Hey does anyone know that the notation $p \mathcal{O}_K$ means in algebraic number theory? Our class is doing a smattering about it, and I can't find it defined in our lecture notes
The reason you use that notation in algebraic number theory is that you can think about $(p)$ as the ideal it generates in $\mathbb{Z}$ as well as in $\mathcal{O}_K$, so this avoids ambiguity
you are fixing a probabiity space $\Omega$, and expectation is the linear operator of measurable functions on $\Omega$ to real/complex/whatever, by integration
In geometry, a hippopede (from Ancient Greek ἱπποπέδη, "horse fetter") is a plane curve determined by an equation of the form
(
x
2
+
y
2
)
2
=
c
x
2
+
d
y
2
{\displaystyle (x^{2}+y^{2})^{2}=cx^{2}+dy^{2}}
,where...
So by implicit differentiation, if $x^2 + y^2 = 1$, then $\frac{dy}{dx} = -\frac{x}{y}$, and then by cancellation, $\frac{y}{x} = -\frac{x}{y}$, by which $y^2 = -x^2$. Substituting into the original equation, $0 = 1$, QED.
In Angelo Margaris's book First Order Mathematical Logic we have the following theorem (see pp. 84),
The equivalence theorem. Let $U$ and $V$ be two formulas (of First Order Predicate Calculus). Let $P_U$ be a formula in which $U$ occurs as a subformula. Let $P_V$ denote the result of replac...
If $f,g$ are paths between $x$ and $y$ in the topological space $X$, $\overline{f}$ and $\overline{g}$ are the reverse paths, and $\overline{f} \star f$ and $\overline{g} \star g$ are path homotopic, does it follow that $f$ and $g$ are path homotopic?
Do you mean $\overline{f} \star f$ and $\overline{g} \star g$?
I think you're right: they are both path homotopic to the constant path...
Hmmm...The problem I'm working on is the following: Prove that for any $x,y \in S^1$, there are countably many homotopy classes in $S^1$ from $x$ to $y$...
If $A$ denotes the class of all such homotopy classes, my thought was to show that $[f] \mapsto [\overline{f} \star f]$ is an injection from $A$ to $\pi_1(S^1,x)$...but if what you're saying is right, this is definitely not injective...
I've been working on this problem for a week and I still can't figure it out...
So $n\alpha\star f$, where $n\alpha$ is any element of $\pi_1(S^1,x)$, works
and we have infinitely many of those
We know these are distinct, because if $n\alpha\star f$ equaled $m\alpha\star f$, then $(n\alpha\star f)\star\bar f$ would equal $(m\alpha\star f)\star\bar f$
meaning $n\alpha\star(f\star\bar f)=n\alpha$ would equal $m\alpha\star(f\star\bar f)=m\alpha$
Okay, so this is a procedure to generate a countable number of such paths. But how do I rule out that the possibility that there are an uncountable number?
That's why I was trying to construct an injection.
Thus, $(F\star\bar f)\star f=n\alpha\star f$, or $F=n\alpha\star f$
@user193319 Basically, the bijection is: given a path from $x$ to $x$, star it by $f$ to get a path from $x$ to $y$. Given a map from $x$ to $y$, star it by $\bar f$ to get a path from $x$ to $x$. These are inverses, and so the two sets ($\{\text{paths from $x$ to $x$}\}$ and $\{\text{paths from $x$ to $y$}\}$) are in bijection.
On unrelated news, I was trying to come up with a smoothed version of the $\min(x,y)$ function
and I ended up with $\dfrac1{\sqrt{\frac1{x^2}+\frac1{y^2}}}$
aka $\dfrac{xy}{\sqrt{x^2+y^2}}$
(This is for positive $x$ and $y$)
I wonder if this is "natural" in some sense
like, maybe it's the only "smoothed" version of $\min(x,y)$ under some natural conditions
I have a question which should be straightforward but I'm not seeing it. If $X$ is a scheme, $x\in X$, $\mathcal O_x$ the local ring at $x$ with maximal ideal $\mathfrak m_x$ why is is $\mathfrak m_x/\mathfrak m_x^2$ a $\mathcal O_x/\mathfrak m_x$-vector space?
