Machine-translated from Chinese.
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
tensor representation
For a general $r$-heavy linear mapping space, we have
$$\mathcal{L}(V_1,\cdots,V_r;Z)=V_1^\ast \otimes \cdots\otimes V_r^\ast \otimes Z.$$
Definition $(r+1)$-heavy linear mapping
$$\begin{aligned}
f:V_1^*\times\cdots\times V_r^*\times Z&\rightarrow \mathcal{L}(V_1,\cdots,V_r;Z),\\ (v_1^*,\cdots,v_r^*,z)&\mapsto v_1^*\otimes \cdots\otimes v_r^*\cdot z,
\end{aligned}$$
That is $f(v_1^\ast ,\cdots,v_r^\ast ,z)(v_1,\cdots,v_r)=v_1^\ast (v_1)\cdots v_r^\ast (v_r) z=v_1^\ast \otimes \cdots \otimes v_r^\ast (v_1,\cdots,v_r) \cdot z.$
Where $\cdot$ is scalar multiplication, determined by the vector space $Z$.
Just use the universal mapping property to show that $g:V_1^\ast \otimes \cdots\otimes V_r^\ast \otimes Z\rightarrow \mathcal{L}(V_1,\cdots,V_r;Z)$ is isomorphism.
First prove
$$\dim V_1^\ast \otimes\cdots \otimes V_r^\ast \otimes Z=\dim V_1^\ast \cdots\dim V_r^\ast \dim Z=\dim\mathcal{L}(V_1,\cdots,V_r;Z).$$
Take the basis $\{\omega_{(1)}^{i_1}\},\cdots,\{\omega_{(r)}^{i_r}\},$ of $V_1^\ast ,\cdots,V_r^\ast ,Z$
$\{\eta_\alpha\}.$ Then any $F\in \mathcal{L}(V_1,\cdots,V_r;Z),$
Can be decomposed into $F^\alpha\eta_\alpha,$
$F^\alpha\in \mathcal{L}(V_1,\cdots,V_r;\mathbb{R})=V_1^\ast \otimes \cdots\otimes V_r^\ast .$
Therefore each $F^\alpha$ can be decomposed into $a^\alpha_{i_1\cdots i_n}\omega_{(1)}^{i_1}\otimes\cdots\otimes \omega_{(n)}^{i_n}.$
That is
$$F=a^\alpha_{i_1\cdots i_n}\omega_{(1)}^{i_1}\otimes\cdots\otimes \omega_{(n)}^{i_n}\cdot\eta_\alpha.$$
By acting on the dual basis $\{e^{(1)}_{i_1}\},\cdots,\{e^{(n)}_{i_n}\},$
From the linear independence of $\{\eta_\alpha\}$,
It can explain the linear independence of $\{\omega_{(1)}^{i_1}\otimes\cdots\otimes \omega_{(n)}^{i_n}\cdot\eta_\alpha\}$,
Therefore it is a set of basis.
noticed,
$$f(\omega^{i_1}_{(1)},\cdots,\omega^{i_r}_{(r)},\eta_\alpha)=\omega_{(1)}^{i_1}\otimes\cdots\otimes \omega_{(n)}^{i_n}\cdot\eta_\alpha=g(\omega^{i_1}_{(1)}\otimes\cdots\otimes\omega^{i_r}_{(r)}\otimes\eta_\alpha),$$
$g$ maps basis to basis, thus it is a linear isomorphism.
The article was last updated on 2021-11-24 22:08:11
