How to describe the Gelfand transform of the Banach algebra of complex Borel measures on the real line?

Question

Let $M$ be the Banach algebra of all complex Borel measures on $\mathbb{R}$. To be clear,

1. Norm: $\| \mu \| = |\mu|(\mathbb{R})$, where $|\mu|(E)$ is the total variance.
2. Product: $(\mu \ast \lambda)(E) = (\mu \times \lambda)(\{(x,y):x+y \in E\})=\int \mu(E-t)d\lambda(t)$. This is commutative and gives a unit, which happens to be the Dirac measure (at $0$).

I'm thinking about the Gelfand transform on $M$ but didn't find any reference on it yet, and I haven't solved it. That's why I'm asking this question. References are welcome but let me explain what I mean by describing the Gelfand transform. For the nonunitary Banach algebra $L^1(\mathbb{R},m)$ we have a nice result

To every non-trivial complex homomorphism $\varphi$ on $L^1$ there corresponds a unique $t \in \mathbb{R}$ such that $\varphi(f)=\hat{f}(t)$. (It can be found on Rudin's Real and Complex Analysis, 9.22-9.23)

Thanks to this, the Gelfand transform on $L^1$ can be identified as the Fourier transform. So I suppose similiar settings can be applied to $M$ in this question as well. Here's my thoughts so far although I didn't figure out how to go in or go further:

1. Fourier transform is still a complex homomorphism in $M$, as one can easily verify. It has to play some role.
2. We may need another Riesz's representation theorem.
3. When thinking about $M$, we can think about the space $C_0(\mathbb{R})$ (continuous functions vanishing at the infinity). But it's not the same as $L^1$, we may need to find a approach parallel to our work on $L^1$. In fact, the ideal of absolutely continuous measures (relative to the Lebesgue measure) is isomorphic to $L^1(\mathbb{R},m)$.
4. Lebesgue decomposition. For any $\mu \in M$, we have an decomposition $\mu = \mu_a+\mu_s$ where $\mu_a$ is absolutely continuous with respect to $m$, and $\mu_s \perp m$. The subspace of $\mu_a$'s is isomorphic to $L^1$, where the Fourier transform is waiting, so perhaps the problem is reduced to the Gelfand transform on the subspace of singular measures.

Answer 1

Many constructions in analysis, such as compactifications, double dual spaces and Gelfald spectra tend to lead to very complicated objects. The spectrum $\widehat{M({\mathbb R})}$ is among these and it is probably very difficult to describe it in simple terms. Nevertheless let me try to offer a perspective that might shed a bit of light into its structure.

The Banach space dual of $C_0(\mathbb R)$ is known to be isomorphic to $M({\mathbb R})$, hence the dual of $M({\mathbb R})$ coincides with the double dual of $C_0(\mathbb R)$, which in turn is a von Neumann algebra, known as the envelopping von Neumann algebra of $C_0(\mathbb R)$, henceforth refered to as $\mathscr M$.

Since a character on $M({\mathbb R})$ is necessarily a continuous linear functional, these are then to be found within $\mathscr M$. Therefore the present question may be interpreted as asking for a characterization of the elements of $\mathscr M$ defining a character of $M({\mathbb R})$.

It may be proved that $\mathscr M$ is a Hopf von Neumann algebra, meaning that it has a canonical co-multiplication, namely a weakly continuous *-homomorphism $$ \Delta :\mathscr M\to \mathscr M\otimes \mathscr M, $$ satisfying some specific hypotheses, among which a crucial co-associativity, which is nevertheless not needed for us here.

Well, so it turns out that an element $a$ in $\mathscr M$ defines a character of $M({\mathbb R})$ if and only if $a$ is nonzero and satisfies the equation $\Delta (a)=a\otimes a$, characterizing it as what the Hopf algebraists call a group-like element.

Should this description be deemed satisfactory I might try to provide further details.