Mixing
Euler characteristic of the product
$begingroup$
I want to prove that Euler characteristic of the product of two compact oriented manifolds is the product of their Euler characteristics.
As always I do, I'm considering Guillemin-Pollack definitions, i.e., the Euler characteristic of M, compact and oriented, $chi(M) = I(Delta,Delta)$ where $Delta$ is the diagonal of $Mtimes M$ and $I(Delta,Delta) = I(i,Delta) =$ sum of orientation numbers of each $pin i^{-1}(Delta)$ using pre image orientation. Here $i:Delta to M times M$ is the inclusion.
Help!
differential-geometry compact-manifolds
asked Feb 13 '14 at 20:20
rseallanrseallan
556
$endgroup$
add a comment |
$begingroup$
I want to prove that Euler characteristic of the product of two compact oriented manifolds is the product of their Euler characteristics.
As always I do, I'm considering Guillemin-Pollack definitions, i.e., the Euler characteristic of M, compact and oriented, $chi(M) = I(Delta,Delta)$ where $Delta$ is the diagonal of $Mtimes M$ and $I(Delta,Delta) = I(i,Delta) =$ sum of orientation numbers of each $pin i^{-1}(Delta)$ using pre image orientation. Here $i:Delta to M times M$ is the inclusion.
Help!
differential-geometry compact-manifolds
asked Feb 13 '14 at 20:20
rseallanrseallan
556
$endgroup$
$begingroup$
Start by looking at the equation $(f_1(x), f_2(x), g_1(y), g_2(y))=(x, x, y, y)$, where $xin X, yin Y$ and $X, Y$ are smooth compact manifolds. Then observe the relation of solutions of this equation to the Euler characteristics of $X$ and $Y$.
$endgroup$
– Moishe Kohan
Feb 13 '14 at 20:36
$begingroup$
Do you mean use Lefschetz Fixed-Point Theory? This is the next chapter, it's supposed to solve without using it.
$endgroup$
– rseallan
Feb 13 '14 at 21:47
add a comment |
$begingroup$
I want to prove that Euler characteristic of the product of two compact oriented manifolds is the product of their Euler characteristics.
As always I do, I'm considering Guillemin-Pollack definitions, i.e., the Euler characteristic of M, compact and oriented, $chi(M) = I(Delta,Delta)$ where $Delta$ is the diagonal of $Mtimes M$ and $I(Delta,Delta) = I(i,Delta) =$ sum of orientation numbers of each $pin i^{-1}(Delta)$ using pre image orientation. Here $i:Delta to M times M$ is the inclusion.
Help!
differential-geometry compact-manifolds
asked Feb 13 '14 at 20:20
rseallanrseallan
556
$endgroup$
I want to prove that Euler characteristic of the product of two compact oriented manifolds is the product of their Euler characteristics.
As always I do, I'm considering Guillemin-Pollack definitions, i.e., the Euler characteristic of M, compact and oriented, $chi(M) = I(Delta,Delta)$ where $Delta$ is the diagonal of $Mtimes M$ and $I(Delta,Delta) = I(i,Delta) =$ sum of orientation numbers of each $pin i^{-1}(Delta)$ using pre image orientation. Here $i:Delta to M times M$ is the inclusion.
Help!
differential-geometry compact-manifolds
differential-geometry compact-manifolds
asked Feb 13 '14 at 20:20
rseallanrseallan
556
asked Feb 13 '14 at 20:20
rseallanrseallan
556
asked Feb 13 '14 at 20:20
rseallanrseallan
556
asked Feb 13 '14 at 20:20
rseallanrseallan
556
asked Feb 13 '14 at 20:20
rseallanrseallan
556
556
$begingroup$
Start by looking at the equation $(f_1(x), f_2(x), g_1(y), g_2(y))=(x, x, y, y)$, where $xin X, yin Y$ and $X, Y$ are smooth compact manifolds. Then observe the relation of solutions of this equation to the Euler characteristics of $X$ and $Y$.
$endgroup$
– Moishe Kohan
Feb 13 '14 at 20:36
$begingroup$
Do you mean use Lefschetz Fixed-Point Theory? This is the next chapter, it's supposed to solve without using it.
