Kevin Knudson: Welcome to My Favorite Theorem. I’m your host, professor of mathematics at the University of Florida Kevin Knudson. This is my cohost.

Evelyn Lamb: Hi! I’m Evelyn Lamb, a math and science writer in Salt Lake City, Utah. Yeah, things are going well here. I went to the mall the other day, and I was leaving—I had to go to get my computer repaired, and I was in a bad mood and stuff, and I was leaving, and there was just, I walked into the parking lot, there was this beautiful view of this mountain. It’s a mall I don’t normally go to, and these mountains: Wow, it’s amazing that I live here.

KK: Is this the picture you put on Twitter?

EL: Yeah, or Facebook.

KK: Yeah, that is pretty spectacular. Well, I had a haircut today, that’s all I can say. Anyway, let’s get to it. We are very pleased in this episode to welcome Laura Taalman. Laura, do you want to introduce yourself and tell people about yourself?

Laura Taalman: Sure. Hi, thank you for having me on this podcast. I am extremely excited to be on it. Thank you.

EL: We’re glad you’re here.

LT: I’m a math professor at James Madison University, which is in Virginia. I’ve been here since 2000. We don’t have graduate students in our department, we only have undergraduate students. So when I got here, straight out of grad school, I had been studying singular algebraic geometry, and I just could not talk about that with students when we were doing undergraduate research. And I switched to knot theory. I’ve since switched to many things. I seem to switch to a new hat every year or so. My new hat is 3D printing. I’ve been doing a lot with mathematical 3D printing, but I think I’m still wearing that math jacket while I’m wearing the 3D printing hat.

EL: That’s a very exciting costume.

LT: Yes, it’s a very exciting costume, that’s true.

KK: And for a while you were the mathematician in residence at the National Museum of Mathematics, right?

LT: MoMath, that’s true. I did a semester at that, and that was the start of me living in New York City for a couple years to solve a two-body problem. I spent a couple years working in industry in 3D printing there. I just recently, last year, came back to the university. I now have the jacket and hat problem.

KK: Well, that’s better than the two-body problem.

LT: It’s better than not having a jacket or a hat.

KK: That too, right. So actually I was just visiting James Madison a couple of months ago. Laura’s department was very nice. Actually, my wife was visiting, and I was just tagging along, so I crashed their colloquium and just gave one. And everybody was really nice. I really, you know, I went to college at Virginia Tech two hours down the road. I’d never really spent any time in Harrisonburg, but it’s a lovely little town.

LT: It is.

KK: It’s very diverse. I had lunch at an Indonesian place.

EL: Oh wow.

KK: It was fantastic. I can’t get that here, you know.

LT: It’s an amazing place.

KK: It is. I thought it was really great. Anyway, so, you’re going to tell us about your favorite theorem. You told us once beforehand, but I’ve kind of forgotten. I remember, but this is pretty great. So Laura, what’s your favorite theorem?

LT: My favorite theorem comes from my knot theory phase. It’s a theorem in knot theory. I don’t know how much knot theory I should assume before saying what this theorem is, but maybe I should just set it up a little bit.

KK: Yeah, set it up a little bit.

EL: That would be great.

LT: In knot theory, you’re in studying, say, you tie a shoelace and you connect the ends, and you do that again with a different piece of string, and you’re wondering if these could possibly be the same knot in disguise, like you could deform one to another. Of course, we don’t study knots in three dimensions like that because no one can draw that. This is, in fact, how I got into 3D printing was trying to print three-dimensional versions of knots that I could look at their conformations.

KK: Very cool.

LT: But really mathematicians study knots as planar diagrams. You’ve got a diagram of a knot with crossings: over crossings and under crossings, a collections of arcs in the plane with crossings. A very old result in knot theory is that if two of those diagrams represent the same knot secretly (they might look very different), there is a sequence of what are known as Reidemeister moves that gets from one to the other. Reidemeister moves are super simple moves, like putting a twist in a strand or moving one strand over another strand, or moving a strand over or under a crossing, right? Super simple. It’s been proved that that’s sufficient, that’s all you need to change one diagram into any other equivalent diagram.

KK: OK.

