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Episode 21 - Jana Rodriguez Hertz

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Content provided by Kevin Knudson and Evelyn Lamb. All podcast content including episodes, graphics, and podcast descriptions are uploaded and provided directly by Kevin Knudson and Evelyn Lamb or their podcast platform partner. If you believe someone is using your copyrighted work without your permission, you can follow the process outlined here https://player.fm/legal.

Evelyn Lamb: Hello and welcome to My Favorite Theorem. I’m one of your hosts, Evelyn Lamb. I’m a freelance math and science writer in Salt Lake City, Utah. And this is your other host.
Kevin Knudson: Hi, I’m Kevin Knudson, professor of mathematics at the University of Florida. How are you doing, Evelyn?
EL: I’m all right. I’m excited because we’re trying a different recording setup today, and a few of our recent episodes, I’ve had a few connection problems, so I’m hoping that everything goes well, and I’ve probably jinxed myself by saying that.
KK: No, no, it’s going to be fine. Positive thinking.
EL: Yeah, I’m hoping that the blips that our listeners may have heard in recent episodes won’t be happening. How about you? Are you doing well?
KK: I’m fine. Spring break is next week, and we’ve had the air conditioning on this week. This is the absurdity of my life. It’s February, and the air conditioning is on. But it’s okay. It’s nice. My son is coming home for spring break, so we’re excited.
EL: Great. We’re very happy today to have Jana Rodriguez-Hertz on the show. So, Jana, would you like to tell us a little bit about yourself?
Jana Rodriguez-Hertz: Hi, thank you so much. I’m originally from Argentina, I have lived in Uruguay for 20 years, and now I live in China, in Shenzhen.
EL: Yeah, that’s quite a big change. When we were first talking, first emailing, I mean, you were in Uruguay then, you’re back in China now. What took you out there?
JRH: Well, we got a nice job offer, and we thought we’d like to try. We said, why not, and we went here. It’s nice. It’s a totally different culture, but I’m liking it so far.
KK: What part of China are you in, which university?
JRH: In Southern University of Science and Technology in Shenzhen. It’s in Shenzhen. Shenzhen is in mainland China in front of Hong Kong, right in front of Hong Kong.
KK: Okay. That’s very far south.
EL: I guess February weather isn’t too bad over there.
JRH: It’s still winter, but it’s not too bad.
EL: Of course, that will be very relevant to our listeners when they hear this in a few months. We’re glad to have you here. Can you tell us about your favorite theorem?
JRH: Well, you know, I live in China now, and every noon I see a dynamical process that looks like the theorem I want to talk to you about, which is the dynamical properties of Smale’s horseshoe. Here it goes. You know, at the canteen of my university, there is a cook that makes noodles.
EL: Oh, nice.
JRH: He takes the dough and stretches it and folds it without mixing, and stretches it and folds it again, until the strips are so thin that they’re ready to be noodles, and then he cuts the dough. Well, this procedure can be described as a chaotic dynamical system, which is the Smale’s horseshoe.
KK: Okay.
JRH: So I want to talk to you about this. But we will do it in a mathematical model so it is more precise. So suppose that the cook has a piece of dough in a square mold, say of side 1. Then the cook stretches the dough so it becomes three times longer in the vertical sense but 1/3 of its original width in the horizontal sense. Then he folds it and puts the dough again in the square mold, making a horseshoe form. So the lower third of the square is converted into a rectangle of height 1 and width 1/3 and will be placed on the left side of the mold. The middle third of the square is going to be bent and will go outside the mold and will be cut. The upper third will be converted to another rectangle of height 1 and width 1/3 and will be put upside down in the right side of the mold. Do you get it?
KK: Yeah.
JRH: Now in the mold there will be two connected components of dough, one in the left third of the square and one in the right third of the square, and the middle third will be empty. In this way, we have obtained a map from a subset of the square into another subset of the square. And each time this map is applied, that is, each time we stretch and fold the dough, and cut the bent part, it’s called a forward iteration. So in the first forward iteration of the square, we obtain two rectangles of width 1/3 and height 1. Now in the second forward iteration of the square, we obtain four rectangles of width 1/9 and height 1. Two rectangles are contained in the left third, two rectangles in the right third. These are four noodles in total.
Counting from left to right, we will see one noodle of width 1/9, one gap of width 1/9, a second noodle of width 1/9, a gap of 1/3, and two more noodles of width 1/9 separated by a gap of width 1/9. Is that okay?
