Chapter 2 Tensors and Differential Forms

1.	(a) Let $l:V\times W\to U$ be a bilinear map. Then by the universal property of free module, $\bar{l}:\mathcal{F}(V\times W)\to U$ is well defined, and by the bilinearity, generators of 2.1(1) are in the kernel of $\bar{l}$. Therefore, the map $\tilde{l}:V\otimes W\to U$ is well defined. Also, it satisfies $\tilde{l}\circ \phi =l$ by the definition of the map. Since $\tilde{l}(v\otimes w)$ is determined by the value of $l$ and elements of the form $v\otimes w$ generate $V\otimes W$, such $\tilde{l}$ must be unique. The rest of the argument is directly followed by the properties of universal objects. (b) Since $V\times W\cong W\times V$, it is true because of the universal property. (c) For $v\in V$, let $f_v:W\times U\to (V\times W)\times U\to (V\otimes W)\times U\to (V\otimes W)\otimes U$ be the map $(w,u)\mapsto ((v,w),u)\mapsto (v\otimes w,u)\mapsto (v\otimes w)\otimes u$. Since it is bilinear, one can define ${\tilde{f}}_v:W\otimes U\to (V\otimes W)\otimes U$. Then, if we set $g:V\times (W\otimes U)\to (V\otimes W)\otimes U,(v,x)\mapsto {\tilde{f}}_v(x)$ for all $x\in W\otimes U$, it is also bilinear, hence one can define $\tilde{g}:V\otimes (W\otimes U)\to (V\otimes W)\otimes U$, and $\tilde{g}(v\otimes (w\otimes u))=(v\otimes w)\otimes u)$. By the same method, one can also define $\tilde{h}:(V\otimes W)\otimes U\to V\otimes (W\otimes U)$, and it is easy to see that $\tilde{g}$ and $\tilde{h}$ are the inverse to each other. Therefore, $V\otimes (W\otimes U)\cong (V\otimes W)\otimes U$. (d) Let $f:V^{\ast }\times W\to \mathrm{Hom}(V,W)$ be a function such that $f(\phi ,w)(v)=\phi (v)\cdot w$. Since it is bilinear, one can define a map $\alpha :V^{\ast }\otimes W\to \mathrm{Hom}(V,W)$. (1) Let $\{v_i\},\{w_j\}$ be a basis of $V,W$, respectively, and $\{v_i^{\ast }\}$ be the dual basis corresponding to $\{v_i\}$, i.e., $v_a^{\ast }(v_b)={\delta }_{ab}$. Then elements of $V^{\ast }\otimes W$ can be represented as ${\sum }_{i,j}{a}_{ij}{v}_i^{\ast }\otimes w_j$. If $\alpha ({\sum }_{i,j}{a}_{ij}{v}_i^{\ast }\otimes w_j)=0$, then $\forall v\in V,{\sum }_{i,j}{a}_{ij}{v}_i^{\ast }(v)\cdot w_j=0$. In particular, if one substitutes $v_i$ for $v$, then ${\sum }_j{a}_{ij}{w}_j=0,\forall i$. Therefore, by the linear independence of the basis, $a_{ij}=0,\forall i,j$, and it means $\alpha$ is injective. (2) Let $\phi \in \mathrm{Hom}(V,W)$. Then if $\phi (v_i)={\sum }_j{a}_{ij}{w}_j$, set $\psi =\alpha ({\sum }_{i,j}{a}_{ij}\otimes w_j)$. Then $\psi$ and $\phi$ have the same value at $v_i$, so they must be the same linear map. Therefore, $\alpha$ is surjective. Therefore, $\alpha$ is an isomorphism. Hence, $\dim V^{\ast }\otimes W=\dim \mathrm{Hom}(V,W)=(\dim V)(\dim W)$, and $\dim V\otimes W=\dim V^{\ast \ast }\otimes W=\left(\dim V^{\ast }\right)(\dim W)=(\dim V)(\dim W)$. (e) By arguments similar to the above, $\left\{e_i\otimes f_j\right\}$ span $V\otimes W$. Since the number of $\left\{e_i\otimes f_j\right\}$ is the same as the dimension of $V\otimes W$ by (d), it must be a basis.
