Consider the following:
two ovaloids $S$ and $S'$
orientations $N$ for $S$ and $N'$ for $S'$
an isometry $f: S \to S'$
while proving Cohn-Vossen's theorem one contructs the following self-adjoint endomorphisms:
$dN_p$
$A(f)_p = -(df_p)^{-1} \circ dN'_p \circ (df_p)$
these are definite endomorphisms because:
$det(dN_p) = K > 0$
$det(A(f)_p) = K' > 0$
where $K,K'$ are the gaussian curvatures of $S,S'$.
However, in Proof of a linear algebra lemma for Cohn-Vossen's theorem I was pointed out that I need to know whether these endormorphisms are positive or negative definite.
In particular, is there a way I can guarantee that $dN_p$ is positive definite and $A(f)_p$ is negative definite? I thought of choosing different $N$ and $N'$ but the problem is that once chosen they stay the same and I the property won't hold for every point $p$.
Let's take $dN_p$. Then, $dN_p$ is an endomorphism which can be diagonalized in the form
\begin{bmatrix} k_1(p) & 0 \\ 0 & k_2(p) \end{bmatrix}
where $k_i(p)$ are the principal curvatures. If we change the point $p$, could we change from a positive definite to negative definite ($dN_p$ is definite because it is a normal of an ovaloid which has positive gauss curvature)? No! Because in that case since $k_i$ is continuous then some $k_i$ would be zero at some point $p$. Therefore, $k_i$ are always positive or negative and $dN_p$ is always definite positive or definite negative. If we need one or the other we can just change the sign. The same reasoning is valid for $dN'_p$ and thus, one can take $A(f)_p$ to be always positive definite or negative definite (it would be the opposite to $dN'_p$).
Finally, I can take the lemma assuming positive definition or any definition for the endomorphisms.