Compact Manifold with odd dimension has Euler numer 0. (Proof using differential topology)












1












$begingroup$


So I am trying to prove this, as an excercise for a course in differential topology, the previous part of the excercise was to find $chi(S^n)$, which is $1+(-1)^n$, it is likely that this fact is meant to use in this part.



I really do not know how to begin, and all proofs I have found, use algebraic topology.



Any help would be appreciated






differential-topology







share|cite|improve this question









$endgroup$












  • $begingroup$
    What definition of Euler characteristic do you use?
    $endgroup$
    – Zircht
    Dec 14 '18 at 21:44










  • $begingroup$
    We just define the Euler characteristic as the Index sum at the zeros of a vector field in $M$, a compact differentiable manifold. (Our Poincare-Hopf just tells you that the index sum is vector field invariant).
    $endgroup$
    – Bajo Fondo
    Dec 14 '18 at 21:50
















1












$begingroup$


So I am trying to prove this, as an excercise for a course in differential topology, the previous part of the excercise was to find $chi(S^n)$, which is $1+(-1)^n$, it is likely that this fact is meant to use in this part.



I really do not know how to begin, and all proofs I have found, use algebraic topology.



Any help would be appreciated






differential-topology







share|cite|improve this question









$endgroup$












  • $begingroup$
    What definition of Euler characteristic do you use?
    $endgroup$
    – Zircht
    Dec 14 '18 at 21:44










  • $begingroup$
    We just define the Euler characteristic as the Index sum at the zeros of a vector field in $M$, a compact differentiable manifold. (Our Poincare-Hopf just tells you that the index sum is vector field invariant).
    $endgroup$
    – Bajo Fondo
    Dec 14 '18 at 21:50














1












1








1





$begingroup$


So I am trying to prove this, as an excercise for a course in differential topology, the previous part of the excercise was to find $chi(S^n)$, which is $1+(-1)^n$, it is likely that this fact is meant to use in this part.



I really do not know how to begin, and all proofs I have found, use algebraic topology.



Any help would be appreciated






differential-topology







share|cite|improve this question









$endgroup$




So I am trying to prove this, as an excercise for a course in differential topology, the previous part of the excercise was to find $chi(S^n)$, which is $1+(-1)^n$, it is likely that this fact is meant to use in this part.



I really do not know how to begin, and all proofs I have found, use algebraic topology.



Any help would be appreciated






differential-topology




differential-topology






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 14 '18 at 20:43









Bajo FondoBajo Fondo

410315




410315












  • $begingroup$
    What definition of Euler characteristic do you use?
    $endgroup$
    – Zircht
    Dec 14 '18 at 21:44










  • $begingroup$
    We just define the Euler characteristic as the Index sum at the zeros of a vector field in $M$, a compact differentiable manifold. (Our Poincare-Hopf just tells you that the index sum is vector field invariant).
    $endgroup$
    – Bajo Fondo
    Dec 14 '18 at 21:50


















  • $begingroup$
    What definition of Euler characteristic do you use?
    $endgroup$
    – Zircht
    Dec 14 '18 at 21:44










  • $begingroup$
    We just define the Euler characteristic as the Index sum at the zeros of a vector field in $M$, a compact differentiable manifold. (Our Poincare-Hopf just tells you that the index sum is vector field invariant).
    $endgroup$
    – Bajo Fondo
    Dec 14 '18 at 21:50
















$begingroup$
What definition of Euler characteristic do you use?
$endgroup$
– Zircht
Dec 14 '18 at 21:44




$begingroup$
What definition of Euler characteristic do you use?
$endgroup$
– Zircht
Dec 14 '18 at 21:44












$begingroup$
We just define the Euler characteristic as the Index sum at the zeros of a vector field in $M$, a compact differentiable manifold. (Our Poincare-Hopf just tells you that the index sum is vector field invariant).
$endgroup$
– Bajo Fondo
Dec 14 '18 at 21:50




