The Dehn Invariant - Numberphile

[@numberphile]

Don't miss our new data analysis playlist over on Computerphile... It goes deep... kzbin.info/aero/PLzH6n4zXuckpfMu_4Ff8E7Z1behQks5ba

[@Jugge83]

You dont have to melt the object. If you play with sand you find that it's possible to fill all kinds of shapes with the same volume of sand. Have I been cheating mathematics my entire childhood?

[@FreeAsInFreeBeer]

@@Jugge83 With more cuts you can get a better approximation. The issue here is doing it with a finite number of cuts. Sand grains are a huge number of cuts - but the approximation is still not perfect (e.g. air between the grains - no smooth surface).

[@sadface7457]

The simpler explination. You can cut flat all polyhedra into a common invarient shape whereas non flat polyherda are formed from multiple invarient shapes. Invariant meaning you cannot change properties by cutting and are unique. By decomposing the planar shaping common shapes and rearranging them all shapes can be made from those of the same volume. The rectangle is useful intermediated because it gives a strategy for achieving these transformations.

[@Jugge83]

@@FreeAsInFreeBeer It sounds like the age old battle between physics and math ;)

[@bluekeybo]

I went through it but wasn't that great. I thought it was too rushed in places it needed to be slow, and slow in places in needed to breeze through. Love Dr. Pound tho but miss Brady's filming for Computerphile

[@lkjadslkfjlaksj]

You should have Daniel on again. He's a really clear expositor.

[@luksus8547]

Right? That was very well presented, really enjoyable

[@Narokkurai]

He reminded me of my favorite TAs in college. Got to the heart of the matter quickly, but explained the process in enough detail that I could still follow along.

[@forloop7713]

Yes

[@General12th]

He's actually really amazing!

[@Bignic2008]

@@ሂድ That's a very rude thing to say. All mathematicians are normal dudes or dudettes, whether they "seem" to be or not.

[@seanm7445]

A major problem for the early mathematicians was being able to pronounce ‘equidecomposability’. Dehn’s proof followed shortly thereafter.

[@gabor6259]

That's my new favourite "mathy" word.

[@Aligartornator13]

The funny thing is "Zerlegungsgleichheit", the german word Hilbert orginally used, is nothing unusual at all.

[@kuro13wolf]

​@@Aligartornator13 I bet dyslexic people from all over the world have nightmares in German.

[@paparomeo4235]

@@Aligartornator13 That would be "decompositionequality", I guess....sounds way easier.

[@gregorywitcher5618]

@@Aligartornator13 Very interesting.

[@EliaIlProfeta]

For people who are wondering about Dehn Invariant for the tetrahedron: For starters, there are other rules in the tensor product, and one of them is that for every integer z, we have that z(a⊗b)=(za)⊗b=a⊗(zb). It follows that the Dehn Invariant for the cube is actually 12⊗(π/2)=1⊗(6π)=1⊗0=0, which is pretty cool. For the tetrahedron, we first compute the dihedral angle. If L is the length of the edge, then the apothem is (L/2)√3, since it's the height of an equilateral triangle. Such a triangle also has the property that its center cuts the height in two pieces, which are exactly 1/3 and 2/3 of the original length (of the triangle's height). So, we get a rectangular triangle which has the tetrahedron's apothem as the hypotenuse, the tetrahedron's height as the longer side and 1/3 of the basis' height as the shorter side. We can compute the latter, which is (L/6)√3. Using trigonometry, if Ѳ is the dihedral angle, then cos(Ѳ) is the ratio between the shorter side and the hypotenuse of the rectangular triangle before mentioned: this ratio is 1/3, thus Ѳ=arccos(1/3). This is an irrational number, and it's NOT a rational multiple of π. For the length of the edge L, we just use the formula L=³√(6V√2), where V is the volume (this isn't too hard to derive, given that we can compute the tetrahedron's height using pythagorean's theorem). Since we need V=1, we get: L=³√(6√2). Since the tetrahedron has 6 edges, we can finally compute its Dehn Invariant: 6(³√(6√2)⊗arccos(1/3))=³√(6√2)⊗6arccos(1/3). We already stated (not proved, since it can be quite a challenge) that arccos(1/3) is not a rational multiple of π, so there's no way to get 0 on any side of the Invariant. This shows that the Dehn Invariant of the tetrahedron is not 0, while the cube one is 0.

[@TheVeryHungrySingularity]

That was way more interesting than I thought it would be

[@Triantalex]

??

[@Hoxbot]

Props to the animator(s?) of these videos

[@dzspdref]

How can I [LIKE] your statement a couple thousand times!? I've thought this same thing many times.

[@Gold161803]

Pete McPartlan

[@sean3533]

This was the ideal amount of "I don't get it". Not too much. Not too little.

