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First, one can resolve the Schubert varieties using Bott-Samelson manifolds, and discover that any two resolutions give the same class upon pushforward. (This good situation ends with K-theory, i.e. i …

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 …

Yes. I assume that your $M_n$ is what is more usually denoted $\overline{M_{0,n}}$. Then the answer is yes, there is a natural map $\overline{M_{0,n}} \twoheadrightarrow M_L$, for each $L$. Specifica …

Things are easiest when the automorphism group of $M$ (with its stratification) acts with finitely many orbits on $T^* M$: see Perverse sheaves on Grassmannians, by Tom Braden.

I'm guessing that, unstated, $M,G$ are finite-dimensional and $G$ is connected Lie. Then $H^*(M/G)$ vanishes for $* \gg 0$, but $H^*_G$ is positively graded, so $H^*(M/G)$ must be a torsion module. …

The connected double cover of $S^1$ (boundary of the Möbius strip) is an $S^0$ bundle that is not the double of the unique $0$-disc bundle over $S^1$.

The construction you describe appears in Tamvakis' The connection between representation theory and Schubert calculus (Enseign. Math. 50 (2004), 267-2860). Basically, instead of working with represent …

The paper by Agnihotri and Woodward, Eigenvalues of products of unitary matrices and quantum Schubert calculus, uses a Narasimhan-Seshadri correspondence between parabolic bundles and unitary connecti …

I assume you mean that $T$ acts preserving each fiber. Then the flatness says that the multigraded Hilbert polynomial is constant. As the Duistermaat-Heckman measure is the leading-order behavior of t …

Check out Almost commuting elements in compact Lie groups by Borel, Freedman, and Morgan. "We describe the components of the moduli space of conjugacy classes of commuting pairs and triples of elemen …

I believe the right reference is Borel-de Siebenthal. A finite-dimensional proof is as follows. The space of conjugacy classes $G/\sim$ can be identified with $T/W = (Lie(T)/\Lambda)/W = Lie(T)/(\Lam …

I would compare at least 2 & 3, if not 4, using maps into classifying spaces. For a space $X$ homotopic to a finite CW-complex (at least), the pullback map $Map(X,Gr_n(\mathbb C^\infty)) \to \{$isomo …

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 …

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 …

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 …