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.
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Here is a problem from Chapter 1 of B. C. Hall. Lie groups, Lie algebras and Representations that I am working on for my REU reading.
Problem: A subset \(E\) of a matrix Lie group \(G\) is called discrete if for each \(A\) in \(E\) there is a neighborhood \(U\) of \(A\) in \(G\) such that \(U\) contains no point in \(E\) except for \(A\). Suppose that \(G\) is a path-connected matrix Lie group and \(N\) is a discrete normal subgroup of \(G\). Show that \(N\) is contained in the center of \(G\). Solution: To solve this problem, I was first immediately reminded of the map \(\Phi_X(A)\colon N\to G\) given by \(\Phi_X(A)=XAX^{-1}\) that Hall introduces earlier in the chapter, where we have \(X\in G\) and \(A\in N\). This map is obviously continuous in both \(X\) and \(A\) due to the nature of matrix multiplication. More motivation for thinking of this map comes from the fact that \(\Phi_X\) actually has its image as a subset of \(N\) since \(N\) is normal. I saw that if \(A\) was in the center, it must be true that \(\Phi_X(A)=XX^{-1}A=A\) for any choice of \(X\in G\). Moreover, the converse is true, since \(XAX^{-1}=A\) implies \(XA=AX\). So it sufficed to show that \(XAX^{-1}=A\). A common construction in path-connected abstract topological spaces is to draw paths connecting a point of interest to an "important point" in the space. I have seen this construction before in algebraic topology, where numerous proofs involving path-lifting requires paths to be drawn from a chosen point in the fiber (which determines the lift), to some other point of interest. In our case, \(G\) is a topological group, and an important element of any group is the identity. So we let \(\lambda_X\colon[0,1]\to G\) to be the path in \(G\) such that \(\lambda_X(0)=X\) and \(\lambda_X(1)=I\). Now define \(f\colon[0,1]\to G\) given by \[f(t)=\Phi_{\lambda_X(t)}(A).\] What I had to show was clear: the function \(\Phi\) evaluated at \(A\) must be constant along the path \(\lambda_X(t)\). In other words, \(f(t)\) needed to be the constant function with image \(A\). At this point, I began fiddling with a literal notion of "closeness". One may put the metric induced by the Hilbert-Schmidt (Frobenius) norm on \(G\) to turn it into a metric space, and do some \(\epsilon\)-\(\delta\) calculations to determine that in fact \(f(t)=A\) for all \(t\). But I later realized that this is not a nice solution, since we are imposing additional (unnecessary) structure on \(G\). It turns out that we can prove the claim entirely topologically. First note that as a composition of continuous functions, \(f\) is continuous. Since \(N\) is discrete, around each \(Y\in\mathrm{Im}\ f\), there exists an open neighborhood \(U_Y\subseteq G\) that intersects \(N\) only at \(Y\). By the continuity of \(f\), the sets of the form \(f^{-1}(U_Y)\) are open in \([0,1]\) (with the subspace topology inherited from \(\mathbb{R}\)) and form an open cover of \([0,1]\), which is compact by the Heine-Borel theorem. So we can extract a finite subcover. But the sets in the cover are preimages, so they are pairwise disjoint. The only way to write \([0,1]\) as a finite union of pairwise disjoint sets that are open in \([0,1]\) is if \([0,1]\) is the only (nonempty) set in the union. So one of the preimages is \([0,1]\), the entire domain of \(f\). That is, there is only one element in the image of \(f\). Since \(f(1)=IAI=A\), this element is \(A\). \(\square\) |
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