Sorry, it seems I am confused about what the holonomy representation is, sorry. It's related but not this. In the Riemannian sense, holonomy group of $M$ at $T_p M$ is the subgroup of $\text{GL}(T_pM)$ of all automorphisms of the tangent space that comes from parallel transporting along loops based at $p$
That's $\text{Hol}_p(M)$. The theorem is if the representation of $\text{Hol}_p(M)$ in $T_pM$ is reducible, $M$ is a product manifold.
@TobiasKildetoft Ah, well, a Riemannian metric is just a choice of an inner product at each tangent space of $M$
Smoothly
If I recall correctly there's theorems about what subgroups of $O(n)$ can appear as $\text{Hol}_p(M)$, and when it does what can be said about $M$. IIRC there are ways to read off if a manifold admits a complex structure from it's holonomy groups.
Actually, the field of fractions of $\Bbb R[x,y]/\langle x^2+y^2-1\rangle$ is isomorphic to $\Bbb R(t)$, through $t=\frac x{y+1}$. (We have $x=\frac{2t}{1+t^2}$ and $y=\frac{1-t^2}{1+t^2}$.)
You essentially think of $x$ and $y$ as $\sin\theta$ and $\cos\theta$. Then $t$ is $\tan(\theta/2)$. It's the tangent half-angle substitution.
Alternatively, it's the parametrization of the circle.
I think it's a general theorem that if $X \subset \Bbb A^n$ is a subvariety of Krull dimension $\ell$, $k(X) :=\text{Frac}\, k[X]$ has transcendence degree $\ell$ over $k$
Well, for characteristic 0 and algebraically closed $k$
Let $R$ be a ring (not necessarily unital) generated by some element $r$; i.e., $R = (r) = \{rx + nr \mid x \in R, n \in \Bbb{Z}\}$. Is the following true:
$R^2 = R$ if and only if $r = \sum_{i=1}^n a_i b_i$ for some $n \in \Bbb{N}$ and $a_i, b_i \in R$, where
$$R^2 = \{ \sum_{i=1}^k x_i ...
Krull dimension is easy to define, but complicated to justify. Krull dimension of a variety $X$ is the maximal length of chains $X_0 \subset X_1 \subset \cdots \subset X_\ell$ of irreducible nonempty subvarieties of $X$.
*Is contemplating a conspiracy to create a field of positive characteristic by killing of certain theorems* https://mathoverflow.net/questions/3551/algebraically-closed-fields-of-positive-characteristic
Question: How can knowing of the automorphism group of a group be useful in studying or determining the structure of that group itself or other groups?
Proof that $(4)$ is maximal in the set $E$ of even integers: Suppose that $(4) \subseteq N \subseteq E$. Since $E$ is a principal ideal ring, $N = (x)$ for some positive$x \in E$. Since $(4) \subseteq (x)$, $4 \in (x)$ which means $x|4$ and therefore either $x = 2$, in which case $(x) = E$, or $x=4$, in which case $(x) = (4)$. Hence, $(4)$ is a maximal ideal in $E$.
@LeakyNun I'm not sure what you mean. I am just using the fact that all ideals in $E$ are principal ideals, which essentially follows from the division algorithm.
let (4) be a proper subset of N. take an element n of N that is not in (4). By division algorithm, n=4q+r. r cannot be 0, 1, or 3. Then, n-4q = 2, so N=E. @user193319
$200 \times 200$ square is colored in chess order.
In each move we can take any $2 \times 3$ rectangle and change the color of all its cells.
Can we make all cells of the square be the same color ?
I still do not have an idea.
Is this construction useful, square with every $2 \times 3$ ...
I imagine the way $H_2$ works in like $S^2$ for example is like, if you integrate a constant function over a nullhomologic(?) thingy it's zero but over the entire thing it's nonzero
And in $\Bbb R^3\setminus\{0\}$, instead of a constant function, you give each tiny rectangle a value equal to the area of its projection onto $S^2$ or something
@Semiclassical Ah, nevermind, there's a simpler way to see $\pi_2(T^2) = 0$. Consider any map $f : S^2 \to T^2 = S^1 \times S^1$. You can write this as $f(x) = (f_1(x), f_2(x))$ where $f_1, f_2$ are the maps $S^2 \to S^1$ given by projecting $f$ to each of the circle components.
