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How to Get to Galois Theory Naturally
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- čas přidán 12. 07. 2024
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Tom Leinster Course Notes
www.maths.ed.a...
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It seems like in general, the video is confused between the group of arbitrary permutations of the roots and the Galois group of the polynomial. The first is not a very useful, as it always gives S_n for any degree n polynomial. The second is much more useful.
At 1:54 you say "all the ways the possible ways the roots can be swapped without changing the equation itself." You eventually say this is known as a symmetry group and it seems like you are talking about the first group I mentioned. How could the equation possibly change when you permute roots?
Around 6:00, you are talking about x^3-1, and say the symmetry group is S_3. And at 6:15 you say "What if I told you that these symmetries can tell you..." But this is the wrong group; the Galois group would tell you this. You go on to say THIS S_3 is an example of a Galois group, but as you noted in another comment, this is not the Galois group of this polynomial.
Note that you do not define Galois groups in this video, but at some point you start talking about it as if the viewer knows what it is.
I see, would really like to see examples of solvable groups and how exactly do we decompose them.
@zcherradi2 ok, we will include more concrete examples in the next videos. Thanks for the constructive criticism 😎
just buy a math book bro 💀💀
@@weirdo911aw sure but the math videos on CZcams are plenty and good. If this CZcamsr adds his own, itd be welcomed by us in the math community.
@@weirdo911aw any suggestions?.
a good first example is C2 (the cyclic group of order 2), which is coincidentally the Galois group of the polynomial x^2 + 1 over Q
Really nice video! Small amount of feedback, part of what makes this video so good is the fact you give some really nice examples as you are going through the video, until you get to talking about solvable groups and unsolvable groups. This really only feels glaring because the rest of the video does a great job of introducing something and elaborating on it and suddenly doesn't at the end when the un-solvability of the quintic is brought up. That being said, I think this is the best introduction to the topic that a person could possibly do in 9 or so minutes.
You're content is great thank you for sharing!
Thank you so much for the detailed feedback and the nice words! Yes, other people commented about this part of this video being a little off… anyway, good to know for next videos. We will focus on leaving only the essential and removing what is extra 😉😎 Please, let us know what kind of content you’d like to see in the channel
One thing that should be noted is that the Galois group of a polynomial does not consist of ALL permutations of the roots, but only of those permutations that lead to an automorphism between the field extensions (you have to explain field extensions to make this point clear).
Roughly speaking it means that swapping the roots, will give rise to the same algebraic structure. It’s very intuitive if you compare it with rotation symmetries: the object should be the same after rotation, for the rotation to be a symmetry.
So in the case of x^3 = 1, the symmetry group is not S3, because you cannot swap 1 with omega and preserve the algebraic structure.
The root x = 1 does not even generate a field extension, because it’s already in the field from which you picked the coefficients (Q, the field of rational numbers).
The only permutation that is an automorphism (a symmetry) is swapping omega and omega^2. But this is the same as complex conjugation and you end up with the same symmetry group (Galois group) as for x^2 = -1.
The Galois group of a polynomial of degree n is always a subgroup of the permutation group Sn, but not necessarily Sn itself.
@@SanderBessels hm… interesting 🤔 thanks for explaining it more precisely
That was the most motivated (natural) explanation of Galois theory I’ve ever seen. Bravo!
@@michaelzumpano7318 Grazie mille!! Let us know what other (natural) explanations you’d like to see in the channel please 😊😎
I've never studied this theory before but those explanations are so good I got interested.
@jeanlucas2834 thanks Jean Lucas!!! I hope you like our other videos as well. 😎 Please let us know what kind of content you’d like to see on the channel 🤩
It is actually pretty advanced, you need a solid foundation in abstract algebra, so if you are not into pure math you are unlikely to see it in your journey
@@sebastiangudino9377 yeah, it is true. However I think group theory and Galois theory are very important anyway. And they do have applications in applied math, physics and chemistry, so even if you are not deep into math it is not crazy to imagine that you may stumble on these concepts eventually 😎
man how did i only found about this channel today! the intro is amazing
@@aziz0x00 I don’t know!!! Anyway, we’re really happy you liked it! Let us know what kind of content interests you so that we can post videos about it here 😎
Great video. I wish you spent more time on solvable groups but this is still a very interesting introduction.
Thanks! Just for us to improve our content for next time: what exactly do you think was missing in the video?
Great timing, ive been trying to understand constructable polygons and this video came up
@@masoncamera273 that’s awesome! I’m glad you liked!! Let us know what content specifically you’d like to see in the channel, please 😎
Great video, grateful to have this information available! It would be nice to delve more into the details of galois groups and solvability.
