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Can someone give me an exemple of a Banach Algebra $A$ which does not have an isometric representation on a Hilbert Space? (With a proof or a reference to the proof.)

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For me, the simplest example would be $A=M_2(\mathbb{C})$ equipped with the operator norm induced by any $\ell^p$ norm ($1\leq p\leq +\infty$, $p\neq 2$) on $\mathbb{C}^2$.

In short: $A$ is naturally a $*$-algebra and an isometric embedding into $B(H)$ for the latter norms is necessarily a $*$-homomorphism, turning $A$ into a $C^*$-algebra. And a $C^*$-norm is unique on a given $C^*$-algebra, as follows from the algebraic characterization of a $C^*$-norm: $\|x\|^2=\|x^*x\|=\rho(x^*x)$.

Here is a detailed elementary proof that $A$ can not be isometrically embedded in any $B(H)$. More precisely, we will see that if such an isometric embedding exists, the norm must be the one induced by the $\ell^2$ norm. Note that this argument can easily be generalized to $M_n(\mathbb{C})$. If you are familiar with matrix units and Peirce decomposition, the idea is simple: if a Banach algebra norm on $M_n(\mathbb{C})$ has $\|e_{jj}\|=1$ for every $j$ and if $\pi:M_n(\mathbb{C})\longrightarrow B(H)$ is an isometric embedding, then $\pi$ is unitarily equivalent to $x\longmapsto x\otimes 1_K$ into $M_n(\mathbb{C})\otimes B(K)$ for some subspace $K$ of $H$. Whence $\|x\|$ must be the unique $C^*$-norm of $M_n(\mathbb{C})$, namely $\|x\|=\sqrt{\rho(x^*x)}$.

Note: with the same argument, we can show that if $\pi:M_n(\mathbb{C})\longmapsto B(H)$ is a nonzero contractive ($\|\pi(x)\|\leq \|x\|$) algebra homomorphism and if the matrix units (canonical basis of $M_n(\mathbb{C})$) and the identity all have norm not greater than $1$ (whence equal to $1$) as it is the case for any induced norm by $\ell^p$, then $\pi$ is actually a $*$-homomorphism (unitarily equivalent to a diagonal embedding) and $\|\pi(x)\|=\sqrt{\rho(x^*x)}$ is the spectral norm of $x$. So the initial norm on $M_n(\mathbb{C})$ can not be induced by an $\ell^p$ norm $(1\leq p\leq \infty)$ other than $p=2$ in this case as well. But note that any Schatten norm, in particular the Hilbert-Schmidt one, is dominated by the spectral norm, giving a contractive non isometric ($n\geq 2$) $*$-representation in $B(H)$.

I will use the following key properties of idempotents in $B(H)$.

Facts: if $p$ is an idempotent ($p^2=p$) is $B(H)$, then $p$ is self-adjoint ($p^*=p$) if and only if $\|p\|\leq 1$. If $1=p_1+p_2$ where $p_1,p_2$ are two projections (=self-adjoint idempotents) such that $p_1p_2=0$, then $K=\mbox{im} p_1=\ker p_2$ and $L=\mbox{im} p_2=\ker p_1$ are orthogonal and $H=K\oplus L$. Finally, $p_1$ and $p_2$ are Murray-von Neumann equivalent (i.e. there exists $u,v\in B(H)$ such that $p_1=uv$ and $p_2=vu$) if and only if their ranges are isometric.

Denote $1=I_2$ the unit and $e_j$ the idempotent matrix of $A$ whose $(j,j)$ coefficient is $1$ and other coefficients are $0$. So we have what is called an orthogonal decomposition $1=e_1\oplus e_2$ of the unit. Note that with the given norm of $A$, $\|e_1\|=\|e_2\|=\|1\|=1$.