KK: Yes.
JRH: So if we iterate n times, we will obtain 2n noodles of width (1/3)n. And if we let the number of iterations go to infinity, that is, if we stretch and fold infinitely many times, cutting each time the bent part, we will obtain a Cantor set of vertical noodles.
KK: Yes.
EL: Right. So as you were saying the ninths with these gaps, and this 1/3, I was thinking, huh, this sounds awfully familiar.
KK: Yeah, yeah.
EL: We’ll include a picture of the Cantor set in the show notes for people to look at.
JRH: When we iterate forward, in the limit we will obtain a Cantor set of noodles. We can also iterate backwards. And what is that? We want to know for each point in the square, that is, for each flour particle of the dough in the mold, where it was before the cook stretched vertically and folded the dough the first time, where it came from. Now we recall that the forward iteration was to stretch in the vertical sense and fold it, so if we zoom back and put it backwards, we will obtain that the backward sense the cook has squeezed in the vertical sense and stretched in the horizontal sense and folded, okay?
EL: Yes.
JRH: Each time we iterate backwards, we stretch in the horizontal sense and fold it and put it in that sense. In this way, the left vertical rectangle is converted into the lower rectangle, the lower third rectangle. And the right side rectangle, the vertical rectangle, is converted into the upper third rectangle, and the bent part is cut. If we iterate backwards, now we will get in the first backward iteration four horizontal rectangles of width 1/9 and the gaps, and if we let the iterations go to infinity, we will obtain a Cantor set of horizontal noodles.
When we iterate forward and consider only what’s left in the mold, we start with two horizontal rectangles and finish with two vertical rectangles. When we iterate backwards we start with two vertical rectangles and finish with two horizontal rectangles. Now we want to consider the particles that stay forever in the mold, that is, the points so that all of the forward iterates and all the backwards iterates stay in the square. This will be the product of two middle-thirds Cantor set. It will look more like grated cheese than noodles.
KK: Right.
JRH: This set will be called the invariant set.
KK: Although they’re not pointwise fixed, they just stay inside the set.
JRH: That’s right. They stay inside the square. In fact, not only will they be not fixed, they will have a chaotic behavior. That is what I want to tell you about.
KK: Okay.
JRH: This is one of the simplest models of an invertible map that is chaotic. So what is chaotic dynamics anyways? There is no universally accepted definition about that. But one that is more or less accepted is one that has three properties. These properties are that periodic points are dense, it is topologically mixing, and it has sensitivity to initial conditions. And let me explain a little bit about this.
A periodic point is a particle of flour that has a trajectory that comes back exactly to the position where it started. This is a periodic point. What does it mean that they are dense? As close as you wish from any point you get one of these.
Topologically mixing you can imagine that means that the dough gets completely mixed, so if you take any two small squares and iterate one of them, it will get completely mixed with the other one forever. From one iteration on, you will get dough from the first rectangle in the second rectangle, always. That means topologically mixing.
I would like to focus on the sensitivity to initial conditions because this is the essence of chaos.
EL: Yeah, that’s kind of what you think of for the idea of chaos. So yeah, can you talk a little about that?
JRH: Yeah. This means that any two particles of flour, no matter how close they are, they will get uniformly separated by the dynamics. in fact, they will be 1/3 apart for some forward or backward iterate. Let me explain this because it is not difficult. Remember that we had the lower third rectangle? Call this lower third rectangle 0, and the upper third rectangle 1. Then we will see that for some forward or backward iterate, any two different particles will be in different horizontal rectangles. One will be in 1, and the other one will be in the 0 rectangle. How is that? If two particles are at different heights, than either they are already in different rectangles, so we are done, or else they are in the same rectangle. But if they are in the same rectangle, the cook stretches the vertical distance by 3. Every time they are in the same horizontal rectangle, their vertical distance is stretched by 3, so they cannot stay forever in the same rectangle unless they are at the same height.
KK: Sure.
JRH: If they are at different heights, they will get eventually separated. On the other hand, if they are in the same vertical rectangle but at different x-coordinates, if we iterate backwards, the cook will stretch the dough in the horizontal sense, so the horizontal distance will be tripled. Each time they are in the same vertical rectangle, they cannot be forever in the same vertical rectangle unless they are in the same, unless their horizontal distance is 0. But if they are in different positions, then either their horizontal distance is positive or the vertical distance is positive. So in some iterate, they will be 1/3 apart. Not only that, if they are in two different vertical rectangles, then in the next backwards iterate, they are in different horizontal rectangles. So we can state that any two different particles for some iterate will be in different horizontal rectangles, no matter how close they are. So that’s something I like very much because each particle is defined by its trajectory.
