I am having some trouble finding any examples of compactly supported exact differential forms on $\mathbb{R}^n$. I have found $e^{\frac{1}{x^2 -1}}$ when taken to be zero everywhere except on the interval $(-1,1)$. What are some other examples?
@JuanSebastianLozano I don't know why you think compactly supported exact forms are so rare. I mean suppose you look at some smooth compactly supported function $f$ (there are a bunch of those). Then $\nabla f$ will also be compactly supported, not? That gives you a compactly supported exact 1-form on $\Bbb R^n$.
@JuanSebastianLozano I just gave you an argument which proves that if you know $N$ many compactly supported functions, you also know at least $N$ many compactly supported exact 1-forms.
@Juan If $f$ is a compactly supported function (you know one, don't you? $\exp(-1/(1-x^2))$ like you said), then so is $\partial f/\partial x_i$. Derivate ad infinitum to get infinitely many compactly supported functions.
I don't know that stuff beyond the basics but I can give you a relatively easy example.
If you have a chain complex, you can dualize that to get a cochain complex.
If you compute the homology $H_\bullet$ of the chain complex and the homology of the cochain complex aka cohomology $H^\bullet$ of the chain complex, it is natural to ask for what's the relation between $H^\bullet$ and $\hom(H_\bullet, \Bbb Z)$ is.
The universal coefficient theorem tells you $H^\bullet$ is essentially an extension of $\hom(H_\bullet, \Bbb Z)$ by a subgroup involving homology of one less degree. That subgroup is described in terms of the $\text{Ext}$ group.
Homological algebra helps you define and compute $\text{Ext}$. That in turn helps in concrete computations in topology.
@BalarkaSen that's amazing, actually. It is something that I kind of had in the back of my mind from seeing these relationships and their similarity to Linear Algebra, where I learned about dualizing and the relationship between $hom(v,f)$ and $v$ and so on, but I guess that shows me how useful those realionships are.
Do you know a good place to learn homological Algebra?
@TobiasKildetoft Yes. They are all called Resident Evil followed by a subtitle. It is a computer game but it generated a series of movies based loosely on the game.
Nice, latest paper went on arXiv today. We managed to do things way more generally than we had expected, leaving only 3 possible exceptions where we had originally not even expected to be able to do more than a few special cases.
Guys, anybody has a clue what spt(f) could mean? For a function $f:\mathbb{R^n} \rightarrow \mathbb{R}$, it will apparently give some sort of subset of the domain of $f$.
hello, one quick question: If all roots of a cubic polynomial are distinct and real then how can we say $f(a)f(b)<0$ , where $a,b$ are two roots of $f'(x)=0$
@ramsay Goes something like: the slope at the roots is nonzero. So, between the largest and smallest roots, the max is positive and the min is negative.
It's used in a lemma called "decomposition of 1" - let $\epsilon \in \mathbb{R}$, there exist functions $f_j$ from $\mathbb{R^n}$ to $\mathbb{R}$, $j \in \mathbb{N}$, that are non-negative, it's partial derivatives are continuous on the domain, $ diam (spt($f$)) < \epsilon$,
and for every $x \in \mathbb{R^n}$ the sum of $f_j(x)$ across $j \in \mathbb{N}$ is equal to $1$, and there exists a neightborhood $U$ of $x$, such that the set $\{j \in \mathbb{N} : spt(f_j) \cap U = \varnothing \}$ , is finite
I'm not really sure Mambo, I wasn't at the last few lectures... There's Gauss' divergence theorem, then there's a lot of lemmas (not sure if these lemmas are used for it though), then there's Green's theorem
@BalarkaSen in amazement you are going to watch captain America ??
Anyways @Balarka do you have a copy if nivan Zuckerman.... I have a few questions I want to ask you in it. I am on my phone so I cannot write stuff in latex
@Hippalectryon Comment vas-tu ? Si tu as un peu de temps, j'ai le problème de Cauchy suivant $y'=f(y)$ avec $y(0)=y_0$, je suppose que $f$ est C^1 et bornée de $\Bbb{R}$ dans $\Bbb{R}$ et $f(y_0)>0$. La question est de montrer que $y(t)$ tend vers $\inf \{z>y_0: f(z)=0\}$.
