ex02

0 (Warmup). For \(U \subseteq \R ^n\) open, let \(f : U \to \R ^n\) be a local diffeomorphism. Show \(f(U)\) is open.

[Solution: Apply the Inverse Function Theorem at each \(x \in U \): this implies the existence of a neighbourhood \(U' \subseteq U\) of \(x\) such that \(f(U')\) is open in \(\R ^n\). Thus \(f(U)\) contains an open neighbourhood of each of its elements.]

1. Let \(U \subseteq \R ^n\) be open and \(f : U \to \R ^m\) be \(C^1\). Suppose \(Df_x\) is surjective for every \(x \in U\). Show that \(f(U)\) be open in \(\R ^m\)?

[Hint: Reduce this problem to the solution of question 1 for each \(x\in U\) by restricting \(f\) to \(\{x+v:v\in W\}\) for a suitable \(m\)-dimensional subspace \(W\leq \R ^n\).]

2. Let \(U\) be an open subset of \(\R ^2 \setminus \{0\}\), and let \(f : U \to \R _{>0} \times \R , \; (x,y) \mapsto (r, \theta )\) be a smooth function such that \(x = r(x,y) \cos \theta (x,y)\) and \(y = r(x,y) \sin \theta (x,y)\) for any \((x,y) \in U\). Compute \(Df_{(x,y)}\) for \((x,y) \in U\).

[Hint: What is the relationship of \(f\) to \(g : \mathbb {R}_{>0} \times \mathbb {R} \to \mathbb {R}^2 \setminus \{0\},\; (r, \theta ) \mapsto (r \cos \theta , r \sin \theta )\) and what does this imply about \(Df\) and \(Dg\)?]

3. Define \(f : \R ^3 \to \R ^2\) by \(f(x,y,z) := (x^2 + y^2 + z, \, xz^2 - yz)\). For which \((x,y,z) \in \R ^3\) is \(Df_{(x,y,z)}\colon \R ^3\to \R ^2\) surjective?

[Hint: When are the two rows of \(Df_{(x,y,z)}\) linearly independent? One approach is to compute the three \(2 \times 2 \) minors. You should find that there are two cases depending on whether \(z=0\) or not.]

4. Let \(GL_n(\R ) \subset M_{n, n}(\R )\) be the subset of invertible matrices in the vector space of real \(n \times n\) matrices. Let \(\inv : GL_n(\R ) \to GL_n(\R )\) be the (continuous) function \(A \mapsto A^{-1}\).
- (i) Explain why \(GL_n(\R )\) is open in \(M_{n,n}(\R )\).
- (ii) Show that the derivative of \(\inv \) at the identity matrix \(I \in GL_n(\R )\) is
  
  \[ D\inv _I = -\Id \nolimits _{M_{n,n}(\R )} .\]
- (iii) Identify the derivative \(D\inv _A : M_{n, n}(\R ) \to M_{n, n}(\R )\) at \(A \in GL_n(\R )\), and deduce that \(\inv \) is smooth.
[Hint: (i) The determinant is a continuous function. (ii) Note that \((I+X)((I+X)^{-1} - I +X) = X^2\). (iii) Define \(L_A\) and \(R_A : \textrm {Mat}_{n,n}(\mathbb {R}) \to \textrm {Mat}_{n,n}(\mathbb {R})\) by \(X \mapsto AX\) and \(X \mapsto XA\). What can you say about \(L_A \circ \mathrm {inv} \circ R_A\)? Now apply the chain rule.]

5.
- (i) Let the function \(\chi : \R \to \R \) be defined by
  
  \[ \chi (t) := \left \{ \begin {array}{ll} 0 & \textrm {for } t \leq 0 \\ e^{-1/t} & \textrm {for } t > 0 \end {array} \right . \]
  
  Show that there is a polynomial \(p_n\) of degree \(\leq 2n\) such that for \(t>0\), the \(n\)th derivative of \(\chi (t)\) is \(p_n(1/t) \chi (t)\). Hence or otherwise, prove that \(\chi \) is a smooth function on \(\R \). [You may assume results about “exponentials dominating polynomials”.]
  
  [Hint: Use induction: you don’t need to find an explicit expression for \(p_n \).]
- (ii) For any \(x\in \R ^n\) and any \(r>0\), show that there is a smooth function \(\rho \colon \R ^n\to \R \) such that \(\{y\in \R ^n: \rho (y)\neq 0\}\) is the open ball \(B_r(x)\). [This is called a “bump function”.]
  
  [Hint: Precompose \(\chi \) with a suitable function of \(\| y- x\| \).]

