Mathematics - 2016-05-04 (page 4 of 5)

Okay, so if we consider the fact that an irrotational field if conservative over a simply connected space $U$, then it seems that all the loops which can exist on the space $U- \{a\}$, where $a$ is some point is a proper subset of the loops that exist on $U$, so then it would still be conservative.

Balarka Sen

Restriction of a conservative vector field is conservative, yes.

So whatever nonconservative vector field over $\Bbb R^2 - 0$ there is, it cannot be extended to $0$.

It must have a singularity at $0$.

Juan Sebastian Lozano

5:17 PM

What about $X = (\ln (y), \frac{x}{y})$?

Semiclassical

you probably want $\ln|y|$ at the very least

Juan Sebastian Lozano

Whoops, of course.

Mike Miller

@JuanSebastianLozano That's not defined on the entire $y$-axis.

$x$-axis, sorry.

Semiclassical

that too

Mike Miller

Try working with rational functions.

Balarka Sen

5:18 PM

^

Semiclassical

if one were going to do a lot of this, i'd suggest going into polar coordinates and assuming rotational symmetry in order to get an example

might be overkill here

Balarka Sen

If one understand what curl = 0 means, I think it becomes easier to construct an example

Semiclassical

sure.

Juan Sebastian Lozano

I'm trying to find functions (other than the obvious ones, which are conservative) where curl = 0, which is how I got the above one.

Balarka Sen

What does curl = 0 really mean? Geometrically?

Juan Sebastian Lozano

5:22 PM

That the vector field looks radial

Mike Miller

@BalarkaSen I suspect you're probably thinking it's too easy to come up with examples. When you first asked about this, you did not understand how the denominator played a role in the story, even given the apparent geometric meaning.

Balarka Sen

No, I don't, not really. I am asking Juan if he knows what curl = 0 means better than just "it looks swirly" - I think it becomes easier to come up with an example then, I am not sure.

Juan Sebastian Lozano

Mike, I like your idea about polar coordinates, but I showing that it's not conservative is harder in polars, I think.

Balarka Sen

I already know an example, so I can only suspect what's easier.

Semiclassical

@JuanSebastianLozano was that directed to MikeM or to me?

Juan Sebastian Lozano

5:25 PM

Oh, whoops, it was to you.

Semiclassical

as it happens, it was correct in any case---my first name is Michael, i just don't use it as my handle.

Juan Sebastian Lozano

That's lucky :P

Okay, so what sort of thing are you meaning by Curl = 0's meaning?

@BalarkaSen

Semiclassical

in any case, if one writes the vector field as $A=A_r \hat{r}+A_\theta \hat{\theta}$

Juan Sebastian Lozano

It's radial if $A_\theta = 0$

Samuel Yusim

looking at an exercise in a book about graph colouring that begins with "Euler's formula implies that every planar graph has a vertex of degree five"

what is being left out/said incorrectly here

Mike Miller

5:29 PM

i am skeptical

can you give some further context, maybe?

Semiclassical

then the 2D curl is $\partial_r(r A_\theta)-\partial_\theta A_r$

Balarka Sen

@JuanSebastianLozano I don't know if I can explain what I mean without explaining what I mean. A vector field has curl = 0 if drawing a small disk on $U$ somewhere it doesn't instantaneously rotate around it's axis under the force of the vector field.

Juan Sebastian Lozano

Every simple planar graph with no cycles has one vertex of degree less than five, as a consequence of Euler's formula

Samuel Yusim

wow.

Hippalectryon

@user1618033 welcome back :-)

Mike Miller

5:31 PM

a little different from the original formula

Samuel Yusim

also, like, that's all it says before it gives part (i) which is proving the 6-colour theorem

Mike Miller

(ii): prove the 5-color theorem

(iii): prove the 4-color theorem

(iv): prove the 1-color theorem

Samuel Yusim

wow

Balarka Sen

1-color theorem!

Juan Sebastian Lozano

(V): prove the -4 color theorem.

Samuel Yusim

5:33 PM

luckily part 2 is some random thing and part 3 is the 5-colour theorem and then it ends

Semiclassical

(VI) prove the 5-senses theorem

Hippalectryon

@BalarkaSen every graph can be colored with one color

Balarka Sen

so deep much math wow

Samuel Yusim

the 1-colour theorem: every planar graph can be coloured with at least 1 colour

Mike Miller

i like the 2-color theorem: every linear graph can be colored with at most two colors

that's a good exercise.

or at least, that's an exercise

Akiva Weinberger

5:36 PM

Does "linear graph" mean what I think it means

Samuel Yusim

I mean if you know induction you're good

Mike Miller

what do you think it means?

