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Stream: learning: questions

Topic: induced automorphisms

David Egolf (Sep 06 2023 at 22:30):

I was reading the start of Leinster's "Galois Theory" (here). On page 15, this example of a group action is provided:

Let $G$ be the 48-element group of isometries (rotations and reflections) of a cube. Then $G$ acts on the 6-element set of faces of the cube: any isometry maps faces to faces.

Intuitively, specifying a way to transform a cube so that it "lands back on itself" provides an induced permutation of the faces of the cube.

This leads me wonder: When (and how) can we induce automorphisms of one object from automorphisms of a different object?

I find it helpful to think about this in the category of sets. Consider the set $\{1,2,3,4\}$ together with the subset $\{1,2\}$. Then there are lots of automorphisms of $\{1,2,3,4\}$ that don't induce automorphisms of $\{1,2\}$ in a nice way. For example, the permutation $1 \mapsto 4, 4 \mapsto 1, 2 \mapsto 2, 3 \mapsto 3$ does not nicely induce a permutation of $\{1,2\}$, because under this permutation $1$ gets mapped to an element outside of $\{1,2\}$.

Motivated by this, I started considering commutative squares like this one:
commutative square

Here, $X$ and $Y$ are objects in some category $C$, $g: X \to X$ is an automorphism, and $m: Y \to X$ is a monomorphism. The intuitive intent is that the square commuting ensures that the $Y$-part of $X$ gets mapped to itself under $g$. We can ask under what conditions a $\bar{g}$ will exist so that the square commutes. When $\bar{g}$ does exist and is an automorphism, we might say it is the automorphism of $Y$ induced by the automorphism $g: X \to X$. By the way, I believe that if $\bar{g}$ exists, it is unique, because if $g \circ m = m \circ \bar{g} = m \circ \bar{g}'$, then $\bar{g} = \bar{g}'$, because $m$ is a monomorphism.

In this context, we can consider a few questions:

• What can we say about the conditions under which $\bar{g}$ exists? At the very least $\overline{1_X} = 1_Y$ always exists.
• If $\bar{g}$ exists, does $\overline{g^{-1}}$ necessarily exist too? Is $\bar{g}$ always an automorphism of $Y$ if $g$ is an automorphism of $X$? On a related note, I think that $\overline{g \circ h} = \bar{g} \circ \bar{h}$ if both $\bar{g}$ and $\bar{h}$ exist.
• If $\bar{g}$ exists for every $g$ in a subgroup $H$ of $\mathrm{Aut}_C( X)$, can we say anything about the structure of $H$? I'm wondering if there's anything special about subgroups of $\mathrm{Aut}_C( X)$ that map $Y$ to itself.

Although I mention some specific questions above for concreteness, it could be interesting to discuss related concepts in this direction too! So please don't hesitate to mention something just because it doesn't directly answer one of my specific questions.

James Deikun (Sep 06 2023 at 22:53):

I think for finite sets in classical foundations, $\overline{g}$ will always be an automorphism if it exists; for infinite sets you probably need something else -- probably for the square to be a pullback.

Mike Shulman (Sep 06 2023 at 22:55):

As a concrete counterexample, let $X=\mathbb{Z}$, $g(x) = x+1$, and $Y = \mathbb{N}$ with $m$ the inclusion. Then $\bar{g}$ exists, but is not an automorphism of $\mathbb{N}$.

James Deikun (Sep 06 2023 at 23:12):

If $m$ pulls back to itself along an isomorphism, then the other projection of the pullback is an isomorphism: take the unique arrow $h$ from Y to Y induced by the arrows $g^{-1}m$ and $\mathrm{id}_Y$, it will be a section of $\overline{g}$ because $\mathrm{id}_Y$ induced it and a retraction of $\overline{g}$ because $\overline{g}h\overline{g} = \overline{g}$ and $mh\overline{g} = g^{-1}gmh\overline{g} = g^{-1}m\overline{g}h\overline{g} = g^{-1}m\overline{g} = m$ so $h\overline{g}$ has the same cone as (and is therefore identical to) $\mathrm{id}_Y$.

James Deikun (Sep 06 2023 at 23:18):

Similarly if a square like this exists with $g$, $\overline{g}$ automorphisms, it is a pullback because given a pullback cone $x : Z \to X, y : Z \to Y$ where $gx = my$, $\overline{g}^{-1}y$ induces the cone ($y$ is immediate, $x$ because $m\overline{g}^{-1}y = g^{-1}my = g^{-1}gx = x$). So being a pullback is exactly enough.

James Deikun (Sep 06 2023 at 23:24):

(Oh, and the inducing morphism is unique because $m$ is a mono. Or because $\overline{g}$ is an iso hence a mono. Take your pick.)

David Egolf (Sep 07 2023 at 16:01):

Mike Shulman said:

As a concrete counterexample, let $X=\mathbb{Z}$, $g(x) = x+1$, and $Y = \mathbb{N}$ with $m$ the inclusion. Then $\bar{g}$ exists, but is not an automorphism of $\mathbb{N}$.

Ah, so sometimes $\bar{g}$ is not an automorphism. I guess that means my plan for inducing automorphisms doesn't always work out!
(In this case $\bar{g}$ I think also adds one to each $\mathbb{N}$, but this is not a bijection and hence not an automorphism of $\mathbb{N}$ in $\mathsf{Set}$.)

