What does isomorphic mean? What is an isomorphism?

What's an isomorphism? It's not an alien warrior race! Isomorphisms are special functions that are used to describe when mathematical objects should be regarded as the same.
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Music:
"Clock Town (First Day)", The Legend of Zelda: Majora's Mask
Composer(s): Koji Kondo, Toru Minegishi, Naoto Kubo
Arranger(s): Mahito Yokota, Naoto Kubo
"Fireside"
Instrumental by Homage
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Outro:
"Lateralus" as performed by Sakis Strigas
Originally by Tool
• Tool - Lateralus (Inst...
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[@Grassmpl]

The map for R+ to R+^(2×2) is x|-> x/2 * J, where J is the all 1s matrix

[@Grassmpl]

Exercise for the viewer. Repeat this problem in the case where 2×2 matrices are replaced with general n×n matrices. Of course n in Z+ is fixed.

[@salvatorezungri6945]

@@Grassmpl x|------> x/n * J where J is the all 1s matrix of order n?

[@MuPrimeMath]

I just finished a lecture on isomorphisms in graph theory one hour ago, and now this video shows up in my recommended. Next week I will try to figure out who in my class works at Google.

[@EpicMathTime]

Hey man, hope you liked it! Congrats on your big integral win. 💪

[@mrnogot4251]

My algebra teacher explained isomorphism as follows “It’s like when a friend of yours shows up to your house wearing a mask and you are kind of scared at first but then you realize: oh it’s just my friend.”

[@boonewalker3973]

We may be the most significant isomorphic sets

[@quantumgaming9180]

I'm going to be honest. I just started college and you explained at the same level, or better maybe, as my teachers. Insanely underrated channel

[@chemistro9440]

0:44 "There is nothing to fear here" [Clock Town music starts] *well played sir*

[@EpicMathTime]

"12:12 Hey, your B is not a group!" *yes it is* "but none of those matrices are inve..." *it's a group tho* "but the identity matrix isn't even in th..." *shhhhhh, only dreams now.*

[@Grassmpl]

Yes I agree. The trick here is the standard identity matrix is NOT the identity of this group. Since none of the matrices have full rank, we are off the hook.

[@poutineausyropderable7108]

+Epic Math Time could you say that The integral of f(x) dx and f(y)dy is isomorphic if the only difference between the two is the symbol?

[@EpicMathTime]

@@poutineausyropderable7108 I mean, those two things are really one-the-nose equal.

[@Grassmpl]

The term "isomorphism" is ambiguous unless the intended underlying equivalence relation is well defined

[@DrKjoergoe]

Maybe I'm not getting the irony but how is this a group?

[@lvl3tensorboi929]

I think a neat example of isomorphism is the application of the concept of euclidean space to continuuous functions which leads to fourier or laplace transforms. It amazed me at first how functions can be seen as basis vectors given a proper scalar product that fulfills the same properties as the euclidean scalar product.

[@BurglarBird]

Just discovered this channel. This topic is exactly what I should be studying atm! It's amazing to be able to procrastinate, but also learn about my courses at the same time. I Subscribed!

[@EpicMathTime]

Welcome :)

[@nadiaarif197]

I’m a math major and this video explained better than both of my math classes combined lol

[@mastercilona]

Legend of zelda background music and tool as outro?? AND a really helpful explanation??? you have secured my sub.

[@kylezs]

Lol I came to the comments to see if there was one on the music, I thought it was runescape music

[@tengsolomon]

As a chem student learning about inorganic chemistry, it just dawned on me that point group operations like rotation and reflection are isomorphic functions. Like you perform rotation about a 180 deg angle, and the properties of the molecule after the rotation is still preserved. The only thing that changed are the matrix representations of the atomic orbitals. This is soo mindblowing.

[@EpicMathTime]

Have you used character tables or learned about group representation theory in your studies? These particular fields are strongly linked. The fact that you are able to see a connection like that at such a foundational level is very impressive.

[@tengsolomon]

@@EpicMathTime Yessss! You determine the point group, represent them as matrices, and every time you perform an operation, the molecular structure is still the same (i.e. geometry, properties, etc) but you reduce the irreducible representations of the molecule. It helps us predict the translational, vibrational, and rotational motions of the molecule.

