Homeomorphism and the group structure on a circle | Algebraic Topology 2 | NJ Wildberger
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- čas přidán 9. 07. 2024
- This is the full second lecture in this beginner's course on Algebraic Topology. We give the basic definition of homeomorphism between two topological spaces, and explain why the line and circle are not homeomorphic.
Then we introduce the group structure on a circle, or in fact a general conic, in a novel way. This gives a gentle intro to the definition of a group. It also uses Pascal's theorem in an interesting way, so we give some background on projective geometry.
This course is given by N J Wildberger at UNSW, also the discoverer of Rational Trigonometry, and a leading advocate of logically correct thinking in mathematics.
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It is true that any topological space homeomorphic to a circle also has a group structure--this is obvious since we can just use the homeomorphism to transfer the multiplication. The mobius strip is not homeomorphic to a circle, but it can be continuously deformed to a circle.
Finally a good camera man! BRAVO, can you have him instruct other camera people?
Very didactic, rich in examples, clear and interesting to follow (which is not so common among math teachers). Thank you for sharing sir.
That's right, it is better to think of the complex numbers as the rational complex numbers, ie both components rational numbers. A surprising amount of classical mathematics can be done in this more careful, yet number theoretically interesting fashion. See my MathFoundations series, around MF90, where I start discussing the weaknesses with the usual approaches to real and complex numbers.
Thank you for making your wonderful lectures available, Dr. Wildberger.
Spectacular classes, very, very nice job. Please, let the classes going on. Greetings from Brazil.
Dr Wildberger, I am thoroughly enjoying your lectures! Thank you.
With regard to the discussed group structure on the circle, I might be missing the point. Isn’t it trivial once we see that the product is actually the sum of circular arcs: A*B=C means arc(OA)+arc(OB)=arc(OC)? The associative rule is particularly trivial. Am I missing something?
circle group The group structure on a circle is the work of F. Lemmermeyer, S. Shirali found in the Mathematical Gazette, Vol 93 #526, March 2009. Given points A, B, and C on the circle their products are also found on the circle giving the group property of closure.
Group properties are the following:
1) Closure property as mentioned above.2) The point O acts as the group identity such that O*A=A*O=A.3) For every point A there is a point B such that A*B=B*A=O existence of inverse.4) (A*B)*C=A*(B*C), associative law.
For the latter two problems, I get the group multiplication rule as alpha(x1)*alpha(x2)=alpha(x1+x2) for the parabola and beta(x1)*beta(x2)=beta(x1x2) for the hyperbola, the fitness of which are not difficult to check. Together with the case of the circle, all the apparently distinct formulas seem to result from the same geometrical interpretation as drawing the parallel intersects, which is really amazing.
Thanks, nice to hear from you.
Wow. Excellent.
Thank you so much for your videos they're really helpful
Thanks for uploading these.
wonderful presentation!! thanks alots !
Thankyou. Dr. Wuldberger
No. There are many different possible group structures that topological spaces might have. If we want the group structure to be continuous, which is only reasonable, then the possible group structures of a given surface is constrained.
A Mobius band cannot be given a continuous group structure if the boundary is included. If the boundary is not included it is a more interesting question, I do not know the answer off hand.
Your playlist is listed so down in search.
thanks for sharing!
Hi, I use a HD Sony Handicam. But I am pretty sure any HD videocam would do just fine; it might be a good idea also to invest in a lapel wireless microphone, which helps improve the sound quality. Some good lighting to remove shadows is also something to think about. Good luck!
I would love to check my thoughts on the final few problems. Is there a posting of the answers anywhere?
I wish I had a teacher like you
Really nice explanation 👍👌👌👌
Good vid!
Its very good to present math like this on youtube, its great, it makes other people want to upload their mathematics on to CZcams too
well, you do it well too !
Thankyou so much.
Fogive me if this is covered later on in the video but I've just learned that apparently my intuitions about homeomorphisms are mistaken and yet I feel fairly confident in my reasoning.
I'm particularly interested in the dinstinction between the topological spaces and the symbols we attach to such spaces because I think it's of vital importance to do so. In a prior video, you discussed the idea of an ant travelling from A -> B, and then when it gets to B we move it onto a separate interval of B -> A, and you said that such a space could be simplified as simply A->A, but I feel that these are completely different because one only has a single distinct point of reference and the other has two.
When you discuss the idea of an interval [0,1) earlier on in this video, you say that it isn't homeomorphic to a circle because 0 and 1 would map to the same point, but again I feel like the devil is in the details because suppose that you ONLY identified the points 0 and 1: by specifying '1)' you're removing it from the domain, meaning that it isn't even a point on the interval to begin with; it's a contradiction. If instead you just had a 'topological segment' or 'topological interval' from 0 --- 0, or just 0 -- , then I would think it would be homeomorphic to a circle with one point 0 along its curve. A topological interval from 0 --- 1, or A --- B however would be its own thing.
