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First, write $M^n \sim M^n$ cobordant, if $M^n$ # $M^n = \partial W^{n+1}$. Where # represent connected sum. Then define

$ \Omega^o_n = \{ \textrm{closed manifolds} \} / \sim$

From $M^n$ # $S^n$ = $M^n$,

This gives \begin{align} \Omega^o_1 = 0, \quad \Omega^o_2 = 0, \quad \Omega^o_3 = 0, \quad \Omega^o_4 \neq 0 \end{align} what is the meaning of $0$ (trivial group) for $\Omega^o_n$?

This topic was coverd in the class of differential topology. professor says that each $0$ meaning special. I want to know what that means.

phy_math
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  • I don't understand the question. Do you want to know the meaning of $\Omega_n^{O} = 0$ for some $n$ ? Do you want to know which manifold is the zero element is this group ? Do you want to know why there is a $O$ exponent in the notation? (which is a letter O, by the way, not a zero) – PseudoNeo May 18 '15 at 07:58
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    I know exponent is letter O, which states orthogonal. I want to know the meaning of $\Omega_n^O=0$ for some $n$. PseudoNeo – phy_math May 18 '15 at 07:59
  • Prof. says that historically, $\Omega^o_3=0$ was proven by Lickorish, Wallece. and he mention that this is related with $3$-manifold – phy_math May 18 '15 at 08:05
  • $\Omega_n^0 = 0$ implies that every compact oriented $n$-manifold bounds. – ljh8372 May 18 '15 at 08:08

1 Answers1

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Just look at the definition. $\Omega_n^O = 0$ means that every $n$-dimensional oriented compact manifold (without boundary) is the boundary of an $(n+1)$-dimensional oriented compact manifold ($0$ is the class of the empty set, so saying that every manifold is in the $0$ class means that every manifold is cobordant to the empty set, which exactly means that the manifold is the boundary of something).

For example, you know that every $1$-manifold is a union of circles. But each circle is the boundary of a disc, so they are zero in $\Omega_1^O$.

For $\Omega_2^O$, you can also use the fact that we know all the orientable $2$-manifolds (= surfaces) and we can prove by hand they are all boundaries of 3-manifolds (if you think of the genus-$g$ surface as sitting in $\mathbb R^3$, it is quite clear that it bounds something...)

For higher dimensions, the result is necessarily less elementary (the next easiest thing is probably the fact that the $\Omega^O_{4n} \neq 0$ for all $n$, which comes from a great invariant: the signature of a manifold) and the full story (the computation of $\Omega_n^O$) is one of the remarkable achievements of 20th-century topology, a tour de force due to René Thom, John Milnor, Sergeï Novikov and C.T.C. Wall.

PseudoNeo
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  • Thanks, PseudoNeo, Your answer is exactly the same context that professor told in the classroom! (which i remind this from your answer!) Wow great! – phy_math May 18 '15 at 08:15