Homework 1:

Question 1: Consider an open set \(U \in \mathbb{R}^k\) and a function \(f: U \rightarrow \mathbb{R}^{n-k}\). The graph of \(f\) is the set of points \((x_1, ..., x_n) \in \mathbb{R}^n\) with \((x_{k+1}, ..., x_n) = f(x_1, ..., x_k)\).

a: The graph of \(f\) was parametrized by \(\begin{equation}\varphi(x_1, ..., x_k) = (x_1, ..., x_k, f(x_1, ..., x_k)) \end{equation}\). Show that the derivative \(D\varphi(x_1, ..., x_k)\) is an injective linear map at all \((x_1, ..., x_k) \in U\).

Answer: Consider the matrix of partial derivatives \(D \varphi\): \(\begin{equation} D\varphi = \begin{bmatrix} \frac{\partial \varphi_1}{\partial x_1} &\ \cdots & \frac{\partial \varphi_1}{\partial x_k} & \frac{\partial f_1}{\partial x_{k+1}} & \cdots & \frac{\partial f_1}{\partial x_n} \\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ \frac{\partial \varphi_k}{\partial x_1} & \cdots & \frac{\partial \varphi_k}{\partial x_k} & \frac{\partial f_k}{\partial x_{k+1}} & \cdots & \frac{\partial f_k}{\partial x_n} \end{bmatrix} \end{equation}\)

Observe that for \(i=j\), we get that \(\frac{\partial \varphi_i}{\partial x_i} = 1\), otherwise, it is \(0\) for \(i \neq j\). So then:

\(\begin{equation} D\varphi = \begin{bmatrix} 1 & \cdots & 0 & \frac{\partial f_1}{\partial x_{k+1}} & \cdots & \frac{\partial{f_1}}{{\partial x_n}} \\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ 0 & \cdots & 1 & \frac{\partial f_k}{\partial x_{k+1}} & \cdots & \frac{\partial f_k}{\partial x_n} \end{bmatrix} \end{equation}\)

Since the matrix of partial derivatives has full rank, we can conclude it is an injective map.

To see why \(D \varphi\) is linear observe that for \( a \in \mathbb{R}\), \(aD\varphi= a * \begin{bmatrix} \frac{\partial \varphi_1}{\partial x_1} &\ \cdots & \frac{\partial \varphi_1}{\partial x_k} & \frac{\partial f_1}{\partial x_{k+1}} & \cdots & \frac{\partial f_1}{\partial x_n} \\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ \frac{\partial \varphi_k}{\partial x_1} & \cdots & \frac{\partial \varphi_k}{\partial x_k} & \frac{\partial f_k}{\partial x_{k+1}} & \cdots & \frac{\partial f_k}{\partial x_n} \end{bmatrix} = \begin{bmatrix} a *\frac{\partial \varphi_1}{\partial x_1} &\ \cdots & a* \frac{\partial \varphi_1}{\partial x_k} & a* \frac{\partial f_1}{\partial x_{k+1}} & \cdots & a* \frac{\partial f_1}{\partial x_n} \\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ a* \frac{\partial \varphi_k}{\partial x_1} & \cdots & a* \frac{\partial \varphi_k}{\partial x_k} & a* \frac{\partial f_k}{\partial x_{k+1}} & \cdots & a* \frac{\partial f_k}{\partial x_n} \end{bmatrix} = D(a\varphi)\)

Now, let \(a,b \in \mathbb{R}^k\), then \(D\varphi(a+b) =\) \(\begin{equation} \begin{bmatrix} \frac{\partial \varphi_1}{\partial x_1}(a+b) &\ \cdots & \frac{\partial \varphi_1}{\partial x_k}(a+b) & \frac{\partial f_1}{\partial x_{k+1}}(a+b) & \cdots & \frac{\partial f_1}{\partial x_n}(a+b) \\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ \frac{\partial \varphi_k}{\partial x_1}(a+b) & \cdots & \frac{\partial \varphi_k}{\partial x_k}(a+b) & \frac{\partial f_k}{\partial x_{k+1}}(a+b) & \cdots & \frac{\partial f_k}{\partial x_n}(a+b) \end{bmatrix} \end{equation}\)