Balarka Sen

graph sitting inside R.

Akiva Weinberger

2-color theorem: If every vertex is degree one, you can color it with two colors where no two adjacent vertices have the same color

(Note: Independant of ZF!)

Juan Sebastian Lozano

$(x-y,x)$?

Huy

5:38 PM

@MikeMiller :(

Semiclassical

i'd assume it's a graph with vertices labelled 1,...,n with edges only between vertices labelled by adjacent numbers

Mike Miller

@BalarkaSen I wouldn't push someone for an example for hours. I would use the example as a good pedagogical example and move on.

@Semiclassical i was just thinking graph inside R

Samuel Yusim

I'll define my graph as having vertex set $\mathbb{R}$ and with adjacency inherited from the euclidean topology

Balarka Sen

Fair enough. @Juan Here goes.

Semiclassical

fair enough

Balarka Sen

5:39 PM

Take $1/\|\mathbf{x}\|^2 (-y, x)$.

Mike Miller

@SamuelYusim is it true that every possibly infinite planar graph can be 4-colored?

Juan Sebastian Lozano

The -4 color theorem: you can uncolor an optimally colored graph by uncoloring only four colors.

Samuel Yusim

I have no idea

Akiva Weinberger

@SamuelYusim How do inherit that from the topology

Balarka Sen

You can show that it integrates to $2\pi$ along the unit circle, so is not conservative.

Mike Miller

5:40 PM

by planar graph here I mean the vertex set should be discrete and there should only be finitely many edges adjacent to a given vertex, and then topological embedding

Samuel Yusim

@akiva: easy, just take consecutive real numbers to be adjacent

Akiva Weinberger

@MikeMiller I'mma go with "yes", though maybe you need Choice?

Mike Miller

why would you need choice?

Semiclassical

to connect with the polar coordinates comment, that corresponds to $A_r = 0$ and $A_\theta = 1/r$

Akiva Weinberger

'Cause infinity be weird.

Semiclassical

5:41 PM

which in accordance with the above formula has zero curl

Mike Miller

great reason

Balarka Sen

Why is it irrotational? You can tell that by looking at the picture: desmos.com/calculator/eijhparfmd. Since all the vectors sticking out are unit vectors, placing a small disk somewhere won't rotate (around it's axis) under the force.

So even though it "looks swirly", curl = 0.

Akiva Weinberger

@MikeMiller And 'cause that other theorem needs it (just a sec)

Samuel Yusim

I'm sitting in on an infinite graph theory class so I'll ask the prof on monday or something

Akiva Weinberger

This one

Juan Sebastian Lozano

5:42 PM

Yeah, that makes complete sense.

Samuel Yusim

in the meantime: is an uncountable disjoint union of copies of $K_{1,3}$ planar?

Juan Sebastian Lozano

I guess the problem is I assumed all fields of that form would be conservative without checking.

Mike Miller

no, @SamuelYusim, since an uncountable set in the plane has an accumulation point

the vertex set is automatically countable if we say planar

Semiclassical

obligatory physics discussion: that's actually a horizontal cross-section of the magnetic vector field produced by a current-carrying wire pointing straight up

Mike Miller

@AkivaWeinberger that's a better reason, thanks

Balarka Sen

5:44 PM

@Juan That there is an irrotational vector field on $\Bbb R^2 - 0$ which is not conservative says that $\Bbb R^2 - 0$ is not simply connected, due to all what we discussed. So there's some topological invariant going on. Let's try to put all these in a different context.

Semiclassical

the fact that the curl is zero where the field is well-defined tells you that there's no source of current away from $r=0$ (i.e. the current runs only through the wire)

Mike Miller

i suspect that when two-three people are trying to teach someone something, the someone learns nothing

Semiclassical

pfft.

Juan Sebastian Lozano

It's okay, I'm learning a lot :P

Balarka Sen

So, now we can introduce the formality of differential forms (about time!)

Juan Sebastian Lozano

5:46 PM

@Semiclassical Yes, that's something we talked about in E&M.

