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Could someone tell me how to compute the cohomology algebra $H^*(K, \mathbb{Z}_2)$ of the three dimensional Klein bottle $K$ defined as follows. Let $S_0,S_1$ be the boundaries of $S^2 \times [0,1]$ with induced orientations from $S^2 \times [0,1]$. Let $h: S_0 \to S_1$ be an orientation reversing homeomorphism. Define $K$ to be the quotient of $S^2 \times [0,1]$ by identifying $S_0$ and $S_1$ via $f$.

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Don't you want an orientation reversing homeo? –  Mariano Suárez-Alvarez Jan 14 '12 at 7:27
    
Many thanks for correction and your detailed answer. –  kelly Jan 14 '12 at 8:01

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$\newcommand\ZZ{\mathbb{Z}}$Look at $S^1$ as $I=[0,1]$ with the endpoints identified. There is an obvious map $\pi:K\to S^1$ which is induced by the second projection $S^2\times I\to I$. Let $0$, $\infty$ be two distinct points in $S^1$ and let $\tilde U=S^1\setminus\{0\}$ and $\tilde V=S^1\setminus\{\infty\}$. Then $\{\tilde U,\tilde V\}$ is an open covering of $S^1$, so defining $U=\pi^{-1}(\tilde U)$ and $V=\pi^{-1}(\tilde V)$, we get an open covering $\{U,V\}$ of $K$.

It is easy to see that $U$ and $V$ have both the homotopy type of a $2$-sphere, and $U\cap V$ has two conected components, each of the homotopy type of a $2$-sphere again. The Mayer-Vietoris exact sequence for $(K;U,V)$ starts with $$ 0\to H^0(K) \to H^0(U)\oplus H^0(V) \to H^0(U\cap V) \to H^1(K) \to H^1(U)\oplus H^1(V) \qquad(\clubsuit) $$ Since $U$ and $V$ "are" $2$-spheres, the rightmost group here is zero. Since $H^0(K)=H^0(U)=H^0(V)=\ZZ_2$ and $H^0(U\cap V)\cong\ZZ_2^2$, and since this is an exact sequence of $\ZZ_2$-vector spaces, we see that $H^1(K)\cong\ZZ_2$.

Likewise, after $H^1(U)\oplus H^1(V)=0$ the M-V sequence is $$ H^1(U\cap V) \to H^2(K) \to H^2(U)\oplus H^2(V) \to H^2(U\cap V) \to H^3(K) \to H^3(U)\oplus H^3(V) $$ Since $U\cap V$ "is" a union of two $2$-spheres, $H^1(U\cap V)=0$; since $U$ and $V$ are $2$-spheres, $H^3(U)\oplus H^3(V)=0$ and $H^2(U)\cong H^2(V)\cong\ZZ_2$; since $U\cap V$ "is" two disjoint $2$-spheres, $H^2(U\cap V)\cong\ZZ_2^2$; finally, since $K$ is a compact manifold, $H^3(K)\cong\ZZ_2$. Replacing all this information in the exact sequence and counting dimensions of vector spaces, we see that $H^2(K)\cong\ZZ_2$.

Remark. One can try to do the same computation with coefficients in $\ZZ$. To get $H^1(K,\ZZ)$ one has to make explicit the map $H^0(U)\oplus H^0(V) \to H^0(U\cap V)$: with a more or less obvious choice of bases for these two free groups of rank $2$ the matrix of the map is $\begin{pmatrix}1&1\\1&1\end{pmatrix}$, and it follows from this and the exact sequence that $H^1(K,\ZZ)\cong\ZZ$. On the other hand, the map $H^2(U)\oplus H^2(V) \to H^2(U\cap V)$ has, in a similar basis, matrix $\begin{pmatrix}1&1\\1&-1\end{pmatrix}$, which is injective, so that $H^2(K,\ZZ)=0$ and has determinant with absolute value $2$, so that its cokernel is cyclic of order $2$, that is, $H^3(K,\ZZ)\cong\ZZ_2$. Here it is wee see fact that the homeo reverses the orientation: if it preserved it, the last matrix would have been $\begin{pmatrix}1&1\\1&1\end{pmatrix}$ which has both kernel and cokernel free of rank $1$.

The exact sequence $(\clubsuit)$ depends covariantly on the coefficients. From the surjection $\ZZ\to\ZZ_2$ we can construct a diagram from it to show that the natural map $H^1(K,\ZZ)\to H^1(K,\ZZ_2)$ is surjective. Since the cup product $H^1(K,\ZZ)\otimes H^1(K,\ZZ)\to H^2(K,\ZZ)=0$ is of course trivial, and since the cup product is natural with respect to ring homomorphisms in the coefficients, this implies that the cup product $H^1(K,\ZZ_2)\otimes H^1(K,\ZZ_2)\to H^2(K,\ZZ_2)$ is also zero.

$$\bullet\bullet\bullet$$

Alternatively, there is a cell decomposition with one cell of each dimension in $\{0,1,2,3\}$ in which, moreover, the $2$-cell is attached with its boundary contracted to the $0$-cell. The cellular complex to compute cohomology is then easily seen to have all coboundary maps equal to zero. Of course, this gives the same result as above.

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Many thanks for your detailed answer. –  kelly Jan 14 '12 at 8:06
    
Notice that I did not compute the product... –  Mariano Suárez-Alvarez Jan 14 '12 at 8:09
    
Thanks again. I hope I would be able to make out the product. –  kelly Jan 14 '12 at 8:36
    
So this gives $H^*(K, \mathbb{Z}_2) \cong \mathbb{Z}_2[a,b]/<a^2,b^2>$ with $deg(a)=1$ and $deg(b)=2$. –  kelly Jan 14 '12 at 9:29
    
The product $H^1\otimes H^2\to H^3$ is missing. –  Mariano Suárez-Alvarez Jan 14 '12 at 10:11

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