ex03

0 (Warmup). Show that \(M:=\{(x,y)\in \R ^2:y=x^2\}\) is a \(1\)-dimensional submanifold of \(\R ^2\).

[Solution: \(M\) is the graph of the smooth function \(h\colon \R \to \R \) with \(h(x)=x^2\). Thus there is a parametrization \(\varphi \colon \R \to M\) with \(\varphi (x)=(x,x^2)\). This is a diffeomorphism because \(F\colon \R ^2\to \R \) with \(F(x,y)=x\) is smooth and \(F|_M=\varphi ^{-1}\). Alternatively, we can apply the Regular Value Theorem: \(f\colon \R ^2\to \R \) with \(f(x,y)=y-x^2\) is smooth and \(Df_{(x,y)}\) is represented by the matrix \([-2x\ 1]\). This is nonzero for all \((x,y)\) so \(0\) is a regular value and hence \(M=f^{-1}(0)\) is a \(1\)-dimensional submanifold of \(\R ^2\). This is related to the first approach, because close to the origin, the proof of the regular value theorem gives a parametrization using the graph of \(h\).]

1. Let \(U \subset \R ^n\) be open, and let \(f : U \to \R ^n\) be a twice differentiable function such that \(Df_x\) is invertible at \(x\in U\). Let \(K\) be the operator norm of \((Df_x)^{-1} : \R ^n \to \R ^n\) and \(N\) the supremum of the operator norm of \(D(Df)_z : \R ^n \to \cL (\R ^n, \R ^n)\) for \(z\in U\) (where \(\cL (\R ^n, \R ^n)\) is itself equipped with the operator norm).

Suppose that \(0 < \delta < 1/(2KN)\) and the open ball \(B_\delta (x)\) is contained in \(U\).
- (i) Let \(\tilde f := (Df_x)^{-1} \circ f\). Show that the image \(\tilde f(B_\delta (x))\) contains \(B_{\delta /4}(\tilde f(x))\).
  
  [Hint: Apply the Mean Value Inequality to \(D f\) and use Lemma 1.22.]
- (ii) Show that the image \(f(B_\delta (x))\) contains the ball \(B_{\delta /(4K)}(f(x))\).
  
  [Hint: What can you say about the image of \(B_{\delta /(4K)}(f(x)) \) under \((Df_x)^{-1} \)?]

2. Consider parametrisations of \(\varphi \) of \(S^n := \{ y \in \R ^{n+1} : \norm {y} = 1 \}\) that are “graphs over a coordinate plane” in the following sense: \(\varphi : B^n \to U\), for \(B^n := \{ x \in \R ^n : \norm {x} < 1\}\) and \(U \subset S^n\) some open subset, and all but one of the \(n+1\) components of \(\varphi (x)\) is equal to a component of \(x\) (e.g., \(\varphi \) could be of the form \((x_1, \ldots , x_n) \mapsto (x_1, \ldots , x_{i-1}, g(x_1, \ldots , x_n), x_i, \ldots x_n)\)). How many parametrisations of this form are required for the images \(U\) to cover \(S^n\)?

[Hint: If \(e_i \in \mathbb {R}^{n+1}\) is one of the \(n+1\) basis vectors, how many parametrisations of the given type have \(e_i\) (or \(-e_i\)) contained in their image?]

3. Define two parametrisations \(\varphi _+, \varphi _- : \R ^n \to S^n\) as follows. For \(x \in \R ^n\), consider the line \(L_{\pm }\) in \(\R ^{n+1}\) that contains \((x,0)\) and \((0,\pm 1)\), and let \(\varphi _\pm (x)\) be the intersection point (other than \((0, \pm 1)\)) of \(L_\pm \) with \(S^n\).

Show that

\[ \varphi _\pm (x) = \frac {1}{1+\norm {x}^2} \left (2x, \, \pm (\norm {x}^2-1) \right ). \]

What are the images \(U_\pm \) of \(\varphi _\pm \)?

