in fact, you can choose a metric on $M$, then $x\mapsto d(x,C)$ is a continuous function vanishing exactly on $C$ and a sufficiently close smooth approximation to this function will be a smooth function only vanishing at $C$
the latter point is technical, but still easier than Whitney extension
for a given continuous function $f\colon M\rightarrow\mathbb{R}$ and a given "threshold function" $\varepsilon\colon M\rightarrow(0,\infty)$, you can always find a smooth function $g\colon M\rightarrow\mathbb{R}$ s.t. $|g(p)-f(p)|<\varepsilon(p)$ for all $p\in M$
this is essentially because you can do it locally by constant functions and then glue together using a partition of unity. making this precise is somewhat technical, though
@TedShifrin I'm curious as to proving all these different what seems like variations of triangle inequality. Are these some of the bed-rock calculations that we will be coming across in Calculus, or is this a really good chance to become intimately familiar with manipulating inequalities? (I'm referring to problems related to problem 12 in Chapter one)
So far, I let $\alpha_i,\beta_i$'s be generators of $H^1(M_h)$ and $\alpha'_i,\beta'_i$'s for $H^1(M_g)$. If I let $f^*(\alpha_i)= \sum_{j=1}^g a_{ij}\alpha'_j+\sum_{j=1}^g b_{ij}\beta_j'$ and $f^*(\beta_i) =\sum_{j=1}^g a'_{ij}\alpha'_j+\sum_{j=1}^g b'_{ij}\beta_j'$ then $f^*(\alpha_i)\smile f^*(\beta_i) = \sum_{j=1}^g a_{ij}b'_{ij}-a'_{ij}b_{ij} = 1$. Is there something I can do more? @TedShifrin
Oh, I only saw the tagged chat. ok for $g = 1,h=2$ case, I get $f^*(\alpha_1)\smile f^*(\beta_1) = (a_{11}b'_{11}-a'_{11}b_{11})\gamma = \gamma$ and $f^*(\alpha_2)\smile f^*(\beta_2) = (a_{21}b'_{21}-a'_{21}b_{21})\gamma = \gamma$ @TedShifrin
just to confirm my observation: when the option is to Sort By Votes (not Active or Oldest), ceteris paribus (e.g., both answers having received exactly 0 upvotes and 0 downviotes), an Accepted answer is now sorted below other same-scored Answers, correct?
@TedShifrin differential bounded manifold M with boundary, submanifold N with codimension 1 contained in some connected component of the boundary, both oriented. N is said to be in-boundary if a positive normal points inwards (??? what does this mean); out-boundary analogous
Still not clear to me. So $N$ is a union of components of the boundary of $M$? I assume they are talking about whether the orientation agrees or disagrees with the orientation on the boundary (component by component).
The boundary of $M$ has a natural boundary orientation. They're taking a COMPONENT of the boundary with an orientation and asking whether its orientation agrees or disagrees with the boundary orientation on (all of) $\partial M$.
either way, i think i won't understand these today, unfortunately. i'm reading because i'm supposed to know them to construct the 2Cob category blah blah and i have a report due today
They're putting random orientations on the components of the boundary. The boundary has a natural orientation. They're saying "in" if the orientation disagrees with the boundary orientation, "out" if it agrees with the boundary orientation. If you take all the boundary components with their boundary orientation, all the components are "out."
yeah, i know. but sincerely i keep getting in this situation because i procrastinate lots of things. i'm pretty sure therapy will help with that... changing bad habits whose toll is always this hell.
@TedShifrin $(f^*(\alpha_1),f^*(\beta_1))$ and $(f^*(\alpha_2),f^*(\beta_2))$ are linearly independent and all linearly dependent for all other cases. But as far as I know, the linear independent and the determinant argument works for vector space.
@TedShifrin nice sentiment, but not always possible. I didn't have many, but I do remember pulling some all-nighters trying to finish something, or cramming for a biology exam.
@BannedUser actually I had an idea with that unrelated to your post
I'm working through another text, the book of numbers by conway. I'm just tinkering a little bit.
$\frac{1}{2}n(n+1)$ gives the nth triangular number.
I kind of had the idea to connect structures together.
to form a chain or molecule of structures kind of.
$\frac{1}{2}n(n+1) + \frac{1}{2}n(n+1) + \frac{1}{2}n(n+1)$ could give a larger more complicated structure.
$3[\frac{1}{2}n(n+1)]$
also connecting different polygonal numbers together like this too.
doesn't have to be $\frac{1}{2}n(n+1)$
n(n+1) is the pronic number, the product of two consecutive integers.