[@ts4gv]

Hah! Agreed. Except I'd swap the term "I don't get it" for "challenge"

[@alexschemm348]

@@ts4gv brilliant way to think!

[@Triantalex]

false.

[@sean3533]

@@Triantalex fight me

[@michaelnovak9412]

Definitely the best Numberphile episode in a while.

[@gregcostanzo4724]

!

[@alainrogez8485]

?

[@Triantalex]

false.

[@staglomagnifico5711]

"This problem was so easy, it only took 2 years to solve!"

[@Life_42]

Lol!

[@staglomagnifico5711]

@@ankushm3t Perhaps if mathematicians are like fantasy wizards and their arcane arts grant them powers of extreme longevity... but last time I checked, they lived about as long as everyone else.

[@trobin]

easy compared to the many of hilberts problems which remain unsolved

[@woonjeng]

@@staglomagnifico5711 better than a 350 years unsolved theorem like fermat's last theorem :l

[@Kartik-yi5ki]

If one of Hilbert's problems takes 2 years to solve, it is easy. Look at the rest of his problems

[@ronilwaslin]

More from this dude. He explains things well, and sounds like he knows cool stuff

[@strider_hiryu850]

MrPainting your profile picture looks like it know's stuff.

[@dimralli4558]

Props to the animator! Without the visuals, I would have been lost 5 minutes in. Great episode.

[@zeeshanmehmood4522]

Please invite Daniel Litt back, this was way interesting and even though he used complicated stuff like tensors I understood everything with his careful building up and his great explaining. New favourite guest on this channel, I think

[@disorganizedorg]

More from Daniel Litt, please! Very clear and engaging explanations.

[@randomaccessfemale]

Agreed. This guy is Litt!

[@pudgebonetv9628]

Straight Litt/e

[@anggalol]

I love how the subtitle is properly formatted, the mathematical expression is stunning

[@donaldasayers]

This is one of those videos that needs something more than just a like. Seriously this is simply one of the best and clearest videos on mathematics on KZbin this year and there is very strong competition out there.

[@fanrco766]

If youre having a hard time understanding this "tensor" business, consider this analogy: when we work with complex numbers, we usually write something of the form "a+bi". This means that we have quantity "a" real numbers and quantity "b" imaginary numbers. We can't simplify "a+bi" because reals and imaginaries are "incompatible" in that they have different rules and properties. Same thing with "L (tensor) Theta". the "L" is a length (which follows the rules of normal real numbers), the "Theta" is an angle (which has special properties. it can only exist between 0 and 2π, its addition is defined mod 2π). Since these two numbers, L and Theta, are "incompatible", we can only write them as a pair of numbers, much like how we write complex numbers as "a+bi". If youd like to learn more about this sort of thing, abstract/linear algebra is a great place to start. I believe this is an example of fields and vector spaces.

[@strider_hiryu850]

fanrco so a tensor is like adding to things that don't add, but not actually adding them, because this is math and we can do that.

[@ceruchi2084]

I'd never seen a "tensor" symbol before this video. Is tensor only used when pairing quantities that can't actually sum (or multiply)? Is it the "apples and oranges" operation?

[@jaktrep]

What you're describing is a direct sum, not a tensor product. To illustrate, consider the following fault in your description of the complex numbers as a tensor product of the reals with the imaginary numbers: Your example suggest that the complex number a+bi is really a⊗bi. This is really the same as (a*1)⊗(b*i) where a and be are taken as elements of the real number field rather than as elements of the real vector space. A property of tensor products implies that this is the same as a*b*(1⊗i), which implies that the complex numbers are a one dimensional vector space over the reals, which we know is not the case. So the direct sum essentially just staples the different objects together while the tensor product entangles them in a more involved way. A big difference between the two is that the direct sum of vector spaces of dimensions a and b has itself the dimension a+b, whereas the tensor product has the dimension a*b. Thinking of the vectors in each vector space as lists of real numbers, e.g. (v1,v2,v3) is in V and (w1,w2,w3) is in W, the direct sum of these vectors would be (v1,v2,v3,w1,w2,w3) whereas the tensor product would be (v1w1,v1w2,v1w3,v2w1,v2w2,v2w3,v3w1,v3w2,v3w3). Conversely, while the direct sum of (a) and (b) would be (a,b), their tensor product would simply be (ab). The tensor product in the video also presents an additional layer of complexity (which was thankfully swept under the rug) because it used a tensor product of Abelian groups (or maybe a tensor product of modules, haven't looked into it) which is a generalisation of vector space tensor products which doesn't have many of the nice properties that vector space tensor products have.

[@satyu131089]

Unfortunately, what you're describing is not a tensor product at all, because a+bi + a+ci is not equal to a+(b+c)i, which is a basic requirement of tensor products. Tensor products are like rectangles: if you add an L*w rectangle and an L*w' rectangle, you get a rectangle of dimensions L*(w+w'). In abstract algebra terms, the tensor product is a certain quotient group of the direct product of two groups.

