Machine-translated from Chinese.
万有映照性质
设$V,W,U$为向量空间, $\psi:V\times W\rightarrow U$为双线性映照.
若$\,\forall\,$向量空间$Z,$ 双线性映照$f:V\times W\rightarrow Z,$
都存在唯一线性映照$g:U\rightarrow Z,$ 使得$f=g\circ\psi,$ 则称
$(U,\psi)$具有万有映照性质.
引理 1. $(U,\psi)$在下述意义下是唯一的: 若$(\widetilde{U},\widetilde{\psi})$具有万有映照性质, 则$\,\exists\,g:U\cong \widetilde{U}$为线性同构, 且$\widetilde{\psi}=g\circ\psi.$
证: 只需说明$\,\exists\,g,\widetilde{g},$ 使得
$$\widetilde\psi=g\circ\psi=(g\circ \widetilde{g})\circ\widetilde\psi=\mathrm{id}\circ\widetilde\psi;$$
$$\psi=\widetilde{g}\circ\widetilde\psi=(\widetilde g\circ g)\circ\psi=\mathrm{id}\circ\psi.$$
利用万有映照性质中的唯一性即可说明$g\circ\widetilde{g}=\mathrm{id}_{\widetilde U},$
$\widetilde{g}\circ g=\mathrm{id}_{U}.$
命题 2. 设$V,W$为向量空间, 则$(V\otimes W,h=\otimes)$具有万有映照性质.
证: 对任意双线性型$f:V\times W\rightarrow Z,$
$f$由基上取值$f(e_i,\sigma_\alpha)$决定. 只需说明使得
$$f(e_i,\sigma_\alpha)=g\circ h(e_i,\sigma_\alpha)=g(e_i\otimes \sigma_\alpha)\tag*{(\ast )}$$
的线性映照$g$存在唯一.
注意到$e_i\otimes \sigma_\alpha$为$V\otimes W$的基,
因此$(\ast )$式决定了$g$, 这说明了唯一性.
至于存在性, 依$(\ast )$式定义$g$即可, 因为在基上满足条件了,
映照又都是线性的, 自然在所有元素上满足$f=g\circ h.$
注 3. 由唯一性, $(\widetilde{U},\widetilde{\psi})\cong (V\otimes W,h),$ 因此总可以用张量积来表示万有映照性质.
推论 4. $\mathcal{L}(V,W;Z)\cong \mathcal{L}(V\otimes W;Z),$ 由性质中的$f\mapsto g$联系. 特别地, $\mathcal{L}(V,W;\mathbb{R})=V^\ast \otimes W^\ast \cong (V\otimes W)^\ast .$
张量积
推广地, 设$\varphi\in \mathcal{L}(V_1,\cdots,V_r;\mathbb{R}),$
$\psi\in \mathcal{L}(W_1,\cdots,W_s;\mathbb{R}),$ 定义
$$(\varphi\otimes \psi)(v_1,\cdots,v_r,w_1,\cdots,w_s):=\varphi(v_1,\cdots,v_r)\cdot\psi(w_1,\cdots,w_s).$$
那么
$$h=\otimes:\mathcal{L}(V_1,\cdots,V_r;\mathbb{R})\times \mathcal{L}(W_1,\cdots,W_s;\mathbb{R})\rightarrow \mathcal{L}(V_1,\cdots,V_r,W_1,\cdots,W_s;\mathbb{R}),$$
满足双线性性, 即张量积满足分配律. 容易验证它也满足结合律. 但是一般的,
张量积不满足交换律.
进一步, 可以对$\varphi_i\in V_i^\ast =\mathcal{L}(V_i;\mathbb{R}),$
定义张量积$\varphi_1\otimes \cdots\otimes \varphi_r\in \mathcal{L}(V_1,\cdots,V_r;\mathbb{R}).$
具体的,
$$(\varphi_1\otimes\cdots\otimes \varphi_r)(v_1,\cdots,v_r)=\varphi_1(v_1)\cdots\varphi_r(v_r),$$
即我们得到了
$$\begin{aligned}
\otimes^r:V_1^*\times\cdots\times V_r^*&\rightarrow\mathcal{L}(V_1,\cdots,V_r;\mathbb{R}),\\ (\varphi_1,\cdots,\varphi_r)&\mapsto \varphi_1\otimes\cdots\otimes \varphi_r,
\end{aligned}$$
为一个$r$-重线性函数, $\otimes$满足分配律.
