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Geometry · Blog Entry
If I'm being honest, today wasn't very interesting. I didn't do much today aside from nap and study and do laundry, so I think I will make this section a bit more of a reflection.
I can't lie, I'm getting a little nervous about college. It's not that I'm nervous about classes; other than the writing seminar, I think I should do alright in most of them. The problem is, I worry about my social life. I thrived before actually coming to Princeton by making friends online through Instagram. This was really easy for me, much easier than socializing in real life. But socializing in real life is a lot harder for me, to be quite honest. I worry that my experience at Princeton will be somewhat cheapened due to my inadequacy in such a skill. And perhaps it is something that I need to learn, as it is something that will serve me well in the future. But still.
To be honest, I always found it difficult to make friends, for whatever reason. Especially close friends. Sometimes I feel strange, perhaps different. Not in a good way, not in a bad way. I mean, who writes a blog and immediately follows it up with graduate-level geometry and topology? My mom says it takes time, it will take time for me to settle into the college life. She says that I just need to be patient. But is that true? What if it never happens to me? What if I never truly settle in? What if I always find myself alone?
I went to Fine Hall today after dinner, to try to take my mind off of things. But I just can't shake the feeling that things are not going to go well. I guess it's true I've made plenty of friends, and tomorrow should actually be a great day for me socially. Maybe it's just a distorted perspective of some kind. But what if it's not? What if it gets worse and worse and worse, until I feel completely alone? Anyway, I started reading more of Bredon. This time, I went over the proofs for the famous Inverse Function Theorem and Implicit Function Theorem. It was a nice change of pace from the point set topology review I was doing before. In a way, it feels more natural to me, which is saying something because topology seems to come much more naturally to me still than other areas of math such as analysis and algebra.
On Tuesday, I will continue classes. There will be a physics lecture, which I believe will cover Newton's laws. Then, I will have Latin class, which should be a bit of a break for me. Then, I have my writing seminar class. I tried to participate often in the first class and share my ideas, which I did not really do in high school English class. Finally, I have a very late night lab for physics, which I've heard mostly consists of Python simulations. But I suppose we will see.
Anyways, you've probably heard enough yap from me, yeah? Let's keep going through Bredon and start our journey through the land of smooth manifolds.
We will skip the rest of Bredon's first chapter on point-set topology and now begin a thorough examination of smooth manifolds. We begin by examining two theorems from multivariable analysis, the implicit function theorem and the inverse function theorem.
Suppose we have a smooth function $f: \mathbb{R}^n \to \mathbb{R}$. If $x=(x_1,\ldots, x_n)$ and $\overline{x}=(\overline{x}_1,\ldots, \overline{x}_n)$, then we have that $$f(x) - f(\overline{x}) = \sum_{i=1}^n \frac{\partial f}{\partial x_i}(\tilde{x})(x_i-\overline{x}_i)$$ where $\tilde{x}$ is on the line segment between $x$ and $\overline{x}$.
First, we consider the function $\mathbb{R}\to \mathbb{R}$ where $t \mapsto f(tx+(1-t)\overline{x})$. It necessarily follows that, by the single-variable mean value theorem, we have that there exists some $t_0$ between 0 and 1 such that $$f(1x+(1-1)\overline{x})-f(0x+(1-0)\overline{x})=f(x)-f(\overline{x})=\frac{df(tx+(1-t)\overline{x})}{dt}\bigg|_{t=t_0}$$ However, using the Chain Rule, we may find that the right hand side becomes $$\frac{df(tx+(1-t)\overline{x}}{dt}\bigg|_{t=t_0}=\sum_{i=1}^n \frac{\partial f}{\partial x_i}(\tilde{x}) \frac{d(tx_i+(1-t)\overline{x}_i)}{dt}\bigg|_{t=t_0}=\sum_{i=1}^n \frac{\partial f}{\partial x_i}(\tilde{x})(x_i-\overline{x}_i)$$ where we let $\tilde{x} = t_0x+(1-t_0)\overline{x}$. The desired expression follows shortly.
Suppose we have a smooth function $f: \mathbb{R}^k \times \mathbb{R}^m \to \mathbb{R}$ where $x\in \mathbb{R}^k$ and $y,\overline{y} \in \mathbb{R}^m$. Then we have $$f(x,y)-f(x,\overline{y})=\sum_{i=1}^m \frac{\partial f}{\partial y_i}(x,\tilde{y})(y_i-\overline{y}_i)$$ where $\tilde{y}$ is on the line segment between $y$ and $\overline{y}$.
