# Why is Kummer's surface a smooth manifold

The $$\mathbb{Z}/2\mathbb{Z}$$ action on $$\mathbb{T}^4$$ has $$16$$ fixed points. Blowing up at these $$16$$ points, we get a new manifold $$\tilde{\mathbb{T}}^4$$. The action extends trivially at each exceptional divisor. Then the quotient space of $$\tilde{\mathbb{T}}^4$$ is called Kummer's surface.

Since the action extends trivially, each exceptional divisor is fixed by the action so that the action is not free there. Thus, we cannot apply quotient manifold theorem. How to see Kummer's suface is a smooth manifold?

• Are you happy with the squaring mapping $\mathcal{O}(-1) \to \mathcal{O}(-2)$, which in effect quotients each fiber by the action $z \mapsto -z$? Jul 4, 2022 at 11:24
• @AndrewD.Hwang Yes I am very happy with that. However, I don't know how to see it. For me, the definition of $\mathcal{O}(-2)$ is $\mathcal{O}(-1)\otimes \mathcal{O}(-1)$. How to obtain your claim by this definition? Could you recommend some references? Jul 4, 2022 at 16:01

$$\newcommand{\Cpx}{\mathbf{C}}\newcommand{\Sheaf}{\mathcal{O}}$$tl; dr: The squaring map $$\Sheaf(-1) \to \Sheaf(-2)$$ is the local model for blowing up the isolated singularity of $$\Cpx^{2}/(\pm1)$$.
Let $$n$$ be an integer. The line bundle $$\Sheaf(n)$$ may be constructed from two copies of $$\Cpx^{2}$$, with respective coordinates $$(z^{0}, \zeta^{0})$$ and $$(z^{1}, \zeta^{1})$$, by identifying (where the formulas are sensible) $$z^{0} = \frac{1}{z^{1}},\qquad \zeta^{0} = \frac{\zeta^{1}}{(z^{1})^{n}}.$$ That is, the transition function is $$g_{01}(z^{0}) = (z^{0})^{-n}$$.
The complement of the zero section of $$\Sheaf(-1)$$ is identified with $$\Cpx^{2}\setminus\{(0, 0)\}$$: In Cartesian coordinates $$(Z^{0} : Z^{1})$$ on $$\Cpx^{2}$$, the identification $$(z^{0}, \zeta^{0}) \simeq (1/z^{1}, z^{1}\zeta^{1})$$ is compatible with the affine coordinates $$z^{0} = Z^{0}/Z^{1}$$ and $$z^{1} = Z^{1}/Z^{0}$$ on the projective line, and the fibre coordinates $$\zeta^{0} = Z^{1}$$ and $$\zeta^{1} = Z^{0}$$.
Now, instead of the involution $$(Z^{0}, Z^{1}) \mapsto (-Z^{0}, -Z^{1})$$ (which fixes the origin and identifies negatives in lines through the origin, a.k.a. fibres of the tautological bundle $$\Sheaf(-1)$$), consider the compatible mappings \begin{align*} &(z^{0}, \zeta^{0}) \mapsto (z^{0}, (\zeta^{0})^{2}) = (z^{0}, \xi^{0}), \\ &(z^{1}, \zeta^{1}) \mapsto (z^{1}, (\zeta^{1})^{2}) = (z^{1}, \xi^{1}) \end{align*} (which fix the zero section and identify negatives in the fibres). We have $$\xi^{0} = (\zeta^{0})^{2} = (\zeta^{1})^{2}(z^{1})^{2} = \xi^{1}(z^{1})^{2},$$ which signifies that $$(z^{0}, \xi^{0})$$ and $$(z^{1}, \xi^{1})$$ are trivializing charts for $$\Sheaf(-2)$$. We have therefore defined a global bundle mapping of total spaces $$\Sheaf(-1) \to \Sheaf(-2)$$ that is the identity on the zero section and is two-to-one elsewhere, identifying pairs $$\pm\zeta$$ in the fibres.
The upshot is, quotienting $$\Cpx^{2}$$ by the antipodal involution then blowing up the image of the origin is the same as blowing up the origin in $$\Cpx^{2}$$ then forming the quotient, which yields the smooth surface $$\Sheaf(-2)$$.
(That's why the Kummer surface as described is smooth, and why it contains sixteen $$-2$$-curves. The other two-dimensional cohomology corresponds to images of the $$\binom{4}{2} = 6$$ real $$2$$-tori in the original real $$4$$-torus.)