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张量积
设$V$是数域$\mathbb{F}=\mathbb{R}$上$n$维向量空间,
$e_1,\cdots,e_n$为$V$的基. $V\ni v=\sum_{i=1}^n a^ie_i.$
那么$V^\ast =\{\text{linear} f:V\rightarrow \mathbb{R}\}$也构成$n$维向量空间,
称为对偶空间. $\,\forall\,f\in V^\ast ,$
$f$由基上取值$\{f(e_1),\cdots,f(e_n)\}$唯一决定.
仅在$e_i$上取$1$的函数记为$\omega^i$, 即$\omega^i(e_j)=\delta^i_j.$
$\{\omega^i\}$构成了$V^\ast $的一组基, 称为$\{e_i\}$的对偶基.
对于有限维向量空间, 显然我们总有$(V^\ast )^\ast =V.$ 记$(f,v)=f(v)$为配对.
设$V_1,\cdots,V_r,Z$为向量空间. 若
$f:V_1\times\cdots\times V_r\rightarrow Z$ 关于每个分量是线性的,
则称$f$是$r$-重线性映照. 当$Z=\mathbb{R}$时, 称为$r$-重线性函数.
内积就是$2$-重线性函数, 称为双线性函数. 向量积(外积)是双线性映照.
行列式也可视为多重线性函数.
记$\mathcal{L}(V_1,\cdots,V_r;Z)=\{\text{multi-linear} f:V_1\times\cdots\times V_r\rightarrow Z\},$
容易验证它也是向量空间. 取$Z$的一组基$\{\eta_\alpha\},$
那么任意$f\in \mathcal{L}(V_1,\cdots,V_r;Z)$可分解成为基上函数$f^\alpha.$
对$r=2$的情形, 设$V,W$是向量空间, $\dim V=n,$ $\dim W=m,$
则$\,\forall\,f\in \mathcal{L}(V,W;\mathbb{R}),$ $\,\forall\,v\in V,$
$w\in W,$
$$f(\cdot,w)\in V^\ast ,\quad f(v,\cdot) \in W^\ast .$$
任取$v^\ast \in V^\ast ,$ $w^\ast \in W^\ast ,$ 引入如下双线性函数:
$$(v^\ast \otimes w^\ast )(v,w):=v^\ast (v)w^\ast (w)\,\Rightarrow\, v^\ast \otimes w^\ast \in \mathcal{L}(V,W;\mathbb{R}).$$
称为$v^\ast $和$w^\ast $的张量积.
$h=\otimes:V^\ast \times W^\ast \rightarrow \mathcal{L}(V,W;\mathbb{R}).$
容易验证, 它是双线性的, 即张量积 $\otimes$ 满足分配律.
定义空间$V^\ast ,W^\ast $间的张量积:
$$V^\ast \otimes W^\ast =\operatorname{span}\{\operatorname{Im}h\}=\left\{\sum_{i=1}^{k}a_i (v_i^\ast \otimes w_i^\ast )\,|\, \,\forall\,a_i\in \mathbb{R}, v_i\in V^\ast , w_i\in W^\ast , k\in \mathbb{Z}_+\right\}.$$
命题 1. $V^\ast \otimes W^\ast =\mathcal{L}(V,W;\mathbb{R}).$
证: 只需说明若$\{\omega^i\}, \{\sigma^\alpha\}$分别是$V^\ast ,W^\ast $的基,
则$\{\omega^i\otimes \sigma^\alpha\}$为$\mathcal{L}(V,W;\mathbb{R})$的基.
特别地由此可得到$\dim \mathcal{L}(V,W;\mathbb{R})=\dim V^\ast \otimes W^\ast =\dim V^\ast \cdot \dim W^\ast =\dim V\cdot \dim W.$
易见$V^\ast \otimes W^\ast =\operatorname{span}\{\omega^i\otimes \sigma^\alpha\},$
证其线性无关, 从而是一组基. 为此, 只需利用对偶基$\{e_i\},$
$\{\xi_\alpha\}$讨论系数即可.
而$\,\forall\,f\in \mathcal{L}(V,W),$ $f$由$f(e_i,\xi_\alpha)$唯一决定.
因此$\dim \mathcal{L}(V,W)=\dim V^\ast \otimes W^\ast .$
由$V^\ast \otimes W^\ast \subset \mathcal{L}(V,W)$即知$V^\ast \otimes W^\ast =\mathcal{L}(V,W).$
由于$V,V^\ast ;$ $W,W^\ast $互为对偶, 类似地可以对$v\in V,w\in W,$ 定义
$$(v\otimes w)(v^\ast ,w^\ast )=v(v^\ast )w(w^\ast )=v^\ast (v)w^\ast (w).$$
此时$h=\otimes:V\times W\rightarrow \mathcal{L}(V^\ast ,W^\ast ;\mathbb{R})$是双线性的,
类似定义$V$和$W$的张量积$V\otimes W=\operatorname{span}\{\operatorname{Im}h\}.$
同理它与空间$\mathcal{L}(V^\ast ,W^\ast ;\mathbb{R})$等同,
$\dim V\otimes W=n\cdot m,$
且$\{e_i\otimes \xi_\alpha\}$为$V\otimes W$的基.