Semiclassical

at the same time, Ampere's law guarantees that there should be a nonzero circulation of the magnetic field around the origin due to the presence of the current

Juan Sebastian Lozano

@BalarkaSen okay

Semiclassical

so if we were to interpret the magnetic field as a force field, then it'd necessarily be a non-conservative field

Juan Sebastian Lozano

(I can see where this is kind of going because you talked about how differential forms can be used to make a 'cohomology theory')

Mike Miller

@Huy every room needs a troll

Semiclassical

5:48 PM

if you wanted a genuine force field, you'd want to replace the magnetic field with an electric field. in that case, one would need to replace the current-carrying wire with a 'string' of magnetic field in its place

moreover that field would need to be time-varying in order to actually produce an effect. not so nice.

Mike Miller

i'm going to bow out of this discussion because i think there are probably already too many people in the pile :)

Semiclassical

i'm done myself, to be clear.

Balarka Sen

@JuanSebastianLozano $U \subset \Bbb R^2$ be an open subset. $0$-forms on $U$ are smooth functions on $U$. 1-forms on $U$ are expressions of the form $f dx + g dy$ where $f, g$ are smooth functions - you should think of them as vector fields $X = (f, g)$, but a version which also enables vector fields to be integrated. 2-forms are expressions of the form $f dx \wedge dy$, where we have the convention that $dx \wedge dy = - dy \wedge dx$.

I am not sure if this is the best way to introduce differential forms but let's move forward and try to understand this better as we proceed.

Note that the collection of i-forms form a $\Bbb R$-vector space, call it $\Omega^i(U)$.

Do you follow me?

Juan Sebastian Lozano

Antisymmetry, like a lie bracket and a cross product? Is that how I should think about them>

Balarka Sen

You should think of it as orientation.

A 2-form is something which can be double integrated (given $f dx \wedge dy$, just define integration as $\iint_D f dx dy$)

And if you switch the order of the $dx$ and $dy$, there is a $-$ sign because orientation gets changed.

Juan Sebastian Lozano

5:59 PM

Okay, yeah, which is the same as the cross product. Okay. I follow so far, but do you mean integration as anti-differentiation? Or do you mean definite integration?

Balarka Sen

I just mean integration over a region, Riemann integral if you prefer. Anti-differentiation doesn't make much sense in this context.

$dx \wedge dy = - dy \wedge dx$ is a signed/oriented version of Fubini's theorem.

Juan Sebastian Lozano

Got it, that makes sense.

Balarka Sen

Alright, we will soon understand this better I hope.

So, we have this vector space $\Omega^i(U)$ (we can add together two $i$-forms, multiply by a scalar, etc...).

Juan Sebastian Lozano

A vector space over $\mathbb{R}$?

Balarka Sen

Yes.

Juan Sebastian Lozano

6:04 PM

Okay

Balarka Sen

OK, so there is a more or less natural map $d : \Omega^0(U) \to \Omega^1(U)$, given by $df = \partial f/\partial x dx + \partial f/\partial y dy$.

This is called the exterior derivative map. As you see, this is just taking $f$ to it's gradient vector field $\nabla f = (\partial f/\partial x, \partial f/\partial y)$.

So in the vector field context it's the same as the map $\nabla$.

Juan Sebastian Lozano

Yea, okay, that makes sense.

Does that mean every $\Omega^i(U)$ is conservative?

Or, just that the image of $d$ is?

Balarka Sen

I don't know what you mean by that. A vector space cannot be conservative :)

Juan Sebastian Lozano

member of the vector space, sorry :P

Balarka Sen

The image of $d : \Omega^0(U) \to \Omega^1(U)$ consists of conservative vector fields, yes, under the identification of the 1-form $fdx + gdy$ with the vector field $X = (f, g)$.

Juan Sebastian Lozano

6:11 PM

Okay, cool, but $\Omega^1(U)$ is just 'the free space' of one forms (meaning all the possible one forms)?

Balarka Sen

Yes, all possible one forms.

Juan Sebastian Lozano

Okay, cool.

Balarka Sen

You see, we're frequently using the identification of a 1-form $fdx + gdy$ with the vector field $X = (f, g)$ to interpret this obscure abstractness. This is useful, but keep in mind a 1-form is more than just it's corresponding vector field. It's something which can be integrated too. But we'll understand more about that soon.

So, we defined exterior derivative map $d : \Omega^0(U) \to \Omega^1(U)$. We can also define an exterior derivative map $d : \Omega^1(U) \to \Omega^2(U)$ given by $d(fdx + gdy) = (\partial g/\partial y - \partial f/\partial x) dx \wedge dy$.

