用户: Solution/ 习题: 陈维桓 微分流形/第一章 预备知识

9.	(1) 代入 ${\text{D}}_{v}f=⟨\nabla f(P),v⟩$ 即可. (2) ${\text{D}}_u(g\cdot h)=⟨\nabla (g\cdot h)(P),u⟩=⟨g(P)\cdot \nabla h(P),u⟩+⟨h(P)\cdot \nabla g(P),u⟩=g(P)\cdot {\text{D}}_{u}h+h(P)\cdot {\text{D}}_{u}g$.
11.	(1) $J_{\tilde{f}}=(\frac{\partial {\tilde{f}}_i}{\partial {\tilde{x}}_j}{)}_{n\times m}=(\frac{\partial {\tilde{f}}_i}{\partial x_k}\frac{\partial x_k}{\partial {\tilde{x}}_j}{)}_{n\times m}=(\frac{\partial (a_l^i{f}_l)}{\partial x_k}\frac{\partial x_k}{\partial (b_s^j{x}_s)}{)}_{n\times m}=(a_j^i\frac{\partial f_i}{\partial x_j}(b_k^j{)}^{-1}{)}_{n\times m}=A{J}_f{B}^{-1}$. (2) 由 $A,B$ 均可逆得 $\mathrm{rk}{J}_f^{\prime }=\mathrm{rk}{J}_f$. (3) 完全同理. 将曲纹坐标系看成坐标变换即可, 此时的坐标变换函数为曲纹坐标变换所对应的 Jacobi 矩阵.
12.	只需证明在所给区域 $\frac{\partial (x^1,\cdots ,x^n)}{\partial (u^1,\cdots ,u^n)}\ne 0$ 即可. 写出行列式并消去第一行并使用数学归纳法能够证明 (具体过程可参考大多数数学分析教材) $$\frac{\partial (x^1,\cdots ,x^n)}{\partial (u^1,\cdots ,u^n)}=r_1{r}_2\cdots r_n(u^n{)}^{n-1}(\cos u^1{)}^{n-2}(\cos u^2{)}^{n-3}\cdots \cos u^{n-2},$$由 $x_1\ne 0$ 得 $u^n,\cos u^1,\cdots ,\cos u^{n-2}\ne 0$, 所以 $\frac{\partial (x^1,\cdots ,x^n)}{\partial (u^1,\cdots ,u^n)}\ne 0$.
17.	$f$ 是线性变换是显然的. 要证明与基底选取无关, 只需取另一组基底 $\widetilde{{\delta }_i}$, 验证这两组基底下的计算结果一致即可. 设另一组基底为 $\widetilde{{\delta }_i}=a^{ij}{\delta }_j$, 则 $\widetilde{{\delta }^i}={\tilde{a}}_{ji}{\delta }^j$, 其中 ${\tilde{a}}_{ij}{a}^{jk}={\delta }_{ik}$, 那么 (已省略对 $i$ 求和) $$\begin{array}{rl}\tilde{f}(u,{\tilde{\delta }}^i){\tilde{\delta }}_i& =\tilde{f}(u,{\tilde{a}}_{ji}{\delta }^j)a^{ik}{\delta }_k\\ & ={\tilde{a}}_{ji}{a}^{ik}\tilde{f}(u,{\delta }^j){\delta }_k\\ & ={\delta }_{jk}\tilde{f}(u,{\delta }^j){\delta }_k\\ & =\tilde{f}(u,{\delta }^i){\delta }_i.\end{array}$$
18.	一方面, 对于每一个 $r$ 重线性映射 $t$, 可以定义 $F(\alpha ,v_1,\cdots ,v_n):=\alpha t(v_1,\cdots ,v_n)$; 另一方面, 对于每一个 $(1,r)$ 型张量 $F(\alpha ,v_1,\cdots ,v_n)$, 可以定义 $t(v_1,\cdots ,v_n)=\sum _{i=1}^{m}F({\delta }^i,v_1,\cdots ,v_n){\delta }_i$. 类似于上一题, 由 $F$ 的线性型可知 $t$ 的定义与基底选取无关.
