Let $\operatorname{Ad}:\operatorname{SL}\_n(\mathbb{R}) \to \operatorname{GL}(\mathfrak{sl}\_n(\mathbb{R}))$ be the adjoint representation (i.e. $\operatorname{Ad}(g)X=gXg^{-1}$) of $SL\_n(\mathbb{R})$. Then is the image of $\operatorname{Ad}$ a Zariski closed subgroup? 


In general, are the any sufficient conditions (e.g. semisimplicity, lack of compact factors) for the image of the adjoint representation of a **real** algebraic group to be Zariski closed?



When is the image of the adjoint representation of a real algebraic group Zariski closed?
Let $G$ be a connected linear algebraic group over the field of real numbers $\mathbb{R}$.
By "connected" I mean "connected over $\mathbb{C}$".
Let $G(\mathbb{R})$ denote the group of $\mathbb{R}$-points of $G$.
It does not have to be connected.
Let $\Gamma\subset G(\mathbb{R})$ be a subgroup of finite index in $G(\mathbb{R})$,
then its Zariski closure is $G(\mathbb{R})$ (because $G$ is connected).
Thus $\Gamma$ is Zariski closed in $G(\mathbb{R})$ if and only if $\Gamma=G(\mathbb{R})$.


Denote by $G^{\mathrm{ad}}$ the image (over $\mathbb{C}$) of the adjoint representation
$$ \mathrm{Ad}\colon G\to \mathrm{GL}(\mathrm{Lie}\, G),$$
this group is defined over $\mathbb{R}$.
The image $\mathrm{Ad}(G(\mathbb{R}))$ is a subgroup of finite index in $G^{\mathrm{ad}}(\mathbb{R})$.
We conclude that $\mathrm{Ad}(G(\mathbb{R}))$ is Zariski closed in $\mathrm{GL}(\mathrm{Lie}\, G)$
if and only if $\mathrm{Ad}(G(\mathbb{R}))=G^{\mathrm{ad}}(\mathbb{R})$.


Now assume that $G$ is simply connected (over $\mathbb{C}$).
Then $G(\mathbb{R})$ is connected, see Onishchik and Vinberg, Lie Groups and Algebraic Groups, 5.2.1, Thm. 3.
It follows that $\mathrm{Ad}(G(\mathbb{R}))$ is connected, hence it is the identity component of $G^{\mathrm{ad}}(\mathbb{R})$.
Thus $\mathrm{Ad}(G(\mathbb{R}))$ is Zariski closed if and only if $G^{\mathrm{ad}}(\mathbb{R})$ is connected.


Now assume that $G=\mathrm{SL}\_n$ (then $G$ is simply connected). In this case $G^{\mathrm{ad}}(\mathbb{R})$ has one connected component when $n$ is odd,
and it has two connected components when $n$ is even, see, e.g.,  [this preprint](http://arxiv.org/abs/1310.7917), Section 9.3.
Thus $\mathrm{Ad}(\mathrm{SL}\_n(\mathbb{R}))$ is Zariski closed if and only if $n$ is *odd*.
In particular, in $n=2$, then $\mathrm{Ad}(\mathrm{SL}\_n(\mathbb{R}))$ is *not* Zariski closed.


Concerning a sufficient condition, it suffices to assume that $G$ is compact (i.e., $G(\mathbb{R})$ is compact), see my second comment to the question.