Note that it is an $\mathcal{O}_x$-module that is annihilated by $\mathfrak{m}_x$
a module structure over a ring $R$ on an abelian group $A$ is a ring homomorphism $R \to \mathrm{End}_{\Bbb Z}(A)$, the kernel of that ring homomorphism is the annihilator $\mathrm{Ann}_R(A)$. If $I$ is an ideal contained in $\mathrm{Ann}_R(A)$, by the homomorphism theorem, it factors through $R/I$, so we get an $R/I$-module structure
So, given a triangle in euclidean space with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$, the area of the triangle is apparently given by $$det\begin{vmatrix} x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1 \end{vmatrix}$$ (which simplifies to $x_1(y_2 -y_3)+x_2(y_3 -y_1)+x_3(y_1 -y_2)$) However, I don't immediately see why this holds.
(Apparently this is very basic stuff, as it's in section 1, chapter 1 of this book. So, I want to make sure I understand why this is.)
I have a simple question when we write $GF(q^m)$ this is a field that is made by modding a polynomial, correct? What do elements look like in extension fields?
The elements are polynomials
so when a polynomial has a splitting field, it means that the roots of that polynomial are POLYNOMIALS?
if you want to compute the value of some polynomial $g \in GF(2)[x]$ at $\alpha$, you can always do long division by $f$ and the remainder of that is the value of $g(\alpha)$
I see, so when we have a polynomial and we want to show that some field element is a root, we just long division the irreducible polynomial of the splitting field and see what it factors to?
@AkivaWeinberger , will you please look at this question? User xbh said that all i need to observe $f^{-1}$ is continuous on each point.
But, then, if I apply same approach to function $f$ from $[0,1)\cup[2,3]\to[0,2]$ defined by $f(x)=x$ if $x\in[0,1)$ and $f(x)=x-1$ if $x\in[2,3]$, i.e. apply Th 4.17 mentioned in link for intervals $[c,d]\subset [0,1)$ and $[e,f]\subset[2,3]$, then I have to deduce that this $f$ has continuous inverse. So, where am I (or xbh) wrong?
@Maximus you can also ignore the polynomial division stuff and work by brue force. All you need to use is that $1+\alpha^2+\alpha^5=0$, which implies that $\alpha^5=-1-\alpha^2$. This allows you to recursively compute all other powers of $\alpha$ as a linear combination of the basis $\{1,\alpha,\alpha^2,\alpha^3,\alpha^4\}$
for example $\alpha^6=\alpha(\alpha^5)=\alpha(-1-\alpha^2)=-\alpha-\alpha^3$
We know that field has only two ideals that is itself and 0 .So if R is a field then R/R , R/{0} are two homomorphic images of it . It is possible that is can have other homorphic images ?
@MatheinBoulomenos I was here during summer a lot and you helped me then too, I dont know if you remember. I noticed you do a lot of abstract algebra and you even have a blog. I am curious. What advice do you have or how do you manage to self study all that stuff on your own.
In terms of knowing whether your solutions are right or wrong.
@Maximus in general for self-studying it's really helpful to try and explain some ideas to other people, that's a test if you truly understand something
@TedShifrin So I still think my claim from yesterday is true. Your example gives a map $F : \Bbb R^2 \to \Bbb R^2$ such that $F^{-1}(0, 0)$ is the $x$-axis and $f$ doesn't fold along it; that's not too hard to visualize. Just squish the $x$-axis, smoothly. Even $F(x, y) = (xy, y)$ works. But I don't think any map $f : \Sigma \to \Sigma'$ with an isolated critical value $p$ such that $f^{-1}(p) \subset \Sigma$ is a circle can locally look like $F$ around any point on $f^{-1}(p)$.
It's just that $(xy, y^3)$ has a particularly bad critical point at $(0, 0)$, which makes your example interesting in it's own. But I don't think it's relevant.
@TedShifrin It's not a counterexample because... $F^{-1}(0, 0)$ is not a circle. You could claim that there exists a function $f$ with $f^{-1}(p)$ a circle such that $f$ looks like $F$ on some chart around the circle. But I do not think that is possible.