20.	取 $V$ 的两个线性无关的向量 $v_1,v_2$, $W$ 的两个线性无关的向量 $w_1,w_2$, $v_1\otimes w_1,v_2\otimes w_2$ 均在此集合中, 但 $v_1\otimes w_1+v_2\otimes w_2$ 并不在其中. 这是因为对于 $v=\sum _{i=1}^m{a}_i{v}_i$ 和 $w=\sum _{j=1}^n{b}_j{w}_j$, $v\otimes w=\sum _{i=1}^m\sum _{j=1}^n{a}_i{b}_j{v}_i\otimes w_j$ 的系数构成的矩阵 $(a_i{b}_j{)}_{\begin{array}{l}1\le i\le m,\\ 1\le j\le n\end{array}}$ 的秩为 $1$, 而 $v_1\otimes w_1+v_2\otimes w_2$ 的系数的矩阵 $\left(\begin{array}{ccc}1& 0& \\ 0& 1& \\ & & \end{array}\right)$ 的秩为 $2$.
21.	若 $e_i=a_i^k{\delta }_k$ 为另一组基底, 且在 $e_i$ 基底下, $f$ 的表示矩阵为 $c_m^j{a}_i^l{b}_l^m$, 其中 $(c_j^i)$ 是 $(a_j^i)$ 的逆矩阵, 则$$\widetilde{b_i^j}\widetilde{b_k^i}\widetilde{b_j^k}=c_m^j{a}_i^l{c}_j^i{a}_k^h{c}_n^k{a}_j^m{b}_l^m{b}_h^l{b}_m^n.=b_l^m{b}_h^l{b}_m^h=B_3$$ 只需对 $(3,3)$ 型张量 $F({\alpha }^1,{\alpha }^2,{\alpha }^3,v_1,v_2,v_3):={\alpha }^1(f(v_2)){\alpha }^2(f(v_3)){\alpha }^3(f(v_1))$ 缩并三次即可表示出 $B_3$.
22.	是 $F(\alpha ,v)=\alpha (v)$, 因为 $F({\delta }^j,{\delta }_i)={\delta }^j({\delta }_i)={\delta }_i^j$.
23.	不存在. 因为该张量无法缩并. 事实上, 任取一个基变换, 容易发现该张量一定与基选取有关.
24.	$$\begin{array}{rl}\sigma (f)(v_1,\cdots ,v_q)& ={\alpha }^1\otimes \cdots \otimes {\alpha }^q(v_{\sigma (1)},\cdots ,v_{\sigma (q)})\\ & ={\alpha }^1(v_{\sigma (1)})\cdots {\alpha }^q(v_{\sigma (q)})\\ & ={\alpha }^{\tau (1)}(v_1)\cdots {\alpha }^{\tau (q)}(v_q)\\ & ={\alpha }^{\tau (1)}\otimes \cdots \otimes {\alpha }^{\tau (q)}(v_1,\cdots ,v_q).\end{array}$$
25.	(1) 由 $\xi =S_q(\xi )$ 容易得到 $\xi$ 对称; 另一方面, 当 $\xi$ 对称时, 有 $S_q(\xi )=\frac{{\sum }_{\sigma \in \mathcal{S}}\sigma (\xi )}{q!}=\frac{{\sum }_{\sigma \in \mathcal{S}}\xi }{q!}=\frac{|\mathcal{S}(q)|\xi }{q!}=\xi$. (2) 由 $\xi =A_q(\xi )$ 容易得到 $\xi$ 反对称; 另一方面, 当 $\xi$ 对称时, 有 $A_q(\xi )=\frac{{\sum }_{\sigma \in \mathcal{S}}\mathrm{sign}(\sigma )\sigma (\xi )}{q!}=\frac{{\sum }_{\sigma \in \mathcal{S}}{\mathrm{sign}(\sigma )}^2\xi }{q!}=\frac{|\mathcal{S}(q)|\xi }{q!}=\xi$.
