# Tag Info

6

In Sur les classes caractéristiques des structures fibrés sphériques, Wu proved that $S^{4n}$ cannot admit an almost complex structure for $n \geq 1$. In Groupes de Lie et puissances réduites de Steenrod, Borel and Serre proved that $S^{2n}$ cannot an admit complex structure for $n \geq 4$. Therefore, the only even-dimensional spheres which can admit an ...

5

Morally speaking, the Lie algebra of vector fields is the Lie algebra of $\text{Diff}(M)$, the diffeomorphism group of $M$. The relationship between these is less tight than in the finite-dimensional case: for example, The exponential map can fail to be defined at any nonzero time (as mentioned by orangeskid in the comments), and Even when defined (say on ...

5

I think Mike's answer works too hard. In fact there are no submersions from an even sphere $S^{2n}$ into any smooth manifold $M$ of strictly smaller dimension. The pullback of the tangent bundle of $M$ along such a submersion would be a nontrivial quotient bundle of the tangent bundle of $S^{2n}$, but the tangent bundle of $S^{2n}$ has no nontrivial ...

4

I do not think there are any. First, we want to apply Ehresmann's theorem. Because $S^m$ is compact any submersion $f: S^m \to S^n$ for is a proper submersion; we need to show that it's surjective. If $f: S^m \to S^n$ was not surjective, you could puncture $S^n$ at one of the points not in the image of $f$ to get a smooth submersion $f: S^m \to \Bbb R^n$. ...

4

If $X \to Y$ is a finite Galois cover with Galois group $G$ then it's known that the induced map $H^{\bullet}(Y, \mathbb{Q}) \to H^{\bullet}(X, \mathbb{Q})$ is injective and induces an isomorphism $$H^{\bullet}(Y, \mathbb{Q}) \cong H^{\bullet}(X, \mathbb{Q})^G.$$ (This has nothing to do with $X$ and $Y$ being compact or manifolds.) So here the answer is ...

3

Let $y$ be the generator of $H^1(\mathbb RP^n, \mathbb Z_2)$. Then $y^n\in H^n(\mathbb RP^n, \mathbb Z_2)$ is the fundamental class of $\mathbb RP^n$. The inverse image $f^*(y^n)$ is equal to either zero or the fundamental class $x\in H^n(M^n,\mathbb Z_2)$. More generally, given the smooth map $f$ between smooth connected closed manifolds $M$ and $N$ of ...

2

Here's another argument: Every regular value $a \in \mathbb{R}P^n$ of $f$ corresponds to a pair of regular values $b,-b \in S^n$ of the lift $g: M^n \to S^n$, and $$|f^{-1}(a)|=|g^{-1}(b)|+|g^{-1}(-b)|.$$ Since the mod-2 degree of a smooth map is given by the cardinality (modulo 2) of the preimage of any regular value, we have $$|f^{-1}(a)| = ... 2 You have things sort of confused. For a 2-form \omega, represented as \omega = dt\wedge\omega_1 + \omega_2 (I prefer to put the dt first to spare sign problems), we define$$I(\omega) = I(dt\wedge\omega_1) = \sum_{i=1}^n\big(\int_0^1 a_i(x,t)dt\big)dx_i.However, to prove your formula, we need to define the operator I, in general, on p-forms, ... 1 A series of hints, which should lead you to the solution 1: Try to show that there exists such a hyperplane, by showing that the set of all such hyperplanes has full measure. 2: Note that a hyperplane corresponds to a line and vice versa. We have found a bijection by taking the orthogonal complement! 3: The projection injects iff the corresponding line ... 1 Let X \subset Y be a submanifold and j : X → Y be the inclusion map. Then ∀x \in X, dj_x : T_x(X) → T_x(Y ) is injective—in fact it is an inclusion. If U is a open subset of a manifold X, T_x(U) = T_x(X) for x \in U. Lemma: If X is a manifold, x \in X, and \phi : U → X is a local parametrization with \phi(0) = x, then (d\phi_0)^{−1} = ... 1 Just look at the definition. \Omega_n^O = 0 means that every n-dimensional oriented compact manifold (without boundary) is the boundary of an (n+1)-dimensional oriented compact manifold (0 is the class of the empty set, so saying that every manifold is in the 0 class means that every manifold is cobordant to the empty set, which exactly means that ... 1 \newcommand{\Reals}{\mathbf{R}}Let k and n be non-negative integers with k \leq n. The simplest example of a k-manifold in \Reals^{n} is \begin{align*} M = \Reals^{k} &= \{(y^{1}, \dots, y^{n}) \text{ in } \Reals^{n} : y^{k+1} = \dots = y^{n} = 0\} \\ &= \{(y^{1}, \dots, y^{k}, 0, \dots, 0) : \text{y^{i} real for each i = 1, \dots, ... 1 Let M be the line x=0, and for any point of M, let U=\Bbb{R}^2, let f be the linear transformationf(x,y)=\pmatrix{0 & 1 \\ 1 & 0}\pmatrix{x \\ y}.$$Then since f is linear it is a diffeomorphism, and$$f(U\cap M)=f(M)=f(\{(0,y):y\in\Bbb{R}\})=\left\{\pmatrix{0 & 1\\ 1 & 0}\pmatrix{0\\y} : y\in\Bbb{R}\right\} = ...

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