Chapter 2: Orbit-Stabiliser Theorem | Essence of Group Theory
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- Опубліковано 22 сер 2024
- An intuitive explanation of the Orbit-Stabilis(z)er theorem (in the finite case). It emerges very apparently when counting the total number of symmetries in some tricky but easy way. This video series continues to develop your intuition towards some fundamental concepts and results in Group theory.
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There is usually also something called Burnside's lemma closely related to this, but this is usually more important in some arguments when proving results in group theory.
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I've been through 2 graduate algebra courses, and it took until this video to realize the Orbit-Stabilizer theorem was so simple/intuitive: it's just understanding a group action by understanding its inverse: its inverse can be broken down into (1) first move x back to itself (x fixed/chosen before), and then (2) then there's exactly one group action left, that fixes x, and moves everyone else back. Just a simple 2 step process. And fantastic special case/example of the dihedral case! It's a perfect instructive and intuitive animation for the Orbit-Stabilizer theorem.
Thank you very much for this - it’s helped me a lot. I’m currently writing a project for my third year undergraduate mathematics degree on enumeration of 4x4 sudoku grids, and reducing the 288 possible enumerations down using group theory. I knew it reduces down to 2 but I struggled to understand why, and when I was pointed in the direction of the orbit-stabiliser theorem I often struggled to understand it properly. You’ve saved the day. Thank you!
Glad to help :)
Hey I just came across your comment and was curious. How did things ultimately turn out?
Hey, I’m curious to how it turned out as well. I have worked on using Groebner basis theory to solving sudokus as well!
Very well explained. This helped me understand orbit stabilizer theorem A LOT.
Thanks! Glad that it helps your understanding!
I was trying to understand the class equation for abstract algebra, and found your video very helpful! Thank you!
I know it's been a few years, but I was just curious. How did the rest of the class end up going?
Thank you so much!! Been having trouble understanding this section in my uni course but your description of stabilizers finally made everything click for me! Had to pause the video and take a minute and just go "OH MY GOD I GET IT NOW" haha
Really glad that it can help!
Out of curiosity, how'd the rest of the class go?
the best one....it was so confusing to understand this theorem ,this video made it so easy, thanks man.
Glad that I made this theorem easier to understand! It was also confusing to me when I first learnt it, so when I thought of this way of understanding the theorem, I just felt compelled to share it with others as well!
You made me finally understand orbits and stabilisers. Thank you!
Glad it can help!
Not sure if you're still replying for this video, but this is my first time studying group theory, and I'm gonna try to use my own words to summarize to see if I understand it correctly.
1. There are only 2 kinds of action. It either moves x, or it doesn't.
2. For every element x in X, there are certain amount of actions that uniquely change the x's position to different places, and there also are unique actions that not change x's position at all.
3. And any one of the former combined with the latter is also a unique action in G.
4. Hence, the product of the actions that would uniquely change x to different places and the unique actions that would not change x at all is all the actions you have in G.
I don't know of any math people can answer this question for me. Any feedback would be much appreciated, thanks ahead.🙏
I feel like the video is kind of unclear.
There is a group T formed by certain transformations(in this context functions from X to itself. Note that this implies closure and we also have to assume identity and inverse, associativity is implied by functions) of a set X with the operation of composition. T is a subgroup of Sym(X) as all functions considered in T are bijective(assuming inverses) and maps X to itself.
For some reason, we want to consider another group G isomorphic to T. we are going to define group action using that abstract group G that has nothing to do with transformations.
A group action F is a function that takes an element of G(representing a transformation) and an element x of X to get a new element y of X. F: G*X->X Basically it defines a way in which you can make a transformation act on X.there are 2 axioms to satisfy. first F(e,x)=x for all x, e being the identity of G. and F(gh,x)=F(g,F(h,x)) (identity transformation fixes everything and transformation composition correspond entirely with the group operation). we often don't want to think about the whole group action, instead we "fix" a g in G and think about what that specific transformation does to the whole set X(this is called currying like the indian spice). so you can think of a group action as a collection of function X to itself each of which can be described as an element of G. then we can show using the axioms that any transformation is bijective. X=F(e,X)=F(g-g,X)=F(g,F(-g,X)). we can then also define a group action as being the homomorphism of a group G to the symmetric group of X sym(X) the group of all bijective maps X to itself(because G is isomorphic to a subgroup of sym(X)). In summary, a group action is just a way to define how a group corresponding to some transformations of a set may act on that set.