Juan Sebastian Lozano

The curl

Balarka Sen

This time one eats a vector field $X$ and spits out it's $\text{curl}(X)$, yes.

So the kernel of this map consists of... what vector fields (identifying 1-forms with vector fields)?

Juan Sebastian Lozano

6:17 PM

Is that map unique? Or is it a clever one informed by vector calculus?

The kernel is the irrotational vector fields.

Balarka Sen

Right, kernel of that map consists of irrotational fields. What do you mean by unique (I think I know what you mean, but I want to clarify before answering)?

Juan Sebastian Lozano

Like, how do we know that that is the choice of map that we want? Is it just that it happens to line up with our notion of curl?

Could we have chosen another map?

Balarka Sen

Very good question :) No, it is in a sense unique. I didn't explain where the map came from because I didn't wedge product in full generality.

I know my definition of $d : \Omega^1 \to \Omega^2$ looks ad hoc, but it's actually quite natural.

But maybe we'll come back and discuss this a bit later? I just want to finish what I am talking about right now before getting to something else, is all.

Juan Sebastian Lozano

Yeah, no problem! Let's continue.

Balarka Sen

OK, so we have a sequence of homomorphisms $\Omega^0(U) \stackrel{d}{\to} \Omega^1(U) \stackrel{d}{\to} \Omega^2(U)$. $d \circ d = 0$, because $\text{curl} \nabla = 0$ as we just proved. This is thus a chain complex. We can look at the 1st homology of this complex $H^1(U) := \text{ker} d/\text{im} d$. This precisely measures how many irrotational vector fields there are modulo conservative ones.

Owatch

6:26 PM

If I'm asked to find the orthogonal for matrix:

1 0
1 0

Balarka Sen

This is called the 1st de Rham cohomology of $U$. We proved about $H^1(U) = 0$ whenever $U$ is simply connected, and that $H^1(\Bbb R^2 - 0)$ is nontrivial since $1/\|\mathbf{x}\|^2 (-y, x)$ is a irrotational field which is not conservative on $\Bbb R^2 - 0$.

This is how all we talked about so far fits into a general context of cohomology.

Owatch

I'm not sure what to do about the second column. It would be a div-zero if I use the formula.

Within: $u_{2} = \frac{1}{||v_{2}||}\times v_{2}$

Tobias Kildetoft

@Owatch What do you mean by the orthogonal?

Owatch

Oh, I'm sorry. I mean I want to normalize.

I have orthogonal basis above, and I need the orthonormal basis.

So I want to normalize the column vectors.

Balarka Sen

@JuanSebastianLozano See if you digest this. We haven't so far introduced anything special: we have just put everything into a philosophically satisfying context of cohomology. In fact we haven't used the extra structure differential forms carry with them either (namely, the ability to be integrated).

Tobias Kildetoft

6:29 PM

@Owatch You don't have any basis there

Owatch

Oh, they have to be nonzero set?

Or the vectors must be nonzero ?

Tobias Kildetoft

@Owatch They have to be a basis (look up the definition if you forgot it)

Juan Sebastian Lozano

Sorry, had to step away for a second.

Balarka Sen

No need to apologize. We all have work to do. :)

Juan Sebastian Lozano

Okay, I see how this is a chain complex, it is a really nice set up. So, for a simply connected open set, the 1st de Rham cohomology group vanishes, just like the first homology group.

Why is this cohomology instead of homology?

Owatch

6:40 PM

I've got "A set of vectors $\{u_{1}...u_{p}\}$ in $R^n$ is said to be an orthogonal set if each pair of distinct vectors from the set is orthogonal. (And the zero vector is orthogonal to everything). That is, if $u_{i} \cdot u_{j} = 0$ when $i \neq j$. Then, "If $S = \{u_{1}...u_{n}\}$ is an orthogonal set of nonzero vectors in $R^{n}$, them S is linearly independent and hence is a basis for the subspace spanned by S".

I guess the vectors are not nonzero.

In my case.

Balarka Sen

@JuanSebastianLozano I lied, it wasn't really a chain complex. It's a cochain complex. The indices increase instead of decreasing.

Anyway, that's a formal thing.

Owatch

So then what is Wolfram doing when it tells me the result is:

1 0
0 0

For the command "orthogonalize {{1,0},{1,0}}" ?

Juan Sebastian Lozano

Okay, yeah, so its because our map goes from $\Omega^0 \to \Omega^1$

Balarka Sen

Mhm.

