《微分流形》讨论稿-线性映照空间

张量表示

对于一般的$r$-重线性映照空间, 我们有

$$\mathcal{L}(V_1,\cdots,V_r;Z)=V_1^\ast \otimes \cdots\otimes V_r^\ast \otimes Z.$$

定义$(r+1)$-重线性映照

$$\begin{aligned} f:V_1^*\times\cdots\times V_r^*\times Z&\rightarrow \mathcal{L}(V_1,\cdots,V_r;Z),\\ (v_1^*,\cdots,v_r^*,z)&\mapsto v_1^*\otimes \cdots\otimes v_r^*\cdot z, \end{aligned}$$

即$f(v_1^\ast ,\cdots,v_r^\ast ,z)(v_1,\cdots,v_r)=v_1^\ast (v_1)\cdots v_r^\ast (v_r) z=v_1^\ast \otimes \cdots \otimes v_r^\ast (v_1,\cdots,v_r) \cdot z.$ 其中$\cdot$为标量乘法, 由向量空间$Z$决定.

只需利用万有映照性质说明$g:V_1^\ast \otimes \cdots\otimes V_r^\ast \otimes Z\rightarrow \mathcal{L}(V_1,\cdots,V_r;Z)$为同构. 首先证明

$$\dim V_1^\ast \otimes\cdots \otimes V_r^\ast \otimes Z=\dim V_1^\ast \cdots\dim V_r^\ast \dim Z=\dim\mathcal{L}(V_1,\cdots,V_r;Z).$$

取$V_1^\ast ,\cdots,V_r^\ast ,Z$的基$\{\omega_{(1)}^{i_1}\},\cdots,\{\omega_{(r)}^{i_r}\},$ $\{\eta_\alpha\}.$ 那么任意$F\in \mathcal{L}(V_1,\cdots,V_r;Z),$ 可分解为$F^\alpha\eta_\alpha,$ $F^\alpha\in \mathcal{L}(V_1,\cdots,V_r;\mathbb{R})=V_1^\ast \otimes \cdots\otimes V_r^\ast .$ 因此每个$F^\alpha$又可分解为$a^\alpha_{i_1\cdots i_n}\omega_{(1)}^{i_1}\otimes\cdots\otimes \omega_{(n)}^{i_n}.$ 即

$$F=a^\alpha_{i_1\cdots i_n}\omega_{(1)}^{i_1}\otimes\cdots\otimes \omega_{(n)}^{i_n}\cdot\eta_\alpha.$$

通过作用对偶基$\{e^{(1)}_{i_1}\},\cdots,\{e^{(n)}_{i_n}\},$ 由$\{\eta_\alpha\}$的线性无关性, 可以说明$\{\omega_{(1)}^{i_1}\otimes\cdots\otimes \omega_{(n)}^{i_n}\cdot\eta_\alpha\}$的线性无关性, 从而其为一组基.

注意到,

$$f(\omega^{i_1}_{(1)},\cdots,\omega^{i_r}_{(r)},\eta_\alpha)=\omega_{(1)}^{i_1}\otimes\cdots\otimes \omega_{(n)}^{i_n}\cdot\eta_\alpha=g(\omega^{i_1}_{(1)}\otimes\cdots\otimes\omega^{i_r}_{(r)}\otimes\eta_\alpha),$$

$g$将基映到基, 从而是线性同构.

文章最后更新于 2021-11-24 22:08:11