$endgroup$
– rseallan
Feb 13 '14 at 21:47
add a comment |
$begingroup$
Start by looking at the equation $(f_1(x), f_2(x), g_1(y), g_2(y))=(x, x, y, y)$, where $xin X, yin Y$ and $X, Y$ are smooth compact manifolds. Then observe the relation of solutions of this equation to the Euler characteristics of $X$ and $Y$.
$endgroup$
– Moishe Kohan
Feb 13 '14 at 20:36
$begingroup$
Do you mean use Lefschetz Fixed-Point Theory? This is the next chapter, it's supposed to solve without using it.
$endgroup$
– rseallan
Feb 13 '14 at 21:47
$begingroup$
Start by looking at the equation $(f_1(x), f_2(x), g_1(y), g_2(y))=(x, x, y, y)$, where $xin X, yin Y$ and $X, Y$ are smooth compact manifolds. Then observe the relation of solutions of this equation to the Euler characteristics of $X$ and $Y$.
$endgroup$
– Moishe Kohan
Feb 13 '14 at 20:36
$begingroup$
Start by looking at the equation $(f_1(x), f_2(x), g_1(y), g_2(y))=(x, x, y, y)$, where $xin X, yin Y$ and $X, Y$ are smooth compact manifolds. Then observe the relation of solutions of this equation to the Euler characteristics of $X$ and $Y$.
$endgroup$
– Moishe Kohan
Feb 13 '14 at 20:36
$begingroup$
Do you mean use Lefschetz Fixed-Point Theory? This is the next chapter, it's supposed to solve without using it.
$endgroup$
– rseallan
Feb 13 '14 at 21:47
$begingroup$
Do you mean use Lefschetz Fixed-Point Theory? This is the next chapter, it's supposed to solve without using it.
$endgroup$
– rseallan
Feb 13 '14 at 21:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
With the notation $Delta_{xy}:={(z,z)~|~zin Xtimes Y}$,
we have to show that
begin{equation*}
I(Delta_{xy},Delta_{xy})
=chi(Xtimes Y)=chi(X)cdotchi(Y) = I(Delta_x,Delta_x) cdot I(Delta_y,Delta_y).
end{equation*}
Now $Delta_{xy}$ is also equal to ${(x,y,x,y)~|~xin X,~yin Y}$.
As a result, if we fix one $y_i$ for which $Delta_y$ intersects with itself, then
$I(Delta_{xy_i},Delta_{xy_i})=I(Delta_x,Delta_x)$.
As we can repeat this for all $y_i$ for which $Delta_y$ intersects with itself, there are then $I(Delta_x,Delta_x) cdot I(Delta_y,Delta_y)$ points of intersection in $I(Delta_{xy},Delta_{xy})$.
The orientation also agrees because a point of the intersection is included with a plus sign if the orientation at the intersection adds up to the orientation of the ambient space (as explained on p. 112 of Guillemin and Pollack's book) which is the product space and thus carries the product orientation as explained on p. 97 of the (same) book.
answered May 26 '17 at 5:29
exchangeexchange
623414
$endgroup$
$begingroup$
What do you mean by "$y_i$ for which $Delta_y$ intersects with itself"? I'm asking since I can't seem to understand why the equality following these words follows. I believe in your notation $Delta_y={(y,y)| yin Y}$, what does it have to do with $y_i$? More generally, I'm confused by your notation since the RHS of your definition of $Delta_{xy}$ is independent of $x,y$.
$endgroup$
– user500094
Mar 29 '18 at 1:55
$begingroup$
Yes with $Delta_y$ I mean ${(y,y)|yin Y}$ and I take $z=(x,y)$ which should explain the label of $Delta_{xy}$. What I mean with the sentence above is that if we fix one $yin Y$ and we call it $y_i$ (we could also call it anything else) such that $I(Delta_{xy_i},Delta_{xy_i})$ is not zero for all $x$, then $y_i$ marks a point where $Delta_{y}$ and $Delta_{y}$ intersect up to homotopy. Then the intersection number only depends on $Delta_x$ and we can find such a $y_i$ as many times as we have non-vanishing intersection numbers $I(Delta_y,Delta_y)$.