26.	设 $\sigma$ 为二阶张量上的对换, 取 ${\xi }_1=\frac{\xi +\sigma (\xi )}{2}$ 与 ${\xi }_2=\frac{\xi -\sigma (\xi )}{2}$. 则 $\xi ={\xi }_1+{\xi }_2$ 且 ${\xi }_1$ 与 ${\xi }_2$ 分别为对称与反对称张量.
27.	通过列举 $\mathcal{S}$ 中元素并利用 $\phi$ 给定性质, 可计算得 $A_3(\phi )=\frac{1}{6}\sum _{\sigma \in \mathcal{S}}\mathrm{sign}(\sigma )\sigma (\phi )=\frac{1}{6}(2-4)\phi =-\frac{1}{3}\phi$. 由 $A_3(\phi )$ 的反对称性可推知 $\phi$ 反对称. 同时, 取任意 $v_1,v_2,v_3\in V$, 由 $\phi (v_2,v_1,v_3)=\phi (v_1,v_2,v_3)=-\phi (v_2,v_1,v_3)$ 得到 $\phi =0$.
28.	任取 $u,v\in V$, 由条件知 $a(u+v,u+v)-a(u,u)-a(v,v)=0$, 化简即得 $a(u,v)+a(v,u)=0$, 即 $S_2(a)=0$.
29.	取定一组基 $\{{\delta }^i\}$. 根据题目条件, 取 $a=E^{ij}+E^{ji}$ 可得 $b_{ij}+b_{ji}=0$, 从而可得 $b$ 反对称.
30.	令 $g:=f-\tilde{f}$, 则对任意 $x\in V$ 有 $g(x,x)=0$, 从第 28 题可得 $g$ 反对称, 但 $g$ 又对称, 因而 $g=0$, 从而 $f=\tilde{f}$.
31.	仅需依据行列式的逆序数定义对于 $i_1\dots i_r$ 是不是 $j_1\dots j_r$ 的置换分别进行讨论即可, 此不赘述.
32.	(1) $$\begin{array}{rl}\frac{1}{r!}{\delta }_{j_1\dots j_r}^{k_1\dots k_r}{\delta }_{k_1\dots k_r{j}_{r+1}\dots j_{r+s}}^{i_1\dots i_{r+s}}& =\frac{1}{r!}{\delta }_{j_1\dots j_r}^{k_1\dots k_r}{\delta }_{k_1\dots k_r}^{i_1\dots i_r}{\delta }_{j_{r+1}\dots j_{r+s}}^{i_{r+1}\dots i_{r+s}}\\ & =\frac{|\mathcal{S}(r)|}{r!}{\delta }_{j_1\dots j_r}^{i_1\dots i_r}{\delta }_{j_{r+1}\dots j_{r+s}}^{i_{r+1}\dots i_{r+s}}\\ & ={\delta }_{j_1\dots j_{r+s}}^{i_1\dots i_{r+s}}.\end{array}$$ (2) 所求即为 $n$ 个数中任取 $r$ 个的排列, 从而 ${\delta }_{i_1\dots i_r}^{i_1\dots i_r}=P_r^n=\frac{n!}{(n-r)!}$.
