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Complementing Ryan's answer: we can assume that $G$ preserves a Riemannian metric on $M$ (since $G$ is compact, we can take just any metric and average it with respect to $G$). Then $V$ from Ryan's po …

I'm not sure what you mean by a line bundle on a real algebraic variety and its Chern classes, but for smooth complex analytic manifolds the Chern class of a line bundle corresponding to a divisor is …

It is a theorem of Whitney's that every closed subset $X$ of a smooth manifold $M$ is the zero locus of a smooth function on $M$. The idea of the proof is as follows: cover the complement $M\setminus …

This depends on what exactly is a smooth mapping from a simplex to the manifold. The standard definition is that the mapping of a non-open subset $X$ of $\mathbf{R}^n$ to a manifold is smooth iff it c …

The wrong-way maps in homology are in fact less mysterious than they look. Suppose that $ f: X \to Y $ is a continuous map of oriented closed topological manifolds. Then there is a composition of maps …

Complementing Andrea's posting: the answer to the question as it is stated is no. Indeed, the proof of the Hodge index formula for Kaehler manifolds uses the strong Lefschetz decomposition, which does …

The answer to the question is positive, due to Wu's formula. See e.g. Milnor-Stasheff, Characteristic classes, lemma 11.13 and theorem 11.14. In fact, all one needs to compute the Stiefel-Whitney clas …

Assume $n>2$. The $SU(n)$-invariant metrics on the sphere $S^{2n-1}$ are in bijection with $SU(n-1)$ invariant metrics on $T_x S^{2n-1}$ where $SU(n-1)$ is realized as the stabilizer of some $x\in S^{ …

This question is similar to Do chains and cochains know the same thing about the manifold? in the sence that both deal with a natural "comparison" quasi-isomorphism that does not preserve the ring str …

This is a triviality, but still: there is a pullback of a differential form, but in general no push-forward of a vector field. As a consequence, one gets e.g. for any smooth map $f:X\to Y$ of smooth m …

If such a complex structure exists, it would weird indeed! For example, as shown by Campana, Demailly and Peternell (Compositio 112, 77-91), if such a thing exists, then $S^6$ would have no non-consta …

The answer in general is no. Take e.g. $M=\mathbb{R}^n$. There are no non-trivial vector bundles on it. So, if $N$ is the zero locus of a transversal section of a bundle $E$ (which is necessarily triv …

This question originated from a conversation with Dmitry that took place here Is there a complex structure on the 6-sphere? The Hirzebruch-Riemann-Roch formula expresses the Euler characteristic of …

Here is a solution in the $C^1$ case [but see upd]. Suppose the vectors $F'(t_1),\ldots, F'(t_n)$ are linearly independent for all $0\leq t_1< \cdots < t_n\leq 1$. Let $L(t_1,\ldots,t_{n-1})$ be vecto …

Let $E=\gamma^1\otimes\mathbb{C}$ be the complexified tautological bundle over $X=\mathbb{P}^6(\mathbb{R})$, and set $F=4E=E\oplus E\oplus E\oplus E$. It is not hard to check that $c(E)=1+a$ and $w(E_ …