By the Mean Value Theorem, we have that $$f(x,y)-f(x,\overline{y})=\sum_{i=1}^k \frac{\partial f}{\partial x_i}(x,\tilde{y}) (x_i-x_i) + \sum_{i=1}^m \frac{\partial f}{\partial y_i}(x,\tilde{y})(y_i-\overline{y}_i)$$ However, note that $x_i-x_i=0$, so $$\sum_{i=1}^k \frac{\partial f}{\partial x_i}(x,\tilde{y}) (x_i-x_i)$$ must therefore be equal to zero. Hence, $$f(x,y)-f(x,\overline{y}) = \sum_{i=1}^m \frac{\partial f}{\partial y_i}(x,\tilde{y})(y_i-\overline{y}_i)$$
One important theorem of note is that of the Banach Contraction Principle. It essentially states that for a function that is a "contraction" (which will be defined later), there is a unique fixed point that is the limit when you apply the contraction infinitely many times to a point.
Suppose $X$ is a complete metric space and $T: X \to X$ is a contraction, which means that for some $K<1$, $\text{dist}(Tx,Ty)\leq K \cdot \text{dist}(x,y)$ for all $x,y\in X$. It follows that $T$ has a unique fixed point $\zeta \in X$, and for any $x\in X$, $\zeta = \lim_{i \to \infty} T^i(x)$.
Suppose we have some $x_0 \in X$, and $x_i = Tx_{i-1}$. If we have that $\delta = \text{dist}(x_0,x_1)$, then we may see by the triangle inequality that $$\text{dist}(x_0,x_k) \leq \text{dist}(x_0,x_1) + \text{dist}(x_1,x_2)+\cdots + \text{dist}(x_{k-1},x_k)$$ which, as distances are always nonnegative, implies $$ \leq \text{dist}(x_0,x_1) + \text{dist}(x_1,x_2) + \cdots + \text{dist}(x_{k-1},x_k) + \cdots$$ and still, since $\text{dist}(Tx,Ty) \leq K \cdot \text{dist}(x,y)$, we have that this implies $$\leq \text{dist}(x_0,x_1)+K \cdot \text{dist}(x_0,x_1) + K^2 \cdot \text{dist}(x_0,x_1)+\cdots$$ which, letting $\delta = \text{dist}(x_0,x_1)$, and combining terms, we obtain $$= \delta (1 + K + K^2 + \cdots)$$ and using the convergence of this specific power series as $K <1$, we obtain $$= \frac{\delta}{1-K}$$ Furthermore, when we have $m \geq n$, $$\text{dist}(x_n,x_m) \leq K \cdot \text{dist}(x_{n-1},x_{m-1})\leq K^2 \cdot \text{dist}(x_{n-2},x_{m-2})$$ $$\leq \cdots \leq K^n \cdot \text{dist}(x_0,x_{m-n}) \leq \frac{\delta K^n}{1-K}$$ As $n\to \infty$, we have that this expression goes to 0, since $K<1$. It therefore follows that $x_0,x_1,x_2,\ldots$ is a Cauchy sequence, and since $X$ is complete, then there exists a $\zeta \in X$ such that $\zeta = \lim_{i\to\infty}x_i$. It is a fixed point, as $T\zeta = \lim_{i\to\infty} T x_i = \lim_{i\to\infty}x_{i+1}=\zeta$. We now prove uniqueness. If $x$ is another fixed point, then we must have that $$\text{dist}(x,\zeta)=\text{dist}(Tx,T\zeta) \leq K \cdot \text{dist}(x,\zeta)$$ Recall that $K<1$, so this implies that the distance is zero, and therefore $x=\zeta$.
We now begin to prove a rather large lemma that will help us prove the Implicit Function Theorem. It is difficult to describe in a way that is non mathematical, and the proof is very long and difficult to understand. But I will try my best.
