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Every smooth manifold with boundary is a smooth manifold with corners.

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1

$begingroup$

Show that every smooth k-manifold with boundary is a smooth manifold with corners

Definitions:

1.If $M subseteq R^n$, M is a smooth k-manifold with boundary if for
every point $pin M$ there exists an open neighbourhood $Vsubseteq M$ of $p$, and an open set $U$ of $mathbb{H}^k$ such that $phi: Uto V$ is a regular embedding.

2.A set $M subseteq R^n$, M is a smooth k-manifold with corners if for every $p in M$ there exist open sets $Vsubseteq M$, $Usubseteq overline {mathbb{R}_+^k}$ and a regular embedding $phi: Uto V$ such that $p in V$

So suppose $M$ is a smooth manifold with boundary and assume that $phi:Uto V$ is a regular embedding which covers the interior points of $M$ and $Pi:Nto M$ is a regular embedding which covers the boundary.

Then for any $pin M$ either $p=phi(x)$ or $p=Pi(y)$ for some $x,y$ in the interrior or boundary of $mathbb{H}^k$

Let $pin int(M)$, then the function $psi:Uto K$, by $psi(x_1,...,x_{k-1},x_k)=(e^{x_1},...,e^{x_{k-1}},e^{x_k})$, is a homeomorphism from ${mathbb{R}_+^k}$ to the interior of $mathbb{H}^k$ and thus $phicircpsi^{-1}:Kto V$ is a regular embedding from an open subset of $mathbb{R}_+^k$ to $M$.

I believe this works for dealing with the interior points. What I am not sure how to deal with is the boundary points.

The only thing I could think of would be to take an orthogonal projection of the points in $partialmathbb{H}^k$ which are not in $partialmathbb{R}^k_+$ say for example in $mathbb{R}^2$, take $(x,0)to(0,vert xvert)$. But I don't think this works in higher dimension.

differential-geometry

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asked Mar 26 at 0:15

AColoredReptileAColoredReptile

408210

$endgroup$

add a comment | 

0

1

$begingroup$

Show that every smooth k-manifold with boundary is a smooth manifold with corners

Definitions:

1.If $M subseteq R^n$, M is a smooth k-manifold with boundary if for
every point $pin M$ there exists an open neighbourhood $Vsubseteq M$ of $p$, and an open set $U$ of $mathbb{H}^k$ such that $phi: Uto V$ is a regular embedding.

2.A set $M subseteq R^n$, M is a smooth k-manifold with corners if for every $p in M$ there exist open sets $Vsubseteq M$, $Usubseteq overline {mathbb{R}_+^k}$ and a regular embedding $phi: Uto V$ such that $p in V$

So suppose $M$ is a smooth manifold with boundary and assume that $phi:Uto V$ is a regular embedding which covers the interior points of $M$ and $Pi:Nto M$ is a regular embedding which covers the boundary.

Then for any $pin M$ either $p=phi(x)$ or $p=Pi(y)$ for some $x,y$ in the interrior or boundary of $mathbb{H}^k$

Let $pin int(M)$, then the function $psi:Uto K$, by $psi(x_1,...,x_{k-1},x_k)=(e^{x_1},...,e^{x_{k-1}},e^{x_k})$, is a homeomorphism from ${mathbb{R}_+^k}$ to the interior of $mathbb{H}^k$ and thus $phicircpsi^{-1}:Kto V$ is a regular embedding from an open subset of $mathbb{R}_+^k$ to $M$.

I believe this works for dealing with the interior points. What I am not sure how to deal with is the boundary points.

The only thing I could think of would be to take an orthogonal projection of the points in $partialmathbb{H}^k$ which are not in $partialmathbb{R}^k_+$ say for example in $mathbb{R}^2$, take $(x,0)to(0,vert xvert)$. But I don't think this works in higher dimension.

differential-geometry

share|cite|improve this question

asked Mar 26 at 0:15

AColoredReptileAColoredReptile

408210

$endgroup$

add a comment | 

$begingroup$

Show that every smooth k-manifold with boundary is a smooth manifold with corners

Definitions:

1.If $M subseteq R^n$, M is a smooth k-manifold with boundary if for
every point $pin M$ there exists an open neighbourhood $Vsubseteq M$ of $p$, and an open set $U$ of $mathbb{H}^k$ such that $phi: Uto V$ is a regular embedding.

2.A set $M subseteq R^n$, M is a smooth k-manifold with corners if for every $p in M$ there exist open sets $Vsubseteq M$, $Usubseteq overline {mathbb{R}_+^k}$ and a regular embedding $phi: Uto V$ such that $p in V$

So suppose $M$ is a smooth manifold with boundary and assume that $phi:Uto V$ is a regular embedding which covers the interior points of $M$ and $Pi:Nto M$ is a regular embedding which covers the boundary.

Then for any $pin M$ either $p=phi(x)$ or $p=Pi(y)$ for some $x,y$ in the interrior or boundary of $mathbb{H}^k$

Let $pin int(M)$, then the function $psi:Uto K$, by $psi(x_1,...,x_{k-1},x_k)=(e^{x_1},...,e^{x_{k-1}},e^{x_k})$, is a homeomorphism from ${mathbb{R}_+^k}$ to the interior of $mathbb{H}^k$ and thus $phicircpsi^{-1}:Kto V$ is a regular embedding from an open subset of $mathbb{R}_+^k$ to $M$.

I believe this works for dealing with the interior points. What I am not sure how to deal with is the boundary points.

The only thing I could think of would be to take an orthogonal projection of the points in $partialmathbb{H}^k$ which are not in $partialmathbb{R}^k_+$ say for example in $mathbb{R}^2$, take $(x,0)to(0,vert xvert)$. But I don't think this works in higher dimension.

differential-geometry

share|cite|improve this question

asked Mar 26 at 0:15

AColoredReptileAColoredReptile

408210

$endgroup$

share|cite|improve this question

asked Mar 26 at 0:15

AColoredReptileAColoredReptile

408210

share|cite|improve this question

asked Mar 26 at 0:15

AColoredReptileAColoredReptile

408210

asked Mar 26 at 0:15

AColoredReptileAColoredReptile

408210

asked Mar 26 at 0:15

AColoredReptileAColoredReptile

408210