2
$\begingroup$

$M$ 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 $$ \int _U \Phi(f(x))\sqrt{g(x)}dx^1...dx^d=\int _U\Phi(y)\sqrt{h(y)}dy^1...dy^d $$

$\endgroup$
  • 1
    $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.