Mixing
Determining induced action on cohomology.
$begingroup$
Consider the map $f:S^3 times S^3 to S^3 times S^3$ given by $(z_0,z_1)mapsto (z_1, -z_0)$. Write $f$ as composition of $g circ h $, where $h$ is $(z_0,z_1)mapsto (z_0, -z_1)$ and $g$ is $(z_0,z_1)mapsto (z_1, z_0)$. Then the corresponding map induced by $h$ on $3^{rd}$ cohomology group with $Bbb Z$-coefficient is given by $A=pmatrix{1&0\0&1}$(as the degree of the antipodal map is $(-1)^{n+1}=1$ in this case) and that of $g$ is given by $B=pmatrix{0&1\1&0}$. So the induced map by $f$ on $3^{rd}$ cohomology group is $AB$ i.e. $pmatrix{0&1\1&0}$. But this must be wrong as then the Lefschetz number $L(f)= 1+ (-1)^3 cdot0+ (-1)^6cdot1=2neq 0$. But $f$ is a free action. So contradiction. Please help me to find my mistake.
I am learning Algebraic Topology by myself. Sorry, incase this a trivial question.
Thank you so much for your help.
algebraic-topology
asked Jan 11 at 18:16
mathstudentmathstudent
394
$endgroup$
add a comment |
$begingroup$
Consider the map $f:S^3 times S^3 to S^3 times S^3$ given by $(z_0,z_1)mapsto (z_1, -z_0)$. Write $f$ as composition of $g circ h $, where $h$ is $(z_0,z_1)mapsto (z_0, -z_1)$ and $g$ is $(z_0,z_1)mapsto (z_1, z_0)$. Then the corresponding map induced by $h$ on $3^{rd}$ cohomology group with $Bbb Z$-coefficient is given by $A=pmatrix{1&0\0&1}$(as the degree of the antipodal map is $(-1)^{n+1}=1$ in this case) and that of $g$ is given by $B=pmatrix{0&1\1&0}$. So the induced map by $f$ on $3^{rd}$ cohomology group is $AB$ i.e. $pmatrix{0&1\1&0}$. But this must be wrong as then the Lefschetz number $L(f)= 1+ (-1)^3 cdot0+ (-1)^6cdot1=2neq 0$. But $f$ is a free action. So contradiction. Please help me to find my mistake.
I am learning Algebraic Topology by myself. Sorry, incase this a trivial question.
Thank you so much for your help.
algebraic-topology
asked Jan 11 at 18:16
mathstudentmathstudent
394
$endgroup$
$begingroup$
Of course, you mean $hcirc g$.
$endgroup$
– Ted Shifrin
Jan 11 at 18:27
$begingroup$
The trace on $H_6$ is $-1$; the map reverses orientation.
$endgroup$
– Lord Shark the Unknown
Jan 11 at 18:29
$begingroup$
@LordSharktheUnknown How to tell a map is orientation preserving or reversing? One way to check I know is to look at the degree of the induced map on top homology. But here how should I know it is orientation reversing assuming I don't know the degree of the map?
$endgroup$
– mathstudent
Jan 13 at 8:20
add a comment |
$begingroup$
Consider the map $f:S^3 times S^3 to S^3 times S^3$ given by $(z_0,z_1)mapsto (z_1, -z_0)$. Write $f$ as composition of $g circ h $, where $h$ is $(z_0,z_1)mapsto (z_0, -z_1)$ and $g$ is $(z_0,z_1)mapsto (z_1, z_0)$. Then the corresponding map induced by $h$ on $3^{rd}$ cohomology group with $Bbb Z$-coefficient is given by $A=pmatrix{1&0\0&1}$(as the degree of the antipodal map is $(-1)^{n+1}=1$ in this case) and that of $g$ is given by $B=pmatrix{0&1\1&0}$. So the induced map by $f$ on $3^{rd}$ cohomology group is $AB$ i.e. $pmatrix{0&1\1&0}$. But this must be wrong as then the Lefschetz number $L(f)= 1+ (-1)^3 cdot0+ (-1)^6cdot1=2neq 0$. But $f$ is a free action. So contradiction. Please help me to find my mistake.
I am learning Algebraic Topology by myself. Sorry, incase this a trivial question.
Thank you so much for your help.
algebraic-topology
asked Jan 11 at 18:16
mathstudentmathstudent
394
$endgroup$
Consider the map $f:S^3 times S^3 to S^3 times S^3$ given by $(z_0,z_1)mapsto (z_1, -z_0)$. Write $f$ as composition of $g circ h $, where $h$ is $(z_0,z_1)mapsto (z_0, -z_1)$ and $g$ is $(z_0,z_1)mapsto (z_1, z_0)$. Then the corresponding map induced by $h$ on $3^{rd}$ cohomology group with $Bbb Z$-coefficient is given by $A=pmatrix{1&0\0&1}$(as the degree of the antipodal map is $(-1)^{n+1}=1$ in this case) and that of $g$ is given by $B=pmatrix{0&1\1&0}$. So the induced map by $f$ on $3^{rd}$ cohomology group is $AB$ i.e. $pmatrix{0&1\1&0}$. But this must be wrong as then the Lefschetz number $L(f)= 1+ (-1)^3 cdot0+ (-1)^6cdot1=2neq 0$. But $f$ is a free action. So contradiction. Please help me to find my mistake.