2.	(a) Let $V={\mathbb{R}}^2$, and consider $e_1\otimes e_2-e_2\otimes e_1\in V\otimes V$. If it is decomposable, say $\left(a{e}_1+b{e}_2\right)\otimes \left(c{e}_1+d{e}_2\right)$, then $ac=bd=0,ad=1,bc=-1$, which is impossible. (b) If the dimension is $1$ or $2$, then it is trivially true. Let $\dim V=3,V=⟨v_1,v_2,v_3⟩$. Then we only need to check that all elements of ${\Lambda }_2(V)$ are decomposable. However, for any element$$x=a{v}_1\wedge v_2+b{v}_2\wedge v_3+c{v}_3\wedge v_1\in {\Lambda }_2(V),$$if $a=0$, then $x=(b{v}_2-c{v}_1)\wedge v_3$, and if $a\ne 0$, then $x=(v_1-\frac{b}{a}{v}_3)\wedge (a{v}_2-c{v}_3)$. Therefore $x$ is always decomposable. (c) If $V={\mathbb{R}}^4$, then $e_1\wedge e_2+e_3\wedge e_4$ is indecomposable. If it is decomposable, say$$(a_1{e}_1+a_2{e}_2+a_3{e}_3+a_4{e}_4)\wedge (b_1{e}_1+b_2{e}_2+b_3{e}_3+b_4{e}_4),$$then$$a_1{b}_2-b_1{a}_2=a_3{b}_4-b_3{a}_4=1,$$$$a_1{b}_3-b_1{a}_3=a_1{b}_4-b_1{a}_4=a_2{b}_3-b_2{a}_3=a_2{b}_4-b_2{a}_4=0.$$If $a_1=0,b_1{a}_2=-1,b_1{a}_3=b_1{a}_4=0$, so $a_3=a_4=0$, which contradicts $a_3{b}_4-b_3{a}_4=1$. If $a_1\ne 0,b_3=\frac{b_1{a}_3}{a_1},b_4=\frac{b_1{a}_4}{a_1}$, which also contradicts $a_3{b}_4-b_3{a}_4=1$. (d) No. Let $\alpha =e_1\wedge e_2+e_3\wedge e_4\in {\Lambda }_2({\mathbb{R}}^4)$. Then $\alpha \wedge \alpha =2e_1\wedge e_2\wedge e_3\wedge e_4\in {\Lambda }_4({\mathbb{R}}^4)$, which is not zero.
3.	(a) That $u\wedge v\in {\lambda }_{k+l}(V)$ is a direct consequence of the associativity of tensor products. Since the last equation is bilinear, we only need to show if $u$ and $v$ are decomposable. However, if $u=u_1\wedge \dots \wedge u_k$ and $v=v_1\wedge \dots \wedge v_l$,$$\begin{array}{rl}v\wedge u& =v_1\wedge \dots \wedge v_l\wedge u_1\wedge \dots \wedge u_k\\ & =(-1{)}^l{u}_1\wedge v_1\wedge \dots \wedge v_l\wedge u_2\wedge \dots \wedge u_k\\ & =\dots \\ & =(-1{)}^{kl}{u}_1\wedge \dots \wedge u_k\wedge v_1\wedge \dots \wedge v_l=(-1{)}^{kl}u\wedge v.\end{array}$$(b) Let us follow the argument in the text. Since an arbitrary element of $I(V)$ has zero determinant, $e_1\otimes \dots \otimes e_n$ cannot be in $I(V)$, i.e., $e_1\wedge \dots \wedge e_n\ne 0$. That $\left\{e_{\Phi }\right\}$ spans $\Lambda (V)$ is obvious. If ${\sum }_{\Phi }{a}_{\Phi }{e}_{\Phi }=0$, since homogeneous parts should be zero, $\forall 0\le k\le n$, ${\sum }_{|\Phi |=k}{a}_{\Phi }{e}_{\Phi }=0$. However, for some ${\Phi }_0$, if we wedge $e_{{\Phi }_0^c}$ on the equation, then we get $a_{{\Phi }_0}{e}_1\wedge \dots \wedge e_n=0$, hence $a_{{\Phi }_0}=0$. Since ${\Phi }_0$ is arbitrary, it means $\{e_{\Phi }\}$ is linearly independent. Therefore, $\{e_{\Phi }\}$ is a basis. The rest of the argument trivially follows. (c) $\tilde{h}$ can be defined because of the universal property of tensor products and $I(V)$ should be in the kernel of the induced map since $h$ is alternating. Its uniqueness is also followed by the universal property of tensor products. If $W=\mathbb{R}$, since $\tilde{h}\in {\Lambda }_k(V{)}^{\ast }$ and $h\in A_k(V)$, ${\phi }^{\ast }:{\Lambda }_k(V{)}^{\ast }\to A_k(V)$ gives an isomorphism.
