2
$\begingroup$

Let $M$ be a submanifold of $\mathbb{R}^n$ show that exists a hyperplane $H$ such that intersects transversaly t $M$

First I define $F:M \times \mathbb{S}^{n-11} \rightarrow \mathbb{R}$ given by $F(x,y)= \langle x | y \rangle$ then $F$ is a submersion because $F$ is the restriction to the submanifold $M \times \mathbb{S}^{n-1}$ of the submersion $G: \mathbb{R}^{2n} \rightarrow \mathbb{R}$ with $G$ the inner product in $\mathbb{R}^n$ then i try to applicated the transversality theorem but I not sure how, because these theorem only gives me to $F_y \pitchfork Z$ f.a.e $y \in \mathbb{R}^{n}$ and $Z$ any submanifold of $\mathbb{R}$ i think to the hyperplane what I am looking for is the inverse image $F^{-1}_{y}(c)$ with $c$ a regular value

Edit: I proved that $0$ is a regular value of $F_y$ and how $\{0\}$ is a submanifold of $\mathbb{R}$ then $F_y \pitchfork \{0\}$, now $F_y^{-1}(0)$ is the set such that the hyperplane $H=\{w \in \mathbb{R}^n ; \langle w,v \rangle =0\}$ intersects to $M$ can i conclude that these intersection is transversaly?

Any hint or suggestions I appreciated

$\endgroup$

1 Answer 1

2
$\begingroup$

We can suppose the origin does not belong to $M$. A hiperplane $H\subset\mathbb R^n$ is generated by $n-1$ independent vectors $u_i$, so we consider the mapping $$ F:\varOmega\times(\mathbb R^{n-1}\setminus\{0\})\to\mathbb R^n:(u,t)\mapsto\sum_it_iu_i $$ Here $u=(u_i), u_i\in\mathbb R^n$, $t=(t_i),t_i\in\mathbb R$ and $\varOmega\subset\mathbb R^{n\times(n-1)}$ is the open set defined by rank$(u)=n-1$. Thus for each $u=(u_i)$ fixed, the image of the partial function $F_u$ is the hyperplane $H_u$ generated by the $u_i$'s, except the origin (because we excluded $t=0$). This mapping $F$ is a submersion. Indeed, fix any $(u^0,t^0)$ with say $t^0_i\ne0$, and for $v\in\mathbb R^n$ denote $u=(0,\dots,v,\dots,0)$. Then $$ d_{(u^0,t^0)}F(u,0)=t_iv, $$ which shows $d_{(u^0,t^0)}F$ is onto. Thus, since $F$ is a submersion, it is transversal to $M$. Consequently, by density of transversality (the basic parametrized version which follows from the Sard-Brown theorem), for almost every $u$, the partial function $F_u$ is transversal to $M$. This means that the hyperplane $H_u$ is transversal to $M$ at every point except the origin, but this point does not belong to $M$. We are done.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .