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 $H$.


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).


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).

  • 2
    $\begingroup$ In other words: equip $M_n$ with operator norm and characterize its linear subspaces as normed spaces. I'm not sure there is much we can say about this; it's certainly a very special property. $\endgroup$
    – Nik Weaver
    Commented Jan 9, 2022 at 13:47
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    $\begingroup$ Any Banach space where the unit ball is a polyhedron is ok just from spaces of diagonal operators. $\endgroup$
    – Will Sawin
    Commented Jan 9, 2022 at 14:50
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    $\begingroup$ I have never seen this questioned addressed. I guess the starting point is to give examples of finite dimensional spaces that do not embed isometrically into $M_n$ for any $n$. Do you know of some? $\endgroup$ Commented Jan 9, 2022 at 18:47
  • $\begingroup$ Thanks. I think that $\ell^1_2$ (that is, $\mathbb{C}^2$ with the $\ell^1$ norm) can't be embedded in $M_n$ for any $n$. I think that I used to know how to prove it but am missing some details (morally speaking because $\ell^1_2$ has a unique operator space structure; and we have a representation of this space as the subspace of $C(S^1)$ generated by the unitaries $1$ and $z$, so $\ell^1_2$ cannot be represented as a unital operator space. This is because unitaries are "hyperrigid"... this getting too long so I'll add stuff to the question. $\endgroup$
    – Orr Shalit
    Commented Jan 9, 2022 at 19:15
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    $\begingroup$ 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. $\endgroup$
    – Terry Tao
    Commented Jan 9, 2022 at 21:01

1 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.)

  • $\begingroup$ Thanks. The fact that there are several recent papers on this problem, together with the fact that having semi-algebraic unit ball is necessary but not sufficient, gives me the feeling that this is going to be the best answer. But I'll wait a few days before accepting it. $\endgroup$
    – Orr Shalit
    Commented Jan 11, 2022 at 5:30

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