Is there an example of an infinite-dimensional $C^\*$-algebra $A$ which admits the following structure:


The $C^\*$ algebra $A$ admits a faithful trace $tr$ such that the multiplication $m: A\otimes A \to A$ would be a bounded operator between pre-Hilbert spaces (with respect to the inner product $tr(ab^\*)$ on $A$ and its natural extension to the algebraic tensor product $A\otimes A$) such that $mm^\*=\lambda\mathrm{Id}$ for some $\lambda \in \mathbb{C}$, where $m^\*$ is the adjoint of $m$ after completion of the above pre Hilbert space? Is there an example of this situation for which $\lambda$ is not an integer number?


**Motivation from finite-dimensional 
 case:** For $A=M\_n(\mathbb{C})$ with the standard trace we have $\lambda=n$ so this process recovers $n$ from $M\_n(\mathbb{C})$, that is $mm^\*=nId$. Namely the matrix multiplication is a rescaled "Partial Isometry".



Recovering "$n$" from $M_n(\mathbb{C})$
As Nik Weaver mentioned, you can change the constant appearing in the equation $m m^{\ast}=\lambda Id$ by simply rescaling the inner product, so we need to assume some normalisation. I will work in the following framework. Let $A$ be a $C^{\ast}$-algebra and let $\varphi$ be a state on $A$. We call $\varphi$ a $\delta$-form if the multiplication map $A\otimes A \to A$ is bounded with respect to the inner product $\langle a, b\rangle = \varphi(a^{\ast}b)$ and $mm^{\ast}=\delta^2 Id$. What we are after are tracial $\delta$-forms.


Let's show that if $A$ is infinite dimensional then it does not admit a tracial state with respect to which the multiplication map is bounded. Indeed, let $(e\_i)\_{i\leq N}$ be an orthonormal set (because $A$ is infinite dimensional, $N$ can be arbitrarily large). Consider the element $x=\sum\_{i=1}^{N} e\_i^{\ast}\otimes e\_i$. Then $m(x) = \sum\_{i=1}^{N} e\_{i}^{\ast}e\_{i}$ and the $L^{2}$-norm of this element is at least $N$; indeed, the state $\varphi$ is contractive on $A$ equipped with its $L^{2}$-norm, and $\varphi(\sum\_{i=1}^{N} e\_i^{\ast}e\_i) = N$. On the other hand, the $L^{2}$-norm of $\sum\_{i=1}^{N} e\_{i}^{\ast}\otimes e\_i$ is equal to $\sqrt{N}$ (because $\varphi$ is tracial). Therefore the norm of $m$ is at least $\sqrt{N}$, but $N$ was arbitrary, so $m$ is not bounded.


So $A$ is finite dimensional, hence of the form $\oplus\_{i=1}^{s} M\_{n\_i}(\mathbb{C})$. Every tracial state is specified by a sequence of weights $(\lambda\_1,\dots,\lambda\_s)$, which add up to $1$. You can now check that it yields a $\delta$-form iff $\lambda\_k = \frac{n\_k^2}{\dim A}$. In this case $\delta^2 = \dim A$. 


