Hugo Chapdelaine
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Registered User
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Mar 27 |
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non-trivial topological line bundles over cartesian product of manifolds not coming from a pullback Ok, I was being idiot |
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Mar 27 |
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non-trivial topological line bundles over cartesian product of manifolds not coming from a pullback I see, so this explains why I was not able to construct such an example. What about complex line bundle? |
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Mar 27 |
revised |
non-trivial topological line bundles over cartesian product of manifolds not coming from a pullback added 90 characters in body; added 52 characters in body |
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Mar 27 |
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non-trivial topological line bundles over cartesian product of manifolds not coming from a pullback Yes, I forgot to mention that I don't want $L$ to be coming from the pullback of a line bundle over $Y$. |
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Mar 27 |
asked | non-trivial topological line bundles over cartesian product of manifolds not coming from a pullback |
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Mar 23 |
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Generalizing the square theorem thanks for the reference! |
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Mar 22 |
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Generalizing the square theorem Hi @Angelo, thanks for the example. I never played with reflexive $\mathcal{O}_X$-modules. They seem to enjoy nice properties like $j_*G=F$. Do you have a reference for this last property? |
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Mar 20 |
asked | Generalizing the square theorem |
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Mar 13 |
asked | On the square theorem for commutative algebraic groups |
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Mar 6 |
awarded | ● Popular Question |
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Jan 31 |
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On Weil’s characters of type (A) Thanks a lot Keerthi! Your proof seems to be correct. |
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Jan 31 |
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On Weil’s characters of type (A) Hi Keerthi, I'm confused about one point. If $Stab(f)=H\leq G_{\mathbf{Q}}$, then you are showing that for all complex conjugation $c\in G_{\mathbf{Q}}$, $cHc^{-1}=H$. But this does not seem to imply that $L:=\bar{\mathbf{Q}}^{H}$ is a CM field. For example, if $L$ is Galois over $\mathbf{Q}$ and not CM, then the condition $cHc^{-1}=H$ is trivially satisfied since $H$ is normal in $G_{\mathbf{Q}}$. Am I right? |
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Jan 31 |
revised |
On Weil’s characters of type (A) edited body |
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Jan 30 |
asked | On Weil’s characters of type (A) |
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Jan 29 |
awarded | ● Popular Question |
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Jan 18 |
revised |
The normalizer a maximal compact subgroup of a semi-simple Lie group added 104 characters in body |
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Jan 18 |
revised |
The normalizer a maximal compact subgroup of a semi-simple Lie group added 30 characters in body |
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Jan 18 |
asked | The normalizer a maximal compact subgroup of a semi-simple Lie group |
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Dec 22 |
awarded | ● Yearling |
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Nov 27 |
awarded | ● Popular Question |
