Explaining the Jacobian conjecture

[00:00:01]If you've been hearing about Fabel solving a math problem, this is the tweet you've probably seen.

[00:00:07]Uh a guy announced that he solved the or proved the Jacobian conjecture to be false thanks to using uh Claude to help him. So, if you're like me, you're probably wondering what the Jacobian conjecture is.

[00:00:17]Here's a little diagram to explain well, a few diagrams. So, first of all, imagine a real number line. As you slide along the number line, you get different values.

[00:00:25]That's not too complicated.

[00:00:27]Now, let's add dimensions to this and create a complex number line. You've got the real axis on the X dimension and the imaginary axis on the Y dimension.

[00:00:37]If you're not sure about imaginary numbers, you should look into that separately. This video is not going to go too deep on it, but it's just going to help you picture it.

[00:00:44]Now, let's imagine we have um a real and a real and an imaginary number as value one.

[00:00:50]And then, we want to get another number involved.

[00:00:53]So, how can we plot that?

[00:00:55]We can create a a third dimension, the Z dimension, and this marks the value of the real uh element of the second value. So, here we have everywhere in this space is corresponding to a real and an imaginary part of the first value and just the real part of the second value.

[00:01:12]So, what happens if we wanted the second value to have an imaginary part as well?

[00:01:16]Then, we could introduce another dimension like color, maybe. As you can see here, as the color changes, the imaginary part of the second value changes. And as we move around the screen, uh different values are selected. So, yeah, this is a way to kind of show the range of different combinations of two imaginary numbers that can be put together.

[00:01:36]Now, let's say that we apply a function.

[00:01:38]So, we take an input that is two real and imaginary numbers, uh two complex numbers, and we apply a function to them, and we'll get an output of two complex numbers. Uh this this diagram kind of shows that. As we move it to different areas and the colors change, then we're getting different inputs of different complex numbers, and we get different outputs of two complex numbers.

[00:01:58]So, that's all a little bit confusing, but let's try and visualize what that looks like.

[00:02:01]So, for example, you have X and Y as the inputs that for the complex numbers and they go through a function here.

[00:02:07]So, an example of a function could be that to get the new X value, you add X + Y squared. And then the Y is just the same.

[00:02:16]So, if you do the math yourself, these two complex numbers when fed in through this system, do a bit of algebra, and they come out as these two numbers on this side.

[00:02:29]So, let's let's visualize that in a different way cuz this doesn't just apply to a single point. This applies to the whole fabric of the of the dimensions in that being changed. So, as you can see here, as the transformation is applied, this is kind of how like the the fabric of the of the graph bends and distorts.

[00:02:47]And then if you take that arrow and you put it on a normalized image of um of the the dimensions, you get this this second graph down here. All right.

[00:02:56]So, that kind of shows you the way that these transformations work.

[00:02:59]They apply across the whole space-time.

[00:03:01]Now, let's say you take a specific point, this green point, and its value is the these three values.

[00:03:07]If you run this through the equation, then you'll get these three output values here.

[00:03:12]So, yeah, it it works everywhere uh the same way the transformation is applied throughout the whole vector space.

[00:03:19]Now, let's talk about the determinant.

[00:03:21]So, let's take the space that we took before. Let's draw a cube around that.

[00:03:27]And if we took we took the area or the volume of this cube, and then we applied the transformation, and we took the new volume of the new shape, and we found the ratio between them, that would be the determinant. It kind of tells you how much the uh shape changes um and and the ratio of how the value increases or decreases based on um the the function. So, that will be different based on whatever this function is up there.

[00:03:55]So, let's talk about the Jacobian.

[00:03:58]So, the Jacobian of a of a function occurs when you find that the partial differentials of each of the elements and it kind of tells you how a particular point how the um how the function is changing at that point. So, if you take any point in the in in the vector space and you find the Jacobian, um you'll you'll find the uh I think it tells you that if you nudge it a little bit, how much it would change in each direction. And that's what this vector represents. So, if you if it changes in the uh X direction, you'll see here and and the Y direction, you get these values here.

[00:04:33]So, yeah, that's the Jacobian.

[00:04:37]You can calculate it by doing these partial derivatives.

[00:04:39]Um and it'll be different at different points.

[00:04:42]But one thing you'll find is that you can then take the the the Jacobian matrix and find the determinant using the some basic linear algebra.

[00:04:56]And if the determinant is constant, then it fits a special case.

[00:05:01]And this special case is the case uh that the conjecture is made about.

[00:05:06]So, what that's saying really is that if if the determinant is constant at every point where you take the Jacobian, then it fits a specific use case.

[00:05:17]Um there are some functions where the the Jacobian is different all over and when you take the determinant, the determinant is therefore different, too.

[00:05:26]Um but then there are some where it's constant. So, you can see in this graph for this function, the determinant is the same all over, no matter where you took the Jacobian, the determinant will be the same value.

