# Tag Info

### Can one glue De Rham cohomology classes on a differential manifolds?

Here is the great answer given by another of my brilliant friends: Let $X$ be $\mathbb C, U_0$ be the open complement in $X$ of the closed disk $\bar D=\{z\in \mathbb C\vert \vert z\vert \leq1 \}$ and ...
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### The convex hull of a manifold whose cobordism class is trivial

Implicit in the other responses is the fact that if $M$ bounds a convex manifold $W$, then $W$ is contractible and so M has the homology of a sphere. So any null-cobordant manifold that is not a ...
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### The convex hull of a manifold whose cobordism class is trivial

There are exotic spheres (which are null cobordant) which do not bound a parallelisable manifold. Since the convex hull is contractible, it would be parallelisable if it were a manifold, so these guys ...
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### Can one glue De Rham cohomology classes on a differential manifolds?

Here is the solution obtained by one of my brilliant geometer friends evoked in the question: Take $X=S^2$, the unit $2$-sphere with equation $x_1^2+x_2 ^2+x_3^2=1$, and cover it by the three open ...
• 44.7k
Accepted

### Are the symmetric spaces $\operatorname{SU}(n)/{\operatorname{SO}(n)}$ always nontrivial in the bordism rings for $n>2$?

There is a fibration $SU(n) \overset p\to SU(n)/SO(n) \overset j\to BSO(n)$, where the $j$ is the classifying map of $p$, viewed as (the projection of) a principal $SO(n)$-bundle. The Stiefel–Whitney ...
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### Can one glue De Rham cohomology classes on a differential manifolds?

This answer provides a positive answer to a refinement of the original question. Recall that two closed differential $k$-forms $ω_0$, $ω_1$ on a smooth manifold $M$ have the same de Rham cohomology ...
• 29.5k
Accepted

### Can one glue De Rham cohomology classes on a differential manifolds?

No. Make $M$ by gluing three strips to two discs to form a thrice-punctured sphere. Take three open sets $U_\lambda$, each made by both discs and two of the strips. Then each $U_\lambda$ is ...
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A manifold $M$ of dimension $>1$ with $H^1(M) \neq 0$ does not have the self-Hopf property. For if so, then there is a map $f: M \to S^1$ that induces a non-trivial homomorphism $f_*: H_1(M) \to \... • 17.1k 17 votes Accepted ### Converse to Hopf degree theorem See the second half of the answer for a complete characterisation of closed orientable manifolds with the Hopf property. Note that$X$having the Hopf property is equivalent to the injectivity of$\...
$\mathbb{CP}^n$ has the self-Hopf property for $n$ odd. See Theorem 2.2 of Self Maps of Projective Spaces C. A. McGibbon Transactions of the American Mathematical Society Vol. 271, No. 1 (May, 1982), ...