Huy

cuz what else do you expect owatch

Balarka Sen

6:44 PM

So, what's next? @Juan.

Owatch

What do you mean, I'm asking here.. It's giving me a result when I apparently don't have any basis at all.

What is the result

Tobias Kildetoft

@Owatch You do not seem to have input the same matrix into WA as you gave us.

Owatch

What do you mean, that's is exactly the same matrix.

I wrote this earlier:

1 0
1 0

Which is: {{1,0},{1,0}}.

And Wolfram Interprets it as such.

Juan Sebastian Lozano

Well, if you want to continue the chain complex you define $\Omega^3$ and a mapping $d:\Omega^2 \to \Omega^3$

Balarka Sen

Yes, we were working with subsets of $\Bbb R^2$, in which case they are zero so I didn't bother defining that. You can do that.

Juan Sebastian Lozano

6:47 PM

Why is that?

(Like I get it at a calculus level, I don't know why that transfers here)

Balarka Sen

Well, first you have to know what 3-forms are.

It's probably worth mentioning that if you work with subsets of $\Bbb R^3$, the map $d : \Omega^2 \to \Omega^3$ in some appropriate way is the divergence map.

Owatch

Another online calculator I found gives me a different result than Wolfram. It just gives back the input..

Juan Sebastian Lozano

Would it just be $dx\wedge dy\wedge dz$

?

In which case, I can see why it would be zero if this product behaves like I think it does, since $dz=0$

user 1618033

@Hippalectryon hheyyyyyyy!!! :-)

Balarka Sen

Yes, 3-forms for subsets of $\Bbb R^3$ are expressions of the form $f dx \wedge dy \wedge dz$, with the convention the switching any two of the three $d(\text{blah})$'s adds a $-$ sign.

Hippalectryon

6:57 PM

@user1618033 I thought you were gone for good and so was your book ;-;

I was wrong yay

user 1618033

@Semiclassical sure. Try it when you have time, it's a nice simple integral.

@Hippalectryon lol :D I'm like a gladiator in this life, I never give up until I have reached all my objectives successfully. :-)

Juan Sebastian Lozano

Okay, that makes sense.

Thanks for explaining all of this. So is there something cool about the integration structure?

user 1618033

@Hippalectryon I can only tell you that I obtained absolutely outstanding results in the last period of time.

Hippalectryon

@user1618033 \oo/ .. and I know I'm always bugging you with this, but how the book doing ?

user147690

Very tired, but must work on forever: I want to talk about taking a point and a line, and taking their span in projective space, how to make this rigorous?

user 1618033

7:01 PM

@Hippalectryon All is fine. It still takes some time for reasons that don't depend on me.

user147690

Do I have to take the cone back into affine space, make sense of it there, and projectivize?

Hippalectryon

Yay :DD

user 1618033

@Hippalectryon I can only tell you that in my book you'll find stuff that no one ever did before (if this seems anything exciting for you).

Hippalectryon

@user1618033 Which is why I keep asking about it :-)

user 1618033

:D

@Hippalectryon How are things there at you? Still in France?

Balarka Sen

7:03 PM

@JuanSebastianLozano Since we're discussing these, let's just define forms in full generality.

Hippalectryon

@user1618033 Well, exams went well this year :P all's going fine for now. Except that I stayed too late working, didn't see the time pass, and only realized it after dinner time was over :(

user 1618033

@Hippalectryon I have the same annoying problem. I don't realize how time goes by, it seems to me that 2, 3 hours pass like 5 minutes. Then I go to bad very late in the night these days, 05:00 am or so.

Balarka Sen

Suppose $U \subset \Bbb R^n$ is an open set. $k$-forms on $U$ are expressions of the form $\sum_I f_I dx_I$ where $dx_I = dx_{i_1} \wedge \cdots \wedge dx_{i_k}$, $I = \{i_1, \cdots, i_k\}$. Index $I$ runs through ordered $k$-tuple subsets of $\{1, \cdots, n\}$

These similarly form a $\Bbb R$-vector space $\Omega^k(U)$.

user 1618033

@Hippalectryon People close to me are really upset because of my schedule, but eventually they want me to be in a good shape. I have many ideas to work on, no time to postpone.

Balarka Sen

We define wedge as $dx_I \wedge dx_J = dx_{(I, J)}$, and extending this billinearly to wedge in general. This product thus takes a $k$-form and a $k'$-form and spits a $k+k'$ form

user 1618033

7:09 PM

@Hippalectryon At some point I was asking myself if one day I'll wake up and realize that I have no new idea, but lately things have improved, I have more and more ideas than ever before.

