OK, @RandoHinn: So you're looking at the region above the $x$-axis and below that crazy $y^2=6(1-x^3)e^{7x}$ and spinning it about the $x$-axis, so, yes, it's what you said if you fill in the stuff.
@RandoHinn: What are the $x$ limits and what is $f$?
I know the fundamental group, don't know how to apply vector calculus here, but have been introduced to it, and I'm trying to use an equivalent trait to simple connectedness (specifically the fact that, in a simply connected space, a loop can be contracted to a point) and general topological traits to prove it.
(I suppose I might be being a little generous when I say I know the fundamental group. We just finished our introduction to the topic during class today. About 6 hours ago.)
@TedShifrin Let me recall your proof: Give $M$ a Riemannian metric, then look at $\text{grad}\, f : M \to TM$. Make this transverse to the zero section to get $X : M \to TM$ such that $X \pitchfork 0$. Being conservative is a closed condition, so $X = \text{grad} g$ for some $g : M \to \Bbb R$. Since $X$ is transverse to the zero section at the critical points of $g$/zeros of $X$, you argue that $g$ has nondegenerate Hessian there.
@TedShifrin Oh, so you just work with $df : M \to T^*M$, and do the same argument I said. That's just dual to what I said though, so it's not very unreasonable. :)
Well, the homework I gave years ago was for vector fields, because we didn't have the cotangent bundle early on. But it's just the transversality theorem to say most perturbations will give you transversality to the zero section (and yes, you have to check that means Morse).
@TedShifrin i know it's very stupid, but for example in the caso $n=2$, $a_2=1$ we have then that $(y^2-x^2)=(y-x)\cap(y+x)$, and the variety associated is a pair of lines in $k^2$, which becomes two in points in the projective space
Note that you are using more than simple transversality. Saying that transverse-to-zero sections are closed in the space of sections of a vector bundle is a baby Thom transversality result
oh and look, a question specifically on this subject that has already been answered, incorrectly, accepted and voted up multiple times math.stackexchange.com/questions/1073541/…
hey I know lets t bag dr suess, great idea, ill hold him down
Model theory studies models of first order theories abstractly (I know this is not exactly informative...), while set theory studies some specific first order theories (ZF and extensions usually) and models of those
@MikeMiller I guess my main question doesn't really concern bounded cohomology (which I don't know about either). I'm curious about representations of the group of $Homeo^{+}(S^1)$ of orientation preserving homeomorphisms of $S^1$. There some connections between some matrices acting on some elements of that group and their rotation number. However, I only really know about the analytic aspects of circle maps, and this connection seems to be algebraic.
Of course, this is very vague right now. Give me a second to provide some more details.
Oh and the reason I mentioned bounded cohomology is because it provides a generalization of the rotation number in a more algebraic setting.
Anyway so here are some more details, I'll keep it brief and if you want to know more let me know!
The maps in $Homeo^{+}(S^1)$ I'm looking at are piecewise affine maps with 2 affine pieces (let me know if this makes sense, I'm not sure if this terminology is standard, I'm coming from a dynamics perspective). Then renormalizations of these maps correspond to actions by elements of $GL_{2}(\mathbb{Z})$. Unsurprisingly, the determinants/eigenvectors of these matrices give information about the rotation number.
This is why I want to know about representations of $Homeo^{+}(S^1)$, and I can't seem to find resources about that online.
@MikeMiller I can define the the renormalization if you are curious about it, it would be much much easier if I could draw some pictures though.
Regarding a paper by Ghys, I'm not sure. However the stuff I skimmed about bounded cohomology and circle maps contains a lot of stuff attributed to him (the document is by Frigieri, I believe).
And regarding your last message, I have no idea. My representation theory is non-existent really, so it's very possible I'm saying non-sense/proposing non-sense. My apologies if that's the case. I'm not getting at anything particularly specific, I just wan…
I just figured having representation of this group would be a natural way of doing this, but since I don't know about rep theory beyond basic definitions, I wanted to ask if this seemed reasonable before I spent significant time learning parts of it.
Thank you for your insight! I'm curious about the idea behind not being able to make this association. I'll check Ghys's paper as well.
And yeah these maps are especially simple, but unfortunately it seems that this class of maps doesn't form a subgroup, which is why I was considering $Homeo^{+}(S^1)$ as a whole.
Thanks! If you get any ideas/are curious about the details of the problem let me know! I'll ask my professor about this when I seem him tomorrow, he will probably have some ideas about this stuff.
The zeroes of $f$ in $\mathbf{A}^{n+1}$ form a cone: scalar multiplication preserves zeroes. (This is immediate from homogeneity.) The zero set in $\mathbf{P}^n$ is the projectivization of this set.
The complementof the zeroes form an open cone: scalar multiplication by non-zero constants preserves the property that $f(x) \neq 0$. We can again projectivize this, and the result is your $D_{\mathbf{P}^n}$.
Put another way: $Z_{\mathbf{A}}$ and $D_{\mathbf{A}}$ are complements of one another, and if you ignore $0$, $$Z_{\mathbf{P}} = Z_{\mathbf{A}} / k^\times$$ and therefore $$D_{\mathbf{P}} = D_{\mathbf{A}}/k^\times.$$
given that you can write any homogeneous element as $\sum_{i=0}^k g_i \cdot y^i$, where $g_i$ is some function of the first $n$ variables with degree $-i\deg f$, you can write this element as $$\left(\sum_{i=0}^k g_i \cdot f^{k-i}\right) \cdot y^k,$$ where the sum is therefore a homogeneous element of degree $k\deg f = -|y^k|$
Problem: Let $T \in \mathcal{L}(V)$ and $p(x)$ and $q(x)$ relatively prime polynomials such that $p(T)q(T) = 0$. Prove that $null(p(T)) \cap null(q(T)) = \{0\}$. Proof: Since $p$ and $q$ are relatively prime, there are polynomials $a$ and $b$ such that $1= a(x)p(x) + b(x)q(x)$ which implies $I = a(T)p(T) + b(T)q(T)$. If $v \in null(p(T)) \cap null(q(T))$, then $Iv = a(T)p(T)v + b(T)q(T)v = 0$ implies $v=0$.
Does this seem correct? I didn't use the fact that $p(T) q(T)=0$, which worries me.