[@keyyyla]

I really like the analogy with the chess game! :)

[@EmilyA-P]

Brilliant vid. Newbie to this area and not a Maths student. You explained it so well

[@cristinapasenelli3720]

Wish my professor had put it in these terms, then I wouldn't have to be scouring the internet for answers. Thanks!

[@gandalf29]

This video has made things very easy for me to grasp. Thanks.

[@timothymoore2197]

I really like this medium of showing math - the clear glass board. It makes these math creatures and what we're saying have this property of "floating around" between us (you and the viewers). I want to create some sort of "math glasses" augmented reality thing where if we put them on, we can have this same effect, but anywhere we want - we wouldn't be bounded by a fixed board placement. I also really like how you speed up the timing when you write down things - it makes the flow of the learning experience better.

[@JM-us3fr]

Yeah I agree. Maybe it would help people understand math concepts better

[@Quasarbooster]

In combinatory calculus, two combinators A and B are isomorphic iff A applied to any x is isomorphic to B applied to x (ie A=B Ax=Bx). For example, the combinator SKK is isomorphic to SKS because when you apply any x, the first reduces to Kx(Kx) and then to x, and the second reduces to Kx(Sx) and then to x. 😊

[@rylanbuck1332]

This is awesome! I just got into my dream college so I can study pure mathematics! I’m currently in calc 2 and it’s very fun! Don’t find it too difficult, just tricky! Of course I’ve started looking ahead at what’s in the future for me (future classes) and I’ve heard this COUNTLESS times! Finally having a clear head on what isomorphic means really makes it easier to understand 😂

[@EpicMathTime]

Great! I'm glad you were able to get something out of it. That's very motivating. Thanks for the comment!

[@fermibubbles9375]

appreciate the sped up writing.. augmented reality for your hand gestures at some point would be wild

[@masontdoyle]

I had to learn about isomorphisms for some undergraduate research I am doing. Your video helped me understand tremendously! Thank you!

[@JM-us3fr]

Good old Category Theory. Someday I will take a class on it.

[@justinotherpatriot1744]

That thumbnail is hilarious AF

[@PhilosophySama]

The kokoroki village music is sending me rn!!

[@halleberry2094]

Yessssss. the lofi in the background love it.

[@FinnBender]

I just wanna say, I love your videos! thx bye

[@EpicMathTime]

Follow me on Instagram to see video previews before they are released: instagram.com/epicmathtime

[@JTan-fq6vy]

In time 2:03, it seems the notion of sameness is a relation , and so the function mentioned in 4:26 is a subset of the relation. So A and B are isomorphic iff (A,B) is in the relation (that we are interested). Please correct me if I am wrong.

[@ashishkumarsharma2584]

Thanks your way of delivering lectures is nice Thanks from India

[@ivan.tucakov]

Thank you! And props for the inverse writing! You must be doing some isomorphic "translating" to pull that off!

[@jacques8277]

I neveer thought I'd get a maths lesson from Bam Margera... in any case great video!

[@RonaldModesitt]

Your presentation was indeed helpful and enlightening!

[@davidk7212]

That hat tells you that, while he may be really into math, he's still hip and sociable.

[@SolidSiren]

Isomorphism is not only used in math!! Its everywhere- chemistry, biology, physics, etc. Did it begin in math? Wonder when this word first was used.

[@TheAAZSD]

Wonderful explanation

[@SAAARC]

Top tier video and explanation

[@stydras3380]

Nice :D Gotta love your structure :P Are you planning on doing a category theory video? :)

[@carlosraventosprieto2065]

man, thank you!!!! amazing video

[@perappelgren948]

Such great a video! 👍👍

[@morgengabe1]

That map diagram at 11:13 seems... raw. Why not make it look more like a function whose domain is some set of order pairs S(x,y) and whose codomain is f(S)? What do continuous functions preserve in topological spaces? Would it be fair to say an Isomorphism is a function that preserves differentiable symmetry?

[@siddid7620]

"we are concerned with their value, not how they look" yeah right

[@Eis461]

Why this channel stopped posting

[@tasmiafatima621]

Really awesome 😍

[@awes0mef4c]

oh hell yeah another video

[@prestondebetaz4300]

Hey that LSU hat makes you 10x cooler. Or, in math terms, the limit as x->cool (EpicMathTime)=infinity

[@temurson]

Just discovered your channel, it is so cool! And I very like the idea of putting problems at the end. I think I solved this one, but don't wanna spoil it for anyone. I tried to think of a function that could have some of the three properties, but not the others (reflexivity, symmetry and transitivity), but I couldn't. I know it's a lazy question, but could you give a couple of examples of those? Thanks.

