Machine-translated from Chinese.
定义
音乐同构(musical isomorphism), 又称典范(canonical)同构,
指黎曼流形切丛和余切丛间的同构, 由黎曼度量给出.
黎曼度量$g=g_{ij}dx^i\otimes dx^j$是一个正定二阶的张量场, 每点有映射
$$
\widehat g_x:T_xM\rightarrow T_x^\ast M,
$$
使得
$$
\widehat g_x(X)(Y)=\left<{}X,Y\right>.
$$
这给出了微分同胚$\widehat g:TM\rightarrow T^\ast M.$
若$X=X^i\partial_i,$ 那么计算得到
$$
\widehat g(X)=g_{ij}X^j\partial_i.
$$
由正定性, 它是一个同构. 上述同构称为降号音乐同构(flat),
记为$\widehat g(X)=X^\flat,$ 表示指标$X^i$下降为下标$g_{ij}X^j.$
逆运算自然称作升号音乐同构(sharp),
记为$\widehat g^{-1}(\omega)=\omega^\sharp,$
将指标$\omega_i$升为上标$g^{ij}\omega_j.$
应用
使用音乐同构可以更简洁地表示一些公式, 如对于一般的向量丛$(E,\nabla,h)$,
计算联络$\nabla:\Gamma(M,E)\rightarrow\Gamma(M,E\otimes T^\ast M)$关于$L^2$内积对偶$\nabla^\ast $的表达式.
对于紧支撑的$f\in\Gamma(M,E),$ $\omega\in \Gamma(M,E\otimes T^\ast M),$
设$\nabla f=f_i\otimes dx^i,$ 那么
$$
\left<{}\nabla f,\omega\right>=\int_M g\otimes h (\nabla f,\omega) d\mathrm{vol}_g=\int_M g^{ij}h(f_i,\omega_j)d\mathrm{vol}_g,\quad \omega=\omega_j \otimes dx^j,
$$
希望得到$\left<{}f,\nabla^\ast \omega\right>=\int_M h(f,\nabla^\ast \omega)d\mathrm{vol}_g$的形式,
因此要利用联络与度量的相容性,
$$
\begin{aligned}
LHS&=\int_M g^{ij} \partial_i h(f,\omega_j) - g^{ij}h(f,\nabla_{\partial_i}\omega_j)d\mathrm{vol}_g\\
&=\int_M (dx^j)^\sharp h(f,\omega_j)-h(f,\nabla_{(dx^j)^\sharp}\omega_j)d\mathrm{vol}_g\\
&=\int_M \operatorname{div}(h(f,\omega_j)(dx^j)^\sharp)-h(f,(\operatorname{div}(dx^j)^\sharp+\nabla_{(dx^j)^\sharp})\omega_j)d\mathrm{vol}_g\\
&=\int_M h(f,\nabla^*\omega)d\mathrm{vol}_g=\left<{}f,\nabla^*\omega\right>.
\end{aligned}
$$
由于等式对所有$f$对,
这就用音乐同构给出了$\nabla^\ast $的表达式. 用降号来表示,
就是对于$\omega=u\otimes X^\flat,$
$$
\nabla^\ast \omega=-\operatorname{div}(X)\omega-\nabla_X\omega.
$$
过程中用到的散度$\operatorname{div}:T_xM\rightarrow \mathbb{R}$指
$$
\operatorname{div}X=\left<{}\nabla_{\partial_i}X,\partial_i\right>=\partial_iX^i+\Gamma_{ij}^iX^j,
$$
满足
$$
\operatorname{div}(fX)=Xf+f\operatorname{div}X.
$$
文章最后更新于 2023-10-30 20:56:03
definition
Musical Isomorphism (musical isomorphism), also known as canonical isomorphism,
Refers to the isomorphism between the tangent bundle and the cotangent bundle of a Riemannian manifold, given by the Riemannian metric.
The Riemannian metric $g=g_{ij}dx^i\otimes dx^j$ is a positive definite second-order tensor field, and each point has a mapping
$$
\widehat g_x:T_xM\rightarrow T_x^\ast M,
$$
make
$$
\widehat g_x(X)(Y)=\left<{}X,Y\right>.
$$
This gives diffeomorphism $\widehat g:TM\rightarrow T^\ast M.$
If $X=X^i\partial_i,$ then calculated
$$
\widehat g(X)=g_{ij}X^j\partial_i.
$$
By positive definiteness, it is an isomorphism. The above isomorphism is called flat musical isomorphism (flat),
Marked as $\widehat g(X)=X^\flat,$, it means that the indicator $X^i$ decreases to the subscript $g_{ij}X^j.$.
The inverse operation is naturally called sharp musical isomorphism (sharp),
Recorded as $\widehat g^{-1}(\omega)=\omega^\sharp,$
Promote indicator $\omega_i$ to superscript $g^{ij}\omega_j.$
Application
Using musical isomorphisms can express some formulas more concisely, such as for the general vector bundle $(E,\nabla,h)$,
Compute the expression of the connection $\nabla:\Gamma(M,E)\rightarrow\Gamma(M,E\otimes T^\ast M)$ with respect to the inner product dual $\nabla^\ast $ of $L^2$.
For compactly supported $f\in\Gamma(M,E),$ $\omega\in \Gamma(M,E\otimes T^\ast M),$
Let $\nabla f=f_i\otimes dx^i,$ then
$$
\left<{}\nabla f,\omega\right>=\int_M g\otimes h (\nabla f,\omega) d\mathrm{vol}_g=\int_M g^{ij}h(f_i,\omega_j)d\mathrm{vol}_g,\quad \omega=\omega_j \otimes dx^j,
$$
I hope to get the form $\left<{}f,\nabla^\ast \omega\right>=\int_M h(f,\nabla^\ast \omega)d\mathrm{vol}_g$,
Therefore, we must take advantage of the compatibility of the connection with the metric,
$$
\begin{aligned}
LHS&=\int_M g^{ij} \partial_i h(f,\omega_j) - g^{ij}h(f,\nabla_{\partial_i}\omega_j)d\mathrm{vol}_g\\
&=\int_M (dx^j)^\sharp h(f,\omega_j)-h(f,\nabla_{(dx^j)^\sharp}\omega_j)d\mathrm{vol}_g\\
&=\int_M \operatorname{div}(h(f,\omega_j)(dx^j)^\sharp)-h(f,(\operatorname{div}(dx^j)^\sharp+\nabla_{(dx^j)^\sharp})\omega_j)d\mathrm{vol}_g\\
&=\int_M h(f,\nabla^*\omega)d\mathrm{vol}_g=\left<{}f,\nabla^*\omega\right>.
\end{aligned}
$$
Since the equation holds for all $f$ pairs,
This gives the expression of $\nabla^\ast $ using musical isomorphism. Expressed with a flat sign,
Just for $\omega=u\otimes X^\flat,$
$$
\nabla^\ast \omega=-\operatorname{div}(X)\omega-\nabla_X\omega.
$$
The divergence $\operatorname{div}:T_xM\rightarrow \mathbb{R}$ used in the process refers to
$$
\operatorname{div}X=\left<{}\nabla_{\partial_i}X,\partial_i\right>=\partial_iX^i+\Gamma_{ij}^iX^j,
$$
satisfy
$$
\operatorname{div}(fX)=Xf+f\operatorname{div}X.
$$
The article was last updated on 2023-10-30 20:56:03
