Two days ago, I had an interesting interaction with some other students in my REU that destroyed two misconceptions I have had in analysis for a long time.
First, we introduce the idea of a bump function. Let \(M\) be a smooth manifold with \(A\subseteq U\subseteq M\), where \(A\) is closed and \(U\) is open. We define a bump function for \(A\) supported in \(U\) to be a continuous function \(\psi\colon M\to\mathbb{R}\) with \(0\leq\psi\leq1\) on \(M\), \(\psi=1\) on \(A\), and \(\mathrm{supp}\ \psi\subseteq U\). So far so good. It is not a particular surprise that such a function exists. We only require that \(\psi\) be continuous. Here is where my intuition had failed me for years up to this point: smooth bump functions exist! This was a complete shock to me. I was not familiar with any smooth functions that were constant in some neighborhood, and non-constant elsewhere. Moreover, I have thought about this scenario many times before and concluded that it ought to be impossible. I figured if a function was of class \(C^{\infty}\), then it could not be constant (so that its derivatives of all orders vanished) and then suddenly get kicked and start moving. I believed that for such a "kick" to occur, one of the functions derivatives must be discontinuous, staying at zero in some neighborhood before suddenly jumping to some nonzero value. Alas, I was wrong. Smooth bump functions do exist. This follows from the existence of smooth partitions of unity. Most of the technical work that goes into proving that smooth bump functions exist is actually hidden under the rug of proving that smooth partitions of unity exist. But taking that for granted, we note that \(U\) and \(M\setminus A\) form an open cover of \(M\). Then, one may find a smooth partition of unity \(\{\psi,\lambda\}\) subordinate to this cover. By definition, \(\lambda\) vanishes on \(A\), so \(1=\psi+\lambda=\psi\) on \(A\). Indeed, this \(\psi\) is the desired bump function! Having crushed one of my childhood misconceptions, another was crushed the other day when I saw the following proposition in Lee's Introduction to Smooth Manifolds. Proposition: Let \(M\) be a smooth manifold with or without boundary, \(p\in M\), \(v\in T_pM\). If \(f,g\in C^{\infty}(M)\) agree on some neighborhood of \(p\), then \(vf=vg\). The statement of this theorem is not particularly relevant, but the hypothesis immediately caught my eye. Why is it stipulated that \(f\) and \(g\) agree just on some neighborhood of \(p\)? I had always figured that the evolution of smooth functions were determined by their behavior in any neighborhood (similar to how continuous functions are determined by their behavior on dense subsets). The motivation here was the uniqueness of solutions to differential equations and dynamical systems. Once one describes a global relationship between a a function and its derivatives, and an initial condition, the evolution of the function past the starting point is determined. I believed that if we strengthen the agreement between two functions to occur not at a single point, but an entire neighborhood, and that both functions were \(C^{\infty}\), then the evolutions of the functions beyond on the neighborhood must also agree. In other words, I did not understand why the proposition said that \(f\) and \(g\) agree on some neighborhood, because I thought that that occurs precisely when they agree everywhere. Alas, this is also wrong. Upon discussing this with some other students at my REU, I found the counterexample I was looking for. Using our previous notation, if one considers \(g=\psi f\), where \(\psi\) is a bump function, both functions agree in any neighborhood contained in \(A\) by construction, but are very free to disagree outside of \(A\)! Somehow, it seems that bump functions allow us to construct strange functions whose derivatives fail to propagate local information globally, even though the functions are infinitely smooth! This observation is further strengthened by the fact that bump functions are actually used in the proof of the proposition written above (and the proposition itself is a statement about how tangent vectors do not care about global behavior—they only care about local behavior). No contradiction is reached with my intuition stemming from unique solutions to differential equations. After all, having a relationship between a function and its derivatives that holds everywhere is quite a strong condition, and it's not any surprise that unique evolution is forced in that scenario. After a little bit of digging, I found the gap in my knowledge. The identity theorem states that analytic functions that agree on some neighborhood must agree everywhere. So it seems that I have stumbled upon a crucial qualitative difference between analytic functions and smooth functions. In particular, analyticity is a strong enough condition for local behavior (at least in a neighborhood) to propagate globally, while smoothness is not. Obviously, any concrete examples that I attempted to think of were analytic. My intuition is still grappling with this. But it's good to know that I was wrong for so long.
0 Comments
Suppose that we have a function \(f\colon U\to\mathbb{R}^m\), where \(U\) is an open subset of \(\mathbb{R}^n\). Then, we define a regular point to be a point \(\vec{c}\in U\) such that the derivative of \(f\) at \(\vec{c}\) is surjective.
