This is another question I have been thinking about, in most differential geometry books, a tensor f is defined as the k-linear map f(...) V-fold into R and a tensor product is essentially the space of these maps (so dual space). In algebra books I am reading the treatment is different, I feel as though they are related by the universal property,
If so, what should I set the map (take k = 2 for simplicity) \phi: V \times W \to M (R-mod).
In the case of vector spaces, it turns out to be the same definition. I'll be glad to talk with you about it, but I have to leave now for the evening. Perhaps another day.
I will try to enter chat when u are available. I feel this is 100% a simple application(s) of universal property (literally every theorem in Dummit following it is an application of the theory....)
@TedShifrin I want to cook up a 5-manifold $M$ (or higher, of odd dimension $2n+1$) which has a hyperplane field $\eta$ such that $\eta$ is not integrable at each point (doesn't admit an integral submanifold of maximal dimension), but has an integral submanifold of dimension 4 (or of $2n\geq d>n$) at each point. I don't know if this is possible, and I don't know how to cook up examples like this. Any ideas?
Task
Proof expression $$ ((\neg C \vee B) \wedge ( A \rightarrow C) \wedge (B \rightarrow D) )\rightarrow \neg A \vee D $$
Attempt to proof
$$ 1. B \text{ (premise) } $$
$$ 2. B \rightarrow D (\text{premise}) $$
$$ 3. A \rightarrow D (\text{premise}) $$
$$ 4. D (\text{modus ponens 1,2}) $$...
> In other words, the pattern of interactions should define the geometry. If a system has many nonlocal connections, then it could be that no geometrical interpretation is possible.
Is false. Geometry proved itself to be so flexible that given any set with some relation R, it can be defined a geometry
It depends on your definition of a geometry. And usually, such a definition would be "A geometry is something that satisfies the following axioms." Of course, when we talk about non-Euclidean geometries, we know what we mean, namely, things that satisfy all axioms for a Euclidean geometry except ...
@TedShifrin Ted you're great! Now the last problem in this issue (I hope): When I substitute you solution of the intermediate equation I got, back I have: $$ \rho(\vec{r})^{g(\vec{r})} \nabla\rho(\vec{r})= \nabla \rho(\vec{r})\times (\nabla\rho(\vec{r})\times \vec{v}(\vec{r}))$$ with your arbitrary function $g$ and another fairly arbitrary scalar function $\rho$ and the $\vec{v}$ in question. Can we give any explicit form to $\vec{v}$ here?
iots like we have to "undo" $\nabla \rho \times$ two times. So I suppose it involves Poincare now?
Well, I think I have a proof for the finite case: Assuming $S$ is a finite semigroup, then fix some element $x\in S$ and consider the set $\{x^{2^n}\mid n\in\mathbb{N}\}$. Since $S$ is finite, there exist $m,n\in\mathbb{N}$ such that $m<n$ and $x^{2^m}=x^{2^n}$. So $x^{2^m}=(x^{2^n})^{2^{n-m}}=x^{2^n}$ and so, since $(x^{2^n})^{2^{n-m}}=x^{2^n}$, if $2^{n-m}=2$ we're done. If not, the multiply both sides by $(x^{2^n})^{2^{n-m-2}}$ to attain $((x^{2^n})^{2^{n-m}-1})^2=(x^{2^n})^{2^{n-m}-1}$
So, if there exist a semigroup with no idempotents, it's infinite.
Typo in there, I believe. It's difficult to tell with the exponents having exponents
(and it's too late to edit)
Yeah, that should be $(x^{2^n})^{2^{n-m}-2}$ and I put $(x^{2^n})^{2^{n-m-2}}$
@Secret I want to apologize once more for bullying you in this chat room earlier. I don't understand most of what you are writing about, but I believe you have your own genius in math. Be well.
I know that X compact, f continuous implies f closed map. (where $Y$ Hausdorff. We may assume all of $X,Y,Z$ are metric spaces). But not that $f$ can be open map.
how to show $f:X\to Y$ continuous from compact $X$ onto Hausdorff $Y$ implies $f$ open map?
well $X$ is compact, which means all its covers can only be finite covers, such covers must be some preimage of some family of open sets in $Z$ as $g \circ f$ is continuous
@Secret right, I was just checking because you said "some open set" rather than a family of such sets (I assume $X$ itself is considered to cover itself)
Continuity is where ch.3 of Munkres begins, which is interpreted because my chemistry PhD started back in 2017, which is why I technically have limits in my background in topology for this area
and I am pretty allergic to talking about metric space in topology, because I hate seeing too many inequality signs in the proofs where I suffer BSOD.