文章最后更新于 2021-11-23 18:00:20
tensor product
Let $V$ be the $n$-dimensional vector space on the number field $\mathbb{F}=\mathbb{R}$,
$e_1,\cdots,e_n$ is the basis of $V$. $V\ni v=\sum_{i=1}^n a^ie_i.$
Then $V^\ast =\{\text{linear} f:V\rightarrow \mathbb{R}\}$ also constitutes a $n$-dimensional vector space,
It is called the dual space. $\,\forall\,f\in V^\ast ,$
$f$ is uniquely determined by the base value $\{f(e_1),\cdots,f(e_n)\}$.
The function that only takes $1$ on $e_i$ is recorded as $\omega^i$, that is, $\omega^i(e_j)=\delta^i_j.$
$\{\omega^i\}$ constitutes a group of bases of $V^\ast $, called $\{e_i\}$ dual basis.
For a finite-dimensional vector space, it is obvious that we always have $(V^\ast )^\ast =V.$ and denote $(f,v)=f(v)$ as a pair.
Let $V_1,\cdots,V_r,Z$ be a vector space. If
$f:V_1\times\cdots\times V_r\rightarrow Z$ is linear with respect to each component,
Then $f$ is called $r$-heavy linear mapping. When $Z=\mathbb{R}$ is, it is called $r$-heavy linear function.
The inner product is the $2$-heavy linear function, called a bilinear function. The vector product (outer product) is a bilinear mapping.
Determinants can also be viewed as multilinear functions.
Remember$\mathcal{L}(V_1,\cdots,V_r;Z)=\{\text{multi-linear} f:V_1\times\cdots\times V_r\rightarrow Z\},$
It is easy to verify that it is also a vector space. Take a set of bases $\{\eta_\alpha\},$ of $Z$
Then any $f\in \mathcal{L}(V_1,\cdots,V_r;Z)$ can be decomposed into a basis function $f^\alpha.$
For the case of $r=2$, let $V,W$ be a vector space, $\dim V=n,$ $\dim W=m,$
Then $\,\forall\,f\in \mathcal{L}(V,W;\mathbb{R}),$ $\,\forall\,v\in V,$
$w\in W,$
$$f(\cdot,w)\in V^\ast ,\quad f(v,\cdot) \in W^\ast .$$
Take $v^\ast \in V^\ast ,$ $w^\ast \in W^\ast ,$ arbitrarily and introduce the following bilinear function:
$$(v^\ast \otimes w^\ast )(v,w):=v^\ast (v)w^\ast (w)\,\Rightarrow\, v^\ast \otimes w^\ast \in \mathcal{L}(V,W;\mathbb{R}).$$
called $v^\ast $ and $w^\ast $ tensor product.
$h=\otimes:V^\ast \times W^\ast \rightarrow \mathcal{L}(V,W;\mathbb{R}).$
It is easy to verify that it is bilinear, that is, the tensor product $\otimes$ satisfies the distributive law.
Define the tensor product between spaces $V^\ast ,W^\ast $:
$$V^\ast \otimes W^\ast =\operatorname{span}\{\operatorname{Im}h\}=\left\{\sum_{i=1}^{k}a_i (v_i^\ast \otimes w_i^\ast )\,|\, \,\forall\,a_i\in \mathbb{R}, v_i\in V^\ast , w_i\in W^\ast , k\in \mathbb{Z}_+\right\}.$$
Proposition 1. $V^\ast \otimes W^\ast =\mathcal{L}(V,W;\mathbb{R}).$
Certificate: It only needs to be stated that if $\{\omega^i\}, \{\sigma^\alpha\}$ is the basis of $V^\ast ,W^\ast $ respectively,
Then $\{\omega^i\otimes \sigma^\alpha\}$ is the basis of $\mathcal{L}(V,W;\mathbb{R})$.
In particular, it follows that $\dim \mathcal{L}(V,W;\mathbb{R})=\dim V^\ast \otimes W^\ast =\dim V^\ast \cdot \dim W^\ast =\dim V\cdot \dim W.$
Easy to see$V^\ast \otimes W^\ast =\operatorname{span}\{\omega^i\otimes \sigma^\alpha\},$
Prove that it is linearly independent, and thus it is a set of basis. To do this, just use the dual basis $\{e_i\},$
$\{\xi_\alpha\}$ Just discuss the coefficients.
And $\,\forall\,f\in \mathcal{L}(V,W),$ $f$ is uniquely determined by $f(e_i,\xi_\alpha)$.
Therefore $\dim \mathcal{L}(V,W)=\dim V^\ast \otimes W^\ast .$
From $V^\ast \otimes W^\ast \subset \mathcal{L}(V,W)$, we know $V^\ast \otimes W^\ast =\mathcal{L}(V,W).$
Since $V,V^\ast ;$ $W,W^\ast $ are dual to each other, $v\in V,w\in W,$ can be defined similarly
$$(v\otimes w)(v^\ast ,w^\ast )=v(v^\ast )w(w^\ast )=v^\ast (v)w^\ast (w).$$
At this time $h=\otimes:V\times W\rightarrow \mathcal{L}(V^\ast ,W^\ast ;\mathbb{R})$ is bilinear,
Similarly define the tensor product $V\otimes W=\operatorname{span}\{\operatorname{Im}h\}.$ of $V$ and $W$
In the same way, it is equivalent to space $\mathcal{L}(V^\ast ,W^\ast ;\mathbb{R})$,
$\dim V\otimes W=n\cdot m,$
And $\{e_i\otimes \xi_\alpha\}$ is the basis of $V\otimes W$.
The article was last updated on 2021-11-23 18:00:20