Mike Shulman (Sep 07 2023 at 16:10):

If what you're after is some way of inducing automorphisms on a subobject, then you may want to consider the stabilizer subgroup of that subobject. In set-theoretic language, if a group $G$ acts on a set $X$, and $S$ is a subset of $X$, for $g\in G$ we define $g\cdot S = \{ g\cdot s \mid s\in S\}$, and then the stabilizer of $S$ is $\mathrm{Stab}_G(S) = \{ g\in G \mid g\cdot S = S\}$. Your hypothesis of the existence of $\bar{g}$ corresponds in this language to asking that $g\cdot S \subseteq S$, which isn't strong enough to yield a subgroup of $G$, but the stronger condition $g\cdot S = S$ is.

In categorical language, $g\cdot S = S$ corresponds to asking that the composite $S \xrightarrow{m} X \xrightarrow{g} X$ is equal to $m$ "as a subobject of $X$", which means that they are isomorphic (necessarily uniquely) in the slice category over $X$. This amounts to explicitly requiring the existence of a $\bar{g}$ that is an isomorphism, so from this point of view we're just solving the problem by fiat. However, the notion of stabilizer of a subobject is a fairly well-studied one, so at least it makes a connection to well-known concepts.

David Egolf (Sep 07 2023 at 16:18):

And thanks @James Deikun for thinking about this! However, I'm still learning the basics of category theory, and it will take me some work to understand your response. If I understand correctly, you are considering the case where $Y$ together with maps $m: Y \to X$ and $\bar{g}: Y \to Y$ is a pullback of the diagram corresponding to the maps $g: X \to X$ and $m: Y \to X$.

Considering the case where the diagram is a pullback wouldn't have occurred to me. I guess I don't have a lot of intuition for when pullbacks are relevant yet, or what they are really like!

I don't understand what this means: "take the unique arrow $h$ from Y to Y induced by the arrows $g^{-1}m$ and $\mathrm{id}_Y$". We have that $g^{-1} \circ m$ is a morphism from $Y$ to $X$, and of course $\mathrm{id}_Y$ goes from $Y$ to $Y$. So I'm assuming there is some standard way of inducing a morphism from $Y$ to $Y$ given a morphism from $Y$ to $X$ and a morphism from $Y$ to $Y$, using a pullback diagram of the form pictured above. However, I don't know what this way is!

EDIT:
Ah, I think I see the idea! The maps $g^{-1} \circ m:Y \to X$ and $1_Y: Y \to Y$ together with $Y$ would provide another cone over the diagram corresponding to the maps $g: X \to X$ and $m: Y \to X$. Then we want to use the universal property of the pullback to induce $h$.

I'll try and see how much further I can get, understanding this.

David Egolf (Sep 07 2023 at 16:27):

Mike Shulman said:

Your hypothesis of the existence of $\bar{g}$ corresponds in this language to asking that $g\cdot S \subseteq S$, which isn't strong enough to yield a subgroup of $G$, but the stronger condition $g\cdot S = S$ is.

I think this was illustrated in that example above, where the action of the $+1$ automorphism of $\mathbb{Z}$ (where $+1: \mathbb{Z} \to \mathbb{Z}$ and $+1(x) = x+1$ for each $x \in \mathbb{Z}$) on $\mathbb{N}$ is such that $+1(\mathbb{N}) \subseteq \mathbb{N}$, but $+1(\mathbb{N}) \neq \mathbb{N}$.

David Egolf (Sep 07 2023 at 18:08):

James Deikun said:

If $m$ pulls back to itself along an isomorphism, then the other projection of the pullback is an isomorphism: take the unique arrow $h$ from Y to Y induced by the arrows $g^{-1}m$ and $\mathrm{id}_Y$, it will be a section of $\overline{g}$ because $\mathrm{id}_Y$ induced it and a retraction of $\overline{g}$ because $\overline{g}h\overline{g} = \overline{g}$ and $mh\overline{g} = g^{-1}gmh\overline{g} = g^{-1}m\overline{g}h\overline{g} = g^{-1}m\overline{g} = m$ so $h\overline{g}$ has the same cone as (and is therefore identical to) $\mathrm{id}_X$.

I think I now follow all of this! The idea that a morphism like $h \overline{g}$ can have a cone confused me for a while. I think the idea is that we can get a new cone from the pullback cone by precomposing at the apex using $h \overline{g}: Y \to Y$. That's pretty cool - I don't think I'd really realized you can get new cones from old ones in this way.

The technique of finding two cone morphisms to the pullback cone, each having the same source cone, and then concluding these two cone morphisms are the same (because there is a unique cone morphism to the pullback cone from any other cone over the appropriate diagram) was also very interesting to see.

(By the way, I think it should be $\mathrm{id}_Y$, not $\mathrm{id}_X$ in the last sentence).

James Deikun (Sep 07 2023 at 18:09):

Ah, yes, you're right. Fixed.

James Deikun (Sep 07 2023 at 18:15):

As for "why a pullback in the first place", a pullback of two monos in $\bold{Set}$ displays the intersection of two subsets, so generalizing from that it seemed likely to work.

David Egolf (Sep 07 2023 at 18:26):

James Deikun said:

Similarly if a square like this exists with $g$, $\overline{g}$ automorphisms, it is a pullback because given a pullback cone $x : Z \to X, y : Z \to Y$ where $gx = my$, $\overline{g}^{-1}y$ induces the cone ($y$ is immediate, $x$ because $m\overline{g}^{-1}y = g^{-1}my = g^{-1}gx = x$). So being a pullback is exactly enough.

And this all makes sense to me as well! Hooray!

So, in summary, the original commutative square is a pullback square if and only if $\bar{g}$ is an automorphism.

David Egolf (Sep 07 2023 at 18:44):

James Deikun said:

As for "why a pullback in the first place", a pullback of two monos in $\bold{Set}$ displays the intersection of two subsets, so generalizing from that it seemed likely to work.

Thanks for explaining! This helps provide some additional intuition (for me) regarding what a pullback is like.