I had no idea. It's like 57: mathematicians are working very hard on determining whether or not it is composite, and they've already gotten partway there by finding out that it's prime.
No, the Zariski-Riemann space of an extension $K/L$ is a scheme whose points are in 1-1 correspondence with valuation rings squashed between the extension.
for example, the one-dimensional ramified covers of $\mathrm{Spec}(\Bbb Z)$ that are included in that can be smooth or non-smooth, Mumford only drew two smooth ones ($V(x)$ and $V(x^2+1)$) Stuff like $V(x^2+4)$ will have a singularity which is related to failure of unique factorization for ideals and failure of integral closedness and you can also use the tangent space of course
for a given, say monic irreducible polynomial, you can read off the ramification points from the discriminant and the singularities from the conductor
I find tom Dieck more readable for a first exposition so I'm working on that now, I think I'll read Hatcher after I'm already familiar with the topics covered
I'd be up for it, but I kinda skipped the homotopy part and started with homology at chapter 9 of tom Dieck (I'm also reading the homotopy part from chapter 2 onward parallel to it though), I don't know what's your background and what kind of AT you're thinking about
@AlessandroCodenotti I'm not sure what kind I'm thinking about either. I'd say my background is a little weak but I've had point-set and algebra courses.
@Fargle My objective is to self study as many topics as I can over the summer from this list to prepare for a couple of grad courses I'd like to take in the winter semester but I don't have the prerequisites for atm
I'm not sure whether math is still allowed in this chatroom but I'm stuck in the proof that for path connected $X$ we have $H_1(X)=\pi_1(X)^{\mathrm{ab}}$
So singular $1$-simplices in $X$ are almost paths in $X$, they just have domain $\Delta^1\subseteq \Bbb R^2$ rather than $[0,1]\subseteq \Bbb R$ but that can be fixed by associating to the simplex $\sigma:\Delta^1\to X$ the path $[0,1]\to X$, $t\mapsto\sigma(1-t,t)$
We have $k:\Delta^1\times I\to X$ an homotopy of paths. We factor $k$ through the quotient map $q:\Delta^1\times I\to\Delta^2$, $(t_0,t_1,t)\mapsto (t_0,t_1(1-t),t_1t)$ to get a map $\sigma:\Delta^2\to X$
We would then like on a manifold $M$ to come up with a recipe that takes as input a function $H$ and spits out a flow which preserves $H$.
How should we do this? We know that the easiest way to come up with a vector field $X_H$ is to take the 1-form $dH$ and contract with some 2-tensor (who knows what the 2-tensor is at this point).
This gives us a recipe to get a map from functions to vector fields, as long as the 2-tensor is a nondegenerate bilinear form at every point, so that $T(X_H, -) = dH$ uniquely defines $X_H$.
No, there is more.
So we see that our 2-tensor should be nondegenerate.
Next, the fact that $X_H(H) = 0$ translates into $dH(X_H) = 0$, or $T(X_H, X_H) = 0$. So $T$ should be skew-symmetric. Thus our theory is defined by a nondegenerate 2-form $\omega$.
@BalarkaSen Right, so if I don't bother with having exactly the standard $n$-simplex in $\Bbb R^{n+1}$ as Dieck does I can take the homotopy of paths to be $k:[0,1]\times [0,1]\to X$ such that $k(x,0)$ and $k(x,1)$ are the two paths I'm dealing with and quotient the side $k(1,t)$ to get a 2-simplex, compute its boundary and conclude that $k_1-k_0$ is a boundary?
Now the flow should also preserve whatever structure we used to define $X_H$; this seems wholly reasonable, lest the recipe change after we apply the flow. Thus $\mathcal L_{X_H} \omega = 0$. But expand with Cartan's magic formula: this is $d \iota_{X_H} \omega + \iota_{X_H} d\omega$. The first term is, by definition of $X_H$, $ddH$, so is zero.
Thus we should demand that $\iota_{X_H} d \omega = 0$. For this to be true for all $X_H$, $d\omega$ had better to be zero.
Thus our theory is defined by a closed nondegenerate 2-form.
This last point I really liked, since the closedness was not a necessary feature of the geometric structure to me - it gives us Darboux charts, but why is it necessary for the dynamics?
