## Archive for the 'quick posts' Category

### Joint Mathematics Meetings in San Francisco -Blog!

My friend and former UT graduate student Adriana Salerno (currently at Bates) will be running the 2010 AMS Joint Math meetings blog. She was also in charge of the blog in previous years (you can check them out here and here). I recommend you check it out in the next few days to see what has been going on at the meetings (specially if, just like me, you don’t happen to be in San Francisco this week).

### New year, new posts

After 4 months of inactivity, I am taking up again the task of updating the blog, which has suffered of neglect due to my terrible time management skills.

I am not going to take off from where I left last time (namely, the posts about the Minkowski problem, which I will finish, someday), but instead will start the year with some shorter, lighter posts. I plant to start with a few posts about varifolds vs currents vs BV sets, and also about the Harnack inequality, maybe later I will write a bit about topics from phase transitions such as the Stefan problem or the Cahn-Hilliard equation

### If you want to kill your productivity, move to a new house!

I know things have been extremely slow lately, but I was moving last week (and well, that means packing everything, moving things to the new house, cleaning the old house, unpacking things at the new one… well, you get the picture).

My goal for this week is presenting Aleksandrov’s solution to  the Minkowski problem (see an earlier post I did introducing this problem). So I am going to leave you a problem as a preview, it is a sort of discrete version of the Minkowski problem:

Let $n_1,...,n_k$  be a family of non-coplanar unit vectors in $\mathbb{R}^3$ and let $\alpha_1,...,\alpha_k$ be positive numbers such that

$\sum \limits_{i=1}^k \alpha_in_i=0$

Then, show that there exists a convex closed polyhedron with exactly $k$ faces with normal vectors given by $n_1,...,n_k$ and corresponding areas $\alpha_1,...,\alpha_k$. Plus, this polyhedron is unique up to translation.

This is fact is not surprising (since it is not hard to check that any polyhedron has this property), but the proof is far from trivial. As you may guess, the proof cannot be constructive, it will use a continuity argument to show that there must be at least one such polyhedron. I will present this in my next post.

Don’t be scared, this has not been turned into a photo blog!  Today I received my copy of Singular Integrals I ordered on-line recently, which is funny given that I have been reading that book for a long time now (at least now I won’t have to borrow a copy from the library or from a friend).

In any case, this reminded me of something I heard once: basically, that throughout the early years of your career as a mathematician, there will be a  list of books (that will depend strongly on your research interests) that you must read completely and in full detail to the point where you are be able to reproduce their contents on command. So, wondering what that list should be for me, I piled up some books from my book shelf and took a picture.

…perhaps I should be reading some of those books instead of writing this blog post

### Quick post: A theorem and an open problem about Gauss curvature

While I finish my next post on the Dirichlet problem for the Monge-Ampere equation, I thought I could mention two neat things I have learned from reading Jerry Kazdan’s survey on Prescribing the curvature of a Riemannian manifold, a short book that I recommend strongly (its a bit out of date, but hey!, geometry has advanced quite a bit in the last couple of years!)

First, a cute solution to an easily-stated problem, as  observed by Wallach and Warner in 1970

Theorem:  Given a compact smooth 2-d manifold and $\Omega$ is a 2-form such that $\int_M \Omega = 2\pi \chi(M)$, then there exists a smooth Riemannian metric $g$ on $M$ such that $\Omega = KdA$, where $K$ is the Gauss curvature of $g$ and $A$ is the area form of $g$.

Proof: Pick your favorite metric $g_0$ on $M$, we will  prove that the metric we are looking for is in fact conformal to $g_0$, for any smooth function $u$, define $g_u = e^{2u}g_0$. Using the well known formulas for the Gauss curvature and area form of $g_u$, one arrives at the identity

$K_udA_u=K_odA_0-(\Delta_0 u) dA_0$

where the sub-indexes $0$ and $u$ refer to the object corresponding to the metric $g_0$ and $g_u$, in particular, $\Delta_0$ is the Laplace operator for $(M,g_0)$. Then one wants to pick $u$ such that

$K_odA_0-\Delta_0 u dA_0=\Omega$

or $(\Delta_0 u) dA_0 = K_0dA_0-\Omega$

but, using the metric $g_0$ we can write $\Omega = f(x)dA_0$ for some $f$ (or equivalently, using the Hodge star operator given by $g_0$) thus the $u$ we want is the solution of the equation

$\Delta_0 u = K_0-f (**)$

but since $\int_M \Omega =\int_MfdA_0= 2\pi\chi(M)=\int_MK_0dA_0$ we have that $\int_MK_0-fdA_0=0$ thus by standard elliptic theory (or Fredholm theory, etc) there exists a smooth function $u$ solving equation (**). Thus the metric $g_u$ is the one we were looking for and the theorem is proved.

Now, an easily stated problem whose solution is not likely to be as short, and has led to a lot of research in the last 2 decades. I  read about it for the first time in an interview with Louis Nirenberg, which can be found here.

Open problem:  Given a two dimensional Riemannian manifold, can we embed it isometrically (even just locally) in three dimensional Euclidean space?

Recall that Gauss curvature solely determines the local geometry of a 2d manifold, so its not surprising that this problem is equivalent to the following problem involving the  (…tataaaaa! )  Monge-Ampere equation:

Open problem restated: Given a $C^\infty$ function $K(x,y)$, find (even just locally) a function $u(x,y)$ which solves the equation

$\frac{u_{xx}u_{yy}-u_{xy}^2}{\left (1+u_x^2+u_y^2 \right)^2} = K(x,y)$

but mind you, this is not your grandpa’s Monge Ampere equation, for it is not an elliptic equation unless $K$ is strictly positive. For general $K$, it is an equation that varies  between hyperbolic or elliptic according to the sign of $K$, so you run into real problems in the set of points where $K$ vanishes .  If on the other hand $K$ is strictly negative, Kazdan says that the equation above is solved using tools from non-linear hyperbolic equations, of which sadly I know nothing. The case $K >0$ is then dealt with the techniques from the last two and the next post, which I should finish in a day or two.

### Pointing out the obvious…

Yes, posting has been extremely low (ahem, nonexistant that is) for the past couple of weeks.  I just got back from my visit at the IAS so hopefully I will have time to post this weekened. In the mean time… ahem Ray, Carl, Orit? (btw Sean, I loved those knot pictures!)

Update: Hat tip to Hector for pointing out a typo in a previous version of this post! (your proof reading skills are unbeatable Hector!)