$endgroup$
– exchange
Mar 30 '18 at 9:12
add a comment |
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1 Answer
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$begingroup$
With the notation $Delta_{xy}:={(z,z)~|~zin Xtimes Y}$,
we have to show that
begin{equation*}
I(Delta_{xy},Delta_{xy})
=chi(Xtimes Y)=chi(X)cdotchi(Y) = I(Delta_x,Delta_x) cdot I(Delta_y,Delta_y).
end{equation*}
Now $Delta_{xy}$ is also equal to ${(x,y,x,y)~|~xin X,~yin Y}$.
As a result, if we fix one $y_i$ for which $Delta_y$ intersects with itself, then
$I(Delta_{xy_i},Delta_{xy_i})=I(Delta_x,Delta_x)$.
As we can repeat this for all $y_i$ for which $Delta_y$ intersects with itself, there are then $I(Delta_x,Delta_x) cdot I(Delta_y,Delta_y)$ points of intersection in $I(Delta_{xy},Delta_{xy})$.
The orientation also agrees because a point of the intersection is included with a plus sign if the orientation at the intersection adds up to the orientation of the ambient space (as explained on p. 112 of Guillemin and Pollack's book) which is the product space and thus carries the product orientation as explained on p. 97 of the (same) book.
answered May 26 '17 at 5:29
exchangeexchange
623414
$endgroup$
$begingroup$
What do you mean by "$y_i$ for which $Delta_y$ intersects with itself"? I'm asking since I can't seem to understand why the equality following these words follows. I believe in your notation $Delta_y={(y,y)| yin Y}$, what does it have to do with $y_i$? More generally, I'm confused by your notation since the RHS of your definition of $Delta_{xy}$ is independent of $x,y$.
$endgroup$
– user500094
Mar 29 '18 at 1:55
$begingroup$
Yes with $Delta_y$ I mean ${(y,y)|yin Y}$ and I take $z=(x,y)$ which should explain the label of $Delta_{xy}$. What I mean with the sentence above is that if we fix one $yin Y$ and we call it $y_i$ (we could also call it anything else) such that $I(Delta_{xy_i},Delta_{xy_i})$ is not zero for all $x$, then $y_i$ marks a point where $Delta_{y}$ and $Delta_{y}$ intersect up to homotopy. Then the intersection number only depends on $Delta_x$ and we can find such a $y_i$ as many times as we have non-vanishing intersection numbers $I(Delta_y,Delta_y)$.
$endgroup$
– exchange
Mar 30 '18 at 9:12
add a comment |
$begingroup$
With the notation $Delta_{xy}:={(z,z)~|~zin Xtimes Y}$,
we have to show that
begin{equation*}
I(Delta_{xy},Delta_{xy})
=chi(Xtimes Y)=chi(X)cdotchi(Y) = I(Delta_x,Delta_x) cdot I(Delta_y,Delta_y).
end{equation*}
Now $Delta_{xy}$ is also equal to ${(x,y,x,y)~|~xin X,~yin Y}$.
As a result, if we fix one $y_i$ for which $Delta_y$ intersects with itself, then
$I(Delta_{xy_i},Delta_{xy_i})=I(Delta_x,Delta_x)$.
As we can repeat this for all $y_i$ for which $Delta_y$ intersects with itself, there are then $I(Delta_x,Delta_x) cdot I(Delta_y,Delta_y)$ points of intersection in $I(Delta_{xy},Delta_{xy})$.
The orientation also agrees because a point of the intersection is included with a plus sign if the orientation at the intersection adds up to the orientation of the ambient space (as explained on p. 112 of Guillemin and Pollack's book) which is the product space and thus carries the product orientation as explained on p. 97 of the (same) book.
answered May 26 '17 at 5:29
exchangeexchange
623414
$endgroup$
$begingroup$
What do you mean by "$y_i$ for which $Delta_y$ intersects with itself"? I'm asking since I can't seem to understand why the equality following these words follows. I believe in your notation $Delta_y={(y,y)| yin Y}$, what does it have to do with $y_i$? More generally, I'm confused by your notation since the RHS of your definition of $Delta_{xy}$ is independent of $x,y$.
