Is sign of $\det(f_*)$ determine how the orientation of the surfaces changes by the isometry?

Question

Let $f:S\rightarrow\bar{S}$ is an isometry. Then the map $f_*:T_pS\rightarrow T_{f(p)}\bar{S}$ is an orthogonal map. Hence $\det(f_*)=\pm1$.

Is sign of $\det(f_*)$ determine how the orientation of the surfaces changes by the isometry?

What is the value of $det(df_*)$?

Answer 1

If $f$ is an isometry then $\langle u,v\rangle_{g_S} = \langle f_* u, f_* v\rangle_{g_\bar S}$. Note that the determinant (including the sign) of $f_*$ depends entirely on the choice of bases for $T_pS$ and $T_f(p)\bar S$. You can't say anything about the orientation just from looking at $f_*$ for some (arbitrary) choice of pair of bases.

But if you choose bases that are oriented consistently with each manifold, you will indeed get that $f$ is orientation-preserving if $\det f_* > 0$. Note that what you're really doing by taking the determinant is pulling back the volume form, and checking that $$\langle dV, f^*d\bar V\rangle > 0.$$