We distribute the partial derivatives over our vectors \(a\) and \(b\) to then get the following:

\(\begin{equation} D\varphi(a+b) = \begin{bmatrix} \frac{\partial \varphi_1}{\partial x_1}(a) &\ \cdots & \frac{\partial \varphi_1}{\partial x_k}(a) & \frac{\partial f_1}{\partial x_{k+1}}(a) & \cdots & \frac{\partial f_1}{\partial x_n}(a) \\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ \frac{\partial \varphi_k}{\partial x_1}(a) & \cdots & \frac{\partial \varphi_k}{\partial x_k}(a) & \frac{\partial f_k}{\partial x_{k+1}}(a) & \cdots & \frac{\partial f_k}{\partial x_n}(a) \end{bmatrix} + \begin{bmatrix} \frac{\partial \varphi_1}{\partial x_1}(b) &\ \cdots & \frac{\partial \varphi_1}{\partial x_k}(b) & \frac{\partial f_1}{\partial x_{k+1}}(b) & \cdots & \frac{\partial f_1}{\partial x_n}(b) \\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ \frac{\partial \varphi_k}{\partial x_1}(b) & \cdots & \frac{\partial \varphi_k}{\partial x_k}(b) & \frac{\partial f_k}{\partial x_{k+1}}(b) & \cdots & \frac{\partial f_k}{\partial x_n}(b) \end{bmatrix} = D\varphi(a) + D\varphi(b) \end{equation}\)

Thus, we can conclude that \(D\varphi\) is an injective linear map at all \((x_1, ..., x_k) \in U\).

b: To do. Derivative of \(g(x_1, ..., x_n) = (x_{k+1}, ..., x_n) - f(x_1,..., x_k) = (x_{k+1} - f_1(x_1,..., x_k) , ..., x_n - f_k(x_1, ..., x_k))\). The associated matrix of partial derivatives is a \(n \times (n-k)\) matrix. When we look at the matrix of partial derivatives of \(g\). We see that we would have an upper block of size \(k\) with the diagonal of the form \(1 - \frac{\partial f_i}{x_i}\), which is zero when \(g=0\) what we have left is then the partials of \(f\) which would give us the appropriate \((n-k)\) linearly indepent vectors of \(Dg\), making it full rank.

Question 2: This uses the Implicit Function Theorem. If we consider the level set \(\{g=0\}\) for some function \(g: \mathbb{R}^n \rightarrow \mathbb{R}^{n-k}\), where \(Dg\) has rank \(n-k\) at every point where \(g=0\). Let \(g(a_1, ..., a_n) = 0\), and apply the implicit function theorem to write some neighborhood of \((a_1, ..., a_n)\), say \(U\) on the level set as the graph of a function \(f: \mathbb{R}^k \rightarrow \mathbb{R}^{n-k}\). What variables in \(\mathbb{R}^n\) act as the domain of \(f\)?

Solution: Let's reimagine \(g\) like this. \(g: \mathbb{R}^n = \mathbb{R} ^k \times \mathbb{R}^{n-k} \rightarrow \mathbb{R}^{n-k}\). With both \(U\), and \(a\) are written in similar form. Since \(g\) has constant rank at every zero, the derivtive of \(g\) is injective. So then by the IFT, we can find neighborhoods \(V_0 \subseteq \mathbb{R}^n\) of \((a_1, ..., a_k)\) and \(W_0 \subseteq \mathbb{R}^{n-k}\) of \((a_{k+1}, ..., a_n)\) with a smooth function \(F: V_0 \rightarrow W_0\) such that \(g^{-1}(a_1, ..., a_n) \cap (V_0 \times W_0)\) is the graph of \(F\). The variables in \(\mathbb{R}^n\) that act as the domain of \(f\) are the \(k\) coordinates that we for \(g: \mathbb{R}^k \times \mathbb{R}^{n-k}\rightarrow \mathbb{R}{n-k}\). Those coordinates will us the appropriate domain of \(f\) that correspond to its graph.

