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### Dual of the Ultraproduct of a Banach Space

Suppose we have a Banach space ultraproduct $(E_i)_U$. I can think of three natural objects which are "dual-like":
$(E_i^*)_U$, the ultraproduct of the duals of the ground spaces.
The space made up ...

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### Double ultrapower of the hyperfinite $II_1$-factor

Let $\omega$ be a free ultrafilter on the natural numbers and $R$ be the hyperfinite $II_1$-factor (the definition of $R$ is recalled in the comments).
Question: Does there exist another free ...

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### Ultrapowers and complemented subspaces

Let $Y$ be a closed subspace of a Banach space $X$, and let $\mathcal{U}$ be a nontrivial ultrafilter on the set $\mathbb{N}$ of all integer numbers.
It is not difficult to see that if $Y$ is ...

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### Which compact topological spaces are homeomorphic to their ultrapower?

It is well known that for any compact metric space $(X, d)$, and any ultrafilter $\mu$ there is a map $i_\mu:\prod_\mu (X, d) \to (X_d)$ in the category of metric spaces and Lipschitz maps where ...

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### Ultraproducts and subobjects of projectives

Usually the question whether the class of projective algebras in a given variety is closed under taking subalgebras seems to be quite hard. In varieties with well understood dual geometry (e. g. ...

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### What is known about the krull dimension of an ultrapower ring?

Let $R$ be a ring, $F$ a free ultrafilter on a set $X$ which is not countably complete, and $R_F$ the ultrapower of $R$ with $R \not\cong R_F$. The following two results are from a masters thesis ...

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66 views

### Reduced products of (abelian and triangulated) categories: references?

For a filter $U$ on a set $X$ and for a family of categories $C_x$ indexed by $X$ I would like to consider the (corresponding categorical version of) reduced product of $C_x$ (for $x\in X$) with ...

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### Ultrapowers of ultrapowers

Suppose that you have some structure $S$, and you want to construct an ultrapower of cardinality $\kappa$ to obtain $S^*_\kappa$. Then, say you want to construct a new ultrapower from $S^*_\kappa$, ...