TITLE: Lie Groups/Lie algebras to algebraic groups
QUESTION [12 upvotes]: I am reading some lie groups/lie algebras on my own..
I am using  Brian Hall's Lie Groups, Lie Algebras, and Representations: An Elementary Introduction
I was checking for some other references on lie groups and found J. S. Milne's notes Lie Algebras, Algebraic Groups,and Lie Groups
It was written in introductory page of algebraic groups chapter that :

Most of the theory of algebraic groups in characteristic zero is
  visible already in the theory of Lie algebras

I would like to know if anybody wants to make it more clear..
I am planning to read some algebraic groups also and I was kind of happy to see that lie groups/lie algebras  and algebraic groups are related.
I had a very basic course in algebraic geometry and I want to learn algebraic groups as well...
I would be happy if one can give some other references  or exposition to comment made by Milne or give some idea of how much algebraic geometry is related to algebraic groups.

REPLY [3 votes]: I state without a complete proof the following interesting theorem.

Theorem 1. Let $G$ be a connected, complex, semisimple Lie group. $G$ admits a unique algebraic group structure.

Proof. In [OV], one can find the proof that for any connected, complex Lie group $G$ such that it is equal to $G^{\prime}$ (the derived subgroup) the holomorphic representations in an algebraic group are polynomial; in particular, this statement holds for $G$ semisimple.
By definition of adjoint representation and by the hypothesis, $\ker\mathrm{ad}=Z(\mathfrak{g})=\{0\}$ and therefore $\mathfrak{g}$ has a faithful representation in $\mathfrak{gl}(\mathfrak{g})$; where $\mathfrak{g}$ is the Lie algebra of $G$ and $\mathfrak{gl}(\mathfrak{g})$ is the general linear group over the vector space $\mathfrak{g}$.
Considered the diagram:
\begin{matrix}
\mathfrak{g} & \stackrel{\mathrm{ad}} {\longrightarrow} & \mathfrak{gl}(\mathfrak{g})\simeq\mathfrak{gl}(\mathbb{V})\\
{\exp}{\downarrow} &  & {\downarrow}{\exp}\\
G & \stackrel{\rho} {\cdots>} & \mathrm{GL}(\mathbb{V})
\end{matrix}
as $G$ is connected then $\mathrm{ad}$ determines a unique holomorphic representation $\rho$ of $G$ on $\mathbb{V}$.
By the previous statement, $\rho$ is a polynomial representation of $G$ on $\mathbb{V}$; as $G$ is a connected group then $\rho$ is a faithful representation.
Summing up, $G$ is a closed subgroup of $\mathrm{GL}(\mathbb{V})$ with respect to the Zariski topology, in particular $G$ is an algebraic group. (q.e.d.) $\Box$
By the proof of previous theorem, we can state the following corollary.

Corollary. Let $G$ be a connected, complex, semisimple Lie group; any closed Lie subgroup of $G$ is an algebraic subgroup of $G$ regarded as an algebraic group.

Remark 1. [BA] proves that an algebraic group (over an algebraically closed field) is affine if and only if it is isomorphic to a closed subgroup of some general linear group (of finite dimension over the same field) thus, any affine algebraic groups can also be called linear algebraic groups.
In particular, any algebraic group is a smooth algebraic variety (for a proof,  one can consult [BS]) so one proves that any complex algebraic group has a structure of complex Lie group. $\Diamond$
Remark 2. The tangent space $T_PM$ of a (differential) manifold $M$ at a point $P$ is isomorphic to the space of (real) linear derivation of the ring of germs of $M$ at $P$ (see [Spk]).
The same construction and the same result hold for the complex manifolds, that is the tangent space $T_PM$ of a complex manifold $M$ at a point $P$ is isomorphic to the space of (complex) linear derivation of $M$ at $P$.
On the other hand, the Zariski tangent space $T_PX$ of an arbitrary algebraic variety $X$ over a field $\mathbb{K}$ at a closed $\mathbb{K}$-valued point $P$ is isomorphic to the space of $\mathbb{K}$-linear derivation of $X$ at $P$ (see [FOAG]).
From all this, the Lie algebra $\mathfrak{g}$ of a connected, complex, semisimple or reductive Lie group $G$ is the Lie algebra of $G$ as algebraic complex group.
Moreover, the dimension of $G$ is the dimension of $\mathfrak{g}$, by the previous statement, the dimension of $\mathfrak{g}$ is the dimension of the Zariski tangent space to $G$ at $1_G$; as proved in [FOAG], the local dimension of $G$ as algebraic group at $1_G$ is the dimension of the Zariski tangent space to $G$ at $1_G$. $\Diamond$
From all this, I can state the following theorem.

Theorem 2. Let $G$ be a connected, complex, semisimple Lie group.
  The Lie algebra $\mathfrak{g}$ of $G$ as complex Lie group is the Lie algebra of $G$ regarding as complex algebriac group; moreover, the categories of holomorphic representations $\mathbf{Rep}_{hol}(G)$ of $G$ is equivalent to the category of algebraic representations $\mathbf{Rep}_{alg}(G)$ of $G$, regarding $G$ as algebraic complex group.

Is it all clear?
Bibliography

[BA] A. Borel (1991) Linear Algebraic Groups, II Edition, Springer-Verlag
[BS] S. Bosch (2011) Algebraic Geometry and Commutative Algebra, Springer-Verlag
[OV] A. L. Onishchik - E. B. Vinberg (1988) Lie Groups and Algebraic Groups, Nauka
[Spk] M. Spivak (1999) A Comprehensive Introduction to Differential Geometry; Volume 1, III Edition, Publish or Perish Inc.
[FOAG] R. D. Vakil (2015) Foundations of Algebraic Geometry, http://math.stanford.edu/~vakil/216blog/FOAGdec2915public.pdf, Stanford University