$f_1, f_2$ are nullhomotopic. So let $h^1_t, h^2_t : S^2 \to S^1$ be a homotopy between $f_1, f_2$ and the constant map $S^2 \to S^1$ to some point $0 \in S^1$
@BalarkaSen In $\Bbb R^2\setminus2$, you have a curve that's nullhomologous but not nullhomotopic. I wonder if in $\Bbb R^3\setminus n$ you could find a surface that's nullhomologous but not nullhomotopic
(By $\Bbb R^3\setminus n$ I mean space minus $n$ points)
It has to be. Because the norm is a linear functional on $L_p$ and then you can always find a function that saturates the Holder bound and then that gives you a connection to $L_q$
Pick out the true statements.
a. Let $f : \mathbb Z\to \mathbb Z^2$ be a bijection. There exists a continuous function from $\mathbb R$ to $\mathbb R^2$ which extends $f.$
b. Let $D$ denote the closed unit disc in $\mathbb R^2.$ There exists a continuous mapping $f : D-\{(0, 0)\}$$\t...
I am having trouble proving that if $(x)$ is prime ideal in $\Bbb{Z}$, then $(x)$ must be maximal ideal (note: I cannot use the equivalence of either of these statements with $(x)=(p)$ for some prime $p$; in fact, I am trying to prove the equivalence of these three statements). I could use a hint.
@Akiva For spheres, the things that are null-homologous but not null-homotopic are the elements in the kernel of the Hurewicz homomorphism, and for (n-1)-connected spaces that map is an isomorphism
I think for connected surfaces in R^3 \ n, null-homologous implies null-homotopic by arguing by hand
So we're asking whether homotopy classes of map from a surface to a wedge of 2-spheres are determined by the degree of the projections on each component of the wedge?
Please verify this: If $A$ is an $m\times 1$ matrix then only row reduced echelon forms possible for $A$ are either $\begin{bmatrix}0&0&\cdots &0 \end{bmatrix}^T$ or $\begin{bmatrix}1&0&\cdots &0 \end{bmatrix}^T$
@PVAL-inactive Yeah...I tried that. Suppose that $(x) \subseteq (y) \subseteq \Bbb{Z}$. If I assume that $(x) \neq (y)$, then the goal is to show $1 \in (y)$ or that $y=1$; if I assume that $(y) \neq \Bbb{Z}$, then the goal would be to show $y \in (x)$, which is when I tried using the division algorithm.
$\Bbb R^n \setminus S$ for a finite set $S$ deformation retracts to wedge of $S^{n-1}$'s, and $\pi_{n-1}$ of that is isomorphic to $H_{n-1}$ by Hurewicz, right?
then you should be able to do an Eckmann-Hilton type argument, so the map composes as a connect sum of maps each hitting each one of the components in the wedge.
Please someone let me know if i am right: If $A$ is an $m\times 1$ matrix then only row reduced echelon forms possible for $A$ are either $\begin{bmatrix}0&0&\cdots &0 \end{bmatrix}^T$ or $\begin{bmatrix}1&0&\cdots &0 \end{bmatrix}^T$
@TedShifrin A map from a genus g surface to a wedge of 2-spheres determine a homology class in the wedge; but how do you know if that class is zero then the map is nullhomotopic?
Hurewicz doesn't tell you that
If the wedge is a single 2-sphere this is Hopf degree theorem
Let $R$ be a ring (not necessarily unital) generated by some element $r$; i.e., $R = (r) = \{rx + nr \mid x \in R, n \in \Bbb{Z}\}$. Is the following true:
$R^2 = R$ if and only if $r = \sum_{i=1}^n a_i b_i$ for some $n \in \Bbb{N}$ and $a_i, b_i \in R$, where
$$R^2 = \{ \sum_{i=1}^k x_i ...
$200 \times 200$ square is colored in chess order.
In each move we can take any $2 \times 3$ rectangle and change the color of all its cells.
Can we make all cells of the square be the same color ?
I still do not have an idea.
Is this construction useful, square with every $2 \times 3$ ...
Pick out the true statements.
a. Let $f : \mathbb Z\to \mathbb Z^2$ be a bijection. There exists a continuous function from $\mathbb R$ to $\mathbb R^2$ which extends $f.$
b. Let $D$ denote the closed unit disc in $\mathbb R^2.$ There exists a continuous mapping $f : D-\{(0, 0)\}$$\t...
