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6

By a couple of normal vector fields, I assume you mean two linearly independent normal vector fields. Let $i : K \to \mathbb{R}^4$ be an embedding. On $K$ we have an exact sequence of vector bundles $$0 \to TK \to i^*T\mathbb{R}^4 \to N \to 0$$ where $N$ is the normal bundle; note that $\operatorname{rank}N = \dim\mathbb{R}^4 - \dim K = 4 - 2 = 2$. If $K$ ...


5

Definitely not. For instance, you could take your example $N\subset\mathbb{R}^3$ and form a connected sum $N\mathbin{\#} P$ where $P$ is any closed oriented 3-manifold. By some simple long exact sequence computations you can find that $H_2(N\mathbin{\#}P)\cong H_2(P)$, so $N\mathbin{\#}P$ cannot be homotopy equivalent to a wedge of circles if $H_2(P)$ is ...


4

A knot is called amphichiral if it is isotopic to its mirror and chiral otherwise. http://mathworld.wolfram.com/AmphichiralKnot.html is perhaps good place to start. Jones' result at the bottom is quite a strong obstruction. Kauffman, Murasugi (independently) and Thistlewaite (also independently) in 1987 proved that for an alternating knot $K$ with an odd ...


4

No. As an obnoxious counterexample, take $g=0$ and $N$ to be the complement of some small open ball in a 3-manifold $M$; then $\pi_1(N) = \pi_1(M)$, so if $M$ was not simply connected (whence $S^3$, thanks to Perelman), then $N$ is not homotopy equivalent to the point. More generally, delete (a small neighborhood of) a wedge of $g$ circles from some ...


3

Probably you can write down an explicit contraction, but recall that $\pi_{k-1}(\Omega_{p,p}(M))\cong \pi_{k}(M)$. This comes from the path space fibration $\Omega_{p,p}(M)\rightarrow P(M)\rightarrow M$. This implies that $\pi_k(M)=0$ for all $k>0$. If $M$ is connected, this says that $M$ is weakly contractible. Manifolds are $CW$-complexes, hence $M$ is ...


2

We can completely characterize the subsets of $\mathbb R^n$ that are the boundaries of open sets -- they are the closed, nowhere dense sets! A set is nowhere dense if it has empty interior. These include Graphs of (continuous!) functions Level sets of polynomials Certain fractals, like the Cantor set Loops that don't intersect themselves. Proof: Suppose ...


2

The following proof is a bit over the top, but here it is: Prove Gauss' isothermal coordinates theorem (every 2-dimensional Riemannian manifold is conformally flat). As a corollary, conclude that every oriented 2-dimensional Riemannian manifold $(M,g)$ has a natural structure of a Riemann surface (the holomorphic atlas is given by charts $\phi: U\subset ...


1

Came up with a bit simpler solution. Let $e_1(x)$ be a tangent vector field on the Klein bottle $K$. Suppose that $n_1(x)$ and $n_2(x)$ - pair of linearly indenpendent normal vector fields. Then we can choose $e_2(x)$ in the $TK_x$ so that four vectors $(e_1(x), e_2(x), n_1(x), n_2(x))$ form a positively oriented basis in $\mathbb{R}^4$. This would give us ...


1

If "intuitive" is what you want, then think of what the graph of a function of two variables looks like. Hills and valleys. If you have $F(x,y) = \text{specified constant}$, then the value of the constant tells you to what depth you fill fill the world with water, and then the curve is the coastline. Why is the coastline a curve? That is the question.


1

Suppose $p(x,y)$ is a polynomial that is not identically $0.$ Let $Z$ be the zero set of $p.$ Then $Z$ is closed, hence $Z = \text { int } Z \cup \partial Z.$ Suppose $\text { int } Z $ is nonempty. Then there is an open disc $D(a,r) \subset Z.$ Then for any nonzero $v\in \mathbb R^2,$ the function $p_v(t) = p(a+tv)$ is a one variable polynomial in $t$ that ...



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