A question about the $2times2$ matrices action on 2-torus.
$begingroup$
Given a $2times2$ matrix $A$ with entries in $mathbb Z$, it acts on
begin{equation}
mathbb T^2=mathbb R^2/mathbb Z^2
end{equation}
and then leads to a map
begin{equation}
A_*colon H_2(mathbb T^2)to H_2(mathbb T^2)
end{equation}
I want to know is this map just the scalar multiplication by $det A$?
Can someone suggest a neat proof?
Thank you in advance.
algebraic-topology
asked Nov 25 '18 at 17:56
Display NameDisplay Name
432212
$endgroup$
add a comment |
$begingroup$
Given a $2times2$ matrix $A$ with entries in $mathbb Z$, it acts on
begin{equation}
mathbb T^2=mathbb R^2/mathbb Z^2
end{equation}
and then leads to a map
begin{equation}
A_*colon H_2(mathbb T^2)to H_2(mathbb T^2)
end{equation}
I want to know is this map just the scalar multiplication by $det A$?
Can someone suggest a neat proof?
Thank you in advance.
algebraic-topology
asked Nov 25 '18 at 17:56
Display NameDisplay Name
432212
$endgroup$
$begingroup$
Very brief idea: $H_2(Bbb T^2)$ is induced by the degree map on cycles.
$endgroup$
– Arthur
Nov 25 '18 at 18:07
1
$begingroup$
Yes. To my mind it's easier to see the corresponding fact for cohomology $H^2$ (because then you can use the cup product), then apply UCT.
$endgroup$
– Qiaochu Yuan
Nov 25 '18 at 19:04
add a comment |
$begingroup$
Given a $2times2$ matrix $A$ with entries in $mathbb Z$, it acts on
begin{equation}
mathbb T^2=mathbb R^2/mathbb Z^2
end{equation}
and then leads to a map
begin{equation}
A_*colon H_2(mathbb T^2)to H_2(mathbb T^2)
end{equation}
I want to know is this map just the scalar multiplication by $det A$?
Can someone suggest a neat proof?
Thank you in advance.
algebraic-topology
asked Nov 25 '18 at 17:56
Display NameDisplay Name
432212
$endgroup$
Given a $2times2$ matrix $A$ with entries in $mathbb Z$, it acts on
begin{equation}
mathbb T^2=mathbb R^2/mathbb Z^2
end{equation}
and then leads to a map
begin{equation}
A_*colon H_2(mathbb T^2)to H_2(mathbb T^2)
end{equation}
I want to know is this map just the scalar multiplication by $det A$?
Can someone suggest a neat proof?
Thank you in advance.
algebraic-topology
algebraic-topology
asked Nov 25 '18 at 17:56
Display NameDisplay Name
432212
asked Nov 25 '18 at 17:56
Display NameDisplay Name
432212
asked Nov 25 '18 at 17:56
Display NameDisplay Name
432212
asked Nov 25 '18 at 17:56
Display NameDisplay Name
432212
asked Nov 25 '18 at 17:56
Display NameDisplay Name
432212
432212
$begingroup$
Very brief idea: $H_2(Bbb T^2)$ is induced by the degree map on cycles.
$endgroup$
– Arthur
Nov 25 '18 at 18:07
1
$begingroup$
Yes. To my mind it's easier to see the corresponding fact for cohomology $H^2$ (because then you can use the cup product), then apply UCT.
$endgroup$
– Qiaochu Yuan
Nov 25 '18 at 19:04
add a comment |
$begingroup$
Very brief idea: $H_2(Bbb T^2)$ is induced by the degree map on cycles.
$endgroup$
– Arthur
Nov 25 '18 at 18:07
1
$begingroup$
Yes. To my mind it's easier to see the corresponding fact for cohomology $H^2$ (because then you can use the cup product), then apply UCT.
$endgroup$
– Qiaochu Yuan
Nov 25 '18 at 19:04
$begingroup$
Very brief idea: $H_2(Bbb T^2)$ is induced by the degree map on cycles.
$endgroup$
– Arthur
Nov 25 '18 at 18:07
$begingroup$
Very brief idea: $H_2(Bbb T^2)$ is induced by the degree map on cycles.
$endgroup$
– Arthur
Nov 25 '18 at 18:07
1
1
$begingroup$
Yes. To my mind it's easier to see the corresponding fact for cohomology $H^2$ (because then you can use the cup product), then apply UCT.
$endgroup$
– Qiaochu Yuan
Nov 25 '18 at 19:04
$begingroup$
Yes. To my mind it's easier to see the corresponding fact for cohomology $H^2$ (because then you can use the cup product), then apply UCT.
$endgroup$
– Qiaochu Yuan
Nov 25 '18 at 19:04
add a comment |
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$begingroup$
Very brief idea: $H_2(Bbb T^2)$ is induced by the degree map on cycles.
$endgroup$
– Arthur
Nov 25 '18 at 18:07
1
$begingroup$
Yes. To my mind it's easier to see the corresponding fact for cohomology $H^2$ (because then you can use the cup product), then apply UCT.
$endgroup$
– Qiaochu Yuan
Nov 25 '18 at 19:04