I have a doubt on the following extract of this book, page 80:
"This action may be exponentiated to the adjoint action of the group $G[[t]]$, where $G$ is the connected simply connected Lie group with Lie algebra $\mathfrak{g}$."
What I don't see is why $G[[t]]$ is a group, given that $G$ is. I think $G[[t]]$ is defined as $G \otimes \mathbb{C}[[t]]$ (that's how $\mathfrak{g}[[t]]$ is defined after all), where $\mathbb{C}[[t]]$ denotes the ring of formal power series with coefficients in $\mathbb{C}$. The multiplication rule should be $$(g\otimes a(t)) \cdot (h \otimes b(t)) = (g\cdot h) \otimes a(t)b(t)$$ But what is, say, the inverse of an element of the form $g \otimes t$?
Moreover, is $G[[t]]$ a (infinite dimensional?) Lie group?
The Lie algebra $\mathfrak{g}$ is assumed in your reference to be a finite-dimensional simple algebra over the field $\mathbf{C}$ of complex numbers. The connected and simply connected group $G$ with $\mathfrak{g}=\mathrm{Lie}(G)$ is therefore an algebraic group (and in fact, defined over $\mathbf{Q}$). Thus it makes sense to consider the set of points $G(A)$ of $G$ with values in $A$, for any commutative $\mathbf{Q}$-algebra $A$.
The notation $G[[t]]$ in this context then means the points of $G$ with values in the ring $\mathbf{C}[[t]]$. For instance, if $G$ is $\mathrm{SL}_n(\mathbf{C})$, then
$$G[[t]]=\{(a_{ij})_{1 \leq i, j \leq n} \in \mathrm{Mat}_{n \times n}(\mathbf{C}[[t]]) \ | \ \mathrm{det}(a_{ij})=1 \}$$ is the group of $n$ by $n$ matrices with entries that are formal power series and with determinant equal to $1$. There are similarly explicit descriptions for all the (simply connected) classical groups. The group $G[[t]]$ is not a Lie group, but rather an Ind algebraic group---in other words, an increasing union of finite-dimensional algebraic varieties equipped with a group structure compatible with this union in a certain sense.