[@bogdanlevi]

@@satyu131089 A⊗B in not allways a quotient of A+B (considering direct sum or direct product, which is the same in this case). Take A=Z^3 and B=Z^3. The direct sum is Z^6 and the tensor product is Z^9.

[@Rohit-ty6hn]

Brady, an intellectual : What if we melt it?

[@xaytana]

I'm hoping this is some kind of 4d chess foreshadowing for a video, or multiple videos, on topology and genus surfaces: Where Genus 0 being a sphere, Genus 1 being a torus, Genus 2 being a double torus, Genus 3 being a triple torus, etc. "Melting," would imply the topological transformation of a sphere to a cube, or any G0 shape into another G0 shape; same with G1 shapes, turning a G1 torus into the typical coffee mug with a closed-loop handle.

[@aoifebakunin1966]

Melting was mentioned earlier in the video as an example of why it was equidecomposibility was possible, so it's not completely out of left field

[@witerabid]

The question I ask myself is: how is melting different from doing infinitely many cuts? Or does this proof only work for a finite amount of cuts?

[@aoifebakunin1966]

@@witerabid equidecomposability only deals with finitely many planar cuts. Anything past that (infinitely many slices, curved cuts, etc) is not covered by the proof

[@witerabid]

@@aoifebakunin1966 Ok, then I must've missed that. Thanks. 😊

[@kenhaley4]

i LOVED this video! I think it's actually somewhat intuitive that the DEHN invariant is indeed an invariant. The genius is coming up with that idea to begin with. Amazing.

[@fanrco766]

Ken Haley Finding these types of invariants must be so crazy hard. Its a case of "when you know it it's obvious". But imagine being a mathematician just sitting there trying to picture these things and coming up with the one thing that is preserved through the cutting of the shapes.

[@Zxv975]

I love this video. It's an interesting, non-trivial result that has been clearly explained and the key points of the maths has been faithfully captured and presented. However, I wish a tiny bit of time was spent explaining the tensor product, to lower its status from "scary mathematical operation" to "a natural intuition that most people watching the video were already doing in their heads without realising". For those interested, forget the word "tensor". Instead, replace it with "pairing". The whole point of that operation was to show there was a pairing between dihedral angles and edges. This was obvious by looking at the animations, yes? Every edge is connected to exactly one dihedral angle, and every dihedral angle corresponds to exactly one edge. The tensor product is just a mathematical way of gluing them together; nothing fancier than this idea is being captured by the use of tensor products in this video (and basically all applications of tensor products boil down to making pairings on some level).

[@eukleidesk6759]

I think this is one of my favourite Numberphile videos to date. Incredibly interesting!

[@numberphile]

Thanks. That’s nice to hear.

[@gevillgar]

There's a moment when I thought: oh man, this is going to get messy. But it didn't. Daniel did a great job explaining the concept.

[@x87-64]

One of the best Numberphile episodes ever.

[@iwersonsch5131]

Applications of tensor products I've heard of: 1 and counting

[@brianlane723]

Does the spin of multiple particle systems count?

[@randomdude9135]

What's a tensor?

[@Ethernet3]

@@randomdude9135 In physics its "something that transforms like a tensor"

[@xario2007]

It's not really a classic tensor though.

[@bogdanlevi]

These are not exactly the same tensors as they use in physics. In physics, a tensor space is usually a tensor product of vector spaces. However, R/(2*pi) is not a vector space, so in this video a tensor product of abelian groups is used. Tensor product can be viewed as a way to multiply things of different origins. You can not simply multiply a real number by a point of a circle (and R/(2*pi) is basically a circle), it just does not make any sense. So you have to invent this new operation called tensor product, with rules that look a lot like multiplication: a*(b+c)=a*b+a*c; (a+b)*c=a*c+b*c.

[@FreeAsInFreeBeer]

Beautiful and simple proof! I hope Daniel Litt will return to the channel often.

[@randomguy263]

Well, if he does it's going to be litt I'm sorry.

[@szmatogowiec3414]

Daniel didn't prove it

[@quickstart-M51]

I’m a physicist and I love being one but I must say that sometimes it seems mathematicians have all the fun.

[@nakulghate9448]

I watched it straight till the end. And I am not the guy who watches a 15 min video without forwarding. Numberphile you rock!!

[@x87-64]

This was one of the most amazing videos on this channel. Daniel Litt is an amazing teacher.

[@JackPitts]

I watched this last night around 2am because I couldn’t sleep. Then today I went back to look again for an invariant among some things I’ve been studying. Found one! I think it’s a good one! Thanks for the inspiration, Dehn and Numberphile!

