I think of localization of spaces completely opposite to how you describe it; instead of localizing the homotopy and homology groups at $p$ and then using some roundabout Hurewicz mod C arguments you just do it at the level of the space. Much more geometric.
This is the perspective I gained from reading Sullivan's original 1970 MIT notes at least
I think you can construct the localization at p pretty explicit, by using $p$-local spheres, which are homotopy colimits of a bunch of $S^n$'s with degree $p$ maps going between, and $p$-local balls instead of spheres and balls when constructing a CW complex
Oh I never got around to reading the completion part in Sullivan
@LeakyNun Yeah, that's what it is. Given a, say simply connected, space $X$ you cook up a new space $X_P$ such that the homotopy groups of $X_{P}$ are $\pi_n(X) \otimes \mathbb{Z}_P$, where $P$ is the multiplicative set generated by some primes.
And it's a fantastic theorem of Sullivan that this happens iff the homology groups are $H_n(X)\otimes \Bbb Z_P$!
To give perspective, you can construct $S^n_P$ where $P = \langle p \rangle$ by taking a bunch of tubes $S^n \times I$ and gluing it front-to-back by a degree $p$ map $S^n \to S^n$
This "kills all $p$-torsion", or "inverts $p$-times any class" in the homotopy groups
This has always been very nonobvious to me - it follows from Sullivan's theorem but it's not clear at all why this should make the homotopy groups $p$-local
@TedE I don't know much about the theory beyond the basics but I like this, yes, because it's one of the few things I can "see"
How do you invert $p$ at the space level? Just do an infinite mapping cylinder construction, like it should be!
Strangely, this is the part that seems least intuitive to me
Taking $\ell$ to be a set of primes, and localisation at $\ell$ means taking an infinite mapping telescope over maps whose degrees form a cofinal set in the set of primes not in $\ell$
The homotopy colimit of $\cdots \to S^n \stackrel{\times p}{\to} S^n \stackrel{\times p}{\to} S^n$ has homology groups $\Bbb Z_p$ in dimension $n$ and $0$ otherwise
If $\ell$ is a multiplicative set, by $X_\ell$ I mean the space whose $\pi_n$ is $\pi_n(X) \otimes \Bbb Z_\ell$, where by $\Bbb Z_\ell$ I mean like everyone does, localizing $\Bbb Z$ at the multiplicative set $\ell$
Your thing is $\Bbb S^n_{\ell}$ for $\ell = primes-\{p\}$, rather than what I thought you meant which had $\ell = \{p\}$ and the degree of the maps in the telescope in the latter case would be products of primes other than $p$
The degree $p$ map $S^2 \to S^2$ acts by multiplication by $p^2$ on $\pi_3 S^2 = \Bbb Z$ because composing with the Hopf fibration $S^3 \to S^2 \to S^2$ you get that the linking number of two generic circle fibers is $p^2$ (by thinking of the configuration as being homologous to $p$ nearby circles linked to $p$ nearby circles, each pair linking with linking number +1)
Yeah for example I think that tells you the generator of $\pi_4 S^2$ is $S^4 \to S^3 \to S^2$ where the first map is suspension of the Hopf fibration and the second map is the Hopf fibration.
But why is that 2-torsion????
No damn clue
(Or generator of $\pi_4 S^3$ is the suspension of the Hopf fibration, even)
The singularity of the cusp, when seen from the $x$-axis, is of order $3$
Also, what's the general way to take intersections? $I, J$ be two ideals in $A$ and look at $V(I) \cap V(J)$ in $\text{Spec}(A)$, which is $V(I \cup J)$. How does one do this intersection scheme-theoretically? Fibered product of the diagram $\text{Spec}(A) \to \text{Spec}(A) \times \text{Spec}(A) \leftarrow \text{Spec}(A/I) \times \text{Spec}(A/J)$?
For closed subschemes $X,Y$ in a scheme $S$, with fixed closed immersions, their intersection is just the fiber product $X\times_ZY$. In the affine case $Z=\text{Spec}(C)$ and $X,Y$ are $\text{Spec}(C/I)$ and $\text{Spec}(C/J)$ respectively. So we are just looking at the fiber product over $\text{Spec}(C/I)\to \text{Spec}(C)\leftarrow \text{Spec}(C/J)$
(where these maps are just Spec of the quotient maps)
General intersections aren't well defined, and there is a whole theory that these notes are building towards
@BalarkaSen How do you define the order of a singularity?
@TedE not sure what you mean by general intersections here - but imo these notes are really about developing chow groups (intersecting subvarieties up to rational equivalence) (and to justify intersection theoretic arguments in the past eg conservation of number etc.)
@loch I mean it doesn't make sense to intersect two arbitrary schemes, and moving to a case where intersections should make sense, I don't think it makes sense to intersect two arbitrary subschemes of a given scheme (and get a scheme back)
Of course, I don't know intersection theory right now, hence why I'm reading a first lecture notes :P
you should be careful with what you mean by an 'arbitrary subscheme' also categorically the fibred product is what you want when you take 'intersections' :p
All I mean to say is this: The definitions I have seen for 'intersections of schemes' always assumed that these schemes were closed subschemes. The fibre product may exist, but in terms of semantics, I haven't seen this called an intersection
I guess with arbitrary subscheme defined in that way, we take the fibre product of the open subschemes, fibred over the closed subscheme, and we're fine
Or maybe not, I guess then I have to first permit intersection of open subschemes :P
@BalarkaSen So in direction $(a,b)$ we get $3ax^2-2yb$ right? What does this tell me? In particular, in the directions $(1,0)$ and $(0,1)$ I'd expect to see a double and a triple point right?
