Search Results

As Johannes Ebert says, the classical areas where sheaves are most important are complex analysis and algebraic geometry. There are two completely different kinds of sheaves one might consider on a c …

The following seems like an extremely basic algebraic topology question, but it's not something I ever learned, nor does it look familiar to the algebraic topologists I've asked. Let $f:X\to Y$ be …

A less interesting but personally checkable answer: consider $f : ({\mathbb C}^2 \setminus 0)/\{\pm 1\} \hookrightarrow {\mathbb C}^3$, $(x,y) \mapsto (x^2,xy,y^2)$. The image is the punctured quadric …

Yes it is simply connected. In general the retraction of $\mathbb C^2$ to $0$ will retract the resolution to the $0$ fiber, which is a tree of $\mathbb{CP}^1$s, hence homotopic to a wedge of $2$-spher …

This is fine for $\mathbb Z$-coefficients; you don't need to go to $\mathbb R$. The part of the sequence you need is $$0\to H_{S^1}^0(S^2) \to H_{S^1}^0(U) \oplus H_{S^1}^0(V) \to H_{S^1}^0(U\cap V …

Start by deformation-retracting $G$ to its maximal compact, so that $G/T$ will be a flag manifold with a Bruhat decomposition, obtainable by Morse theory as in Agol's answer. If what you want anyway …

Let $Y = P^2 \setminus ${two of its fixed points}, and $X = Y$ with $f$ the identity. Then the fiber over $y$ is a point, but $Y$ is not contractible, I don't think. I'm pretty sure your "has one fix …

Given an $n$-manifold $M$ (say), we can talk about its Euler characteristic $\chi(M)$. This can be generalized to the Euler number of any $n$-dimensional bundle ${\mathcal V}$. Or indeed, the Euler …

Such an endomorphism of $M$ gives an automorphism of the cohomology ring that acts by $-1$ on top cohomology. The cohomology ring of your example $M = {\mathbb C \mathbb P}^{2n}$ doesn't have such aut …

Note that your statement is only true in rational cohomology. For example, $H^\ast(SO(5)/T)$ is not generated in degree $2$ (though it is rationally). The easiest proof I know starts from equivariant …

One. The Poincaré-Hopf theorem is usually stated as a formula for the Euler characteristic of the tangent bundle TM. Is there a version for Euler classes, of oriented real vector bundles? It seems l …

Let $B = Gr_{\dim E}(\infty)$ denote the classifying space for $\dim E$-bundles. Your $P(c_i(E))$ defines a class on $B$, which (if $P$ has integer coefficients, which you better've meant it to!) can …

I think it's because we have well-developed techniques with which to prove that this condition holds, and when those fail, people don't put that much effort into trying to describe the (more difficult …

The direct sum over $n$ of the total homology of the Hilbert scheme of $n$ points in the plane. (Reference: Nakajima's book.) The (stably-almost-)complex cobordism ring of a point. I expect that's pr …

Well, there's going to be some map $T^* \to H^2_T(pt)$ taking a weight $\lambda$ to the equivariant Euler class of the corresponding line bundle over a point. If you add weights, that tensors the line …