[@semm174]

I’m a third year university mathematics student and one of my modules is a really interesting one that looks at the mathematics of hilbert’s problems and their applications. His third problem is one of them! Could have done with knowing this video existed before taking the exam though, took me a while to understand tensor products.

[@heimdall1973]

The cool thing Dehn invariant: although you can't turn a tetrahedron into a cube, if you chop of the corners (just right), you get a shape the can be turned into a cube. If you scale a solid by factor 1/2 (with volume 1/8 of the original) and use 2 of them, that will have the same Dehn invariant. Or have one that's 1/3 and one that's 2/3. Or you have several, as long as all scale factors add up to 1. Take that original solid and cut off the scaled down solids. For example, have a dodecahedron with side 3 and cut off a dodecahedron with side 1 and a dodecahedron with side 2. Or start with a tetrahedron with side 1 and cut off 2 tetrahedra each with side 1/2. Or instead cut off 4 tetraedra each with side 1/4. Or... whatever. Because you cut off something with the same Dehn invariant, you end up with something with Dehn invariant zero, same as the cube. That means you should then be able to cut it up and reassemble into a cube.

[@MrSigmaSharp]

Extremely simple description. I read this proof before and had a real hard time understanding that. But Daniel explained it very simple and elegant. I'm amazed.

[@gcewing]

I'm sure if you gave Banach and Tarski an infinite number of razor blades they would be able to do it.

[@irrelevant_noob]

Might not even need an "infinite" number of razor blades, the result worked with a finite number of subsets. ;)

[@gcewing]

True, but then the razor blades might need to have some very strange shapes.

[@strider_hiryu850]

Nah, we just need razor blades of infinidecimal thiccness

[@NSwaper]

Somebody watches VSauce!

[@matthewryan4844]

They could do it twice!

[@benjaminnelson5455]

I'm glad you addressed the idea of melting one, I was going to ask how that differs from making an enormous, but still finite, number of cuts. It seems that the answer is that any cut must also result in a piece with edges and angles, so that you can still compute the Dehn invariant?

[@tom7]

Loved this one! Enough detail to grok the proofs!

[@zlaja111]

I keep watching this channel and I understand like 3% of stuff they're covering. I wish my school maths were as engaging as Numberphile content

[@Damathematician]

The strategy at 3:30 is so gangster...boy have I gotten rusty with proofs.

[@willnewman9783]

I agree. That is one of my favorite proof techniques: showing all of a certain thing are equivalent to each other by showing that they are all equivalent to a specific thing.

[@jacobcasey28]

More of this harder abstract stuff ! Brilliant episode

[@subh1]

This filled my heart with an indescribable amount of joy that's quite hard to explain. Not sure if it was the Dehn's clever invariant or Dr. Litt's smile and cheerful nature. Perhaps a combination of both! :)

[@atmosphericSkull]

as someone who has never read about the tensor product, I feel like I got a good sense of it from his description of that combined mutant set of reals and reals mod 2pi

[@francescomussin]

Even with the tensor product that looks a bit shady it's still a fascinating proof and your video managed to capture that. Congrats!

[@Alkis05]

since when I studied differential geometry, I got a great appreciation for invariants. It is amazing the kind of generalization and abstraction that you got with them.

[@bogdanlevi]

Invariants truly shine in general topology. There are lots of them, but not only that, there are lots of relations between them, examples and counterexamples showing that relations are exact and so on.

[@Israel2.3.2]

I first learned about this in Ch. 27 of Robin Hartshorne's Geometry: Euclid and Beyond. Fantastic book.

[@jonhillery7736]

TIL Hartshorne wrote more than the algebraic geometry book

[@jonhillery7736]

Lol this is the second video I've seen you on today. The first being "David Nadler: 2016 Breakthrough Prize in Mathematics Symposium"

[@Israel2.3.2]

@@jonhillery7736 Wow, forgot I left that comment lol. Still have a long way to go ^_^

[@jonhillery7736]

@@Israel2.3.2 Don't we all. I'm excited to be taking Nadler's algebraic geometry course in the fall

[@theflaggeddragon9472]

@@jonhillery7736 Try his book on deformation theory ;)

[@maxkoller6315]

wow I'm just so amazed now and I finally accepted that there are people like Dehn or this mathematician or Einstein if you will whose skill of mathematical deduction I'll never reach, not in my lifetime

[@waffamoto]

"I'm not going to pretend to totally understand that but... I believe you" I need that on a t-shirt

[@takatotakasui8307]

Definitely have Daniel on this show again.

[@nvgirishs]

This was amazing! Can we have more from Daniel Litt?

[@zubin8010]

One of my favorite Numberphile videos!

[@IainHendry]

This was absolutely fantastic. I love your channel!