4.	Since (2), (3), and (4) are bilinear forms, we only need to consider when $f$ and $g$ are both induced by decomposable elements. Therefore if $\phi :\Lambda (V{)}^{\ast }\to A(V)$ is the isomorphism, and $f=\phi \circ \alpha (w_1^{\ast }\wedge \dots \wedge w_p^{\ast }),g=\phi \circ \alpha (w_{p+1}^{\ast }\wedge \dots \wedge w_{p+q}^{\ast })$. Then$$\begin{array}{rl}f{\wedge }_{\alpha }g(v_1,\dots ,v_{p+q})& =w_1^{\ast }\wedge \dots \wedge w_p^{\ast }\wedge w_{p+1}^{\ast }\wedge \dots \wedge w_{p+q}^{\ast }(v_1\wedge \dots \wedge v_{p+q})\\ & =\det (w_i^{\ast }{v}_j{)}_{1\le i,j\le p+q}\\ & =\sum _{\sigma \in S_{p+q}}\mathrm{sgn}(\sigma )\prod _{i=1}^{p+q}{w}_i^{\ast }{v}_{\sigma (i)}\\ & =\sum _{\pi :p,q\text{ shuffles}}\sum _{{\sigma }_1\in S_p}\sum _{{\sigma }_2\in S_q}\mathrm{sgn}(\pi )\mathrm{sgn}({\sigma }_1)\mathrm{sgn}({\sigma }_2)\prod _{i_1=1}^p{w}_{i_1}^{\ast }{v}_{\pi ({\sigma }_1(i_1))}\prod _{i_2=1}^q{w}_{p+i_2}^{\ast }{v}_{\pi (p+{\sigma }_2(i_2))}\mathrm{sgn}(\pi )\det (w_i^{\ast }{v}_{\pi (i)}{)}_{1\le i\le p}\det (w_j^{\ast }{v}_{\pi (j)}{)}_{p+1\le j\le p+q}\\ & =\sum _{\pi :p,q\text{ shuffles}}\mathrm{sgn}(\pi )f(v_{\pi (1)},\dots ,v_{\pi (p)})g(v_{\pi (p+1)},\dots ,v_{\pi (p+q)}).\end{array}$$On the other hand, let $f=\phi \circ \beta (w_1^{\ast }\wedge \dots \wedge w_p^{\ast }),g=\phi \circ \beta (w_{p+1}^{\ast }\wedge \dots \wedge w_{p+q}^{\ast })$. Then$$\begin{array}{rl}f{\wedge }_{\beta }g(v_1,\dots ,v_{p+q})& =w_1^{\ast }\wedge \dots \wedge w_p^{\ast }\wedge w_{p+1}^{\ast }\wedge \dots \wedge w_{p+q}^{\ast }(v_1\wedge \dots \wedge v_{p+q})\\ & =\frac{1}{(p+q)!}\det (w_i^{\ast }{v}_j{)}_{1\le i,j\le p+q}\\ & =\frac{1}{(p+q)!}\sum _{\sigma \in S_{p+q}}\mathrm{sgn}(\sigma )\prod _{i=1}^{p+q}{w}_i^{\ast }{v}_{\sigma (i)}\\ & =\frac{1}{(p+q)!}\sum _{\pi :p,q\text{ shuffles}}\sum _{{\sigma }_1\in S_p}\sum _{{\sigma }_2\in S_q}\mathrm{sgn}(\pi )\mathrm{sgn}({\sigma }_1)\mathrm{sgn}({\sigma }_2)\prod _{i_1=1}^p{w}_{i_1}^{\ast }{v}_{\pi ({\sigma }_1(i_1))}\prod _{i_2=1}^q{w}_{p+i_2}^{\ast }{v}_{\pi (p+{\sigma }_2(i_2))}\\ & =\frac{1}{(p+q)!}\sum _{\pi :p,q\text{ shuffles}}\mathrm{sgn}(\pi )\det (w_i^{\ast }{v}_{\pi (i)}{)}_{1\le i\le p}\det (w_j^{\ast }{v}_{\pi (j)}{)}_{p+1\le j\le p+q}\\ & =\frac{p!q!