33.	对于任意的 $\sigma \in \mathcal{S}$ 有$$\det a=\mathrm{sign}(\sigma )a_{i_1}^{\sigma (i_1)}\cdots a_{i_n}^{\sigma (i_n)},$$因此$$\det a=\frac{1}{n!}\sum _{\sigma \in \mathcal{S}}\mathrm{sign}(\sigma )a_{i_1}^{\sigma (i_1)}\cdots a_{i_n}^{\sigma (i_n)}=\frac{1}{n!}{\delta }_{j_1\cdots j_n}^{i_1\cdots i_n}{a}_{i_1}^{j_1}\cdots a_{i_n}^{j_n}.$$
34.	构造 $(r,r)$ 型张量 $f({\alpha }^1,\cdots ,{\alpha }^r,v_1,\cdots ,v_r):={\sum }_{\sigma \in \mathcal{S}(r)}\mathrm{sign}(\sigma ){\alpha }^1(v_{\sigma (1))})\cdots {\alpha }^r(v_{\sigma (r)})$ 取定 $V$ 的一组基 $\{e_i\}$ 和对偶基 $\{e^i\}$, 则 $f$ 在这组基下的分量$$\begin{array}{rl}{f}_{j_1\cdots j_r}^{i_1\cdots i_r}& =f(e^{i_1},\cdots ,e^{i_r},e_{j_1},\cdots ,e_{j_r})\\ & =\sum _{\sigma \in \mathcal{S}(r)}\mathrm{sign}(\sigma )e^{i_1}(e_{j_{\sigma (1))}})\cdots e^{i_r}(e_{j_{\sigma (r)}})\\ & =\sum _{\sigma \in \mathcal{S}(r)}\mathrm{sign}(\sigma ){\delta }_{j_{\sigma (1)}}^{i_1}\cdots {\delta }_{j_{\sigma (r)}}^{i_r}={\delta }_{j_1\cdots j_r}^{i_1\cdots i_r}.\end{array}$$
35.	不能, 因为该分量不是协变的. 事实上, 任取一个基变换, 容易发现该张量一定与基选取有关.
36.	设 $\{e_i\}$ 是另一组基, 满足 $e_i=a_i^j{\delta }_j$. 设 $({\bar{b}}_i^j)$ 是 $f$ 在基 $\left\{e_i\right\}$ 下的表示矩阵, 则 ${\bar{b}}_i^j=a_i^m{c}_n^j{b}_m^n$. 其中 $(c_i^j)$ 是 $(a_i^j)$ 的逆矩阵. 在基底 $\{e_i\}$ 下计算得$$\begin{array}{rl}{\tilde{B}}_r^{\prime }& ={\delta }_{j_1\cdots j_r}^{i_1\cdots i_r}{\bar{b}}_{i_1}^{j_1}\cdots {\bar{b}}_{i_r}^{j_r}={\delta }_{j_1\cdots j_r}^{i_1\cdots i_r}(a_{i_1}^{m_1}{c}_{n_1}^{j_1}{b}_{m_1}^{n_1})\cdots (a_{i_r}^{m_r}{c}_{n_r}^{j_r}{b}_{m_r}^{n_r})\\ & =\sum _{\sigma \in \mathcal{S}(r)}{\delta }_{i_{\sigma (1)}\cdots i_{\sigma (r)}}^{i_1\cdots i_r}(a_{i_1}^{m_1}{c}_{n_1}^{i_{\sigma (1)}}{b}_{m_1}^{n_1})\cdots (a_{i_r}^{m_r}{c}_{n_r}^{i_{\sigma (r)}}{b}_{m_r}^{n_r})\\ & =\sum _{\sigma \in \mathcal{S}(r)}\mathrm{sign}(\sigma )(a_{i_1}^{m_1}{c}_{n_{{\sigma }^{-1}(1)}}^{i_1})\cdots (a_{i_r}^{m_r}{c}_{n_{{\sigma }^{-1}(r)}}^{i_r})b_{m_1}^{n_1}\cdots b_{m_r}^{n_r}\\ & =\sum _{\sigma \in \mathcal{S}(r)}\mathrm{sign}(\sigma ){\delta }_{n_{{\sigma }^{-1}(1)}}^{m_1}\cdots {\delta }_{n_{{\sigma }^{-1}(r)}}^{m_r}{b}_{m_1}^{n_1}\cdots b_{m_r}^{n_r}\\ & =\det ({\delta }_{n_j}^{m_i})b_{m_1}^{n_1}\cdots b_{m_r}^{n_r}={\delta }_{n_1\cdots n_r}^{m_1\cdots m_r}{b}_{m_1}^{n_1}\cdots b_{m_r}^{n_r}={\tilde{B}}_r.\end{array}$$
37.	见 P40,41.