[@lesterdavepaguio4680]

Wow! I have been wondering about this since 7th grade. I just tried to figure it out by cutting the pieces in real life using a board! That was one of the most amazing things that happened in my life! This is one of the reasons why I love math so much!!!😊😊

[@levitheentity4000]

13:40 actually he's right you can turn a cube into a smaller tetrahedron, with a bit of leftover

[@alxjones]

Or a lot of leftover, depending on how small you make it.

[@luisrosano3510]

I don`t understand why people hate matematics, it is beautiful, and this solution in particular is gorgeous. Thank you for the video guys. I really enjoy it.

[@atzuras]

Dehn: solves one of the Hilbert problems Numberphile: it was clever..

[@nbrader]

9:44 I think you glossed over some of the subtle significance of this equivalence. As I can see so far, it's not so much that angles generally are equivalent after a full turn, it's that when slicing a polyhedron, new edges that aren't where edges were before would have had, in a sense, a dihedral angle of half a turn and that needs to be preserved in our prospective invariant.

[@kameronbriggs235]

This reminds me of stories ive heard, about some massive amount of work on amazons algorithm for loading bins and trucks to maximize space and unloading time and whatnot.

[@ceruchi2084]

I like this guy's style! Would love more videos with him.

[@KevinHorecka]

In higher dimensions, are the number of invariants always one fewer than the number of dimensions as we see in 2D and 3D? If we don't know, how would you even go about trying to prove such a thing?

[@matteogauthier7750]

I second this! I would add, if not n-1, can the number of invariants of nth-dimensional polygons be expressed as a function of n?

[@Nukestarmaster]

You would need to prove that by adding a dimension you also add one invariant. The proof would follow through induction.

[@nothayley]

I think you could combine volume and Dehn into one with clever application of the tensor product... Possibly.

[@diligar]

My guess is that in 4D the new invariant might have something to do with the shape’s 2D faces? Similar to how Dehn’s 3D invariant looked at the shapes 1D edges perhaps?

[@johnburnham6239]

Aren’t there the same number of invariants from 2 to 3 dimensions?

[@paulpantea9521]

Great video. Brady, I have maybe one suggestion: Before shooting the video, ask the interviewee to explain the notions that might be a bit more tricky, e.g. the tensor product. This way you don't get confused when he explains stuff on camera. Those at home can replay the video, search on Google etc., but you are put on the spot. Daniel is right, the tensor product is nothing complicated or mysterious, although it can be confusing when you see it for the first time.

[@SalixAlba256]

Interestingly the Dehn invariant of a cube is 0 ⊗0. Starting with 12⊗π/2 = 6 ⊗ π = 3 ⊗ 2π, but as the second term is calculated mod 2π, its just zero. The dihedral angle of a tetrahedron is arccos(1/3), not a rational multiple of pi and the edge length is cuberoot(3) sqrt(2).

[@SoWe1]

1 ⊗ 0 not 0⊗0 but a bunch more representations of course, eg 3*n⊗0

[@SoWe1]

1 ⊗ 0 not 0⊗0 but a bunch more representations of course, eg 3*n⊗0

[@SoWe1]

1 ⊗ 0 not 0⊗0 but a bunch more representations of course, eg 3*n⊗0

[@SoWe1]

1 ⊗ 0 not 0⊗0 but a bunch more representations of course, eg 3*n⊗0

[@SoWe1]

1 ⊗ 0 not 0⊗0 but a bunch more representations of course, eg 3*n⊗0

[@TheTrevorS1]

Best Numberphile in a long time! Do more involved ones like this, please.

[@grumpyparsnip]

The Dehn invariant of a cube can be simplified to 0 since 12 tensor pi/2 =1 tensor 6pi = 1 tensor 0 = 0.

[@p.singson3910]

O Beard of knowledge, enchant me more.

[@grumpyparsnip]

@@spagetychannel5070 Indeed. Well spotted.

[@Etothe2iPi]

It's actually great fun to find the cuts and pieces of two polyhedra with the same Dehn invariant, like the cube and the rhombic dedecahedron.

[@tedbo1819]

I'm surprised that they didn't mention one of the nice applications of the theorem: there is no proof that the volume of a cone is (1/3) x (base area) x (height) based entirely on geometry, you must use calculus.

[@oofusmcdoofus]

Yeah in textbooks they just show the activity of putting sand in a cylinder using a cone

[@rosiefay7283]

But there *are* formulae for the volumes of a cube and regular tetrahedron, and their Dehn invariants are different, so how do you conclude that there is no geometrical proof for the volume of a cone?

[@davide1913]

Watched this many times, Would love to see another video about Dehn Invariants or similar. Thank You guys for this brilliant work! fantastic !

