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Restricted one statement to the complex base field
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Torsten Ekedahl
Torsten Ekedahl
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What does make sense is to speak of an algebraic hull of an embedding of $\mathfrak g\subseteq \mathfrak{gl}_m$, i.e., of a (faithful) $\mathfrak g$-representation. In that case one may consider the intersection of all algebraic subgroups of $\mathrm{GL}_m$ whose Lie algebra contains $\mathfrak g$. In terms of Zariski closures (when the base field is $\mathbb C$) it is the Zariski closure of the exponentials of all elements of $\mathfrak g$ (inside of $\mathfrak{gl}_m$). From the Tannakian point of view this is the group that corresponds to the tensor subcategory of the category of $\mathfrak g$-representations generated by the given representation.

What does make sense is to speak of an algebraic hull of an embedding of $\mathfrak g\subseteq \mathfrak{gl}_m$, i.e., of a (faithful) $\mathfrak g$-representation. In that case one may consider the intersection of all algebraic subgroups of $\mathrm{GL}_m$ whose Lie algebra contains $\mathfrak g$. In terms of Zariski closures it is the Zariski closure of the exponentials of all elements of $\mathfrak g$ (inside of $\mathfrak{gl}_m$). From the Tannakian point of view this is the group that corresponds to the tensor subcategory of the category of $\mathfrak g$-representations generated by the given representation.

What does make sense is to speak of an algebraic hull of an embedding of $\mathfrak g\subseteq \mathfrak{gl}_m$, i.e., of a (faithful) $\mathfrak g$-representation. In that case one may consider the intersection of all algebraic subgroups of $\mathrm{GL}_m$ whose Lie algebra contains $\mathfrak g$. In terms of Zariski closures (when the base field is $\mathbb C$) it is the Zariski closure of the exponentials of all elements of $\mathfrak g$ (inside of $\mathfrak{gl}_m$). From the Tannakian point of view this is the group that corresponds to the tensor subcategory of the category of $\mathfrak g$-representations generated by the given representation.

Added what may be an actual answer to the question
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Torsten Ekedahl
Torsten Ekedahl
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[Added] I somewhat rudely hijacked the question by taking up things that maybe weren't that pertinent to the question so let me give an answer which I think is more on track.

The problem is that one can not always define the algebraisation of an abstract finite dimensional Lie algebra $\mathfrak g$ even if some algebraisation exists. As an examples consider a $2$-dimensional Lie algebra with basis $x,y$ and $[x,y]=y$. This is the Lie algebra of an infinite number of algebraic groups: Let the $1$-dimensional torus $\mathbb G_m$ act on the additive group $\mathbb G_a$ by $(t,v) \mapsto t^nv$, where $n\not=0$ and let $G_n$ be the semi-direct product of this action. These groups all have $\mathfrak g$ as Lie algebra but the only isomorphisms between them is that $G_n$ is isomorphic to $G_{-n}$.

What does make sense is to speak of an algebraic hull of an embedding of $\mathfrak g\subseteq \mathfrak{gl}_m$, i.e., of a (faithful) $\mathfrak g$-representation. In that case one may consider the intersection of all algebraic subgroups of $\mathrm{GL}_m$ whose Lie algebra contains $\mathfrak g$. In terms of Zariski closures it is the Zariski closure of the exponentials of all elements of $\mathfrak g$ (inside of $\mathfrak{gl}_m$). From the Tannakian point of view this is the group that corresponds to the tensor subcategory of the category of $\mathfrak g$-representations generated by the given representation.

However, if one wants something that is independent of a particular representation one has to pass to an inverse limit of groups coming from different representations. This leads to an infinite dimensional monster even in the case when $\mathfrak g$ is $1$-dimensional.

[Added] I somewhat rudely hijacked the question by taking up things that maybe weren't that pertinent to the question so let me give an answer which I think is more on track.