38.	设 $\left\{e_i\right\},\{e^i\}$ 是另一组基和对偶基. 设 $e_i=a_i^j{\delta }_j$, 则 $e^i=c_i^j{\delta }^j$, 其中 $(c_i^j)$ 是 $(a_i^j)$ 的逆矩阵. 那么$$\begin{array}{rl}& \sum _{1\le i_1<\cdots <i_r\le n}\left(e_{i_1}\wedge \cdots \wedge e_{i_r}\right)\otimes \left(e^{i_1}\wedge \cdots \wedge e^{i_r}\right)\\ & =\frac{1}{r!}\left(e_{i_1}\wedge \cdots \wedge e_{i_r}\right)\otimes \left(e^{i_1}\wedge \cdots \wedge e^{i_r}\right)\\ & =\frac{1}{r!}{a}_{i_1}^{j_1}\cdots a_{i_r}^{j_r}{c}_{k_1}^{i_1}\cdots c_{k_r}^{i_r}\left({\delta }_{j_1}\wedge \cdots \wedge {\delta }_{j_r}\right)\otimes \left({\delta }^{k_1}\wedge \cdots \wedge {\delta }^{k_r}\right)\\ & =\frac{1}{r!}{\delta }_{k_1}^{j_1}\cdots {\delta }_{k_r}^{j_r}\left({\delta }_{j_1}\wedge \cdots \wedge {\delta }_{j_r}\right)\otimes \left({\delta }^{k_1}\wedge \cdots \wedge {\delta }^{k_r}\right)\\ & =\frac{1}{r!}\left({\delta }_{j_1}\wedge \cdots \wedge {\delta }_{j_r}\right)\otimes \left({\delta }^{j_1}\wedge \cdots \wedge {\delta }^{j_r}\right)\\ & =\sum _{1\le j_1<\cdots <j_r\le n}\left({\delta }_{j_1}\wedge \cdots \wedge {\delta }_{j_r}\right)\otimes \left({\delta }^{j_1}\wedge \cdots \wedge {\delta }^{j_r}\right).\end{array}$$
39.	这是第 43 题的特殊情况.
41.	由第 37 题, ${\delta }^{i_1}\wedge \cdots {\delta }^{i_r}(u_1,\cdots ,u_r)=\det (u_l^{i_k})$. 设 ${\alpha }^k={\alpha }_i^k{\delta }^i$, 则$$\begin{array}{rl}{\alpha }^1\wedge \cdots {\alpha }^r(u_1,\cdots ,u_r)& =\sum _{i_1,\cdots ,i_r}{\alpha }_{i_1}^1\cdots {\alpha }_{i_r}^r{\delta }^{i_1}\wedge {\delta }^{i_r}(u_1,\cdots ,u_r)\\ & =\sum _{i_1,\cdots ,i_r}{\alpha }_{i_1}^1\cdots {\alpha }_{i_r}^r\det (u_l^{i_k})\\ & =\sum _{i_1,\cdots ,i_r}\det ({\alpha }_{i_k}^k{u}_l^{i_k})\\ & =\det ((\sum _{i=1}^n{\alpha }_i^k{\delta }^i)u_l)\\ & =\det ({\alpha }^k(u_l)).\end{array}$$