$endgroup$
– user500094
Mar 29 '18 at 1:55
$begingroup$
Yes with $Delta_y$ I mean ${(y,y)|yin Y}$ and I take $z=(x,y)$ which should explain the label of $Delta_{xy}$. What I mean with the sentence above is that if we fix one $yin Y$ and we call it $y_i$ (we could also call it anything else) such that $I(Delta_{xy_i},Delta_{xy_i})$ is not zero for all $x$, then $y_i$ marks a point where $Delta_{y}$ and $Delta_{y}$ intersect up to homotopy. Then the intersection number only depends on $Delta_x$ and we can find such a $y_i$ as many times as we have non-vanishing intersection numbers $I(Delta_y,Delta_y)$.
$endgroup$
– exchange
Mar 30 '18 at 9:12
add a comment |
$begingroup$
With the notation $Delta_{xy}:={(z,z)~|~zin Xtimes Y}$,
we have to show that
begin{equation*}
I(Delta_{xy},Delta_{xy})
=chi(Xtimes Y)=chi(X)cdotchi(Y) = I(Delta_x,Delta_x) cdot I(Delta_y,Delta_y).
end{equation*}
Now $Delta_{xy}$ is also equal to ${(x,y,x,y)~|~xin X,~yin Y}$.
As a result, if we fix one $y_i$ for which $Delta_y$ intersects with itself, then
$I(Delta_{xy_i},Delta_{xy_i})=I(Delta_x,Delta_x)$.
As we can repeat this for all $y_i$ for which $Delta_y$ intersects with itself, there are then $I(Delta_x,Delta_x) cdot I(Delta_y,Delta_y)$ points of intersection in $I(Delta_{xy},Delta_{xy})$.
The orientation also agrees because a point of the intersection is included with a plus sign if the orientation at the intersection adds up to the orientation of the ambient space (as explained on p. 112 of Guillemin and Pollack's book) which is the product space and thus carries the product orientation as explained on p. 97 of the (same) book.
answered May 26 '17 at 5:29
exchangeexchange
623414
$endgroup$
With the notation $Delta_{xy}:={(z,z)~|~zin Xtimes Y}$,
we have to show that
begin{equation*}
I(Delta_{xy},Delta_{xy})
=chi(Xtimes Y)=chi(X)cdotchi(Y) = I(Delta_x,Delta_x) cdot I(Delta_y,Delta_y).
end{equation*}
Now $Delta_{xy}$ is also equal to ${(x,y,x,y)~|~xin X,~yin Y}$.
As a result, if we fix one $y_i$ for which $Delta_y$ intersects with itself, then
$I(Delta_{xy_i},Delta_{xy_i})=I(Delta_x,Delta_x)$.
As we can repeat this for all $y_i$ for which $Delta_y$ intersects with itself, there are then $I(Delta_x,Delta_x) cdot I(Delta_y,Delta_y)$ points of intersection in $I(Delta_{xy},Delta_{xy})$.
The orientation also agrees because a point of the intersection is included with a plus sign if the orientation at the intersection adds up to the orientation of the ambient space (as explained on p. 112 of Guillemin and Pollack's book) which is the product space and thus carries the product orientation as explained on p. 97 of the (same) book.
answered May 26 '17 at 5:29
exchangeexchange
623414
answered May 26 '17 at 5:29
exchangeexchange
623414
answered May 26 '17 at 5:29
exchangeexchange
623414
answered May 26 '17 at 5:29
exchangeexchange
623414
623414
$begingroup$
What do you mean by "$y_i$ for which $Delta_y$ intersects with itself"? I'm asking since I can't seem to understand why the equality following these words follows. I believe in your notation $Delta_y={(y,y)| yin Y}$, what does it have to do with $y_i$? More generally, I'm confused by your notation since the RHS of your definition of $Delta_{xy}$ is independent of $x,y$.
$endgroup$
– user500094
Mar 29 '18 at 1:55
$begingroup$
Yes with $Delta_y$ I mean ${(y,y)|yin Y}$ and I take $z=(x,y)$ which should explain the label of $Delta_{xy}$. What I mean with the sentence above is that if we fix one $yin Y$ and we call it $y_i$ (we could also call it anything else) such that $I(Delta_{xy_i},Delta_{xy_i})$ is not zero for all $x$, then $y_i$ marks a point where $Delta_{y}$ and $Delta_{y}$ intersect up to homotopy. Then the intersection number only depends on $Delta_x$ and we can find such a $y_i$ as many times as we have non-vanishing intersection numbers $I(Delta_y,Delta_y)$.
