4
$\begingroup$

I'm stuck to prove that the singular cohomology groups with coefficients in $\mathbb{Z}$, $H^*(S^1\times S^3;\mathbb{Z})$ and $H^*(S^1\vee S^3 \vee S^4;\mathbb{Z})$, are not isomorphic as rings.

What I've done so far: First I calculated the cohomology groups:

$H^n(S^1\vee S^3 \vee S^4)\cong H^n(S^1\times S^3)\cong\begin{cases} \mathbb{Z}, & \text{if }n=0,1,3,4\\ 0, & \text{else. } \end{cases}$

I used that $H^n(S^1\times X)\cong H^n(X)\oplus H^{n-1}(X)$, the isomorphismus $$f:H^n(X)\oplus H^{n-1}(X)\to H^n(S^1\times X)$$is given by $$(\eta_1,\eta_2)\mapsto (\pi_X^*(\eta_1)+\xi \times \eta_2),$$ with $\xi\in H^1(S^1)$ a generator and $\pi_X:S^1\times X\to X$ is the projection. Furthermore, I used that for the reduced singular cohomology it is : $H_{red}^n(S^1\vee S^3 \vee S^4)\cong H_{red}^n(S^1)\oplus H_{red}^n(S^3)\oplus H_{red}^n(S^4)$ for all $n\ge 0$.

Then we defined the cup-product in singular cohomology $$\cup: H^p(X,A;R)\otimes H^q(X,B;R)\to H^{p+q}(X,A\cup B;R)$$ by $$\cup([\alpha],[\beta]):=[\alpha\cup\beta].$$ See here cup-product for singular homology? for the definition.

The cup-product is the ring-multiplication of the cohomology groups, in our cases is $A=B=\emptyset$. Next I considered

  1. $\cup: H^1(S^1\times S^3)\otimes H^3(S^1\times S^3)\rightarrow H^4(S^1\times S^3)$ and we can identify: $H^1(S^1\times S^3)\cong H^0(S^3)$, $H^3(S^1\times S^3)\cong H^3(S^3)$, $H^4(S^1\times S^3)\cong H^3(S^3)$.

  2. $\cup: H^1(S^1\vee S^3 \vee S^4)\otimes H^3(S^1\vee S^3 \vee S^4)\rightarrow H^4(S^1\vee S^3 \vee S^4)$ and we can identify: $H^1(S^1\vee S^3 \vee S^4)\cong H^1(S^1)$, $H^3(S^1\vee S^3 \vee S^4)\cong H^3(S^3)$ and $H^4(S^1\vee S^3 \vee S^4)\cong H^4(S^4)$.

But now I'm stuck. I think I have to choose generators of the cohomology groups, multiply them and find out what the difference in both rings is. But I didn't know how to do this exactly, how to choose generators an so on (because I don't know, what the generators are). Could you help me to complete the proof? Best, mrs.mop

Edit: Meanwhile I considered the cup product on $H^*(S^n)$ and for the generators $1\in H^0(S^n)$, $x\in H^n(S^n)$ it is $1\cup x=1$, $x\cup 1=x$ and $x\cup x=0$. Therefore the product of the two nonzero generators 1 and x under $\cup: H^1(S^1\times S^3)\otimes H^3(S^1\times S^3)\rightarrow H^4(S^1\times S^3)$ is nonzero.

If you consider $\cup: H^1(S^1\vee S^3 \vee S^4)\otimes H^3(S^1\vee S^3 \vee S^4)\rightarrow H^4(S^1\vee S^3 \vee S^4)$, in the answer is claimed that the product of two generators is zero (I don't know how to prove this).

Can I conclude that the cohomologys are not isomorphic as rings?

|cite|improve this question
$\endgroup$
  • $\begingroup$ Do you know Kunneth formula?? $\endgroup$ – Anubhav Mukherjee Nov 7 '15 at 12:57
  • $\begingroup$ no, I don't know this formula from lecture. (But it seems we will discuss the theorem in lecture in the next few weeks) $\endgroup$ – user197416 Nov 7 '15 at 13:04
  • $\begingroup$ What do you know about the cup product? Just the definition? $\endgroup$ – user99914 Nov 7 '15 at 13:24
  • $\begingroup$ I know the properties (acssociative property, distributivity, the "commutativity"..), the cross product in singular cohomology (we had the the same definition as you can see here math.stackexchange.com/questions/759969/… ), which is in strong relation with the cup-product I think. That's all. $\endgroup$ – user197416 Nov 7 '15 at 13:36
0
$\begingroup$

I am assuming cohomology with $\mathbb{Z}$ coefficient. If $\alpha \in H^1(S^1\times S^3), \beta \in H^3(S^1\times S^3)$ be two generators , then by kunneth formula, $\alpha \cup \beta \in H^4(S^1\times S^3)$ is a generator.

On the other hand if $a\in H^1(S^1\vee S^3\vee S^4)$ and $b\in H^3(S^1\vee S^3\vee S^4)$ be two generators, then $a\cup b$ is zero.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ thanks. how to choose the generators in the special case above? does then the multiplication on $H^*(S^1\vee S^3\vee S^4)$ send everything to zero ? $\endgroup$ – user197416 Nov 7 '15 at 17:53
  • $\begingroup$ I'm not sure what you mean or what exactly to do, sorry. For $H^*(S^n)$ we have the cup products $1\cup x=x$, $x\cup 1=x$, $x\cup x=0$, with $1$ is a generator of $H^0(S^n)$ and $x$ is a generator of $H^n(S^n)$.But I still don't understand why $a\cup b$ is zero for 2. Could you elaborate your answer a littlebit? $\endgroup$ – user197416 Nov 7 '15 at 22:09
  • $\begingroup$ consider the inclusion map $S^1\vee S^3 \to S^1\vee S^3\vee S^4$, then this map induced an isomorphism in 1st and 3rd co-homology group... now for $S^1\vee S^3$ $H^4$ is zero, so $a\cup b$ is zero. $\endgroup$ – Anubhav Mukherjee Nov 8 '15 at 8:09
  • $\begingroup$ oh, thank you very much! $\endgroup$ – user197416 Nov 8 '15 at 8:10
  • $\begingroup$ And sorry for late reply :) $\endgroup$ – Anubhav Mukherjee Nov 8 '15 at 8:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy