@JackDon It's an inside joke understood only by fans of Brian Lumley and the Necroscope himself. But I would say Harry used his higher maths to trap the math students in the void.
@JackDon Ignore that I couldn't resist. I'm just here cause I am terrible at math and I seek a numbers wizard.
@JackDon Harry is the protagonist in a series of books about a guy that can speak to the dead, hence learning math from past geniuses (Mobius especially) to travel through space and time to hunt vampires.
Anyway, I was looking for help to figure out the progression of degrees to turn and how often based on speed to cover a grid of squares of a set size. I realize I could go in a linear fashion, turn 90 degrees for one grid and then 90 again. . im looking to spiral outward instead
Suppose $A_{i}$, $B_{i}$, $C_{i}$ are chain complexes such that the sequence $0 \rightarrow A_i \rightarrow^{f} B_i \rightarrow^{g} C_i \rightarrow 0$ is exact for all i. Does this imply that $f,g$ are chain maps?
I would expect it to be false in general, but because of my lack of examples of chain complexes I couldnt find one
That property has nothing to do with the differential, so certainly not
As a quick example, take $B_0 = B_1 = \Bbb Z$, with differential $d(b_1) = b_0$, with $A_1 = \Bbb Z$ and $C_0 = \Bbb Z$. Let $f$ be the obvious inclusion and $g$ be the obvious projection
@Mike This is actually what I wanted to ask. Actually I am proving that if such a short exact sequence exists then it induces a long exact sequence of the homology groups. But you have the map $H_p(A_i) \rightarrow^{f*} H_p(B_i)$ in the long exact sequence which can only happen if $f$ is a chain map, but my notes don't mention f being a chain map.
Another thing, in the long exact sequence $\cdots \rightarrow H_p(A) \rightarrow^{f}* H_p(B) \rightarrow^{g}* H_p(C) \rightarrow{\partial}* H_{p-1}(A) \rightarrow^{f}* \cdots$ is the map $\partial *$ unique?
In the sense that (atleast from what I have been given) you define $\partial*[c] = [f^{-1} \circ \partial \circ g^{-1}(c)]$, but looking at the commutative diagram there are many other ways you can define this map
@Sayan the snake lemma produces long exact sequence of homology groups out of a short exact sequence of chain complexes in a natural way. if you have a map of exact sequence of chain complexes (0->A->B->C->0) -> (0->A'->B'->C'->0) then it gives a map of long exact sequences (H(A)->H(B)->H(C)) -> (H(A')->H(B')->H(C'))
if you have two recipes of producing a connecting homomorphism such that naturality is satisfied, you can use it on the identity map (0->A->B->C->0)->(0->A->B->C->0) to get that they are the same
@TedShifrin Suppose I have a Riemannian surface $(M, g)$ with a triangulation of $M$. Can you give me a homotopy $g_t$, $t \in [0, 1)$ of the metric such that the curvature function $K_t$ has a well defined limit $K_1 = \lim_{t \to 1} K_t$ to a function which vanishes everywhere on $M$ except on the vertices of the triangulation $M$, on which it agrees with the angle defect?
That is, are simplicial surfaces just limits of Riemannian manifolds?
@BS If we think of this as embedded, you can deform a polyhedron to the manifold. But you have singularities on edges of faces, not just at vertices, no?
In probability and statistics, Student's t-distribution (or simply the t-distribution) is any member of a family of continuous probability distributions that arises when estimating the mean of a normally distributed population in situations where the sample size is small and the population standard deviation is unknown. It was developed by William Sealy Gosset under the pseudonym Student.
The t-distribution plays a role in a number of widely used statistical analyses, including Student's t-test for assessing the statistical significance of the difference between two sample means, the construction...
@Mathphile according to wiki, it has one parameter, namely $\nu \in (0,\infty)$
@LeakyNun I really don't think about the $t$-distribution with non-integral degrees of freedom. To me it's just $t_{n-1}$ where $n$ is a natural number.
@BalarkaSen Let $C \subset M$ be a Morse-Bott critical locus, i.e. $\nabla f = 0$ on $C$ whereas $\text{Hess}^N f$, the normal Hessian, is everywhere nondegenerate on $C$. Is there anything special about the map $\nabla f: M \to TM$ along $C$ that we can see just at the level of manifolds, not using special structure of $M$ or $TM$?