EL: Right, so you can tell exactly what you are by where you’ve been.
JRH: Yeah, two particles are defined by what they have done and what they will do. That allows something that is very interesting in this type of chaotic dynamics, which is symbolic dynamics. Now you know that any two points in some iterate will have distinct horizontal rectangles, so you can code any particle by its position in the horizontal rectangles. If one particle is in the beginning in the 0 rectangle, you will assign to them a sequence so that its zero position is 0, a double infinite sequence. If the first iterate is in the rectangle 1, then in the first position you will put a 1. In this way you can code any particle by a bi-infinite sequence of zeroes and ones. So in dynamics this is called conjugation. You can conjugate the horseshoe map with a sequence of bi-infinite sequences. This means that you can code the dynamics. Anything that happens in the set of bi-infinite sequences, happens in the horseshoe and vice versa. This is very interesting because you will find particles that describe any trajectory that you wish because you can write any sequence of zeroes and ones as you wish. You will have all Shakespeare coded in the horseshoe map, all of Donald Trump’s tweets will be there too.
KK: Let’s hope not. Sad!
JRH: Everything will be there.
EL: History of the world, for better and worse.
KK: What about Borges’s Library of Babel? It’s in there too, right?
JRH: If you can code it with zeroes and ones, it’s there.
EL: Yeah, that’s really cool. So where did you first run into this theorem?
JRH: When I was a graduate student, I ran into chaos, and I first ran into a baby model of this, which is the tent map. A tent map is in the interval, and that was very cool. Unlike this model, it’s coded by one-sided sequences. And later on, I went to IMPA [Instituto de Matemática Pura e Aplicada] in Rio de Janeiro, and I learned that Smale, the author of this example, had produced this example while being at IMPA in Rio.
KK: Right.
JRH: It was cool. I learned a little more about dynamics, about hyperbolic dynamics, and in fact, now I’m working in partially hyperbolic dynamics, which is very much related to this, so that is why I like it so much.
KK: Yeah, one of my colleagues spends a lot of time in Brazil, and he’s still studying the tent map. It’s remarkable, I mean, it’s such a simple model, and it’s remarkable what we still don’t know about it. And this is even more complicated, it’s a 2-d version.
EL: So part of this show is asking our guests to pair their theorem with something. I have an idea of what you might have chosen to pair with your theorem, but can you tell us what you’ve chosen?
JRH: Yeah, I like this sensitivity to initial conditions because you are defined by your trajectory. That’s pretty cool. For instance, if you consider humans as particles in a system, actually nowadays in Shenzhen, it is only me who was born in Argentina, lived in Uruguay, and lives in Shenzhen.
EL: Oh wow.
JRH: This is a city of 20 million people. But I am defined by my trajectory. And I’m sure any one of you are defined by your trajectory. If you look at a couple of things in your life, you will discover that you are the only person in the world who has done that. That is something I like. You’re defined, either by what you’ve done, or what you will do.
EL: Your path in life. It’s interesting that you go there because when I was talking to Ami Radunskaya, who also chose a theorem in dynamics, she also talked about how her theorem related to this idea of your path in life, so that’s a fun idea.
JRH: I like it.
KK: Of course, I was thinking about taffy-pulling the whole time you were describing the horseshoe map. You’ve seen these machines that pull taffy, I think they’re patented, and everything’s getting mixed up.
EL: Yeah.
JRH: All of this mixing is what makes us unique.
EL: So you can enjoy this theorem while pondering your life’s path and maybe over a bowl of noodles with some taffy for dessert.
KK: This has been fun. I’d never really thought too much about the horseshoe map. I knew it as this classical example, and I always heard it was so complicated that Smale decided to give up on dynamics, and I’m sure that’s false. I know that’s false. He’s a brilliant man.
JRH: Actually, he’s coming to a conference we’re organizing this year.
EL: Oh, neat.
KK: He’s still doing amazingly interesting stuff. I work in topological data analysis, and he’s been working in that area lately. He’s just a brilliant guy. The Fields Medal was not wasted on him, for sure.
EL: Well thanks a lot for taking the time to talk to us. I really enjoyed talking with you.
JRH: Thank you for inviting me.
