The double cover of $SL(2,\mathbb R)$ is not algebraic.

This can be blamed on the fact that the map $$\pi_1\big(SL(2,\mathbb R)\big)\cong \mathbb Z\quad\longrightarrow\quad \pi_1\big(SL(2,\mathbb C)\big)=0$$ is not injective.

If the double cover of $\pi_1(SL(2,\mathbb R))$ was algebraic, it would have a complexification, which would be a double cover of $SL(2,\mathbb C)$. But $SL(2,\mathbb C)$ doesn't have any double covers since its fundamental group is trivial.

Using that method, you can actually detect which covers are algebraic:
Let $G$ be a real algebraic Lie group, and let $A$ be a finite abelian group. A central extension of $G$ by $A$ determines a homomorphism $\pi_1(G)\to A$. The cover is algebraic iff that homomorphism extends to a homomorphism $\pi_1(G_{\mathbb C})\to A$.

The double cover of $SL(2,\mathbb R)$ is not algebraic.

This can be blamed on the fact that the map $$\pi_1\big(SL(2,\mathbb R)\big)\cong \mathbb Z\quad\longrightarrow\quad \pi_1\big(SL(2,\mathbb C)\big)=0$$ is not injective.

If the double cover of $\pi_1(SL(2,\mathbb R))$ were algebraic, it would have a complexification, which would be a double cover of $SL(2,\mathbb C)$. But $SL(2,\mathbb C)$ doesn't have any double covers since its fundamental group is trivial.

Using that method, you can actually detect which covers are algebraic:
Let $G$ be a real algebraic Lie group, and let $A$ be a finite abelian group. A central extension of $G$ by $A$ determines a homomorphism $\pi_1(G)\to A$. The cover is algebraic iff that homomorphism extends to a homomorphism $\pi_1(G_{\mathbb C})\to A$.