irreducible representations of $GL_2$ over $p$-adic field Let $E$ be a finite extension of $\mathbf{Q}_p$. In class we stated the following fact :
Every irreducible algebraic representation of $GL_2(E)$ is of the form 
$$
\mathrm{Sym}^{k-2}(E^2) \otimes_E {\det}^m
$$
for $k \geq 2$ and $m \in \mathbf{Z}$
Questions : 
1) What does algebraic mean here ?
2) do you know a reference ?
 A: To confirm what Tobias said, "algebraic" in this case means a homomorphism of algebraic groups $GL_2(E)\rightarrow GL_N(E)$ for some $N$, i.e. a group homomorphism defined via polynomial functions. A slight word of warning: in the earlier years of the subject, an "algebraic representation" of a $p$-adic group such as $GL_2(E)$ meant something very different to this (what's now usually called a smooth representation).
As for the most straightforward proof that I can think of, let me sketch out how you can do this by first considering $SL_2(E)$. Let me cover myself by saying that I haven't thought very carefully about the details, but I don't think that anything should go wrong; maybe some things will just be a bit harder than I'm claiming.
A fairly well-known result is the classification of the finite-dimensional irreducible complex representations of the Lie algebra $\frak{sl}_2(\Bbb{C})$, which is via looking at highest weights. This leads to the result that the irreducible representations are precisely the representations $Sym^{k-2}(\Bbb{C}^2)$. In fact, this holds more generally: one obtains the same classification of the $k$-representations of ${\frak{sl}}_2(k)$ for any field $k$ of characteristic $0$: see here, for example.
The algebraic group $SL_2(E)$ has a Lie algebra defined in a way reasonably analogous to the usual definition (the tangent space at the identity identifies with derivations and inherits a Lie bracket), which leads to the usual result that it's Lie algebra is ${\frak{sl}}_2(E)$ as above, namely the Lie algebra of 2x2 traceless matrices over $E$. Then by the usual argument of "differentiating" an irreducible algebraic representation of $SL_2(E)$, one obtains an irreducible representation of ${\frak{sl}}_2(E)$. So one needs only to classify the irreducible representations of ${\frak{sl}}_2(E)$, and then note that every such representation arises via "differentiation", in order to classify the irreducible algebraic representations of $SL_2(E)$.
It then remains to extend this classification to $GL_2(E)$ (in a general setting, this is fairly common: $GL_2$ is a reductive group with derived subgroup $SL_2$. The group $SL_2$ is semisimple, and so a bit easier to work with as one doesn't really have to worry about the centre, and the representation theories of the two groups are closely related). Let me write $V(k)$ for the representation $Sym^{k-2}(E^2)$ of $SL_2(E)$. Since the action of $SL_2(E)$ on $E^2$ clearly extends to an action of $GL_2(E)$ (which obviously remains algebraic), $V(k)$ extends to a representation of $GL_2(E)$ for any $k\geq 2$. This means that any irreducible representation of $GL_2(E)$ which contains $V(k)$ upon restriction to $SL_2(E)$ must be isomorphic to $V(k)\otimes\chi$, where $\chi$ is some irreducible representation of $GL_2(E)/SL_2(E)\simeq G_m(E)$, viewed as a representation of $GL_2(E)$ via inflation. It's easy to see that the only such $\chi$ are the powers of $\det$: use the classification of characters of $G_m(E)$ (it's a split torus of rank 1 so they are precisely the characters $x\mapsto x^m$ for $m\in\Bbb{Z}$) and then note that inflating to $GL_2(E)$ means pulling back through $\det$, so $\chi$ is of the form $x\mapsto\det(x^m)$, i.e. $\chi=\det^m$.
