# Riemann surfaces

It is known that $\mathbb{H}$ (upper hyperplane), $\mathbb{C}$, $\mathbb{C} \cup \infty$ (riemann sphere) are riemann surfaces. $\mathbb{H} \cup i\infty$ is not, but what is the principal difference between this case and $\mathbb{C} \cup \infty$?

• Boundary of $\mathbb H$ is homeomorphic with $S^1$, but in this case you are only adding a single point in the boundary. – Anubhav Mukherjee Oct 24 '16 at 16:53
• $\mathbb{H}$ is homeomorphic with disk. anyway, thanks – Sasha Mayer Oct 24 '16 at 17:30
• so what?? I didn't get the meanng of your comment? Do you know the boundary of of upper half plane? – Anubhav Mukherjee Oct 24 '16 at 18:02
• @Anubhav, i've missed the word "boundary " in your previous comment, sorry. i agree with you, that the problem is with neighbourhood of that point. – Sasha Mayer Oct 24 '16 at 18:06
• The issue is the same with adding $0$ to the open upper half-plane (no neighborhood of $0$ in $\mathbf{H} \cup \{0\}$ is biholomorphic to an open disk) versus adding $0$ to the punctured plane. – Andrew D. Hwang Oct 25 '16 at 14:14

To answer this, one must understand that there is no isomorphism between an annulus $A_a=\{a<\vert z\vert <1\}$ and the punctured disk $\{0<\vert z\vert <1\}$.
Let us think about $\bf H$ as a disk ${\bf D}=\{ \vert z\vert <1\}$, and consider the annulus $A= A_{1/2}$ ; assume that there exists a complex structure on ${\bf D}\cup \infty$ which restricts to the given structure on $A$. Then $A\cup \infty$ would be a simply connected Riemann surface, hence either $\bf C$ of $\bf D$.
It cannot be $\bf C$ as $\bf C\setminus \{1pt\}$ is not isomorphic to a punctured disk (the universal cover of $\bf C\setminus \{1pt\}$ is $\bf C$). So it must be $\bf D$, and (as $:bf D$ is homogeneous) removing a point would provide an isomorphism between the punctered disk and $A_{1/2}$.