Shadowing Practice: The category of G-Sets - Learn English Speaking with Video

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Hi everyone, I'm John Turula, one of the authors of Topology, a Categorical Approach.

In this video, I'm going to talk about the category of G-sets for a fixed group G.

My motivation for talking about the category of G-sets is topological because the category

of covering spaces of a topological space B is isomorphic to the category of G sets, where G is the fundamental group of the base space.

So you can understand everything about covering spaces in topology by understanding the category of G sets for a fixed group G.

Now what is this category?

Well, the objects are sets together with a function from S cross G into S that's compatible with the group operations.

Here I've written out what it means.

It means that if I act on an element S in my set by a group element G and then by H, it's the same as first multiplying the group elements G, H, and then acting on the set.

And the second condition is that the identity element of the group acts as the identity on the set.

And it's worth pointing out that here I'm considering right group actions.

Those are the ones relevant for the theory of covering spaces.

You could also consider the parallel discussion for left group actions.

Now, what are morphisms between G sets?

Well, a morphism from a G set S to a G set T is a function from S to T that's compatible with the actions, and I've written here what that means.

Now there are a couple of important definitions to know when you have a G set.

So suppose S is a G set.

One is the stabilizer of an element of that set.

That's equal to the set of all elements in the group that fix the element in the set.

Now you can check that the stabilizer of any element is a subgroup of your group G.

And the other important definition is what is the orbit of an element A from your set S.

The orbit of A is just the set of all elements in your set S that can be obtained by acting on that one element A by an element of the group G.

So we have these two important definitions associated to an element in your set in your G set S.

One is the stabilizer of that element, it's a subgroup of G, and the other is the orbit of that element, which is a subset of S.

There's one more piece of terminology I want to share, which is what it means for G to act transitively.

This is the same as saying that the entire set S consists of one orbit.

And more generally, the orbits of S partition the set S, which proves this theorem, that every G set decomposes into the disjoint union of sets on which G acts transitively.

And so now we can ask the question, what is a typical G set on which G acts transitively look like?

Well, every G set S on which G acts transitively is isomorphic to G mod H as a G set,

where G mod H is the set of right cosets of a subgroup H of the group G, and in fact H is related to the set because it's the stabilizer of an element in that set.

How do you prove it?

Well, start by taking any element, little s, from your G set capital S, and let H be the stabilizer subgroup of that element.

Then I'm going to define a map from

the cosets of the stabilizer, a map from G mod H, to the set S by sending the coset H times a little G to the element A acted on by G.

Now, you have to check that this map is well-defined, it's not hard,

from which it will follow pretty easily that it's a bijection, and then you have to check that it's equivariant with respect to the g action, where the g action on the set of cosets is just defined by right multiplication.

Here in my notes, I've written hg goes to ag.

I should have written hg goes to sg.

Anyway, there's a detail in this proof that I think is worth looking a little closer at, which is that we picked h to be the stabilizer of one element s.

The stabilizer of two different elements in a G set on which G acts transitively, let's say A and A prime, are conjugate subgroups.

In other words, if A prime can be obtained from A by the action of little g, then the stabilizer of A prime is G inverse times the stabilizer of A times little g.

Stabilizer subgroups are conjugate subgroups in the group G, which implies that as G sets, G mod H and G mod G inverse HG are isomorphic as G sets.

Now there's another detail that's worth pointing out, which is namely that if you have a G set S on which G acts transitively, we know that it's isomorphic to G mod h where h is the stabilizer subgroup.

But we know how many elements are in g mod h.

We know how many cosets there are.

And so you get a theorem which is known as the orbit stabilizer theorem.

A great but really useful theorem in mathematics.

And what it says is when gx transitively on a set s, then the cardinality of the orbit of any element times the cardinality of the stabilizer of that element equals the cardinality of the group G.

So we're starting to get a really good picture of what the category of G sets looks like.

So the first observation is that every G set decomposes into the disjoint union of G sets on which G acts transitively.

In this category, disjoint union are co-products and sets on which G act transitively are sometimes called homogeneous G sets.

So all the objects of this category are co-products of homogeneous G sets.

then each homogeneous G set itself is isomorphic to one of the form G mod H for a subgroup H of G.

So that tells us what the objects are in this category.

And now we just need to look a little closer at the morphisms, which we know are just equivariant set maps.

But then the question is, does that have a nice description in terms of the group structure of G?

and we'll find out that yes indeed it does.