[@steveashkarian3201]

Amazing!

[@Zeegoner]

Thanks for this one

[@nftnick7815]

Awesome you should do a video on "Hylomorphism" $DAG Constellation network uses this on their Directed Acyclic Graph as they redefine the internet with the HGTP

[@bobtom1243]

awesome!!!

[@huseyinhalitince4404]

What does it mean to be the same as itself .is it possible to think of wise versa? Would you give an example of wise versa

[@NightmareCourtPictures]

Same as itself is supposed to be an obvious fact…like for example say I took you into a cloning machine and made a perfect clone of you…you and this clone are isomorphic because there is a mapping you can do, atom by atom where that mapping is bijective. Say now that I took your clone and I smushed it into just a lumpy pile of goo. This lumpy pile of goo is still isomorphic to you because there is still a bijective mapping you can make with every atom of your smushed clone and you. You can think of a bijective mapping as me taking pieces of your smushed clone, and placing them somewhere new…so a -1 from the clone and a +1 over here…and slowly but surely from the pile of mush I build you up again into a human being. These operations (+1 and -1) is the bijective function in that case…but the function can be any mathematical function, so long as there is an inverse operation to it.

[@DynestiGTI]

Man it's so reassuring to hear other people found the term isomorphism scary, I was just getting to grips with homomorohism and bijection 😅 same thing happened with real analysis, I was so dumbfounded about where these random epsilon/3's were coming from but so was everyone else.

[@parepidemosproductions4741]

I have subscribed and the notifications bell is on for when the inverse of a function is not isomorphic. thanks

[@benjaminbrady2385]

5:20 🅱️®️⛎🏨, that's a cop right there

[@supermarc]

I have a question: are there also mathematical objects for which an isomorphism is more than just a map f : A -> B on the underlying sets? For example, if the mathematical object has additional properties whose preservation cannot be expressed in terms of the function immediately?

[@EpicMathTime]

In an abstract categorical sense, certainly, but that isn't enlightening because category theory puts an abstraction on isomorphisms to the point that it may not capture the intuition that I wanted to address here. But in a more concrete mathematical sense, I think a good example is as follows: instead of vector spaces over a field F, with isomorphisms in the usual way (bijective linear transformations), we can look at all vector spaces over any field isomorphic to F, with isomorphisms taken to be bijective semilinear maps. Semilinear maps are more than just a map on the underlying sets because it must not only translate the vectors in the vector space, but also the scalars to the new field. So it effectively contains two maps, translating the vectors, and translating the field elements in that external field acting on the vector space. In the standard study of linear algebra, this is a nonissue and not needed because we lock interacting vector spaces into being over the same field. Once we broaden the class of objects in the way described, one map on the underlying vectors is not enough to preserve what we want.

[@supermarc]

@@EpicMathTime Nice one! If we cheat a little bit, though, then we could see such an object as an ordered pair (F,V), where F is a field and V is a vector space over F. Then a morphism could still be seen a map f: (F_1,V_1) -> (F_2,V_2).

[@EpicMathTime]

@@supermarc There's nothing wrong with describing V over F as a single object (which it is) and denoting it (V,F), but the set-wise meaning of this is unclear. If we want to have a map f:(V,F)->(V',F'), we have to know what (V,F) means as a set. Are elements of (V,F) ordered pairs (v,c)? If so, it's hard to make this agree with the notion of our vector space (though I'd have to think about that more). If the elements of (V,F) are just vectors, then we of course have the same issue as before. Now, if we instead are associating (V,F) to (V',F') not by a map between them as sets, but in a way that views them as single mathematical objects in which (V,F) is the "input" and (V',F') is the "output" (as opposed to the domain and target) then we aren't describing a morphism at all, but a functor.

[@PolarSky]

This is great! Thank you :)

[@valeriobertoncello1809]

Are you by any chance related to Paul from the LangFocus channel? You two really look alike

[@EpicMathTime]

Haha nope, although someone else did mention that recently.