Hippalectryon

@user1618033 Well, dont put your future health at risk though :-)

user 1618033

@Hippalectryon Yeap, good to consider that.

Balarka Sen

Then the exterior derivative map becomes much more apparent: $d : \Omega^0(U) \to \Omega^1(U)$ is defined as usual by $df = \sum_i \partial f/\partial x_i dx_i$.

user 1618033

@Hippalectryon are you working in the same place? You did some stuff related to chemistry, didn't you?

Balarka Sen

And we define $d : \Omega^k(U) \to \Omega^{k+1}(U)$ by $d(f dx_I) = df \wedge dx_I$ and extending linearly.

Hippalectryon

7:11 PM

@user1618033 I haven't moved

Balarka Sen

You can check that this definition agrees with curl for subsets of R^2 and $d : \Omega^1(U) \to \Omega^2(U)$.

user 1618033

@Hippalectryon I see.

Hippalectryon

@user1618033 Maths, physics, chemistry. I have been way more on the chemistry chat lately than in the math chat, since you weren't there (or at least I didn't see you)

user 1618033

@Hippalectryon You seemed pretty busy in some of the past days.

Owatch

I'm getting bloody nowhere. I want to QR factorize a 2x2 matrix of 1's. To do this, I perform the Gram-Schmidt procedure to find the orthonormal basis for it. I get the orthogonal set of vectors {{1,0},{1,0}}. And then I need to find the orthonormal vectors. So I normalize each column vector, which is then impossible because you can't normalize a zero vector.

Hippalectryon

7:12 PM

@user1618033 Ah yeah the last two weeks were especially busy

user 1618033

@Hippalectryon I wasn't around for a good while, I tried to focuse only on my mathematical research. It was a great time to me.

@Hippalectryon I see.

Balarka Sen

@Adeek How did the algebraic topology exam go?

Juan Sebastian Lozano

Wait, so there are $\binom{k}{n}$ terms?

Balarka Sen

$\binom{n}{k}$, but yeah.

Juan Sebastian Lozano

Okay, and what do you mean by $dx_{(I,J)}$?

Balarka Sen

7:16 PM

$I$ is a $k$-tuple, $J$ is a $k'$-tuple. $(I, J)$ here is concatenation: it's the $k+k'$ tuple by concatenating $I$ and $J$.

So ({1, 2}, {3}) = {1, 2, 3}

user 1618033

@Hippalectryon Just a curiosity about what you said before: did I ever look like someone that might give up? Referring to my book?

:-)

Hippalectryon

@user1618033 No, but you could have not come again to this chat, and I'd have no way to contact you to know when your book is out

user 1618033

@Hippalectryon OK :-)

Juan Sebastian Lozano

So $dx_{1,2,3,...}= dx_1 \wedge dx_2 \wedge ...$, and $dx_{1,2} \wedge dx_{3,4} = dx_{1,2,3,4}$?

Balarka Sen

Yepper.

Juan Sebastian Lozano

7:24 PM

Okay, and the negative from the curl comes from the order of the differentials.

Okay, yeah, I got it, that makes sense. It was a little hard to digest at first.

Balarka Sen

@JuanSebastianLozano Yep.

The formality in setting up differential forms can be intimidating. But there's more geometry going on than one sees at first glance.

Juan Sebastian Lozano

Does this extend to a more general context than $\mathbb{R}^n$?
Also, can you set up a chain complex by integrating? Or do the mappings not work like that.

Balarka Sen

Yes, it extends to arbitrary smooth manifolds.

Integrating a form over something results a scalar (I mean, it's just $\iint_D f dx_1 dx_2 \cdots dx_n$). Not sure how you'd construction a chain complex out of that.

Owatch

If the normalized form of the zero vector is the zero vector. Then how is it that normalizing the vectors of the Matrix:

1 0
1 0

Gives:

.707 .707
.707 .707

When I get:

Juan Sebastian Lozano

I was thinking of a single integration, but I guess that only makes sense in special contexts, when the order integration doesn't matter and the domain is nice.

Owatch

7:38 PM

.707 0
.707 0

Clearly. The first column vector is: $\frac{1}{\sqrt{1^{2} + 1^{2}}} \times v_{1}$.

But how on earth is the second vector the same? It should be zero. Why does the online calculator say differently

Mike Miller

@JuanSebastianLozano There's an interesting idea here I won't say anything about.

