Right, orthonormal vectors — unit vectors that are orthogonal. So now maybe you should think about how to make an orthonormal basis for $\Bbb R^4$ with the first vector $(1/2,1/2,1/2,1/2)$.
This is a question about studying for the Putnam examination (and, secondarily, other high-difficulty proof-based math competitions like the IMO). It is not about the history of the competition, the advisability of participating, the career trajectories of former participants, or other such thing...
@Mathphile: Start in $\Bbb R^3$. If I give you one unit vector, how to find two other vectors so that the matrix with those columns is orthogonal? You already said you need pairwise orthogonal vectors that all have length 1.
@skullpatrol Thanks :) I'm thinking about doing it next school year. I don't want to be too ambitious, but I want to see if I'll get more than 0 points XD
Practice up on earlier exams. The average score is usually 0, so it's no shame to try and not get any. They don't give much partial credit. I think they give 0, 1, 3, 10 points on a problem when they score, or used to anyhow.
Now the remaining standard basis vectors can go to any three orthonormal vectors all orthogonal to $(1/2, 1/2, 1/2, 1/2)$, so you have to find one possibility.
Hello @José Carlos Santos , could you possibly reopen the question here (math.stackexchange.com/questions/3640960/…) I have the answer ready, have edited it, and the OP has confirmed in comments how they wish it to be answered.
Use an appropriate bijection to prove that each of the following sets has cardinality $\boldsymbol{c}$
(a) $(0, \infty)$
(c) $\mathbb{R}-\{0\}$
I found this question on internet while reading cardinality stuff. What does c stands over here and How to show there is bijection between those two...
can someone explain to me the last step that 5Xum did in his proof? the one where he says $(\bar a + \bar b)\delta - \delta ^2<\epsilon$ why did the signs change here to minus?
und why is $\epsilon$ bigger than the given number, why its not just equal?
I have an exercise that's asking me to prove $\iota_v\circ\iota_v=0$, where $\iota_v$ denotes interior product of an alternating multilinear form with a vector. I'm not exactly sure how to interpret this composition. Am I just supposed to note that $(\iota_v(\iota_v\omega))(v_3,...,v_k)=(\iota_v\omega)(v,v_3,...,v_k)=\omega(v,v,v_3,...,v_k)$ for an alternating $k$-tensor $\omega$, $k\ge3$?
The answer shows that $\pi\not\in \Bbb Q(\pi^3)$, which is enough to conclude that $\pi$ is algebraic of degree $3$ over $\Bbb Q(\pi^3)$ since it is a root of $x^3-\pi^3$
note that the same argument shows that if $p$ is any polynomial over $F$ and $c$ is transcendental over $F$, then $p(c)$ is also transcendental over $F$
Also @Balarka being torsion free is obviously not a quasi isometry invariant, since it's easy to build groups that are not torsion free but have a finite index torsion free subgroup, but more surprisingly being virtually torsion free is also not a quasi isometry invariant!
@Alessandro That's interesting. Having torsion forces infinite cohomological dimension, so that's a little surprising. Not sure how cohomological dimension interacts with quasi-isometries though
You say a group $G$ has $M$-valued cohomological dimension $n$ if its group cohomology $H^*(G; M)$ vanishes above degree degree $n$ for some $G$-module $M$
Hi @Ted!
