# The Poincare sphere

Following on from my previous post, I would like to add a little introduction to the Poincare homology sphere early on in Chapter 1.  This gives an opportunity to introduce the $$E_8$$ plumbing in a reasonably down-to-earth context, so it will not come as such a surprise when we bring it up with the exotic spheres.

So the historical introduction goes: classification of 2-manifolds; Poincaré and 3-manifolds, and the history of the Poincare conjecture; the Fifth Complement to Analysis Situs and the Poincaré homology sphere.

Poincaré’s discussion is phrased in terms of (what became known as) a Heegard splitting, but without too much difficulty this can be related to the calculations (using van Kampen’s theorem) to compute the fundamental group below…

We start with 2-disk bundles over the 2-sphere.  These are classified by an Euler number.  We take 8 copies of this bundle and plumb them together according to the $$E_8$$ Dynkin diagram: Each node of the Dynkin diagram represents a copy of our bundle, and two bundles are plumbed together if the two nodes are linked by an edge.  The result (after rounding the corners) is a smooth 4-manifold with boundary; call the boundary  $$M$$, a closed 3-manifold.

The homology of $$M$$ is computed from a complex whose only non-trivial differential is the $$E_8$$ matrix.  Because this matrix is unimodular, $$M$$ is a homology sphere.  This part of the computation is the same as in higher dimensions.

However, $$M$$ is not simply connected.   This can be seen as follows.  Using van Kampen’s theorem, one computes that the fundamental group of $$M$$ has one generator for each node of the Dynkin diagram, and also one relator for each node.  Let $$x_k$$ be the generator corresponding to node $$k$$. The relator corresponding to node $$k$$ is of the form

$x_k^2 x_{j_1}x_{j_2}\ldots = 1 ,$

where $$j_1,j_2,\ldots$$ are the nodes adjacent to $$k$$ in the Dynkin diagram.  In the case of the diagram above this yields

$1 = x_1^2x_2 = x_2^2x_1x_3 = x_3^2x_4x_5x_2 = x_4^2x_3 = x_5^2x_3x_6 = x_6^2x_5x_7=x_7^2x_8x_6 = x_8^2x_7.$

Writing $$x=x_1$$, $$y=x_8$$ allows us to present the group $$\pi_1(M)$$ as

$\langle x,y | x^3 = y^5 = (xy)^2 \rangle.$

The permutations $$x = (531)$$, $$y = (12345)$$ generate the alternating group $$A_5$$ and satisfy the above relations (with $$x^3 = y^5 = (xy)^2 = 1$$.)  Thus $$\pi_1(M)$$ surjects onto $$A_5$$ and in particular is not trivial.  (In fact, it is known that the kernel is of order 2; $$\pi_1(M)$$ is the binary icosahedral group of order 120.

Thus $$M$$ is a homology sphere but not a homotopy sphere.

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