[@ytseberle]

Except Paul speaks with a Canadian accent and Epic sounds southern U.S., probably Louisiana based on his LSU hat :-)

[@thebiber9401]

I'm a physicist. Can someone verify that the following terms are correct? The isomorphisms for... - ... geometry are the Euclidean Transformations (e.g. translations and rotations) - ... vector spaces are linear maps - ... groups are homomorphisms (like representations in quantum mechanics) I don't know about topology.

[@EpicMathTime]

These are all correct as morphisms. They are isomorphisms with the added stipulation they are also bijective, with inverses that are also morphisms. (that is, an isomorphism is a special kind of morphism). For topology, the morphisms are continuous functions. Sometimes the steps for a morphism to become an isomorphism get satisfied automatically, depending on category. For Euclidean geometry, all morphisms are automatically isomorphisms (every Euclidean transformation is bijective and has an inverse that is also a Euclidean transformation). For most algebraic categories, any bijective morphism is automatically an isomorphism (the inverse of any linear map is a linear map). This isn't true of topology though (there are bijective continuous functions whose inverses are not continuous). So the statement "bijective morphism whose inverse is a morphism" is meant to be an all-encompassing general description.

[@muckchorris9745]

Don't miss arrows on coordinate systems.

[@kalekaleb8148]

I suddenly love 😍 mathematics.

[@illiztDesignsHD]

Please start a math podcast!

[@rohitbhosle6521]

5:54 so funny 😂😂😂😂

[@Grassmpl]

Exercise. let G be a group consisting of a subset of n×n matrices over a field F wrt the standard multiplication. Prove that the elements of G are either ALL invertible or ALL singular. Furthermore, prove that in the invertible case, the identity of G MUST BE the standard identity matrix.

[@EpicMathTime]

:) Proof: First, we show if A in G is invertible, then the identity of G is the identity matrix. Let E denote the identity matrix of G, let A* denote the inverse of A in G (let A^(-1) be its standard inverse, and I the standard identity matrix). Then we have that AA* = E and AA^(-1) = I, hence AA*AA^(-1) = EI = E => AEA^(-1) = E => AA^(-1) = E => E = I. Hence E = I. Now, we show if B in G is not invertible, then the identity of G cannot be the identity matrix. B is in G, therefore it has an inverse B* such that BB* = B*B = E. Since B is not invertible, E =/= I. This shows that A and B cannot lie in the same group (as A's membership requires E = I, while B's requires E =/= I).

[@Grassmpl]

@@EpicMathTime Well done. Your grade is 5/5. Here is another trivial exercise which I know won't stump you. Let G consist of invertible matrices and H consist of singular matrices. Is it possible that G is isomorphic to H? Either give an example or a disproof as appropriate.

[@Grassmpl]

@@EpicMathTime btw these questions are from my own head which I know the answer to. Ie they are NOT from textbooks, internet, home works, exams, etc.

[@EpicMathTime]

@@Grassmpl Yes, it is possible, let E be an idempotent matrix, then {I} and {E} are isomorphic. Or do we want to ban that?

[@Grassmpl]

@@EpicMathTime this is good. As u can see very trivial. Although since I is idempotent u should mention that E is singular as well. Better yet let E=0

[@nourhanramadan7487]

you save me thanks

[@raffypongcol5731]

😍😍😍

[@euclidselements9522]

hey i have this monopoly, let's finish our chess game

[@vojislavbelic896]

What i really want explained is why some guys wear beanies all the time

[@tomkerruish2982]

Ask Tim Pool.

[@captainsal7074]

Clocktown Majora mask

[@sollinw]

niiiiiiiiiiiiiice

[@lettersfromanihilist9092]

this is sorta niche, but is anyone else here because they're reading Wittgenstein's tractatus, and was told that one of the ideas is that "language is isomorphic to reality"

[@NonTwinBrothers]

Ooooh here I am thinking he learned how to draw backwards, nice lol

[@captainsal7074]

Lateralus Tool

[@AR-vb4xy]

Bro what's your math qualifications?

[@thehippievan1288]

You mean my fortnite skins are isomorphic to the standard ones??? Damn. . .

[@FooodConfusion]

lovely brother you nailed it dumb like me in maths can even understand this thing

[@Anteino]

Kudos for writing backwards so fast

[@Anteino]

No wait, you wrote like normally and mirrored the footage. That must be it.

[@inigo8740]

Equally handsome, equally smart.

[@wwtorm510]

This music is distracting

[@beastslayer9691]

You are cute