[00:05:39]Um and here we we changed the equation.

[00:05:44]As you can see, we've got now X squared and Y, [clears throat] and the determinant is is different all over. So, it depends on where you decide to take the Jacobian to find the partial differentials in that area, the determinant will be different, which means that when you do the transformation, the shape will scale in a different way at a different rate.

[00:06:03]So, those are the kind of like things we got to determine between here, whether the function is giving you a constant determinant throughout or if it's giving you a variable determinant.

[00:06:18]Again, we're trying a different thing here.

[00:06:20]We're showing that in this version, the the colors are still showing the second part of the imaginary number. As the colors are changing, the values are changing of the determinant. So, you can see here that this function wouldn't be in the club, as you call it.

[00:06:44]So, if we take two points, we're going to take two equations here.

[00:06:47]Uh this is the first equation, where the determinant ends up being two, so it's constant throughout. And the second equation, the determinant ends up being 2x, so it changes based on where you took the Jacobian.

[00:07:00]So, we're going to show you something there. If you take two inputs in both of these both of these functions, one function is in this special club uh that the conjecture is made about, and the other one is not.

[00:07:11]And you take two values, the the red and orange squares as just two specific input points, and you apply the transformation, what you'll find is that in the in the ones where the it's in the club, where the determinant is constant throughout, the two inputs will usually go to two different outputs. And in fact, that's what the conjecture actually states is that in the instance where the determinant is constant throughout, the inputs will always be corresponding to to a singular output, and the outputs will correspond to a singular input.

[00:07:41]Um whereas in this graph on the right, you can see that two of the same input have ended up in the same output. So, you couldn't really reverse it cuz it could go any other direction.

[00:07:51]Um and the conjecture basically tells us that if the if the graph has a determinant as constant, you'll be able to it's possible to uh map each output to one particular input point, which is obviously impossible on the graph on the right. And that's that's the conjecture. That's what was stated. And that's what was proven wrong in this tweet. Um so, we'll go over it in a minute, but before we go deeper into that, let's talk a little bit about the complex plane in more depth.

[00:08:22]So, if you imagine a complex number of uh of one dimension, um you've got the real axis and the imaginary axis.

[00:08:31]And when you apply function to that, that changes.

[00:08:34]Uh you can see the the graph on the right shows this, and you get a new output.

[00:08:38]And then, what we've been looking at so far is a complex dimension uh the complex number in this kind of squared, so there's two different complex numbers.

[00:08:47]Um and so, you get this axis that we've been looking at that uses three dimensions plus a dimension of color.

[00:08:53]Uh you can see that when the function's applied to it, the whole of the the dimension of space kind of bends and and warps.

[00:09:01]And now, let's look at the the instance when there's three complex numbers. With it's kind of tricky for this one, so what we've done is we've put every point with put a vector a new vector in that has a magnitude and a phase, and that magnitude and phase is encoding the third uh complex number. Which is it's quite complicated, so it's it's hard to visualize those. It's you know, cuz this is a six-dimensional graph at basically this point.

[00:09:27]So, we show that this is what it looks like before the transformation.

[00:09:33]Transformation gets applied.

[00:09:35]The the changes, and now we go through all the different color options that there are after this. You could see how how this changes.

[00:09:42]So, this is just going to help you visualize what it looks like when uh uh three complex numbers are transformed.

[00:09:52]If there's two points in this complex number graph and they end up meeting, um then that would be something that that people would predict that the determinant must be variable.

[00:10:04]So, let's take a look at this function.

[00:10:06]So, this is the actual function that the Twitter user uh said he thinks is going to be break the conjecture that does break the conjecture. So, this is the visualization of what happens when you apply the transform to this uh to these three complex numbers and and you apply it in this way. You see that this this formula here, here's what happens to the to the transform of the graph area.

[00:10:30]So, the vector space bends and three points ends up connecting them one point, so three inputs have the same output, which means it's not reversible.

[00:10:39]So, you know, you think, "Well, I'm sure it's got to have a variable determinant, and that would explain it."

[00:10:45]But when you actually calculate the determinant of it, you take the Jacobian, um you find that everything actually perfectly cancels out, and if you want to do the math yourself, you can, and the determinant ends up just being a constant of minus two.

[00:10:57]So, this disproves the uh the Jacobian conjecture.

[00:11:03]And um yeah, it's pretty interesting for using AI to just kind of teach you math.

[00:11:09]Um I'll put out another video explaining exactly how this guy actually managed to find this example, cuz what he's really done there is he's found a function um and he's found specific inputs that map to a specific output. That's quite a computationally hard thing to do, so there's been some method to it. Um I think it's [clears throat] all coming out more about how he's done this now, but this is more about the the problem of what problem he's really solved here, as this is like a very old problem that people thought was insoluble. No one thought you could disprove it, but yeah, there you go. Thanks for watching.