@@kallek9645 hi! Thanks for the nice words and the suggestion. We will definitely delve into these subjects since other people asked us the same. We will include in the list of next videos in the channel 😎
For those of you not inderstanding well enough to their faste: the link between group and polynomial is clearer once you learn about Fields, homomorphism and Galois correspondance
Great invitation otherwise, it is thanks to content like yours I learned deeply about Galois
@@saltydemon7107 thanks for the nice comment and the insight!! We are planning to make another video where we show clearer the connection between Galois theory and symmetries, since many people in the comments seemed confused. Thanks!!!
@@dibeos REALLY??? I shall wait for it then! The truth is, I am planning to make my own documentary about Galois Theory, I will wait yours, just to make sure I can bring another point of view and another way of seing Galois!! That is the main problem with the Galois documentaries.... All of them tend to explain in the exact same maner....
Can't wait to see how you'll manage to explain it!! You do have something to bring !!
@@saltydemon7107 yes, and please put the link of your documentary as soon as it’s ready in a comment here in my channel so that I can watch it 🥸
Great video, and it is nice to see that you studied Physics in USP, tamo junto mano!
@@Leo-if5tn I’m glad you liked!!! Oh yeah, tamu juntu! 😎
A majestic presentation of this topic!
@@jonathanlister5644 thanks for your majestic comment, Jonathan 😎 please, tell us what kind of videos you’d like to see in the channel, this way we can search, create a cool explanation and post a video about it
Thanks for the great videos guys. I can already see your content improving, and that's really exciting.
Something I'd like to add is that we can easily conclude that degree n equations are not solvable by radicals for n >= 5. This is because S_5 embeds into S_n, when n>=5, and thus S_n will contain a copy of A_5. Hence S_n is not solvable.
@@thedude882 thanks for the nice comment! It really encourages us to keep going. Interesting remark about S_5. It does contain a subgroup isomorphic to A_5. I would really like to see a rigorous proof though of this and of the fact that it cannot be solved by radicals. I might even make a video about it… 🤔
@@dibeos A_5 is defined as the set even permutations of S_5, so it's going to be a subgroup.
S_5 -> S_n, where n>=5, is an injection, so A_5 is isomorphic to its image in S_n, and the image is a subgroup in S_n.
If a group is solvable, so are its subgroups. But then S_n is not solvable for n >= 5, as it contains a subgroup isom to A_5 which is not solvable.
This channel deserve Million subscriber
@@Satisfiyingvideo-uu9pw we will get there. We will grow together with our audience (you guys) learning and doing math 😎
Correct me if I am wrong, but isn’t the Galois group of x^3-1 supposed to be Abelian? In which case it’s symmetry group can’t be S_3? 🤔 great video in any case! 😎
Hi CurlBro, thanks for the comment. The Galois group of x^3-1 is Abelian (C_2, because it consists of only two automorphisms (the identity and complex conjugation) that map omega to omega^2 and vice versa, forming a cyclic group of order 2).
The symmetric group of the permutations of the roots of x^3-1 is S_3.
I hope this answers your question 😎
@@dibeos Hi dibeos, yes it does! Thanks for the quick reply. 😎
@@CurlBro15 😎
Another confusion: complex numbers are not necessarily imaginary numbers. Please check your definitions.
@@YQ-jerzy you’re right. I just used the same term in order to simplify the concepts taught in the video. If I were to define every word I said the video, it would have turned out to be too long and complicated… unfortunately it was a necessary sacrifice. But thanks for reminding everybody 😌
Sir,
If I place my hand flush against the wall and try to push the wall, does it mean I am applying force on the wall.
Since FORCE = MASS X ACCELERATION
ACCELERATION = VELOCITY / TIME
VELOCITY = DISPLACEMENT / TIME
Since there is no displacement between my hand and the wall therefore velocity becomes 0 and as a consequence Acceleration becomes 0 and hence
Force becomes 0. So does that mean I am applying 0 force on the wall ?
Am I right or wrong Sir, please explain where I've gone wrong if I did.
I searched a lot on the Internet and CZcams but didn't find suitable answer for situation mentioned above
Because of Newton’s third law. When you push on the wall, your hand will also experience a force back onto it (normal force). The force is there, but since there is an equal force opposing it, they cancel out and thus no velocity. An object will only move if the forces are not balanced. For example, when you sit in a chair, you are not moving (relative to the chair), because the normal force and your weight force cancel out (balanced forces). If the normal force did not exist,(i.e the force making you sit upright), you would fall through the chair and the ground. You can use F=ma to find the ‘winning’ force if an object is moving, but you would still need to take into account other forces acting on the object. I hope this makes sense.
