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No. In fact compact Riemannian manifolds can have zero sectional curvature; take flat tori $\mathbb{R}^n/\Gamma$ where $\mathbb{R}^n$ has the usual Euclidean metric and $\Gamma$ is a lattice. Apparently every compact Lie group is the isometry group of some compact Riemannian manifold, but I don't know how cavalier one can be about specifying its curvature. ...


Every closed 4-manifold is homeomorphic to a CW complex if and only if every compact 4-manifold with nonempty boundary is homeomorphic to a CW complex.


Yes. Since the product of a form and a smooth function is again a form, extend the form to a small tubular neighborhood of your submanifold and multiply by a bump function. Georges Elencwajg has suggested I explain further. To extend a $p$-form $\omega\in\Omega^p(M)$ to a $p$-form on $\mathbb{R}^n$, we need to first define a $p$-form on ...


Drawing a phase diagram would probably have shown you right away that the nonnegative quadrant $Q=(X\geqslant0,Y\geqslant0)$ is stable by the dynamics, that the equilibria in the quadrant $Q$ are the point $(X,Y)=(0,0)$ and the segment $S=(X\geqslant0,Y\geqslant0,X+Y=N)$, and that $(0,0)$ is an unstable node. At every point $(X_0,Y_0)$ on $S$, an eigenvalue ...


The space $H^1(K) = \mathbb{Z}^2$ is generated by the Poincaré duals $\alpha = A^*$ and $\beta = B^*$ to $A$ and $B$, respectively. (I'm working over $\mathbb{Z}_2$ throughout, so that $H^*(K)$ is actually $H^*(K, \mathbb{Z}_2)$. For reasons of dimension, the only products you need to compute in the ring $H^*(K)$ are $\alpha^2, \alpha \beta$, and $\beta^2$. ...


Just extend the frame arbitrarily to an open subset $U\subset N$, and then apply Gram-Schmidt to it. The Gram-Schmidt process won't change the vector fields on $U\cap N$ where they're already orthonormal.


I'm reading his introduction to topological manifolds :)! I'll solve (b) and (c) (since you have found (a) yourself): (b) Let's look at the real line $\mathbb{R}$ and as a cover we take: $$ \mathcal{U} = ]0,\infty[ \cup \{]-n-2,-n[ \cup ]n,n+1[ \mid n \in \mathbb{N}\} $$ this is clearly an open cover of $\mathbb{R}$ and $]0,\infty[$ has a nonempty ...

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