$S^2 times S^4$ is not homotopy equivalent to $mathbb{C}P^3$ using cohomology rings
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I am trying to show that $S^2 times S^4$ is not homotopy equivalent to $mathbb{C}P^3$ using cohomology rings.
I know that $H^*{mathbb{C}P^3} simeq mathbb{Z}[lambda]/(lambda^4)$ as a graded ring with $|lambda|=2$.
By the Künneth formula, we have
$H^*(S^2 times S^4) simeq H^2(S^2) otimes H^4(S^4) simeq mathbb{Z}[alpha]/(alpha^2) otimes mathbb{Z}[beta]/(beta^2)$ where $|alpha|=2$ and $|beta|=4$.
What I am stuck at now is showing how they have a different cup product structure. I think I can somehow use the fact that multiplication on the basis of the graded tensor product is defined by $(a otimes b)(c otimes d) = (-1)^{|b||c|}(ac otimes bd)$.
algebraic-topology homology-cohomology
asked Sep 16 '18 at 20:43
TuoTuoTuoTuo
1,776516
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add a comment |
$begingroup$
I am trying to show that $S^2 times S^4$ is not homotopy equivalent to $mathbb{C}P^3$ using cohomology rings.
I know that $H^*{mathbb{C}P^3} simeq mathbb{Z}[lambda]/(lambda^4)$ as a graded ring with $|lambda|=2$.
By the Künneth formula, we have
$H^*(S^2 times S^4) simeq H^2(S^2) otimes H^4(S^4) simeq mathbb{Z}[alpha]/(alpha^2) otimes mathbb{Z}[beta]/(beta^2)$ where $|alpha|=2$ and $|beta|=4$.
What I am stuck at now is showing how they have a different cup product structure. I think I can somehow use the fact that multiplication on the basis of the graded tensor product is defined by $(a otimes b)(c otimes d) = (-1)^{|b||c|}(ac otimes bd)$.
algebraic-topology homology-cohomology
asked Sep 16 '18 at 20:43
TuoTuoTuoTuo
1,776516
$endgroup$
$begingroup$
Can't you just notice that the generator of $H^4(Bbb CP^3)$ is the square of the generator of $H^2(Bbb CP^3)$? Is that true of $S^2times S^4$?
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– Ted Shifrin
Sep 16 '18 at 20:59
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@TedShifrin Thanks Ted, I corrected the typos. Let me think about what you said about the generators.
$endgroup$
– TuoTuo
Sep 16 '18 at 21:01
add a comment |
$begingroup$
I am trying to show that $S^2 times S^4$ is not homotopy equivalent to $mathbb{C}P^3$ using cohomology rings.
I know that $H^*{mathbb{C}P^3} simeq mathbb{Z}[lambda]/(lambda^4)$ as a graded ring with $|lambda|=2$.
By the Künneth formula, we have
$H^*(S^2 times S^4) simeq H^2(S^2) otimes H^4(S^4) simeq mathbb{Z}[alpha]/(alpha^2) otimes mathbb{Z}[beta]/(beta^2)$ where $|alpha|=2$ and $|beta|=4$.
What I am stuck at now is showing how they have a different cup product structure. I think I can somehow use the fact that multiplication on the basis of the graded tensor product is defined by $(a otimes b)(c otimes d) = (-1)^{|b||c|}(ac otimes bd)$.
algebraic-topology homology-cohomology
asked Sep 16 '18 at 20:43
TuoTuoTuoTuo
1,776516
$endgroup$
I am trying to show that $S^2 times S^4$ is not homotopy equivalent to $mathbb{C}P^3$ using cohomology rings.
I know that $H^*{mathbb{C}P^3} simeq mathbb{Z}[lambda]/(lambda^4)$ as a graded ring with $|lambda|=2$.
By the Künneth formula, we have
$H^*(S^2 times S^4) simeq H^2(S^2) otimes H^4(S^4) simeq mathbb{Z}[alpha]/(alpha^2) otimes mathbb{Z}[beta]/(beta^2)$ where $|alpha|=2$ and $|beta|=4$.
What I am stuck at now is showing how they have a different cup product structure. I think I can somehow use the fact that multiplication on the basis of the graded tensor product is defined by $(a otimes b)(c otimes d) = (-1)^{|b||c|}(ac otimes bd)$.
algebraic-topology homology-cohomology
algebraic-topology homology-cohomology
asked Sep 16 '18 at 20:43
TuoTuoTuoTuo
1,776516
asked Sep 16 '18 at 20:43
TuoTuoTuoTuo
1,776516
edited Sep 16 '18 at 21:00
TuoTuo
asked Sep 16 '18 at 20:43
TuoTuoTuoTuo
1,776516
asked Sep 16 '18 at 20:43
TuoTuoTuoTuo
1,776516
asked Sep 16 '18 at 20:43
TuoTuoTuoTuo
1,776516
1,776516
$begingroup$
Can't you just notice that the generator of $H^4(Bbb CP^3)$ is the square of the generator of $H^2(Bbb CP^3)$? Is that true of $S^2times S^4$?
$endgroup$
– Ted Shifrin
Sep 16 '18 at 20:59
$begingroup$
@TedShifrin Thanks Ted, I corrected the typos. Let me think about what you said about the generators.
