I am slightly confused with the definition of a virtual diagonal for a Banach algebra $A$. For instance in Runde; Amenable Banach algebras they are defined as follows. Consider the diagonal operator $\triangle:A\otimes^{\gamma}A\to A$, where $\otimes^{\gamma}$ denotes the projective tensor product. Furthremore we make $A\otimes^{\gamma}A$ an Banach $A$ module by defining $a(x\otimes y)=ax\otimes y$ and $(x\otimes y)a=x\otimes ya$. A virtual diagonal for $A$ is an element $D\in (A\otimes^{\gamma}A)^{\ast\ast}$ such that $a\cdot D=D\cdot a$ and $a\cdot \triangle ^{\ast\ast}D=a$, for each $a\in A$.
At this point I am not sure how to make sense of the second multiplication; $a\cdot \triangle ^{\ast\ast}D=a$. I read in Virtual diagonals and $n$-amenability for Banach algebras, by Paterson that the second statement should be read as $\hat{a}\cdot \triangle ^{\ast\ast}D=\hat{a}$, where $\hat{a}$ is the embedded version of $a$ in to the second dual. Now my real question is, is there a canonical multiplication on $A^{\ast\ast}$, how should I interpret the multiplication $\hat{a}\cdot \triangle ^{\ast\ast}D$?
Let $A$ be a Banach algebra. For every $a,x\in A$, $f\in A^{*}$, $F,G\in A^{**}$ the following multiplications are defined \begin{array}{ccc} fa(x) = f(ax), & Gf(a) = G(fa), & F*_1 G(f) = F(Gf) \\ af(x) = f(xa), & fG(a) = G(af), & G*_2 F(f) = F(fG) \\ \end{array} The products in the last column are called the first & second Arens products, respectively. Generally these two products are not the same. $A$ is called Arens regular if both products are equal. However, when $a\in A$ and $F\in A^{**}$, then it is not difficult to show that $$F*_1 a = F*_2 a\hspace{9mm}\textrm{and}\hspace{9mm} a*_1 F = a*_2 F $$
In fact, for each $b\in A$, $$\hat{a}f(b) = \hat{a}(fb) = fb(a) = f(ba) = af(b).$$ Thus, $\hat{a}f = af$. Second, for each $f\in A^{*}$, $$F*_1\hat{a}(f) = F(af) = fF(a) = \hat{a}(fF) = F*_2 \hat{a}(f).$$ Thus, $F*_1\hat{a} = F*_2 \hat{a}$. The other equality can be shown to hold similarly.