@anakhro: I just taught precalculus last year in AoPS. I was reasonably happy with the standard approach to trig. You switch over to the circle (and then to $\Bbb R$), but I think it's fine to start with first quadrant only.
Not that much, but I demanded leeway when I taught calculus. In both courses I wrote a number of my own problems for them to work on — especially the better students — because several of the students weren't challenged enough. But I couldn't get the kids to spend any time to write up stuff for me to criticize, which to me is the essential in learning mathematics — not just on-line homework.
I did a bit more in the way of proofs in calculus and even changed the exams a little bit. But it probably wasn't worth all my effort.
Whence my quitting.
One of the kids in my precalc class was one of the junior olympiad winners. He was amazing. A few of the calc kids could have been tremendous if I'd had them in an actual class with a grade ...
Various sorts. In calculus, I gave them some of the problems I used to assign in my calc theory class. In trig, I had them set up some of the word problems that would show up in max/min problems using trig. Fill in steps of different proofs of the addition formulas, etc.
Oh yeah, and a proof of Heron's formula relating to the inscribed circle, etc.
Well, now the trend in the US is to flipping classrooms, so it's doing away with lectures and putting the responsibility on the students to watch some videos and work some practice problems before doing more serious groupwork in class the next day. I think that's an improvement over having dull lectures with passive students sitting in large lecture halls. But I never tried it myself. Had I not retired, I would have tried it in my multivariable math class using my videos. ...
I don't know what you mean by compartmentalizing. 95% of calculus students are there from science and engineering (and business — different clientele). The curriculum is driven by what their faculty want, not what pure mathematicians want.
It shouldn't just be passive watching, but sadly, that's what most classes are.
However, I don't think the way stuff is organized is optimal. Particularly in multivariable calculus, I hate a lot of the way books are written/organized.
I have bitched for 40 years that most of the people who write calculus books are not "comfortable" with multivariable and don't think about it right.
You trivialize over two intervals and look at the overlap (which is two intervals). So it depends on how many connected components you have in $GL(2)$ (or $O(2)$).
This is classically called the "clutching construction" for bundles on spheres.
If you integrate it $dt$ you'll get a nowhere-vanishing $2$-form on the Möbius strip, I do believe.
Anyhow, the simple answer to your question is to look at $S^1$-bundles over $S^1$. You either have a torus or you have a Klein bottle. You were forgetting about the latter.
The third (or second for some people) isomorphism theorem says if $I ⊆ J ⊆ R$ are ideals in $R$ then $(R/I)/(J/I) \cong R/J$.
In the proof, we show that the map $\psi: R/I \to R/J$ given by $\psi(r+I) = r+J.$ We show that $\psi$ is a well-defined surjective ring homomorphism; and by the first isomorphism theorem $(R/I)/\ker(\psi) \cong R/J$.
It seems to me to be a thickened (and rather heavily stylized) Möbius strip, i.e., a torus with square cross section that is given a one-half twist.
I made this image just now using the code from my math.SE question, Drawing a thickened Möbius strip in Mathematica
@TedShifrin what is this shape?
the mobius strip is $M = \Bbb R^2 / \Bbb Z$ where $n \cdot (x, y) = (x + n, (-1)^n y)$
this feels like a semidirect product
let's see why the candidate 2-form on $M$ isn't well-defined
it's just $\mathrm dx \land \mathrm dy$
oh well-defined <=> invariant under the pullback of the action
one thing I think will really help me along with what ive been reading is if someone can explain the logic behind the choice of the term algebraic integer, it has to be significant
to compensate for the confusion the choice makes, is it because algebraic integers are roots of monatomic polynomials with integer coefficents? I mean they use the term quadratic integer for quadratic fields
@SohamChowdhury I been thinking about your hint. Would the map be $f: K \to J$ given by $f(I+r) = J$? Maybe the same can I say $J/I$ and $\ker(\psi)=K$ are either equal or disjoint, as cosets. But $I+r \in K \implies I+r=J+0 \implies r \in J$, meaning $J/I = \ker(\psi).$
whoever wrote the wiki for algebraic integer deserves many thanks I mean look at how many definitions this forces you to learn in order to see the equivalence of the one definition
@Adam Well, let $K$ be a subfield of a field $L$. Let $a\in L$. Do you know what is meant by '$a$ is algebraic over $K$'? Do you know what it means for an element to be integral? Maybe you would prefer integral integer :P.
