Machine-translated from Chinese.
对称张量与反称张量
对$\Phi\in V^\ast \otimes \cdots\otimes V^\ast =L(V,\cdots,V;\mathbb{R}),$ 若
$$\Phi(v_1,\cdots,v_i,\cdots,v_j,\cdots,v_k)=\pm \Phi(v_1,\cdots,v_j,\cdots,v_i,\cdots,v_k),\quad \,\forall\,1\le i,j\le k,$$
则称$\Phi$是对称(反称)的.
用$\odot^k(V^\ast )$表示$V^\ast \otimes\cdots\otimes V^\ast $中对称张量全体,
$\Lambda^k(V^\ast )$表示反称张量全体.
令$J(k)$表示$k$阶置换群. 对于$\sigma\in J(k),$
定义$\operatorname{sgn}:J(k)\rightarrow\{\pm 1\},$ 表示奇偶排列.
易验证$\operatorname{sgn}(\sigma\tau)=\operatorname{sgn}(\sigma)\operatorname{sgn}(\tau).$
定义作用$\sigma\Phi(v_1,\cdots,v_k):=\Phi(v_{\sigma(1)},\cdots,v_{\sigma(k)}),$
易见$\sigma\Phi\in V^\ast \otimes\cdots\otimes V^\ast ,$ 且$\sigma$是线性映照.
那么$\Phi$是对称(反称)的$\Leftrightarrow$
$\sigma\Phi=(\operatorname{sgn}\sigma)\Phi.$
取$V$的一组基$\{e_i\},$ 对偶基$\{\omega^i\}.$
则对$\Phi=\Phi_{i_1\cdots i_k}\omega^{i_1}\otimes\cdots\otimes \omega^{i_k},$
可以得到$\Phi$对称(反称)$\Leftrightarrow$
$\Phi_{i_1\cdots i_k}$关于指标$\{i_1,\cdots,i_k\}$是对称(反称)的.
定义如下两个线性变换:
$S_k:\Phi\mapsto \frac{1}{k!}\sum_\sigma \sigma \Phi,$
称为对称化算子;
$A_k:\Phi\mapsto \frac{1}{k!}\sum_{\sigma} (\operatorname{sgn}\sigma)\sigma \Phi,$
称为反称化算子. 记$V_k=V^\ast \otimes\cdots\otimes V^\ast ,$ 我们有如下命题:
命题 1. $\odot^k(V^\ast )=S_k(V_k),$ $\Lambda^k(V^\ast )=A_k(V_k),$ 且$\Phi$是对称(反称)的$\Leftrightarrow$ $S_k(\Phi)=\Phi$($A_k(\Phi)=\Phi.$)
证: 取$\tau\in J(k),$ 首先证明$\tau(S_k(\Phi))=S_k(\Phi),$
$\tau(A_k(\Phi))=(\operatorname{sgn}\tau)A_k (\Phi).$
其次再说明$A_k|_{\Lambda^k(V^\ast )}=\mathrm{id},$
$S_k|_{\odot^k(V^\ast )}=\mathrm{id}$即可.
拉回映照
设$f:V\rightarrow W$为线性映照, 定义$f^\ast :W^\ast \rightarrow V^\ast ,$
$(f^\ast w^\ast )(v)=w^\ast (f(v)),$ 称为拉回映照.
可以推广到$f^\ast :W_k\rightarrow V_k,$
$(f^\ast \Phi)(v_1,\cdots,v_k)=\Phi(f(v_1),\cdots,f(v_k)).$
命题 2. $S_kf^\ast =f^\ast S_k,$ $A_kf^\ast =f^\ast A_k.$
证: 利用$\sigma f^\ast =f^\ast \sigma$即可.
外代数
$\Lambda^r(V^\ast ):=A_r(V_r),$ 称为$r$阶反称张量空间.
