I stumbled upon Problem 13-2 on p.344 in John Lee's Introduction to Smooth Manifolds (2nd Edition) where Lee explains that the proof for the existence of a Riemannian Metric on a manifold is done by a "Partition of Unity"-Argument and he emphasizes that a crucial part in the proof was that the set of inner products on a given tangent space is a convex subset of the vector space of all symmetric $2$-tensors.
Now it seems that the convexity property can not be omitted in "usual partition of unity" arguments but unfortunately, I don't understand why it's so important. Where exactly does the convexity of the (respective) subset comes into play if we want to do such partition of unity argument (in a much more general sense)?
So, to make my question a bit more precise:
My question: Say we want to patch together local objects to a global one (e.g. local sections of a vector bundle to a global section) by a usual partition of unity argument. Why and where do we need to use convexity of the subset containing the images of the objects we want to patch together? (in the case of local sections we require that there is an open set $V\subset E$ (total space) so that $V\cap E_p$ is a convex subset of the fiber $E_p$. Why do we need convexity of $V\cap E_p$ for example?
Thanks in advance for any help!
A convex subset of a vector space $C\subseteq V$ is a set which is closed under all convex combinations, which are wieghted sums whose wights are nonnegative and sum to $1$: $$ \lambda_1u_1+\lambda_2u_2+\cdots\lambda_nu_n \\ u_i\in U,\ \ \ \lambda_i\ge 0,\ \ \ \sum_i\lambda_i=1 $$ When combining together local sections $\sigma_i$ of vector bundles using a partition of unity $\psi_i$ in an expression like $\sum_i\psi_i\sigma_i$, we are effectively taking a convex combination at each fiber. This means that the combination retains any fiberwise properties which hold on a convex subset of each fiber (such as positive definiteness and symmetry of $(0,2)$ tensors).
The same argument doesn't work for indefinite metrics of arbitrary signature because these tensors do not form a convex subset, and so the wighted sum $\sum_i\psi_ig_i$ may fail to be a nondegenerate metric of the same signature..