$endgroup$
– TuoTuo
Sep 16 '18 at 21:01
add a comment |
$begingroup$
Can't you just notice that the generator of $H^4(Bbb CP^3)$ is the square of the generator of $H^2(Bbb CP^3)$? Is that true of $S^2times S^4$?
$endgroup$
– Ted Shifrin
Sep 16 '18 at 20:59
$begingroup$
@TedShifrin Thanks Ted, I corrected the typos. Let me think about what you said about the generators.
$endgroup$
– TuoTuo
Sep 16 '18 at 21:01
$begingroup$
Can't you just notice that the generator of $H^4(Bbb CP^3)$ is the square of the generator of $H^2(Bbb CP^3)$? Is that true of $S^2times S^4$?
$endgroup$
– Ted Shifrin
Sep 16 '18 at 20:59
$begingroup$
Can't you just notice that the generator of $H^4(Bbb CP^3)$ is the square of the generator of $H^2(Bbb CP^3)$? Is that true of $S^2times S^4$?
$endgroup$
– Ted Shifrin
Sep 16 '18 at 20:59
$begingroup$
@TedShifrin Thanks Ted, I corrected the typos. Let me think about what you said about the generators.
$endgroup$
– TuoTuo
Sep 16 '18 at 21:01
$begingroup$
@TedShifrin Thanks Ted, I corrected the typos. Let me think about what you said about the generators.
$endgroup$
– TuoTuo
Sep 16 '18 at 21:01
add a comment |
1 Answer
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You should first decipher the isomorphism $$H^*(S^2 times S^4) cong mathbb{Z}[alpha, beta]/(alpha^2, beta^2, (alpha cup beta)^2)$$ where $|alpha|=2$ and $|beta|=4$. Thus, if you take a generator of $H^*(S^2 times S^4)$, then its cube is clearly zero.
But in the cohomology ring $H^*(mathbb C P^3)$ it is not the case.
answered Dec 15 '18 at 11:43
M. Giovanni LucarettiM. Giovanni Lucaretti
797
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add a comment |
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You should first decipher the isomorphism $$H^*(S^2 times S^4) cong mathbb{Z}[alpha, beta]/(alpha^2, beta^2, (alpha cup beta)^2)$$ where $|alpha|=2$ and $|beta|=4$. Thus, if you take a generator of $H^*(S^2 times S^4)$, then its cube is clearly zero.
But in the cohomology ring $H^*(mathbb C P^3)$ it is not the case.
answered Dec 15 '18 at 11:43
M. Giovanni LucarettiM. Giovanni Lucaretti
797
$endgroup$
add a comment |
$begingroup$
You should first decipher the isomorphism $$H^*(S^2 times S^4) cong mathbb{Z}[alpha, beta]/(alpha^2, beta^2, (alpha cup beta)^2)$$ where $|alpha|=2$ and $|beta|=4$. Thus, if you take a generator of $H^*(S^2 times S^4)$, then its cube is clearly zero.
But in the cohomology ring $H^*(mathbb C P^3)$ it is not the case.
answered Dec 15 '18 at 11:43
M. Giovanni LucarettiM. Giovanni Lucaretti
797
$endgroup$
add a comment |
$begingroup$
You should first decipher the isomorphism $$H^*(S^2 times S^4) cong mathbb{Z}[alpha, beta]/(alpha^2, beta^2, (alpha cup beta)^2)$$ where $|alpha|=2$ and $|beta|=4$. Thus, if you take a generator of $H^*(S^2 times S^4)$, then its cube is clearly zero.
But in the cohomology ring $H^*(mathbb C P^3)$ it is not the case.
answered Dec 15 '18 at 11:43
M. Giovanni LucarettiM. Giovanni Lucaretti
797
$endgroup$
You should first decipher the isomorphism $$H^*(S^2 times S^4) cong mathbb{Z}[alpha, beta]/(alpha^2, beta^2, (alpha cup beta)^2)$$ where $|alpha|=2$ and $|beta|=4$. Thus, if you take a generator of $H^*(S^2 times S^4)$, then its cube is clearly zero.
But in the cohomology ring $H^*(mathbb C P^3)$ it is not the case.
answered Dec 15 '18 at 11:43
M. Giovanni LucarettiM. Giovanni Lucaretti
797
edited Dec 15 '18 at 14:16
answered Dec 15 '18 at 11:43
M. Giovanni LucarettiM. Giovanni Lucaretti
797
answered Dec 15 '18 at 11:43
M. Giovanni LucarettiM. Giovanni Lucaretti
797
answered Dec 15 '18 at 11:43
M. Giovanni LucarettiM. Giovanni Lucaretti
797
797
add a comment |
add a comment |
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Can't you just notice that the generator of $H^4(Bbb CP^3)$ is the square of the generator of $H^2(Bbb CP^3)$? Is that true of $S^2times S^4$?
$endgroup$
– Ted Shifrin
Sep 16 '18 at 20:59
$begingroup$
@TedShifrin Thanks Ted, I corrected the typos. Let me think about what you said about the generators.
$endgroup$
– TuoTuo
Sep 16 '18 at 21:01