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.
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.