Let $\zeta \in \mathbb{R}^n$ and $\eta \in \mathbb{R}^m$. If $f: \mathbb{R}^n \times \mathbb{R}^m \to \mathbb{R}^m$ is smooth, and $f = (f_1,\ldots, f_m)$, and $f(\zeta, \eta) = \eta$ and $$\frac{\partial f_i}{\partial y_j}(\zeta,\eta)=0$$ where $x_i$ and $y_i$ are defined as before, then there exist some $a>0$ and $b>0$ such that there exists a unique continuous function $\phi: A \to B$, where $A = \{x \in \mathbb{R}^n \mid \|x-\zeta\|\leq a\}$ and $B = \{y \in \mathbb{R}^m \mid \| y - \eta \| \leq b\}$ such that $\phi(\zeta)=\eta$ and $\phi(x)=f(x,\phi(x))$ for all $x \in A$.
Let us assume that $\zeta = 0$, $\eta = 0$, and $f(0,0)=0$. We may be able to do so by transforming our coordinates appropriately. Now, apply Corollary 1 to each $f_i$, and using the assumption that since $\partial f_i/\partial y_i = 0$ at the origin, it is "small" in some neighborhood of the origin, there exists some $a>0$ and $b>0$ such that $$\| f(x,y) - f(x,\overline{y})\| < K \|y - \overline{y}\|$$ where $\|x\|\leq a$, $\|y\|\leq b$, and $\|\overline{y}\|\leq b$. Furthermore, if $a$ is even smaller, we may be able to have that $$Kb + \|f(x,0)\| \leq b$$ since $Kb < b$. Now consider the set $\textbf{F}$ of all functions $\phi: A \to B$ where $\phi(0)=0$. Let $\text{dist}(\phi,\psi) = \sup\{\|\phi(x)-\psi(x)\|\mid x \in A\}$. This metric space is complete. Furthermore, define $T: \textbf{F}\to \textbf{F}$ where $(T\phi)(x)=f(x,\phi(x))$. Note that $(T\phi)(0)=f(0,\phi(0))=f(0,0)=0$, and $$\|(T\phi)(x)\|=\|f(x,\phi(x))\|\leq \|f(x,\phi(x))-f(x,0)\|+\|f(x,0)\|$$ $$ \leq K \|\phi(x)\| + \|f(x,0)\|$$ $$ \leq Kb + \|f(x,0)\| \leq b$$ and so $(T\phi)(x) \in B$, as its norm is less than $b$. $T$ is a contraction, but I am too lazy to go through all the calculations for that. Therefore, by the Banach Principle, we may find a unique $\phi$ such that $\phi(0)=0$ and $T\phi= \phi$, or $f(x,\phi(x))=\phi(x)$ for all $x \in A$. Furthermore, $\phi = \lim \phi_i$ where $\phi_0$ is arbitrary, and $\phi_i$ is defined like in the Banach Contraction Principle. If we have $\phi_0(x)=0$, then all $\phi_i$ will be continuous because $f$ is continuous. Therefore, $\phi$ is a uniform limit of continuous functions, and is therefore continuous.
I do not think that I will be providing the entire proof of the Implicit Function Theorem here, because it is two pages long and I do not want to write all of that down. However, I will state it in full:
Suppose we have a differentiable $g: \mathbb{R}^n \times \mathbb{R}^m \to \mathbb{R}^m$, and $g(\zeta, \eta)=0$. Suppose that the Jacobian determinant of the composition $$y \mapsto (\zeta, y) \mapsto g(\zeta,y)$$ is nonzero. Then there are some $a>0$, $b>0$ such that there exists a unique function $ \phi: A \to B$ where $\phi(\zeta) = \eta$, and $g(x,\phi(x))=0$ for all $x \in A$. Additionally, if $g$ is $C^p$, which means it has up to $p$ continuous derivatives, then so is $\phi$. This includes the case where $p = \infty$.
The Implicit Function Theorem essentially tells us under which conditions does there exist an appropriate function $y(x)$ such that at some point, there is a neighborhood around which $y$ may satisfy some relation (that is, $g(x,y(x))$).
Suppose that we have $\theta: \mathbb{R}^m \to \mathbb{R}^m$ be $C^1$, where $\theta(\eta)=\zeta$ and the differential at $\eta$ is nonsingular, so the Jacobian determinant is nonzero at that point. Then there exist $a>0$ and $b>0$ such that there is a function $\phi: A \to B$, defined in the previous manner, where $\phi(\zeta)=\eta$ and $\theta(\phi(x))=x$ for all $x \in A$. Also, $\phi$ is $C^p$ when $\theta$ is $C^p$.