Suppose \(f(\vec{c})=\vec{b}\). Then, the implicit function theorem states that in a neighborhood of \(c\), the equation \(f(\vec{x})=\vec{b}\) can be expressed as an implicit function \(g\) acting on \(n-m\) active variables (components of \(\vec{x}\)), yielding the remaining \(m\) components (passive variables). In particular, we may write \[g\begin{pmatrix} x_{i_1}\\ \vdots\\ x_{i_{n-m}}\end{pmatrix}=\begin{pmatrix}x_{j_1}\\ \vdots\\ x_{j_m}\end{pmatrix},\] where \(\{x_{j_1},...,x_{j_m}\}=\{x_1,...,x_n\}-\{x_{i_1},...,x_{i_{n-m}}\}\). The regularity condition is an interesting one. I'm not sure I quite get it. When the derivative of \(f\) fails to be surjective, that means that the rank of the Jacobian matrix is not equal to \(m\). This means that at least one of the column vectors of the Jacobian can be written as a linear combination of the others. That is, one of the partial derivatives of \(f\) is a linear combination of some of the others. This is a strange condition. The partial derivatives of \(f\) must be linearly independent for an implicit function to be guaranteed to exist. If I think really hard, I can come up with the following heuristic. Suppose a partial derivative of \(f\) can be written as a linear combination of other partial derivatives of \(f\). This means that there are two directions one can move from \(\vec{c}\) in \(U\) so that the change in \(f\) is the same. One is along a standard basis vector, and the other is along some linear combination of other standard basis vectors. This leads to a problem. If there are two points in a small neighborhood of \(\vec{c}\) that have the same image through \(f\), there must be a small neighborhood of \(f(\vec{c})\) (by continuity) that has a point that is the image of more than one point in the domain. This immediately implies that a local inverse doesn't exist here. But apparently, a local implicit function doesn't exist here either. Anyway, perhaps it's a lot more meaningful to think about what the implicit function allows us to do, which is to talk about a rudimentary definition of a manifold. To do this, first we define the graph of a function. I know, I know, we all know what a graph of a function looks like. But what is a graph? Definition: Consider a function \(g\colon\mathbb{R}^{k}\to\mathbb{R}^{n-k}\). The graph of \(g\), denoted by \(\Gamma(g)\), is the set of points \(\begin{pmatrix}\vec{x}\\\vec{y}\end{pmatrix}\in\mathbb{R}^n\) such that \(f(\vec{x})=\vec{y}\). This makes a lot of sense, especially if you think about visualizable cases like \(k=2\) and \(n=3\). In this case, the graph of the function lives in \(\mathbb{R}^3\) and is, well, literally the graph of the function we are thinking about. A k-dimensional manifold is then some set that locally resembles \(\mathbb{R}^k\). Before we get more precise, let us think of an example. The saddle surface of a potato chip is a two-dimensional manifold since if we zoomed in on it, it looks like \(\mathbb{R}^2\). The union of the coordinate axes in \(\mathbb{R}^2\) is not a one-dimensional manifold, since at the origin, no matter how much we zoom, it does not look like \(\mathbb{R}\). Oh, and it should be mentioned that most manifolds that we think about are actually submanifolds that are embedded in a higher-dimensional space. But for our purposes, we'll ignore this and call what we think of just manifolds. Suppose that \(\vec{b}\) is a regular value of \(f\colon\mathbb{R}^n\to\mathbb{R}^m\). Then, consider the subset of the graph \(\Gamma(f)\) where the only value \(f\) is mapping to is \(\vec{b}\). Denote this subset as \(f^{-1}(\vec{b})\). In particular, \(f^{-1}(\vec{b})\) lives in \(\mathbb{R}^{m+n}\). By the implicit function theorem, in a small neighborhood of any point in \(f^{-1}(\vec{b})\), the intersection of \(f^{-1}(\vec{b})\) and that neighborhood is the graph of some local implicit function \(g\colon U\to\mathbb{R}^m\), where \(U\subset\mathbb{R}^{n-m}\). The idea is that \(g\) is an isomorphism between a subset of \(\mathbb{R}^{n-m}\) and a part of \(f^{-1}(\vec{b})\). So, for small sections, \(f^{-1}(\vec{b})\) is isomorphic to \(\mathbb{R}^{n-m}\). This is what we mean when we say "resembles". In our particular case, \(f^{-1}(\vec{b})\) is a \((n-m)\)-dimensional submanifold of \(\mathbb{R}^{m+n}\). I'll eventually get around to learning the proof of the implicit function theorem. It's pretty messy, and involves Kantorovich's theorem and Lipschitz ratios, neither of which I am familiar with. Perhaps I'll start a blog series on my progress treading through those topics! |
Categories
All
Archives
July 2023
|