[@willfuller3689]

Great explainer: clear and to the point

[@HexRey]

Been hoping for a video on this forever! Awesome

[@michaurbanski5961]

Isn't that true for the cube then: 12 tensor pi/2 = 12*(1 tensor pi/2) = 1 tensor 6*pi = 1 tensor 0?

[@Dominikbeck12]

That's the question!

[@UlrichPennig]

This is true. However, the Dehn invariant of the tetrahedron is non-zero.

[@shakedel]

Was thinking the exact same thing

[@shakedel]

@@UlrichPennig but, if there is more than one representation for the invariant, namely 12×π/2 = 1×0, how can you tell that there aren't other representations if the same "invariant" that match the tetrahedron's. It feels like something is missing.

[@UlrichPennig]

@@shakedel As mentioned in the video, some things are identified in the tensor product. This means that you have to carefully check whether two invariants are the same or not, which is a bit more complicated than the video makes it seem. In the example, a tensor product of the form a x b is only zero if you can divide a by an integer c, such that c*b is divisible by 2pi. This follows from the observation that (a x b) = (a/c x cb). Notice the little Z below the tensor product in the video? That tells you exactly that you can move integers from the left side of the tensor product to the right in the way described in the last formula.

[@IFearlessINinja]

I really enjoyed this video! I was going to give credit to the interviewee, but Brady's animations also play a huge part in the quality of the video. You both did a great job!

[@Jodabomb24]

I was kinda hoping to see a Numberphile2 video about why it actually is invariant.

[@bastiaanabcde]

Probably one shows this by showing that for any way of cutting and pasting, if you introduce some extra intermediate cuts, then you can make sure all cuts are of this special shape. For cuts in this special shape, this invariant is an invariant by definition.

[@TIO540S1]

Joseph McGowan I’d like to see a proof, or at least the idea of a proof that the Dean invariant is invariant.

[@markorezic3131]

@@TIO540S1 well it should be enough that you show every type of cutting you can do(which he does in the video) and show that no matter how you cut it, these values(lengths, angles) sum up to the starting, unchanging value. Therefore no amount/type of cuts can ever change that value, its invariant

[@NotaWalrus1]

he explained it in the video. Each cut can only affect an edge in one of three ways, it either cuts along the edge, cuts the edge, or doesn't affect it. In either case the Dehn invariant for that edge is the same, so the sum of the invariants for all the edges must also be the same.

[@technowey]

Thank you. That was an excellent and clear presentation of complicated concepts.

[@rickseiden1]

This is the first time I've ever had a Numberphile video go over my head.

[@tristrumandrewsfisho339]

This was great. instead of confusing me with symbology it taught me some gently. Thank you! i understood all of it because we were stepped through a piece at a time. i guess my brain's dehn variant is the same as this video and the information was cut up just right.

[@lioneldamtew9533]

13:03 Couldn't you use the same logic to show that the invariant of the cube is 1⊗(12*(π/2)) = 1⊗0 ? I'm using the third rule: a⊗b1+a⊗b2=a⊗(b1+b2). Does that mean, that the invariants 1⊗0 and 12⊗(π/2) are equivalent?

[@EliaIlProfeta]

Yes, they are the same element in the tensor product, and that element is 0

[@alxjones]

@@EliaIlProfeta 1⊗0 is not 0. 1⊗0 is 1⊗0 and 0 is 0. The operation ⊗ is not like multiplication in that we can simplify it down to a single element, it's more like forming a special kind of ordered pair, a⊗b = (a,b) with (a,b) + (x,b) = (a+x,b) and (a,b) + (a,y) = (a,b+y) -- the first property shown in the video is actually a property of ℝ/2π and *not* of ⊗. However, you *can* show that 0⊗0 is an additive identity in the set ℝ⊗(ℝ/2π), so if there's any element that you might write as simply 0, it would be that one.

[@EliaIlProfeta]

I'm not sure what your objection is about, considering that 1⊗0=0⊗0, which is pretty standard to just write down as 0 when it's clear that we're working in the context of tensor products. I might very well be missing the point, but I'm fairly certain that there's nothing weird or incorrect about writing 1⊗0=0 here.

[@alxjones]

@@EliaIlProfeta In a proper tensor product of commutative monoids, (a,0)~(0,b)~0 must be taken explicitly because it doesn't follow from distributivity (as opposed to abelian groups, where it is a consequence). The operation shown in the video is exactly that tensor product *without* that property given. I haven't read the original paper, so maybe he does use the proper tensor product, but as demonstrated in the video it's just not true.

[@beardedboulderer2609]

Daniel said it very well: tensor is the most brutal way to combine two objects things (behind formal product).