With this in mind, I want to say that I share your aforementioned criticism of the real number line and I think that 0, negative infinity, and positive infinity can each be considered to be such points of interest along a topological interval like A --- B --- C, but I find great peace of mind in the idea that infinity should be treated as a single point much like zero, where its sign is really just representative of the direction from which it's approached, this means that the real number line or the positive and negative integers (not strictly the real number line but if confusing I can clarify later) is homeomorphic to a circle with two distinct points of interest at 0 and infinity. In other words, we can treat it like the pseudo-interval A --- B --- A and hence map it to a circle.
What this means is that homeomorphism is heavily dependent upon these 'points of interest', and is not so much dependent on the shape of the interval.
To illustrate this point: suppose you had a unit circle with two points of interest on it A and B -- for clarity's sake, place them on opposite sides (I apologise for not having a diagram). If we characterise two transformations A and B on such a space that creates the new space with two points of interest on it AA and AB for example (like matrix operations) on the same unit circle. and we define a AA = A, AB = BA = B, and BB = A, then we see that a transformation of A creates the same space with the same positioning, and a transformation of B flips the points of interest. This, I feel, is really what defines negative and positive numbers. We can extend this notion to four points, A,B,C,D, define a similar matrix-like relationships and where AX=X, BB = A, AB = BA = B, DD = B, etc. ensuring that all possible operations collapse into one of the original 4 points, and there we have defined the complex numbers -- we can observe that multiplying by C or D 'rotates' it by 90 degrees. Again we can extend such a space to higher degrees of notable points and occupy a topological surface instead and define the quaternions, etc. I imagine doing so ad nausium could create an arbitrarily precise rotatable space, but I'm yet to find that out.
Anyway, I just wanted to convey my thoughts on the matter; thank you for such great lectures
Well, I didn't manage to figure out how to apply the Pascal's THM at the special parallel case into our proof of the associativity. I thought we might first have to pick out 6 from the totally 7 points (O, A, B, C, A*B, B*C, A*B*C), in order to use the THM. Could someone please show which 6 should be selected?
Thank you Dr. Wildberger. What I mean in the first part is, Does every topological space homeomorphic to circle should have a algebraic group structure. The reason behind my thought is there will definitely exist an inverse to every element on a circle and any point on the circle can be taken as reference(identity) and associativity and closed set property are valid for a circle.i believe mobius strip will not be homeomorphic to a circle as it cannot be continously deformed to a circle.
It's really interesting ...and the way you teach the subject is very nice
And nice to here 16:30 Indian name...(highschool teacher).
circle algebra - to do A*A technically you don't need this assumption with tangent, you can set any point B and use the rule (A*A)*B = A*(A*B) :)
@njwildberger True. I didn't take into account the background. I was just surprised that it turns out to be the same thing!
Hi sir, just a little question at around 43:00, you drew the parallel hexagon to show the projective plane lets parallel lines meet at infinity. Why did the lines now meet externally rather than like the previous example where the intersections are inside the conic?
Since I, personally, had trouble making the connection between associativity and Pascal's Theorem, here is the corollary:-
Proof that (AxB)xC = Ax(BxC)
In the (possibly) intersecting hexagon A, B, C, AxB, O, BxC, A we have
1. AB//O(AxB) by construction
2. BC//O(BxC) by construction and since 2 pairs of // lines exist then
3. (AxB)C//A(BxC) the third pair must also be // by Pascal’s Theorem hence
O(AxB)xC//OAx(BxC) by construction. But both these lines pass through O
So they must be coincident lines, hence the points (AxB)xC and Ax(BxC) must also be coincident on the circle. Hence (AxB)xC=Ax(BxC) QED
@maximuspower2 That's right, but this way only involves a bare minimum of geometry. Think of how much work would be required to define properly the exponential function, angles etc (rarely done)... there is a big difference.
At 47:57, what about the case h_1=1/h_2? Anyway, after algebrizing the geometrical group structure myself with basic Cartesian geometry I've got that h_3(h_1,h_2)=|(1-h_1h_2)/(h_1+h+2)| if h_1+h_2
eq 0 and h_3(h_1,h_2)=0 if h_1+h_2=0. And, with this, I think that the correct approach to define algebrically this circle group would be to firstly define your function ''e'', take its image as being the set/realm of the group and on this, define the operation A*B=e(h_3(e^{-1} (A) ,e^{-1} (B))), show that this definition is well defined (which is equivalent with showing ''e'' is bijective), and verify the group laws. EDIT: this would be a great setup if we would have a proper theory of functions of realms.
@maximuspower2 That's a good question, I don't know the answer.
The problem 4 at 48:13, shouldn't be $h3 = \frac{h1*h2-1}{h1+h2}$ instead of $h3 = \frac{h1+h2}{1-h1*h2}$ ?
Actually, I was wondering-- can you think of a simple way adapt this way of thinking about the group structure to S^3, S^7, etc?