I know that the average of $3$ numbers is $88$ and the number is between $0$ and $100$. I need to find a new number that with it, the average will be $90$. This should be very easy but I don't find a way to solve it in a fast way.
@Hippalectryon Car sinon le TVI nous donne l'annulation en un point et le théorème de Cauchy (unicité) nous donne que la solution est la solution nulle.
C'est peut-être tout bête, mais je ne vois pas pourquoi c'est strictement croissant. Si on suppose l'inverse, alors en effet par le TVI on a un point $x$ tel que $y'(x)=0$. Q'est-ce qu'on en déduit ?
@Hippalectryon si deux solution coincident en un point de l'intervalle "solution" avec $f$ localement lipschitzienne alors elles coincident partout sur l'intervalle.
@Hippalectryon la contradiction vient du fait que si j'ai une solution qui s'annule alors l'unicité me dit que la solution est identiquement nulle, c'est impossible car on demande d'avoir $f(y_0)>0$
@JeSuis Bah avec les nouveaux programmes, ça tape dans le grand classique. C'était de la diffusion thermique genre effect de peau avec de la génération de chaleur par radioactivité suivi de méca Q bateau (effet tunnel, radioactivité alpha) pour phys 2. Phys 1 c'était d'abord des équations de sup suivies d'ondes dans les milieux matériels
@Shadock J'ai découvert en 3/2 que tout l'effort que tu passe à bien faire un TIPE paye assez peu en fait. Et que le faire (bien) rapidement ça donne le même résultat (en physique en tout cas; en chimie il y a des manips qui demandent du temps)
@robjohn I want to show the embedding $W^{1,p}(0,1) \subset C^0 [0,1]$.
So we pick a $u \in W^{1,p}(0,1)$ and want to show that $u \in C^0 [0,1]$.
Let $x_n \to x$. We want to show that $u(x_n) \to u(x)$.
Since $u \in W^{1,p}(0,1)$ we have that $u \in L_p$ and $u' \in L_p$.
And you told me to use the fact that $u(x+h)-u(x)=\int_{x}^{x+h} u'(t) dt$.
We know that a function $f$ is continuous in $x_0$ if $\forall \epsilon>0 \exists \delta>0$ such that $\forall y \in [x,x+h]$ with $|y-x_0|< \delta$ it holds that $|f(y)-f(x_0)|< \epsilon$.
par l'absurde, tu supposes $y$ non monotone, donc c'est à dire qu'il existe trois réels sur l'intervalle solution $I'$ tels que $a<b<c$ et que $y(b)$ est 1) y(b)>y(a) et y(b)>y(c) 2)y(b)<y(a) et y(b)<y(c)
pour chaque cas tu peux supposer, par le TVI blabla, que y(a)=y(c).
Ensuite tu prends une primitive de $f$ notée $F$,tu as $(F\circ y)'=f(y)y'=y'^2$
et le reste découle du fait qu'un final $y'^2=0$ sur [a,c] donc constante, absurde car $y(b)\ne y(a)$
For any $p \in S$, $T_pS$ has a canonical choice of basis $\{\partial g/\partial x, \partial g/\partial y\}$. I want to know the area of the parallelogram spanned by $\partial g/\partial x$ and $\partial g/\partial y$ in $\Bbb R^3$.
This is given by the Gram's formula: if $u, v$ are two vectors in $\Bbb R^n$ ($\Bbb R^n$ is equipped with the dot product here), take the orthogonal complement of the span of $\{u, v\}$. Choose an orthogonal basis of that, say $v_3, \cdots, v_n$. Then $\text{det}(u, v, v_3, \cdots, v_n)$ is the volume of the parallelopiped spanned by these, which is the same as the area of the parallelogram spanned by $u, v$ since $v_i$'s are all orthogonal to $u, v$.
But if $A = [u, v, v_3, \cdots, v_n]$, $\text{det}(A^2) = \text{det}(A^{\mathsf{T}}A)$ which is the determinant of the block matrix consisting of 4 blocks, with the first diagonal block $[u \cdot u, u \cdot v; v \cdot u, v \cdot v]$, the second diagonal block identity, and the rest of the blocks $0$. So $\text{det}(A)$ is $\sqrt{|u \cdot u, u \cdot v; v \cdot u, v \cdot v|}$.