@@gant-arro I don't think this is correct. Newton's third law always applies. So, by your arguement, pushing a book on a table would not move, because Newton's third law would give an equal and opposite force.
I believe your confusion is on what is the opposite force applied to. It is applied to the object creating the initial force. So in the original example, the opposite force is acting on the hand, not the wall.
The actual reason the wall does not move would be due to structural internal factors. Just like you need a lot of force to push a tough spring, you need a lot of force to move a wall.
@@willnewman9783 when you push a book on a table (for the sake of the argument assume that the force is perpendicular to the table), the force you are pushing against comes from friction (and the weight force, but assume that it is light so we can ignore it). If the coefficient of friction is insanely large, then yes you would not be able to push the book and you would experience the normal force. It is also important to note that you do experience a normal force when you slide a book along the table (if you punch it really hard it will still hurt), but the force you are pushing the book with will still be greater than the normal force, hence the movement. The reason why it is difficult to push a wall down IS because of the internal structure, giving it enough resistance to withstand a small amount of force. Look up ‘resolving forces’ for more info
With respect, I don't like this video. The obvious problem is that I don't see how Galois groups can be something not in the form S_n for some n. If you say that "for any polynomial its Galois group is the group of symmetries of swapping the roots" then obviously if a polynomial of degree n has distinct root, its Galois group is S_n! If you look at the Tom Leinster Course Notes, in definition 1.2.1 there is a condition that such a permutation must satisfy something, i.e., the permuted tuple must be a conjugate of the original one. (Which is maybe not so easy to explain but I think it's worth mentioning that there is such condition -- not all permutations work)
This brings us to another obvious problem: I don't see anything natural in Galois theory. Why do we consider structures like symmetric groups? Why does solvability of groups have anything to do with solvability of polynomial equations? So many questions are left unanswered and unmotivated. (To be honest I only click this video because I want to know a natural motivation for Galois theory, which I failed to recognize while studying)
So, yes... this is my review. 🙂
Thank you for your feedback! It is really thoughtful. I appreciate you taking the time to watch the video and share your concerns.
To address your first point, the Galois group of a polynomial doesn't always have to be the symmetric group S_n. For instance, the Galois group can be a proper subgroup of S_n, depending on the specific symmetries of the polynomial's roots. It is true that for a generic polynomial of degree n with distinct roots, the Galois group can be S_n, but there are many interesting cases where this is not true, and the Galois group can be something else.
Example: x^4-2 => roots: (2)^(1/4), -(2)^(1/4), i(2)^(1/4) and -i(2)^(1/4). The Galois group of this polynomial is actually the Klein four-group V_4 (which is a proper subgroup of S_4).
Regarding the second point about the natural motivation for Galois theory, it comes from solving polynomial equations and understanding their symmetries. The connection to solvability by radicals and the use of symmetric groups is a crucial part of the theory, but I agree that it is not easy to grasp at first (at least for me it was not). I highly recommend looking into other specific examples where the Galois group is smaller and seeing how this affects the solvability of the polynomial.
I'll keep your points in mind for future videos though to better address these foundational questions (maybe with more examples to illustrate the intuition behind this connection between symmetries and polynomials). Thanks again for your comment! 😊😎
Can you consider doing a video on category theory.
@@user-wr4yl7tx3w we already made one on Category Theory 😎
Here you go! @user-wr4yl7tx3w czcams.com/video/mKixqJ9xnRM/video.htmlsi=aY3l6VeuNzChzKGj
I have a physics question. Suppose that we have a pendulum. Now it is a well-known fact that due to the conservation of energy the bob cannot swing higher than it's initial height. Let the bob be charged with some charge on it. Then if we let the pendulum swing and make sure the charge doesn't go anywhere it will swing normally. But then once it started swinging and we apply an electric field, then won't the pendulum bob get attracted to the electric field and actually get more energy to go higher than its initial height. If so where was that energy before the electric field was applied?
One thing I forgot to mention is that the electric field is applied by bringing another charge of opposite sign to that on the bob.
@@hazimahmed8713 The energy for the pendulum bob to swing higher than its initial height comes from the work done by the electric field on the charged bob. When you bring another charge of opposite sign close to the pendulum bob, you create an electric field. This field exerts a force on the charged bob, doing work on it and thereby increasing its potential energy. This additional energy allows the pendulum bob to swing higher.
In summary, the energy was introduced by the external work done by the electric field. Conservation of energy still holds because the increase in mechanical energy of the pendulum comes from the energy supplied by the electric field 😎
Thank you
@@dibeos did you just use chatgpt to answer???