}{(p+q)!}\sum _{\pi \in S_{p+q}}\mathrm{sgn}(\pi )\det (w_i^{\ast }{v}_{\pi (i)}{)}_{1\le i\le p}\det (w_j^{\ast }{v}_{\pi (j)}{)}_{p+1\le j\le p+q}\\ & =\frac{1}{(p+q)!}\sum _{\pi \in S_{p+q}}\mathrm{sgn}(\pi )f(v_{\pi (1)},\dots ,v_{\pi (p)})g(v_{\pi (p+1)},\dots ,v_{\pi (p+q)}).\end{array}$$Therefore,$$\begin{array}{rl}f{\wedge }_{\alpha }g(v_1,\dots ,v_{p+q})& =\sum _{\pi :p,q\text{ shuffles}}\mathrm{sgn}(\pi )f(v_{\pi (1)},\dots ,v_{\pi (p)})g(v_{\pi (p+1)},\dots ,v_{\pi (p+q)})\\ & =\frac{1}{p!q!}\sum _{\pi \in S_{p+q}}\mathrm{sgn}(\pi )f(v_{\pi (1)},\dots ,v_{\pi (p)})g(v_{\pi (p+1)},\dots ,v_{\pi (p+q)})\\ & =\frac{(p+q)!}{p!q!}f{\wedge }_{\beta }g.\end{array}$$
5.	(where $\gamma$ is an integral curve of $X$)$$\begin{array}{rl}{L}_X(f){|}_m& =\underset{t\to 0}{\lim }\frac{\delta X_t(f_{X_t(m)})-f_m}{t}=\underset{t\to 0}{\lim }\frac{f_{X_t(m)}-f(m)}{t}\\ & =\underset{t\to 0}{\lim }\frac{f({\gamma }_m(t))-f(m)}{t}=d\gamma \left(\frac{d}{dt}{|}_0\right)(f)\\ & =X_m(f).\end{array}$$
6.	(Since $\sqrt{t}$ is defined on $[0,\epsilon )$, the limit should be calculated when $t\to 0^{+}$.) We will use L’Hôpital’s rule. Let $g(x,y,z,w)=f(Y_{-x}{X}_{-y}{Y}_z{X}_w)(m)$. Then,$$\begin{array}{rl}\frac{\partial g}{\partial x}(x,y,z,w)& =-Y_{Y_{-x}{X}_{-y}{Y}_z{X}_w(m)}(f)\\ \frac{\partial g}{\partial y}(x,y,z,w)& =-X_{X_{-y}{Y}_z{X}_w(m)}(f\circ Y_{-x})\\ \frac{\partial g}{\partial z}(x,y,z,w)& =Y_{Y_z{X}_w(m)}(f\circ Y_{-x}\circ X_{-y})\\ \frac{\partial g}{\partial w}(x,y,z,w)& =X_{X_w(m)}(f\circ Y_{-x}\circ X_{-y}\circ Y_z)\\ \frac{{\partial }^{2}g}{\partial x^2}(x,y,z,w)& =Y_{Y_{-x}{X}_{-y}{Y}_z{X}_w(m)}(Y(f))\\ \frac{{\partial }^{2}g}{\partial y^2}(x,y,z,w)& =X_{X_{-y}{Y}_z{X}_w(m)}(X(f\circ Y_{-x}))\\ \frac{{\partial }^{2}g}{\partial z^2}(x,y,z,w)& =Y_{Y_z{X}_w(m)}(Y(f\circ Y_{-x}\circ X_{-y}))\\ \frac{{\partial }^{2}g}{\partial w^2}(x,y,z,w)& =X_{X_w(m)}(X(f\circ Y_{-x}\circ X_{-y}\circ Y_z))\\ \frac{{\partial }^{2}g}{\partial x\partial y}(x,y,z,w)& =X_{X_{-y}{Y}_z{X}_w(m)}(Y(f)\circ Y_{-x})\\ \frac{{\partial }^{2}g}{\partial x\partial z}(x,y,z,w)& =-Y_{Y_z{X}_w(m)}(Y(f)\circ Y_{-x}\circ X_{-y})\\ \frac{{\partial }^{2}g}{\partial x\partial w}(x,y,z,w)& =-X_{X_w(m)}(Y(f)\circ Y_{-x}\circ X_{-y}\circ Y_z)\\ \frac{{\partial }^{2}g}{\partial y\partial z}(x,y,z,w)& =-Y_{Y_z{X}_w(m)}(X(f\circ Y_{-x})\circ X_{-y})\\ \frac{{\partial }^{2}g}{\partial y\partial w}(x,y,z,w)& =-X_{X_w(m)}(X(f\circ Y_{-x})\circ X_{-y}\circ Y_z)\\ \frac{{\partial }^{2}g}{\partial z\partial w}(x,y,z,w)& =X_{X_w(m)}(Y(f\circ Y_{-x}\circ X_{-y})\circ Y_z)\end{array}$$Therefore,$$\begin{array}{rl}\underset{t\to 0^{+}}{\lim }\frac{f(\beta (t))-f(\beta (0))}{t}& =\underset{t\to 0^{+}}{\lim }\frac{f(\beta (t^2))-f(\beta (0))}{t^2}\\ & =\underset{t\to 0^{+}}{\lim }\frac{g(t,t,t,t)-g(0,0,0,0)}{t^2}\\ & =\underset{t\to 0^{+}}{\lim }\frac{dg(t,t,t,t)/dt}{2t}\\ & =\underset{t\to 0^{+}}{\lim }\frac{d^{2}g(t,t,t,t)/d{t}^2}{2}.\end{array}$$Since$$dg(t,t,t,t)/dt{|}_{t=0}=-Y_m(f)-X_m(f)+Y_m(f)+X_m(f)=0,$$we have$$\begin{array}{rl}\underset{t\to 0^{+}}{\lim }\frac{d^{2}g(t,t,t,t)/d{t}^2}{2}& =\frac{1}{2}\left(Y_m(Y(f))+X_m(X(f))+Y_m(Y(f))+X_m(X(f))\right)\\ & +X_m(Y(f))-Y_m(Y(f))-X_m(Y(f))-Y_m(X(f))\\ & -X_m(X(f))+X_m(Y(f))\\ & =X_m(Y(f))-Y_m(X(f))=[X,Y]{|}_m(f).\end{array}$$
7.	For $m\in M$, let $x_1,\dots ,x_n$ be local coordinates around $m$. Since the equation is linear with respect to $\omega$, it is enough to consider the case when $\omega$ is decomposable, i.e., $\omega =gd{x}_{m_1}\wedge \dots \wedge d{x}_{m_p}$. Then,$$\begin{array}{rl}{L}_{Y_0}(\omega (Y_1,\dots ,Y_p))& =Y_0(g\det (Y_1(x_{m_i}),\dots ,Y_p(x_{m_i}){)}_{1\le i\le p})\\ & =Y_0(g)\det (Y_1(x_{m_i}),\dots ,Y_p(x_{m_i}){)}_{1\le i\le p}\\ & +\sum _{j=1}^{p}g\det (Y_1(x_{m_i}),\dots ,Y_0(Y_j(x_{m_i})),\dots ,Y_p(x_{m_i}){)}_{1\le i\le p},\end{array}$$$$\begin{array}{rl}{L}_{Y_0}(\omega )(Y_1,\dots ,Y_p)& =Y_0(g)\det (Y_1(x_{m_i}),\dots ,Y_p(x_{m_i}){)}_{1\le i\le p}\\ & +\sum _{j=1}^{p}g\det (Y_1(x_{m_i}),\dots ,Y_j(Y_0(x_{m_i})),\dots ,Y_p(x_{m_i}){)}_{1\le i\le p},\end{array}$$$$\begin{array}{rl}\omega (Y_1,\dots ,L_{Y_0}{Y}_j,\dots ,Y_p)& =\det (Y_1,\dots ,Y_0(Y_j(x_{m_i})),\dots ,Y_p{)}_{1\le i\le p}\\ & -\det (Y_1,\dots ,Y_j(Y_0(x_{m_i})),\dots ,Y_p{)}_{1\le i\le p}.\end{array}$$(e) follows by combining these equations.