约定$\Lambda^0(V^\ast )=\mathbb{R},$ $\Lambda^1(V^\ast )=V^\ast .$
设$\Phi\in \Lambda^r(V^\ast ),$ $\Psi\in \Lambda^s(V^\ast ),$
则$\Phi\wedge \Psi:=\binom{r+s}{r,s}A_{r+s}(\Phi\otimes \Psi)\in \Lambda^{r+s}(V^\ast )$称为$\Phi$与$\Psi$的外积,
满足如下性质:
分配律:
$(a\Phi_1+b\Phi_2)\wedge\Psi=a\Phi_1\wedge \Psi+b\Phi_2\wedge\Psi;$
$\Phi\wedge(a\Psi_1+b\Psi_2)=a\Phi\wedge \Psi_1+b\Phi\wedge \Psi_2.$
反交换律: $\Phi\wedge \Psi=(-1)^{rs}\Psi\wedge \Phi;$
结合律: $(\Phi\wedge \Psi)\wedge\eta=\Phi\wedge(\Psi\wedge \eta).$
证明基本利用张量积的性质即可, 外面套上线性的反称化算子$A.$ 对于反交换律,
利用置换
$$\tau=\begin{pmatrix}
1&\cdots&r&r+1&\cdots&r+s\\
s+1&\cdots&s+r&1&\cdots&s
\end{pmatrix}$$
即可, 满足$\operatorname{sgn}\tau=(-1)^{rs}.$
特别地, 对于$r=s=1$的情形, $\Phi\wedge\Psi=-\Psi\wedge\Phi,$
$\Phi\wedge\Phi=0.$
对于结合律, 引入置换
$$\widetilde\tau=\begin{pmatrix}
1&\cdots&r+s&r+s+1&\cdots&r+s+t\\
\tau(1)&\cdots&\tau(r+s)&r+s+1&\cdots&r+s+t
\end{pmatrix}$$
进行常规的定义讨论即可. 特别地,
可以在证明中看到,
$\Phi_1\wedge\cdots\wedge\Phi_k=\binom{r_1+\cdots+r_k}{r_1,\cdots,r_k}A(\Phi_1\otimes \cdots\otimes \Phi_k).$
推论 3. 设$\theta^1,\cdots,\theta^k\in \Lambda^1(V^\ast )=V^\ast ,$ $x_1,\cdots,x_k\in V,$ 则$\theta^1\wedge\cdots\wedge\theta^k(x_1,\cdots,x_k)=\det(\theta^i(x_j))_{k\times k}.$
证: 展开发现式子恰为行列式的组合定义即可.
即$\sum_\sigma \operatorname{sgn}(\sigma) \theta^1(x_{\sigma(1)})\cdots\theta^k(x_{\sigma(k)}).$
推论 4. 设$\theta^1,\cdots,\theta^k\in \Lambda^1(V^\ast )=V^\ast ,$ 则$\theta^1,\cdots,\theta^k$线性相关$\Leftrightarrow$ $\theta^1\wedge\cdots\wedge\theta^k=0.$
证: 利用反交换律与反证法即可.
文章最后更新于 2021-12-02 11:36:31
Symmetric and antisymmetric tensors
For $\Phi\in V^\ast \otimes \cdots\otimes V^\ast =L(V,\cdots,V;\mathbb{R}),$ if
$$\Phi(v_1,\cdots,v_i,\cdots,v_j,\cdots,v_k)=\pm \Phi(v_1,\cdots,v_j,\cdots,v_i,\cdots,v_k),\quad \,\forall\,1\le i,j\le k,$$
Then $\Phi$ is said to be Symmetry (Anti-symmetry) of.
Use $\odot^k(V^\ast )$ to represent all symmetric tensors in $V^\ast \otimes\cdots\otimes V^\ast $,
$\Lambda^k(V^\ast )$ represents the entire inverse tensor.
Let $J(k)$ represent the permutation group of order $k$. For $\sigma\in J(k),$
Definition $\operatorname{sgn}:J(k)\rightarrow\{\pm 1\},$ represents odd-even arrangement.
Easy to verify$\operatorname{sgn}(\sigma\tau)=\operatorname{sgn}(\sigma)\operatorname{sgn}(\tau).$
Define role$\sigma\Phi(v_1,\cdots,v_k):=\Phi(v_{\sigma(1)},\cdots,v_{\sigma(k)}),$
It is easy to see that $\sigma\Phi\in V^\ast \otimes\cdots\otimes V^\ast ,$ and $\sigma$ are linear mappings.
Then $\Phi$ is symmetric (antisymmetric) $\Leftrightarrow$
$\sigma\Phi=(\operatorname{sgn}\sigma)\Phi.$
Take a set of basis $\{e_i\},$ dual basis $\{\omega^i\}.$ of $V$
Then for $\Phi=\Phi_{i_1\cdots i_k}\omega^{i_1}\otimes\cdots\otimes \omega^{i_k},$
You can get $\Phi$ symmetry (antisymmetry) $\Leftrightarrow$
$\Phi_{i_1\cdots i_k}$ is symmetric (anti-symmetric) about the indicator $\{i_1,\cdots,i_k\}$.
