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Exercise verification in John Lee's Introduction to Smooth Manifolds, Part 1

0

$begingroup$

Let $pi: Eto M$ be a smooth vector bundle, and $V$ an open subset of $E$ with the property that $V cap E_p$ is non empty and convex for all $pin M$.

By "a section of $V$" we mean a (local or global) section of $E$ whose image lies in $V$.

Show that there exists a smooth global section of $V$.

My Argument

Let $pin M$ and $(sigma_i^p:U_pto E)_i=1^k$ a smooth local frame with $p in U_psubseteq M$ open subset.

Since $E_p cap Vneemptyset$ then $v^i_psigma_i^p|_pin V cap E_p$ for a certain $v_p in mathbbR^k$.

Define a smooth local section $sigma^p:U_pto E, qmapsto v^i_psigma^p_i|_q$, and take $(sigma^p)^-1(V)$ which is open in $M$.

Then $(sigma^p)^-1(V):pin M$ is an open cover of $M$. Take a smooth partition of unity subordinate to it, say $(psi_p:pin M)$.

Consider $sigma=sum_ppsi_psigma^p$. Then $sigma$ is a smooth global section of $E$.

It remains to show that the image of $sigma$ lies in $V$.

Let $qin M$, then I have $sigma(q)=psi_p(q)sigma^p|_q$ for finite indices $p$, and I know that $sigma^p|_q$ lies in $V$ for each $p$. It follows that $sigma(q)=psi_p(q)sigma^p|_q$ lies in $V$ by the convexity of $Vcap E_q$. $qquadsquare$

Is my proof correct? It is possible to simply it?

P.S: this is the first part of Problem 13.2 in Lee's book: "Introduction to smooth manifolds, 2 Edition"

Here Exercise verification in John Lee's Introduction to Smooth Manifolds, Part 2 is the link to part 2 of the question.

differential-geometry smooth-manifolds vector-bundles smooth-functions

share|cite|improve this question

edited Mar 18 at 17:24

Minato

asked Mar 18 at 12:30

MinatoMinato

593314

$endgroup$

add a comment | 

0

$begingroup$

Let $pi: Eto M$ be a smooth vector bundle, and $V$ an open subset of $E$ with the property that $V cap E_p$ is non empty and convex for all $pin M$.

By "a section of $V$" we mean a (local or global) section of $E$ whose image lies in $V$.

Show that there exists a smooth global section of $V$.

My Argument

Let $pin M$ and $(sigma_i^p:U_pto E)_i=1^k$ a smooth local frame with $p in U_psubseteq M$ open subset.

Since $E_p cap Vneemptyset$ then $v^i_psigma_i^p|_pin V cap E_p$ for a certain $v_p in mathbbR^k$.

Define a smooth local section $sigma^p:U_pto E, qmapsto v^i_psigma^p_i|_q$, and take $(sigma^p)^-1(V)$ which is open in $M$.

Then $(sigma^p)^-1(V):pin M$ is an open cover of $M$. Take a smooth partition of unity subordinate to it, say $(psi_p:pin M)$.

Consider $sigma=sum_ppsi_psigma^p$. Then $sigma$ is a smooth global section of $E$.

It remains to show that the image of $sigma$ lies in $V$.

Let $qin M$, then I have $sigma(q)=psi_p(q)sigma^p|_q$ for finite indices $p$, and I know that $sigma^p|_q$ lies in $V$ for each $p$. It follows that $sigma(q)=psi_p(q)sigma^p|_q$ lies in $V$ by the convexity of $Vcap E_q$. $qquadsquare$

Is my proof correct? It is possible to simply it?

P.S: this is the first part of Problem 13.2 in Lee's book: "Introduction to smooth manifolds, 2 Edition"

Here Exercise verification in John Lee's Introduction to Smooth Manifolds, Part 2 is the link to part 2 of the question.

differential-geometry smooth-manifolds vector-bundles smooth-functions

share|cite|improve this question

edited Mar 18 at 17:24

Minato

asked Mar 18 at 12:30

MinatoMinato

593314

$endgroup$

add a comment | 

$begingroup$

Let $pi: Eto M$ be a smooth vector bundle, and $V$ an open subset of $E$ with the property that $V cap E_p$ is non empty and convex for all $pin M$.

By "a section of $V$" we mean a (local or global) section of $E$ whose image lies in $V$.

Show that there exists a smooth global section of $V$.

My Argument

Let $pin M$ and $(sigma_i^p:U_pto E)_i=1^k$ a smooth local frame with $p in U_psubseteq M$ open subset.

Since $E_p cap Vneemptyset$ then $v^i_psigma_i^p|_pin V cap E_p$ for a certain $v_p in mathbbR^k$.

Define a smooth local section $sigma^p:U_pto E, qmapsto v^i_psigma^p_i|_q$, and take $(sigma^p)^-1(V)$ which is open in $M$.

Then $(sigma^p)^-1(V):pin M$ is an open cover of $M$. Take a smooth partition of unity subordinate to it, say $(psi_p:pin M)$.

Consider $sigma=sum_ppsi_psigma^p$. Then $sigma$ is a smooth global section of $E$.

It remains to show that the image of $sigma$ lies in $V$.

Let $qin M$, then I have $sigma(q)=psi_p(q)sigma^p|_q$ for finite indices $p$, and I know that $sigma^p|_q$ lies in $V$ for each $p$. It follows that $sigma(q)=psi_p(q)sigma^p|_q$ lies in $V$ by the convexity of $Vcap E_q$. $qquadsquare$

Is my proof correct? It is possible to simply it?

P.S: this is the first part of Problem 13.2 in Lee's book: "Introduction to smooth manifolds, 2 Edition"

Here Exercise verification in John Lee's Introduction to Smooth Manifolds, Part 2 is the link to part 2 of the question.

differential-geometry smooth-manifolds vector-bundles smooth-functions

share|cite|improve this question

edited Mar 18 at 17:24

Minato

asked Mar 18 at 12:30

MinatoMinato

593314

$endgroup$

share|cite|improve this question

edited Mar 18 at 17:24

Minato

asked Mar 18 at 12:30

MinatoMinato

593314

share|cite|improve this question

edited Mar 18 at 17:24

Minato

asked Mar 18 at 12:30

MinatoMinato

593314

asked Mar 18 at 12:30

MinatoMinato

593314

asked Mar 18 at 12:30

MinatoMinato

593314