42.	$$\begin{array}{rl}{x}_1\wedge \cdots \wedge x_r& =\sum _{i_1,\cdots ,i_r}{a}_1^{i_1}\cdots a_r^{i_r}{e}_{i_1}\wedge \cdots \wedge e_{i_r}\\ & =\sum _{i_1<\cdots <i_r}\sum _{\sigma }{a}_1^{i_{\sigma (1)}}\cdots a_r^{i_{\sigma (r)}}{e}_{i_{\sigma (1)}}\wedge \cdots \wedge e_{i_{\sigma (r)}}\\ & =\sum _{i_1<\cdots <i_r}\sum _{\sigma }\mathrm{sign}(\sigma )a_1^{i_{\sigma (1)}}\cdots a_r^{i_{\sigma (r)}}{e}_{i_1}\wedge \cdots \wedge e_{i_r}\\ & =\sum _{i_1<\cdots <i_r}\det (a_l^{i_k})e_{i_1}\wedge \cdots \wedge e_{i_r}.\end{array}$$
43.	$$\begin{array}{rl}\phi \wedge \psi (v_1,\cdots ,v_{r+s})& =\frac{1}{r!s!}\sum _{\sigma }\mathrm{sign}(\sigma )\phi \otimes \psi (v_{\sigma (1)},\cdots ,v_{\sigma (r+s)})\\ & =\frac{1}{r!s!}\sum _{i_1,\cdots ,i_r}\phi (v_{i_1},\cdots ,v_{i_r})\sum _{\sigma (k)=i_k}\mathrm{sign}(\sigma )\psi (v_{\sigma (r+1)},\cdots ,v_{\sigma (r+s)})\\ & =\sum _{\begin{array}{c}{i}_1<\cdots <i_r\\ i_{r+1}<\cdots <i_{r+s}\end{array}}\mathrm{sign}({\sigma }_{i_1,\cdots ,i_r})\phi (v_{i_1},\cdots ,v_{i_r})\psi (v_{i_{r+1}},\cdots ,v_{i_{r+s}}),\end{array}$$其中 ${\sigma }_{i_1,\cdots ,i_r}(1,2,\cdots ,r+s)=(i_1,\cdots ,i_r,i_{r+1},\cdots ,i_{r+s})$, 故 $\mathrm{sign}({\sigma }_{i_1,\cdots ,i_r})={\delta }_{1,\cdots ,r+s}^{i_1,\cdots ,i_{r+s}}$, 代入即证.
44.	(1) 易证. (2) $$\begin{array}{rl}i(u_1)(\phi \wedge \psi )& (u_2,\cdots ,u_{r+s})=\sum _{\begin{array}{c}{i}_1<\cdots <i_r\\ i_{r+1}<\cdots <i_{r+s}\end{array}}{\delta }_{1,\cdots ,r+s}^{i_1,\cdots ,i_{r+s}}\phi (u_{i_1},\cdots ,u_{i_r})\psi (u_{i_{r+1}},\cdots ,u_{i_{r+s}})\\ & =\sum _{\begin{array}{c}{i}_1<\cdots <i_r\\ i_{r+1}<\cdots <i_{r+s}\end{array}}[\sum _{1\in \{i_1,\cdots ,i_r\}}{\delta }_{1,2,\cdots ,r+s}^{1,i_2,\cdots ,i_{r+s}}\phi (u_1,u_{i_2},\cdots ,u_{i_r})\psi (u_{i_{r+1}},\cdots ,u_{i_{r+s}})\\ & +\sum _{1\notin \{i_1,\cdots ,i_r\}}{\delta }_{1,\cdots ,r,r+1,r+2,\cdots ,r+s}^{i_1,\cdots ,i_r,1,i_{r+2},\cdots ,i_{r+s}}\phi (u_{i_1},\cdots ,u_{i_r})\psi (u_1,u_{i_{r+2}},\cdots ,u_{i_{r+s}})]\\ & =\sum _{\begin{array}{c}{i}_2<\cdots <i_r\\ i_{r+1}<\cdots <i_{r+s}\end{array}}{\delta }_{2,\cdots ,r+s}^{i_2,\cdots ,i_{r+s}}(i(u_1)\phi )(u_{i_2},\cdots ,u_{i_r})\psi (u_{i_{r+1}},\cdots ,u_{i_{r+s}})\\ & +(-1{)}^r\sum _{\begin{array}{c}{i}_1<\cdots <i_r\\ i_{r+2}<\cdots <i_{r+s}\end{array}}{\delta }_{2,\cdots ,r+1,r+2,\cdots ,r+s}^{i_1,\cdots ,i_r,i_{r+2},\cdots ,i_{r+s}}\phi (u_{i_1},\cdots ,u_{i_r})(i(u_1)\psi )(u_{i_{r+2}},\cdots ,u_{i_{r+s}})\\ & =((i(u_1)\phi )\wedge \psi +(-1{)}^r\phi \wedge (i(u_1)\psi ))(u_2,\cdots ,u_{r+s}).\end{array}$$