So I think the first observation to make is that if you have an equivariant function between two homogeneous G sets,

because G acts transitively on those sets, I only need to tell you what happens to one element.

So if you have a morphism of G sets from G mod H to G mod K, I just need to tell you what F does to one coset and I might as well do that for the coset h.

So h is going to have to go to some coset of k.

Let's say it's k times g for some g in the group g.

So this entire equivariant map f is determined by one element little g in the group.

Now I'm actually defining a map on these equivalence classes and so not every element g is going to actually give a well-defined map.

You can check that in order for this map f determined by this element little g to be well defined, g is going to have to satisfy a condition related to the subgroups h and k.

And in fact, the condition, the way I like to write the condition, is that if you conjugate the subgroup h by g, you have to land inside the subgroup k.

And there's some redundancy in this description of equivariant g maps between g mod h and g mod k by elements of G.

In particular, you can check that if the coset KG and the coset KG' are the same, then the two maps you get by sending H to KG and H to KG' are identical.

Now, let's just put that redundancy aside for a moment and make the following conclusion,

that there exists a morphism of G sets from G mod H to G mod K, if and only if there's an element, little G, in the group so that G inverse H, G sits inside the subgroup K.

Now let's consider the case that the subgroup K equals the subgroup H, so that we're considering the group of automorphisms of G mod H as a G set.

Now considering the redundancy, what we find is a very nice theorem,

namely that the group of automorphisms of g mod h as a g set is isomorphic to the normalizer of the subgroup h modulo h.

And the proof is just an application of what we've already described as the morphisms between two homogeneous G sets of the form G mod H and G mod K.

So in other words,

substituting K equals H into those discussions tells us that the map from G mod H into G mod H is determined by one element in the group G,

and that element in the group G has to satisfy G inverse HG sits inside of H.

Moreover, two such maps will be equal if the coset HG equals the coset HG prime,

which is the same as saying that G times G prime inverse is in the subgroup H.

And the conclusion then is that the set of morphisms divided by the ones that are equal give you the normalizer of H modulo h.

And then you can check that this is an isomorphism of groups.

The automorphisms of a g set always form a group, and this quotient of the normalizer of h by h is a group, and these are isomorphic groups.

And so this completes all the details that I wanted to give that gives a nice description of the category of G sets.

First, every G set decomposes into the co-product of homogeneous G sets.

The homogeneous G sets look like the quotient of G mod a subgroup H.

At the very beginning, G mod the identity, just G, maps to every other G set because the identity sits inside of every subgroup H.

All of these G sets map to the trivial G set, which is what happens when the stabilizer subgroup is all of G.

And then you'll want to know what the automorphisms of one of these homogeneous G sets are.

And so if you fix a subgroup H, the automorphisms of G mod H are just isomorphic to the group, the normalizer of H mod H.

So that finishes the details of what I wanted to say about the category of G sets.

You can read about group actions in most any algebra book.

I wanted to pull together all the just the little pieces in a simple way that give you this picture of the category of G sets because this is equivalent to the category of covering spaces where you apply where G is the fundamental group of the base space.

And so to really understand the theory of covering spaces you have to understand this category of G sets.

That's all for this video.

Thanks very much for your attention.

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Woordenschat en spreektips bij deze les

Deze spreekles op niveau C2 is gebaseerd op de video “The category of G-Sets”. Deze woorden komen het vaakst terug: element, mod, subgroup, stabilizer, category. Deze video bevat 95 zinnen en 1917 woorden om na te spreken. Het gesproken deel duurt 12:20. De spreker praat in een natuurlijk tempo van ongeveer 155 woorden per minuut, dicht bij een alledaags gesprek. Slechts 71% van de woorden hoort bij de 3.000 meest gebruikte Engelse woorden, dus de woordenschat is pittig.

Belangrijke woorden in deze video

De 15 moeilijkste woorden uit de video, met uitspraak en betekenis:

