The Brieskorn Spheres And Their Fundamental Groups

The Brieskorn spheres $\Sigma(p, q, r)$ are important examples of homology 3-dimensional spheres, i. e., manifolds with the homology being like one of a 3-dimensional sphere. They were extensively studied by Egbert Brieskorn and John Milnor. Henry Poincaré described the first one, $\mathbb{S}^3 / 2I\cong \Sigma(2, 3, 5)$ (see the proof below) in his “Cinquiéme complément à l’Analysis situs” (1904), where

is the binary icosahedral group. Now, the manifold $\mathbb{S}^3 / 2I$ is known to be the Poincaré sphere.

Definition 1

A Brieskorn manifold $\Sigma(p, q, r)$, (with positive integers $p, q, r$) is a manifold

\begin{equation*} \Sigma(p, q, r) := \{ z\in\mathbb{C}^3 \ | \ z_1^p + z_2^q + z_3 ^r = 0\}\cap \{ z\in\mathbb{C}^3 \ | \ |z| = 1\}. \end{equation*}

Remark 1. By analogy, one can define manifolds $\Sigma(a_1, …, a_n)$ of higher dimensions. There are 28 well-known 7-dimensional exotic spheres (i. e., 27 of them have non-standard smooth structures) discovered by Milnor:

Remark 2. In dimension greater than 3, Brieskorn manifolds $\Sigma(a_1, …, a_n)$ are simply connected. So, in this case, any such a homology sphere will be homeomorphic to the standard sphere. Brieskorn formulated the criteria of $\Sigma(a_1, …, a_n)$ being trivial. This one relies on the vanishing of some polynomial $P(a_1 ,…, a_n)$.

The Poincaré Sphere

Recall the construction of Poicaré’s sphere. We can look at $\mathbb{R}^4$ as the quaternion algebra $\mathbb{H}$. There is a 2-fold cover $p: \mathbb{S}^3\to \mathrm{SO}(3)$ sending each unit quaternion $q$ (i. e., a quaternion with the standard Euclidean norm 1) to the inner automorphism $\rho_q$ (i. e., $\rho_q(q_1) = qq_1q^{-1}$ for any quaternion $q_1$). The icosahedra rotations group $A_5$ sits inside $\mathrm{SO}(3)$. Consider its inverse image under $p$, then the corresponding quotient space $\mathbb{S}^3 / p^{-1}(A_5)$ will be the desired Poincaré sphere.

When Does A Brieskorn Manifold Become A Homology Sphere?

It turns out that under some easy conditions, manifolds $\Sigma(p, q, r)$ happen to be homology spheres:

Theorem 1 (Milnor’75)

A manifold $\Sigma(p, q, r)$ is a homology sphere if and only if $p, q$ and $r$ are pairwise coprime. If at lest one of them equals 1, then $\Sigma(p, q, r)$ homeomorphic to the standard sphere $\mathbb{S}^3$.

Sketch of proof. The commutator subgroup of a group

acts freely on the universal covering space $\widetilde{\mathrm{Isom}^{+}}(P)$ of the plane $P$ isometries group $\mathrm{Isom}^{+}(P)$. P denotes a plane with the constant curvature $K = \mathrm{sign}(1/p + 1/q + 1/r - 1)$. As it was shown by Milnor, the resulting quotient space $\widetilde{\mathrm{Isom}^{+}}(P) / \Gamma’(p, q, r)$ is diffeomorphic to $\Sigma(p, q, r)$. Recall, if $K = 0, -1$ or $1$, then we will have $\widetilde{\mathrm{Isom}^{+}}(P)$ to be $\mathbb{S}^3$, $\widetilde{\mathrm{PSL}(2, \mathbb{R})}\cong \mathbb{R}^3$ or $\mathbb{R}^2\rtimes \mathrm{SO}(2)\cong\mathbb{R}^3$ respectively.

So, it suffices to prove $\Gamma’(p, q, r)$ being perfect (i. e., the first homology vanishes). It can be done by noticing that a manifold $\Sigma(p, q, r)$ is a Seifert fibration over $\mathbb{S}^3$ and it is obtained by some sequence of Dehn surgeries. Let $p, q$ and $r$ be denoted by $a_i$, $i=1, 2, 3$ for simplicity. A calculation of the fundamental group via the Seifert-Van Kampen theorem concludes

Now, it is easy to see the desired property of $\Gamma(p, q, r)$.

