For sake of brevity let $A$ denote the Banach algebra formed by equipping $\ell^1({\mathbb N})$ with pointwise multiplication. This algebra is clearly not isomorphic as a Banach algebra to any uniform algebra (since it has idempotents of arbitrarily large norm).

It has been known since the 1970s that there exists a Hilbert space $H$ and a continuous injective algebra homomorphism $\theta:A\to B(H)$ which has (norm-)closed range. That is, $A$ is isomorphic to a non-self-adjoint operator algebra. However, the proof is non-explicit; the only argument I have seen uses the fact that multiplication $A \otimes A \to A$ extends continuously to the injective tensor product of $A$ with itself, and then uses abstract GNS-flavoured techniques to build the required space $H$. See

Varopoulos, N. Th. Some remarks on $Q$-algebras. Ann. Inst. Fourier (Grenoble) 22 (1972), no. 4, 1–11.

Davie, A. M. Quotient algebras of uniform algebras. J. London Math. Soc. (2) 7 (1973), 31–40.

For each $n$ let $A_n$ be the subalgebra of $A$ generated by the first $n$ minimal idempotents (so $A_n$ is just ${\mathbb C}^n$ with the $\ell^1$-norm and pointwise product).

UPDATE I originally mis-stated part of the question, as Andreas, Bill and Gideon were all quick to point out; my apologies for the confusion. UPDATE, ENCORE My apologies for having omitted yet another condition in my haste to get the original edits done - thanks to fedja for catching this

Taking the existence of $\theta$ as read, it is straightforward to show that given $a\in A_n$ there exists a subspace $H_n\subset H$ of dimension $n+1$ and a homomorphism $\theta_n A_n \to B(H_n)$ such that

$$ c\Vert a\Vert_1 < \Vert\theta_n(a)\Vert \leq C \Vert a\Vert_1 $$ where the constants $c$ and $C$ are strictly positive and independent of $n$ and $a$.

By a crude compactness argument, this implies the following:

There exist constants $0 < c < C$ such that, for each $n\geq 1$, we can find $m(n)$ and idempotents $E_1,\dots, E_n$ in $M_{m(n)}({\mathbb C})$, which satisfy $$ E_jE_k=0 \quad\hbox{for $j\neq k$} $$ and $$ c\sum_{j=1}^n |a_j| \leq \left\Vert \sum_{j=1}^n a_j E_j \right\Vert \leq C \sum_{j=1}^n |a_j| $$ for all $a_1,\dots, a_n\in {\mathbb C}$.

My question is this: does anybody know of an explicit construction of such $E_1,\dots, E_n$? I would be interested just to know of a construction which works for a sequence $n_1 < n_2< \dots$ Also, how small can we make the containing matrix algebra? (The compactness argument referred to above produces an embedding into the block-diagonal matrices with block size $n+1$ -- I haven't tried to work out just how `long' the diagonal has to be.)

These seem to me like questions that must have been looked at before, so perhaps the answer would be as simple as supplying an appropriate reference. (The choice to work over complex scalars here is not important; a solution just for real scalars would be just as welcome.)

Remark: perhaps it might be useful to use the fairly well-known equivalence between the $\ell^1$-norm and the norm $$ \Vert x \Vert_S := \sup_{F \subseteq {\mathbb N}} \left\Vert \sum_{i\in F} x_i \right\Vert $$

  • $\begingroup$ How do you get $2n+2$? $\endgroup$ Dec 17, 2010 at 10:31
  • $\begingroup$ Yemon, are you sure about the dimension $2n+2$? I think it is wrong. Maybe you mean $2^n$? $\endgroup$ Dec 17, 2010 at 10:56
  • $\begingroup$ Gideon: thanks for catching that ($2n+2$ is very probably too small). I was over-hastily transcribing from some of my notes and misread them. I will have a think about what the correct number should be and update my post $\endgroup$
    – Yemon Choi
    Dec 17, 2010 at 19:36
  • $\begingroup$ Isn't the type 2 constant of $M_n$ something like $\log n$ (by comparing to $S_p^n$ with $p=\log n$? (I am too busy [or lazy] to look it up or think about it right now.) The type 2 constant of $\ell_1^n$ is $n^{1/2}$. That should give the right order up to a multiplicative $\log$ factor even for an embedding that is only linear. $\endgroup$ Dec 17, 2010 at 20:23
  • $\begingroup$ Hi Bill. The type 2 constant of $M_m$ is of order $\sqrt{\log m}$, by comparing with that of $S_p^n$ (which is of order $\sqrt p$). This gives that if $\ell_1^n$ nicely embeds in $M_m$ then m in exponential in $n$. $\endgroup$ Dec 17, 2010 at 20:56

2 Answers 2


You can manage to get $m(n)=(n+2)2^{n}$ (this can certainly be made a bit better). By your remark, it is enough to construct projections $E_1,\dots,E_n$ in $M_{n+2}(\mathbb C)$ such that:

  1. $E_i E_j=0$ for $i \ne j$

  2. $\|E_i\|\leq 3$

  3. $\| \sum s_i E_i\| \geq |\sum s_i|$ for any complex numbers $s_1,\dots s_n$.