Torsion free hyperbolic groups should have finite cohomological dimension
Haha. Well, if $G$ is discrete group then group cohomology with trivial $\Bbb Z$-valued coefficients is the singular cohomology group $H^*(K(G, 1); \Bbb Z)$ where $K(G, 1)$ is the unique space upto homotopy equivalence which has fundamental group $G$ and universal cover contractible
Man I have so much to read, my reading list is exploding
$\lim_{z \to \infty} |C_b(z)|^2 = 1.$ where z is a complex variable and $C_b$ is the firing rate of neurons in the anterior occipital. Under certain environmental stressors and assuming the SOR model of course, it's pretty straightforward to show that the limit exists (with enough data), and tends to one. So in summary, the stress in the brain (under "attack") heightens and firing increases rapidly. It's good to be able to model the decay rate of the stress and show that it goes to a constant
I misspoke; Weinstein's theorem says oriented isometries of positively curved even dimensional manifolds have a fixed point. This implies the theorem I quoted, because any nontrivial element in the fundamental group will give a deck transformation of the universal cover, which an oriented isometry and will thus have a fixed point - but deck transformations which have fixed points are identity
Doesn't explain where I am using even dimensional; need to read Weinstein's theorem's proof
Or maybe I can try to cook a proof up if it's not too hard
The basic idea should be if $p \in M$ is a point, $(U, \varphi)$ is a chart around $p$ on $M$, then let $B$ be a neighborhood of $p$ in $\Bbb R^{k+n} = \Bbb R^k \times \Bbb R^n$ such that $B \cap M = U$, and consider $B \to \Bbb R^{k+n}$, $(x, y) \mapsto (x, y + \varphi(x))$. Then image of $B \cap M$ under this is the graph of $\varphi$... graph of functions are locally flat submanifolds, right?
@Balarka: But where did you fundamentally use the $y$ in that construction? With smoothness, I would say that you're slicing a diffeomorphism, but I guess I don't get it.
Yeah what I said strictly speaking doesn't work because $\varphi$ isn't defined on all of $\Bbb R^k$; I can't seem to extend it in a way that gives me a homeomorphism $B \to \Bbb R^{k+n}$ either.
Trying to fix
@TedShifrin My idea is the following picture; take a wild arc in $\Bbb R^3$, and embed it in $\Bbb R^3 \times \Bbb R^1$. Then you should be able to find an ambient homeomorphism of $\Bbb R^3 \times \Bbb R^1$ which shifts the wild arc vertically along the $\Bbb R^1$ direction as you traverse it, making it a graph
It's a category $E$ with some additonal axioms such as finitely complete and cofinitely complete such that the category acts abstractly like $\textbf{Set}$.
@TedShifrin I think this is actually OK. $\varphi : U \subset M \subset \Bbb R^k \to \Bbb R^n$ be my chart homeo around $p \in U$. Extend this to a map $\tilde{\varphi} : V \to \Bbb R^n$ where $V$ is a neighborhood of $p$ in $\Bbb R^k$ such that $V \cap M = U$; of course, this is no longer a homeomorphism or anything, but $\tilde{\varphi}|_U = \varphi$.
Now re-embed $M \subset \Bbb R^k \subset \Bbb R^{k+n}$ and consider a ball $B$ around $p$ in $\Bbb R^{k+n}$, define $F : B \to \Bbb R^{k+n}$, $F(x, y) = (x, y + \tilde{\varphi}(x))$. This is a homeomorphism (to an open subset of $\Bbb R^{k+n}$), as it is obviously injective and the inverse is given by $G(a, b) = (a, b - \tilde{\varphi}(b))$.
@enjoys My ultimate research question is: How to classify certain algebraic function fields in Hadamard spaces that are enriched with co-complete fibered CAT(0) spaces. So right now I don't know the answer but I think I can figure it out
So my claim is if I have a function $f : \Bbb R^n \to \Bbb R^k$ and $\Gamma_f = \{(f(x), x) : x \in \Bbb R^n\} \subset \Bbb R^{k + n}$ is the graph, then there is a homeomorphism $\Bbb R^{k+n} \to \Bbb R^{k+n}$ taking $\Gamma_f$ to $\Bbb \{0\} \times \Bbb R^n$. Just consider $(a, b) \mapsto (f(b) - a, b)$
but I have a big problem. I don't know how to fiber CAT(0) spaces in the proper way. Also I don't know how to "place" the function fields in Hadamard space. And finally, I don't know how to enrich the co-complete things.
Use an appropriate bijection to prove that each of the following sets has cardinality $\boldsymbol{c}$
(a) $(0, \infty)$
(c) $\mathbb{R}-\{0\}$
I found this question on internet while reading cardinality stuff. What does c stands over here and How to show there is bijection between those two...
