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

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).

• 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. Commented Jan 9, 2022 at 13:47
• Any Banach space where the unit ball is a polyhedron is ok just from spaces of diagonal operators. Commented Jan 9, 2022 at 14:50
• 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? Commented Jan 9, 2022 at 18:47
• 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. Commented Jan 9, 2022 at 19:15
• 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. Commented Jan 9, 2022 at 21:01

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 $$\ell_2$$.