I'm not sure. $k[x, y]/(x^3 - y^2, x - y)$ gets you $k[x]/(x^2) \oplus k$ for example, which tells the line $x - y = 0$ vanishes with order $2$ at $(0, 0)$ and order $1$ at $(1, 1)$.
If you parametrize the curve using $(t^2, t^3)$ and the line using $(t, t)$, it does tell that the "osculation index" of the curves at $(0, 0)$ is $2$.
The $0$-th derivative agrees, the first derivative of either is $(0, 0)$ and $(1, 1)$ - parallel. Second derivative is $(2, 0)$ and $(1,1)$ - not parallel anymore
So upto the first two derivatives the curves match
Also, I think that the $x$ axis is the unique line through the origin that gives a fat point of order 3, since otherwise you always get an order $2$ intersection at the origin, and an order $1$ intersection elsewhere $k[x]/(x^2)\oplus k[x]/(x-a^2)$ for $a\in k$
Has to because total intersection multiplicity cannot exceed $3$, and any other line than the x-axis which passes through origin hits the curve somewhere else
Taking away a +1 contribution of intersection multiplicity
It is order of contact, like I said, I think. Given two curves $\gamma_1, \gamma_2$, asking what's the maximal $k$ such that $\gamma_1^{(k)}(t) = \gamma_2^{(k)}(t)$
alternatively / also what intersection multiplicity *should* suggest - is to perturb your line a little bit - you'll see if you move x=0 a little bit, the double point becomes two points
whereas if you move y=0 a little bit (and put on your complex number lens) the triple point becomes three points
if $V(I)$ and $V(J)$ are complementary Krull-dimensional affine sets in $\Bbb C[x_1, \cdots, x_n]$, should I expect the intersection multiplicity to be dimension of $\Bbb C[x_1, \cdots, x_n]/(I + J)$? Why's that even a $0$-dimensional algebra over $\Bbb C$?
Looks like you guys are in the middle of something, but if you can help me clarify this it would be greatly appreciated. I'm currently getting used to the formality of dual spaces, but I'm not seeing how my text arrives at a dual vector being represented by a matrix product with the vector it's dual to. I'm working with the Minkowski metric, here it is:
Define $\mathcal{B} = \{e_i\}$ as the basis, and $\mathcal{B^*} = \{e^i\}$ as the dual basis, where $e^i(e_j) = \delta_j^i$. A dual vector $\tilde{v}$ can be represented by its components with $\tilde{v} = \tilde{v}_ie^i$, where $\tilde{v}_i = \tilde{v}(e_i)$ (repeated index summation notation). Now, I'm looking at how these components are represented by the vectors $v$ of which they are dual to. So...
But even if $V(I)$ and $V(J)$ are smooth complementary dimensional subvarieties of $\Bbb C^n$ which are topologically transverse, does it imply $\Bbb C[x_1, \cdots, x_n]/(I + J)$ is a $0$-dimensional $\Bbb C$-algebra
I think topological transversality forces $I + J$ to be a radical ideal
Historical publications bear their old name and this can be problematic (both for the trans person themself and for people who don't "understand" their situation)
if you take a point in the intersection and compute its tangent space you should expect to see dim 0 which is good enough to tell you that the thing you had was a 0-dimensional algebra (since dim of tangent space >= dim of local ring)
@Soham I feel like the conversation would go "Oh I got married so my name is different on some of my old publications" and that's it lol, in the trans case the conversation might be a lot more difficult
So in the second set of notes, a weil divisor is defined to be an irreducible codimension $1$ variety. I imagine this is already wrong, and it should be a formal sum of such things
Since the first example is given by $(x-1)^2(x^2-2)/(x-3)$ over $\Bbb A^1_{\Bbb C}$ and $\Bbb{A}^1_{\Bbb Q}$
Dual spaces are cool. They pop up a bit in geometry, but I feel like I never truly understood what they're for. Either way, I can't deal with physics notation either.
This comes from Jech Lemma 8.19 (page 99).
If $\kappa$ is a regular uncountable cardinal, $A, B \subset \kappa$ are stationary subsets,
$A<B \ \ \iff \ A \cap \alpha \ \text{ is stationary for almost all } \alpha \in B \iff \exists \text{ closed unbounded }C: \forall \alpha , \alpha \in C\cap ...
Hey! How do we know that a bilinear form is positive definite (without knowing the eigenvalues and without calculation)? and if we only know that the constant elements of the matrix (that represents the bilinear form) look like this: $a_ij*a_uv$.
This comes from Jech Lemma 8.19 (page 99).
If $\kappa$ is a regular uncountable cardinal, $A, B \subset \kappa$ are stationary subsets,
$A<B \ \ \iff \ A \cap \alpha \ \text{ is stationary for almost all } \alpha \in B \iff \exists \text{ closed unbounded }C: \forall \alpha , \alpha \in C\cap ...
