Fall 2014


Let \(X = {\mathbf{R}}^3 - \Delta^{(1)}\), the complement of the skeleton of regular tetrahedron, and compute \(\pi_1(X)\) and \(H_*(X)\).

Lay the graph out flat in the plane, then take a maximal tree - these leaves 3 edges, and so \(\pi_1(X) = {\mathbf{Z}}^{\ast 3}\).

Moreover \(X \simeq S^1\vee S^1 \vee S^1\) which has only a 1-skeleton, thus \(H_*(X) = [{\mathbf{Z}}, {\mathbf{Z}}^3, 0\rightarrow]\).


Let \(X = S^1 \times B^2 - L\) where \(L\) is two linked solid torii inside a larger solid torus. Compute \(H_*(X)\).



Let \(L\) be a 3-manifold with homology \([{\mathbf{Z}}, {\mathbf{Z}}_3, 0, {\mathbf{Z}}, \ldots]\) and let \(X = L \times\Sigma L\). Compute \(H_*(X), H^*(X)\).

Useful facts:

  • \(H_{k}(X\times Y) \cong \bigoplus _{{i+j=k}}H_{i}(X)\otimes H_{j}(Y) \bigoplus_{i+j=k-1}\operatorname{Tor}(H_i(X), H_j(Y))\)
  • \(\tilde H_i(\Sigma X) = \tilde H_{i-1}(X)\)

We will use the fact that \(H_*(\Sigma L) = [{\mathbf{Z}}, {\mathbf{Z}}, {\mathbf{Z}}_3, 0, {\mathbf{Z}}]\).

Represent \(H_*(L)\) by \(p(x, y) = 1 + yx + x^3\) and \(H_*(\Sigma L)\) by \(q(x,y) = 1 + x + yx^2 + x^4\), we can extract the free part of \(H_*(X)\) by multiplying

\begin{align*}p(x,y)q(x,y) = 1 + (1+y)x + 2yx^2 + (y^2+1)x^3 + 2x^4 + 2yx^5 + x^7\end{align*}

where multiplication corresponds to the tensor product, addition to the direct sum/product.

So the free portion is \begin{align*}H_*(X) = [{\mathbf{Z}}, {\mathbf{Z}}\oplus {\mathbf{Z}}_3, {\mathbf{Z}}_3\otimes{\mathbf{Z}}_3, {\mathbf{Z}}\oplus {\mathbf{Z}}_3\otimes{\mathbf{Z}}_3, {\mathbf{Z}}^2, {\mathbf{Z}}_3^2, 0, {\mathbf{Z}}] \\ =[{\mathbf{Z}}, {\mathbf{Z}}\oplus {\mathbf{Z}}_3, {\mathbf{Z}}_3, {\mathbf{Z}}\oplus {\mathbf{Z}}_3, {\mathbf{Z}}^2, {\mathbf{Z}}_3^2, 0, {\mathbf{Z}}] \end{align*}

We can add in the correction from torsion by noting that only terms of the form \(\operatorname{Tor}({\mathbf{Z}}_3, {\mathbf{Z}}_3) = {\mathbf{Z}}_3\) survive. These come from the terms \(i=1, j=2\), so \(i+j=k-1 \implies k = 1+2+1 = 4\) and there is thus an additional torsion term appearing in dimension 4. So we have

\begin{align*}H_*(X) = [{\mathbf{Z}}, {\mathbf{Z}}\times {\mathbf{Z}}_3, {\mathbf{Z}}_3, {\mathbf{Z}}\times {\mathbf{Z}}_3, {\mathbf{Z}}^2 \times {\mathbf{Z}}_3, {\mathbf{Z}}_3^2, 0, {\mathbf{Z}}] \\ = [{\mathbf{Z}}, {\mathbf{Z}}, 0,{\mathbf{Z}},{\mathbf{Z}}^2,0,0,{\mathbf{Z}}] \times [0,{\mathbf{Z}}_3,{\mathbf{Z}}_3,{\mathbf{Z}}_3,{\mathbf{Z}}_3,{\mathbf{Z}}_3^2,0,0]\end{align*}

and \begin{align*}H^*(X)= [{\mathbf{Z}}, {\mathbf{Z}}, 0,{\mathbf{Z}},{\mathbf{Z}}^2,0,0,{\mathbf{Z}}] \times [0, 0,{\mathbf{Z}}_3,{\mathbf{Z}}_3,{\mathbf{Z}}_3,{\mathbf{Z}}_3,{\mathbf{Z}}_3^2,0] \\ = [{\mathbf{Z}}, {\mathbf{Z}}, {\mathbf{Z}}_3,{\mathbf{Z}}\times {\mathbf{Z}}_3,{\mathbf{Z}}^2\times {\mathbf{Z}}_3,{\mathbf{Z}}_3,{\mathbf{Z}}_3^2,{\mathbf{Z}}].\end{align*}


Let \(M\) be a closed, connected, oriented 4-manifold such that \(H_2(M; {\mathbf{Z}})\) has rank 1. Show that there is not a free \({\mathbf{Z}}_2\) action on \(M\).

Useful facts:

  • \(X \twoheadrightarrow_{\times p} Y\) induces \(\chi(X) = p\chi(Y)\)
  • Moral: always try a simple Euler characteristic argument first!

We know that \(H_*(M) = [{\mathbf{Z}}, A, {\mathbf{Z}}\times G, A, {\mathbf{Z}}]\) for some group \(A\) and some torsion group \(G\). Letting \(n=\mathrm{rank}(A)\) and taking the Euler characteristic, we have \(\chi(M) = (1)1 + (-1)n + (1)1 + (-1)n + (1)1 = 3-2n\). Note that this is odd for any \(n\).

However, a free action of \({\mathbf{Z}}_2 \curvearrowright M\) would produce a double covering \(M \twoheadrightarrow_{\times 2} M/{\mathbf{Z}}_2\), and multiplicativity of Euler characteristics would force \(\chi(M) = 2 \chi(M/{\mathbf{Z}}_2)\) and thus \(3-2n = 2k\) for some integer \(k\). This would require \(3-2n\) to be even, so we have a contradiction.


Let \(X\) be \(T^2\) with a 2-cell attached to the interior along a longitude. Compute \(\pi_2(X)\).

Useful facts:

  • \(T^2 = e^0 + e^1_1 + e^1_2 + e^2\) as a CW complex.
  • \(S^2/(x_0 \sim x_1) \simeq S^2 \wedge S^1\) when \(x_0, x_1\) are two distinct points. (Picture: sphere with a string handle connecting north/south poles.)
  • \(\pi_{\geq 2}(\tilde X) \cong \pi_{\geq 2}(X)\) for \(\tilde X \twoheadrightarrow X\) the universal cover.

Write \(T^2 = e^0 + e^1_1 + e^1_2 + e^2\), where the first and second 1-cells denote the longitude and meridian respectively. By symmetry, we could have equivalently attached a disk to the meridian instead of the longitude, filling the center hole in the torus. Contract this disk to a point, then pull it vertically in both directions to obtain \(S^2\) with two points identified, which is homotopy-equivalent to \(S^2 \vee S_1\).

Take the universal cover, which is \({\mathbf{R}}^1 \cup_{{\mathbf{Z}}} S^2\) and has the same \(\pi_2\). This is homotopy-equivalent to \(\bigvee_{i\in {\mathbf{Z}}}S^2\) and so \(\pi_2(X) = \prod_{i\in {\mathbf{Z}}} {\mathbf{Z}}\) generated by each distinct copy of \(S^2\). (Alternatively written as \({\mathbf{Z}}[t, t^{-1}]\)).