So I'm thinking one should be able to construct the path explicitly.
Let $p,q\in \Bbb R^n-E$. Let $L\subset\Bbb R^n$ be the straight line from $p$ to $q$. If $E$ does not intersect $L$ between $p$ and $q$, they are trivially connected by a path. Thus, we assume that $L\cap E$ is nonempty between $p$ and $q$. Now, $L$ cannot actually be a subset of $E$ because this would imply $p,q\in E$, contrary to assumption. Thus $L$ and $E$ intersect at precisely one point $y$.
We set up coordinates $x^1,\dotsc, x^n$ with $y$ the origin and $E$ spanning the first $n-2$ coordinates. That is, $E$ is the level set $x^{n-1}=x^n=0$. Since $p,q\notin E$, at least one of the two last coordinates are nonzero.
I don't really have an intuition for this problem.
@0celo7 Ohh. Then it's much simpler. Change coordinates so that $E$ is spanned by the basis vectors. Now it's linear algebra to try to write down a path (generically it'd always be a straightline, try to find conditions where it's not a straightline and fix accordingly)
But I see you did approach along those lines. Not sure where you got stuck.
Alright, let's just be a little clean about this. Assume your $p, q$ are random and you don't really know if the line through them intersects $E$ or not.
You need to learn high school algebra before learning linear algebra. $q_n (q_{n-1} - p_{n-1}) = q_{n-1}(q_n - p_n)$. Aka, $q_n q_{n-1} - q_n p_{n-1} = q_{n-1} q_n - q_{n-1} p_n$.
So, given fixed $p$ and $q$ this time, can you always find a $w$ such that $\ell(p, w)$ and $\ell(w, q)$ are both disjoint from $E$? Using the condition above?
If you can, you're done by picking the obvious path which goes from $p$ to $w$ along $\ell(p, w)$ and $w$ to $q$ along $\ell(w, q)$.
No problem. But the lesson from this is that you shouldn't try to get away from computation saying logic stuff like "If it doesn't, there's nothing to prove, so why not just assume it does".
Because that didn't really help.
General nonsense is arguments like this, but in a broader sense involving category theory. So many people do this (even I do) while trying to get away from computations (mostly in algebraic topology), and it's always the same rut.
The two equations above says $(q_n - p_n) t = q_n$ and $(q_{n-1} - p_{n-1}) t = q_{n-1}$ after re-arranging. Multiply the first by $q_{n-1} - p_{n-1}$ and the second by $q_n - p_n$.
in this topic http://math.stackexchange.com/questions/1913728/probability-that-no-couple-sits-together-in-a-circle there are a few derivations of the nevessary result of counting desired possibilities and eventually they arrive at the same explicit formula. The problem I'm having is understanding the $(-1)^k$, I know it's related to the inclusion-exclusion principle
Hello!! I want to use the Chinese remainder theorem for the following: $\mathbb{F}_2[X]$, $x\equiv 1\pmod X \\ x\equiv 0\pmod {X+1}$ Could you give me some hints how we could do that?
@MAFIA36790 Have you tried doing that? It doesn't work. Try clicking on the hyperlink. It directs me to the main site, not the meta site. Is there any way to tag meta tags?
@robjohn according to your estimations of my contibutions in math.se, would u think i would chase nice notes if ever i repass my bachelor degree, that is math specialized ?
Hey guys, I wanted to crowdsource a question out to you: If you could talk to Abel Prize and Fields Medal winners like Atiyah, Bhargava, Faltings, Hironaka, Mori, Ngo, Szemeredi, Voevodsky, and Wiles what would you want to ask them?
I have to prove that $7$ divides $3^{2n+1}+2^{n+2}$. I have done a proof which involves induction . But I want a proof just based on congruences . Could you help me with that @robjohn
@robjohn i think that if i take $a\in R\setminus Q$ and $I=Q\cap (-\infty, a)$ and $J=Q\cap (a,+\infty) $ it i good , an A=N\cap (-\infty,a) and B=N\cap (a,+\infty)
Hi guys, a small linear algebra question here, If I have a square matrix $M$ that is not full rank, is it always possible to use jordon decomposition with a certain basis to show that the triangular matrix $M'$ will always have number of zeros at the diagonal matching the nullity of the matrix?
Or more directly, for M of not full rank, is it always true that there exists at least one basis set which the diagonal component of the matrix will have number of zero eignevalues matching that of the nullity?
@robjohn For M of not full rank, is it always true that there exists at least one basis set which the diagonal component of the matrix will have the number of zero eigenvalues matching that of the nullity of M?
I see, that's a counterexample, thanks! (Background: So that means I need to actually ask the physics guys in order to work out what it means when M has a time depedent rank in that system...)
So, is it normal to define the dual of the tautological bundle over $\Bbb P^n$ to be the one with fibers equal to the lines that the elements of $\Bbb P^n$ define??
Because the whole big deal about the tautological line bundle is that you build all the isomorphism classes out of it, (using tensor product) giving this bijection to $\Bbb Z$
That's only in the complex line bundle category though. Isomorphism of complex vector bundles is slightly different than isomorphism of the underlying real bundles.