@TedShifrin I think if you projectivize your example, you'll get a map $f$ with $f^{-1}(0)$ a circle, but a tubular neighborhood around it will be a Moebius strip, and $f$, restricted to that, will be the map $M \to M/C \cong D^2$, blowdown of the central circle of the Moebius strip.
$C \subset M$ is the central circle in the Moebius strip
At least, that's the case with $f(x, y) = (xy, y)$. With $(xy, y^3)$ it might be ickier, there may be a cusp involved.
Let's see. If we have $\Bbb RP^2$ as the domain, the normal bundle of the preimage will be determined by the degree of the polynomials.
So you'll get $k$ half-twists if the degree is $k$, I think. I know this is true for a curve in $\Bbb P^2$ defined by a homogeneous polynomial of degree $k$, so this requires thought.
@MatheinBoulomenos yes but so is not $(a,b)$ here, but xbh says that considering $[c,d]\subset(a,b)$ and deducing that $f$ when restricted to $[c,d]$, has continuous inverse is enough to conclude f has continuous inverse on $(a,b)$. But when I apply that approach to $[0,1)\cup[2,3]\to[0,2]$ example, then I am in trouble!
@TedShifrin That sounds right. But I don't think you need to work that hard; $f^{-1}(0, 0)$ in $\Bbb R^2$ is just $y = 0$. If you projectivize it's a $\Bbb{RP}^1 \subset \Bbb{RP}^2$.
But you get a metric for free in this case; distance between two self-homeomorphisms of the circle is the sup norm of the two homeomorphisms, in the standard metric on S^1.
In general if X is a compact space and Y is a metric space, then the "topology of uniform convergence" on C(X, Y) is given by the metric d(f, g) = sup_x d_Y(f(x), g(x)). The topology induced by this is the same as the compact-open topology on C(X, Y).
@Maximus first of all, the pasting lemma doesn't apply to infinite closed coverings (without assuming something like locally finite), if you try to apply the approach that xbh gave to your example, then you run into trouble, because if you take the compact interval $[0,1] \subset [0,2]$, then the inverse image $f^{-1}([0,1])=g([0,1])=[0,1) \cup \{2\}$ is not compact, so you can't apply the theorem
@TedShifrin On a related note, I was reading a little bit about singularity theory properly. It seems the stable singularities for maps between surfaces are just folds and cusps.
I.e., maps with folds and cusps are dense in the space of maps between two given surfaces.
$O(2,1)$ has four connected components. First divide it into two parts, first block acting of 2 dim subspace and and the rest on 1-dim subspace. Then consider two more parts by determinant {+-1} in each of these. These are the four components. But I don't know the explicit continuous map that will help me see there are 4 connected comp0nents.
@Silent A big difference is that (most) closed subsets of $(a,b)$ are compact
Try to show that the inverse image in $f$ of a sufficiently small closed neighborhood around a point is compact.
(Using the fact that $f$ is strictly increasing.)
@Silent (Note that, in your counterexample function, the inverse image of a small closed neighborhood of $1$ will look like $[t_0,1)\cup[2,t_1]$, which is closed in $[0,1)\cup[2,3]$ but not compact.)
Weird question. When we say that $X \subset F$ where $F$ is some family of subsets, then we are essentially saying that $X$ can be a bunch of elements from $F$? But when we say $X \in F$ we mean that $X$ is only a single element taken from $F$?
What does the function $f(x,y)$ reduce to?
$$f(x,y) = \lim_{n \to \infty} \sum_{i = 0}^{n} (((x \bmod 2^{i-1})-(x \bmod 2^i)) \cdot ((y \bmod 2^{i-1})-(y \bmod 2^i)) \cdot \frac {1}{2^i})$$
In Angelo Margaris's book First Order Mathematical Logic we have the following theorem (see pp. 84),
The equivalence theorem. Let $U$ and $V$ be two formulas (of First Order Predicate Calculus). Let $P_U$ be a formula in which $U$ occurs as a subformula. Let $P_V$ denote the result of replac...