@@completo3172 no…
Another video, I love it❤
Hi, Dibeo's)
@@SobTim-eu3xu hi!! We are glad you like it!!! 😎
your feelings are irrational
@@dibeos as I always say:"You make it, I like it"
With luck and more power to you.
hoping for more videos.
@@Khashayarissi-ob4yj oh yeah, more videos are definitely coming!!! But please let me know what you would like to watch here in the channel 😎
Great video, subscribed
@@jacksonstenger thanks!!!! Let us know what kind of content you’d like to see in the channel please 😎
Nice video. Though I do not understand why it seems surprising that if 'i' was an answer then '-i' would be also, especially since if you had x^2=1, the answers are both 1 and -1.
Yes, it is not that surprising but we were trying to show this symmetry between complex and complex conjugate 😎
@@dibeos The proof that if you have real coefficients (required) and you have a complex root of a polynomial, then its conjugate is also a root is the 'best' showing I have seen for the symmetry
he sounds like fundy
@@runnow2655 the Minecraft gamer?
@@dibeos yea
@@runnow2655 thanks, I’m flattered 😎
I need to understand what are groups before I can grasp this concept
Groups are basically sets that fall under a binary operation with 4 properties: closure, associativity, having an identity, and having an inverse for every element within the set. So, for example, take the integers under addition (addition being our binary operator). Every two integers' sum is also within the set of integers (e.g. 4+7 = 11) (closure). If I have a integers a, b, c, then (a+b) + c = a + (b+c) , so grouping does not matter (associativity). The set of integers has a number a such that any number b added to it gives a sum equal to b (the number 0 in this case) which is the identity of the integers under addition (identity). Lastly, every number within the integers has an additive inverse, a number such that if we have a number a and b is said to be its inverse, a + b = 0 , in this case b = -a (e.g. 2 + -2 = 0, 3 + -3 = 0, etc). Since the integers under addition has all of these properties, we say that the integers under addition are a group.
@@felipefred1279 well, the definition of a group was already explained in the comment below (by William Arcor), but I would add a few (intuitive) things. Think of groups as a way to capture the idea of symmetry and structure. For example, the rotations of a square that leave it looking the same form a group, illustrating the concept of symmetry. Groups are crucial in solving polynomial equations, as they show the symmetries in the solutions through the Galois group. They help classify and understand different mathematical structures, with applications in physics and chemistry to describe symmetries in molecules and physical laws. Groups are important in abstract algebra, and form the foundation for understanding more complex structures like rings and fields, which have applications in number theory, cryptography, and beyond. So as you can see, groups are a powerful concept in both pure and applied mathematics 😎 do these explanations help you ?
@@dibeos it helps a lot, thanks!
@@williamarcor251 great explanation. Short and precise. I think that something that helps to grasp the concept of a group as well is showing an example of something that is not a group. For example, the natural numbers under addition do not form a group (it lacks an inverse element) and the 2x2 matrices under subtraction do not form a group (it is not associative), etc. 😎
ok but why do solvable groups have anything to do with radicals?
@@ekxo1126 Solvable groups are connected to radicals because a polynomial's roots can be expressed in terms of radicals if and only if its Galois group is solvable. This relates to solving polynomial equations using algebraic operations, which involve radicals
@@dibeos ok sorry my question was a bit rude and unclear. I mean the video is very good and likable but I didn't like that the fact that the central point of the video, "a polynomial's roots can be expressed in terms of radicals if and only if its Galois group is solvable" is just said because it's probably very difficult to prove or even intuitively show. In that sense, the end of the video it's not at all "natural", as the video wanted to seem. I understood the definition and motivation behind the Galois group, but it doesn't seem obvious at all why having it be solvable would be equivalent to the existence of a closed radical form for the solutions.
You never exactly answered her question. Why would you specifically need the capability to utilize complex numbers. You simply provided a generic equation. Under what specific context/s would one want the capabilities to solve for a complex number ? You can not provide the innumerable scenerios, but a few examples for the non-mathematically inclined would be nice. Not to harp on specifics, but I assume you to be a mathematician, as apparently all you can perceive are numbers. The contextualizations under which such mathematics are applied, are actually what is most relevant. 😮 The math is the easy part, where, when & why is far more crucial. 😮
@@mikeolsze6776 Thank you for your comment! 😎
Imaginary numbers are not just abstract concepts but have practical applications in various fields. For example, in electrical engineering, they are used to analyze and design circuits. Specifically, they are important in the study of alternating current (AC) circuits, where they help in calculating impedance and phase differences.