I am learning Algebraic Topology by myself. Sorry, incase this a trivial question.
Thank you so much for your help.
algebraic-topology
algebraic-topology
asked Jan 11 at 18:16
mathstudentmathstudent
394
asked Jan 11 at 18:16
mathstudentmathstudent
394
edited Jan 13 at 8:23
mathstudent
asked Jan 11 at 18:16
mathstudentmathstudent
394
asked Jan 11 at 18:16
mathstudentmathstudent
394
asked Jan 11 at 18:16
mathstudentmathstudent
394
394
$begingroup$
Of course, you mean $hcirc g$.
$endgroup$
– Ted Shifrin
Jan 11 at 18:27
$begingroup$
The trace on $H_6$ is $-1$; the map reverses orientation.
$endgroup$
– Lord Shark the Unknown
Jan 11 at 18:29
$begingroup$
@LordSharktheUnknown How to tell a map is orientation preserving or reversing? One way to check I know is to look at the degree of the induced map on top homology. But here how should I know it is orientation reversing assuming I don't know the degree of the map?
$endgroup$
– mathstudent
Jan 13 at 8:20
add a comment |
$begingroup$
Of course, you mean $hcirc g$.
$endgroup$
– Ted Shifrin
Jan 11 at 18:27
$begingroup$
The trace on $H_6$ is $-1$; the map reverses orientation.
$endgroup$
– Lord Shark the Unknown
Jan 11 at 18:29
$begingroup$
@LordSharktheUnknown How to tell a map is orientation preserving or reversing? One way to check I know is to look at the degree of the induced map on top homology. But here how should I know it is orientation reversing assuming I don't know the degree of the map?
$endgroup$
– mathstudent
Jan 13 at 8:20
$begingroup$
Of course, you mean $hcirc g$.
$endgroup$
– Ted Shifrin
Jan 11 at 18:27
$begingroup$
Of course, you mean $hcirc g$.
$endgroup$
– Ted Shifrin
Jan 11 at 18:27
$begingroup$
The trace on $H_6$ is $-1$; the map reverses orientation.
$endgroup$
– Lord Shark the Unknown
Jan 11 at 18:29
$begingroup$
The trace on $H_6$ is $-1$; the map reverses orientation.
$endgroup$
– Lord Shark the Unknown
Jan 11 at 18:29
$begingroup$
@LordSharktheUnknown How to tell a map is orientation preserving or reversing? One way to check I know is to look at the degree of the induced map on top homology. But here how should I know it is orientation reversing assuming I don't know the degree of the map?
$endgroup$
– mathstudent
Jan 13 at 8:20
$begingroup$
@LordSharktheUnknown How to tell a map is orientation preserving or reversing? One way to check I know is to look at the degree of the induced map on top homology. But here how should I know it is orientation reversing assuming I don't know the degree of the map?
$endgroup$
– mathstudent
Jan 13 at 8:20
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The mistake is that the degree of the map is $-1$, not $1$. If $alpha$ is the generator of the top cohomology of the first factor and $beta$ that of the second, then note that $pi_1^*alphacuppi_2^*beta = (-1)^{3cdot 3}pi_2^*betacuppi_1^*alpha$.
answered Jan 11 at 18:28
Ted ShifrinTed Shifrin
63.7k44591
$endgroup$
$begingroup$
Thanks. Just to clarify myself. So on the $3^{rd}$ cohomology the induced action of $f$ is correct and given by the above matrix $AB=pmatrix{0&1\1&0}$. And on the $H^6$ it is $-1$. Please tell whether this is correct or not?
$endgroup$
– mathstudent
Jan 11 at 19:41
$begingroup$
Yes, @mathstudent.
$endgroup$
– Ted Shifrin
Jan 11 at 19:48
add a comment |
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$begingroup$
The mistake is that the degree of the map is $-1$, not $1$. If $alpha$ is the generator of the top cohomology of the first factor and $beta$ that of the second, then note that $pi_1^*alphacuppi_2^*beta = (-1)^{3cdot 3}pi_2^*betacuppi_1^*alpha$.
answered Jan 11 at 18:28
Ted ShifrinTed Shifrin
63.7k44591
$endgroup$
$begingroup$
Thanks. Just to clarify myself. So on the $3^{rd}$ cohomology the induced action of $f$ is correct and given by the above matrix $AB=pmatrix{0&1\1&0}$. And on the $H^6$ it is $-1$. Please tell whether this is correct or not?
