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Let's consider closed simply connected manifold $M^n$ and a $a\in H^k(M)$ and $a*\in H^{n-k}(M)$ is the dual to $a$.

Is it true that dual to $a$ is realisable as a immersed sphere or $ a*=bc $ for some $b,c\in H^*(M)$ ?

Edit: it is more natural to ask about possibility to decompose dual to $a$ as a product, see example in the answer below.

Edit2: let's assume that $M$ is not decompasable, so there is no $X,Y$ such that $M = X\times Y$

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Maybe I'm missing something easy here: why would you expect the statement to be true? –  Marco Golla Oct 22 '11 at 15:48
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What if your $M=X\times Y$ of dimension $i+j$. and take $a$ to be the generator of $H^i(X)$. This may not be realisable by an immersed sphere. On the other hand, the dual of $a$ will be the generator of $H^j(Y)$ which needn't split! –  Somnath Basu Oct 22 '11 at 17:04
    
to Somnath Basu: you are right, thank you, I change my question once more, sorry) –  Nikita Kalinin Oct 22 '11 at 17:53
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I no longer understand what you are trying to ask. now you are looking at Poincare duality betwwen cohomology and cohomology (so presumably with torsion free coefficients?). what does it have to do with immersed spheres then? are you trying to ask if the chomology ring is spherically generated? such question does make sense once stated appropriately. it has nothing to do with poincare duality though so maybe that's not what you want. –  Vitali Kapovitch Oct 23 '11 at 2:10
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to Vitali Kapovitch: the Sergey's answer below is exactly about what I mean. Sorry for ununderstandable question. –  Nikita Kalinin Oct 23 '11 at 11:31
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2 Answers 2

up vote 4 down vote accepted

I will construct a closed simply-connected $8$-manifold $M$ and an $a\in H^3(M;\Bbb Z)$ such that the Poincare dual $b$ of $a$ is not realizable by a map $S^5\to M$, and a Hom-dual element in $H^5(M;\Bbb Z/2)$ to the $\bmod 2$ reduction of $b$ is not a nontrivial product.

Let $K$ be the suspension over $\Bbb C P^2$. Then there is an $\alpha\in H^3(K;\Bbb Z/2)$ with $Sq^2(\alpha)\ne 0$, but the $\bmod 2$ cohomology ring of $K$ is trivial. Let $N$ be a regular neighborhood of a PL copy of $K$ in some $\Bbb R^m$. So $N$ is homotopy equivalent to $K$. A loop in $\partial N$ bounds a disk in $N$, which can be pushed off $K$ as long as $5+2\le m-1$. Thus $\partial N$ is simply-connected. Let $M$ be the double of $N$, i.e. $M=\partial (N\times I)$. So $M$ is a closed $m$-manifold, it is simply-connected by Seifert-van Kampen, and the inclusion $N\subset M$ is split by the projection $\phi:M\subset N\times I\to N$. So the cohomology of $N\simeq K$ is a direct summand in the cohomology of $M$. Let $\beta=\phi^\ast(\alpha)$, then $\gamma:=Sq^2\beta\ne 0$, and $\gamma$ is not a nontrivial product. Since the nonzero element of $H^5(S^5;\Bbb Z/2)$ is not in the image of $Sq^2$, there is no map $f:S^5\to M$ such that $f^*(\gamma)\ne 0$.

Let $b\in H_5(M;\Bbb Z)$ be such that $\gamma(b)\ne 0$, and let $a\in H^{m-5}(M;\Bbb Z)$ be the Poincare dual of $b$. If $b$ is realized by an immersion, or just a map, $f:S^5\to M$, then $0\ne\gamma\smallfrown f_\ast[S^5]=f_\ast(f^\ast(\gamma)\smallfrown[S^5])$, contradicting $f^*(\gamma)=0$.

As for $m$, $\Bbb C P^2$ is the mapping cone of the Hopf map $h:S^3\to S^2$. The mapping cylinder of $h$ embeds in $S^2*S^3=S^6$, so $\Bbb CP^2$ embeds in $\Bbb R^7$ and $K$ in $\Bbb R^8$.

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thank you for detailed answer! –  Nikita Kalinin Oct 22 '11 at 16:24
    
Sorry, it is not my question. It is more natural to ask about decomposition as a product of dual to $a$, see edit. –  Nikita Kalinin Oct 22 '11 at 16:36
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The above example also works rationally, because $H^*(K)$ is torsion-free. A simpler example, which does not work rationally: take $K$ to be the suspension over $\Bbb R P^2$. Then the same construction (using $Sq^1$, i.e. the Bockstein homomorphism) produces a simply-connected closed 5-manifold $M$ and an $a\in H^2(M)$ with similar properties. –  Sergey Melikhov Oct 23 '11 at 1:51
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Hi Sergey, nice answer! One question - why do you need to take the double of $N$, and not just $\partial N$, for your argument? It seems to me that if $m$ is large enough the cohomology of $N$ injects into that of $\partial N$ by Lefschetz duality. Or do you really need it to be a direct summand? –  Mark Grant Oct 23 '11 at 7:27
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Hi Mark, thanks. I now see: a cocycle with support in the interior of $N$ is a coboundary in $\Bbb R^m$ and hence in $N$. So $\partial N$ gives a $7$-dimensional rational example (now using that $\partial N$ is simply-connected, which was not really used in the above argument). If we start from $K=\Sigma(\Bbb R P^2)$, this does not help to lower the dimension because we would have to embed this $K$ in $\Bbb R^6$ to make sure that $\partial N$ is simply-connected. –  Sergey Melikhov Oct 23 '11 at 20:04
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Consider $M=SU(4)$. Rationally, the cohomology ring $H^\ast(M;\mathbb{Q})$ is the product $H^\ast(S^3;\mathbb{Q})\otimes H^\ast(S^5;\mathbb{Q})\otimes H^\ast(S^7;\mathbb{Q})$. Consider $a$ to the product of the generators of $S^3$ and $S^5$. The dual is the generator of $S^7$ which doesn't split. However, $\pi_8(SU(4))=\mathbb{Z}_{4!}$ which is torsion and hence a sphere cannot possibly generate $a$.

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