For simplicity, I will be talking only about connected groups over an algebraically closed field of characteristic zero.


The basic theorem of affine algebraic groups is that they all admit faithful, finite-dimensional representations. The *fundamental* theorem for semisimple groups is that these representations are all completely reducible, but unfortunately there is no reason that any irreducible summand of a faithful representation should be faithful, only that the kernels of all these representations intersect trivially.



> 
> My question is whether such a representation does, in fact, exist.
> 
> 
> (Answered: iff the center is cyclic.)
> 
> 
> 


This does not hold of general reductive groups for the following reason: if $T$ is any torus of rank $r > 1$, then its irreducible representations are all characters $\chi \colon T \cong \mathbb{G}\_m^r \to \mathbb{G}\_m$, which therefore have nontrivial kernels. More generally, any reductive group $G$ has connected center a torus of some rank $r$, so by Schur's lemma this center acts by a character $\chi$ in any irreducible representation of $G$ and if $r > 1$, therefore does not act faithfully.


The exceptional case $r = 1$ *does* have an example, namely $\operatorname{GL}\_n$, whose standard representation is faithful and irreducible and whose center has rank 1. A more general version of this question might be, then:



> 
> Does any reductive group whose center has rank at most 1 have a faithful irreducible representation?
> 
> 
> (Answered: when not semisimple, iff the center is connected.)
> 
> 
> 


Another special case is that if $G$ is simple and of adjoint type, then its adjoint representation is irreducible and faithful by definition (or, depending on your definition, because the center is trivial). A constructive version of this question for any $G$ (semisimple or reductive of central rank 1) is then:



> 
> Can we give a construction of a faithful, irreducible representation of $G$ from its adjoint representation?
> 
> 
> (Not yet answered!)
> 
> 
> 


This is deliberately a little vague since I don't want to restrict the possible form of such a construction, only that it not start out with "Throw away the adjoint representation and take another one such that..."


Finally, suppose the answer is "no".



> 
> What is the obstruction to such a representation existing?
> 
> 
> (Answered: for $Z$ the center, it is the existence of a generator for $X^\*(Z)$.)
> 
> 
> 



Do semisimple algebraic groups always have faithful irreducible representations?
*Edit*: I now give the argument for general reductive $G$.


Let $G$ be a reductive algebraic group over an alg. closed field $k$ of char. 0. Fix a max
torus $T$ and write $X = X^\*(T)$ for its group of characters. Write $R$ for the
subgroup of $X$ generated by the roots of $G$. Then the center $Z$
of $G$ is the diagonalizable subgroup of $T$ whose character group is $X/R$. 



> 
> **Claim:** $G$ has a faithful irreducible representation if and only if the character 
>  group $X/R$ of $Z$ is cyclic.
> 
> 
> 


Note for semisimple $G$, the center $Z$ is finite. Since the characteristic of $k$ is 0, in this case the group of $k$-points of $Z$ is (non-canonically) isomorphic
to $X/R$. Thus $Z$ is cyclic if and only $X/R$ is cyclic.


In general, the condition that $X/R$ is cyclic means either that the group of points $Z(k)$ is finite cyclic, or that $Z$ is a 1 dimensional torus.


As to the proof, for $(\implies)$ see Boyarsky's comment following reb's answer.


For $(\Leftarrow)$ let me first treat the case where $G$ is *almost simple*; i.e. where the root system $\Phi$ of $G$ is irreducible. Supopse that the class of $\lambda \in X$ generates the cyclic group $X/R$. Since
the Weyl group acts on $X$ leaving $R$ invariant, the class of any $W$-conjugate of $\lambda$ is also a generator of $X/R$. Thus we may as well suppose $\lambda$ to be dominant
and non-0 [if $X=R$, take e.g. $\lambda$ to be a dominant root...]
Now the simple $G$-module $L=L(\lambda) = H^0(\lambda)$ with "highest weight $\lambda$"
will be faithful. To see this, note that since $\lambda \ne 0$, $L$ is not the trivial representation. Since $G$ is almost simple, the only proper normal subgroups
are contained in $Z$. Thus it suffices to observe that the action of $Z$ on the
$\lambda$ weight space of $L$ is faithful.


The general case is more-or-less the same, but with a bit more book-keeping.
Write the root system $\Phi$ of $G$ as a disjoint union $\Phi = \cup \Phi\_i$
of its irreducible components. There is an isogeny
$$\pi:\prod\_i G\_{i,sc} \times T \to G$$
where $T$ is a torus and $G\_{i,sc}$ is the simply connected almost simple group
with root system $\Phi\_i$. Write $G\_i$ for the image $\pi(G\_{i,sc}) \subset G$.


The *key fact* is this: 
a representation $\rho:G \to \operatorname{GL}(V)$ has
$\ker \rho \subset Z$ if and only if the restriction $\rho\_{\mid G\_i}$ is non-trivial
for each $i$.


Now, as before pick $\lambda \in X$ for which the coset of $\lambda$ generates
the assumed-to-be cyclic group $X/R$. After replacing $\lambda$ by a Weyl group
conjugate, we may suppose $\lambda$ to be dominant. After possibly repeatedly replacing
$\lambda$ by $\lambda + \alpha$ for dominant roots $\alpha$, we may suppose
that $\lambda$ has the following property:
$$(\*) \quad \text{for each $i$, there is $\beta\_i \in \Phi\_i$ with $\langle \lambda,\beta\_i^\vee \rangle \ne 0$}$$


Now let $L = L(\lambda)$ be the simple module with highest weight $\lambda$. Condition
$(\*)$ implies that $G\_i$ acts nontrivially on $L$ for each $i$, so by the "key fact",
the kernel of the representation of $G$ on $L$ lies in $Z$. but since $\lambda$
generates the group of characters of $Z$, the center $Z$ acts faithfully on the
$\lambda$ weight space of $L$.


