Which finite dimensional Banach spaces can be represented isometrically as spaces of bounded operators on a finite dimensional Hilbert space?

Question

Background:

It is known that every Banach space $X$ can be embedded isometrically as a subspace in the space $C(K)$ of continuous functions on a compact Hausdorff space $K$. Indeed, one can take $K$ equal to the closed unit ball in the dual $X^*$, and embed $X$ in $C((X^*)_1)$ using the duality mapping $x \mapsto \hat{x}\big|_{(X^*)_1} \in C((X^*)_1)$, where $\hat{x} \in X^{**}$ is given by $\hat{x}(f) = f(x)$, for $f \in X^*$. By the Gelfand-Naimark theorem we know that $C(K)$ can be faithfully represented as a sub-C*-algebra of the algebra $B(H)$ of bounded linear operators on some Hilbert space $H$, and so we conclude:

Every Banach space can be represented isometrically as a closed subspace of $B(H)$ for some Hilbert space $H$.

Question:

Which (finite dimensional) Banach spaces can be represented isometrically as closed subspaces of $B(H)$ for a finite dimensional Hilbert space $H$?

A related questions is: do you know any work related to this question?

I usually have spaces of the complex numbers in mind, but I'll also be happy to hear about real spaces.

Why did I ask myself this question?

Nothing fancy. I was giving a lecture about operator systems to undergraduates and I started from operator spaces, but since they didn't all take a course in functional analysis I worked mostly in $M_n$. I made a point that every banach space is an operator space in $B(H)$ for some $H$, and showed them how $\ell^2_2$ and $\ell^\infty_2$ (the space $\mathbb{C}^2$ with the $\ell^2$ and $\ell^\infty$ norms, respectively) can be isometrically embedded in $M_2$. My older notes said that $\ell^1_2$ cannot be embedded in $M_n$ for finite $n$ (I forgot the complete details of the proof but I'll write the idea below). Speaking about this I got curious about which finite dimensional spaces can be embedded in $M_n$ for finite $n$.

The simplest question is:

Can $\ell^1_2$ be embedded in $M_n$?

For this simple question I think that the answer is: no. The reason I think this is that $\ell^1_2$ has a unique operator space structure. We have an isometric representation as the operator subspace of the C-algebra $C(S^1)$ generated by the unitaries $1$ and $z$. Since unitaries are hyperrigid, any unital isometric map from $span\{1,z\}$ onto a subspace of $M_n$ would extend to $*$-isomorphism between the generated C-algebras, and this is impossible. Thus, $\ell^1_2$ is not a unital operator space. I haven't worked much in the nonunital setting, I think that in this particular case the technical difference between unital and nonunital won't change the end result. But then I got stuck (Gupta and Reza's linked paper below gives an elementary proof that $\ell^1_1$ cannot be embedded in $M_n$, and one of the steps is a nice trick that shows how indeed you can reduce to the case of unital operator space).

Update:

Bunyamin Sari, in a comment to Bill Johnson's answer, put a reference to a paper by Samya Kumar Ray showing that $\ell^p_n$ cannot be embedded isometrically as compact operators. That papers contains a reference to a paper by Gupta and Reza that provides an elementary proof that $\ell^1_2$ cannot be embedded in $M_n$. It follows, of course, that no $\ell^1_m$ can be embedded in $M_n$.

Terry Tao commented below: "In order to embed for a finite dimensional Banach space to embed isometrically into some $M_n$ it is necessary that the unit ball be a semi-algebraic set."

To see this, note that the unit ball of an operator space is the intersection of a linear subspace with the unit ball of $M_n$, which can be given as the set of matrices satisfying the matrix inequality $A^*A \leq I$. However, having a semi-algebraic unit ball is not a sufficient condition for embeddability as an operator subspace of $M_n$; for example consider $\ell^p_2$ for $p=4$ and see the paper by Ray above.

(Thanks to Guy Salomon and Eli Shamovich for some offline discussions).

Answer 1

In Ray - On Isometric Embedding $\ell_p^m \to S_\infty^n$ and Unique operator space structure it is shown that for $p\in (2,\infty)\cup \{1\}$ and $n\ge 2$, $\ell_p^n$ does not isometrically embed into the space of compact operators on $\ell_2$.

(This was previously in the comments, now deleted for housekeeping.)

Answer 2

Proposition: A real Banach space $X$ is isometrically isomorphic to a real subspace of some $M_n(\mathbb{C})$ if and only if the unit ball $X_1$ is a spectrahedron.

This is not particularly deep and almost surely already known to the OP. Nevertheless, I think that it's relevant to the question since it brings the abundant literature on spectrahedra to bear on the problem. For example, using the negative resolution of the Helton-Nie conjecture, it should be possible to give examples of finite-dimensional real Banach spaces $X$ with semialgebraic unit ball such that $X$ is not even a quotient of any finite-dimensionally representable Banach space.

The proposition is not particularly deep because the very definition of spectrahdron is in terms of a similar notion of embeddability: with $\mathsf{Her}_n \subseteq M_n(\mathbb{C})$ denoting the subspace of hermitian matrices and $\mathsf{PSD}_n \subseteq \mathsf{Her}_n$ the positive semidefinite cone, a set $S \subseteq \mathbb{R}^m$ is called a spectrahedron if and only if there is an affine map $f : \mathbb{R}^m \to \mathsf{Her}_n$ such that $S = f^{-1}(\mathsf{PSD}_n)$.

Proof of the proposition: if $g : X \to M_n(\mathbb{C})$ is an isometric embedding, then $$f(x) := \begin{pmatrix} 1 & g(x) \\\\ g(x)^* & 1 \end{pmatrix}$$ is positive semidefinite if and only if $\|g(x)\| \le 1$, and thus $f$ realizes the unit ball of $X$ as a spectrahedron.

Conversely, if the unit ball of $X$ is a spectrahedron realized by $f : \mathbb{R}^m \to \mathsf{Her}_n$, then the support of $f(x)$ is maximal for $x = 0$ since the origin is in the interior of the unit ball. Multiplying by the square root of the pseudoinverse of $f(0)$ from both sides and restricting to the support achieves $f(0) = 1_n$. Then we have $\|f(x)\| \le 2$ for all $x \in X_1$, since otherwise $f(-x) = 2\cdot 1_n - f(x)$ would not be positive semidefinite. Thus defining $$g(x) := 1_n - f(x)$$ makes $g$ linear and such that $g(x)$ has spectrum in $[-1,1]$ if and only if $x \in X_1$, which is as required. $\square$