Sorry to resurrect, but we leave a $($detailed$)$ proof here that $TM$ has the structure of an oriented $2n$-manifold, even if the $n$-manifold $M$ is non-orientable. Let $\{(U_\alpha, \phi_\alpha)\}... View answer Accepted answer 19 votes Sorry for reviving such an old problem... Anyways, what is important here is that$\text{arctan}$is a bijection from$\mathbb{R}$to$( -\pi/2, \pi/2 )$, and it is an isometry if we give$\left(-\pi/...

This is not quite true, since, for example, the open unit disk is embedded in $\mathbb{R}^2$ by the inclusion, and the inverse image of the closed unit disk is the open unit disk. What we will show is ...

We will show when we have a normed space $E$ and $r>0$, $E$ and its subset $B_r(0) = \{x \text{ }|\text{ }\|x\| < r\}$ are homeomorphic. $($Wikipedia gives a pretty okay primer on normed spaces, ...

Sorry to resurrect such an old post, but I would like to supply a proof of this without invoking Jordan normal form or Schur decomposition. So we want to show the following statement. Theorem. Let $... View answer 11 votes Some good$($introductory$)$sources, in general, for all things smooth manifolds: Topology from the Differentiable Viewpoint, by Milnor Differential Topology, by Guillemin-Pollack Differential Forms ... View answer Accepted answer 11 votes Suppose$f(z) \neq 0$. Then $${{f(z+h)^8 - f(z)^8}\over{h}} = {{f(z+h) - f(z)}\over{h}}\left(f(z+h)^7 + f(z+h)^6f(z) + \dots + f(z)^7\right).$$ Using continuity, we see the expression in parentheses ... View answer Accepted answer 10 votes It turns out$p_n(x)$is reducible only when it is "obvious." More precisely, polynomial$p_n(x)$is reducible over$\mathbb{Z}$if and only if at least one of the following two conditions hold. ... View answer 8 votes We know that if$X \subset \mathbb{R}^N$is a$k$-dimensional manifold, then$T_x(X)$is a$k$-dimensional linear subspace of$\mathbb{R}^N$. Moreover,$\partial X$is a$(k-1)$-dimensional manifold ... View answer Accepted answer 8 votes 1. We first talk a bit about the underlying method of the proof. We want look for$(I + \epsilon T)^{-1}$as an infinite series, similar to the way we would expand$1/(1+x)$into a power series for$|...

Sorry for resurrecting such an old post, but Section §1.6 of Hartshorne annoys me in the sense that the concept of "abstract nonsingular curve" is not used very much (if at all) in algebraic geometry. ...

Let $K$ be a global field with a distinguished place $w$. Let $\mathbb{A}^w$ be the restricted direct product of all the $K_{v\ne w}$ with respect to the subgroups $\mathcal{O}_v$ whenever $v$ is ...

This is essentially Problem #7 from Tournament of the Towns, Fall 2011, Junior A-Level and Problem #2 from USAMO 2012. Tournament of the Towns, Fall 2011, Junior A-Level, Problem #7. Each vertex of ...

Going off user26857's comment, we provide a counterexample for Proposition 3.14 in the case that $M$ is not finitely generated. Hopefully you can use this to construct a counterexample for Corollary 3....

Have $\mu$ be a $\sigma$-finite measure on the $\sigma$-algebra $\mathscr{M}$. Have $\mu^*$ be the outer measure induced by $\mu$, $\mathscr{M}^*$ the $\mu^*$-measurable sets and $\overline{\mu}$ the ...

The answer is $n = 1$, $n = 4$, or $n$ is any prime number. Let $S$ be the set of integers less than or equal to $n$ that are relatively prime to $n$, and let $T$ be the set of integers that are ...

The first sentence of the problem's statement says that $(\mathbb{Z}/2\mathbb{Z})^7$ is a simple $A$-module; hence, by Schur's Lemma, the ring $A' = \text{End}_A((\mathbb{Z}/2\mathbb{Z})^7)$ is a skew ...

In the interests of completeness, we provide a complete solution. Lemma. Let $X_1, X_2, \dots$ be a sequence of Bernoulli random variables $($not necessarily independent, not necessarily with same $p$...

We claim that given the conditions set forth in the problem, $\text{Ind}_{Z(G)}^G \rho$ is always not irreducible. Set $Z = Z(G)$. Let $W$ be an irreducible $Z$-representation. We would like to ...

Pick coordinate charts around $p \in M$ and $f(p) \in N$; assume that $M$ and $N$ are of dimension $m$, $n$ respectively, and $($by continuity$)$ that the points of $M$ corresponding to our chart ...

We have $$x^3 = y^2 = (xy)^2 = 1,$$ so $$yx = (yx)^{-1} = x^2y,\text{ }yx^2 = x^2yx = x^4y = xy.$$ a. Fix a nonzero $u\in V$, and let $$v = u+xu+x^2u = (1+x+x^2)u,\text{ }w=u+yu = (1+y)u.$$ Since $\{1,... View answer 5 votes A simpler solution than the one of Lee Mosher is as follows. Let$z_1, \dots, z_n$be the roots of$p$inside the unit circle, and let$w_1, \dots, w_m$be the roots of$p$outside the unit circle. So$...

If I had to give a tl;dr to my incoherent drivel above, it would probably be as follows. $V \otimes V^*$ has a canonical basis if and only if $V$ is $1$-dimensional. We have a circle. We draw the ...

b. We compute $\nabla(\alpha)$ for an arbitrary $1$-form $\alpha(x) = \sum_{i=1}^n \alpha_i(x)\,dx_i$. We have$$\partial \hspace{.5mm} d\alpha = \partial\left(\sum_{i \neq j} D_j\alpha_i\,dx_j \wedge ... View answer 5 votes First of all, it is a worthwhile exercise to identify the set H of hyperfaces (which are cubes of side length 2) and verify that |H| = 8. Choose any hyperface of T. This hyperface can be ... View answer 4 votes Let U be a nonempty open subset of an irreducible topological space X. Denote by \overline{U} the closure of U in X. Then (X - U, \overline{U}) is a decomposition of X. Because X is ... View answer Accepted answer 4 votes We give a completely elementary proof simply working in the ring \mathbb{C}[t] and using that it is a UFD. Suppose there are some solutions of$$a(t)^3 + b(t)^3 = c(t)^3$$in \mathbb{C}[t]. Choose a ... View answer Accepted answer 4 votes Here is a visualization for the regular singular homology theory with coefficients in \mathbb{R}. The generator for H_2 is going to be the cycle that is "the entire manifold" (this is called the ... View answer 4 votes Solution 1. Let$$g: X \to S^n, \text{ }(p, q) \mapsto p.We see that $g \circ f = \text{Id}_{S^n}$. Now, let $(p, q) \in X$. There is a well-defined shortest path $\lambda_{p, q}: [0, 1] \to S^n$ ...
The prime ideals of $\mathbb{R}[x]$ have the form $\langle f\rangle$, where $f$ is one of: $f(x) = 0$; $f(x) = x - \lambda$, where $\lambda \in \mathbb{R}$; $f(x) = x^2 + bx + c$, where \$b, c \in \...