Some exercise solutions for John M. Lee Introduction to Smooth Manifolds Second Edition

1 Smooth Manifold

Exercise 1.18

• Suppose ${\mathcal{A}}_{\mathcal{1}},{\mathcal{A}}_2$ determine the same smooth structure $\mathcal{A}$ in $M$.

  By the definition of maximality, ${\mathcal{A}}_1\subseteq \mathcal{A}$ and ${\mathcal{A}}_2\subseteq \mathcal{A}$. Since every pair of charts in $\mathcal{A}$ is smoothly compatible, the same holds for any pair of charts taken from ${\mathcal{A}}_1\cup {\mathcal{A}}_2\subseteq \mathcal{A}$. Thus ${\mathcal{A}}_1\cup {\mathcal{A}}_2$ is a smooth atlas.

• Suppose ${\mathcal{A}}_1\cup {\mathcal{A}}_2$ is a smooth atlas.

  Let $\mathcal{A}$ be the unique maximal smooth atlas containing ${\mathcal{A}}_1\cup {\mathcal{A}}_2$. Since ${\mathcal{A}}_1\subseteq {\mathcal{A}}_1\cup {\mathcal{A}}_2\subseteq \mathcal{A}$, the maximal smooth atlas determined by ${\mathcal{A}}_1$ must be contained in $A$. But a maximal smooth atlas cannot be properly contained in another. Thus the maximal smooth atlas determined by ${\mathcal{A}}_1$ is $\mathcal{A}$. The same argument applies to ${\mathcal{A}}_2$, so ${\mathcal{A}}_1$ and ${\mathcal{A}}_2$ determines the same smooth structure.

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12 Tensor

Exercise 12.3

Let $F,F^{\prime }\in L(V_1,\dots ,V_k;\mathbb{R}),G,G^{\prime }\in L(W_1,\cdots ,W_l;\mathbb{R})$, and scalars $a,b\in \mathbb{R}$.

• Bilinearity

  For the first variable,

  $$\begin{array}{rl}& ((aF+b{F}^{\prime })\otimes G)(v,w)\\ =& (aF+b{F}^{\prime })(v)G(w)\\ =& (aF(v)+b{F}^{\prime }(v))G(w)\\ =& aF(v)G(w)+b{F}^{\prime }(v)G(w)\\ =& a(F\otimes G)(v,w)+b(F^{\prime }\otimes G)(v,w)\end{array}$$

  For the second variable,

  $$\begin{array}{rl}& (F\otimes (aG+b{G}^{\prime }))(v,w)\\ =& F(v)(aG+b{G}^{\prime })(w)\\ =& F(v)(aG(w)+b{G}^{\prime }(w))\\ =& aF(v)G(w)+bF(v)G^{\prime }(w)\\ =& a(F\otimes G)(v,w)+b(F\otimes G^{\prime })(v,w)\end{array}$$

  Thus the bilinearity holds.

• Associativity

  Let $H\in L(U_1,\dots ,U_m;\mathbb{R})$.

  Left-hand side:

  $$((F\otimes G)\otimes H)(v,w,u)=(F\otimes G)(v,w)H(u)=F(v)G(w)H(u)$$

  Right-hand side:

  $$(F\otimes (G\otimes H))(v,w,u)=F(v)(G\otimes H)(w,u)=F(v)(G(w)H(u))$$

  Left-hand side equals right-hand side by the associativity of multiplication.

Thus the operation $\otimes$ is bilinear and associative.

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