $f\in G$ that fix some $x\in X$, i.e. send $x$ to itself. So we have a subgroup $F_x$ of $G$ associated to each element of $X$. Now if $k$ sends $x$ to $y$ then we can think of $F_x$ and $F_y$ like our $H(X)$ and $H(Y)$ from before, the important question is, does $k$ give a natural map from $F_x$ to $F_y$ or vice versa?

If we have an element $f \in F_x$ so $f(x) = x$ and some $k\in G$ with $k(x) = y$ then $k^{-1}(y) = x$ and we can see that $ k(f(k^{-1}(y))) = y$ so the element $kfk^{-1} \in F_y$. So this map coming from $k$ sends $F_x \to F_y$, it is pushing forward along $k$ as it is taking things about $x$ to things about $y$ and $k$ took $x$ to $y$.

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What about $k^*$?

If we continue thinking about $k$ and the relationship between $F_x$ and $F_y$ some more, we see that we can also map $F_y$ to $F_x$, if we send $g\in F_y$ to $k^{-1}gk$! The definition looks very similar, but it is a different map. So what should we call this map, well we are using $k$ to send these subgroups backwards compared to the way $k$ sent the points, so this map should be called $k^*$, it is the pullback along $k$.