MA40254 Differential and geometric analysis : Solutions 2

1. For any \(x \in U\), the hypothesis implies there is a subspace \(W \leq \R ^n\), with \(\R ^n=W\oplus \ker Df_x\), such that the restriction of \(Df_x\) to \(W\) gives a linear isomorphism \(W \to \R ^m\). Now consider the translation map \(T_x : W \to \R ^n, \; z \mapsto x + z\). Then \(\tilde U := T_x^{-1}(U)\) is an open subset of \(W\), containing the origin. Let \(\tilde f := f \circ T_x : \tilde U \to \R ^m\). This is a smooth function, with \(D\tilde f_0 = Df_{x}|_W\) by the chain rule. Since that is an isomorphism, the image \(\tilde f(\tilde U)\) contains an open neighbourhood of \(\tilde f(0) = f(x)\) by the Inverse Function Theorem. Since \(\tilde f(\tilde U) \subseteq f(U)\), therefore \(f(U)\) too contains an open neighbourhood of \(f(x)\). Applying this argument for each \(x \in U\) shows that \(f(U)\) is open.

2. Let \(g : \mathbb {R}_{>0} \times \mathbb {R} \to \mathbb {R}^2 \setminus \{0\},\; (r, \theta ) \mapsto (r \cos \theta , r \sin \theta )\). Then \(g \circ f = \Id _U\), so \(Df_{(x,y)}\) is the inverse of \(Dg_{f(x,y)}\). The matrix representing \(Dg_{(r,\theta )}\) is \(\bigl (\begin {smallmatrix} \cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta \end {smallmatrix}\bigr )\), with inverse

\[ \frac {1}{r} \begin {pmatrix} r \cos \theta & r\sin \theta \\ -\sin \theta & \phantom {r} \cos \theta \end {pmatrix} \; = \; \begin {pmatrix} \dfrac {x}{\sqrt {x^2+y^2}} & \dfrac {y}{\sqrt {x^2+y^2}} \\[4mm] -\dfrac {y}{x^2+y^2} & \dfrac {x}{x^2+y^2} \end {pmatrix} . \]

3. \(Df_{(x,y,z)} : \R ^3 \to \R ^2\) is represented by the matrix

\[ \pmat {2x & 2y & 1 \\ z^2 & -z & 2xz - y} . \]

\(Df_{(x,y,z)}\) is not surjective if and only if this matrix has less than full rank—equivalently its rows are linearly dependent, or its three minors are all zero. The first minor is

\[ 2x(-z) - 2yz^2 = -2z(x+yz), \]

so vanishes if and only if \(x = -yz\) or \(z = 0\). When \(x = -yz\), the vanishing of the other two minors

\[ 2x(2xz-y) - z^2, \qquad 2y(2xz-y) + z, \]

reduces to \(y = \pm \sqrt {\frac {z}{4z^2+2}}\). On the other hand, if \(z = 0\) then the vanishing of the other two minors reduces to \(x = 0\) or \(y = 0\). The case \(x = 0\) is already covered by \(x = -yz\) as before. On the other hand, if \(y = 0\) then the same formula for \(y\) as before is still valid. All in all, we can say that the derivative fails to be surjective precisely when
\(\seteqnumber{0}{}{0}\)
\begin{align*} y &= z = 0, \textrm { or}\\ y &= \pm \sqrt {\frac {z}{4z^2+2}} \textrm { and } x = -yz . \end{align*}

4.
- (i) \(GL_n(\R )\) is the complement to the zero set of \(\det : M_{n,n}(\R ) \to \R \). The determinant function is continuous, so its zero set is closed.
- (ii) We claim that \(D\inv _I = -\Id _{M_{n, n}(\R )}\). This is equivalent to saying that
  
  \[ \frac {\norm {(I+X)^{-1} - I + X}}{\norm {X}} \to 0 \]
  
  as \(X \to 0\) in \(M_{n,n}(\R )\). We can use any norms we like, but lets use the operator norm. Then, since \((I+X)((I+X)^{-1} - I +X) = I + (X^2 - I) = X^2\),
  
  \[ \norm {(I+X)^{-1} - I + X} \leq \norm {(I+X)^{-1}}\norm {X^2} \leq \norm {(I+X)^{-1}}\norm {X}^2 . \]
  
  Because \(\inv \) is continuous, certainly \(\norm {(I+X)^{-1}} < 2\norm {I} = 2\) for \(\norm {X}\) small, so
  
  \[ \frac {\norm {(I+X)^{-1} - I + X}}{\norm {X}} < 2 \norm {X} \to 0 \]
  
  as \(X \to 0\), thus proving that \(D\inv _I(X) = -X\).
- (iii) Given \(A \in GL_n(\R )\), define linear maps
  \(\seteqnumber{0}{}{0}\)
  \begin{align*} L_A &: M_{n,n}(\R ) \to M_{n,n}(\R ), X \mapsto AX,\\ R_A &: M_{n,n}(\R ) \to M_{n,n}(\R ), X \mapsto XA. \end{align*} The fact that \(B^{-1} = (AB)^{-1}A\) for any \(B \in GL_n(\R )\) means that \(\inv = R_A \circ \inv \circ L_A\). By the chain rule (and using that \(L_A\) and \(R_A\) are linear maps)
  
  \[ D\inv _I = R_A \circ D\inv _A \circ L_A . \]
  
