4

Let $(H_n)_n$ be a sequence of separable Hilbert spaces and $\omega$ be a free ultrafilter. Is it true that $\Pi_{\omega}H_n$ is also separable? Note that $\Pi_{\omega}H_n$ is the quotient space $\mathcal{l}^{\infty}(\mathbb{N},H_n)/I$, where $I=\{(x_n)_n\in\mathcal{l}^{\infty}(\mathbb{N},H_n):\ \lim_{n\rightarrow\omega}\|x_n\|=0 \}.$ If we take each $H_n=\mathbb{C}$, Then $\mathcal{l}^{\infty}(\mathbb{N},H_n)=\mathcal{l}^{\infty}(\mathbb{N})$, which is non separable. What about $I= \{(x_n)_n\in\mathcal{l}^{\infty}(\mathbb{N}):\ \lim_{n\rightarrow\omega}|x_n|=0 \}$. Is it separable?

David C. Ullrich
  • 89,985
Rajeev ranjan
  • 83
  • Yes, it is a Hilbert space. The inner product on $\Pi_{\omega}H_n$ is given by $\left\langle (x_n)n,(y_n)_n \right\rangle=\lim{n\rightarrow \omega}\left\langle x_n,y_n \right\rangle$. – Rajeev ranjan Aug 06 '21 at 05:21
  • My bad. $ \ \ $ – Martin Argerami Aug 06 '21 at 05:34
  • 1
    $\ell^\infty(\mathbb N)\cong C(\beta\mathbb N)$ and $\omega$ corresponds to a point $x\in\beta\mathbb N\setminus \mathbb N$ so that $I\cong {f\in C(\beta\mathbb N\mid f(x)=0}$ under the isomorphism above. But this is a maximal ideal, hence $\ell^\infty(\mathbb N)/I\cong \mathbb C$. – MaoWao Aug 06 '21 at 09:32

1 Answers1

2

I will write $(\xi_n)^\bullet$ for the equivalence class of the sequence $(\xi_n)$ in the ultraproduct. I will also write $H^\omega$ for the ultraproduct $\prod_{n\to\omega}H$.

Since we can embed every separable Hilbert space into $\ell_2$, the question is really if $\ell_2^\omega$ is separable. To show this, let $(e_k)$ denote the canonical ONB of $\ell_2$ and let $$ T\colon \ell_\infty(\ell_2)\to\bigoplus_{k=1}^\infty\mathbb C^\omega,\,(\xi_n)\mapsto ((\langle \xi_n,e_1\rangle)^\bullet,(\langle \xi_n,e_2\rangle)^\bullet,\dots). $$ By a Fatou-type argument, $$ \lVert T((\xi_n))\rVert^2=\sum_{k=1}^\infty\lim_{n\to\omega}\lvert\langle \xi_n,e_k\rangle\rvert^2\leq \lim_{n\to\omega}\sum_{k=1}^\infty \lvert \langle \xi_n,e_k\rangle\rvert^2=\lim_{n\to\omega}\lVert \xi_n\rVert^2=\lVert (\xi_n)^\bullet\rVert^2. $$ Thus $T$ can be decomposed as $$ \ell_\infty(\ell_2)\overset{q}{\to}\ell_2^\omega\overset{\iota}\to \bigoplus_{k=1}^\infty\mathbb C^\omega, $$ where $q$ is the canonical projection and $\iota$ is a contraction.

As discussed in the comments, $\ell_\infty=C_b(\mathbb N)\cong C(\beta \mathbb N)$ and $I$ is a maximal ideal in $C(\beta \mathbb N)$ under this identification, so that $\mathbb C^\omega=\ell_\infty/I\cong \mathbb C$. Thus $\ell_2^\omega$ embeds into a separable Hilbert space, which means that it itself must be separable. In fact, since $\ell_2\hookrightarrow \ell_2^\omega$ as constant sequences, it follows that $\ell_2^\omega\cong\ell_2$.

MaoWao
  • 15,139
  • 1
  • 18
  • 36
  • Just a comment: the Fatou type argument appearing is a direct application of Fatou's lemma w.r.t integration over the counting measure of $\mathbb{N}$, i.e. summation. Nice answer! – Just dropped in Nov 22 '21 at 20:32
  • @JustDroppedIn Well, more or less. One has to be a little careful because these are not limits of sequences, but really limits of nets. – MaoWao Nov 22 '21 at 20:35
  • Oh-oh I didn't notice that; isn't this going to be a problem? related: https://math.stackexchange.com/questions/141198/a-net-version-of-dominated-convergence – Just dropped in Nov 22 '21 at 20:38
  • @JustDroppedIn For general measure spaces this is indeed a problem. But in this case it's ok: For every finite $N$ we have $\sum_{n=1}^N\lim_{n\to\omega}=\lim_{n\to\omega}\sum_{n=1}^N\leq \lim_{n\to\omega}\sum_{n=1}^\infty$, where we already know that the last limit exists. Then you can let $N\to\infty$ on the left side. – MaoWao Nov 22 '21 at 20:42