I'm also about to begin the Introduction to Number Theory by Apostol.
conway does this actually in ch2 of the book. it's kind of like combining polygonal numbers into a jigsaw puzzle. I'm just tinkering with extending the idea a bit.
if inserting a diagonal on $I\times I$ between endpoints b and d (call diagonal bd) gives us a triangulation of the torus, then we must have a topological embedding $h:bd->h(bd)$ such that $h(b)=h(d)$ and this is a contradiction since $h$ is injective
@TedShifrin $f^*(\alpha_1)$ and $f^*(\beta_1)$ are lineary independent. But $f^*(\alpha_2)$ and $f^*(\alpha_1)$ are $\Bbb Z$-linearly dependent and $f^*(\alpha_2)$ and $f^*(\beta_1)$ are $\Bbb Z$-linearly dependent so $f^*(\alpha_1)$ and $f^*(\beta_1)$ are $\Bbb Z$-linearly dependent which is a contradiction.
but I wonder if this determinant argument holds for general case
Btw I found in Dummit foote chatper 11.4 that I can still use the result (relation between det and lineary dependency) for general commutative ring with 1 $R$.
@love_sodam Well, this is the real reason I love differential forms as opposed to singular cochains. Wedge product is skew-commutative. Cup product isn't quite. I haven't thought that part through.
P.S. before I take the cat to the vet at an ungodly hour. @love_sodam These are matrices of integers, so why are you talking about determinants with non-commutative rings?
If $M_g$ denotes the closed orientable surface of genus $g$, show that degree $1$ maps $M_g\to M_h$ exist iff $g\geq h$.
Construction of degree $1$ map for $g\geq h$ is easy. I want to prove the converse using cup product. From now on, the basic idea is given by @TedShifrin. First I consider $(...
to focus my brain on something. suppose i have some rank-1 projector $P$ on $\mathbb{C}^d$. If I apply a unitary transformation, i get another projector $P'=U P U^\dagger$. Suppose I apply a random unitary transformation and average over $U(n)$, i.e., $\int_{U(d)} U P U^\dagger \,d\mu(U)$ where $\mu(U)$ is the left-invariant Haar measure on $U(n)$. I think this should just be the identity matrix $1_d$, up to a multiplicative constant.
well, the argument by invariance i'm seeing is that for any unitary $V$ we have $$\int_{U(d)} VU P U^\dagger\,d\mu(U) = \int_{U(d)} U P (V^\dagger U)^\dagger\,d\mu(U)=\int_{U(d)} U P U^\dagger V\,d\mu(U)$$
@TedShifrin its recurrent :-). the son of an irish friend is visiting and just set off on a week long california trip, so we had a brief 'son' for a night and then a goodbye.
OK, @Semiclassic, I've convinced myself the $1/2$ is right. Start with projecting on the line spanned by $v$ and then on the orthogonal line. Average those. As we rotate, the average stays the same.
$$\sum_{n=1}^{\infty}\frac{1 \cdot 4 \cdot 7 \cdot ... \cdot (3n+1)}{n^2 \cdot 3^n \cdot n!}$$
This is my solution:
Let's rewrite the question:
$$\sum_{n=1}^{\infty}\frac{(3n+1)!!!}{n^2 \cdot 3^n \cdot n!}$$
If we use D'Alembert's criterion:
$$\lim_{n \to \infty} \frac{\frac{(3n+4)(3n+1)!!!}{(n+1...
right multiplication by a fixed element in a ring allows us to view the ring as a subring of endomorphisms on its underlying abelian group. is there a name for rings with the property that every such endomorphism corresponds to right multiplication by an element?
in some sense I might haven been better of working in terms of the projectors $P_{\pm}=\frac12(1_d\otimes 1_d\pm S)$, but i didn't want to use $P_{\pm}$ for that
guess I could have done $P_S$ for + and $P_A$ for - (symmetric and antisymmetric)
this is converging on the results i expected, at any rate
the lecture notes we lifted that problem from are more advanced, yeah
the solution is sorta neat, though
you're looking for functions $f_A(\sigma\cdot a),f_B(\sigma\cdot b)$ on the sphere for unit vectors $a,b$
such that both take values in $[0,1]$ on the sphere, their averages on the sphere are both 1/2, and the average of their product should be $\frac14(1-\lambda\,a\cdot b)$ for some $\lambda$
first way that doesn't work: Pick $f_A=1$ when $\sigma\cdot a>0$ and 0 otherwise, and $f_B=1$ when $\sigma\cdot b<0$
those both average to 1/2 as needed, but the average of the product is computed via the overlap of these two hemispheres
There's the $\sin\phi$ or $\sin\theta$ or whatever you have that you need to integrate. This is why zones of spheres have area dependent only on height.
because if we choose it like this, then the average value of $$f_A(\sigma\cdot a)f_B(\sigma\cdot b)=\frac14(1+\sigma\cdot a-\sigma\cdot b-(\sigma\cdot a)(\sigma\cdot b))$$
is just $\frac14(1-\frac13 a\cdot b)$
if we pick the same sign, we'd have gotten the wrong sign
and the lesson will end up being that you can do it, if $\lambda\leq 1/2$
what the present example does is show that you can make it work for $\lambda=1/3$, which is actually enough to establish it for $0\leq \lambda \leq 1/3$
@TedShifrin Been attending a student present rational homotopy theory from Griffths-Morgan. Quite a beautiful fact that rational differential forms on simplicial complexes can be extended off subcomplexes, despite being piecewise polynomial.
