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I have the following problem in my homework for algebraic topology: Does there exist a compact 2-dimensional manifold $M$ without boundary such that $M\times M$ is homotopically equivalent to $M$? I was thinking that no, because, there are two cases:

  1. The manifold is orientable of genus $g$: $M_g$. Then it has a CW-complex structure given by one $0$-cell, $2g$ $1$-cells $\{a_1, b_1,\dots,a_g,b_g\}$ and one $2$-cell attached by product of commutators $[a_1,b_1][a_2,b_2]\dots[a_g,b_g]$ with the homology groups: $H_n(M_g)=\begin{cases}\mathbb{Z}, \text{ if }n=0,2\\\mathbb{Z}^{2g},\text{ if }n=1\\0,\text{ else}\\\end{cases}$. Using Kunneth's formula it follows that $H_4(M_g\times M_g)\neq 0$, therefore $M_g\times M_g$ is not homotopically equivalent to $M_g$.
  2. The manifold is not orientable of genus $g$: $N_g$. Then it has a CW-complex given by one $0$-cell, $g$ $1$-cells $\{a_1,\dots,a_g\}$ and one $2$-cell $e$ atached by the word $a^2_1\dots a^2_g$ with the homology groups: $H_n(N_g)=\begin{cases}\mathbb{Z}, \text{ if }n=0\\\mathbb{Z}^{g-1}\oplus\mathbb{Z}/2,\text{ if }n=1\\0,\text{ else}\\\end{cases}$. Using Kunneth's formula it follows that $H_2(N_g\times N_g)\neq 0$, therefore $N_g\times N_g$ is not homotopically equivalent to $N_g$.

Is my reasoning correct or am I missing something?

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  • $\begingroup$ Can I use $M = \mathbb R^2$? $\endgroup$
    – user99914
    Commented Feb 12, 2015 at 6:22
  • $\begingroup$ I guess the problem reffers to closed manifolds. $\endgroup$
    – Iulia
    Commented Feb 12, 2015 at 6:29

1 Answer 1

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Looks ok to me.

You can simpy observe that $M\times M$ is a closed $4$-manifold, so that $H_4(M\times M,\mathbb Z_2)\neq0$.

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  • $\begingroup$ I'm using the fact that every connected manifold has is $\mathbb Z_2$-orientation. $\endgroup$ Commented Feb 12, 2015 at 6:34

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