An $\epsilon$-neighborhood of the Moebius strip in $\Bbb R^3$ is indeed the solid torus. It cannot be $M \times (0, 1)$ because any open subset of $\Bbb R^3$ is necessarily orientable.
The point is the "normal bundle" to the Moebius strip along the center circle is also a Moebius strip
That's why you get Moebius $\oplus$ Moebius as your bundle
@LeakyNun This picture is exactly Mobius $\oplus$ Mobius (you imagine taking a cross X and then twisting it around and coming back to the same point), hence a solid torus :)
(I guess I meant half-twist as opposed to quarter-twist, but w/e)
@MikeMiller I always forget the definition of $K$-theory. $\tilde{K}^0(S^1)$ is $\Bbb Z_2$, right? Generated by the 2-torsion class of the Mobius strip
@TedE no can you give me an example to work from I get how to write out all algebraic integers of a particular degree in set builder notation but yeah can you give an example
I suppose $\tilde{K}^0(S^9)$ is also $\Bbb Z_2$, because of Bott periodicity?
$\widetilde{K}^0(X)$ is $[X, BO \times \Bbb Z]_{pt}$ I think
I can forget about the formalism and check this directly; rank $n$ vector bundles over $S^9$ are classified by $\pi_8(O(n))$ by clutching construction, and taking direct sum with trivial bundle correspond to pushing forward this class by $O(n) \to O(n+1)$, including an orthogonal $n \times n$ matrix as a block consisting of 1 in top-left corner and the $n \times n$ block below
So the stable class lies in $\pi_8(O)$ as promised, and homotopy groups of $O$ are $8$-periodic is the classical Bott periodicity theorem
I wonder what the vector bundle over $S^9$ corresponding to the nontrivial stable class looks like
If $\Bbb F$ is a field and $R = \Bbb F[x,y]$ I'm being told that $\Bbb F$ has an R module structure induced by the quotient map $\Bbb F[x,y]\twoheadrightarrow \Bbb F$. I fail to understand what this means, would anyone be able to shed some light on this?
So if I look at $\phi : p \mapsto (xp, yp)$ and $\psi : (p,q) \mapsto yp-xq$ this gives me a projective resolution of $\Bbb F$ of the form $0 \to R \to R\oplus R\to R$ right ?
Note that your resolution can also be thought as a free resolution of the module (x, y) over R = F[x, y], as 0 -> R -> R oplus R -> (x, y) -> 0
Which says (x, y) is not a flat F[x, y]-module
The algebro-geometric interpretation of that is that (x, y) is like the skyscraper sheaf at the point (0, 0) in A^2, so the fibers "vary drastically" point-to-point
Spec F -> Spec F[x, y] given by dualzing F[x, y] -> F[x, y]/(x, y) = F is not a flat morphism
(One can of course alternately argue that any f.g. flat module is projective and projective modules over polynomial rings are free by Quillen-Suslin, and (x, y) is not a free F[x, y]-module :))
@Astyx Tensoring $0 \to R \to R^2 \to R \to F = R/M \to 0$ with $F$ says that $0 \to F \to F^2 \to F \to F \to 0$ is our chain complex where all maps except the last $F \to F$ becomes zero I think, so $\text{Tor}^2_R(F, F) \cong F$.
And $\text{Tor}^1(M, F) \cong \text{Tor}^2(F, F)$, so that means $\text{Tor}^1(M, F) \cong F$, which is the witness of $M = (x, y)$ being not flat over $R = F[x, y]$
@AlessandroCodenotti the Prof replied, he asked if i was attending his lecture course because he likes to get to know students better before supervising their theses. I‘m not attending his course atm because I don’t have the background for it but I would do so in the future lol, otherwise he didn’t say no! He invited me to come and speak to him first
I'm not sure what your map is @anakhro. Try Z/6 otimes Z/4. Then the map x otimes 1 to x, Z/6 (x) Z/4 -> Z/6 is not a real Mao. The domain is isomorphic to Z/2.
If you already showed somehow that there's a unique x so that an elementary tensor is equal to x (x) 1 in Q (x)_Z Q then fine. But how did you show this?
