@TedShifrin a proposition is a sentence with a truth value. $p\wedge q$ is a proposition since it can be true or false depending on the values of $p$ and $q$
@TedShifrin my advisor is the head of the board of examiners (not sure if they're responsible, but okay) and I know him quite well, so I'll try to talk to him on monday
I distributed and ended up with $[(\neg p\wedge\neg r)\vee\neg q]\wedge(q\vee\neg q)$, the last part is always true, and true with another thing is that thing, so we end up with $(\neg p\wedge\neg r)\vee\neg q$
Yeah, I guess you're right you can't get rid of $q$ entirely. If $q$ is false, the proposition holds no matter what. If $q$ is true, you need $\neg (p\vee r)$ to hold.
So I guess we could rewrite this as an implication, but $q$ is still there.
@TedShifrin ah! I never understood the meaning of "the proposition holds no matter what" and "$\neg (p\vee r)$ to hold". What do you mean by "hold"? Thanks!
@TedShifrin I'm helping a friend, and he told me that in the textbook, the $q$ disappeared magically. That is why I ask you, although it is very possible that the book is wrong
It is not a famous textbook, it's a local one, don't worry $\ddot\smile$
@BalarkaSen I have a proof that if $P$ is a closed oriented 3-manifold which fibers over an oriented surface, and $P$ also fibers over a circle with fiber a closed oriented surface of genus $g>1$, then $P$ is isomorphic a product $S^1 \times \Sigma_g$ and the bundles are isomorphic to the projection maps. This is also true for $g=0$ and I suspect for $g=1$ but the proofs don't work.
@TedShifrin if we can count them, then your sentence "It can only happen, pretty much, formally if you have $∨(q∨¬q)$ or $∧(q∨¬q)$... or something similar" is true, but if we cannot count them, I think that your sentence is not true at all. In Logic, we don't like "not true at all" (?)
It used to annoy me over and over that colloquium talks would start with five/ten minutes reviewing what projective space is. WTF. Can't we assume second-year grad students and faculty know this?
@Ryan: Yes, of course. At MIT and Princeton that doesn't happen. But it really shouldn't have at UGA, either. Is every number theory talk going to start by reviewing modular arithmetic? :D
@TedShifrin @RyanUnger any good quick summary/reference that just does all the important definitions and theorems from multivariable/vector calculus, but on general (abstract) manifolds?
Eisenstein's criterion seems relevant to things Scholze might talk about, since if a polynomial is $p$-Eisenstein, it is irreducible over $\Bbb Q_p$ and defines a totally ramified extension and conversely, totally ramified extensions always have a primitive element that is $p$-Eisenstein
maybe some hyperbolic stuff, @Mathein. One grad student overruled his NT adviser and stayed in my grad diff geo course for the year because he thought it might be useful.
can someone explain to me the Laplace-Beltrami operator and why its spectrum is that important? Apparantly the Selberg trace formula is important for NT, but I never worked with the spectrum of a differential operator before
ok you probably know functional analysis so if we want to solve some abstract problem like $x'(t)=Ax(t)$ where $x(t)$ is a curve in some Banach space and $A$ is a linear operator then we can write $x(t)=e^{At}x_0$, where $x_0$ is the "initial condition"
i was going to write an integral kernel in a second
now $e^{t\Delta}:L^2\to L^2$ is actually a trace-class operator for $t>0$
and we have the nice formula $$\mathrm{tr}\,e^{t\Delta}=\sum_{j=0}^\infty e^{-t\lambda_j}$$
now the problem is that actually computing the RHS is kind of impossible from what I just said, so we have to consider another representation of the heat kernel
To fix notation let $\{\phi_j,\lambda_j\}_{j=0}^\infty$ be the spectral resolution for the Laplacian (on a closed Riemannian manifold)
this means that $\Delta \phi_j+\lambda_j\phi_j=0$, $\phi_j\in C^\infty$, $\{\phi_j\}$ is an orthonormal basis for $L^2$, and $0=\lambda_0<\lambda_1\le \lambda_2\le\cdots\to \infty$