It will not work in the form you're thinking, though.

Balarka Sen

I am curious to hear what's the idea.

Mike Miller

That's why I'm not going to talk about it.

Owatch

Wolfram seems to agree this time. But I don't know why all these tools are so inconsistent.

Balarka Sen

Sure. But can I hear what's the goal? To construct a map $\Omega^n \to \Omega^{n-1}$?

Mike Miller

7:42 PM

No.

Ask me when you're done with the book.

Balarka Sen

OK, thanks.

Mike Miller

Also, one can of course construct a map ($d^*$). It's not even interesting that one can write down maps in the other direction that form a chain complex. The point is that you can do so by integrating.

Balarka Sen

I see.

Mike Miller

I guess I should say something trivial about it. If you have a (compactly supported say) $n$-form on $\Bbb R^n$, what would you try to do to get a form of smaller degree?

Balarka Sen

If it's a closed form, then it's exact since it lives in $\Bbb R^n$. I can try to find out what is $\omega$ such that my form is $d\omega$. Seems rather far-fetched, I suppose.

Mike Miller

7:47 PM

That doesn't seem in the spirit of what Juan suggested at all.

Owatch

Well looks like the tool I was using is in-fact wrong.

Yet more evidence I am always right.

And that the internet is trying to undermine me.

And I fixed the problem.

Good. All is fine for now.

Balarka Sen

I don't know what I want to do, I suppose.

Vrouvrou

Hello, if g is continuous on a compact [0,1 is g^2 still continuous on [0,1]

?

Mike Miller

ok

Akiva Weinberger

@Vrouvrou Yes. The product of continuous functions is continuous.

Unless you meant to write "uniformly continuous" (which is different), the domain [0,1] doesn't matter.

Vrouvrou

7:59 PM

so if i have $f,g\in (C([0,1],\mathbb{R})$ then i can say that $\int_0^1 (gf)^2 dt=\int_0^1 g^2(t) f^2(t) dt \leq \sup_{x\in[0,1]} g^2(x)\int_0^1 f^2(t)dx$ @AkivaWeinberger

Akiva Weinberger

I believe so.

Though, now, the domain [0,1] is relevant because it assures us that the functions are bounded and thus the supremum isn't infinity

Vrouvrou

yes

Owatch

Well I sent an email about the tool mistake to the owner.

Hopefully he fixes it.

Vrouvrou

g^2 must be continuous on [0,1] to say that $sup g^2 <\infty$ @AkivaWeinberger

Balarka Sen

@MikeMiller I guess I have an idea to "antiderivate" a compactly supported n-form on $\Bbb R^n$ to get an $n-1$ form but it looks far fetched and not useful. Take $f dx_1 \wedge \cdots \wedge dx_n$ and spit out a signed sum of $(\int_0^t f(x) dx_i) dx_1 \wedge \cdots \hat{dx_i} \cdots \wedge dx_n$. Doesn't look promising.

Mike Miller

8:09 PM

Nah.

Balarka Sen

Perhaps we should really discuss this when I have done the book and have learnt more about differential forms. Thanks though.

Mike Miller

let's write $\Bbb R^n$ has $\Bbb R \times \Bbb R^{n-1}$. Write every form as $\sum g_I dt \wedge dx_I$. then 'integration over the first factor' gives you a $(k-1)$ form on $\Bbb R^{n-1}$

Vrouvrou

I need help to solve this problem: in $X=C([0,1],\mathbb{R})$ with the norme $||f||_2=\sqrt{\int_0^1 f^2(x)dx$ we define $T:X\rightarrow X$ by $Tf=gf$ for $g\in X$ How to prove that $T$ is continuous and how to find $||T||$ ? I find the constante of continuity $\sup g^2$ but i can't prove $||T||=\sup g^2$ have an idea @AkivaWeinberger ?

Balarka Sen

Oh yeah that's true. I was close, why the hell didn't I think of that.

$f$ is compactly supported. Duh.

Mike Miller

how does this play with closed forms?

Vrouvrou

8:18 PM

@DanielFischer hello

Balarka Sen

8:29 PM

Sorry, I was away. So let me try to understand your map a bit. Suppose $fdx + gdy$ is a 1-form, $f, g$ are compactly supported. What does integration give me? $\int_\Bbb R f(x, y) dx$? I guess I am confused about if your $t$ in $dt \wedge dx_I$ is fixed : in this case $dy$ cannot be written as $dx$ wedge something.