[@patrickwienhoft7987]

I think I'm missing somthing in the 3D case... So I fully understand that when cutting an edge either the resulting lengths or the angles add up to the original. But what about the new edges that are created? E.g. cut a cube into two 1x1x0.5 cuboids. Then still all angles are 90° but all lengths of the edges summed up are 20 instead of 12 as for the cube. Is it because you have two objects now? Will the invariant be the same again once we put the two cuboids together again? Also that gives rise to another interesting question: What Dehn invariants can two 3D objects have that are the result of cutting another 3D object once?

[@burnheart123]

I pretty much have the same question. You can put them together again in a way, where not much edge length is lost, for example when there is only a tiny overlap at one corner.

[@lukebaczynskyj9353]

Patrick Wienhöft when they are two different polyhedron it doesn’t work because the Dehn Invariant is for one complete polyhedron. If you take your two 1x1x0.5 prisms and put them together again in a new configuration, say creating a 1x2x0.5 rectangular prism, you would find that the Dehn Invariant is again 12 as it was originally.

[@prdoyle]

It's because the lengths are not what is invariant. When you cut a cube, the overall edge length invariably increases, and that's fine and expected. What is invariant is the sum of the tensor-products, which they didn't actually define.

[@skyscraperfan]

@@lukebaczynskyj9353 But weren't we looking for an invariant that does not change if you make a cut? That was the whole idea of the proof.

[@patrickwienhoft7987]

@@lukebaczynskyj9353 Well but is that true for all ways to put them back together? For example if they are offset, you then have some 270° angles.

[@emorgan0085]

I think an important part was left out of the proof, which is how to deal with the new edges you create when you cut a face, since you are creating length out of nowhere. This might be wrong, but I think the missing piece is to point out that A tensor pi is always 0 for any A, which causes any newly created edge pairs to cancel each other out.

[@nimulamin]

I demand a video from @3blue1brown going deep into this topic.

[@PhilipSmolen]

Not yet. I want to see where he's going with his series on differential equations.

[@javierantoniosilva8477]

When is my man Daniel making a comeback? Cause this video is Litt🔥🔥

[@simonsallen]

It is rare for me to be able to claim the first comment. I had pondered this problem some years ago but I was unaware that it had been proved. After some musing, I came to the view that it was probably not possible. How to prove it I had no idea but just for once, my intuition led me to conclude the correct answer. Usually, my intuition leds me astray. Another great video from Numberphile.

[@sparky9705]

suicide train comes for all

[@randomdude9135]

Are you a patreon?

[@Anklejbiter]

how tf was this comment made 2 days ago?

[@tooru]

Why would you even point out that your comment is first? Do such things make you happy?

[@cate01a]

@@Anklejbiter Must've been an unlisted video for patreons only

[@madagaskar8162]

Awesome. I like how the newer episodes deal with the convoluted, crazy stuff that makes your head spin. But you feel so much smarter once you got the gist if it. :D

[@prdoyle]

Started off great--possibly my favourite Numberfile yet--and then didn't describe what the "tensor" operator actually is. Kind of unsatisfying. Will there be a Numerphile2 video going into that?

[@edelopo]

They kind of described it, right? Maybe they missed multiplication by scalars but apart from that it's fine.

[@randomdude9135]

Yeah, not all viewers know the def of a tensor. Inc me

[@randomdude9135]

It's already a 15min vid. 2min wouldn't have made a difference

[@ruinenlust_]

They've shown what it does and how it operates - that alone is enough to implicitly define an operator.

[@edelopo]

@@randomdude9135 My point is that what they said is one possible definition of the tensor product: formal sums of those weird looking symbols (basically pairs of objects) with some funky rules for the sum and scalar product. I agree that they left out the multiplication part and could have covered it, but i think that it wouldn't have made things more understandable. They just used (1 tensor pi/2) added 12 times equals (1 added 12 times) tensor pi/2 equals 12 tensor pi/2.

[@recklessroges]

I liked how Daniel grew up with Sesame Street counting. "1,2,3,4,5; 6,7,8,9,10; 11. 12."

[@greencornucopia3298]

This is one of those videos you watch at 2:59 am

[@therogue1542]

This was beautifully explained

[@ivarkrabol]

This was really interesting, but really frustrating to watch. I'm left with too many questions that I feel should have been at least hinted at, like: What about new edges? What if you use rule 3 instead of rule 2 to calculate the invariant of the cube, and get a different result? Am I supposed to interpret the rules to mean that for example a⊗(2b) = (2a)⊗b, somehow, since they are both the sum of a⊗b + a⊗b? Does that generalize to n(a⊗b) = (na)⊗b = a⊗(nb) for all real numbers n? Does that mean a⊗b = b/2π(a⊗2π) = (ab/2π)⊗2π = (ab/2π)⊗0 = (ab/2π)(1⊗0) = 1⊗0? Probably not, but I don't know why it wouldn't. Maybe I'm not allowed to have a tensor product whose dihedral angle is 0 (mod 2π).