The map $\text{Hess} f: TM/TC \to TM$ is recoverable just using manifold-level constructions (if $M = X$ and $TM = Y$ with the $0$-section being $Z \subset Y$, this is the composite $TM \to TY \to TY/TZ$, factored as $TM/TC \to TY/TZ$) but the normal Hessian requires that you kill off $TC \subset TM$, and knowing that subspace of $TM$ requires that you know the codomain is a tangent bundle (so we can take $TC$ for submanifolds $C$ of the domain)
It's not totally obvious to me whether that $TC$ is very important. Here, let's try this. Maybe the abstract notion we're getting at (Morse : transversality :: Morse-Bott : ?) is a map $F: X \to Y$ and a submanifold $S \subset Y$ so that a) $f^{-1}(S)$ is a submanifold $C$, and b) $DF\big|_C: TX/TC \to TY/TS$ is injective.
i dont want to use the normal hessian since that requires special structure
i do not think there is a good set of assumptions that gets around assumption (a) --- I think you always need to assume the inverse image is a submanifold and then state the transversality property
@TedShifrin I'm trying to take the following situation: $f$ is Morse-Bott, and $g$ is Morse along the critical submanifolds of $f$, and I consider $F_t: M \to TM$ given by $\nabla(f+tg)$. I would like to phrase this setup abstractly, not using that the codomain is the tangent bundle of the domain. That's where I run into trouble --- I already don't know quite how to encode "$f$ is Morse-Bott".
@TedShifrin I'm gonna do like another week or so of logic/proof practice and then move on. I feel like I have a better understanding of the techniques and stuff now.
Write $\psi = DF|_C: TX/TC \to TY/TS$. Then the condition I decided was the right encoding of "$f$ is Morse-Bott" above was just that $\psi$ is fiberwise injective. But $g$ itself is not apparent in this abstract setting, since we can't add elements of $Y$.
Only $F'_0$ is, and that's a map $X \to TY$, not a map $X \to Y$. It's tough.
I'm probably making my life harder than it has to be
@BalarkaSen OK, "a pseudotransverse map" is all the data of: $(X,C)$, $(Y,S)$, a map $f: X \to Y$ so that $F^{-1}(S) = C$, and such that the map $\psi = DF|_C: TX/TC \to TY/TS$ is injective.
I think there's a way to write it but I'm not going to bother
I've forgotten how to prove the Morse lemma
Oh is this like a Moser trick
If $Q(v)$ is your function with $Q(0) = 0$ and $D_0 Q = 0$ but $H(v) = \langle (D^2 Q) v, v\rangle$ nondegenerate, then set $Q_t(v) = Q(v) + t(Q(v)-H(v))$; then $Q_t$ has its first two derivatives constant in time
The Hessian is nondegenerate, so find coordinates where it diagonalizes. You immediately have the desired expression of $f$ in those coordinates, given by writing $f$ in Taylor expansion w/ reminder, no?
Here's the key lemma. Take a smooth function $f : \Bbb R^n \to \Bbb R$, $f(0) = 0$. I claim you can find functions $g_i : \Bbb R^n \to \Bbb R$ such that $g_i(0) = \partial f(0)/\partial x_i$ such that $f(x) = \sum x_i g_i(x)$
The proof is to integrate: $f(x) = \int_0^1 \nabla f(tx) \cdot x dt$ like you were writing earlier, I think.
Last expression is coming from $f(x) = \int_0^1 \partial_t f(tx) dt$
I wrote the Taylor's theorem with remainder term wrong, it is $f(x) = f(a) + Df(a) (x - a) + 1/2 (x - a)^T Hf(\xi) (x - a)$. $\xi$ is a function of $x - a$ which vanishes at $0$.
I forget how to diagoalize it as a function of $x - a$. I can diagonalize it as a function of $\xi$, but that's not what we want.
How would you describe the geometry of $\Bbb H^n$ at infinity? Would you say it has a "conformal structure", in the sense that angles still makes sense at infinity.
I suppose that's why if you have a finite group $G$ acting by isometries on $\Bbb H^n$ then the quotient, if noncompact, can always be "cusp compactified"; the only way it can be noncompact is if the fundamental hyperbolic polyhedron has vertices at infinity, in which case you can just put that point in
I suppose those manifest as orbifold points downstairs because of this "conformal structure at infinity"
Let's do $n = 2$ because higher dimension is too hard for me
Yeah it's basically the reference point to study negatively curved spaces. Alexandrov curvature is defined by comparing a space with $\Bbb H^n$ and see if triangles in the former are more deflated than triangles in the latter
There's people who do ggt that are pretty much topologists, but some are pretty much group theorists, some are actually analysts in disguise, and there's even some that are model theorists
@JackOhara you have to use something, i mean. you can "prove" triangle inequality by knowing triangles in R^n lie on a 2d subspace, and on 2d you "know" it by euclidean geometry
A language question: If someone says an element is 4-torsion, do they specifically mean that the element has order $4$, or just that the order divides $4$?
Someone has asked similiar question before, but the answer to that question omits the proof.
And I couldn't find the proof on Rudin's Real and Complex Analysis or Stein's Real Analysis.
Prove that :$
\int_a^b |f'(x)|\,dx \leq V(f,[a,b])$, where f is a function of bounded variation.
I tried to ...