They asked to what extent does $H(x,y,t)$ look like $e(x,y,t)$ as $x\to y$ and $t\searrow 0$
Recall that on a Riemannian manifold we have a distance function $d$. So we consider the "naive heat kernel" $$\mathcal E(x,y,t)=(4\pi t)^{-n/2}e^{-d(x,y)^2/4t}$$
their idea is that we should be able to add correction terms to this and obtain the real heat kernel
since this thing obviously is like $e$ for short distances and times, and if the correction terms are small, then $H$ looks like $e$ too
it turns out the right thing to look at is $$H_k(x,y,t)=\mathcal E(x,y,t)\sum_{j=0}^kt^ju_j(x,y)$$
and we want to compute $u_j$ so that they satisfy $$(\partial_t-\Delta_x)H_k=-\mathcal Et^k\Delta_x u_k$$
I don't have a great reason for why this equation, but it works out in the end
one can show that for $k$ sufficiently large, $H_k$ approximates $H$ very well and we have $$\sum e^{-\lambda_jt}=(4\pi t)^{-n/2}\left(\sum_{j=0}^k t^j\int u_j(x,x)\,dx\right)+O(t^\alpha),$$
where $\alpha$ is some power
I actually wanted $u_0(x,x)=1$ haha
so the first term gives the volume, always
and when $n=2$ you get the Euler characteristic in the second term
@BalarkaSen Let $P$ be a closed oriented 3-manifold, and suppose $p_1: P \to \Sigma$ is an $S^1$ bundle, where $\Sigma$ is an oriented surface, and furthermore $p_2: P \to S^1$ is a fiber bundle with fiber $\Sigma_g$, where $g > 1$.
I claim that the map $p_1 i_2: \Sigma_g \to \Sigma$ has nonzero degree; in fact, it is homotopic to a covering map. For consider the induced map $gf:\pi_1 \Sigma_g \to \pi_1 P \to \pi_1 \Sigma$; I claim this composite is injective and its image is finite index. First observe that $f: \pi_1 \Sigma_g \to \pi_1 P$ is injective, because it's the inclusion map of a fibration whose base is aspherical.
Now consider the short exact sequence $\Bbb Z \to \pi_1 P \to \pi_1 \Sigma$; if $gf$ is not injective, then $\text{ker}(gf)$ is the preimage of $\Bbb Z \subset \pi_1 P$, and in particular is either zero or a normal subgroup isomorphic to $\Bbb Z$.
In the latter case, we have the action $\pi_1 \Sigma_g \looparrowright \Bbb Z$, which comes from a homomorphism $\pi_1 \Sigma_g \to \Bbb Z/2$. Clearly this homomorphism has nonzero kernel; an element of this kernel commutes with $\Bbb Z$, and so we have constructed an embedding $\Bbb Z^2 \to \pi_1 \Sigma_g$. This contradicts the assumption $g>1$ by a hyperbolic geometry argument. So $\text{ker}(gf) = 0$ and $p_1 i_2$ is homotopic to a covering map.
Every noncompact surface has free fundamental group, so it's a finite index covering.
Now $P$ has two special cohomology classes, $S$ and $C$, the pullbacks of the fundamental classes of the Circle and the Surface. Then $S \smile C = \text{deg}(gf)$, by dualizing to intersection products of fibers. But if the circle bundle over $\Sigma$ has Euler class $n$, then $nS = 0$ by the Gysin sequence. Thus the Euler class is zero and the circle bundle is trivial, and hence $P \cong S^1 \times \Sigma$.
I do not know if we have $\Sigma = \Sigma_g$. I guess no. I picture you could cover $\Sigma$ non-trivially as the "$n$th root bundle" of the $U(1)$-bundle.
@user76284 they are simple poles (order 1) by combining formula (10) on this page mathworld.wolfram.com/PrimeZetaFunction.html with the fact that the zeta function pole at $s=1$ is also simple (of order 1).
@pre-kidney But PrimeZetaP[s] + 1/(s - 1/2) doesn't converge at $s=1/2$? I can cancel the logarithmic singularity at $s=1$ by using PrimeZetaP[s] + Log[s - 1].
What do you call this process of drawing things by circles? editor.p5js.org/natureofcode/full/bxog_Xt6O I know that this is related to fourier but I wanted to know the name of this process so I can search and read about it.
Maybe you can also try to check pictures with similar Fourier series animations to see whether somebody uses specific terminology for this "cumulative" diagram.
Google search also suggest the term harmonic circles. Although I haven't heard that term before.
Anyone know how Mathematica may have gotten this Borel sum for the harmonic series? mathoverflow.net/a/335061/74578 I keep getting something that diverges.