45.	(1) 易证. (2) $$\begin{array}{rl}e(\alpha )(\phi \wedge \psi )& =\alpha \wedge (\phi \wedge \psi )\\ & =(\alpha \wedge \phi )\wedge \psi \\ & =(e(\alpha )\phi )\wedge \psi ,\end{array}$$ $$\begin{array}{rl}(\alpha \wedge \phi \wedge \psi )(u_1,\cdots ,u_{r+s+1})s& =(\phi \wedge \alpha \wedge \psi )(u_2,\cdots ,u_{r+1},u_1,u_{r+2},\cdots ,u_{r+s+1})\\ & =(-1{)}^r(\phi \wedge \alpha \wedge \psi )(u_1,\cdots ,u_{r+s+1}),\end{array}$$故 $e(\alpha )(\phi \wedge \psi )=(-1{)}^r\phi \wedge (\alpha \wedge \psi )=(-1{)}^r\phi \wedge (e(\alpha )(\psi )).$
46.	由第 44、45 题, $$\begin{array}{rl}i(u)\circ e(\alpha )\phi +e(\alpha )\circ i(u)\phi & =(i(u)\alpha )\wedge \phi -\alpha \wedge (i(u)\phi )+\alpha \wedge (i(u)\phi )\\ & =\alpha (u)\phi .\end{array}$$即证.
47.	(1)(2) 我们直接证明 (2) 的 $n\ge 1$ 的情形. $n=1$ 显然成立. 对 $n$ 归纳: 假设 $n$ 已成立, 对于 $\dim V=n+1$ 中的 $n$ 次外形式$$\Omega =\sum _{i=1}^{n+1}{a}_i{\delta }^1\wedge \cdots \wedge \widehat{{\delta }^i}\wedge \cdots \wedge {\delta }^{n+1},$$$a_i$ 不全为 $0$, $\{{\delta }_i\}$ 为 $V$ 的一组基. 考虑 $\psi =\sum b_i{\delta }^i$, 则 $\psi \wedge \Omega =(\sum a_i{b}_i(-1{)}^{i+1}){\delta }^1\wedge \cdots \wedge {\delta }^{n+1},$ 易见存在 $\{b_i\}$ 使得 $\sum a_i{b}_i(-1{)}^{i+1}=0$. 即存在 $\psi$ 使得 $\psi \wedge \Omega =0$. 由定理 5.8 (P43) , 存在 $\phi$ 使得 $\Omega =\psi \wedge \phi$. 注意到存在 $W,\dim W=n$ 使得 $V^{\ast }=\mathbf{R}\psi \oplus W^{\ast }$, 考虑 $\phi$ 在 $W^{\ast }$ 的投影 $\tilde{\phi }$, 可见 $\Omega =\psi \wedge \tilde{\phi }$. 由归纳假设, $\tilde{\phi }$ 可分解, 于是 $\Omega$ 可分解. (3) 当 $\dim V=4$ 时 ${\delta }^1\wedge {\delta }^2+{\delta }^3\wedge {\delta }^4$ 不可分解, 否则存在 $\sum _{i=1}^4{a}_i{\delta }^i\ne 0$ 使得 $\left(\sum _{i=1}^4{a}_i{\delta }^i\right)\wedge ({\delta }^1\wedge {\delta }^2+{\delta }^3\wedge {\delta }^4)=0$, 展开得 $a_1=a_2=a_3=a_4=0$, 矛盾.