[Hint: Parametrize \(L_\pm \) by \(t\in \R \) and find \(y\in L_\pm \) with \(\norm {||y||^2=1}\). This should give a quadratic equation for \(t\)—one solution gives the point \((0,\pm 1)\) and the other is \(\varphi _\pm (x)\).]

4. Let \(k\) be a positive integer, and let \(M = \{(x,y,z) \in \R ^3 : x^k + y^k + z^k = 1 \}\).
- (i) Show that \(M\) is a submanifold of \(\R ^3\).
  
  [Hint: Show that \(1\) is a regular value of \((x,y,z) \mapsto x^k + y^k + z^k \).]
- (ii) Show that if \(k\) is even then \(M\) is diffeomorphic to \(S^2\).
  
  [Hint: To see the difference between when \(k\) is even or odd, it may be helpful to draw a sketch of the set \(\{(x,y) \in \mathbb {R}^2 : x^k + y^k = 1\} \) for \(k = 1,2,3,4\). Now you need to find a smooth function \(f:M \to S^2\) with a smooth inverse \(g:S^2 \to M\). To do this, find explicitly smooth maps \(F\colon \mathbb {R}^3 \setminus \{0\} \to S^2\) and \(G\colon \mathbb {R}^3 \setminus \{0\} \to M\) such that the restriction of \(F\) to \(M\) and the restriction of \(G\) to \(S^2\) are inverses.]

5. Fix \(k > 0\), and define \(f : \R ^3 \to \R \) by

\[ f(x,y,z) = \frac {x^2 + y^2}{(x^2 + y^2 + z^2 + k)^2} . \]

What are the regular values of \(f\)? For each regular value \(q\) of \(f\), describe \(f^{-1}(q)\).

[Hint: First identify the points where \(\frac {\partial f}{\partial z} = 0 \), then identify which of those have \(\frac {\partial f}{\partial x} = \frac {\partial f}{\partial y} = 0 \) too.]

MA40254 Differential and geometric analysis : Solutions 3

1.
- (i) By the chain rule \(D\tilde f_z = (Df_x)^{-1} \circ Df_z\) for all \(z\). Hence
  
  \[ \norm {D\tilde f_z - \Id \nolimits _{\R ^n}}=\norm {(Df_x)^{-1}}\norm {D f_z - D f_x} \leq K \norm {D f_z - D f_x} \leq KN \norm {z - x} \]
  
  by the Mean Value Inequality for \(Df\), which is \(< 1/2\) for \(z\in B_\delta (x)\). Lemma 1.22 thus ensures that the image \(\tilde f(B_\delta (x))\) contains \(B_{\delta /4}(\tilde f(x))\).
- (ii) Since \((Df_x)^{-1}\) is linear with \(\norm {(Df_x)^{-1}}_{op}=K\), the image of \(B_{\delta /(4K)}(f(x))\) under \((Df_x)^{-1}\) is contained in \(B_{\delta /4}(\tilde f(x))\), and hence the image of \(B_{\delta /4}(\tilde f(x))\) under \(Df_x\) contains \(B_{\delta /(4K)}(f(x))\). Hence \(f(B_{\delta }(x)) = (Df_x)(\tilde f (B_\delta (x)))\) contains \(B_{\delta /(4K)}(f(x))\).

2. One can cover \(S^n\) by \(2n+2\) parametrisations of this form, two for each of the \(n+1\) coordinates. For example, writing the last coordinate as a function of the other \(n\) we get

\[ (x_1, \ldots , x_n) \mapsto (x_1, \ldots , x_n, \sqrt {1 - x_1^2 - \cdots x_n^2}) \]

and

\[ (x_1, \ldots , x_n) \mapsto (x_1, \ldots , x_n, -\sqrt {1 - x_1^2 - \cdots x_n^2}) . \]

Now, each of the \(2n+2\) vectors \(\pm e_1, \ldots , \pm e_{n+1}\) is contained in the image of only one parametrisation of this form, so \(S^n\) cannot be covered by less than \(2n+2\) such parametrisations.