Heh. You should think of this as analogous to working mod 2 for unoriented manifolds, while for complex manifolds you always have a canonical orientation so you can work in integers.
@Danu Well, it suffices to find a section of $L \otimes L$. But that's exactly what a Riemannian metric on $L$ does, so you don't do anything new (it's the same proof that $L$ is isom to it's dual).
Exactly. The reason doing anything like that is garbage for complex line bundles is because a continuous choice of nondegenerate billinear pairing $L \otimes L \to \Bbb C$ (NOT the same as a Hermitian inner product!! That was garbage-talk on my behalf) might as well not exist.
Partition of unity won't work because that doesn't exist on the holomorphic category.
Yes, tautological is the one which assigns to every point the line it corresponds to.
@Danu Ah, now that's a good question.
$O(-1)$ admits no holomorphic global section except the zero section (which you might know already), but admits a nontrivial global smooth section, yes?
Take such a section and make it intersect with the zero section.
Then you'd get intersection number (after making these two sections transverse) $-1$.
For $O(1)$ doing the same construction would give you $1$.
I have always believed that scalars are just 1-dim vectors.
Yestoday, I however realized that units of measurements may be the compontnts of the vector but they cannot be scalars. That is, 1 meter can be 1-dim vector but it cannot be a scalar because you can add meters to meters (vectors must be additive) but you can multiply an object with number only. If you multiply an object with 1 meter, you obtain an object that cannot be added to the original vector.
The bottom line is that scalars can only be numbers. As long as you deal with pure numbers, you can say that scalars are 1-dim vectors. The difference comes up when units of measurement or more complex objects enter the picture.
Is it right? I never liked the bullshit that vectors have a direction and relative whereas scalars have no direction (and absolute?).
Because the latter definition is not informative (every 1-dim object has a direction on its 1-dim axis) and makes sutdents to believe that scalars are another name for the length of the vectors.
@Secret +/- 1 meter does not work with units? Stop thinking about it as a scalar and start thinking as a vector on an axis and see the difference. It is much easier now, right?
If you want to see whether something is a vector, it has to obey a transformation law under coordinate transformaitons. On 1D, the only possible coordinate transfomration is a shift $x\rightarrow x+ a$. Check if units obey that first
Little Alien: Ok consider a general coordinate transformation $x'\rightarrow ax+b$. Suppose your 1D vector measured from the origin is $\vec{v}=k-0 \text{ units }$. Then your new vector $\vec{v}'=((ak+b )-(a0+b)) \text{ units}=ak\text{ units}$. Now $\vec{v}'=a\vec{v}=\frac{dx'}{dx}\vec{v}$ thus vector components in 1D do obey the required transformation law. Now consider
in general, if you multiply out (z-u)(z-v)(z-w)...(more terms), the coefficients will go: sum of roots, sum of pairs of roots, sum of triples off roots, etc., except with minus signs introduced every other coefficient
this fact is called vieta's formulas
anyway, if z^3-(u+v+w)z^2+(uv+vw+wu)z-(uvw) = z^3+0z^2+az+b, then what can we say?
Vieta's formulas (the relationships between the roots and the polynomial coefficients) are very useful
I instantly recognized 2p and p^2+q^2 as sum and products of two roots, then saw the 0 coefficient meant the third root could be expressed using the other two
[Ok it turns out whether something is scalar or not depends on the type of transformation you do](http://physics.stackexchange.com/questions/279522/in-what-sense-can-a-complex-number-be-a-scalar/279542#279542).
However one issue of treating pure numbers as vectors in 1D is that now you have 2 (for real scalars) and a circle of unit vectors for each direction (for complex scalars) that you need to take account of.
Suppose, as measured at the origin you have a scalar k=+1, a 1D vector v=+1 and a 1D vector u=+a. If you are just computing the scalar multiplication k*u, you get +a as expected …
even if you restrict to 1 dimensional spaces
Danu, I guess that will also answer you question earlier on what does it mean to be a 1D-vector
Hi, I have a question for any here who majored in computer science, and then decided to pick up mathematics in grad school, or as a second major in undergrad.
Well, if we look at this in terms of abstract algebra, nobody stops one from defining a binary operator that allow you to map a 1 dimensional vector space and a n dimensional vector space to a n dimensional vector space (We have done similar things when complexify vector spaces).
What I showed above is that even if you try to hack the 1D vectors in so that they behave like the scalars we knew, the result will still differ because they are not scalars (and thus affectedb y coordinate transformations)
there is no issue treating numbers as vectors. vectors are just elements of vector spaces, and vector spaces are things satisfying a list of axioms. the fact the complex numbers are 2D over the real numbers is no issue at all.
vectors in a 1D space can be treated as numbers, but there is no canonical way of doing so. also you're confusing affine coordinate changes with coordinate changes for a vector space (which must fix the origin). not every vector is even a coordinate vector, so it doesn't always make sense to talk about a vector's "components."
everything about this discussion strikes me as trivial and uninteresting
Request for definition: "Locally free $\mathcal O_X$-module" ($\mathcal O_X$ is the sheaf of holomorphic functions on $X$)---in the context of complex geometry.
