**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](https://arxiv.org/pdf/1911.00241.pdf) by Samya Kumar Ray showing that $\ell^p\_n$ cannot be embedded isometrically as compact operators. That papers contains a reference to a [paper](https://arxiv.org/pdf/1605.03769.pdf) 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).



Which finite dimensional Banach spaces can be represented isometrically as spaces of bounded operators on a finite dimensional Hilbert space?
In [Ray - On Isometric Embedding $\ell\_p^m \to S\_\infty^n$ and Unique operator space structure](https://arxiv.org/abs/1911.00241) 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.)