I can also quotient both the $k(0,t)$ side and the $k(1,t)$ side to get rid of the constant term in the boundary actually, the quotient is a disk so still homeomorphic to the standard 2-simplex and it's all fine
How much of this relationship between $H_1$ and $\pi_1$ works for $H_n$ and $\pi_n$? It seems to me that the same trick can be used to show that if $f,g:[0,1]^n\to X$ are in the same class in $\pi_n(X,x_0)$ then they should be in the same class in $H_n(X)$ as well, since $[0,1]\cong\Delta^n$
If I knew how to edit for shit I'd draw a peaceful simplex with the buildup, and then when it changes to the heavier music, then boom it becomes Concise
There are some restrictive conditions that tell you something about that map (this more general statement is often called the Hurewicz isomorphism theorem, and that map the Hurewicz map: the n = 1 case is subsumed)
Ah, I see. Allow me to rethink my exposition briefly
Given any "tensor" $T$ on a manifold (a tensor is in a sense an object that takes as input $r$ vector fields and spits out $s$ vector fields, in a smoothly varying way, and linear at each point - a vector field is a basic kind of tensor that takes 0 pieces of input data and spits out 1 vector field, while a form is a kind of tensor that takes in $k$ vector fields and spits out $0$ vector fields - just a function)
And a vector field $X$, you can define the "derivative of $T$ in the direction of $X$", called the Lie derivative, $\mathcal L_X T$. The idea is this. A vector field has "integral curves" $\gamma$, with the rule that $\gamma'(t) = X_{\gamma(t)}$ (the tangent to the curve is just the vector field, at that point)
Now on an integral curve $\gamma$ for the vector field $X$, we can restrict $T$ to $\gamma$. Because $\gamma$ is identified with $\Bbb R$, we can take the usual kind of derivative from calculus - the time derivative. If $\gamma(0) = p$, we define $$(\mathcal L_X T)_p = \frac{d}{dt} T_{\gamma(t)} \big|_{t = 0}.$$
I apologize for the mess of notation, but the idea is supposed to be that given a vector field $X$, we have a direction in which we can take the derivative given at each point by the vector field itself, and we take the derivative of the tensor in that direction
I suspect I am dropping on you a mess of concepts you haven't seen before which probably make this more scary
If that is the case, then ignore me for a bit, and let's table this for some period of time :) These will all seem quite natural eventually
@Alessandro Given $\alpha \in \pi_n(X)$, there is a representative $f : S^n \to X$ such that $[f] = \alpha$. Triangulate $S^n$ so that it's a simplicial complex. Then $f$ naturally gives rise to a singular cycle in $X$ by considering the formal sum of the restrictions of $f$ to each of those simplices, upto sign according to identification. Let $\xi$ be that $n$-cycle. $\pi_n(X) \to H_n(X)$ is given by $\alpha \mapsto [\xi]$.
This is well defined because of what you said, difference of two of those $\xi$'s arising from two different representative $f$'s is a boundary given by the homotopy.
For $n = 1$, this is a homomorphism $\pi_1(X) \to H_1(X)$ of a generally nonabelian group to an abelian group. So the commutator subgroup dies in the kernel of this map. By universal theorem of quotients it descends to a homomorphism $\pi_1(X)/[\pi_1(X), \pi_1(X)] \to H_1(X)$
The reason is living in the paragraph of Hatcher I wanted you to read. Classes in $H_n(X)$ can be represented as maps from $n$-manifolds with at least(most??? codimension is weird, linguistically) codim 2 singularities.
For $H_1(X)$ there is no issue as codimension 2 doesn't exist in dimension 1
So every element of $H_1(X)$ can be represented as a map from a 1-dimensional closed manifolds (simply circles) to $X$.
So every element of $H_1(X)$ is indeed in the image of the homomorphism, as they can be represented as a loop.
Now if you remember Hatcher, you'd know that there are no issues in codimension 2 as well. Every homology class is represented as a map from a pseudomanifold with singularity living in codimension at least (nudge @Mike) 3. So why can't this same argument push for $\pi_2(X) \to H_2(X)$?
@BalarkaSen hmmm well for $n=1$ we need path connectdness to make it work, so I guess we need some stronger condition for $n=2$ to have any hope of it working, but it sure looks like it should work!
@MikeMiller Here's a question pulled out of my ass. Say I pick a homology class $\alpha \in H_n(X)$ arbitrarily, where $X$ is some nice space (manifold if you want). Suppose I pick a representative $f : Z \to X$ from a pseudomanifold with codim 3 singularities for $\alpha$. The singularity set in $Z$ is going to be a subcomplex itself. So restrict $f$ to that to get a cycle in $H_{n-3}(X)$. What is this boio? Is this independent of the choice of $f$?
I should really look for what happens when $\alpha$ is representable as a map from a stratified space. I can restrict it to each stratum-closure to get a long chain of elements in $\bigoplus H_k(X)$.
Has Kreck thought about this phenomenon? Do you get some filtration on $H_\bullet(X)$ or something?