Woord	Uitspraak	Betekenis
stabilizer zelfstandig naamwoord	/ˈsteɪbəlaɪzɚ/	stabilisator
homogeneous bijvoeglijk naamwoord	/ˌhoʊ.moʊˈd͡ʒi.ni.əs/	homogeen
isomorphic bijvoeglijk naamwoord	/ˌaɪ.soʊˈmɔɹ.fɪk/	isomorf
morphism zelfstandig naamwoord	/ˈmɔɹfɪzəm/	morfisme
orbit zelfstandig naamwoord	/ˈɔɹ.bɪt/	baan
inverse bijvoeglijk naamwoord	/ɪnˈvɝs/	tegengesteld, tegenovergesteld
theorem zelfstandig naamwoord	/ˈθiərəm/	theorema, stelling
cardinality zelfstandig naamwoord	/kɑɹdɪˈnælɪti/	kardinaliteit
conjugate werkwoord	/ˈkɒn.d͡ʒə.ɡeɪt/	vervoegen
decompose werkwoord	/ˌdiːkəmˈpəʊz/	ontbinden
redundancy zelfstandig naamwoord	/ɹɪˈdʌndən(t)si/	overtolligheid, redundantie
quotient zelfstandig naamwoord	/ˈkwoʊʃənt/	quotiënt
topology zelfstandig naamwoord	/təˈpɑləd͡ʒi/	topologie
compatible bijvoeglijk naamwoord	/kəmˈpæt.ə.bəl/	verenigbaar
satisfy werkwoord	/ˈsætɪsfaɪ/	voldoen, voldoende zijn

Phrasal verbs die je zult horen

Woord	Betekenis
point out werkwoord	aanwijzen, aanduiden
find out werkwoord	ontdekken, onderzoeken

Grammatica in deze video

De structuren die de spreker het meest gebruikt, met de exacte woorden uit de video:

Structuur	In de video
Lijdende vorm be + voltooid deelwoord — het gaat om wat er gebeurt, niet om wie het doet	can be obtained · is just defined · is known
Betrekkelijke bijzinnen who / which + zin — extra informatie over een persoon of ding	element, which is · share, which is · out, which is
Present perfect have/has + voltooid deelwoord — iets uit het verleden dat nu nog telt	I've written · we've already described

Uitspraak om op te letten

De spreker gebruikt 13 samentrekkingen en verkorte vormen, zoals I'm, I've, we're. Spreek ze kort uit, zoals je ze hoort.

De klanken “sh” en “zh”: quotient /ˈkwoʊʃənt/, bijection /baɪˈd͡ʒɛk.ʃən/, multiplication /ˌmʌltɪplɪˈkeɪʃən/, partition /pɑɹˈtɪ.ʃən/
Lange woorden — let op de klemtoon: stabilizer /ˈsteɪbəlaɪzɚ/, homogeneous /ˌhoʊ.moʊˈd͡ʒi.ni.əs/, isomorphic /ˌaɪ.soʊˈmɔɹ.fɪk/, cardinality /kɑɹdɪˈnælɪti/, redundancy /ɹɪˈdʌndən(t)si/

Klanken die Nederlandstaligen lastig vinden:

/æ/ — opener dan de Nederlandse “e”: cardinality /kɑɹdɪˈnælɪti/, compatible /kəmˈpæt.ə.bəl/, satisfy /ˈsætɪsfaɪ/, algebra /ˈæl.d͡ʒɪ.bɹə/
/g/ — een harde plofklank, geen Nederlandse “g”: subgroup /ˈsʌbˌɡɹuːp/, conjugate /ˈkɒn.d͡ʒə.ɡeɪt/
/w/ — met ronde lippen: quotient /ˈkwoʊʃənt/, equivalence /ɪˈkwɪv.ə.ləns/

Zo oefen je met deze video

Luister de hele video één keer zonder te spreken en noteer de woorden die je niet kent.
Begin op 0,75× snelheid, spreek zin voor zin na en ga terug naar normale snelheid zodra het makkelijk gaat.
Neem jezelf op en vergelijk met het origineel; let daarbij op woorden als stabilizer, homogeneous, isomorphic.

Wat is de Shadowing-techniek?

Shadowing is een wetenschappelijk onderbouwde taalleermethode die oorspronkelijk is ontwikkeld voor professionele tolkentraining en gepopulariseerd door polyglot Dr. Alexander Arguelles. De methode is eenvoudig maar krachtig: je luistert naar native Engelse audio en herhaalt het onmiddellijk hardop — als een schaduw die de spreker volgt met slechts 1–2 seconden vertraging. In tegenstelling tot passief luisteren of grammaticadrills, dwingt shadowing je hersenen en mondspieren om echte spraakpatronen tegelijkertijd te verwerken en te reproduceren. Onderzoek toont aan dat het de uitspraaknauwkeurigheid, intonatie, ritme, verbonden spraak, luisterbegrip en spreekvaardigheid aanzienlijk verbetert — waardoor het een van de meest effectieve methoden is voor IELTS Speaking-voorbereiding en echte Engelse communicatie.

Shadowing-techniek: lees de volledige stap-voor-stap-gids →