I actually understood many of the real analysis theorems a lot more ever since I knew about the more general framework of point set topology (which includes everything including non hausedoff topologies) as putting everything in terms of nets helps to avoid a lot a lot of inequality and absolute value signs
and makes it more intuitive since I am more comfortable with infinities
Have not followed up on much of his works since our meeting in 2014, back then he shared a similar sentiment as I am on the issue of inexactness since computationally speaking, all transcendental functions are power series, thus all we ever get are approximations
by redefining distances as distance squared, he basically includes a lot of constructible numbers into measurements of geometric figures, hence sidestepping the sin cos power series problem as everything is at least expressible in terms of combinations of roots, and polynomials
though technically speaking, there is a limit
for one, not all algebraic numbers are in the constructible numbers, such as solutions of 5th degree polynomials cannot be expressed in terms of elementary functions due to Galois theory (the corresponding Galois group is not solvable)
We have not discussed anything about his view on the reals, we only have discussed about rational trigonometry
I feel really ignorant in asking this question but I am really just don't understand how a compact set can be considered closed.
By definition of a compact set it means that given an open cover we can find a finite subcover the covers the topological space.
I think the word "open cover" i...
> It is true, however, that compact sets in Hausdorff spaces are closed, though a bit of work is required to establish the result.
one does need hausedoff to establish compact sets are closed. This helps to continue the above proof as by taking complements, $Y-f(X)$ is open and $Z-(g\circ f)(X)$ is also open
Thus $g(Y-f(X))$ is some subset in $Z-(g\circ f)(X)$
meaning it is open thus $g$ maps open sets to open sets, hence $g$ is also open map and hence continuous
Let $A $ be an $m \times n$ matrix of rank $n$ with real entries. Choose the correct statement.
1.$Ax=b$ has a solution for any $b$.
2.$Ax=0$ does not have a solution.
3.if $Ax=b$ has a solution then it is unique.
4.$y'A=0$ for some nonzero $y'$ where $y'$ is transpose of vector $y$
Now opti...
Wow! it seems correct. How do you come up with such clear cut reasoning? When i solve these kind of questions the i solely rely on intuition, but that feels unsatisfying. Which book are you usin for linear algebra? @N.Maneesh
Is it true that $\Bbb Z[1/2]$ is infinite dimensional over $\Bbb Z$?
@AlessandroCodenotti I mean how do we see that $\Bbb Z[1/2]$ has finite basis over $\Bbb Z$? I can't think other than $\{1,1/2,1/4,1/8,...\}$ as basis.
Hi, I made two measurements of one object's mass using different techniques and there are the results I got: First method: m1 = (80,0 ± 1,0) g Second method: m2 = (77,6 ± 7,2) g
You can treat the first method as a reference one. And I have a question: Is there a way to express these two results in terms of percentage? I mean "in how many percent the second result is coherent with the reference one"?
I tried to make this using a standard ratio Îľ = (m1/m2) and also the error of ratio calculated using exact differential
In the first method it should be m1 = (80,0 ± 0,1) g. Sorry, the page won't let me change it.
@anakhro Yes I noticed. Your answer is much appreciated. I see you made some efforts trying to explain my question.
So the thing is I'am not taking course in logic. I'am however taking course in discrete mathematics which is aimed at first year computer science students.
@Tuki yes I figured your skill level was around there. The Fitch-style notation there might not be helpful for your course, but it explains how you can save the half of the argument you made.
The first method is probably the way your professor intended.
But your try wasn't bad. You just needed to finish it, that's all.
I am trying to get the other answerer to elaborate on his answer for you. He's using a more formal style of proof where you use the $\vdash$ to denote "proves" and then use sets of formulas. So $\{P,Q\}\vdash R$ means that $P,Q$ prove $R$.
@BalarkaSen a problem I was working on yesterday: "I want to cook up a 5-manifold $M$ (or higher, of odd dimension $2n+1$) which has a hyperplane field $\eta$ such that $\eta$ is not integrable at each point (doesn't admit an integral submanifold of maximal dimension), but has an integral submanifold of dimension 3 (or of $2n>d>n$) at each point. I don't know if this is possible, and I don't know how to cook up examples like this."
The significance of such an example is kind of cute. Contact geometry is the case where you can only find integral submanifolds of dimension $d\leq n$.
And integrable hyperplane fields would be where you can find ones of $d=2n$.
So the example I am looking for would be between integrable and completely nonintegrable.
I have to sit down and work at it sometime today. For now I am looking at mr. Darboux & Pfaff. Trying to organize in my head the reasoning behind Moser's trick.
Trying to make my thesis intro a little bit more robust than current literature on the subject. I think I did a good job on Frobenius's theorem so far, everything being at least in one spot rather than 50 references.
I'm not sure why you wouldn't expect lots of examples to exist. Eg, a contact 3-manifold usually has a lot of embedded Legendrian knots. Why is $d > n$ special? Am I missing something?
