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Geometry · Blog Entry
So, today, I woke up and got breakfast (pancakes). I was pretty excited, for today, there would be the second lecture for topology.
I first attended a precept for the physics class on Newton's laws. I really need to practice with these problems. At its most simple level, the problems are trivial to me. However, they can get increasingly complicated, and even though they may be easily done for someone who still has those skills freshly developed, I have not touched mechanics in a long while, and have not adequately practiced, either. So I will have to practice and re-learn all the techniques now.
I had lunch afterwards, which was pizza, and then attended my Latin class. My teacher is seriously proposing the idea that I skip out of this Latin class if it is too slow for me. I want the easy A. Anyways, we covered the first and second declensions of nouns today. It was all very straightforward. I also did the exercises due tomorrow (as of now) before class (today) even began. There is a quiz tomorrow, but I am not worried. I got this in the bag, to be honest.
I had little time for a break afterwards, for fifteen minutes after Latin ended, I had my topology lecture. Today, we covered some topological spaces, the notion of a basis, the product topology, the order topology, and the subspace topology. This is comparatively more interesting than set theory that was covered in the first lecture, but still very familiar material for me. I imagine that the amount of familiar content will dwindle as the semester goes on and we move into the fundamentals of algebraic topology.
After topology, I studied at the Frist Campus Center with two of my friends, Darby and Jessica. I lent Darby my algebra textbook (which was required for her class on abstract algebra) and I began to work on the topology problem set, with 3 out of 7 problems completed so far. After that, nothing more is of interest to the reader. I took a nap and missed dinner, which kind of sucks.
I still have writing seminar work to do, which I will do tomorrow. I essentially have to "criticize" a source that I read before, pointing out potential tensions in the text compared to a map that I see. Essentially, I need to discuss the limitations of the text, and where my "motive" diverges from the literature. Then I need to do a 3 and a half page reading which is no big deal. All of this should be easily done in the morning. I will update you guys later.
We begin with the notion of a critical point, which is a generalization of the concept of a critical point from single-variable calculus. The image of such is called a critical value, and any point that is not a critical value is considered regular.
Suppose $\phi: M^m \to N^n$ is a smooth map. We say a point $p \in M^m$ is a critical point of $\phi$ if $\phi_*: T_p(M) \to T_{\phi(p)}(N)$ has rank less than $n$. The image of a critical point in $N^n$ is called a critical value. Any point in $N^n$ that is not a critical value is called a regular value, even if it is not in the image of $\phi$.
We now approach one of the main results regarding regular values, the proof of which requires some measure theory, and so will not be provided here. The interested reader is directed to Appendix C of Bredon's Topology and Geometry for further information.
If $\phi: M^m \to \mathbb{R}^n$ is a smooth map, then the set of critical values has measure zero; that is, we can cover the set of critical values with a countable number of intervals of arbitrarily small length.
A corollary swiftly follows from Sard's Theorem.
Suppose that $\phi: M^m \to N^n$ is smooth. Then the set of regular values is residual in $N^n$.
Suppose $C$ is the critical set of $\phi$ (that is, the set of critical points), and $K \subset M$ is a compact set. It naturally follows that $\phi(C \cap K)$ is compact and has empty interior, due to Sard's Theorem. It naturally follows from elementary properties in real analysis that $\phi(C \cap K)$ is nowhere dense in the codomain $N$. $M$ may be covered by a countable number of such compact $K$, so $\phi(C)$ is of first category (that is, it can be covered by a countable number of nowhere dense sets), and so its complement is residual (I still have no idea what residual means).
Interestingly enough, Sard's theorem can be used to prove the Fundamental Theorem of Algebra. We shall do so now, with an argument straight from Milnor's Topology from the Differentiable Viewpoint.
Suppose $p(z)$ is a complex polynomial, with degree not zero. Then $p(z)$ has a zero.
Suppose $p(x+iy) = u(x,y)+iv(x,y)$. Differentiating, we get $p'(z)=(u_x+iv_x)=-i(u_y+iv_y)$, as we may be able to differentiate the expression with respect to $x$ and $y$ and equate the two. It follows that the Jacobian $J(u,v;x,y) = u_x^2 + u_y^2$ is zero if and only if $p'(z)=0$. It follows that there are only a finite number of points (which are zeros of $p'$) that can satisfy this, so $p: \mathbb{R}^2 \to \mathbb{R}^2$ has only a finite number of critical points, the set of which we shall denote by $F$. Now, suppose $p(z) = a_0 z^n + a_1 z^{n-1}+ \cdots + a_n$ such that $a_0 \not = 0$. We have that $$ |p(z)| = |z|^n |a_0 + a_1 z^{-1}+\cdots + a_n z^{-n}|$$ shows that as $|z| \to \infty$, then $|p(z)| \to \infty$. So, we may extend $p$ continuously to the one-point compactification $S^2$ of $\mathbb{R}^2$. This allows each compact subset of $S^2$ to have a compact image, and so $p$ is by definition a proper map. It is an earlier result from the textbook that it follows that $p$ is a closed mapping. Now, we have that $p^{-1}(c)$ consists of the zeros of the polynomial $p(z)-c$, and so may contain at most $n$ points (I guess it is a result we may use that a polynomial has at most its degree in terms of roots). Suppose $k = k(c)$ be the number of points in $p^{-1}(c)$. Then, if $c$ is regular and $p^{-1}(c) = \{z_1,\ldots, z_k\}$, then we may say that each $z_i$ has a neighborhood $U_i$ which is diffeomorphic to a neighborhood $V_i \subset \mathbb{R}^2 - F$ of $c$. Note that $\mathbb{R}^2$ is Hausdorff, so let $U_i$ be disjoint. Consider that there exists an open connected neighborhood $V$ of $c$ inside the very complicated looking open set $V_1 \cap \cdots \cap V_k - f(\mathbb{R}^2 - (U_1 \cup \cdots \cup U_k))$. Let $W_i = U_i \cap p^{-1}(V)$. This is diffeomorphic by $p$ onto $V$. We also have that $p^{-1}(V) = W_1 \cup \cdots \cup W_k$, which are each disjoint. Therefore, $k(c)$ is locally constant on $\mathbb{R}^2 - F$, but $\mathbb{R}^2 - F$ is connected, so $k(c)$ is constant on $\mathbb{R}^2 - F$. It is nonzero, but the image of $p$ must contain $\mathbb{R}^2$, and to be honest this book has completely lost me. Why is Bredon like this? I think even Milnor's treatment of the Fundamental Theorem of Algebra was more clear.