Intuition behind centralizers of Langlands parameters In the description of the Langlands correspondence for $\mathbb{Q}_p$, we consider admissible representations of $G(\mathbb{Q}_p)$ for $G$ a reductive group defined over $\mathbb{Q}_p$, and admissible Langlands parameters $\phi: W \to ^LG$, where $W$ denotes the absolute Weil group of $\mathbb{Q}_p$. In particular, to each irreducible admissible representation $\pi$ of $G(\mathbb{Q}_p)$, we (to some extent conjecturally) assign an $L$-parameter $\pi_{\phi}$. The map $\pi \mapsto \pi_{\phi}$ satisfies a number of compatibility conditions and is finite to one. The set theoretic fiber of an $L$-parameter is called an $L$-packet and denoted $\Pi_{\phi}$. Various descriptions of the contents of $\Pi_{\phi}$ have been suggested (in particular in the tempered case), typically involving some variant of the set of irreducible representations of the centralizer of the image of $\phi$, denoted $S_{\phi}$.
My question asks for an intuitive (or otherwise) explanation of why representations of this centralizer group should be involved in parametrising the contents of the corresponding $L$-packet.
 A: The simplest non-trivial case for this is for $G = SL_2$. If you take an unramified principal series representation of $\tilde G = GL_2$ and restrict it to $G$, then it will almost always be irreducible; but there is a bad case where it's a direct sum of 2 things, which is when the ratio of the Satake parameters is $-1$, and these two summands form a non-singleton $L$-packet. This is all worked out very beautifully in Labesse--Langlands.
Now, since the Langlands parameter for $\tilde G$ is just a diagonal matrix, you can explicitly compute its centraliser in ${}^L \tilde G = GL_2(\mathbf{C})$ and in ${}^L G = PGL_2(\mathbf{C})$. [*] You will see that the former is always connected, but the latter has either 1 or 2 components, and it has 2 components exactly when the ratio of Satake params is $-1$. 
This, at least, shows that the component group of the centraliser is giving the right answer in this simple case.
([*] I am lying a bit here: this is the dual group $\hat G$, not the $L$-group ${}^L G$ which is the semidirect product of $\hat G$ with the local Weil group. However, in our case the semidirect product is actually direct, because both groups are split, so we can ignore the second factor.)