$begingroup$
We just define the Euler characteristic as the Index sum at the zeros of a vector field in $M$, a compact differentiable manifold. (Our Poincare-Hopf just tells you that the index sum is vector field invariant).
$endgroup$
– Bajo Fondo
Dec 14 '18 at 21:50










1 Answer
1






active

oldest

votes


















2












$begingroup$

HINT: Let $vec v$ be a "good" vector field on your compact manifold $M$. (So it has isolated zeroes, each with index $pm 1$, although we really don't need this.) If $p$ is a zero of $vec v$ with index $k$, what is the index of $-vec v$ at $p$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ok.. I get it... Since $v$ and $-v$ share the same zeros you have that if $v$ has index $k$ at $p$, then $deg(dg^{-1}ovog)$ is $k$ ($g:U to V subset M$ parametrization). Then the index of $-v$ at $p$ is $deg(dg^{-1}o-vog)=deg(-dg^{-1}ovog)$. Take $f(x)=-x$, then the index of $-v$ is $deg(fo(dg^{-1}o-vog))=deg(f)deg(dg^{-1}o-vog)=(-1)^nk=-k$. Then add up all of then and you will have that $sum iota = -sum iota$, hence $sum iota =0$. Thank you.
    $endgroup$
    – Bajo Fondo
    Dec 14 '18 at 23:40








  • 2




    $begingroup$
    Looks right, except that you need to learn some MathJax :)
    $endgroup$
    – Ted Shifrin
    Dec 14 '18 at 23:51











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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

HINT: Let $vec v$ be a "good" vector field on your compact manifold $M$. (So it has isolated zeroes, each with index $pm 1$, although we really don't need this.) If $p$ is a zero of $vec v$ with index $k$, what is the index of $-vec v$ at $p$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ok.. I get it... Since $v$ and $-v$ share the same zeros you have that if $v$ has index $k$ at $p$, then $deg(dg^{-1}ovog)$ is $k$ ($g:U to V subset M$ parametrization). Then the index of $-v$ at $p$ is $deg(dg^{-1}o-vog)=deg(-dg^{-1}ovog)$. Take $f(x)=-x$, then the index of $-v$ is $deg(fo(dg^{-1}o-vog))=deg(f)deg(dg^{-1}o-vog)=(-1)^nk=-k$. Then add up all of then and you will have that $sum iota = -sum iota$, hence $sum iota =0$. Thank you.
    $endgroup$
    – Bajo Fondo
    Dec 14 '18 at 23:40








  • 2




    $begingroup$
    Looks right, except that you need to learn some MathJax :)
    $endgroup$
    – Ted Shifrin
    Dec 14 '18 at 23:51
















2












$begingroup$

HINT: Let $vec v$ be a "good" vector field on your compact manifold $M$. (So it has isolated zeroes, each with index $pm 1$, although we really don't need this.) If $p$ is a zero of $vec v$ with index $k$, what is the index of $-vec v$ at $p$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ok.. I get it... Since $v$ and $-v$ share the same zeros you have that if $v$ has index $k$ at $p$, then $deg(dg^{-1}ovog)$ is $k$ ($g:U to V subset M$ parametrization). Then the index of $-v$ at $p$ is $deg(dg^{-1}o-vog)=deg(-dg^{-1}ovog)$. Take $f(x)=-x$, then the index of $-v$ is $deg(fo(dg^{-1}o-vog))=deg(f)deg(dg^{-1}o-vog)=(-1)^nk=-k$. Then add up all of then and you will have that $sum iota = -sum iota$, hence $sum iota =0$. Thank you.
    $endgroup$
    – Bajo Fondo
    Dec 14 '18 at 23:40








  • 2




    $begingroup$
    Looks right, except that you need to learn some MathJax :)
    $endgroup$
    – Ted Shifrin
    Dec 14 '18 at 23:51














2












2








2





$begingroup$

HINT: Let $vec v$ be a "good" vector field on your compact manifold $M$. (So it has isolated zeroes, each with index $pm 1$, although we really don't need this.) If $p$ is a zero of $vec v$ with index $k$, what is the index of $-vec v$ at $p$?