1. obviously an element can either be moved or not moved. in general, an action g permutes all elements of X(because of closure, inverse transformations), i.e. g is a bijective function from X to itself. either gx=x(x is fixed, not moved) or gx exists in X and isn't x(so x has been moved somewhere in X)
2. there may or may not be any actions that change x. for example, just consider the symmetry of permuting a set a,b,c that also fixes a. there may also be more than one actions that fix(don't change) x. everything depends on what set you are considering and what transformations you allow
3 and 4. we want to show the orbit stabilizer theorem. first if F is a set of functions f that all fix x and G a set of functions that don't fix x, then clearly F and G are disjoint. same is true if you consider it for a specific place x can go thus they are all distinct. to show that we didn't miss any, clearly x has to go somewhere(orbit). then we have to place the rest of them(stabilizer) so any transformation has to have an orbit and a stabilizer for x which means that it must have already been considered.
Awesome video. So glad I found this channel! I don't even have any math background and I am able to understand the series so far. Well explained for sure.
Thanks so much for the compliment! Great to hear!
I'm not exactly sure how the graphics package works, but at 0:31 the word "structure" is surrounded by two left quotation marks; if it's being rendered by LaTeX, you'll want to type it as ``structure'' instead (with the last two being apostrophes) to get the quotation marks to face the right way
I’m pretty sure this is Manim, initially created by Grant Sanderson. It’s open source now.
Thank you so much for making video on orbit stabilizer theorem in such a fashion.
Orbit-Stabiliser Theorem? More like "Oh man, superb videos, thanks man!" I've tried to tackle this material several times before, but seeing things illustrated like this was pretty much better than anything else I've ever done. Thanks so much!
If I have an octagon, there will be:
1 identity element
8-1=7 rotations (R1 is just the identity)
8/2=4 simmetry planes that go through the vertices
1 inversion centre
I just don't get how multiplying the number of possible rotations by 2 (identity and simmetry plane inside the stabiliser) gives out the total number of elements (so it takes in consideration the inversione centre and the other 3 remaining simmetry planes).
Also...7*2=14 and I have counted 1+7+4+1=13 elements 😅. Maybe I'm missing the plane that lies on the octagon, but that is equivalent to the identity (just like the R1 that I've not counted).
There's the identity, 7 rotations, and 8 flips. 4 of the flips go through two opposite vertices, and the other 4 go through the center of two opposite edges. The inversion centre is just a 180° rotation in this case.
You sir , now have a loyal subscriber. GReat work
Welcome aboard!
Really clear! Much appreciate!
Glad that it helps!
This concept is a lot easier to learn on the symetries of a cube. 6*4=24
Really good video, great job!
Thanks so much!
It's very inspiring to see the productive followers of the great 3blue1brown.
so symmetry groups Aut(X), card(X)=n is using the orbit stabilizer multiple time where fixing an element is n and the orbit is the same question but of n-1. we get the relation |X_n| = n*|X_n-1| which is the definition of a factorial. for a proof, the orbit stabilizer clearly reaches every possible element of G as every permutation (multiplication by g in G) has to put x somewhere and then count what can happen to the other elements. It can't overcount neither because all permutations are different depending on where x is and the stabilizer.
Certainly my favo urite math channel. Love u mathemaniac. icey =]
Thanks for this educational contribution.
As a beginner, I'm not sure I understand 'fix'? is that mean that under any sigma in G, some vertexes would not change？
Boy, you know how to make something simple... so so complicated
Thank you
Thank you so much!!!
THANK YOU SO MUCH!!!!!!!!
Glad to help!
I don't know if I'm understanding this correctly, but are you counting the order of the group by counting the number of possible inverses, decoding the action of the inverse into that of bringing x_k back to x (though the other elements under the action may still fail to be themselves - i.e., the group element bringing x_k back to x may not be the group element corresponding to the inverse of our original action), then applying the symmetry which fixes x and brings everything else back to its original place, so that the composition gives the unique inverse of the original symmetry?
Now that I think of it, depending on what set of "return" actions you choose for each possible x_k in the orbit, we get a unique decomposition for every element in the group as a composition of something in that "return" set and something in the stabilizer
thank you very much
I have a question! When describing the orbit of the vertex 1 of the octagon, it seemed like we only considered rotations. But I can think of reflections that also fix the vertex! Why aren't these considered?
EDIT: I think my question boils down to: What are r_2, r_3, ..., r_n? There are multiple ways that move, for example, vertex 2 to vertex 1.
Yes, there are lots of ways to move, and we just choose one of them - the argument wouldn't change anyway.
@@mathemaniac Thanks!
The goat.
Bless your soul
thanks from an Indian , much love from India ....