$endgroup$
– exchange
Mar 30 '18 at 9:12
add a comment |
$begingroup$
What do you mean by "$y_i$ for which $Delta_y$ intersects with itself"? I'm asking since I can't seem to understand why the equality following these words follows. I believe in your notation $Delta_y={(y,y)| yin Y}$, what does it have to do with $y_i$? More generally, I'm confused by your notation since the RHS of your definition of $Delta_{xy}$ is independent of $x,y$.
$endgroup$
– user500094
Mar 29 '18 at 1:55
$begingroup$
Yes with $Delta_y$ I mean ${(y,y)|yin Y}$ and I take $z=(x,y)$ which should explain the label of $Delta_{xy}$. What I mean with the sentence above is that if we fix one $yin Y$ and we call it $y_i$ (we could also call it anything else) such that $I(Delta_{xy_i},Delta_{xy_i})$ is not zero for all $x$, then $y_i$ marks a point where $Delta_{y}$ and $Delta_{y}$ intersect up to homotopy. Then the intersection number only depends on $Delta_x$ and we can find such a $y_i$ as many times as we have non-vanishing intersection numbers $I(Delta_y,Delta_y)$.
$endgroup$
– exchange
Mar 30 '18 at 9:12
$begingroup$
What do you mean by "$y_i$ for which $Delta_y$ intersects with itself"? I'm asking since I can't seem to understand why the equality following these words follows. I believe in your notation $Delta_y={(y,y)| yin Y}$, what does it have to do with $y_i$? More generally, I'm confused by your notation since the RHS of your definition of $Delta_{xy}$ is independent of $x,y$.
$endgroup$
– user500094
Mar 29 '18 at 1:55
$begingroup$
What do you mean by "$y_i$ for which $Delta_y$ intersects with itself"? I'm asking since I can't seem to understand why the equality following these words follows. I believe in your notation $Delta_y={(y,y)| yin Y}$, what does it have to do with $y_i$? More generally, I'm confused by your notation since the RHS of your definition of $Delta_{xy}$ is independent of $x,y$.
$endgroup$
– user500094
Mar 29 '18 at 1:55
$begingroup$
Yes with $Delta_y$ I mean ${(y,y)|yin Y}$ and I take $z=(x,y)$ which should explain the label of $Delta_{xy}$. What I mean with the sentence above is that if we fix one $yin Y$ and we call it $y_i$ (we could also call it anything else) such that $I(Delta_{xy_i},Delta_{xy_i})$ is not zero for all $x$, then $y_i$ marks a point where $Delta_{y}$ and $Delta_{y}$ intersect up to homotopy. Then the intersection number only depends on $Delta_x$ and we can find such a $y_i$ as many times as we have non-vanishing intersection numbers $I(Delta_y,Delta_y)$.
$endgroup$
– exchange
Mar 30 '18 at 9:12
$begingroup$
Yes with $Delta_y$ I mean ${(y,y)|yin Y}$ and I take $z=(x,y)$ which should explain the label of $Delta_{xy}$. What I mean with the sentence above is that if we fix one $yin Y$ and we call it $y_i$ (we could also call it anything else) such that $I(Delta_{xy_i},Delta_{xy_i})$ is not zero for all $x$, then $y_i$ marks a point where $Delta_{y}$ and $Delta_{y}$ intersect up to homotopy. Then the intersection number only depends on $Delta_x$ and we can find such a $y_i$ as many times as we have non-vanishing intersection numbers $I(Delta_y,Delta_y)$.
$endgroup$
– exchange
Mar 30 '18 at 9:12
add a comment |
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$begingroup$
Start by looking at the equation $(f_1(x), f_2(x), g_1(y), g_2(y))=(x, x, y, y)$, where $xin X, yin Y$ and $X, Y$ are smooth compact manifolds. Then observe the relation of solutions of this equation to the Euler characteristics of $X$ and $Y$.
$endgroup$
– Moishe Kohan
Feb 13 '14 at 20:36
$begingroup$
Do you mean use Lefschetz Fixed-Point Theory? This is the next chapter, it's supposed to solve without using it.
$endgroup$
– rseallan
Feb 13 '14 at 21:47