(Then consider $\oplus_{F \subset (1,\dots,n)} M_{|F|+2}$ inside $M_{m(n)}(\mathbb C)$).

We use the following elementary fact: there exists $\Omega$ a unit vector in the euclidean space $\mathbb C^{n+2}$ and $x_1,\dots,x_n$ and $y_1,\dots,y_n$ of norm $\sqrt 3$ in $\mathbb C^{n+2}$ such that $\langle x_i,y_j\rangle=1$ if $i=j$ and $0$ if $i\ne j$, and such that $\langle x_i,\Omega\rangle=\langle y_i,\Omega\rangle=1$. (if $\Omega,e_0,\dots,e_n$ is an orthonormal basis, take $x_i=\Omega+e_0+e_i$ and $y_i=\Omega-e_0+e_i$).

Then the matrices $E_i=x_i^* \otimes y_i$ (the rank one operator mapping $x$ to $\langle x_i,x\rangle y_i$ satisfy the conditions above. The only point perhaps not completely evident is the third point, but this is because $\langle \Omega,\sum_i s_i E_i \Omega \rangle = \sum_i s_i$.

  • $\begingroup$ Simple and nice, Mikael. $\endgroup$ Dec 18, 2010 at 17:30
  • $\begingroup$ Thanks Mikael - I'm afraid I don't have time to check the details right now but on a quick glance it looks good. If this holds up, would you have any objection to me citing this as part of something I'm writing? (I have a section where I needed some "counterexamples for contrast", and originally I used the construction outlined in my question, but it would clearly make for a better paper if I could just refer people to a direct construction such as yours.) $\endgroup$
    – Yemon Choi
    Dec 19, 2010 at 0:20
  • $\begingroup$ Having a second look at this, I am now a bit unsure how exactly Mikael's embedding works, although I think something like this is on the right lines and should yield an example. If $E_1, \dots, E_n$ are as in my question, then what does the compression of each of them to $M_{|F|+2}$ look like when $F=\{1,2,3\}$ say? $\endgroup$
    – Yemon Choi
    Dec 19, 2010 at 7:46
  • $\begingroup$ No objection, Yemon. $\endgroup$ Dec 19, 2010 at 7:47
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    $\begingroup$ Let me clarify, and to avoid confusion denote by $P_1,\dots,P_n \in M_{n+2}(\mathbb C)$ the rank one projections in my answer. The $E_i$'s as in your question are defined by $E_i=\oplus_{F \subset (1,\dots,n} P_i 1_{i \in F} \in \oplus_F M_{n+2}(\mathbb C)$. $\endgroup$ Dec 19, 2010 at 7:54

Here is simple construction for $m(n)=2^n$. As is indicated in the remarks above one can't get an essentially lower dimension.

The $E_i$-s will be diagonal matrices. $E_1$ will have the first half diagonal entries equal to 1 the second half 0. $E_2$ the first and third fourths 1-s the rest zeroes, ..., $E_n$ alternatively 1 and zero. (These are $(r_i+1)/2$ where $r_i$ are the Rademacher functions). The point is that for each subset $A$ of $\{1,\dots,n\}$ there is a $j$ such that $\{E_1(j,j),\dots,E_n(j,j)\}$ is exactly the indicator function of $A$. The $E_i$ are idempotents.

Using the equivalent $\ell_1$ norm $\|\cdot\|_S$ at the end of the question, it is easy to see that these satisfy the inequalities. They satisfy them with constants 1 for the $\|\cdot\|_S$ norm.

Update: Disregard this answer. It relates to a previous version of Yemon's question.

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    $\begingroup$ OP forgot to mention the condition $E_iE_j=0$ for $i\ne j$, of course. $\endgroup$
    – fedja
    Dec 18, 2010 at 6:57
  • $\begingroup$ Je suis devenu senile ... thank you, fedja, and my apologies to Gideon. I've changed the question to rectify this embarrassing lapse. $\endgroup$
    – Yemon Choi
    Dec 18, 2010 at 7:39
  • $\begingroup$ Gideon's linear embedding looks to me like embedding $\ell_1^n$ into $\ell_\infty^m$, and I have seen something like this particular embedding before. (Of course any finite-dimensional Banach space embeds with small BM-distance into $\ell_\infty^m$ for $m$ suitably large.) In fact one knows by my initial remark (about $\ell^1$ not being a uniform algebra) that we can't achieve the embedding I'm after using only diagonal matrices. $\endgroup$
    – Yemon Choi
    Dec 18, 2010 at 7:40

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