Define the following two linear transformations:
$S_k:\Phi\mapsto \frac{1}{k!}\sum_\sigma \sigma \Phi,$
called Symmetrization operator;
$A_k:\Phi\mapsto \frac{1}{k!}\sum_{\sigma} (\operatorname{sgn}\sigma)\sigma \Phi,$
called desymmetrization operator. Note $V_k=V^\ast \otimes\cdots\otimes V^\ast ,$ we have the following proposition:
Proposition 1. $\odot^k(V^\ast )=S_k(V_k),$ $\Lambda^k(V^\ast )=A_k(V_k),$ and $\Phi$ are symmetrical (antisymmetrical) $\Leftrightarrow$ $S_k(\Phi)=\Phi$($A_k(\Phi)=\Phi.$)
Certificate: Take $\tau\in J(k),$ and first prove $\tau(S_k(\Phi))=S_k(\Phi),$
$\tau(A_k(\Phi))=(\operatorname{sgn}\tau)A_k (\Phi).$
Next, explain $A_k|_{\Lambda^k(V^\ast )}=\mathrm{id},$
$S_k|_{\odot^k(V^\ast )}=\mathrm{id}$ is enough.
Pullback map
Let $f:V\rightarrow W$ be a linear mapping, define $f^\ast :W^\ast \rightarrow V^\ast ,$
$(f^\ast w^\ast )(v)=w^\ast (f(v)),$ is called Pullback map.
Can be generalized to $f^\ast :W_k\rightarrow V_k,$
$(f^\ast \Phi)(v_1,\cdots,v_k)=\Phi(f(v_1),\cdots,f(v_k)).$
Proposition 2. $S_kf^\ast =f^\ast S_k,$ $A_kf^\ast =f^\ast A_k.$
Certificate: Just use $\sigma f^\ast =f^\ast \sigma$.
Exterior Algebra
$\Lambda^r(V^\ast ):=A_r(V_r),$ is called $r$ order inverse tensor space.
Agreement $\Lambda^0(V^\ast )=\mathbb{R},$ $\Lambda^1(V^\ast )=V^\ast .$
Let $\Phi\in \Lambda^r(V^\ast ),$ $\Psi\in \Lambda^s(V^\ast ),$
Then $\Phi\wedge \Psi:=\binom{r+s}{r,s}A_{r+s}(\Phi\otimes \Psi)\in \Lambda^{r+s}(V^\ast )$ is called the combination of $\Phi$ and $\Psi$ outer product,
Satisfy the following properties:
Distributive law:
$(a\Phi_1+b\Phi_2)\wedge\Psi=a\Phi_1\wedge \Psi+b\Phi_2\wedge\Psi;$
$\Phi\wedge(a\Psi_1+b\Psi_2)=a\Phi\wedge \Psi_1+b\Phi\wedge \Psi_2.$
Anti-commutative law: $\Phi\wedge \Psi=(-1)^{rs}\Psi\wedge \Phi;$
Associative law: $(\Phi\wedge \Psi)\wedge\eta=\Phi\wedge(\Psi\wedge \eta).$
The proof is basically based on the properties of tensor products, and the linear desymmetry operator $A.$ is put outside it. For the inverse commutative law,
Use substitution
$$\tau=\begin{pmatrix}
1&\cdots&r&r+1&\cdots&r+s\\
s+1&\cdots&s+r&1&\cdots&s
\end{pmatrix}$$
That’s it, satisfying $\operatorname{sgn}\tau=(-1)^{rs}.$
In particular, for the case of $r=s=1$, $\Phi\wedge\Psi=-\Psi\wedge\Phi,$
$\Phi\wedge\Phi=0.$
For the associative law, a permutation is introduced
$$\widetilde\tau=\begin{pmatrix}
1&\cdots&r+s&r+s+1&\cdots&r+s+t\\
\tau(1)&\cdots&\tau(r+s)&r+s+1&\cdots&r+s+t
\end{pmatrix}$$
Just have the usual definition discussion. In particular,
As can be seen in the proof,
$\Phi_1\wedge\cdots\wedge\Phi_k=\binom{r_1+\cdots+r_k}{r_1,\cdots,r_k}A(\Phi_1\otimes \cdots\otimes \Phi_k).$
Corollary 3. Assume $\theta^1,\cdots,\theta^k\in \Lambda^1(V^\ast )=V^\ast ,$ $x_1,\cdots,x_k\in V,$ then $\theta^1\wedge\cdots\wedge\theta^k(x_1,\cdots,x_k)=\det(\theta^i(x_j))_{k\times k}.$
Certificate: Expand it and find that the expression is exactly the combinatorial definition of the determinant.
That is $\sum_\sigma \operatorname{sgn}(\sigma) \theta^1(x_{\sigma(1)})\cdots\theta^k(x_{\sigma(k)}).$
Corollary 4. Assume $\theta^1,\cdots,\theta^k\in \Lambda^1(V^\ast )=V^\ast ,$, then $\theta^1,\cdots,\theta^k$ is linearly related $\Leftrightarrow$ $\theta^1\wedge\cdots\wedge\theta^k=0.$
Certificate: Just use the anti-commutative law and proof by contradiction.
The article was last updated on 2021-12-02 11:36:31
