Hey @Mike, if I have a space $X$ and subspaces $A,B\subset X$, is there a LES $...\rightarrow\overline{H}^k(X,A)\rightarrow\overline{H}^k(X,B)\rightarrow \overline{H}^k(A,A\cap B)\rightarrow...$? I think I have such a thing, but it looks so bizarre that I'm doubting.
The thing to do is try to find a lower bound on |T|_{p -> q} |T^{-1}|_{q -> p}. The infimum of this over all linear isos gives a metric on the space of normed spaces mod isometry.
Well, log of that. That's 1 iff the normed spaces are isometric.
Then find d(R^n_2, R^n_p) = log n |1/p - 1/2|. The identity map gives the best possible, and then you find a lower bound by playing with the parallelogram law and watching what happens to T(sum pm e_i).
From there, for p,q both on the same side of 2, you get d(R^n_p, R^n_q) = log n |1/p-1/q|.
You get a lower bound on distance for non conjugate exponents but not an equality when they're on opposite sides of 2. Now the identity map is no longer the best possible T.
Doing it for conjugate exponents other than 1, infty is hard. I don't know how to do it.
What are situations in which one uses naturality of the isomorphism $\tilde{H}_k(X) \rightarrow \tilde{H}_{k+1}(\Sigma X)$? It is useful in one of our exercises, finding a degree $k$ map for for every $n$-sphere, where this quickly yields the induction step
It all feels too much like a bag of tricks to me, so I'm starting to look for heuristics
R^n with the p-norm. Geometry of finite dimensional.normed spaces.
@user2103480 Uh I can probably come up with examples but that's probably not necessary. The point is that it's a construction on spaces. Any decent construction on spaces is functorial, so you get naturality for free.
@Thorgott Yes, this is a good definition of orientability.
w_1(E) can be constructed as the obstruction to trivializing a bundle over the 1-skeleton
I have an inductive argument for finite 1-dim CW complexes. The infinite case isn't as clear to me. Is it some limiting argument (the complex is the colimit of its finite subcomplexes, so the bundle should be the colimits of its restrictions to the finite subcomplexes and then a colimit of trivial bundles will be trivial)?
also, algebraic construction and not a topological one, but the snake lemma construction turning an SES of chain complexes into LES of homology groups is functorial
if you ever write down a diagram and wanna argue it commutes, this fact takes care of most of those cases
Okay nice these I could've thought of as well. Also, more in the algebraic spirit of what thor mentioned, the SES from the universal coefficient theorem
I am having problems with the solving this integration.May be I am making some mistake in a certain step.The r and W values are constants.The answer reads 16.48
Limits are given as required.There is no unit balancing that needs to be done in y understainding
@user586228 I recognise some of the marks on the paper but there is little by way of coherency or context. It is not surprising that you are having issues.
Is the improper integral $$\int_{-1}^{2} \frac{\mathrm d x}{x}$$ divergent? I don't think so. We can break the given integral in the three following integrals: $$\underbrace{\int_{-1}^{2} \frac{\mathrm d x}{x}}_{I} = \lim_{h\to 0^{+}} \left(\underbrace{\int_{-1}^{-h} \frac{\mathrm d x}{x}}_{I_1} + \underbrace{\int_{h}^{1} \frac{\mathrm d x}{x}}_{I_2} + \underbrace{\int_{1}^{2} \frac{\mathrm d x}{x}}_{I_3}\right)$$ In the above sum, $I_1 = -I_2$, thus we obtain...
Hi, I have a question. Does every(compact or non-compact) surface without boundary has the property: there is a Riemannian metric on the surface such that every non-trivial element of the fundamental group can be represented by a shortest loop.
For compact I know this.
But what about non-compact I don't know
shortest in the sense of least length in free homotopy class
Let me tell you what I know far more: every non-compact surface without boundary cover(continuous sense) closed surface of negative Euler characteristic.
I think you are talking about: think annulus as suspension type object on S^1 with vertex removed, and then circles in the there are infinitely many circles has a decreasing length converging to vertecies.
I'm certain that the calculus of variations proof will apply in the noncompact case given some basic assumptions (complete metric, maybe some lower bound on sectional curvature?)
What the natural assumptions are should become clear as one chases through a proof
We do not know because we do not know the proof. I am saying that if you carefully read a "calculus of variations'' style proof in the case of closed hyperbolic surfaces you will be able to identify exactly what assumptions are used and figure out the generality in which it is true.
There's no value in having a bunch of facts without understanding why they are true. That's why we do mathematics, to understand
Of course it is just my taste to do a calculus of variations style proof, it seems most natural to me personally. I don't know a reference. There are probably references that do this using hyperbolic geometry, but I do not know much about hyperbolic geometry.
Which should work for any surface without boundary by the uniformization theorem, which asserts that by (pointwise) scaling the metric you can make it hyperbolic.
Oh, my bad, of course this argument doesn't work. Take a punctured hyprebolic surface. Then the ends look like Gabriel's horn. So the length of the curve goes to 0 as you slide along the cusp.
Even constant curvature -1 isn't sufficient, seems like you need flatness or something on these kinds of ends. Seems like something you can do by hand. Whatever!
Indeed I never thought about isometries of R^n with different p norms, and although it seems very plausible that no such isometries exist, I also faceplanted trying to prove it @Mike
Makes sense. I wanted to take preimage of a point, but I don't have $T_1$, so I'll have to find another closed set instead
What does a vector bundle$\pi:E\to B$ such that no matter which finite open cover of $B$ I pick, $E$ is not locally trivial on a member of the cover even look like? I guess I could have something ugly over $(0,1)$ with more and more "twistiness" accumulating toward $0$ or something like that
Ok my question concerns with the threshold for different properties. As we increase the number of nodes, the probability of edge formation decreases. But for properties like connectedness, wouldn't the low edge probability act in a opposite direction?
@Alessandro you can always pick some finite open cover such that the bundle is trivial over at least some members of that open cover, simply by appending finitely many open sets on which it is trivial to the cover
yesterday and this morning were kind of busy- I am only confused with why you wrote $f = u/v h$ in the beginning, but assuming it's ok the rest makes sense.
It concerns with Erd\"os R\'enyi random graphs. For example, the threshold for property like "diameter two" to appear is $sqrt(2\frac{lnn}{n})$ where $n$ is the number of nodes. As we increase the number of nodes, the threshold gets pushed to left..that seems somewhat counterintuitive to me.
I understand the threshold is related to expected value. And if expected value goes with 0 for a certain probability then the probability of finding that property is zero. And at the threshold expected value of the property shift other way.. and hence the threshold.
@BigSocks So $f\in (h)$ implies that there is an element $a\in k(x)[y]$ such that $f = ah$. You can write $a = u/v$ where $u\in k[x,y]$ and $v\in k[x]$
My question was very poorly phrased, here is an understandable version. What is an example of a bundle $\pi\colon E\to B$ such that all covers $B_i$ of $B$ with $\pi^{-1}(B_i)\simeq F\times B_i$ for all $i$, are infinite?
h is a polynomial in $k[x,y]$ that generate (f) + (g) in $k(x)[y]$ (which is a PID)
the reason we can take a polynomial is that polynomials in k[x] are "constants" of k(x)[y], in the sense that you can invert them, so $(fu) = (f)$ for any $u\in k[x]$