In physics, particularly in quantum mechanics, imaginary numbers play a key role in describing the behavior of particles at a microscopic level. They are part of the complex numbers used in the Schrödinger equation, which predicts how quantum systems evolve over time.
Also, in signal processing, imaginary numbers are used in the Fourier transform, which transforms signals between time and frequency domains. This is used for analyzing frequencies in signals and is widely applied in communications and audio engineering.
So, while the equations themselves might seem abstract, their applications in these and other fields show their practical importance.
A concrete example:
In analyzing an RLC circuit (a circuit with a resistor, inductor, and capacitor in series), solving the characteristic quadratic equation can yield complex roots. Given an RLC circuit with resistance R = 10 ohms, inductance L = 1 Henry, and capacitance C = 0.01 Farads, the characteristic equation s^2 + 10s + 100 = 0 results in complex roots s = -5 plus or minus 5*sqrt{3}i. These roots indicate the natural frequencies of the circuit, where the real part -5 represents the exponential decay due to resistance, and the imaginary parts plus or minus 5*sqrt{3} indicate oscillations. This understanding is very important for designing circuits, such as in radio receivers, to ensure they function correctly by tuning to specific frequencies.
I hope I answered your question. If not, let me know 😎
Girls also like maths wow nice to see❤
@@Cooososoo oh yeah, there are some cases
@@dibeos I think in future your subs will cross 1 milli
@@Cooososoo we hope so!!! But honestly even if we don’t we will continue to make these videos, because it’s just so fun to make them. And we are learning A LOT by making them. Also, the fact that people comment correcting our mistakes and giving us feedback helps a lot 😎
@@dibeos is she single 😂
@@evin4899 Sofia is my wife 😎 yeah, I’m really lucky, huh?!
(x+1)(x-1) + 2 = 0 🧐
@ValidatingUsername sorry, we didn’t understand. Could you explain better?
@@dibeos the equation reduces to x²+1=0
@@kirthankamble95 yes, and that’s what is shown in the beginning of the video, which has no real solutions
@@dibeos Again, a little math humor to add to the content and spur some discussion about other ways to solve the problem to help with your algorithm metrics 🥺
@@ValidatingUsername Hahahha ok, next time I’ll get it right away. When I see your comments from now on I know that it’s a math joke 😅
As good as the presentation is, I could not more fully despise this story of i. I see it everywhere, but as an avid researcher of maths history I know its _completely false_ and is literally remnant propaganda from the Quaternion wars. Ping me if you want the real story.
@@tinkeringtim7999 we do want the real story!!! Please, tell us
@dibeos I'm tired of dumping it all in a compressed form in YT comments which makes it look like I just don't understand - so many things have to be untwisted together.
I'm looking for someone to collaborate with to get the story out properly.
@tinkeringtim7999 well, you have my attention. How do you plan to expose the real story?
@@dibeos Still undecided if I'm going to go the PhD -> publish route or just work on YT materials. This piece ties together with fluxions, which were also terribly misunderstood.
I have a remarkably well stocked library of original books from the last few hundred years, including every major work on quaternions.
Geometric algebra and topology has _almost_ caught up with everything you can do using fluxions and quaternions, but still haven't.
The key is a line through the narrative is formed by people with a particular (synethstetic) neurodiversity. They made a framework, then others coloured it in, then when they broke it someone with the same class of neurodiverse traits puts it back together in the modern language. Then everyone piles on and breaks it again.
That's literally the overarching story, but the specifics of the Quaternion wars are really juicy. Look up PG Tait's anger at the "hermaphrodite monsters" the formalists made of the new system.
Hamilton was not trying to extend complex numbers, that was just a loose end he was tying off in a much larger framework.
Hamiltons system had division, Gibbs/Heaviside didn't because they didn't care - they didn't seem to see any need for their equations to be reversible, because they only thought of maths in one direction (to give credibility to the ideas they wanted to sell).
I'm going into YT comment essay mode again, I digress. I should spend time on materials rather than YT comments.
@@dibeos I just wrote a rather long comment and it seems to have disappeared on my app. Did you see it? Have you got an email address we can exchange in more depth?
still not good enough
It is ok, we will get better. Do you want to tell us how to improve the content please? 😎
i agree
@@baobin82 do you agree that it is not good enough? If yes, would you mind telling us so that we can improve for next time? 🤔
You guys are doing well!
Dont take tips from an account with less than 500 subs, who never did quality video editing
@@alextrebek5237 hi Alex! Thanks for the nice comment. Would you like to give us some advice? We are constantly trying to improve our content 😎 any suggestion is welcome!