  Hence
  
  \[ D\inv _A = (R_A)^{-1} \circ (-\Id \nolimits _{M_{n,n}(\R )}) \circ (L_A)^{-1} = R_{-A^{-1}}\circ L_{A^{-1}}, \]
  
  i.e.,
  
  \[ D\inv _A(X) = - A^{-1}XA^{-1}. \]
  
  An alternative proof is to observe that \(m\circ (\Id ,\inv ): M_{n,n}(\R ) \to M_{n,n}(\R ); A\mapsto AA^{-1}=I\) is constant, so
  
  \[ 0=Dm_{A,A^{-1}}\circ (D\Id \nolimits _A, D\inv _A)(X)=Dm_{A,A^{-1}}(X,D\inv _A(X))=XA^{-1}+AD\inv _A(X) \]
  
  (by the product rule) and again we get \(D\inv _A(X)=- A^{-1}XA^{-1}\).
  
  In other words \(D\inv =RL\circ \inv \) where \(RL(B)=R_{-B}\circ L_B\) is a product of the linear maps \(B\mapsto R_{-B}\) and \(B\mapsto L_B\), and hence is differentiable by the product rule with derivative \(D(RL)_B(Z) = R_{-Z} \circ L_B + R_{-B}\circ L_Z\). Since \(\inv \) is differentiable, \(D\inv \) is differentiable by the chain rule with \(D(D\inv )_A(X)= R_{A^{-1}XA^{-1}} \circ L_{A^{-1}} + R_{-A^{-1}}\circ L_{-A^{-1}XA^{-1}}\). Applying this to \(Y\) we get
  
  \[ D^2\inv _A(X,Y)= A^{-1}Y (A^{-1} X A^{-1}) + (A^{-1}X A^{-1}) Y A^{-1}. \]
  
  In general, induction on \(k\), using linearity, the chain rule, product rule and \(D\inv \), shows that
  
  \[ D^k\inv _A(X_1,\ldots X_k)=(-1)^k \sum _{\sigma \in S_k} A^{-1}X_{\sigma (1)}A^{-1}X_{\sigma (2)}\cdots A^{-1} X_{\sigma (k)} A^{-1}. \]
  
  Alternatively, observe that \(A^{-1} = \mathrm {adj}(A)/\det (A)\) where \(\mathrm {adj}(A)\) and \(\det (A)\) are polynomial in the entries of \(A\). Hence \(\mathrm {adj}(A)\) and \(1/\det (A)\) have partial derivatives of all orders (by induction in the latter case, since \(\det (A)\) is nonvanishing), and now the iterated product rule shows \(A^{-1}\) is smooth.

5.
- (i) We use induction, as the claim clearly holds for \(n=0\) with \(p_{n}(x)=1\). Suppose the claim holds for \(n = k\). Then for \(t > 0\)
  
  \[ \chi ^{(k+1)}(t) = \frac {d}{dt} \left ( \textstyle p_k\left (1/t\right ) e^{-1/t}\right ) = \left (-\frac {p_k'(1/t)}{t^2} + \frac {p_k\left (1/t\right )}{t^2} \right )e^{-1/t} , \]
  
  so if we set \(p_{k+1}(x) := x^2( -p'_k(x) + p_k(x))\) we are done.
  
  For \(t < 0\) it is clear that the \(n\)th derivative of \(\chi (t)\) is \(0\). We prove \(\chi ^{(n)}(t)\) is differentiable at \(t = 0\) with derivative \(\chi ^{(n+1)}(0) = 0\) by induction on \(n\): \((\chi ^{(n)}(0+\epsilon )-\chi ^{(n)}(0))/\epsilon = \chi ^{(n)}(\epsilon )/\epsilon \) which is \(p_n(1/\epsilon )e^{-1/\epsilon }/\epsilon \) for \(\epsilon >0\) and zero for \(\epsilon <0\). Thus it has limit \(0\) as \(\epsilon \to 0\) since exponentials dominate polynomials.
- (ii) Define \(\rho : \R ^n \to \R \) by
  
  \[ \rho (y) := \chi \left (r^2 - \norm {y - x}^2\right ). \]
  
  This is a smooth function (by the chain rule) that is non-zero precisely on \(B_r(x)\).