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## Hot answers tagged differential-topology

3

The result should extend to submanifolds of an orientable manifold. The key point is that a bundle is orientable precisely if its top wedge power is trivial. (And, by definition, a manifold is orientable iff its tangent bundle is orientable.) You should be able to combine this with what you've already proved to finish.

3

Yes, as we can take singular homology (which we can for all topological spaces), then we define the Betti numbers to be the ranks of the individual homology groups and then if the alternating sum of Betti numbers of the space is finite, then we define the Euler characteristic to be this alternating sum. The Euler characteristic is undefined if this ...

3

The argument is valid, but you don't need Nash's theorem to come up with the canonical measure on a nonorientable Riemannian manifold $M$. There are other ways: $M$ is a metric space, and thus carries the $m$-dimensional Hausdorff measure. This measure (up to normalization) agrees with what you get from the volume form when $M$ is orientable. See ...

2

Let $\partial$ be the vector differentiation operator on some region $M$ that is $n$-dimensional. In the calculus of clifford algebra, this is a single operator that incorporates both the exterior derivative $d$ and the interior derivative $\delta$. That is, given a $k$-vector field (or equivalently, a $k$-form) $F$, we can arrive at a $k+1$-vector field ...

1

I first give a solution in the context of CW-complexes, topological bundles and continuous sections. In this setting, using contractibility of the fibers and induction on skeleta, one constructs a (continuous) section $$s: M\to E$$ of the bundle. Next, again using contractibility of the fibers and induction on skeleta, one constructs a (continuous) map $$... 1 Hint: Think about what the map f does to points around x. Also, recall what a submersion means for the dimensions of your two spaces. Addendum: Everything you did was correct. You just need to push the conclusion out of what you already know. And that comes from the submersion theorem. So we can assume that these open sets you mentioned line up in ... 1 Say that X is of dimension m and Y is of dimension n. If y is a regular value of f, then f^{-1}(y) is a properly embedded submanifold of X with dimension m-n. Call this submanifold S. Given a point p in S, this means that there are coordinate charts (U, \varphi) in X containing p and (V, \psi) in Y containing f(p) such ... 1 Let U = S^n\setminus A and V = S^n\setminus B then by hypothesis U and V cover S^n so we can apply Mayer-Vietoris:$$ H_1(S^n) \overset{\partial_*}\longrightarrow H_0(U\cap V) \overset{(i_*,j_*)}{\longrightarrow} H_0(U)\oplus H_0(V) \overset{g}{\longrightarrow} H_0(S^n) \longrightarrow 0  If we assume that $n\geq 2$ then $H_1(S^n) = \{0\}$, ...

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I hope this answer might still be useful after almost two months. These notes contains some nice pictures of examples of the Whitney trick. But here is a very low-dimensional example which really captures the whole idea. Suppose you have two oriented simple closed curves $c_1,c_2$ in the sphere $S^2$. You should visualize this by drawing two such curves ...

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The key idea is to come up with coordinate charts; that is, one-one maps between open sets $V\subset M$ and open subsets of $\mathbb{R}^n$. The two-dimensional special case is enough to illustrate the general idea. If we have a chart on $V\subset M$, then we have coordinates $(x,y)$ in the patch $V$. There are a bunch of ways to define $TM$ formally, but in ...

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None. By Moser, the only invariants of a symplectic surface are its topological type and its total volume. Suppose we can embed some surface of genus $g$ and $b$ boundary components as a symplectic manifold of area $A$. Then, you can embed it as a submanifold of arbitrary area by scaling $\mathbb{R}^4$. Now suppose we can embed a surface of genus $G$ and ...

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