[outro]

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Episode 21 - Jana Rodriguez Hertz

My Favorite Theorem

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Manage episode 210442319 series 1516226
Content provided by Kevin Knudson and Evelyn Lamb. All podcast content including episodes, graphics, and podcast descriptions are uploaded and provided directly by Kevin Knudson and Evelyn Lamb or their podcast platform partner. If you believe someone is using your copyrighted work without your permission, you can follow the process outlined here https://player.fm/legal.

Evelyn Lamb: Hello and welcome to My Favorite Theorem. I’m one of your hosts, Evelyn Lamb. I’m a freelance math and science writer in Salt Lake City, Utah. And this is your other host.
Kevin Knudson: Hi, I’m Kevin Knudson, professor of mathematics at the University of Florida. How are you doing, Evelyn?
EL: I’m all right. I’m excited because we’re trying a different recording setup today, and a few of our recent episodes, I’ve had a few connection problems, so I’m hoping that everything goes well, and I’ve probably jinxed myself by saying that.
KK: No, no, it’s going to be fine. Positive thinking.
EL: Yeah, I’m hoping that the blips that our listeners may have heard in recent episodes won’t be happening. How about you? Are you doing well?
KK: I’m fine. Spring break is next week, and we’ve had the air conditioning on this week. This is the absurdity of my life. It’s February, and the air conditioning is on. But it’s okay. It’s nice. My son is coming home for spring break, so we’re excited.
EL: Great. We’re very happy today to have Jana Rodriguez-Hertz on the show. So, Jana, would you like to tell us a little bit about yourself?
Jana Rodriguez-Hertz: Hi, thank you so much. I’m originally from Argentina, I have lived in Uruguay for 20 years, and now I live in China, in Shenzhen.
EL: Yeah, that’s quite a big change. When we were first talking, first emailing, I mean, you were in Uruguay then, you’re back in China now. What took you out there?
JRH: Well, we got a nice job offer, and we thought we’d like to try. We said, why not, and we went here. It’s nice. It’s a totally different culture, but I’m liking it so far.
KK: What part of China are you in, which university?
JRH: In Southern University of Science and Technology in Shenzhen. It’s in Shenzhen. Shenzhen is in mainland China in front of Hong Kong, right in front of Hong Kong.
KK: Okay. That’s very far south.
EL: I guess February weather isn’t too bad over there.
JRH: It’s still winter, but it’s not too bad.
EL: Of course, that will be very relevant to our listeners when they hear this in a few months. We’re glad to have you here. Can you tell us about your favorite theorem?
JRH: Well, you know, I live in China now, and every noon I see a dynamical process that looks like the theorem I want to talk to you about, which is the dynamical properties of Smale’s horseshoe. Here it goes. You know, at the canteen of my university, there is a cook that makes noodles.
EL: Oh, nice.
JRH: He takes the dough and stretches it and folds it without mixing, and stretches it and folds it again, until the strips are so thin that they’re ready to be noodles, and then he cuts the dough. Well, this procedure can be described as a chaotic dynamical system, which is the Smale’s horseshoe.
KK: Okay.
JRH: So I want to talk to you about this. But we will do it in a mathematical model so it is more precise. So suppose that the cook has a piece of dough in a square mold, say of side 1. Then the cook stretches the dough so it becomes three times longer in the vertical sense but 1/3 of its original width in the horizontal sense. Then he folds it and puts the dough again in the square mold, making a horseshoe form. So the lower third of the square is converted into a rectangle of height 1 and width 1/3 and will be placed on the left side of the mold. The middle third of the square is going to be bent and will go outside the mold and will be cut. The upper third will be converted to another rectangle of height 1 and width 1/3 and will be put upside down in the right side of the mold. Do you get it?
KK: Yeah.
JRH: Now in the mold there will be two connected components of dough, one in the left third of the square and one in the right third of the square, and the middle third will be empty. In this way, we have obtained a map from a subset of the square into another subset of the square. And each time this map is applied, that is, each time we stretch and fold the dough, and cut the bent part, it’s called a forward iteration. So in the first forward iteration of the square, we obtain two rectangles of width 1/3 and height 1. Now in the second forward iteration of the square, we obtain four rectangles of width 1/9 and height 1. Two rectangles are contained in the left third, two rectangles in the right third. These are four noodles in total.
Counting from left to right, we will see one noodle of width 1/9, one gap of width 1/9, a second noodle of width 1/9, a gap of 1/3, and two more noodles of width 1/9 separated by a gap of width 1/9. Is that okay?
