Let $K(\mathbb P^1)$ be the Grothendieck group of sheaves on $\mathbb P^1$. I want to show that the map $K^{{\rm PGL}(2)\times \{\pm 1\}}(\mathbb P^1) \to K(\mathbb P^1)$ is not onto. I read somewhere that the class of a vector bundle of odd degree on $\mathbb P^1$ is not equal to the class of a sheaf equivariant under ${\rm PGL}(2)\times \{\pm 1\}$. Could you explain me why this is the case?



Equivariant sheaves on $\mathbb P^1$
Let me explain why the line bundle $\mathcal{O}(1)$ does not admit a $\mathrm{PGL}(2)$-equivariant structure. Indeed, if it does, then the vector space
$$
\mathrm{Hom}(\mathcal{O}, \mathcal{O}(1))
$$
would have a structure of a $\mathrm{PGL}(2)$-representation, compatible with the standard $\mathrm{SL}(2)$-representation structure. But this is not true, because the central element $-1 \in \mathrm{SL}(2)$ acts on this space non-trivially.