且设$\omega_{(i)}^{j_{i} }$是$V_i^\ast $的基,
那么$\{\omega_{(1)}^{j_1}\otimes\cdots\otimes \omega_{(r)}^{j_r}\}$构成$V_1^\ast \otimes \cdots\otimes V_r^\ast =\operatorname{span}\{\operatorname{Im}\otimes^r\}$的基.
因此$\dim V_1^\ast \otimes \cdots\otimes V_r^\ast =\dim V_1\cdots\dim V_r.$
归纳地, 同理可证
$$V_1^\ast \otimes \cdots\otimes V_r^\ast =\mathcal{L}(V_1,\cdots,V_r;\mathbb{R}).$$
由互为对偶性, 也可定义$v_1\otimes\cdots\otimes v_r,$ $v_i\in V_i,$
以及$V_1\otimes\cdots\otimes V_r=\mathcal{L}(V_1^\ast ,\cdots,V_r^\ast ;\mathbb{R}).$
若$\{e^{(i)}_{j_i}\}$为$V_i$的基,
则$\{e^{(1)}_{j_1}\otimes\cdots\otimes e^{(r)}_{j_r}\}$为$V_1\otimes\cdots\otimes V_r$的基,
$\dim V_1\otimes \cdots\otimes V_r=\dim V_1\cdots \dim V_r.$
对于万有映照性质, 也有多重形式的推广: 即设$V_1,\cdots,V_r,U$为向量空间,
$h:V_1\times \cdots \times V_r\rightarrow U$为$r$-重线性映照.
若$\,\forall\,r$-重线性映照$f:V_1\times \cdots\times V_r\rightarrow Z,$
存在唯一$g:U\rightarrow Z,$ 使得$f=g\circ h.$
则称$(U,h)$具有万有映照性质.
类似地, 这样的$(U,h)$在线性同构意义下是唯一的,
且$(V_1\otimes \cdots\otimes V_r,h)$具有万有映照性质.
由此推出$(V_1\otimes \cdots\otimes V_r)^\ast \cong V_1^\ast \otimes \cdots\otimes V_r^\ast .$
而对于一般的$r$-重线性映照, 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
$$\begin{aligned}
f:V_1^*\times\cdots\times V_r^*\times Z&\rightarrow \mathcal{L}(V_1,\cdots,V_r;Z),\\ (v_1^*,\cdots,v_r^*,z)&\mapsto v_1^*\otimes \cdots\otimes v_r^*\cdot z,
\end{aligned}$$
利用万有映照性质说明$g$为同构即可.
张量
设$V$是$\mathbb{R}$上$n$维向量空间, 令
$$V^r_s=(V\otimes \cdots\otimes V)^r\otimes (V^\ast \otimes\cdots\otimes V^\ast )_s=\mathcal{L}(V^\ast ,\cdots,V^\ast ,V,\cdots,V;\mathbb{R}),$$
称为$(r,s)$-型张量空间, 其中元素称为$(r,s)$型张量. 如$V_0^1=V,$
$V_1^0=V^\ast ,$ $V_0^0=\mathbb{R}.$
设$\{e_i\}$为$V$的基, $\{\omega^i\}$为对偶基,
则$\{e_{i_1}\otimes\cdots\otimes e_{i_r}\otimes \omega^{j_1}\otimes\cdots\otimes \omega^{j_s}\}$构成$V^r_s$的基,
$\dim V_s^r=n^{r+s}.$
$\,\forall\,\Phi\in V^r_s,$
$\Phi$的分量$\Phi^{k_1\cdots k_r}_{l_1\cdots l_s}=\Phi(\omega^{k_1},\cdots,\omega^{k_2},e_{l_1},\cdots,e_{l_s}).$
若另取基$\{\widetilde{e}_i\}$与对偶基$\{\widetilde{\omega}^i\},$
那么有$\widetilde{e}_i= a_i^je_j,$ $\widetilde{\omega}^i=b_j^i\omega^j,$
其中$a_i^jb_j^k=\delta_i^k.$ 那么变换后, 分量上的变化为:
$$\widetilde{\Phi}^{i_1\cdots i_r}_{j_1\cdots j_s}=b_{k_1}^{i_1}\cdots b_{k_r}^{i_r} a_{j_1}^{l_1}\cdots a_{j_s}^{l_s} \Phi_{l_1\cdots l_s}^{k_1\cdots k_r}.$$
称它是$r$-阶反变, $s$-阶共变的.
文章最后更新于 2021-11-24 16:14:25
Universal Mapping Property
Let $V,W,U$ be a vector space and $\psi:V\times W\rightarrow U$ be a bilinear mapping.