48.	设 $V^{\ast }$ 的基底为 $\{{\delta }^i\}$, $2$ 次外形式 $\alpha$ 可表为 $$\alpha =\sum _{i<j}{\alpha }_{ij}{\delta }^i\wedge {\delta }^j.$$ 假定 ${\alpha }_{ij}$ 不全为 $0$ , 不妨设 ${\alpha }_{12}\ne 0$ , 于是 $$\begin{array}{rl}\alpha & =({\alpha }_{12}{\delta }^1+{\alpha }_{32}{\delta }^3+\cdots +{\alpha }_{n2}{\delta }^n)\wedge ({\delta }^2+\frac{{\alpha }_{13}}{{\alpha }_{12}}{\delta }^3+\cdots +\frac{{\alpha }_{1n}}{{\alpha }_{12}}{\delta }^n)+\sum _{3\le i<j\le n}{\alpha }_{ij}^{\prime }{\delta }^i\wedge {\delta }^j\\ & ={\sigma }^1\wedge {\sigma }^2+\sum _{3\le i<j\le n}{\alpha }_{ij}^{\prime }{\delta }^i\wedge {\delta }^j,\end{array}$$其中 ${\sigma }^1={\alpha }_{12}{\delta }^1+{\alpha }_{32}{\delta }^3+\cdots +{\alpha }_{n2}{\delta }^n,{\sigma }^2={\delta }^2+\frac{{\alpha }_{13}}{{\alpha }_{12}}{\delta }^3+\cdots +\frac{{\alpha }_{1n}}{{\alpha }_{12}}{\delta }^n$. 若 ${\alpha }_{ij}^{\prime }$ 不全为零, 不妨设 ${\alpha }_{34}^{\prime }\ne 0$, 继续上面的过程. 因 $n$ 是有限数, 故上述步骤必在有限步之后终止, 即 $\alpha$ 可表成$$\alpha ={\sigma }^1\wedge {\sigma }^2+\cdots +{\sigma }^{2r-1}\wedge {\sigma }^{2r},$$易见 ${\sigma }^1,\cdots ,{\sigma }^{2r}$ 是线性无关的 $1$ 次形式. 由 ${\sigma }^{2i-1}\wedge {\sigma }^{2i}\wedge {\sigma }^{2j-1}\wedge {\sigma }^{2j}={\sigma }^{2j-1}\wedge {\sigma }^{2j}\wedge {\sigma }^{2i-1}\wedge {\sigma }^{2i}$ 可得 ${\alpha }^r=r!{\sigma }^1\wedge {\sigma }^2\wedge \cdots \wedge {\sigma }^{2r-1}\wedge {\sigma }^{2r}\ne 0$, 由 ${\alpha }^{r+1}$ 为高于 $2r$ 次的外形式得 ${\alpha }^{r+1}=0$, 所以 $2r$ 只与 $\alpha$ 有关. Remark: 可以直接通过反对称二次型的合同标准型说明.
49.	(1) 对称性和双线性是显然的. 由于 $(⟨v_{\alpha },v_{\beta }⟩{)}_{\alpha ,\beta }$ 为内积的 Gram 矩阵, 于是正定性成立. (2)$⟨{\delta }_{i_1}\wedge \cdots \wedge {\delta }_{i_p},{\delta }_{j_1}\wedge \cdots \wedge {\delta }_{j_p}⟩=\det ({\delta }_{i_{\alpha }}^{j_{\beta }})={\delta }_{j_1\cdots j_p}^{i_1\cdots i_p}.$
50.	(1) 注意到映射 ${\wedge }^{p}A:\prod _{i=1}^{p}V\to W$ 为交错多线性映射, 由 ${\wedge }^{p}V$ 的泛性质, ${\wedge }^{p}A:\prod _{i=1}^{p}V\to W$ 诱导了 ${\wedge }^{p}A:{\wedge }^{p}V\to W$. (2)${\wedge }^{p}A({\delta }_{i_1}\wedge \cdots \wedge {\delta }_{i_p})={\delta }_{{\alpha }_1\cdots {\alpha }_p}^{i_1\cdots i_p}{a}_{i_1}^{{\alpha }_1}\cdots a_{i_p}^{{\alpha }_p}.$ (3)${\wedge }^p(AB)(z_1,\cdots ,z_p)=(AB{z}_1)\wedge \cdots \wedge (AB{z}_p)={\wedge }^{p}A(B{z}_1,\cdots ,B{z}_p)={\wedge }^{p}A\circ {\wedge }^{p}B(z_1,\cdots ,z_p).$