 The corresponding eigenvalue is sum of the squares. And with this choice of eigenfunctions, once again, you will see very nice picture of zero set. Zero set of cosine in one direction, zero set of cosine in different direction will give you a very beautiful, simple tonalized picture. But the truth is that both on the sphere and on the torus, much more happens. We have eigenvalues of high multiplicity. You see here, there are many harmonic, homogeneous polynomials of degree n. So there are many functions that you can combine and look at their combinations. These beautiful pictures correspond to a very specific choice of a basis there. The same with the torus. Sorry, there's some of the squares. We know that there are many numbers that you can write as sum of the squares in many different ways. If you pick such eigenvalue, you will have lots of eigenfunctions. And instead of taking these nice products, you can try to take a linear combination of those and look at the corresponding picture of the zero set or nodal domains. I'll show you some of the examples, hopefully. This is an example of two nodal domains for torus or square. So you can combine eigenfunctions such that the number of your nodal domains is just two. The eigenvalue is high. You can do it for very high, for very large eigenvalues. The number of eigendomains is bounded by the number of this eigenvalue in the list by current nodal domain theorem. But on the other hand, it could be very small. It could be two. Same is true on the sphere. I don't have a nice picture, but it's a very old result that there are eigenfunctions of high-frequency, large eigenvalue on the sphere with just two nodal domains. If on the torus you go to really high frequency and look what happens there when you take random combination of these crosshines, you will see a picture like that. It's blue is where the function is negative, white is where it's positive. So there are lots of lines in between where the function is zero, and they're not like that at all. And finally, the last illustration, it's a very beautiful picture. There are many, yeah? You take these elementary eigenfunctions and you take a linear combination with random coefficients. So normalize it in L2, for example, so I think this is illustration when you try to do it on your computer with randomized coefficients. The last picture is nodal domains on the sphere. Many beautiful pictures if you go to this site, it's just one of them, and you see that the nodal set, the lines on the sphere are not so nice structured as before. And in this case, your sphere is divided into several nodal domains. So there is no precise current nodal domain theorem as we go to high dimensions. The eigenvalue doesn't control the number of nodal domains, but still you, looking at the picture, you can guess what your eigenvalue is. Looking at this picture, you can say that eigenvalue is quite large, but probably if you compare this one and that one, the eigenvalue here is smaller than eigenvalue there. How do we know that? There is the following statement, it's called, yeah, what is this? That tells you that if you take an eigenfunction on, and you look at the zero set in dimension D, it's D minus one dimensional thing, and three two-dimensional sphere, it's union of curves. High dimension, the dimension of the zero set is D minus one, so it tells you that this is bounded by square root of lambda with two constants that depend on the manifold. And the hypothesis was proved in the end of 80s by Donald and Fiferman for the case where when metric is real analytic, we'll discuss some of the ideas behind it. Tomorrow, hopefully, but it will give you that on the torus and the sphere with standard metric, looking at the picture, if you able to estimate the zero, going to the zero set, you know approximately what is your eigenvalue. What I'm going to prove is much more mild result, but it already tells you that looking at the picture, you should be able to say something about eigenvalue. What I want to show is that we know the density of the zero set. So we have our compact manifold and claim that there is a constant C such that if we take an eigenfunction corresponding to eigenvalue lambda, then the zero set of this eigenfunction is dense on scale one over square root of lambda. In other words, the distance from any point on the manifold to the zero set is larger than or equal to constant over square root of lambda. The larger is the frequency or the eigenvalue of your eigenfunction, the denser are the zero sets. If you see really big gaps here, you see some balls on this nodal domain and you look at the size of this ball, if you know your constant for this sphere, you can estimate the eigenvalue say that eigenvalue is not very large because we have large balls on nodal domains. It's for any acts on the manifold. Sorry, thank you, sure. There is one classical result on elliptic PD that will need to prove this. It's Harnack inequality that probably should be called Moser inequality for elliptic PDs. It was known to Harnack for harmonic functions and it's the following. If I have solution to my elliptic PD that is positive in some ball, then in the half of this ball the minimum of the function is larger than the maximum of the function times some constant C. The constant depends only on the elliptic constant of the operator. For this result, you don't need elliptic coefficients. You need that you have