Is the integration on a compact Lie group invariant from left translation?

Let G be a compact Lie group of dimension n and $\Omega$ be a left-invariant n-form,$f:G \to \textbf{R}$ be a smooth map.

Is the equation below true? $$\int\limits_{\rm{G}} {f \circ \mu \left( {h, \cdot } \right)\Omega } = \int\limits_{\rm{G}} {f\Omega }$$ where $\mu(h,g)=hg$ is the multiplication in G.

If is,how to prove it?

• Pull back the $n$-form on the right by the diffeomorphism $L_h$. – Ted Shifrin Mar 6 '17 at 16:55
• I got it!Thank you.So if we pull back a n-form by a diffeomorphism,the value of integration dose not change,right?I just wonder what it happen if we replace the diffeomorphism by any smooth maps. – Huanghow Mar 7 '17 at 2:17
• In general, you get a factor of the degree of the map. Note that $L_h$, being homotopic to the identity (assuming your Lie group is connected or $h$ is in the identity component), is orientation-preserving. – Ted Shifrin Mar 7 '17 at 3:24