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Dave Benson
Dave Benson
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Lemma. If $A$ is an abelian group satisfying $A\otimes A=0$ and $\mathop{\rm Tor}(A,A)=0$ then $A=0$.

Proof. Since $\mathop{\rm Tor}$ is left exact on abelian groups, an inclusion of a finite cyclic group $C$ in $A$ gives an injection $\mathop{\rm Tor}(C,C)\to \mathop{\rm Tor}(A,A)$. So if $\mathop{\rm Tor}(A,A)=0$ then $A$ is torsion free. Then $A$ embeds in $\mathbb{Q}\otimes A$, so if $A\otimes A=0$ then $(\mathbb{Q}\otimes A)\otimes(\mathbb{Q}\otimes A)=\mathbb{Q}\otimes(A\otimes A)=0$. So $\mathbb{Q}\otimes A=0$ and then $A=0$.

Now given your manifold $M$, since it is smooth, simply connected, and finite dimensional, it has the homotopy type of a finite dimensional CW complex. So by the Whitehead theorem, if it's not contractible then it has some non-vanishing homology group in degree $\geqslant 2$. Let $H_k(M)\ne 0$ with $k\geqslant 2$ as large as possible (so $k$ is at most the dimension of $M$). Then by the Künneth theorem and the lemma, either $H_{2k-1}(M\times M)\ne 0$$H_{2k}(M\times M)\ne 0$ or $H_{2k}(M\times M)\ne 0$$H_{2k+1}(M\times M)\ne 0$. But $2k-1>k$, soSo we have a contradiction if $M\times M\simeq M$.

Lemma. If $A$ is an abelian group satisfying $A\otimes A=0$ and $\mathop{\rm Tor}(A,A)=0$ then $A=0$.

Proof. Since $\mathop{\rm Tor}$ is left exact on abelian groups, an inclusion of a finite cyclic group $C$ in $A$ gives an injection $\mathop{\rm Tor}(C,C)\to \mathop{\rm Tor}(A,A)$. So if $\mathop{\rm Tor}(A,A)=0$ then $A$ is torsion free. Then $A$ embeds in $\mathbb{Q}\otimes A$, so if $A\otimes A=0$ then $(\mathbb{Q}\otimes A)\otimes(\mathbb{Q}\otimes A)=\mathbb{Q}\otimes(A\otimes A)=0$. So $\mathbb{Q}\otimes A=0$ and then $A=0$.

Now given your manifold $M$, since it is smooth, simply connected, and finite dimensional, it has the homotopy type of a finite dimensional CW complex. So by the Whitehead theorem, if it's not contractible then it has some non-vanishing homology group in degree $\geqslant 2$. Let $H_k(M)\ne 0$ with $k\geqslant 2$ as large as possible (so $k$ is at most the dimension of $M$). Then by the Künneth theorem and the lemma, either $H_{2k-1}(M\times M)\ne 0$ or $H_{2k}(M\times M)\ne 0$. But $2k-1>k$, so we have a contradiction if $M\times M\simeq M$.

Lemma. If $A$ is an abelian group satisfying $A\otimes A=0$ and $\mathop{\rm Tor}(A,A)=0$ then $A=0$.

Proof. Since $\mathop{\rm Tor}$ is left exact on abelian groups, an inclusion of a finite cyclic group $C$ in $A$ gives an injection $\mathop{\rm Tor}(C,C)\to \mathop{\rm Tor}(A,A)$. So if $\mathop{\rm Tor}(A,A)=0$ then $A$ is torsion free. Then $A$ embeds in $\mathbb{Q}\otimes A$, so if $A\otimes A=0$ then $(\mathbb{Q}\otimes A)\otimes(\mathbb{Q}\otimes A)=\mathbb{Q}\otimes(A\otimes A)=0$. So $\mathbb{Q}\otimes A=0$ and then $A=0$.

Now given your manifold $M$, since it is smooth, simply connected, and finite dimensional, it has the homotopy type of a finite dimensional CW complex. So by the Whitehead theorem, if it's not contractible then it has some non-vanishing homology group in degree $\geqslant 2$. Let $H_k(M)\ne 0$ with $k\geqslant 2$ as large as possible (so $k$ is at most the dimension of $M$). Then by the Künneth theorem and the lemma, either $H_{2k}(M\times M)\ne 0$ or $H_{2k+1}(M\times M)\ne 0$. So we have a contradiction if $M\times M\simeq M$.

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Dave Benson
Dave Benson
  • 16.2k
  • 2
  • 42
  • 95

Lemma. If $A$ is an abelian group satisfying $A\otimes A=0$ and $\mathop{\rm Tor}(A,A)=0$ then $A=0$.

Proof. Since $\mathop{\rm Tor}$ is left exact on abelian groups, an inclusion of a finite cyclic group $C$ in $A$ gives an injection $\mathop{\rm Tor}(C,C)\to \mathop{\rm Tor}(A,A)$. So if $\mathop{\rm Tor}(A,A)=0$ then $A$ is torsion free. Then $A$ embeds in $\mathbb{Q}\otimes A$, so if $A\otimes A=0$ then $(\mathbb{Q}\otimes A)\otimes(\mathbb{Q}\otimes A)=\mathbb{Q}\otimes(A\otimes A)=0$. So $\mathbb{Q}\otimes A=0$ and then $A=0$.

Now given your manifold $M$, since it is smooth, simply connected, and finite dimensional, it has the homotopy type of a finite dimensional CW complex. So by the Whitehead theorem, if it's not contractible then it has some non-vanishing homology group in degree $\geqslant 2$. Let $H_k(M)\ne 0$ with $k\geqslant 2$ as large as possible (so $k$ is at most the dimension of $M$). Then by the Künneth theorem and the lemma, either $H_{2k-1}(M\times M)\ne 0$ or $H_{2k}(M\times M)\ne 0$. But $2k-1>k$, so we have a contradiction if $M\times M\simeq M$.