3. Points on the two lines in question can be written as \((1-t)(0,\pm 1)+ t(x,0)\) with \(t \in \R \). The intersection points with \(S^2\) are given by

\[ 1 = \norm {(1-t) (0, \pm 1) + t(x, 0)}^2 = \norm {(1-t)(0, \pm 1)}^2 + \norm {tx}^2 = 1 - 2t + t^2(\norm {x}^2 + 1) , \]

with solutions \(t = 0\) and \(t = \frac {2}{1+\norm {x}^2}\). The former corresponds to \((0,0,\pm 1)\), while the latter gives the desired expression for \(\varphi _\pm (x)\).

\(U_\pm \) is \(S^n \setminus \{ (0, \pm 1) \}\). (Thus we have covered \(S^n\) by two parametrisations.)

4.
- (i) The derivative of the function \(\R ^3 \to \R , (x,y,z) \mapsto x^k + y^k + z^k\) is represented by
  
  \[ \pmat {kx^{k-1} & ky^{k-1} & kz^{k-1}} , \]
  
  which vanishes only at the origin. In particular, \(1\) is a regular value of this function, so its preimage \(M\) is a submanifold.
- (ii) Define \(F : \R ^3 \setminus \{0\} \to S^2\) and \(G : \R ^3 \setminus \{0\} \to M\) by
  
  \[ F(x,y,z) = \frac {(x,y,z)}{\sqrt {x^2 +y^2 +z^2}}, \quad G(x,y,z) = \frac {(x,y,z)}{(x^k +y^k +z^k)^{1/k}} \]
  
  These are both well-defined smooth functions (\(k\) even ensures that \(x^k + y^k + z^k\) never vanishes), so the restrictions \(f = F|_{M} : M \to S^2\) and \(g = G|_{S^2} : S^2 \to M\) are smooth too, and they are inverse to each other.

5. We first identify the points \((x,y,z) \in \R ^3\) where \(Df_{(x,y,z)} : \R ^3 \to \R \) fails to be surjective. Since the codomain is \(\R \), that just means checking when all the partial derivatives are zero. Now \(\partial {f}/\partial {z} = \frac {-4(x^2+y^2)z}{(x^2 + y^2 + z^2 + k)^3}\) vanishes only when either \(z = 0\) or \(x = y = 0\).

For fixed \(z\), we have \(f(x,y,z)=g(R)\) where \(g(R) := \frac {R}{(R + z^2 + k)^2}\) and \(R=x^2+y^2\). Hence

\[ \frac {\partial f}{\partial x} = 2x g'(R), \quad \frac {\partial f}{\partial y} = 2y g'(R) , \]

so both vanish if and only if \(x = y = 0\) or

\[ 0 = g'(R) = \frac {-R + z^2 + k}{(R+z^2 + k)^3} . \]

Hence the set of points where \(Df_{(x,y,z)}\) fails to be surjective is the union of the line \(\{x = y = 0\}\) and the circle \(\{z = 0, \, x^2 + y^2 = k\}\). The image of this set is \(\{0, \frac {1}{4k}\}\), so the set of regular values is \(\R \setminus \{0, \frac {1}{4k}\}\). For \(q > 0\)

\[ f(x,y,z) = q \; \Leftrightarrow \; x^2 + y^2 + z^2 + k = \tfrac {1}{\sqrt {q}} \sqrt {x^2 + y^2} \; \Leftrightarrow \; (\sqrt {x^2 + y^2} - \tfrac {1}{2\sqrt {q}})^2 + z^2 = \tfrac {1}{4q} - k. \]

If \(0 < q < \frac {1}{4k}\), then this is an equation defining a torus in \(\R ^3\), while if \(q < 0\) or \(q > \frac {1}{4k}\) then \(f^{-1}(q)\) is empty.