# Fall 2014

## 1

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 $$H_*(X) = [{\mathbf{Z}}, {\mathbf{Z}}^3, 0\rightarrow]$$.

## 2

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

\todo[inline]{Solution}

## 3

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*}

## 4

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.

## 5

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 $${\mathbf{Z}}[t, t^{-1}]$$).