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fixed de Rham's surname
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Take, for example, the Klein bottle K. Its Dede Rham cohomology with coefficients in $\mathbb{R}$ is $\mathbb{R}$ in dimension 1, while its singular cohomology with coefficients in $\mathbb{Z}$ is $\mathbb{Z} \times \mathbb{Z}_2$ in dimension 1. It is in general true that Dede Rham cohomology ignores the torsion part of singular cohomology. This is not a big surprise since Dede Rham cohomology really just gives the dimensions of the spaces of solutions to certain PDE's, but I'm wondering if there is some other way to directly use the differentiable structure of a manifold to recover torsion. I feel like I should know this, but what can I say...

Thanks!

Take, for example, the Klein bottle K. Its De Rham cohomology with coefficients in $\mathbb{R}$ is $\mathbb{R}$ in dimension 1, while its singular cohomology with coefficients in $\mathbb{Z}$ is $\mathbb{Z} \times \mathbb{Z}_2$ in dimension 1. It is in general true that De Rham cohomology ignores the torsion part of singular cohomology. This is not a big surprise since De Rham cohomology really just gives the dimensions of the spaces of solutions to certain PDE's, but I'm wondering if there is some other way to directly use the differentiable structure of a manifold to recover torsion. I feel like I should know this, but what can I say...

Thanks!

Take, for example, the Klein bottle K. Its de Rham cohomology with coefficients in $\mathbb{R}$ is $\mathbb{R}$ in dimension 1, while its singular cohomology with coefficients in $\mathbb{Z}$ is $\mathbb{Z} \times \mathbb{Z}_2$ in dimension 1. It is in general true that de Rham cohomology ignores the torsion part of singular cohomology. This is not a big surprise since de Rham cohomology really just gives the dimensions of the spaces of solutions to certain PDE's, but I'm wondering if there is some other way to directly use the differentiable structure of a manifold to recover torsion. I feel like I should know this, but what can I say...

Thanks!

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Paul Siegel
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Can analysis detect torsion in cohomology?

Take, for example, the Klein bottle K. Its De Rham cohomology with coefficients in $\mathbb{R}$ is $\mathbb{R}$ in dimension 1, while its singular cohomology with coefficients in $\mathbb{Z}$ is $\mathbb{Z} \times \mathbb{Z}_2$ in dimension 1. It is in general true that De Rham cohomology ignores the torsion part of singular cohomology. This is not a big surprise since De Rham cohomology really just gives the dimensions of the spaces of solutions to certain PDE's, but I'm wondering if there is some other way to directly use the differentiable structure of a manifold to recover torsion. I feel like I should know this, but what can I say...

Thanks!