It all seems a bit silly when multiplication is an obvious homomorphism from one to the other, which is exactly the map you want to describe
@LeakyNun derived functors of the tensor product functor :3
@LeakyNun A duck and a goat are homeomorphic, doesn't mean they are the "same pictures"
Mike set up the right context: there's a natural flat connection that depends on how many twists you use to identify the sides of the square. So they are not isomorphic as flat bundles
@LeakyNun I would call it Moebius strip x [0, 1]. You will not find a picture because open Moebius strip x (0, 1) does not embed as a submanifold of R^3 because any open subset of R^3 is orientable
One of my favourite professors always would throw out a disclaimer when he said he can't think in >3 dimensions, "but perhaps one of you are an alien from a distant planet and can do so".
Why can $p(z_i)$ be written as $\prod_i p(y_i \mid x_i)$? I would like to note that I understand that, given that the observations are independently drawn, then the joint distribution of $p(z_i)$ could be written as $\prod_i p(z_i)$, but why does it become a conditional?
In fact to prove this you have to use universal property with the multiplication map Q o_Z Q -> Q anyway rendering what you were doing a bit roundabout
@BalarkaSen sorry, I am confused over the term "multiplication map". It was advised that there was the map $x\otimes y \mapsto xy$. Is this the one you are meaning, or which one?
Forget simple tensors. There is a map $\Bbb Q \times \Bbb Q \to \Bbb Q$, $(r, s) \mapsto rs$. The ring multiplication of $\Bbb Q$. This is billinear, yes?
I was just confused what map it was you were discussing.
So how would I finish it via my method? I have a map $x\otimes y \mapsto xy$, but I have to show that this is well-defined to finish off with the surjectivity/injectivity (thus showing the isomorphism)?
I mean the map you just stated isn't your method, it's the linear map induced by multiplication. It's how we are suggesting thinking about the map from the tensor product.
I thought your method was to write everything as x o 1 and forget the 1.
@anakhro You wan to prove $1 \otimes x = 0$ implies $x = 0$. To do this, consider the bilinear map $\Bbb Q \times \Bbb Q \to \Bbb Q$ which factors through the linear map $\Bbb Q \otimes \Bbb Q \to \Bbb Q$, and $(1, x)$ gets sent to $x$. Image of $(1, x)$ by $\Bbb Q \times \Bbb Q \to \Bbb Q \otimes \Bbb Q$ is $1 \otimes x$, so $1 \otimes x = 0$ would force $x = 0$.
If you want the element based defn of tensor product, it's freely generated by symbols v o w, modulo the relations (cv) o w = v o (cw) and (v+w) o u = v o u + w o u and v o (w + u) = v o w + v o u. These relations are satisfied for any bilinear map, like multiplication, so that multiplication defines a map on Q o_Z Q to Q
From chapter 1 we have the following definition of space $\mathcal{D}_K$.
If $K$ is a compact set in $\mathbb{R}^n$ then $\mathcal{D}_K$ denotes the space of all $f \in C^{\infty}(\mathbb{R}^n)$ whose support lies in $K$.
The topology on such space is introduced by the family of seminorms
...
Let $A,B,C > 0 $
$a_1 = A$
$a_2 = B$
For integer $ n > 2 : $
$$a_n = \frac{a_{n-1}(a_{n-1} + C)}{a_{n-2}}.$$
$$ T = \lim_{k \to \infty} \frac{a_k}{ a_{k - 1}}.$$
Notice the limit seems to depend on 3 variables, but actually only depends on 2 ;
$A/C,B/C$.
You can check that if you multiply...
@anakhro how can I determine all elements of the subset of $\mathbb Z[x]$ if I am given a presumed to be algebraic integer? like all the possible monic polynomials with integer coeffs
or what is the most efficient algorithm for doing so i mean
the subset of all monic polynomials with integer coeffs that a given algebraic integer is a root of like you know how you can do with Einsteinian integers by getting the first polynomial, and then generalizing by introducing two variables and making a linear combination
From chapter 1 we have the following definition of space $\mathcal{D}_K$.
If $K$ is a compact set in $\mathbb{R}^n$ then $\mathcal{D}_K$ denotes the space of all $f \in C^{\infty}(\mathbb{R}^n)$ whose support lies in $K$.
The topology on such space is introduced by the family of seminorms
...
Let $A,B,C > 0 $
$a_1 = A$
$a_2 = B$
For integer $ n > 2 : $
$$a_n = \frac{a_{n-1}(a_{n-1} + C)}{a_{n-2}}.$$
$$ T = \lim_{k \to \infty} \frac{a_k}{ a_{k - 1}}.$$
Notice the limit seems to depend on 3 variables, but actually only depends on 2 ;
$A/C,B/C$.
You can check that if you multiply...