In 10:55 you said this is only true for finite groups, so the orbit-stabiliser theorem doesn't work for circles, cause their |org_g(x)| is infinite, right? So my question is, can a infinite group have finite orbits?
It depends on what your action really is! For the circular symmetry group, if the group only acts on one point - the centre of the circle, then clearly the orbit can only have size 1.
could you not put text in the same space where go the captions? it is very annoying
Will take more care of that in the future.
@@mathemaniac thank you so much! love your video
4:22
For of Group Theory, starting from the discussion of orbit stabilizer thm seems not very usual
This video series is supposed to look at group theory in an unusual perspective though :)
Why is it that we only get 2 stabilizers in the case of |stabG(1)|? Are there not more reflections possible such that 1 gets fixed? For instance a reflection line between 1 and 4...
The reflection along the line joining 1 and 4 doesn't fix the octagon, so it is not a symmetry of the octagon in the first place.
@@mathemaniac thanks!
During the "translation to the right by 2 units" the number line went to the left. Is it just me? I apologize if it is just me. Maybe I'm having a dyslexic moment. By the way, excellent video and thank you very much for this production.
Do you see the "translation to the right by 1 unit" as moving to the left? I think they are going to the same direction though... it might be possible due to frame rate that you view the video in, but anyway, thanks for liking the video!
Why 2 goes to 8 ??
2 was adjacent to 1, so after any symmetry, it also has to be adjacent to 1. There are two options to be adjacent to 1: being at 2 or 8.
Correct me @mathemaniac we move with stabilizer and in free rotation keeping distance same 2 moves to 8
@@naturalchillman Yes: 2 can only move to 2 or 8 because these symmetries keep their distance.
If x goes to the same vertex through 2 different symmetries, then are those symmetries equal? I know this is a stupid and naïve question but it would be great if you replied. Thanks in advance!
Not stupid at all! The answer is not necessarily, but if every point is like the x that you described, not just a particular vertex, then the two symmetries have to equal.
For example, in the octagon discussed in the video, we can consider the identity symmetry, and the reflectional symmetry along the horizontal axis. Vertex 5 goes to vertex 5 in both symmetries, but I think we can all agree that the identity and the reflection are not the same thing. This is because other vertices do not go to the same place under the two symmetries!
@@mathemaniac sorry man but I am kind of new here. I have another question. You say all the symmetries can be fixed through a rotation and a reflection or no reflection. Let's say through R1 and a reflection a transformation created by a symmetry is fixed. So any symmetry's transformation that gets fixed in the same way are equal ( I am assuming this since the number of symmetries are 2n). This implies if through 2 symmetries a vertex is fixed in the same way those symmetries are equal. Maybe I don't understand fully what's going on, please explain me what I am struggling to understand. Thanks!
@@meetjoshi9853 You are in the right track. There is no need to assume the number of symmetries here: as long as two symmetries goes to the identity in exactly the same process (r_1 then reflection in your example), they are the same! This is because we can always reverse the symmetry: the two symmetries are actually just a reflection then r_1 but in reverse!
@@mathemaniac Thank you very much! I understand it now.
After 100 views I might have an idea.
sadly fix in English has 2 meanings
The animation are cool but he needs to be better explained. First of all, what do you mean by "fixing"? Couldn't I just do another symmetry operation and move 1?
Thanks for your compliment about the animation. Fixing means that the vertex stays unchanged after all these symmetries. The point of the whole thing is to identify what the symmetry really is, through reverting back to identity, which does nothing to the figure, i.e. fixing every vertex. So we have got to start somewhere, and so we need to fix 1 first, then fix the others. Of course, you can fix other vertices. I just chose vertex 1 for the video.
@@mathemaniac Thank you, don't take my comment as a bad review, I think this video is well made. But it would be nice if you added more of "why we are doing this", "what's our goal", "why things are in this way".... etc
More generally, aiming for an intuitive explanation
Well done, cheers
This is a nice comment! I didn't take this as bad comment at all! I think that I had sort of implied the things in my comment, but it might not be clear enough. Definitely will look out for this in the future.
@@mathemaniac In the video when you show arrows with word "fixes" above it, do you still mean "keep it where it is", or do you mean "corrects"?
@@orangus01 Fixing a point in this context means that after the symmetries, it stays where it is.
I like your robotic voice.
good video, hate the pacing of your voice
Can you be a bit more specific? Is it too fast or too slow or something else about the voice? It will make it easier to improve.
@@mathemaniac Sounds fine to me. Thanks for making these!
@@mathemaniac perfect video
@@naturalchillman Thanks!