Question 3: The points of the projective plane \(\mathbb{RP}^2\) correspond to the lines \(ax +by +cz =0\) through the origin in \(\mathbb{R}^3\). We have 3 open sets with corresponding charts. \(U_1= \{a \neq 0 \}\), \(U_2 = \{b \neq 0\}\), \(U_3 = \{c \neq 0 \}\)

\(\begin {equation} \varphi_1(ax+by+cz=0) = (\frac{b}{a}, \frac{c}{a}) \end{equation}\)

Define \(\varphi_2, \varphi_3\). Compute the transition function \(\psi = \varphi_2 \circ \varphi ^{-1}_1\) and show that it is smooth.

Solution: The corresponding maps are \(\varphi_2: U_2 \rightarrow \mathbb{R}^2\), \(\varphi_3: U_3 \rightarrow \mathbb{R}^2\). Where

\(\begin{equation} \varphi_2(ax+by+cz=0) = (\frac{a}{b}, \frac{c}{b}) \end{equation}\)

\(\begin{equation} \varphi_3(ax+by+cz=0) = (\frac{a}{c}, \frac{b}{c}) \end{equation}\)

Now, let \((a,b) \in \mathbb{R}^2\) and consider the the transition function \(\varphi_2 \circ \varphi^{-1}_1: \varphi_1(U_1 \cap U_2) \rightarrow \varphi_2(U_1 \cap U_2)\). Upon evaluation, \(\varphi_2 ( \varphi^{-1}_1(b,c)) = \varphi_2(x+by+cz=0) = (\frac{1}{b}, \frac{c}{b})\). Since \(\frac{1}{x}\) is smooth for \(x\neq0\), we can see that the transition function is smooth on its components and thus, a smooth function.

Problem 4: Let \(\mathcal{A}\) be an atlas of charts on a smooth manifold \(M\). Let \((U_1, \varphi _1)\) and \((U_2, \varphi_2)\) be charts on \(M\) that are smoothly compatible with the atlas \(\mathcal{A}\). Show that \(\varphi_1\) and \(\varphi_2\) are smooth compatile with each other.

Solution: Let's recall the definition of smoothly compatible. Two charts \((V, \psi) , (W, \kappa)\) are called smoothly compatible whenever \(V \cap W = \emptyset\) or \(\psi \circ \kappa ^{-1}\) is a diffeomorphism. If we are in the first case, then we are already finished. The empty intersection implies that \(\varphi_1 , \varphi_2\) are smoothly compatible. If we are in the second case, consider the following. Since each of the maps is smoothly compatible with \(\mathcal{A}\), let \(\nu\) be a chart in \(\mathcal{A}\) that is smoothly compatible with both \(\varphi_1\) and \(\varphi_2\) in the latter sense. From this, we can observe that \(\varphi_1 \circ \nu ^{-1}\) and \(\nu \circ \varphi_2 ^{-1}\) are diffeomorphisms. So then \(\varphi_1 \circ \nu ^{-1} \circ \nu \circ \varphi_2^{-1} = \varphi_1 \circ \varphi_2^{-1}\) is also a diffeomorphism. Thus, our two charts are also smoothly compatible.

Problem 5: Let \(M\) be a manifold with boundary. Let \(x \in M\) be a point and \((U_1, \varphi_1)\) a chart on \(M\) with \(x\in U_1\) and \(\varphi_1(x) \in \partial \mathbb{H}^n = \{(x_1, ..., x_n) | x_n=0 \}\). If \((U_2, \varphi_2)\) is any other chart on \(M\) with \(x\in U_2\), show that \(\varphi_2(x) \in \partial \mathbb{H}^n\) as well.

Solution: Since \(U_1 \cap U_2 \neq \emptyset\). The transition function \(\varphi_2 \circ \varphi^{-1}_1: \varphi_1(U_1 \cap U_2) \rightarrow \varphi_2(U_1 \cap U_2)\) is a homeomorphism. \(\varphi_1(x) \in \mathbb{H}^n\) guarantees our homeomorphism \(\varphi_2(x) \in \partial \mathbb{H}^n\), any open set \(V \cap \mathbb{H}^n\) containing \(\varphi_(x)\) would be mapped to another open \(W \cap \mathbb{H}^n\).