@anakhro I tried hard to understand the Moser trick once. It's still a trick to me
The key fact I remember is that if $\omega_t = \omega_0 + d\alpha_t$ is a homotopy of symplectic forms lying in the same cohomology class, then there is a canonical way to cook up a family of vector fields $X_t$ such that $\mathcal{L}_{X_t} \omega_t = \dot{\omega_t}$
Then the isotopy between $\omega_t$ and $\omega_0$ is just given by flowing along $X_t$.
Choosing $X_t$ so that $\omega(X_t, -) = \dot{\alpha_t}$ works. I don't know why
Maybe physicists understand this better; dualizing by the symplectic form corresponds to the Hamiltonian and all.
@BalarkaSen it's somewhere in between contact and integrable. I want something strictly in between. Not something which is integrable. Not something which is contact.
Legendrian submanifolds are integral manifolds of maximal dimension in contact manifolds. By the magic of symplectic geometry (dimensions of isotropic submanifolds), this dimension is always $n$ (where the contact manifold is $2n+1$).
For $n=1$, your (hyper)plane fields can admit integral submanifolds of dimensions 1 or 2. There is no middle guy, but as soon as you increase $n$, you get those middle guys.
So "completely non-integrable" coincides with "non-integrable" for probably only $n=1$.
There are probably lots of examples for $n\geq 2$, but I couldn't cook up any last time I tried. Maybe I just didn't have the right idea.
If I come up with something to explain Moser, I will let you know. :P
@anakhro Oh, ok, so integrable submanifolds of dimension $d > n$ is just not possible in the contact world. I should have seen that. If $(M^{2n+1}, \alpha)$ is contact and $\xi = \ker \alpha$, $d\alpha|_\xi$ is symplectic. Integrable subbundles of $\xi$ are precisely the ones for which $d\alpha = 0$ along them.
u can show that minimal dimension for integrable submanifolds <=> u have a contact form so you just need to find a form $\alpha$ so that $d\alpha$ is non-degenerate but not too non-degenerate i.e. its rank isnt full
@ÉricoMeloSilva Intuitively I'd think of taking direct sum of a completely integrable subbundle with a completely nonintegrable one instead of cooking up a form. I'm not sure if that's a correct way of going about it.
I don't really have good intuition for the complete nonintegrability condition
i like saying it formy bc it means finding a 1-form $\alpha$ with $(d\alpha\vert_{\eta})^{k} \neq 0$ but $(d\alpha\vert_{\eta})^{k + 1} = 0$ for $0 < k < n$ in ambient $2n + 1$ which is like finding a bunch of matrices of certain rank which sounds like something i can do despite being dumb
well semester's over; i took analysis-2 (metric spaces+integration+bit of multivariable), algebra-2(linear algebra), probability-2(continuous distributions, ending with limit theorems) and classical mechanics this sem
Yeah if $\omega$ is the canonical symplectic form on the phase space $T^*M$, the Hamiltonian of a function $f : M \to \Bbb R$ is the $\omega$-dual of $df$
@anakhro I think they should just pick up a book and go through it than doing it piecemeal. Eg, I think linear algebra was quite sub-par this semester. I don't think the real point of various theorems were explained to us
There's also Hamilton-Jacobi theory, which formally takes the Hamiltonian $H=p^2/2m+V$ (for a one-coordinate system) and replaces $H\to E,p\to \partial S/\partial x$
and therefore you get a PDE $(\partial S/\partial x)^2+V(x) = E$
a @Balarka: I disagree. I like Artin's treatment, and the canonical form stuff is very natural (as opposed to ad hoc like the "advanced" linear algebra books that don't have modules).
@Semiclassical I like thinking about it in terms of DEs. Lagrangian is the second order equations in $T\mathbb R^n$, and Hamilton gives first order DEs in $T^*\mathbb R^n$
The instructor told me and a friend of mine to cover Jordan normal form and rational canonical form at the end of the semester, otherwise nobody would have seen it
@TedShifrin That was also a joint set of two talks with a friend actually. We have a growing community of highly motivated first years consisting of... 4 people :)
Task
Proof expression $$ ((\neg C \vee B) \wedge ( A \rightarrow C) \wedge (B \rightarrow D) )\rightarrow \neg A \vee D $$
Attempt to proof
$$ 1. B \text{ (premise) } $$
$$ 2. B \rightarrow D (\text{premise}) $$
$$ 3. A \rightarrow D (\text{premise}) $$
$$ 4. D (\text{modus ponens 1,2}) $$...
I feel really ignorant in asking this question but I am really just don't understand how a compact set can be considered closed.
By definition of a compact set it means that given an open cover we can find a finite subcover the covers the topological space.
I think the word "open cover" i...