uniformly elliptic equation and then the constant here depends only on the elliptic constant. It also tells you that you can use Harnack inequality on each scale, small one or large ones. If you control elliptic constant of your operator you have this inequality with the same constant for small and large balls. Excuse me? Let us put it this way because there is a constant function, right? But still for the constant function if I take this small it will be... You want this to be non... Okay. We have maximum principle as well for elliptic PDs so if it's non-negative it's positive in a slightly smaller ball. We will need one more useful trick to prove our density result. The truth is we like to work with harmonic functions or solutions to elliptic PDs. We know a lot about them. Eigenfunctions, they are almost there but you have this annoying term with coefficients that you don't have control on so you prefer to make an eigenfunction into a harmonic function in a sense and you can always do it by introducing a new variable and looking at a function that is now defined on your manifold or your domain times r in the following way you take the value of the eigenfunction at point x and multiply by exponential to the power squared of lambda t. Now for this function on the product manifold we have the Laplace operator that you get when you take the usual metric the product here this is zero. So h is... this function is a solution of standard second-order elliptic PD in divergence form with no lower-order coefficients and I want to explain how hardening quality together with this lifting argument give you the result on the density of the zero-set. Assume that my eigenfunction doesn't change sign for example it's positive on some ball of radius r another way of writing that the distance from x to the zero-set is larger than r my function has the same sign on this ball. Look at the function h that is the lift of the eigenfunction this is positive on the product of this ball times the whole real line it will take just a small part of it so I had a ball and I have a positive function here I'll just scribe a ball into this one and look at the half of the ball and try to look at minimum and maximum of this lifting function in the smaller ball there I will need one more blackboard but remember what lifting is it's written there so the maximum of h over this now d plus one dimensional ball let's write one half of... let me take one of those okay sorry so I had a ball b my initial one in there I constructed a large ball of dimension plus one that I called b prime and half of the ball is where we can apply the Harlequin equality the maximum of this function is larger than the maximum of the function h on this ball I'll write one fourth of the initial b take even smaller ball inside and the maximum of h is definitely larger than maximum of sorry phi lambda on the small one on this plane we multiply by one so the function h is equal to phi on this plane where t is equal to zero on the other hand the minimum of function h if I take a point in the small ball I can still go down and being in this one half of b prime I can go down to a distance that is comparable to r and when I go down I will multiply my function by e to the power square root of lambda times the distance I can travel so the minimum is less than or equal to the maximum over this ball phi lambda but now multiplied to e to minus square root of lambda cr if what the constant c is it's an absolute constant there but compared to the Harlequin equality we see that the ratio between maximum and minimum should be bounded under this term here so we know that e to minus square root of lambda cr is bounded by c lambda let me see this was the minimum this was the maximum so this ratio is bounded by the constant and it tells you that the exponent is bounded or r is bounded by constant over the square root of lambda so the zero set is one over square root of lambda dense there it's very far from the hypothesis it will not give you this result but at least it gives you some feeling that when you change the eigenvalue the nodal set changes and it becomes denser and denser when you increase the eigenvalue I have some more minutes to tell you about part of the truth about your hypothesis there was another very nice conjecture of Derashvili that was initially formulated for harmonic functions harmonic functions, say in R3 suppose that we have a harmonic function in the unit ball of Rd so that the value of the function is zero is zero look at the zero set of this function and the question was is it true that the length of the zero set in this ball is bounded from below by some constant that depends on dimension only in dimension two we're looking at unit circle and harmonic function that is zero in the center we know the maximal principle so it's not a lonely zero there should be a curve through the zero point and this curve doesn't bound anything here so this curve should leave the circle so the length of this curve is definitely at least two but in dimension three you're now looking at a surface that goes through this point and you can draw a surface of even a surface that is zero set of a harmonic function that is very narrow and the area of this surface is very small so now in dimension three you can touch the zero by a very small finger here and there is no simple argument showing that this inequality is true however this was open for many years