Corollary 1

A Brieskorn homology sphere $\Sigma(p, q, r)$ is an aspherical space (i. e., all except $\pi_1$ homotopy groups vanish) if and only if

\begin{equation*} 1/p + 1/q + 1/r < 1 \end{equation*}

Note here the following fun fact:

Theorem 2 (Milnor’57)

The Poincaré sphere is the only homology sphere with a finite non trivial fundamental group.

The Geometric Meaning Of $\Gamma(p, q, r)$

The groups $\Gamma(p, q, r)$ from the proof have the following geometric interpretation. Consider the Coxeter group $\Delta(p, q, r)\subset \mathrm{Isom}(P)$ generated by side reflections of a triangle with angles $\pi/p, \pi/q$ and $\pi/r$ on the plane $P$ with a metric of a constant Gaussian curvature $K = 0, \pm 1$. The group $\Delta(p, q, r)$ contains a subgroup $\Delta^{+}(p, q, r)$ of proper isometries. On can show that the von Dyck group

is isomorphic to $\Delta^{+}(p, q, r)$.

There is the universal cover

over the plane isometries group. Then,

The inverse image

is a central extension $\Gamma(p, q, r)$ of $D(p, q, r)$ by $C$

As it was mentioned in the proof of the theorem above, Brieskorn’s spheres are the quotient spaces: $\mathbb{S}^3 / \Gamma’(p, q, r)$ if the curvature of a triangle is positive, and $\mathbb{R}^3 / \Gamma’(p, q, r)$ otherwise.

Example 1. The Poincaré sphere $\mathbb{S}^3/2I$ is a Brieskorn homology sphere $\Sigma(2, 3, 5)$. Indeed, the group $D(2, 3, 5)$ is isomorphic to $A_5$ (see the presentation above). Its inverse image $\Gamma(2, 3, 5)$ under $p$ will be $2I$. But $2I$ is perfect, hence, $(2I)’ = 2I$, and $\Sigma(2, 3, 5)\cong \mathbb{S}^3/2I$.

The Deficiency Of $\Gamma(p, q, r)$

The deficiency $\mathrm{def}(G)$ of a group $G$ is an important invariant in combinatorial group theory. It is defined as the maximal value of $|F| - |R|$ over all possible presentations of

Example 2. The deficiency of any finite group $G$ with finite group presentation equals 0. Indeed, for such a group, its first homology $H_1(G) = G/G’$ is finite. So, $|F|\leqslant |R|$ for any presentation. Particularly, the trivial group and the binary icosahedral group have zero deficiency.

In this work, the following inequality is proved for any group $G = \langle F\ | \ R\rangle$:

where $s(H_2(G))$ denotes a minimal number of generators of $H_2(G)$.

Applying this inequality and using the fact that $H_2(G)$ is finitely generated¹, we obtain

Proposition 1

For any finitely presented group $G$, its deficiency is a finite number, i. e., $\mathrm{def}(G)<\infty$.

From the inequality ($*$), one can easily derive the following folklore result:

Proposition 2

Suppose that a finitely presented group $G$ has zero the first and the second homologies (with trivial $\mathbb{Z}$ coefficients) — such groups are called super perfect. Then its deficiency equals 0.

Now, let us prove

Proposition 3

A group $\Gamma(p, q, r)$ has zero deficiency for all $p, q, r > 1$.

Proof. Since $\Gamma(p, q, r)$ has balanced presentation, it suffices to prove that $\mathrm{def}(\Gamma(p, q, r))\leqslant 0$. As we we saw earlier, $\Gamma(p, q, r)$ is a fundamental group of the closed 3-dimensional manifold

Now, consider the following well-known exact sequence

By the Poincaré duality, $H_2(M)\cong H^1(M)$. And the universal coefficient theorem gives

So, from our favourite inequality ($*$) then follows $\mathrm{def}(M)\leqslant 0$.

References

1. M. H. Poincaré. Cinquième complément a l’Analysis situs. Rendiconti del Circolo Matematico di Palermo (1884-1940), 18:45–110, 1904.

2. John Milnor. On the 3-dimensional Brieskorn manifolds $M(p, q, r)$, pages 175–226. Princeton University Press, Princeton, 1975

3. E. Brieskorn. Beispiele zur Differentialtopologie von Singularitäten. Inventiones mathematicae, 2:1–14, 1966
4. Milnor J. Groups Which Act on Sn Without Fixed Point. American Journal of Mathematics, 79(3):623–630, 1957
5. John Milnor. Singular Points of Complex Hypersurfaces. (AM-61). Princeton University Press, 1968
6. R.V. Gamkrelidze, N. Saveliev, and A. Vassiiev. Invariants of Homology 3-Spheres. Encyclopaedia of Mathematical Sciences. Springer Berlin Heidelberg, 2002