8.	$(\Rightarrow )\forall (m,n)\in M\times N,\forall Y_2,\dots ,Y_p\in (M\times N{)}_{(m,n)}$, $$\begin{array}{rl}i(X){\omega }_{(m,n)}(Y_2,\dots ,Y_p)& =i(X)\delta \pi (\alpha {)}_{(m,n)}(Y_2,\dots ,Y_p)\\ & =\delta \pi (\alpha {)}_{(m,n)}(X_{(m,n)},Y_2,\dots ,Y_p)\\ & ={\alpha }_m(d\pi (X_{(m,n)}),d\pi (Y_2),\dots ,d\pi (Y_p))=0.\end{array}$$Also,$$\begin{array}{rl}i(X)d{\omega }_{(m,n)}(Y_2,\dots ,Y_p)& =i(X)d(\delta \pi (\alpha ){)}_{(m,n)}(Y_2,\dots ,Y_p)\\ & =d(\delta \pi (\alpha ){)}_{(m,n)}(X_{(m,n)},Y_2,\dots ,Y_p)\\ & =\delta \pi (d\alpha {)}_{(m,n)}(X_{(m,n)},Y_2,\dots ,Y_p)\\ & =d{\alpha }_m(d\pi (X_{(m,n)}),d\pi (Y_2),\dots ,d\pi (Y_p))=0.\end{array}$$Since $L_X=i(X)\circ d+d\circ i(X)$, $L_X\omega =0$. $(\Leftarrow )$ Let $(m,n)\in M\times N$, and choose local coordinates of $m\in M,n\in N$, say $\{x_1,\dots ,x_s\},\{y_1,\dots ,y_t\}$, such that local coordinates of $(m,n)\in M\times N$ consist of $\{x_1,\dots ,x_s,y_1,\dots ,y_t\}$. Then, $\omega$ can be described as$$\sum _{|\Phi |+|\Psi |=p}{g}_{\Phi ,\Psi }d{x}_{\Phi }\wedge d{y}_{\Psi }.$$And we can take a vector field $X$ on $M\times N$ as $\frac{\partial }{\partial x_i}\times ({\mathcal{C}}^{\infty }$ function which is zero outside the local chart and $1$ at $(m,n))$. Then $X_{(m,n)}=\frac{\partial }{\partial x_i}{|}_{(m,n)}$ and $d\pi (X)=0$. Therefore, by assumption, $L_X\omega =0$. But,$$L_X\omega =\sum _{|\Phi |+|\Psi |=p}\frac{\partial g_{\Phi ,\Psi }}{\partial x_i}d{x}_{\Phi }\wedge d{y}_{\Psi }.$$Therefore $\forall x_i,\frac{\partial g_{\Phi ,\Psi }}{\partial x_i}=0$. If we let ${\iota }_{m^{\prime }}:N\to M\times N,n^{\prime }\mapsto \left(m^{\prime },n^{\prime }\right)$,$$(\delta {\iota }_m\omega {)}_n=\sum _{|\Psi |=p}{g}_{\Phi ,\Psi }(m,n)d{y}_{\Psi }.$$Therefore, by the result above, on the small neighborhood of $m\in M$ and for all $m^{\prime }$ in the neighborhood, ${\left(\delta {\iota }_m^{\prime }\omega \right)}_n$ is the same as ${\left(\delta l_m\omega \right)}_n$. Since $(m,n)$ is arbitrary, we can say that $\delta {\iota }_m\omega$ is locally the same regardless of the choice of $m$. Also, since $M$ is connected, we conclude that $\delta {\iota }_m\omega$ is always the same regardless of the choice of $m\in M$. Let $\alpha$ be this $p$-form on $N$. Then, $\omega =\delta \pi (\alpha )$. Indeed, for all $(m,n)\in M\times N$, and $X_1,\dots ,X_p\in (M\times N{)}_{(m,n)}$,$$\begin{array}{rl}(\omega -\delta \pi (\alpha ){)}_{(m,n)}(X_1,\dots ,X_p)& ={\omega }_{(m,n)}(X_1,\dots ,X_p)-\delta \pi (\delta {\iota }_m\omega {)}_{(m,n)}(X_1,\dots ,X_p)\\ & ={\omega }_{(m,n)}(X_1,\dots ,X_p)-{\omega }_{(m,n)}(d({\iota }_m\circ \pi )(X_1),\dots ,d({\iota }_m\circ \pi )(X_p))\\ & ={\omega }_{(m,n)}(X_1-d({\iota }_m\circ \pi )(X_1),\dots ,X_p-d({\iota }_m\circ \pi )(X_p)).\end{array}$$The last term is zero since if we set $X$ as a vector field of $M\times N$ such that$$X_{(m,n)}=X_1-d({\iota }_m\circ \pi )(X_1)$$and $d\pi X=0$, (it is possible since$$d\pi (X_1-d({\iota }_m\circ \pi )(X_1))=0,$$the last term is$$(i(X)\omega {)}_{(m,n)}(X_2-d({\iota }_m\circ \pi )(X_2),\dots ,X_p-d({\iota }_m\circ \pi )(X_p))$$which should be zero by the assumption.