KK: Yes.
JRH: So if we iterate n times, we will obtain 2n noodles of width (1/3)n. And if we let the number of iterations go to infinity, that is, if we stretch and fold infinitely many times, cutting each time the bent part, we will obtain a Cantor set of vertical noodles.
KK: Yes.
EL: Right. So as you were saying the ninths with these gaps, and this 1/3, I was thinking, huh, this sounds awfully familiar.
KK: Yeah, yeah.
EL: We’ll include a picture of the Cantor set in the show notes for people to look at.
JRH: When we iterate forward, in the limit we will obtain a Cantor set of noodles. We can also iterate backwards. And what is that? We want to know for each point in the square, that is, for each flour particle of the dough in the mold, where it was before the cook stretched vertically and folded the dough the first time, where it came from. Now we recall that the forward iteration was to stretch in the vertical sense and fold it, so if we zoom back and put it backwards, we will obtain that the backward sense the cook has squeezed in the vertical sense and stretched in the horizontal sense and folded, okay?
EL: Yes.
JRH: Each time we iterate backwards, we stretch in the horizontal sense and fold it and put it in that sense. In this way, the left vertical rectangle is converted into the lower rectangle, the lower third rectangle. And the right side rectangle, the vertical rectangle, is converted into the upper third rectangle, and the bent part is cut. If we iterate backwards, now we will get in the first backward iteration four horizontal rectangles of width 1/9 and the gaps, and if we let the iterations go to infinity, we will obtain a Cantor set of horizontal noodles.
When we iterate forward and consider only what’s left in the mold, we start with two horizontal rectangles and finish with two vertical rectangles. When we iterate backwards we start with two vertical rectangles and finish with two horizontal rectangles. Now we want to consider the particles that stay forever in the mold, that is, the points so that all of the forward iterates and all the backwards iterates stay in the square. This will be the product of two middle-thirds Cantor set. It will look more like grated cheese than noodles.
KK: Right.
JRH: This set will be called the invariant set.
KK: Although they’re not pointwise fixed, they just stay inside the set.
JRH: That’s right. They stay inside the square. In fact, not only will they be not fixed, they will have a chaotic behavior. That is what I want to tell you about.
KK: Okay.
JRH: This is one of the simplest models of an invertible map that is chaotic. So what is chaotic dynamics anyways? There is no universally accepted definition about that. But one that is more or less accepted is one that has three properties. These properties are that periodic points are dense, it is topologically mixing, and it has sensitivity to initial conditions. And let me explain a little bit about this.
A periodic point is a particle of flour that has a trajectory that comes back exactly to the position where it started. This is a periodic point. What does it mean that they are dense? As close as you wish from any point you get one of these.
Topologically mixing you can imagine that means that the dough gets completely mixed, so if you take any two small squares and iterate one of them, it will get completely mixed with the other one forever. From one iteration on, you will get dough from the first rectangle in the second rectangle, always. That means topologically mixing.
I would like to focus on the sensitivity to initial conditions because this is the essence of chaos.
EL: Yeah, that’s kind of what you think of for the idea of chaos. So yeah, can you talk a little about that?
JRH: Yeah. This means that any two particles of flour, no matter how close they are, they will get uniformly separated by the dynamics. in fact, they will be 1/3 apart for some forward or backward iterate. Let me explain this because it is not difficult. Remember that we had the lower third rectangle? Call this lower third rectangle 0, and the upper third rectangle 1. Then we will see that for some forward or backward iterate, any two different particles will be in different horizontal rectangles. One will be in 1, and the other one will be in the 0 rectangle. How is that? If two particles are at different heights, than either they are already in different rectangles, so we are done, or else they are in the same rectangle. But if they are in the same rectangle, the cook stretches the vertical distance by 3. Every time they are in the same horizontal rectangle, their vertical distance is stretched by 3, so they cannot stay forever in the same rectangle unless they are at the same height.
KK: Sure.
JRH: If they are at different heights, they will get eventually separated. On the other hand, if they are in the same vertical rectangle but at different x-coordinates, if we iterate backwards, the cook will stretch the dough in the horizontal sense, so the horizontal distance will be tripled. Each time they are in the same vertical rectangle, they cannot be forever in the same vertical rectangle unless they are in the same, unless their horizontal distance is 0. But if they are in different positions, then either their horizontal distance is positive or the vertical distance is positive. So in some iterate, they will be 1/3 apart. Not only that, if they are in two different vertical rectangles, then in the next backwards iterate, they are in different horizontal rectangles. So we can state that any two different particles for some iterate will be in different horizontal rectangles, no matter how close they are. So that’s something I like very much because each particle is defined by its trajectory.