LT: So my favorite theorem is by Joel Haas and Jeffrey Lagarias, I think is his name. Haas is from UC Davis, and Lagarias is at Michigan. And in 2001, they proved an upper bound for the number of Reidemeister of moves that it takes to turn a knot diagram that’s secretly unknotted and turn it into basically a circle, the unknot. So they wanted to answer this question.

We know we can, if it’s unknotted, turn it into a circle. The question is how many of these Reidemeister moves are you going to need, and even worse than that, if you start with a diagram that has, like, 10 crossings, you might actually have to increase the number of crossings along the way while simplifying the knot. It’s not necessarily true that the number of crossings will be monotonically decreasing throughout the Reidemeister move process. You might increase the number, you might have to increase the number of crossings by a lot. So this is a nontrivial question of how many Reidemeister moves. So they said, OK, look. We want to find this one constant that will give you an upper bound for any knot that’s trivial to unknot it, the number of Reidemeister moves, and they said that the bound would be of the form 2 times [ed note: Taalman misspoke here and meant to the power instead of times, as is clear from the rest of the conversation] a constant times n, where n is the number of crossings. So if it’s a 10-crossing knot, it would be like 2^10 times this constant, right?

KK: OK.

LT: I was playing around with some numbers, so for example, if you had a 6-crossing knot, right, and if the constant happened to be 10, this would be 2^60, which is over a quintillion.

KK: That’s a lot.

LT: If that constant were 10, and your knot started out with just 6 crossings, that’s a big number. But that is not the bound that they found.

KK: It’s not 10.

LT: Their theorem, my favorite theorem, is that they came up with a bound that the maximum number of Reidemeister moves that would be needed to unknot a trivial knot, that constant is 2^10^11 times n. The constant is 10^11, so 2^(10^11) times n. So I put this into Wolfram Alpha with n=6. So say you have a 6-crossing knot. It’s not so bad. I put in 2^10million [ed note: Taalman misspoke here and meant hundred billion; 10^7 or 10 million comes up as a bound in a different part of the paper], and then also times 6 in the exponent. I just did this this afternoon, and do you know what Wolfram Alpha said?

KK: It couldn’t do it?

LT: I’ve never seen this. It said nothing.

EL: You broke it?

LT: It didn’t spin and think about it, and it didn’t attempt to say something. It literally just pretended that I did not press the button. This is really a big number.

KK: I’m surprised. You know what it should have done? It should have given you the shrug emoji.

LT: Yeah, that would be great if it had that. That would be perfect. So the reason it’s my favorite theorem, I guess there are a lot of reasons, but the primary reason is: this is ridiculous, right? If you have a 6-crossing knot, there’s no way you’re going to need even a quintillion Reidemeister moves in reality. If I actually give you a 6-crossing knot in reality, you’re not going to need a quintillion Reidemeister moves, let alone this number of silence that Wolfram Alpha can’t even calculate. So to me, it’s just really funny. And I could talk a little more about that. But it’s an important result because it’s the first upper bound, which is great, but also, it’s just, it’s ridiculous.

KK: It’s clearly not sharp. They didn’t cook up an example.

LT: It’s clearly not sharp.

KK: They didn’t cook up an example where they had to use that many moves.

LT: Right, no, they did not. It’s kind of like what happened with the twin prime conjecture, and people online were looking at the largest gap you could guarantee, I don’t know if I’m going to say this right, the largest gap.

KK: Right, it was 70 million.

LT: And eventually primes would have to appear with that gap. That gap started out being huge, I don’t remember what it was, but it was really big, and it ended up getting better and better and better and better.

KK: Right.

LT: So this is like the first shot in that game for Reidemeister moves, is 2 to the 10 to the 11th times the number of crossings.

KK: Has anybody made that better yet?

LT: They have. So that was in 2001, this exponential upper bound with very large exponent, and in 2011, two different mathematicians, Coward and Lackenby, I think, proved a different bound that involved an exponential tower. That gives you an idea of just how big that first bound was, if this bound is an exponential tower.

EL: And it’s better?

LT: Actually, let me say that slightly differently because this is not necessarily better. Their result was actually a little bit different. Their result wasn’t taking a knot to the unknot. It was taking any knot to any other knot it was equivalent to.