49:45 Is that particular group structure connected to the lemma you used for the derivation of the Triple quad formula? Namely that Q_1*Q_2=Q^2 - meaning that multiplication of quadrances is closed under multiplication. I still have some trouble with this lemma - because I tried to understand it without the obvious use of lengths requiring square roots, namely sqrt(Q_1)*sqrt(Q_2)=Q iff Q_1*Q_2=Q^2. You said the lemma is obvious showing a picture - but what's a quadrance squared supposed to mean?
Thank you Prof. Wildberger for all of your great and very interesting videos! I'm going through these lectures now, and I'm wondering what your opinion about category theory is. You might touch on that in this series but I'm not sure at this point.
And on a totally unrelated note, I wonder what your opinion of geometric algebra and geometric calculus is.
Finally, an alternative approach to the group structure of a circle could use rotation matrices expressed in terms of the rational parametrization of the circle. And now I see that is exactly what you used for Problem 4!
Hi Connor, I think category theory is a useful meta language in which to discuss large scale mathematical structures, but it does not yet hold as a logical theory within maths on account of the set theoretical ambiguities. As for geometric algebra, that is also tricky. I will eventually discuss that.
Wow very nice presentation of the group structure on the unit circle. However, I think it's the same as the regular group structure: namely, if e^{i\theta_1}, e^{i\theta_2} are on the circle, that group structure will have the same multiplication, e^{i\theta_1}e^{i\theta_2}=e^{i(\theta_1+\theta_2}.
@andresgoens Keep going, you might find that they are actually reasonably mathematical after all.
dear N j wildberger does this mean only sets with group structure are homeomorphic to circle does a mobius have a group structure if yes then is a mobius strip is homeomorphic to circle(disc).
Thanks for posting these video, if one wanted to follow this course is there a Text that goes along with the lectures?
Colin Merrick Sorry there is not at this point.
Thanks
hey professor can you please explain me what would happen to the inverse if i take A the point on the circle just opposite to the O like we can get two tangent on a circle which are parallel to each other and let say that point is A is that point then what will be its inverse of A may be it is the same point, am i right?
Yep! In other words... That Point has order two.
Do I understand that the circle space is not contractible because, if you fix the initial point of the string and take up the slack on the free (running) end at the origin, it will not collapse, whereas if you do the same for the interval it will do so?
That's the right idea.
The circle analogy for a group structure works only for Abelian groups as the line AB has no directionality, or am I missing something?
Later on in the video he specifies that all groups mentioned in the lecture are commutative
great lectures, but do you have a book where you emphasize those courses? thank you
@maximuspower2 Yes, they're the same. It's easy to show (i) If C is the complex product of A and B, then OC || AB, and the converse i.e. (ii) If OC || AB, then C must represent the complex product of A and B, which should complete the proof of equivalence between the two approaches.
However, it's interesting that the professor wants to avoid 'irrationalities' while sitting on one i.e the circle that 'revolves' around pi. A pi-less 'perspective' on circle, though an illusion, is interesting.
In the given (A*B)*C=(A*B)*C example, the group structure does not look like to be convex, but then, what are conditions to have a convex group structure ?
Olivier Hault There is no notion of convex group structure that I know of.
i think you did point that Circle -[1 point] is meromorphic to line A1 in the previous video
very very usuful for a beginer as me
interesting
They in in shrinkable is the question.
@abstractUser Yes I already said that they are the same. However, you probably shouldn't judge his views on rationals unless you fully heard his reasoning (or any other ideas anyone has).
Around 18:00, what does it even mean to lines be parallel in this context? If the circle is defined as circle only in topological sense does it (necessarily) lie in another space? And if not, how can a line occupy the points not defined the topological circle, that is the 'space' that is not there. I'm sorry I can't formulate my question more rigorously.
Bit late, but: they are parallel in the context of Euclidean geometry. You can find an algebraic expression for A*B by defining the unitary circle S^1 as the set of (x,y) coordinates in R^2 that satisfies x^2+y^2=1, and then using the standard techniques of cartesian geometry to find the point that satisfies the property. Q^2 works too: take two rational points in the unit circle, and then the line that passes through them both will have a rational slope, and cut the y axis in a rational point as well. Then if O is also rational, A*B is rational.
You are a wonderful professor
Excuse me, in fact I have two questions
1. What is the name of Topologic property that we used for
comparison between the circl and line (the worm path).
2. in any way be a space of linear lines homeomorfic
with circle because the line does notthe same as point topologicly
Hani Mechri
it is called contractibility.
R is contractible but S1 is not.
Note however a unit disk is contractible.
the Indian high school teacher thing got me.
I second that!
intetesting..
12:20
the natural question from 43:08 is provable without pascal theorem;) it only requires math from the 6th grade, no need for points at infinity:) also aren't there suppose to be 2 lines at infinity, for this particular case??:) (43:54)
Depends on what you mean by the meaning of is, I mean same. Bill Clinton?