9.	$(\Leftarrow )$ If $a_1{v}_1+\dots +a_r{v}_r=0$ and some of $a_i$’s are nonzero, then without loss of generality, one can assume $a_1\ne 0$. Dividing the equation by $a_1$, one can also assume $a_1=1$. Then$$v_1\wedge \dots \wedge v_n=(-a_2{v}_2-\dots -a_r{v}_r)\wedge v_2\wedge \dots \wedge v_r=0.$$ $(\Rightarrow )$ Suppose $v_1,\dots ,v_p$ are linearly independent, then there is a linear map $\phi :V\to k^r,v_i\mapsto e_i$. By the universal property, it induces a map ${\Lambda }_r(\phi ):{\Lambda }_r(V)\to {\Lambda }_r(k^r)$. Since $\phi (v_1\wedge \dots \wedge v_p)=e_1\wedge \dots \wedge e_p\ne 0$ by 2.6, $v_1\wedge \dots \wedge v_p\ne 0$.
10.	$(\Rightarrow )$ Let $v_i={\sum }_j{a}_{ij}{w}_j$. Then by direct calculation, $v_1\wedge \dots \wedge v_r=(\det A)w_1\wedge \dots \wedge w_r$. $(\Leftarrow )$ If $w\in ⟨w_1,\dots ,w_r⟩\setminus ⟨v_1,\dots v_r⟩$, $v_1\wedge \dots \wedge v_r\wedge w\ne 0$ by Exercise 9, which contradicts $w_1\wedge \dots \wedge w_r\wedge w=0$. Therefore $⟨w_1,\dots ,w_r⟩\subset ⟨v_1,\dots v_r⟩$ and vice versa.
11.	(Condition$\Rightarrow$(a)) Since $d_{{\omega }_i}\in \mathfrak{I}$ by definition of differential ideals. ((a)$\Rightarrow$(b))$$d\omega =d{\omega }_1\wedge \dots \wedge {\omega }_r+\dots +(-1{)}^{r+1}{\omega }_1\wedge \dots \wedge d{\omega }_r=\alpha \wedge \omega$$by using (a). ((b)$\Rightarrow$(a))$$d\omega =d{\omega }_1\wedge \dots \wedge {\omega }_r+\dots +(-1{)}^{r+1}{\omega }_1\wedge \dots \wedge d{\omega }_r=\alpha \wedge \omega .$$Since $\omega \wedge {\omega }_i=0$, by wedging ${\omega }_i$ on the equation, we get $d{\omega }_i\wedge \omega =0$. Therefore, $d{\omega }_i\in ⟨{\omega }_1,\dots ,{\omega }_r⟩$. ((a)$\Rightarrow$condition)$$d⟨{\omega }_1,\dots ,{\omega }_r⟩\subset ⟨d{\omega }_1,\dots ,d{\omega }_r,{\omega }_1,\dots ,{\omega }_r⟩$$by the property of derivations, and the result directly follows.
12.	For another basis $w_1,\dots ,w_n$, let $w_i={\sum }_j{a}_{ij}{v}_j$. Then$$\begin{array}{rl}A{w}_1\wedge \dots \wedge A{w}_n& =\sum _j{a}_{1j}A{v}_j\wedge \dots \wedge \sum _j{a}_{nj}A{v}_j\\ & =\det (a_{ij})A{v}_1\wedge \dots \wedge A{v}_n.\end{array}$$Therefore by Exercise 10, $\det A$ does not depend on the choice of basis. If we choose the standard basis $e_1,\dots ,e_n$, the equation of $\det A$ follows directly. Also, for two matrices $A,B$,$$B{v}_1\wedge \dots \wedge B{v}_n=(\det B)v_1\wedge \dots \wedge v_n,$$so$$\begin{array}{rl}AB{v}_1\wedge \dots \wedge AB{v}_n& =(\det B)A{v}_1\wedge \dots \wedge A{v}_n\\ & =(\det B)(\det A)v_1\wedge \dots \wedge v_n,\end{array}$$which proves $\det A\det B=\det AB$.
13.	The orthonormality comes from direct calculations. (5) Since $\ast \ast$ is linear, we only need to show this property for the basis elements. If $\Phi =\left\{i_1,\dots ,i_p\right\}\subset \{1,\dots ,n\}$ and let $f$ be the element which satisfies $e_{\Phi }\wedge f=e_1\wedge \dots \wedge e_n$, then$$f\wedge e_{\Phi }=(-1{)}^{p(n-p)}{e}_1\wedge \dots \wedge e_n.$$Therefore $\ast \ast e_{\Phi }=\pm \ast f=(-1{)}^{p(n-p)}{e}_{\Phi }$. (6) Since the equation is bilinear in $v$ and $w$, we only need to show this for the basis elements. Therefore we can assume that $v=e_{{\Phi }_1},w=e_{{\Phi }_2}$. If $v\ne w$,$$⟨v,w⟩=\ast (w\wedge \ast v)=\ast (v\wedge \ast w)=0$$by definition of star and inner product. If $v=w$,$$⟨v,w⟩=\ast (w\wedge \ast v)=\ast (v\wedge \ast w)=1$$also by definition. Hence (6) is proved.