share|cite|improve this answer









$endgroup$



HINT: Let $vec v$ be a "good" vector field on your compact manifold $M$. (So it has isolated zeroes, each with index $pm 1$, although we really don't need this.) If $p$ is a zero of $vec v$ with index $k$, what is the index of $-vec v$ at $p$?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 14 '18 at 22:24









Ted ShifrinTed Shifrin

64.2k44692




64.2k44692












  • $begingroup$
    Ok.. I get it... Since $v$ and $-v$ share the same zeros you have that if $v$ has index $k$ at $p$, then $deg(dg^{-1}ovog)$ is $k$ ($g:U to V subset M$ parametrization). Then the index of $-v$ at $p$ is $deg(dg^{-1}o-vog)=deg(-dg^{-1}ovog)$. Take $f(x)=-x$, then the index of $-v$ is $deg(fo(dg^{-1}o-vog))=deg(f)deg(dg^{-1}o-vog)=(-1)^nk=-k$. Then add up all of then and you will have that $sum iota = -sum iota$, hence $sum iota =0$. Thank you.
    $endgroup$
    – Bajo Fondo
    Dec 14 '18 at 23:40








  • 2




    $begingroup$
    Looks right, except that you need to learn some MathJax :)
    $endgroup$
    – Ted Shifrin
    Dec 14 '18 at 23:51


















  • $begingroup$
    Ok.. I get it... Since $v$ and $-v$ share the same zeros you have that if $v$ has index $k$ at $p$, then $deg(dg^{-1}ovog)$ is $k$ ($g:U to V subset M$ parametrization). Then the index of $-v$ at $p$ is $deg(dg^{-1}o-vog)=deg(-dg^{-1}ovog)$. Take $f(x)=-x$, then the index of $-v$ is $deg(fo(dg^{-1}o-vog))=deg(f)deg(dg^{-1}o-vog)=(-1)^nk=-k$. Then add up all of then and you will have that $sum iota = -sum iota$, hence $sum iota =0$. Thank you.
    $endgroup$
    – Bajo Fondo
    Dec 14 '18 at 23:40








  • 2




    $begingroup$
    Looks right, except that you need to learn some MathJax :)
    $endgroup$
    – Ted Shifrin
    Dec 14 '18 at 23:51
















$begingroup$
Ok.. I get it... Since $v$ and $-v$ share the same zeros you have that if $v$ has index $k$ at $p$, then $deg(dg^{-1}ovog)$ is $k$ ($g:U to V subset M$ parametrization). Then the index of $-v$ at $p$ is $deg(dg^{-1}o-vog)=deg(-dg^{-1}ovog)$. Take $f(x)=-x$, then the index of $-v$ is $deg(fo(dg^{-1}o-vog))=deg(f)deg(dg^{-1}o-vog)=(-1)^nk=-k$. Then add up all of then and you will have that $sum iota = -sum iota$, hence $sum iota =0$. Thank you.
$endgroup$
– Bajo Fondo
Dec 14 '18 at 23:40






$begingroup$
Ok.. I get it... Since $v$ and $-v$ share the same zeros you have that if $v$ has index $k$ at $p$, then $deg(dg^{-1}ovog)$ is $k$ ($g:U to V subset M$ parametrization). Then the index of $-v$ at $p$ is $deg(dg^{-1}o-vog)=deg(-dg^{-1}ovog)$. Take $f(x)=-x$, then the index of $-v$ is $deg(fo(dg^{-1}o-vog))=deg(f)deg(dg^{-1}o-vog)=(-1)^nk=-k$. Then add up all of then and you will have that $sum iota = -sum iota$, hence $sum iota =0$. Thank you.
$endgroup$
– Bajo Fondo
Dec 14 '18 at 23:40






2




2




$begingroup$
Looks right, except that you need to learn some MathJax :)
$endgroup$
– Ted Shifrin
Dec 14 '18 at 23:51




$begingroup$
Looks right, except that you need to learn some MathJax :)
$endgroup$
– Ted Shifrin
Dec 14 '18 at 23:51


















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