1.What is the definition of hyperfinite II$_1$-factor? Can anyone show me concrete examples?
2.If $R$ is a hyperfinite II$_1$-factor ,we can define the ultraproduct of $R^{\omega}=l^\infty(R)/I_{\omega}$,where $\omega $ is any free ultrafilter.Is $I_{\omega}$ nuclear?
A II$_1$ factor is an infinite-dimensional von Neumann algebra that has trivial centre and a faithful trace. It is hyperfinite (or AFD, or amenable) if it has a (sot, wot, $\sigma$-wot, $\sigma$-sot, etc.) dense subalgebra that is an increasing union of finite-dimensional C$^*$-algebras.
There is a unique hyperfinite II$_1$, factor, often denoted by $R$. Common ways to construct it are, among others,
As a group algebra, $$ (\mathbb C \, \mathbb S_\infty)''\subset B(H_\lambda), $$ where $\lambda:S_\infty\to B(H_\infty)$ is the left regular representation and $S_\infty$ is the group of finite permutations of $\mathbb N$.
As $\overline{\bigotimes_{n\in\mathbb N} M_2(\mathbb C)}^{sot}\subset B(H)$, where $H$ comes from the GNS representation of the trace.
$\overline{{\rm UHF}(2^\infty)}^{sot}\subset B(H)$, where $H$ comes from the GNS representation of the trace.
$L^\infty[0,1]\rtimes \mathbb R$, where $\mathbb R$ acts by translation, i.e. $s\cdot f=f(s\dotplus\cdot)$, where $\dotplus$ is addition modulo $1$.
As for $I_\omega$, it is non-separable as a C$^*$-algebra (it is an ideal in a von Neumann algebra), so asking about its nuclearity is weird. I would say that it is not nuclear (because I would expect its double dual to not be semidiscrete), but I'm not entirely sure how to argue about it.