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$X = \Bbb{CP}^4 \#\Bbb{CP}^4$ does not support an almost complex structure. I suspect this is true for $\Bbb{CP}^{2n} \# \Bbb{CP}^{2n}$ for any $n>0$; I can't prove it. The general idea is to obtain a "higher-dimensional" Wu theorem on which manifolds admit complex structures in terms of various characteristic classes of your manifold. I think this is ...


1

The space $F(p)$ is an open submanifold of $X^p$, so it has the same dimension*, namely $np$. The group $S_p$ acts freely and properly discontinuouly on $F(p)$ and is discrete, thus $B(p) = F(p) / S_p$ has the same dimension as $F(p)$, again $np$. * If you want to be convinced, each $x \in F(p)$ has an open neighborhood contained entirely in $F(p)$; $X^p$ ...


1

Since $\mu_{*}$ is linear, $$\mu_{*,(e,e)}(X,Y) = \mu_{*,(e,e)}(X,0) + \mu_{*,(e,e)}(0,Y).$$ Let $\alpha(t)$, $\beta(t)$ be integral curves of $X$ and $Y$, respectively, starting at $e$. Then $c_1(t) := (\alpha(t), e)$ and $c_2(t) := (e, \beta(t))$ are integral curves through $(X,0)$ and $(0,Y)$, respectively, starting at $(e,e)$. Thus ...


1

Let $p$ be a point in $M$. Since $M$ is Hausdorff and $G$ is finite, there is an open set $U$ such that $p\in U$ and $gp\not\in U$ for all $g\in G\setminus\{1\}$. Since $M$ is regular, there is an open set $V$ such that $p\in V\subseteq\overline V\subseteq U$. Now the set $W=V\setminus\bigcup_{g\in G\setminus\{1\}}g(\overline V)$ is open, contains $p$. What ...



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