Now assume there exists an isometric algebra homomorphism $\pi:A\longrightarrow B(H)$ for some Hilbert space $H$. Note $\pi$ is injective and that $\pi(1), \pi(e_1),\pi(e_2)$ are norm $1$ idempotents of $B(H)$ with norm $1$, whence projections. Since $e_1e_2=0$, we have $\pi(e_1)\pi(e_2)=0$. Since $\pi(1)$ commutes with $\pi(x)$ for every $x\in A$, $\pi(A)$ leaves the image of $\pi(1)$ invariant and is null on its nullspace. So without loss of generality, we can assume that $\pi(1)=1$ and $1=\pi(e_1)+\pi(e_2)$. Since there exist $u_1,u_2\in A$ such that $e_1=u_1u_2$ and $e_2=u_2u_1$ (take $u_1=e_{12}$ and $u_2=e_{21}$ from the canonical basis), we see that $\pi(e_1)$ and $\pi(e_2)$ are Murray-von Neumann equivalent. So their ranges $K$ and $L$ give $H=K\oplus L$ an orthogonal decomposition of $H$ with $K$ and $L$ isometric. Thanks to this decomposition, $B(H)$ is $*$-isomorphic to $M_2(B(K))=M_2(\mathbb{C})\otimes B(K)$. With respect to this $2\times 2$ decomposition of $B(H)$, the embedding $\pi$ is simply the natural extension of the scalar embedding $\mathbb{C}\longmapsto B(K)$. Indeed, every $x\in A=M_2(\mathbb{C})$ can be uniquely written $x=e_1xe_1+e_1xe_2+e_2xe_1+e_2xe_2$ (the Peirce decomposition, which is just decomposition with respect to the canonical basis of $M_2(\mathbb{C})$. Applying $\pi$ yields the corresponding $2\times 2 $ decomposition in $B(H)\simeq M_2(B(K))$. So under this identification, $\pi$ is simply $$ \pi:x=\pmatrix{a&b\\c&d}\longmapsto \pmatrix{a 1_K&b1_K\\c1_K&d1_K}=x\otimes 1_K. $$ Therefore $$ \|x\|_A=\|\pi(x)\|=\|x\otimes 1_K\|=\|x:(\mathbb{C}^2,\|\cdot\|_2)\longmapsto (\mathbb{C}^2,\|\cdot\|_2)\|=\sqrt{\rho(x^*x)} $$ That is, the norm on $A$ must be the norm induced by the $\ell^2$ norm on $\mathbb{C}^2$. And this is obviously not equal to any norm induced by an $\ell^p$ norm for $p\neq 2$.

Precision: given $p_j=\pi(e_j)$ and $v_j=\pi(u_j)$, the identification between $B(H)=B(K\oplus L)$ and $M_2(B(K))$ is given by $$ y=\pmatrix{a&b\\ c&d}\longmapsto\pmatrix{1_K&0\\0&v_1}\pmatrix{a&b\\ c&d}\pmatrix{1_K&0\\0&v_2}=V_1yV_2 $$ with $V_1:K\oplus L\longmapsto K\oplus K$ and $V_2:K\oplus K\longmapsto K\oplus L$ are such that $V_1V_2=1$ and $V_2V_1=1$. Now with the assumptions, $\|V_j\|\leq 1$. So both both are invertible contractions, with contractive inverse. This implies that the $V_j$ are surjective isometries whence unitaries, and $V_2=V_1^*$. So the identification is a unitary equivalence, in particular a $*$-isomorphism.

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    $\begingroup$ thank you for the time that you spent on this wonderful answer ! $\endgroup$
    – thetruth
    Jun 18 '13 at 19:33
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Could I answer this way ? (half inspired by an answer i got on MathOverflow)

We know that a C$^\star$-algebra is Arens Regular. We know that for a locally compact $\textit{infinite}$ group $G$, $L^{1}(G)$ is not Arens Regular. Therefore, if there was a representation of $L^{1}(G)$ in some $H$, we would have a contradiction, since a closed subalgebra of a Arens Regular algebra (such as $B(H)$, since it's a C$^{\star}$ algebra) is still Arens Regular (but $L^{1}(G)$ is not).

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  • $\begingroup$ Yes, that's fine. This is the classical counter-example. $\endgroup$ Oct 6 '14 at 14:04
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I should share an easy example of a Banach algebra that an entry level student can verify easily. On $\mathbb{C}^2$, with point wise linear operations, define the multiplication $$(x_1,x_2)(y_1,y_2) = (x_1y_1,0).$$ One verifies that with sup norm this makes $\mathbb{C}^2$ a Banach algebra. Naturally the sup norm does not satisfy the parallelogram law. Hence this cannot be isometrically isomorphic to any Hilbert space or it's subspace.

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    $\begingroup$ The question asks whether there is an isometric representation on a Hilbert space, i.e. an isometric homomorphism $A \to B(H)$ to the bounded operators on a Hilbert space, not an isometric embedding $A \to H$ to the Hilbert space itself. Your example is the same as the continuous functions on a $2$-point space and you have the diagonal embedding $\mathbb{C}^2 \to M_{2}(\mathbb{C}) = B(\mathbb{C}^2)$ which is isometric. $\endgroup$
    – Martin
    Jun 21 '13 at 7:46
  • $\begingroup$ First of all this is not the example of the continuous functions on a 2-point space as on that the multiplication would be $(x_1,x_2)(y_1,y_2) = (x_1y_1,x_2y_2)$, whereas in my multiplication second component is always zero. Consequently, the diagonal embedding you mentioned does not work. Anyway, rating down anything requires a responsibility and understanding. Thanks. $\endgroup$ Jul 12 '13 at 8:12
  • $\begingroup$ I didn't vote on your answer and I overlooked the zero in the second coordinate, sorry about that. Anyway, my objection still applies: the parallelogram law only shows that the algebra itself is not isometric to a Hilbert space, not that it does not admit an isometric representation on a Hilbert space. You can realize your algebra isometrically in $M_3(\mathbb{C})$ via $\begin{bmatrix} x_1 & 0 & 0 \cr 0 & 0 & x_2 \cr 0 & 0 & 0\end{bmatrix}$. $\endgroup$
    – Martin
    Jul 15 '13 at 11:50

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