Right. But why is that relevant? I thought stein structures were complex structures? Where do we get the symplectic structure?
Ok, I googled it, you pick a plurisubharmonic function $\varphi$ and let $\omega$ be $dJ^*d\varphi$. Is there a nicer way of thinking of the symplectic structure? Would one prefer to think of this as a special kind of Kahler manifold to begin with?
A Stein manifold (one of many definitions) $(M,J)$ is a complex manifold which admits an exhausting proper morse function $f: M \to [0, \infty)$ such that the form $-d(J^*(df))$ is symplectic.
@MikeMiller "Is there a nicer way to think about their symplectic structure?" No. The $J$-convex function is exactly what is manipulated to give nice characterizations of which smooth manifolds admit these structures (MUCH nicer characterizations than Kahler).
The definition that makes them "intrinsically interesting" is that they are exactly the smooth affine analytic varieties but the functions are what you use to study them (along with the symplectic structure coming from them and the the contact level sets associated to the function). I just wrote a rather trivial application of Stein structures here math.stackexchange.com/a/1410952/83337
Is the end of an open Stein manifold a product end automatically? I guess it must because "varieties can't have complicated ends". I don't know why j believe that but I do.
I guess because every affine bariety actually had a compactification in the projectivization. Ok.
Any infinite 2-handlebody (one needs to be somewhat precise as what is meant here I mean a handlebody admitting a proper morse function with crit points $\leq 2$) is homeomorphic (definitely not diffeomorphic) to a Stein manifold by a theorem of my advisor.
A lot of the time you just deal with the compact version called Stein domains which are just Stein manifolds with $J$-convex boundary obtained by attaching finitely many handles to $D^4$
Lots of examples. Some non-existence results from the adjunction inequality. Eliashberg showed that every 2n dimensional n-handlebody admits a smooth structure except when $n=2$ and in that dimension the framing on the $2$-handles must be $tb(K)-1$, which reduces a lot of the filling questions to Kirby calculus.
Okay I thought of the short exact sequence problem which you gave me. So we have $N\to G\to \Bbb{Z}^n$ a SES . If I define a map $s: \Bbb{Z}^n \to G$ by the map $s(x)=\text{ker}(m)$ where $m:G \to \Bbb{Z}^n$ . Will this map do the job. By that I mean to say that if this map is an homomorphism (which I guess it is) by the splitting lemma $G \cong \Bbb{Z}^n \oplus N$. Right?
I am not asking you to count all the homomorphisms, keep that in mind. I want a simple, easy way to construct a homomorphism between $\Bbb Z$ and $A$, given an arbitrary abelian group $A$.
Can I define a homomorphism $f : \Bbb Z \to A$ by sending $1 \in \Bbb Z$ to $a \in A$ and linearly extending? That is, $n \mapsto a + a + \cdots + a$, where the sum is done $n$ times?
What you can do, on the other hand, is to define an action of $\Bbb Z$ on $A$. That, in short, is written $na$. But anyway, you'll learn this when you do modules.
@Rememberme huh?
I am not sure if you're quite clear on the notation $na$.
We define the homomorphism $f : \Bbb Z \to A$ by $1 \mapsto a$ for some fixed $a \in A$, and then linear extending, i.e., $n \mapsto a + \cdots + a$
This is a situation similar to vector spaces, where you define a homomorphism between vector spaces by mapping basis elements to basis elements and linearly extending, right?
OK. So it turns out this is indeed a kind of generalization to what we do in vector spaces. $\Bbb Z$ here is a free module over itself, one might say. But nevermind that.
@Balarka Can I define a map from $A \to \Bbb{Z}$ by just the inverse idea that is $a\mapsto 1$ that is $a+a+\cdots +a \mapsto n$ ? Now if I compose the previous map with this one then I will get $1\mapsto 1$ and since $\Bbb{Z}$ is the cyclic group this will be the identity map.
Didn't we just discuss that what a map which is an homomorphism from $\Bbb{Z}\to G$ will look like for an abelian G... So shouldn't I be finding a map from $G\to \Bbb{Z}$ which cancels out with the map we had discussed
@Remember I'd recommend you write down the statement of the problem, what you want to do, and all we discussed 'till now. You've started messing up everything.
You have an exact sequence $N \stackrel{f}{\to} G \stackrel{g}{\to} \Bbb Z$. You are already given the map $g : G \to \Bbb Z$, you don't need to construct it anymore!
You have to construct a map $s : \Bbb Z \to G$ such that the composition $\Bbb Z \stackrel{s}{\to} G \stackrel{g}{\to} \Bbb Z$ is the identity map $\Bbb Z \to \Bbb Z$.
Say you have a function f: R -> R. Is there a standard way to denote the function R^2 -> R^2 defined by applying f to the first element of an ordered pair?
The second element in each ordered pair is left alone.
@Balarka Now as we had discussed that homomorphisms from $\Bbb{Z} \to G$ should look like $n \mapsto g+g+\cdots+g$. So when I think of this map $N \stackrel{f}{\to} G \stackrel{g}{\to} \Bbb Z$ since the last map is surjective for every $g+g+\cdots g \in G$ there should be some value for it in $\Bbb{Z}$ which has to be n(because if it is some other value the map cannot be surjective). Therefore the composition will $g \circ f$ will become the identity map
But you haven't quite thrashed out the detail yet. Why don't you tell me what your section $s : \Bbb Z \to G$ does to the element $1 \in \Bbb Z$? Where does it send it?
