Excercise: Show that if two Riemannian metrics $g_1$ $g_2$ are such that $g_1=fg_2$ for some positive definite $f \in C^{\infty}(M)$ for Riemannian Manifold $M$, then $g_1$ and $g_2$ have the same angle measurements but not length.
My attempt:
$\forall X,Y \in T_pM$, $g_1(X,Y)=f(p)g_2(X,Y)$
Thus $g_1(X,X)=f(p)g_2(X,X)$
And so by definition
$f(p)|X|_2|X|_2cos(0)=|X|_1|X|_1cos(0)$
Since $f$ is positive definite, we have that $|X|_2 \neq |X|_1$ and so the length measurements made by $g_1$ and $g_2$ are not equal.
I'm trying now to show that $g_1$ and $g_2$ do indeed preserve angle measurements. My first observation is that these metrics preserve orthogonality. Can somebody help me rigorously finish the proof? Or at least give me a hint? Thanks.
On a side note, what's the easiest way to explicitly prove that a change of basis on an inner product space preserves angles between vectors?
Best Answer
If you define the cosine of the angle $\theta$ between $X_1$ and $X_2$ as $$\cos\theta_g = \frac{g(X_1, X_2)}{\sqrt{g(X_1, X_1,) g(X_2, X_2)}} $$ then if $g_1(X_1, X_2) = f(p)g_2(X_1, X_2)$ $\forall X, Y \in T_p M$, you can show that $$\cos\theta_{g_1} = \cos\theta_{g_2}.$$