$endgroup$
– mathstudent
Jan 11 at 19:41
$begingroup$
Yes, @mathstudent.
$endgroup$
– Ted Shifrin
Jan 11 at 19:48
add a comment |
$begingroup$
The mistake is that the degree of the map is $-1$, not $1$. If $alpha$ is the generator of the top cohomology of the first factor and $beta$ that of the second, then note that $pi_1^*alphacuppi_2^*beta = (-1)^{3cdot 3}pi_2^*betacuppi_1^*alpha$.
answered Jan 11 at 18:28
Ted ShifrinTed Shifrin
63.7k44591
$endgroup$
$begingroup$
Thanks. Just to clarify myself. So on the $3^{rd}$ cohomology the induced action of $f$ is correct and given by the above matrix $AB=pmatrix{0&1\1&0}$. And on the $H^6$ it is $-1$. Please tell whether this is correct or not?
$endgroup$
– mathstudent
Jan 11 at 19:41
$begingroup$
Yes, @mathstudent.
$endgroup$
– Ted Shifrin
Jan 11 at 19:48
add a comment |
$begingroup$
The mistake is that the degree of the map is $-1$, not $1$. If $alpha$ is the generator of the top cohomology of the first factor and $beta$ that of the second, then note that $pi_1^*alphacuppi_2^*beta = (-1)^{3cdot 3}pi_2^*betacuppi_1^*alpha$.
answered Jan 11 at 18:28
Ted ShifrinTed Shifrin
63.7k44591
$endgroup$
The mistake is that the degree of the map is $-1$, not $1$. If $alpha$ is the generator of the top cohomology of the first factor and $beta$ that of the second, then note that $pi_1^*alphacuppi_2^*beta = (-1)^{3cdot 3}pi_2^*betacuppi_1^*alpha$.
answered Jan 11 at 18:28
Ted ShifrinTed Shifrin
63.7k44591
answered Jan 11 at 18:28
Ted ShifrinTed Shifrin
63.7k44591
answered Jan 11 at 18:28
Ted ShifrinTed Shifrin
63.7k44591
answered Jan 11 at 18:28
Ted ShifrinTed Shifrin
63.7k44591
63.7k44591
$begingroup$
Thanks. Just to clarify myself. So on the $3^{rd}$ cohomology the induced action of $f$ is correct and given by the above matrix $AB=pmatrix{0&1\1&0}$. And on the $H^6$ it is $-1$. Please tell whether this is correct or not?
$endgroup$
– mathstudent
Jan 11 at 19:41
$begingroup$
Yes, @mathstudent.
$endgroup$
– Ted Shifrin
Jan 11 at 19:48
add a comment |
$begingroup$
Thanks. Just to clarify myself. So on the $3^{rd}$ cohomology the induced action of $f$ is correct and given by the above matrix $AB=pmatrix{0&1\1&0}$. And on the $H^6$ it is $-1$. Please tell whether this is correct or not?
$endgroup$
– mathstudent
Jan 11 at 19:41
$begingroup$
Yes, @mathstudent.
$endgroup$
– Ted Shifrin
Jan 11 at 19:48
$begingroup$
Thanks. Just to clarify myself. So on the $3^{rd}$ cohomology the induced action of $f$ is correct and given by the above matrix $AB=pmatrix{0&1\1&0}$. And on the $H^6$ it is $-1$. Please tell whether this is correct or not?
$endgroup$
– mathstudent
Jan 11 at 19:41
$begingroup$
Thanks. Just to clarify myself. So on the $3^{rd}$ cohomology the induced action of $f$ is correct and given by the above matrix $AB=pmatrix{0&1\1&0}$. And on the $H^6$ it is $-1$. Please tell whether this is correct or not?
$endgroup$
– mathstudent
Jan 11 at 19:41
$begingroup$
Yes, @mathstudent.
$endgroup$
– Ted Shifrin
Jan 11 at 19:48
$begingroup$
Yes, @mathstudent.
$endgroup$
– Ted Shifrin
Jan 11 at 19:48
add a comment |
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$begingroup$
Of course, you mean $hcirc g$.
$endgroup$
– Ted Shifrin
Jan 11 at 18:27
$begingroup$
The trace on $H_6$ is $-1$; the map reverses orientation.
$endgroup$
– Lord Shark the Unknown
Jan 11 at 18:29
$begingroup$
@LordSharktheUnknown How to tell a map is orientation preserving or reversing? One way to check I know is to look at the degree of the induced map on top homology. But here how should I know it is orientation reversing assuming I don't know the degree of the map?
$endgroup$
– mathstudent
Jan 13 at 8:20