So area of the parallelogram spanned by $\partial g/\partial x, \partial g/\partial y$ is $\sqrt{EG - F^2}$ where $E = \|\partial g/\partial x\|^2, G = \partial g/\partial x \cdot \partial g/\partial y$ and $F = \|\partial g/\partial y\|^2$. As a corollary area of the surface $S$ is $\int_U \sqrt{EG - F^2} dA_{xy}$.
@MikeMiller Not sure if you're listening, thus not sure if I should continue.
Right, so an interesting question would be how much about $S$ and about the embedding and about $\Bbb R^n$ did we actually need to define "area of $S$"?
We see that Gram's formula was an essential tool in this. To use the Gram's formula we needed an inner product on $\Bbb R^n$ to begin with, namely, a dot product.
In particular did we actually need dot product on the $\Bbb R^n$ my surface is embedded in? No, we just need a inner product on $T_pS$ for each $p$ to compute the area of the parallelogram given by $\partial g/\partial x_1(p), \partial g/\partial x_2(p)$. To do this for each $p$ in a consistent manner we need a smoothly varying inner product.
Alright, ok. So $M$ be a smooth manifold. If I choose some top dimensional form, $\omega$, on $M$, I can define the volume of $M$ with respect to that form as $\int_M 1 \omega$.
If I have a Riemannian metric $g$ on $M$, I can choose a "canonical volume form", by analogizing what we did for a surface above.
Without using Taylor’s Theorem show similarly that if $f$ is twice continuously differentiable in a neighbourhood of $x_0$ then $f$ can be written in the form: $f(x)=f(x_0)+f'(x_0)(x-x_0)+\frac{\psi(x)}{2}(x-x_0)^2$ where $\psi$ is continuous at $x_0$ and $\psi(x_0) = f′′(x_0)$.
Volume of codimension one submanifolds of a given manifold, I suppose. But since the submanifold gets a natural Riemannian metric from the top, we can just do the same computation, which is why I am confused by what you mean because I don't see why it's going to be distinct from getting volume of manifolds.
Sure, that's more or less the same thing if you have a Riemannian metric on the total space, which is of course a lot more structure than just a volume form.
I just didn't realize what you were trying to do.
Note that you never actually needed Gram's formula to define the top form on your surface...
\end{matrix}\right.$ And then showed that it is continuous at $x_0$, (I don't think I needed L'Hopital but the Questions says I might need it.) Then I rearranged and subbed in $f'(x) =\frac{f(x)-f(x_0)}{x-x_0}$
@MikeMiller Right, for my surface one just needs the normals. Then I can define the area 2-form on the surface by $n_3 dx \wedge dy - n_2 dx \wedge dz + n_1 dy \wedge dz$.
If $(M, g)$ is a Riemannian manifold, then I can define the volume form on $M$ in local coordinates by $\sqrt{|g_{ij}|} dx_1 \wedge \cdots \wedge dx_n$.
Ah, I have to show it's independent of the charts on $M$ we choose, so typing it takes some time.
I don't really like your definition, though it's perfectly correct. Can you give me a coordinate free one by specifying what it does to tangent vectors?
The idea is that if $f : V \to U$ is some diffeomorphism with $x_i = f_i(y_1, \cdots, y_n)$ then $dx_1 \wedge \cdots \wedge dx_n = |Df| dy_1 \wedge \cdots \wedge dy_n$ and $g_{ij}$ w.r.t. the new coordinates is $(Df) (g_{ij}) (Df)^T$. Straightforward calculation shows that my volume form leaves unchanged.
@MikeMiller Well, it eats the orthonormal basis of $T_pM$ and spits out the volume of the box spanned by them. That much is clear by Gram's formula.
I am having some trouble finding any examples of compactly supported exact differential forms on $\mathbb{R}^n$. I have found $e^{\frac{1}{x^2 -1}}$ when taken to be zero everywhere except on the interval $(-1,1)$. What are some other examples?
