$M$ is a Riemannian manifold ,and $g_{ij}$ is Remannian metric. Let $x=(x^1...x^d)$ $(i.e. x:U_x\rightarrow R^d)$ be a local coordinates ,and $v,w\in T_pM$ with coordinate representations $(v^1...v^d)$ and $(w^1...w^d)$$(i.e.v=v^i\frac{\partial}{\partial x^i},w=w^j\frac{\partial}{\partial x^j})$.

Let $y=f(x)$ $(i.e. y:U_y\rightarrow R^d)$ define different local coordinates.In these coordinates,$v$ and $w$ have representations $(\tilde{v}^1...\tilde{v}^d)$ and $(\tilde{w}^1...\tilde{w}^d)$. Let the metric in the new coordinates be given by $h_{kl}(y)$,then,we have: $$ h_{kl}(f(x))\frac{\partial f^k}{\partial x^i}\frac{\partial f^l}{\partial x^j}=g_{ij}(x) $$

Let $g=det(g_{ij}),h=det(h_{kl})$,and $\Phi$ is a function on $M$, $U=U_x\bigcap U_y$

Show that the integral of a function $\Phi$ on $M$ is invariant ,namely $$ \int _U \Phi(f(x))\sqrt{g(x)}dx^1...dx^d=\int _U\Phi(y)\sqrt{h(y)}dy^1...dy^d $$

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    $\begingroup$ Looks familiar... Haven't you asked that already? $\endgroup$ – Amitai Yuval Sep 18 '15 at 8:42
  • $\begingroup$ @AmitaiYuval Yes, because the previous one is not good ,I have delete it .and ask again. $\endgroup$ – lanse2pty Sep 18 '15 at 8:45
  • $\begingroup$ Do you know how to integrate an $n$-form on a $n$-dimensional orientable manifold? $\endgroup$ – user99914 Sep 18 '15 at 9:14
  • $\begingroup$ @JohnMa integrate it by pulling back to $R^n$ ? $\endgroup$ – lanse2pty Sep 18 '15 at 10:49

If $ \phi : {\bf R}^n \rightarrow M$ is coordinate, then $$ \int_{U:=\phi ({\bf R}^n) } \Phi (p)\ d {\rm Vol}_p := \int_{{\bf R}^n} \Phi (\phi(x)) \sqrt{{\rm det}\ g_{ij}} dx^1\cdots dx^n $$ where $$ g_{ij}:=g(d\phi e_i,d\phi e_j ) $$

If $ f : x\in {\bf R}^n\rightarrow y\in {\bf R}^n $ is a diffeomorphism, we have another coordinate $$ \phi\circ f^{-1} : {\bf R}^n\rightarrow M $$

That is by above definition we have $$ \int_{U} \Phi (p)\ d {\rm Vol}_p = \int_{{\bf R}^n} \Phi (\phi\circ f^{-1} (y) ) \sqrt{{\rm det}\ g_{\alpha\beta}} dy^1\cdots dy^n $$ where $$ g_{\alpha\beta}:= g(d(\phi\circ f^{-1}) e_\alpha,d(\phi\circ f^{-1} ) e_\beta ) $$

Here $$ f^\ast ( \Phi (\phi\circ f^{-1} (y) ) \sqrt{{\rm det}\ g_{\alpha\beta}} dy^1\cdots dy^n ) $$ $$= \Phi (\phi (x)) \sqrt{ {\rm det}\ (f^{-1})^k_\alpha (f^{-1})^l_\beta g_{kl} }\ {\rm det} f^\alpha_k\ dx^1\cdots dx^n $$ $$= \Phi (\phi (x)) \sqrt{ {\rm det}\ g_{kl} }\ dx^1\cdots dx^n$$

That is since two integrals are equal (cf. change of variables), the above definition is well defined.


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