It can't send it to an arbitrary $g$. It has to send it to a $g$ such that after composing with the map $G \to \Bbb Z$ it's sent back to $1 \in \Bbb Z$.
@Rememberme Not really. What's the identity element of $\Bbb Z^n$?
@Rememberme Yes. See, you defined the map $s : \Bbb Z \to G$ by tracing out the diagram : you're sending $1$ to some element of $G$ which is in turn sent to $1$ by $q$.
You can trace out the arrows on the short exact sequence by your fingers to "see" where the map came from.
@Tobias Sanity-check : $V, V'$ be affine varieties. A $k$-algebra isomorphism $k[V] \to k[V']$ induces a Zariski isomorphism $V' \to V$, right? i.e., the correspondence between affine $k$-algebras and affine varities is functorial, correct?
I guess I should be specific and say I want an analogue of the fact that (for $1 \leq i, j \leq n$) $\sum_{i < j} x_ix_j = \dfrac{\left(\sum_{\ell = 1}^n x_i \right)^2 - \sum_{\ell = 1}^n x_i^2}{2}$
@columbus By $G$ acting on $k[x_1, \cdots, x_n]$, I mean a pairing $G \times k[x_i] \to k[x_i]$ $(g, f) \mapsto f'$, where $f, f'$ are polynomials in $k[x_i]$
Well, a pairing which satisfies a few desirable things
@TobiasKildetoft $g \in GL_m(k)$ acts on the vector space $V/k$ generated by $x_i$ by linear transformations $V \to V$ matrices of whose entries are rational functions of the entries of $g$.
yeah, yeah, I see what you mean. You defined a representation for which the invariant algebra is trivial.
But Eisenbud considers arbitrary rational representations on the space.
OK, that clears a lot up. I'll try to give that section another try today.
@TobiasKildetoft Another interesting thing Eisenbud talks about is geometric invariant theory.
If you have a variety $X$ such that a group $G$ acts on $X$, then you have a natural morphisms (of sets) $X \to X/G$. However, in general, $X/G$ cannot be made into a variety so that the morphism is a regular map.
For example, consider $k^{\times}$ acting on $\Bbb A_k^1$ by multiplication. The quotient is a two-point set (corresponding to the class $[0]$ and $[1]$ of all nonzero elements). However, if you try to give it the Zariski topology compatible with the morphism, then $[0]$ has to be contained in the closure of $[1]$. Impossible.
ok. now if you restrict attention to the open set of $\Bbb A^1_k$ consisting of the nonzero elements, then the restriction of the quotient morphism to this set is a perfectly good regular map between varieties (quotient is a singleton - an ok variety)
This is in general what you try to do in geometric invariant theory : approximate the $X/G$ by algebraic varieties. to be rigorous, if $X/G$ can be given a structure of a variety such that $X \to X/G$ is a regular map, then the coordinate ring of $X/G$ should be the invariant affine $k$-algebra $k[X]^G$.
But in general, as this doesn't happen, you look for subvarieties of $X$ for which the restriction of the quotient map is a regular map between varieties with the base variety having coordinate ring $k[X]^G$
I find this theory beautiful. Eisenbud says this was developed by Mumford and Fogarty in order to define moduli spaces.
@BalarkaSen I have only heard very little about it. But would it not make more sense to start with schemes than varieties for this? You would get a lot more quotients "for free" to start with, plus you have some idea of where you can take the quotients even when they are not schemes (being faiscaeu instead, or however that is spelled)
I am not sure if I understand the reason you want to do this with schemes, mostly because of my ignorance. Can you elaborate? (and the word's faisceaux, I think)
@BalarkaSen Right, that spelling looks more correct (for the plural). I don't actually understand this nearly as well as I would like. There is a chapter on it in Jantzen's book, but that is somewhat brief
@TobiasKildetoft We can just describe the adjoint functor of $V \to k[V]$ from the category of affine varieties to affine $k$-algebras to be the functor $A \to \text{mSpec} \,A$, the maximal spectrum of the $k$-algebra, can't we?
@BalarkaSen Well, just like an inverse function. But I suppose in our case we only get something isomorphic to what we started with, not necessarily the same thing
Week 5 just ended and I get things done just in time, spending all of my time working on uni without pretty much any breaks and getting 6 hours sleep a night :P. Starting to think I am just bad at math
user147690
Does everyone feel like they are bad at math sometimes?
hey, I have a probability question from "a first course in probability" by Sheldon Ross. He provides a solution but I have certain concerns about the solution. Shall I post it here?
Suppose V is C-vector space of finite dimension with inner product say $f$ and say $f^*$ its conjugate operator and $λ$ eigenvalue of $f^*f$ and $v$ non zero eigenvector of $λ$
a) express $||f||^2$ subject to $||v||$ and $λ$ and prove that $λ$ is a real non negative number and that is true for ...
Suppose V is C-vector space of finite dimension with inner product say $f$ and say $f^*$ its conjugate operator and $λ$ eigenvalue of $f^*f$ and $v$ non zero eigenvector of $λ$
a) express $||f||^2$ subject to $||v||$ and $λ$ and prove that $λ$ is a real non negative number and that is true for ...
