I hope this is not too elementary!


Let $G$ be a algebraic reductive group over $\mathbb{C}$.
The group $G(\mathbb{C}[[t]])$ can be given the structure of a pro algebraic group as follows.


Let $l\in \mathbb{N}$ and $J^l:=\mathbb{C}[[t]] /t^l\mathbb{C}[[t]])$. Then $G(J^l)$ is known to be an algebraic group and $G(\mathbb{C}[[t]])$ can be identified with 
$\varprojlim\_l G(J^l)$.


So my question is:



> 
> Why is $G(J^l)$ an algebraic group.
> 
> 
> 


I mean I can prove this with dumb calculations. However, I wonder if there is an nice argument to see this?



About the pro-algebraic group structure of $G(\mathbb{C}[[t]])$
As hinted by S. Carnahan, this is a particular case of a general construction, known as Weil restriction, Greenberg functor, or arc space, depending on the context.


Let $X$ be a scheme of finite type over a field $k$. Then there is a scheme $\mathcal L(X)$, which is a projective limit of schemes $\mathcal L\_m(X)$ of finite type over $k$, such that $\mathcal L(X)(F)=X(F[[t]])$ and $\mathcal L\_m(X)(F)=X(F[[t]]/(t^m))$, for every $k$-algebra $F$.


Most of the construction is, in fact, relatively easy. 


Begin with $X=\mathbf A^1$. Then, it suffices to take $\mathcal L\_m(X)=\mathbf A^m$, the identification of $\mathcal L\_m(X)(F)$ with $X(F[[t]]/(t^m))$
begin given by $(x\_0,\dots,x\_{m-1})\mapsto x\_0+x\_1t+\dots+x\_{m-1}t^{m-1}$.
Then $\mathcal L(X)=\mathbf A^\infty=\mathop{\mathrm {Spec}}(T\_0,T\_1,T\_2,\dots)$.


This generalizes readily to $X=\mathbf A^n$ (take the $n$th power of
the preceding schemes).


Now, if $X$ is a closed subscheme of $\mathbf A^n$, with ideal $I=(P,\dots)$,
one can expand
$P(x\_0+x\_1t+\dots+x\_{m-1}t^{m-1})=P\_0(x)+P\_1(x)t+\dots+P\_{m-1}(x)t^{m-1}
\pmod {t^m}$ and $\mathcal L\_m(X)$ is viewed as a closed subscheme of $\mathcal L\_m(\mathbf A^n)$ by adding the equations $P\_0=\dots=P\_{m-1}=0$ for every polynomial $P\in I$ (or in a generating subset of $I$).
In your particular case, where $X$ is an (affine) algebraic group, this is all you need.


If $U$ is an affine open subscheme of $X$, then $\mathcal L\_m(U)$ identifies as an open affine subscheme of $\mathcal L\_m(X)$. This will allow to define $\mathcal L\_m(X)$ in general by gluing $\mathcal L\_m(U)$, for affine open subschemes $U$ of $X$.


One then takes the limit $\mathcal L(X)=\varprojlim\_m \mathcal L\_m(X)$,
which exists as a scheme, because the transition morphisms $\mathcal L\_{m+1}(X)\to \mathcal L\_m(X)$ are affine. 


Finally, the formula $\mathcal L(X)(F)=X(F[[t]])$ is easy if $F$ is a field, or if $X$ is affine (this is all you need), and is relatively easy if $X$ is quasiprojective. The general case is due to B. Bhatt (private communication). His proof used techniques of derived algebraic geometry (a theorem of Lurie/Brandenburg-Chiravasitu) and existence of enough perfect complexes (Thomason-Trobaugh).


EDIT: I realize that I did not fully answer your question, which was *why* one gets a pro-algebraic group when one begins with an algebraic group. The point is that this construction is functorial and commutes with products. Consequently, if $X$ is a group scheme, then one gets morphisms $\mathcal L\_m(X)\times\_k \mathcal L\_m(X)\to \mathcal L\_m(X)$ and $\mathcal L(X)\times\_k\mathcal L(X)\to\mathcal L(X)$ which give $\mathcal L\_m(X)$ a structure of algebraic group, and $\mathcal L(X)$ a structure of group scheme, projective limit of algebraic groups.