If $\,\forall\,$ vector space $Z,$ bilinear mapping $f:V\times W\rightarrow Z,$
There exists a unique linear mapping $g:U\rightarrow Z,$ such that $f=g\circ\psi,$ is said to
$(U,\psi)$ has Universal Mapping Property.
Lemma 1. $(U,\psi)$ is unique in the following sense: if $(\widetilde{U},\widetilde{\psi})$ has universal mapping properties, then $\,\exists\,g:U\cong \widetilde{U}$ is a linear isomorphism, and $\widetilde{\psi}=g\circ\psi.$
Certificate: Just state that $\,\exists\,g,\widetilde{g},$ makes
$$\widetilde\psi=g\circ\psi=(g\circ \widetilde{g})\circ\widetilde\psi=\mathrm{id}\circ\widetilde\psi;$$
$$\psi=\widetilde{g}\circ\widetilde\psi=(\widetilde g\circ g)\circ\psi=\mathrm{id}\circ\psi.$$
It can be explained by using the uniqueness of the universal mapping property$g\circ\widetilde{g}=\mathrm{id}_{\widetilde U},$
$\widetilde{g}\circ g=\mathrm{id}_{U}.$
Proposition 2. Assuming $V,W$ is a vector space, then $(V\otimes W,h=\otimes)$ has universal mapping properties.
Certificate: For any bilinear type $f:V\times W\rightarrow Z,$
$f$ is determined by the value $f(e_i,\sigma_\alpha)$ on the basis. Just state that such
$$f(e_i,\sigma_\alpha)=g\circ h(e_i,\sigma_\alpha)=g(e_i\otimes \sigma_\alpha)\tag*{(\ast )}$$
The linear mapping $g$ of exists uniquely.
Note that $e_i\otimes \sigma_\alpha$ is the basis of $V\otimes W$,
Therefore, formula $(\ast )$ determines $g$, which illustrates the uniqueness.
As for existence, just define $g$ according to the formula $(\ast )$, because the conditions are satisfied on the basis,
The mapping is linear, so it naturally satisfies $f=g\circ h.$ on all elements.
Note 3. Due to uniqueness, $(\widetilde{U},\widetilde{\psi})\cong (V\otimes W,h),$ can always use tensor product to express the universal mapping property.
Corollary 4. $\mathcal{L}(V,W;Z)\cong \mathcal{L}(V\otimes W;Z),$ is related by $f\mapsto g$ in the property. In particular, $\mathcal{L}(V,W;\mathbb{R})=V^\ast \otimes W^\ast \cong (V\otimes W)^\ast .$
tensor product
Promotion area, let $\varphi\in \mathcal{L}(V_1,\cdots,V_r;\mathbb{R}),$
$\psi\in \mathcal{L}(W_1,\cdots,W_s;\mathbb{R}),$ Definition
$$(\varphi\otimes \psi)(v_1,\cdots,v_r,w_1,\cdots,w_s):=\varphi(v_1,\cdots,v_r)\cdot\psi(w_1,\cdots,w_s).$$
Then
$$h=\otimes:\mathcal{L}(V_1,\cdots,V_r;\mathbb{R})\times \mathcal{L}(W_1,\cdots,W_s;\mathbb{R})\rightarrow \mathcal{L}(V_1,\cdots,V_r,W_1,\cdots,W_s;\mathbb{R}),$$
It satisfies bilinearity, that is, the tensor product satisfies the distributive law. It is easy to verify that it also satisfies the associative law. But generally,
The tensor product does not satisfy the commutative law.
Further, $\varphi_i\in V_i^\ast =\mathcal{L}(V_i;\mathbb{R}),$ can be
Define tensor product $\varphi_1\otimes \cdots\otimes \varphi_r\in \mathcal{L}(V_1,\cdots,V_r;\mathbb{R}).$
specific,
$$(\varphi_1\otimes\cdots\otimes \varphi_r)(v_1,\cdots,v_r)=\varphi_1(v_1)\cdots\varphi_r(v_r),$$
i.e. we get
$$\begin{aligned}
\otimes^r:V_1^*\times\cdots\times V_r^*&\rightarrow\mathcal{L}(V_1,\cdots,V_r;\mathbb{R}),\\ (\varphi_1,\cdots,\varphi_r)&\mapsto \varphi_1\otimes\cdots\otimes \varphi_r,
\end{aligned}$$
is a $r$-heavy linear function, $\otimes$ satisfies the distributive law.