The problem is that one can not always define the algebraisation of an abstract finite dimensional Lie algebra $\mathfrak g$ even if some algebraisation exists. As an examples consider a $2$-dimensional Lie algebra with basis $x,y$ and $[x,y]=y$. This is the Lie algebra of an infinite number of algebraic groups: Let the $1$-dimensional torus $\mathbb G_m$ act on the additive group $\mathbb G_a$ by $(t,v) \mapsto t^nv$, where $n\not=0$ and let $G_n$ be the semi-direct product of this action. These groups all have $\mathfrak g$ as Lie algebra but the only isomorphisms between them is that $G_n$ is isomorphic to $G_{-n}$.

What does make sense is to speak of an algebraic hull of an embedding of $\mathfrak g\subseteq \mathfrak{gl}_m$, i.e., of a (faithful) $\mathfrak g$-representation. In that case one may consider the intersection of all algebraic subgroups of $\mathrm{GL}_m$ whose Lie algebra contains $\mathfrak g$. In terms of Zariski closures it is the Zariski closure of the exponentials of all elements of $\mathfrak g$ (inside of $\mathfrak{gl}_m$). From the Tannakian point of view this is the group that corresponds to the tensor subcategory of the category of $\mathfrak g$-representations generated by the given representation.

However, if one wants something that is independent of a particular representation one has to pass to an inverse limit of groups coming from different representations. This leads to an infinite dimensional monster even in the case when $\mathfrak g$ is $1$-dimensional.

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Torsten Ekedahl
Torsten Ekedahl
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After some thought my pessimism (as expressed in my concurrence with the answer of Milne) has abated somewhat. If I were bold enough I would conjecture the following (assuming that the characteristic zero base field is algebraically closed): Let $\mathfrak g$ be a finite dimensional Lie algebra over $k$ and let $G$ be the pro-algebraic group whose representation tensor category is equivalent to the tensor category of finite dimensional $\mathfrak g$-modules. Then if $S$ is the (pro-)radical of $G$ and $U$ the (pro-)unipotent radical $U$ and $G/S$ are algebraic groups (unipotent and semi-simple respectively). Furthmore, the pro-torus $T:=S/U$ has as character group $\mathfrak{u}/[\mathfrak{g},\mathfrak{u}]$ considered as an additive group. Hence the only infinite-dimensional part is $T$ but its character group, \emph{à priori} only an abstract group, is reasonably well controlled. This is analogous to the case of irreducible infinite-dimensional representations of a semi-simple Lie group where the center of the enveloping algebra acts by a character and the set of characters as a set is very large. However it is the set of $k$-points of an algebraic variety which means that it is under control. The analogy goes further as the category of $G$-representations (assuming $U$ is finite dimensional) splits up into a direct product of categories parametrised by cosets of the character group of $T$ with respect to the subgroup generated by the characters occurring in the action of $T$ on the Lie algebra of $U$.

Here are some comments on the conjecture (I do not vouch for the complete veracity of my claims).

We can get a picture of $G/U$ by looking at the irreducible $\mathfrak g$-representations (as they correspond exactly to the irreducible $G/U$-representations). All such representations factor through $\mathfrak g/[\mathfrak{g},\mathfrak{u}]$ which is the product of $\mathfrak{u}/[\mathfrak{g},\mathfrak{u}]$ and $\mathfrak g/\mathfrak{u}$. Hence, the irreducible representations are parametrised by pairs of a $1$-dimensional representation of $\mathfrak{u}/[\mathfrak{g},\mathfrak{u}]$ and an irreducible representation of the semi-simple algebra $\mathfrak g/\mathfrak{u}$. This gives the prediction that $G/U$ should be the product of a torus with character group $\mathfrak{u}/[\mathfrak{g},\mathfrak{u}]$ and the simply-connected semi-simple group with Lie algebra $\mathfrak g/\mathfrak{u}$.

As for $U$, the idea is that the category of unipotent representations (i.e., successive extensions of the trivial representation) of $\mathfrak g$ is equivalent to the category of representations of the unipotent group with Lie algebra $\mathfrak g'$, the maximal unipotent quotient of $\mathfrak g$. Something similar ought to be true for successive extension of the same irreducible representation and there shouldn't be too much "intermixing" between different irreducibles.