Vrouvrou

someone help me with my question ?

Mike Miller

i guess i should be more careful. let's work with $\Bbb R^2$, like you say. Define the 1-form on $\Bbb R$ by $\alpha'(x) = \int_{\Bbb R \times \{*\}} \alpha(x,\cdot)$.

so the $dy$ just dies.

Hippalectryon

@Vrouvrou What is ||T|| ? The subordinate norm of T with respect to ||_|| ?

Juan Sebastian Lozano

@MikeMiller doesn't this map give you a family of functions, then, since the $dy$ dies?

Hippalectryon

The continuity of T is rather clear via the definition of continuity btw @Vrouvrou

Juan Sebastian Lozano

8:38 PM

Which is consistent with regular antidifferentiation.

Balarka Sen

@MikeMiller so I suspected.

Juan Sebastian Lozano

But not great for our purposes, because then nothing maps to zero uniquely, for example

Balarka Sen

@Juan we're not antidifferentiating. it's just integrating with respect to the first coordinate over all of $(-\infty, \infty)$.

Vrouvrou

@Hippalectryon $||T||=\sup_{||f||_2=1}||Tf||_2$

Balarka Sen

which we can do as out form is compactly supported.

Mike Miller

8:39 PM

how am I differentiating?

Juan Sebastian Lozano

I don't really understand what is going on, now that re-read that question.

So, disregard it.

Vrouvrou

@Hippalectryon for the continuity i do: $||Tf||^2_2=\int_0^1 (gf)^2 dt=\int_0^1 g^2(t) f^2(t) dt \leq \sup_{x\in[0,1]} g^2(x)\int_0^1 f^2(t)dx$

Balarka Sen

I have to leave for now. I'll think about what's the deal with this. I'll be back later.

Thanks, @MikeMiller. See ya @Juan.

Juan Sebastian Lozano

Bye!

Hippalectryon

@Vrouvrou Well it's clear that $||T||\le\sup g^2$ right ?

Vrouvrou

8:48 PM

@Hippalectryon no $\leq \sqrt{\sup g^2}$

Mary Star

Hello!!

Suppose that $R$ is a commutative ring and $(x,y)$ is an ideal.
We have that $(x,y)^2=(x,y)(x,y)=\{ij \mid i\in (x,y) , j\in (x,y)\}=\{x^2, xy, yx, y^2\}=\{x^2, xy, y^2\}$.
Is the ideal $(x^2,xy, y^2)$ equal to the set $\{x^2, xy , y^2\}$ ? Or is it maybe $\{x^2, xy , y^2\}\subseteq (x^2, xy , y^2)$ ?

Juan Sebastian Lozano

@MikeMiller wait so the mappings from $\Omega^{i+1} \to \Omega^i$ stayed within $\mathbb{R}^n$, but your notion of integration creates a new form in a lower dimension, right?

Vrouvrou

@Hippalectryon yes

Hippalectryon

@Vrouvrou Yeah sorry that's what I meant (I can't edit the message above anymore)

Mike Miller

@JuanSebastianLozano Yes, it's a map $\Omega^{i+1}(\Bbb R^n) \to \Omega^i(\Bbb R^{n-1})$.

Vrouvrou

8:49 PM

ok @Hippalectryon

Hippalectryon

@Vrouvrou Now, can you find a function $f$ such that $||Tf||^2=\sup g^2,||f||=1$ ?

Vrouvrou

@Hippalectryon i tryed $f=1$ but it gives me $||Tf||\leq \sqrt{\sup g^2}$

Juan Sebastian Lozano

So you could make a diagram that looks like this, where $\psi$ is your map?

Vrouvrou

@Hippalectryon are you there ?

Hippalectryon

@Vrouvrou yes, writing down some ideas. It's a bit less "trivial" than I thought indeed. We probably can't find $f$ such that $||Tf||^2=\sup g^2$, but we can construct a sequence $f_n$ such that $\lim||Tf_n||^2=\sup g^2$

Vrouvrou

8:58 PM

@Hippalectryon that's a good idea

Hippalectryon

@Vrouvrou Let $a$ be such that $g(a)=\max g$. Now let $\epsilon>0$ and $f_n(x)=\begin{cases}\sqrt{\frac{n}{2\epsilon}},x\in[a-\epsilon/n,a+ \epsilon/n]\\0\text{ otherwise}\end{cases}$

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$Mathematics$ Mathematics