[@ElTurbinado]

That's funny I just asked exactly the same question lol. It's weird, right? 12⊗π/2 = 1⊗6π = 1⊗0 according to those rules and I have no idea what that means for polyhedra.

[@BillTheManiac]

You've got it right for 2, and for 3, 4, 5, and so on, and all integers, but not for real numbers. That's a big part of what's clever about it - most of the time in mathematics where you have a construction that behaves well under multiplication by integers, it behaves well under multiplication by real numbers as well. But Dehn saw that to solve this problem it was crucial to make an invariant with nice properties under integer multiplication but not real number multiplication.

[@paxmaniac1]

Yeah, this. Although on the upside it forced me to read up on it, so mission accomplished for Numberphile.

[@Jesin00]

I love proofs based on invariants like this. I would appreciate more videos like this one, if you can find more!

[@AlwinMao]

I thought it was interesting that he chose to show 12 copies of 1 x pi/2 = 12 x pi/2 but I guess you could also use the other rule (a x b1) + (a x b2) = a x (b1+b2) so that 12 copies of 1 x pi/2 = 1 x 6 pi = 1 x 0 I'm guessing that's what wikipedia means by "The Dehn invariant is zero for the cube"

[@markorezic3131]

Yes that is correct, and the angles for the tetrahedron do not sum up to a multiple of 2pi, however they do not sum up to a rational multiple of pi and thats why we know for sure that their invariants arent the same

[@BrokenSofa]

Love this feeling of interest I get from watching these videos

[@LucasPreti]

incredibly satisfying sound effects

[@tedchirvasiu]

Math ASMR

[@Life_42]

@@tedchirvasiu ^^^ (:

[@VanVlearMusic]

More Daniel! Love his presentation

[@TheOfficialCzex]

Ooo... My discrete math knowledge is rushing back to me as he speaks.

[@Tekay37]

This is the kind of video I subscribed for years ago. Very interesting and insightful. Thank you!

[@johnredberg]

Whoa whoa whoa "famous mathematician _at the time"_ ?

[@samuelluria4744]

This is extremely intuitive, if you imagine the following analogy: Imagine a cylinder - a soda can, for example - it's volume is measured by a set of lines, along the height, and across the diameter (specifically, r²). If we crush this soda can with our foot, the "lines" will all still completely EXIST within the crushed can; if we shrank down to the size of an ant, we could walk all along the lines, and tracing them, realize that the full lengths of them are in fact still very much "there". But, obviously, the crushed can can by no means CONTAIN the same volume as it did before.

[@donaldasayers]

An utterly brilliant video, even I understood it. We need an extra video to explain the details of tensor multiplication in this context, please?

[@xario2007]

There has been no tensor multiplication. Better just call it "special plus" or "lazy plus", because adding a length and an angle doesn't compute to something new, it only gives you a "length-and-angle"-object. It does however distribute, so in that regard it's more like a "lazy product".

[@smort123]

@@xario2007 " "length-and-angle"-object"? A vector?

[@windowslogo3577]

definitely

[@alimanski7941]

It's a form of operation that has special rules, and given those rules, you can create a space for objects that behaves nicely. You can see the rules that define a "nice behaviour" in the wikipedia article for Vector spaces, under the section "definition".

[@donaldasayers]

@@xario2007 But without an actual rigorous definition it could be hard to show that two values are different.

[@maxhaibara8828]

10 minutes of explanation Brady: *cool*

[@grill-surf-bust]

Crystal clear explanation. Would love to see more.

[@StephenFarthing]

This would have been more enlightening if he had explained what a tensor is.

[@killerbee.13]

I think that the only issue was when he said it was like multiplication. My understanding is that it's not like multiplication, because multiplication takes two numbers and makes one number, while the tensor product takes two numbers and combines them into one "quantity" which is still two numbers, they just have a special relationship.

[@menachemsalomon]

With a background in OOP, it's quite easy to conceive of an object with two fields and a limited number of operations combining such objects. (In many ways, learning programming made learning math later on much easier.)

[@LeToastMeister]

I was feeling so smart up until the part where you start messing around with symbols.

[@Royvan7]

to my understand it's kinda just a formal way to take the two math objects and make them into one more complicated math object.

[@bryantames3716]

It feels kinda like how you go from the reals to the complex but with different rules.

[@natlyon6437]

All he did was define an operation using rules that are pretty straight forward if you know any trig. If you don't know what the mod operation means again it's quite simple and you'll see that R mod 2(pi) makes sense because of the way the unit circle works.

[@kevina5337]

Ya I feel he should have explained what that x inside a circle meant lol

[@natlyon6437]

@@kevina5337 he did at 9:43 The symbol is defined by those rules

[@sajanjohn6300]

Everybody gangsta until the "tensor" comes in.