Path of nondegenerate forms still nondegenerate. Good morning,
i was reading a text where was defined a parametrized two-form on a compact manifold M by $$w_t = (1-t)w + tf^{*}w$$ where $~f:M \to M$ is some diffeomorphism of M and $w$ is the volume form of M. Now it was stated that (fix a point $p \in M$) $(w_t)_p$ again is a nondegenerate form. When $w$ is a volume form then this is equivalent of $w_p$ being nondegenerate for all $p\in M$. As $w_p$ is nondegenerate and f a bijection it seems to be clear that also $(f^{*}w)_p$ is nondegenerate. By the way, I use the following definition of nondegeneracy: For any fixed point $p \in M$
$$w_p\ \text{nondegenerate} \Leftrightarrow \forall v \in T_pM: w_p(u,v)=0 \Rightarrow u = 0 $$
$$\Leftrightarrow \forall v \neq 0 \in T_pM\ \exists u\in T_pM: w_p(u,v) \neq 0$$
But why is the sum of them nondegenerate, too? For example take $t=0.5$ and
Assume $(w_t)_p(u,v)=0$ for all $v \in T_pM$. Then this yields $$w(u,v) = -w_{f(p)}((df)_p(u), (df)_p(v)).$$ If $f$ was the identity, then it is clear to me that $u$ must be zero, but what if $f$ is not the identity. I'm sure one can build a diffeo such that this equality is fullfilled for all $v \in T_pM$ without u being zero...
 A: For this statement to be true, it is necessary and sufficient to assume that $M$ is orientable and that the diffeomorphism $f$ preserves (one and thus any choice of) orientation.
As an easy example of the necessity of this condition on $f$, consider $(M, w) = (\mathbb{R}, dx)$ and consider the orientation reversing diffeomorphism $f(x) = -x$. Then $f^{\ast}w = -dx$ and $w_{1/2} \equiv 0$.
Let $M$ be an orientable $n$-dimension manifold. Let $w \in \Omega^n(M)$ be a volume form on $M$, that is a no-where vanishing section of the (real) line bundle $\Lambda^nM$. The volume form $w$ determines an orientation of the manifold $M$.
A diffeomorphism $f : M \to M$ sends volume forms to volumes forms, but it might change the orientation. More precisely, given any smooth map $f : M \to M$, we have a well-defined map $q_f : M \to \mathbb{R} : p \mapsto \frac{f^{\ast}w_p}{w_p}$ (observe that the map $\Lambda^nM \to M \times R : (p, \lambda \, w_p) \to (p, \lambda)$ is a global trivialisation of the bundle, so this quotient is well-defined) ; If $f$ is a diffeomorphism, then $q_f$ is non-vanishing ; $f$ is orientation preserving if and only if $q_f$ is (strictly) positive.
Notice then that
$$ \left. w_t \right|_p := (1-t) \left. w \right|_p + t \left. f^{\ast}w \right|_p = (1-t) \left. w \right|_p + t \, \left. q_f \cdot w \right|_p = \left( 1-t + t \, q_f(p)  \right) \, \left. w \right|_p \, , $$
so that $w_t$ vanishes at $p$ if and only if $q_f(p) = - (1-t)/t < 0$ (understood that we only consider $t \in [0,1]$). Hence, a sufficient and necessary condition for $w_t$ to never vanish on $M$, whichever the value of $t$, is that $q_f > 0$ i.e. that $f$ be orientation preserving.
