# Regarding the representation theory of $SL_2(\mathbf{R})$.

Dear friends of mathematics, I have the following question for you.

(a) According to Wikipedia there is a unique irreducible (real??) $2$-dimensional representation of $SL_2(\mathbf{R})$, which must be the standard representation $i$ (for it is irreducible).

(b) If we take an automorphism $\phi$ of $SL_2$, then the representation $i\circ\phi$ must thus be isomorphic to $i$, which implies that $\phi$ is an inner automorphism. However, the group of outer automorphisms of $SL_2(\mathbf{R})$ is supposed to have order $2$, namely conjugation by the matrix $\mathrm{diag}(1,-1)$ should be an outer automorphism.

I think (a) and (b) contradict each other. How is this possible? Is the information given in Wikipedia wrong? Or does it refer to complex representations? For Wikipedia also says that the representation theory of $SL_2(\mathbf{R})$ is "the same" as the one of $SU_2$, in which case it would be true for complex representations.

If this is true, how many real $2$-dimensional irreducible representations does $SL_2$ have? References are much welcome. How is the general relationship between complex and real representations? Also I wonder how the 2-dimensional real irreducible repreentations of $SL_2(R)$ are related to the 2-dimensionl complex irred reps of $SL_2(\mathbf{C})$

The fact that $i\circ\phi$ is isomorphic to $i$ doesn't imply that $\phi$ is an inner automorphism.
• Thanks for the answer. But it is true that it implies that $\phi$ is of the form $\phi_R=R(-)R^{-1}$ right? Your point is that (unless $R$ has determinant $>0$) $R$ need not be (up to a scalar multiple) an element of $SL_2$. Over the complex numbers it would be an inner automorphism, though? – Herr Harry Heiner Jan 23 '16 at 19:10
• But is it true that $SL_2$ has only one $2$-dim irred rep? – Herr Harry Heiner Jan 23 '16 at 19:12
• In the above $R$ is a member of $GL_2$ – Herr Harry Heiner Jan 23 '16 at 20:34
• I would like to accept your answer, but I would like to know whether I understand it correctly. I mean if $i\circ\phi$ is isomorphic to $i$ implies that there is $R\in GL_2$ such that for all $A\in SL_2$ we have $\phi(A)\circ R=R\circ A$. – Herr Harry Heiner Jan 24 '16 at 10:26
• @HerrHarryHeiner Yes, that's correct. And to answer the question in your previous comment, yes, $SL_2(\mathbf{R})$ has only one $2$-dimensional real irreducible representation: the classification of the finite-dimensional irreducible representations is the same whether you're looking at real or at complex representations. – Jeremy Rickard Jan 24 '16 at 10:32