The Dehn Invariant - Numberphile

It was #3 on Hilbert's list of the most important problems in mathematics - until his student solved it.
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NUMBERPHILE

Videos by Brady Haran
Editing and animation by Pete McPartlan

[numberphile]

Don't miss our new data analysis playlist over on Computerphile... It goes deep...

[lkjadslkfjlaksj]

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

[seanm]

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

[sean]

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

[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:
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

[staglomagnifico]

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

[michaelnovak]

Definitely the best Numberphile episode in a while.

[Hoxbot]

Props to the animator(s?) of these videos

[dimralli]

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

[ronilwaslin]

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

[disorganizedorg]

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

[Rohit-tyhn]

Brady, an intellectual : What if we melt it?

[zeeshanmehmood]

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

[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 Youtube this year and there is very strong competition out there.

[anggalol]

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

[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.

[kenhaley]

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.

[fanrco]

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.

[iwersonsch]

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