KK: OK.

EL: OK.

LT: This could well be worse, actually. And to tell you the truth, I was not entirely certain how to type this number into Mathematica, into Wolfram Alpha. It could be a lot worse. Their bound for the maximum number of Reidemeister moves that you need to take one knot to another knot that it’s ambient isotopy equivalent to in 3-space, if you had that knot. I’ve got to get my piece of paper to look at this. Their number is what they call exp^c^n(n), so the n is the sum of the crossing numbers of the two knots. The c^n: c is some constant to be determined. It could be laughably large, right? And what exp means is that it’s 2^n iterated that many times. So exp^k, or exp(k)(n) would be 2^n iterated k times.

KK: Right. 2 to the 2 to the 2 to the…

LT: …2 to the n. So this number is 2 to the 2 to the 2 to the…tower, and the height of this tower is c^n, where n is the number of crossings, and then there’s an n at the top. And the number c is 10 to the one millionth power.

KK: Wow.

EL: Wow. So this is bad news.

LT: This is very bad. So the tower is 10 to the one million high. I’m sure this is worse than the other one.

KK: It’s got to be worse.

LT: They didn’t try at all to make that low. I did a small example: what if the tower was only length 2 and there was a 6 on the top, so 2^2^6. And you’re doing your brackets from the top down, so 2 to the quantity 2^6.

EL: Right.

LT: That is over a quintillion.

KK: Sure.

EL: Yeah, like this is Graham’s number stuff.

LT: Yeah, Graham’s number, all that stuff with the arrows. All that stuff with the arrows.

EL: Yeah, basically you can’t even tell someone how big Graham’s number is because you don’t have the words to describe the bigness of this number.

LT: Yeah, and even with a tower of 2, I’m getting a quintillion. Their length is 10 to the one million. I already don’t understand what 10 to the one million is.

KK: No. You know this thing where you pack the known universe with protons, do you know how many there’d be?

LT: No. Not many?

KK: 10^126.

LT: Oh my God.

KK: So 10 to the one million. You’ve surely seen Powers of 10, this old Eames movie, right?

LT: Yeah, yeah.

KK: The known universe just isn’t that big, you know? It’s what, 10 to the 30th across or whatever. It’s nothing.

EL: You definitely can’t come up with an example that needs this because the heat death of the universe would occur well before we proved this example needed this many steps.

KK: Yeah.

LT: I think that these mathematicians know how funny their result it. It’s definitely, it’s not just funny. The proofs are very complicated and have to do with piecewise linear 3-manifolds and all this. I don’t understand the proofs. This is very sophisticated, so I’m not besmirching them by saying it’s funny. But I think they understand how crazy this sounds. They’ll say things like, this Coward-Lackenby paper has a line in there like, notice that this solves the problem of figuring out if two knots are Reidemeister equivalent because all you have to do is look at every sequence of Reidemeister moves of that length, look at them all, and then see if any two of them turn out to be the same knot. Boom, you’ve solved your problem.

KK: All you have to do.

LT: All you have to do! Problem solved.

EL: Yes.

LT: Or that, so earlier you asked if the result has been improved upon, and it has, but that wasn’t the reference I wanted to say for that. It has been improved just three years ago by Lackenby, one of the authors of that other result, and their result is polynomial. They found a polynomial bound, not an exponential bound. It’s much better. They found that if n is the number of crossings to go from a trivial knot to the trivial circle, this is back to that problem, it’s 236 times n to the 11th power.

KK: OK.

LT: It’s not so bad.

KK: Right.

LT: Not so bad. It is actually pretty bad. But it’s something that Wolfram could calculate. So I did it for example with n equals 3. So say you have a 3-crossing trivial knot. What’s the most number of Reidemeister moves that you would need according to this bound to unknot it? That would be 236 times 3 to the 11th power. That is 2 times 10^31 power, which is 10 nonillion.

KK: Right, OK.

LT: 10 nonillion.

EL: So this isn’t great.

LT: But it had a name! Dressed in scientific notation. Positive change.

EL: It didn’t cause Wolfram Alpha to run away in fright.

LT: No. I think this is the best one so far, this 2014 result by Lackenby. I think it’s the best one.