And suppose $\omega_{(i)}^{j_{i} }$ is the basis of $V_i^\ast $,
Then $\{\omega_{(1)}^{j_1}\otimes\cdots\otimes \omega_{(r)}^{j_r}\}$ forms the basis of $V_1^\ast \otimes \cdots\otimes V_r^\ast =\operatorname{span}\{\operatorname{Im}\otimes^r\}$.
Therefore $\dim V_1^\ast \otimes \cdots\otimes V_r^\ast =\dim V_1\cdots\dim V_r.$
Inductively, the same argument can be proved
$$V_1^\ast \otimes \cdots\otimes V_r^\ast =\mathcal{L}(V_1,\cdots,V_r;\mathbb{R}).$$
Due to mutual duality, $v_1\otimes\cdots\otimes v_r,$ $v_i\in V_i,$ can also be defined
and $V_1\otimes\cdots\otimes V_r=\mathcal{L}(V_1^\ast ,\cdots,V_r^\ast ;\mathbb{R}).$
If $\{e^{(i)}_{j_i}\}$ is the basis of $V_i$,
Then $\{e^{(1)}_{j_1}\otimes\cdots\otimes e^{(r)}_{j_r}\}$ is the basis of $V_1\otimes\cdots\otimes V_r$,
$\dim V_1\otimes \cdots\otimes V_r=\dim V_1\cdots \dim V_r.$
For the universal mapping property, there are also multiple forms of generalization: that is, assuming $V_1,\cdots,V_r,U$ is a vector space,
$h:V_1\times \cdots \times V_r\rightarrow U$ is a $r$-heavy linear mapping.
If$\,\forall\,r$-heavy linear mapping$f:V_1\times \cdots\times V_r\rightarrow Z,$
There is a unique $g:U\rightarrow Z,$ such that $f=g\circ h.$
Then it is said that $(U,h)$ has Universal Mapping Property.
Similarly, such $(U,h)$ is unique in the sense of linear isomorphism,
And $(V_1\otimes \cdots\otimes V_r,h)$ has the property of reflecting everything.
From this it follows $(V_1\otimes \cdots\otimes V_r)^\ast \cong V_1^\ast \otimes \cdots\otimes V_r^\ast .$
For the general $r$-heavy linear mapping, 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
$$\begin{aligned}
f:V_1^*\times\cdots\times V_r^*\times Z&\rightarrow \mathcal{L}(V_1,\cdots,V_r;Z),\\ (v_1^*,\cdots,v_r^*,z)&\mapsto v_1^*\otimes \cdots\otimes v_r^*\cdot z,
\end{aligned}$$
Just use the universal mapping property to show that $g$ is isomorphic.
Tensor
Assume $V$ is a $n$-dimensional vector space on $\mathbb{R}$, let
$$V^r_s=(V\otimes \cdots\otimes V)^r\otimes (V^\ast \otimes\cdots\otimes V^\ast )_s=\mathcal{L}(V^\ast ,\cdots,V^\ast ,V,\cdots,V;\mathbb{R}),$$
is called a $(r,s)$-type tensor space, where the elements are called $(r,s)$-type Tensor. Such as $V_0^1=V,$
$V_1^0=V^\ast ,$ $V_0^0=\mathbb{R}.$
Let $\{e_i\}$ be the basis of $V$, $\{\omega^i\}$ be the dual basis,
Then $\{e_{i_1}\otimes\cdots\otimes e_{i_r}\otimes \omega^{j_1}\otimes\cdots\otimes \omega^{j_s}\}$ constitutes the basis of $V^r_s$,
$\dim V_s^r=n^{r+s}.$
$\,\forall\,\Phi\in V^r_s,$
Component $\Phi^{k_1\cdots k_r}_{l_1\cdots l_s}=\Phi(\omega^{k_1},\cdots,\omega^{k_2},e_{l_1},\cdots,e_{l_s}).$ of $\Phi$
If we take another basis $\{\widetilde{e}_i\}$ and dual basis $\{\widetilde{\omega}^i\},$
Then there are $\widetilde{e}_i= a_i^je_j,$ $\widetilde{\omega}^i=b_j^i\omega^j,$
Among them $a_i^jb_j^k=\delta_i^k.$ After transformation, the change in components is:
$$\widetilde{\Phi}^{i_1\cdots i_r}_{j_1\cdots j_s}=b_{k_1}^{i_1}\cdots b_{k_r}^{i_r} a_{j_1}^{l_1}\cdots a_{j_s}^{l_s} \Phi_{l_1\cdots l_s}^{k_1\cdots k_r}.$$
It is called $r$-order inverse variation and $s$-order covariation.
The article was last updated on 2021-11-24 16:14:25