14.	It suffices to show that for all $v\in {\Lambda }_{p+1}(V)$ and $w\in {\Lambda }_p(V)$, $$⟨\gamma (v),w⟩=⟨v,\xi \wedge w⟩.$$But it is equivalent to (using Exercise 13):$$\begin{array}{rl}& (-1{)}^{np}⟨w,\ast (\xi \wedge \ast v)⟩=⟨\xi \wedge w,v⟩\\ \iff & (-1{)}^{np}\ast (w\wedge (-1{)}^{p(n-p)}\xi \wedge \ast v)=⟨\xi \wedge w,v⟩\\ \iff & (-1{)}^p\ast (w\wedge \xi \wedge \ast v)=⟨\xi \wedge w,v⟩\\ \iff & \ast (\xi \wedge w\wedge \ast v)=⟨\xi \wedge w,v⟩\end{array}$$which is true by Exercise 13.
15.	(It is true only if $\xi \ne 0$) Obviously $(\xi \wedge {)}^2=0$. Conversely, if $\xi \wedge w=0$, then$$w=\xi \wedge \gamma \left(\frac{w}{⟨\xi ,\xi ⟩}\right).$$To prove this, it is enough to show that for all $v\in {\Lambda }_{p+1}(V)$,$$⟨\xi \wedge \gamma (w),v⟩=⟨\xi ,\xi ⟩⟨w,v⟩.$$But$$\begin{array}{rl}⟨\xi \wedge \gamma (w),v⟩& =⟨\gamma (w),\gamma (v)⟩\\ & =⟨(-1{)}^{np}\ast (\xi \wedge \ast w),(-1{)}^{np}\ast (\xi \wedge \ast v)⟩\\ & =\ast (\ast (\xi \wedge \ast w)\wedge (-1{)}^{p(n-p)}\xi \wedge \ast v)\\ & =\ast (\xi \wedge \ast v\wedge \ast (\xi \wedge \ast w))\\ & =⟨\xi \wedge \ast v,\xi \wedge \ast w⟩.\end{array}$$Let us consider a basis $\{v_i\}$ of $V$ which contains $\xi$ and an expression of $w$ in terms of the basis derived by $\left\{v_i\right\}$, say ${\sum }_{|I|=p+1}{a}_I{v}_I$ where $I=\{v_{i_1},\dots ,v_{i_{p+1}}\}\subset \{v_i\}$ and $v_I=v_{i_1}\wedge \dots \wedge v_{i_{p+1}}$ (to clarify the definition, it is assumed that orders of elements of all finite subset of $\{v_i\}$ are determined, possibly by using the Axiom of Choice if necessary). Since $\xi \wedge w=0$, it follows that all $I$ which $a_I\ne 0$ should contain $\xi$. Therefore, considering an expression of $\ast w$ derived by $w$, we see that there is no $\xi$ in the expression of $\ast w$. It means that if calculating $⟨\xi \wedge \ast v,\xi \wedge \ast w⟩$ by splitting matrices and getting determinants, the first column of matrices should be $(⟨\xi ,\xi ⟩,0,\dots ,0)$. It means that $⟨\xi \wedge \ast v,\xi \wedge \ast w⟩$ should be $⟨\xi ,\xi ⟩⟨\ast v,\ast w⟩=⟨\xi ,\xi ⟩⟨v,w⟩$.
16.	${\sum }_i{\theta }_i\wedge {\omega }_i=0$. If we wedge ${\omega }_1\wedge \dots \wedge \widehat{{\omega }_i}\wedge \dots \wedge {\omega }_p$, we get$${\theta }_i\wedge {\omega }_1\wedge \dots \wedge {\omega }_p=0.$$Therefore ${\theta }_i\in ⟨{\omega }_1,\dots ,{\omega }_p⟩$. So it is possible to write ${\theta }_i={\sum }_j{A}_{ij}{\omega }_j$, where $A_{ij}$ are ${\mathcal{C}}^{\infty }$ functions. One can derive $A_{ij}=A_{ji}$ by just putting these expressions into the given equation and noting that ${\omega }_i\wedge {\omega }_j$’s are linearly independent.