EL: Well that’s interesting, because you know, just for the example of 3, if you try, like, 10 Reidemeister moves, that’s gotta be it. It feels like that has to be so much lower. It’ll be interesting to see if it’s possible to shrink this down more to meet some more realistic bound.

LT: Honestly, 3 is a ridiculous example. I used it because it was the smallest, but you’re right. If you think about it, there’s really not that many three-crossing diagrams that one can draw.

KK: Right.

LT: Of the ones that are trivial, I’m sure you could find a path of Reidemeister moves. This result isn’t made for low-crossing knots, really, I think. Or at least not three. But you’re right, it’s got to be way better than this.

KK: This is where mathematicians and computer scientists are never going to see eye to eye on something. A computer scientist will look at this and say, that’s ridiculous. You have not solved the problem.

LT: I agree. It’s not good enough. They did have one result in this 2014 paper. Remember I said that you may have to increase the number of crossings? Well back in the original 2001 paper, Haas and Lagarias were like, hey, here’s a fun corollary: you only have to increase the number of crossings by 2 to the power of 10 to the 11th times n at most, because you can’t have more crossings than what it would take for the number of Reidemeister moves. So that’s their corollary. In 2014, that bound is super significantly improved. They just say it’s (7n) squared. That’s not bad at all. They’re saying it doesn’t have to get worse than that on your way to making it the unknot.

KK: You might have to go up and down and up and down and up and down, right?

LT: Right. I guess then they’re saying the most it would ever have to go up is to that.

KK: Yeah.

LT: So things are getting better.

KK: All the time getting better. So part of the fun of this podcast, aside from just learning about absurd numbers, is that we ask our guests to pair their theorem with something. So what have you chosen to pair your theorem with?

LT: That one is actually harder to answer than what is your favorite theorem.

KK: Sure.

LT: I could answer that right away. But I’ve thought about it, and I’ve decided that the best thing to pair it with is champagne.

KK: OK.

LT: Here’s why. First of all, you should really celebrate that a first upper bound has been found.

EL: Yeah.

LT: Especially in terms of when you have undergraduates who are doing research, this kind of meta question of what does it mean to have a first upper bound, a completely non-practical upper bound. The fact that that’s worthy of celebration is something I want them to know. It doesn’t have to be practical. The theory of having an upper bound is very important.

KK: Right.

LT: So champagne is to celebrate, but it’s also to get you over these numbers. I don’t know, maybe it represents how you feel when you’re thinking about the numbers, or what you need to do when you have been thinking about the numbers, is you need a stiff drink. It can be for both.

EL: And champagne is kind of funny, too. It’s got the funny little bubbles, and you’re always happy when you have it. I think it goes very well with the spirit. It’s not practical either.

KK: No.

LT: Yeah.

EL: As drinks go, it’s one of the less practical ones.

KK: And if you get cheap champagne, it will give you a headache, just like these big numbers.

LT: It’s very serious if you had a tower of exponential champagne, this would be a serious problem for you.

KK: Yeah.

EL: Yeah.

KK: Oh wow. We always like to give our guests a chance to plug anything they’re working on. You tweet a lot. I enjoy it.

LT: I do tweet a lot. If you want to find me online, I’m usually known as mathgrrl, like riot grrl but for math. If you’re interested in 3D printable mathematical designs, I have a ton of free math designs on Thingiverse under that name, and I also have a shop on Shapeways which makes you great 3D printed mathematical jewelry and stuff.

EL: It’s all really pretty. You also have a blog, is Hacktastic still going?

LT: Hacktastic is still there. A lot of it has been taken over by these tutorials I’ve been writing about 3D printing with a million different types of software. If you go to mathgrrl.com, Hacktastic is one of the tabs on that.

EL: I like that one.

KK: All over the internet.

EL: Yeah. She will definitely bring some joy to your life on Twitter and on 3D printing worlds. Yeah, thank you so much for being on here. I’m definitely going to look up these papers and try to conceptualize these numbers a little bit.

LT: These are very big numbers. Thank you so much. It’s been really fun talking about this, and thank you for asking what my favorite theorem is.

KK: Thanks, Laura.

[outro]