"If I were to be stranded on an island, wanted to understand the beauty and intricacies of the modern study of smooth 4-manifolds, and were allowed to take only one reference, the book under review would most certainly be my choice."
Let $x>1$ be a real variable.
$$f(x) = \lim_{n = \infty} \sum_1^n \ln(n)^2 + n \ln(x)^2 - \sum_1^n \ln(x+n)^2$$
Is there an integral representation for $f(x)$ that uses no $\sum$ , named polynomials nor non-analytic functions ?
( so no " cheap" Floor function or Sum in the integrand )
Notice t...
Assume we have a displacement field $u=u(x,z)=(u_x(x,z), u_z(x,z)))$ which takes the point $R=(x,z)$ to $r=R+u=(x+u_x(x,z), z+u_z(x,z))$. We want to find the tangent and normal to the curve which was $z=0$ before applying the displacement. This new curve is given by: $$r=(x+u_x(x,0),0+u_z(x,0))$$...
Well by definition a linear map $f : \Delta^n \to Y$ satisfies $f(a + b) = f(a) + f(b)$ and $f(ca) = cf(a)$. Now write down any $v \in \Delta^n$ as linear combination of the vertices ('cause $\Delta^n$ is just the convex hull on the vertices) and work it out.
This is algebra, should be up on your line of thought. Not sure what's unclear about it.
Well, as Hatcher explains, $D_{m(\sigma)}(\partial \sigma) - D(\partial \sigma)$ is a linear combination of $D_{m(\sigma)}(\sigma_i) - D_{m(\sigma_i)}(\sigma_i)$ where $\sigma_i$ are restrictions of $\sigma$ to the faces of $\Delta^n$. Now if $\sigma$ fits into $U$ after $m(\sigma)$ barycentric subdivisions, so will $\sigma_i$. Thus $m(\sigma_i) \leq m(\sigma)$.
So if you write out $D_{m(\sigma)}(\sigma_i) - D_{m(\sigma_i)}(\sigma_i)$ explicitly (use the definition), you'll get terms of the form $TS^k(\sigma_i)$ with $k \geq m(\sigma_j)$, because things below that will cancel out. These fit in $U$.
I wrote a problem, it asks for something with $n=2016$. Unfortunately a lot of calculations are needed, how can we reduce the calculations without losing style?
I thought of changing 2016 for 216 but that looks desperate
Let $V$ be a finite dimensional vector space, and $H$ a subspace of $V$. How do I show that $H$ has a basis? This seems obvious but I am having trouble proving it.
@AkivaWeinberger You aren't halfway through chapter 2 of Hatcher yet, I don't think. To be honest I felt it was not natural to me as fundamental groups and covering spaces even when I finished learning the theory in chapter 2.
@rorty: Why are you interested in subspaces specifically? It's true that every vector space has a basis (with some Choice required for infinite dimensions)
At the moment I'm actually skimming the rest of Chapter 2, with the intent of going back in more detail, mostly because of that frustration I was talking about
Let $x>1$ be a real variable.
$$f(x) = \lim_{n = \infty} \sum_1^n \ln(n)^2 + n \ln(x)^2 - \sum_1^n \ln(x+n)^2$$
Is there an integral representation for $f(x)$ that uses no $\sum$ , named polynomials nor non-analytic functions ?
( so no " cheap" Floor function or Sum in the integrand )
Notice t...
("What familiar space is the quotient ∆ complex of a 2 simplex [v0, v1, v2] obtained by identifying the edges [v0, v1] and [v1, v2], preserving the ordering of vertices?")
Not sure what you mean. You can see this by taking a rectangle, gluing opposite sides (which gives the band) and then let length of one side tend to 0. This pinches of a bit of the boundary (which is a circle) of the band, so no harm is done.
Anyway, it's good fact to keep in one's head that R-orientability means top homology with R-coeffs is R, but remember that this is nontrivial and won't be proved until chapter 3.3.
Sure. All of this will be done in 3.3. Anyway, did you get why I was being pedantic? Having the right answer is not the point: what's the proof that H_2(nonorientable surface) = 0?
I feel like I could do it if I really wanted to — I suppose I know what $C_2$ and $C_1$ are — but it'll just turn into a big jumble of mindless algebra, I think.
(Interestingly, that other book I have calls them $S_n$ instead of $C_n$.)
True, but not quite the reason. There's an interesting concept one stumbles upon while solving this which is actually a general phenomenon for 3-manifolds. But nevermind that.
So, my guess is that I identify $A$ with $-B$, and $C$ with $-D$, where those are the faces, but I can't visualize that so it might not be $S^3$ at all
Hmm akiva, I think I can actually make a plausible construction of the factorial for simplicial complexes; although it seems kind of useless and I don't have any intuition for it.
@CarryonSmiling We were talking about Hatcher's Algebraic Topology, but we got sidetracked by my believe that we should be able to take the factorial of anything.
Here's the deal: if you can arrange $H_n(X^n, X^{n-1})$ in a chain complex and compute it's homology, it'd be computable entirely from the cell structure of $X$.
Let $x>1$ be a real variable.
$$f(x) = \lim_{n = \infty} \sum_1^n \ln(n)^2 + n \ln(x)^2 - \sum_1^n \ln(x+n)^2$$
Is there an integral representation for $f(x)$ that uses no $\sum$ , named polynomials nor non-analytic functions ?
( so no " cheap" Floor function or Sum in the integrand )
Notice t...