EL: Right, so you can tell exactly what you are by where you’ve been.
JRH: Yeah, two particles are defined by what they have done and what they will do. That allows something that is very interesting in this type of chaotic dynamics, which is symbolic dynamics. Now you know that any two points in some iterate will have distinct horizontal rectangles, so you can code any particle by its position in the horizontal rectangles. If one particle is in the beginning in the 0 rectangle, you will assign to them a sequence so that its zero position is 0, a double infinite sequence. If the first iterate is in the rectangle 1, then in the first position you will put a 1. In this way you can code any particle by a bi-infinite sequence of zeroes and ones. So in dynamics this is called conjugation. You can conjugate the horseshoe map with a sequence of bi-infinite sequences. This means that you can code the dynamics. Anything that happens in the set of bi-infinite sequences, happens in the horseshoe and vice versa. This is very interesting because you will find particles that describe any trajectory that you wish because you can write any sequence of zeroes and ones as you wish. You will have all Shakespeare coded in the horseshoe map, all of Donald Trump’s tweets will be there too.
KK: Let’s hope not. Sad!
JRH: Everything will be there.
EL: History of the world, for better and worse.
KK: What about Borges’s Library of Babel? It’s in there too, right?
JRH: If you can code it with zeroes and ones, it’s there.
EL: Yeah, that’s really cool. So where did you first run into this theorem?
JRH: When I was a graduate student, I ran into chaos, and I first ran into a baby model of this, which is the tent map. A tent map is in the interval, and that was very cool. Unlike this model, it’s coded by one-sided sequences. And later on, I went to IMPA [Instituto de Matemática Pura e Aplicada] in Rio de Janeiro, and I learned that Smale, the author of this example, had produced this example while being at IMPA in Rio.
KK: Right.
JRH: It was cool. I learned a little more about dynamics, about hyperbolic dynamics, and in fact, now I’m working in partially hyperbolic dynamics, which is very much related to this, so that is why I like it so much.
KK: Yeah, one of my colleagues spends a lot of time in Brazil, and he’s still studying the tent map. It’s remarkable, I mean, it’s such a simple model, and it’s remarkable what we still don’t know about it. And this is even more complicated, it’s a 2-d version.
EL: So part of this show is asking our guests to pair their theorem with something. I have an idea of what you might have chosen to pair with your theorem, but can you tell us what you’ve chosen?
JRH: Yeah, I like this sensitivity to initial conditions because you are defined by your trajectory. That’s pretty cool. For instance, if you consider humans as particles in a system, actually nowadays in Shenzhen, it is only me who was born in Argentina, lived in Uruguay, and lives in Shenzhen.
EL: Oh wow.
JRH: This is a city of 20 million people. But I am defined by my trajectory. And I’m sure any one of you are defined by your trajectory. If you look at a couple of things in your life, you will discover that you are the only person in the world who has done that. That is something I like. You’re defined, either by what you’ve done, or what you will do.
EL: Your path in life. It’s interesting that you go there because when I was talking to Ami Radunskaya, who also chose a theorem in dynamics, she also talked about how her theorem related to this idea of your path in life, so that’s a fun idea.
JRH: I like it.
KK: Of course, I was thinking about taffy-pulling the whole time you were describing the horseshoe map. You’ve seen these machines that pull taffy, I think they’re patented, and everything’s getting mixed up.
EL: Yeah.
JRH: All of this mixing is what makes us unique.
EL: So you can enjoy this theorem while pondering your life’s path and maybe over a bowl of noodles with some taffy for dessert.
KK: This has been fun. I’d never really thought too much about the horseshoe map. I knew it as this classical example, and I always heard it was so complicated that Smale decided to give up on dynamics, and I’m sure that’s false. I know that’s false. He’s a brilliant man.
JRH: Actually, he’s coming to a conference we’re organizing this year.
EL: Oh, neat.
KK: He’s still doing amazingly interesting stuff. I work in topological data analysis, and he’s been working in that area lately. He’s just a brilliant guy. The Fields Medal was not wasted on him, for sure.
EL: Well thanks a lot for taking the time to talk to us. I really enjoyed talking with you.
JRH: Thank you for inviting me.
[outro]

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