@Astyx I found this definition while going through MIT reading material of graph theory. According to the reading material, If the graph is a bipartite degree constrained, there must exist a matching that covers L.
@Astyx A matching in a graph, G, is a set of edges such that no two edges in the set share a vertex. A matching is said to cover a set, L, of vertices iff each vertex in L has an edge of the matching incident to it
@Astyx Okay so. I was considering it to be an equivalence and I had few examples in my mind to contradict it. Yeah, I did not read it well and mistakenly considered a implication to be an equivalence.
But it means, Degree constrained condition could not be used to check if a graph has matching or not. It means I will have to use Hall's theorem ( that eventually checks every subset of graph and thus time consuming ).
I think all they mean to say, If a graph has degree constrained condition - matching exist to cover L. But If a graph is not degree constrained, There may or not may not exist matching that cover L.
I think taking Guillemin's differential topology course hooked me. Then I took some several complex variables and complex manifolds as a senior and it fascinated me.
ah, that's the deep one — that's equality of mixed partials ... but that's why $d$ is so important. From the perpective of homological algebra, it gives you a complex. From the perspective of PDE, it gives you integrability conditions.
I don't usually think it about axiomatically. I define it in coordinates and then check it's well-defined. You can do this directly (but it's hard) or you can verify the properties from the definition and then argue uniqueness.
@LucasHenrique it was in reference to a previous conversation I had with Mike where I expressed distaste at definitions which are axiomatic when others exist that give more intuition as to why you would be interested in the object to begin with.
So while axioms play a large role in mathematics, the point I would try to get across is that so does "intuition" and "inspiration".
This is not really an answer, but you want the wedge to be a cup product for De Rahm cohomology. But then one asks why is there a $(-1)^p$ in cup products and I guess that fixes some orientation
@TedShifrin I don't think I know how I should think about differential forms to begin with. Thinking about them as multilinear forms that eat tangent vectors doesn't seem very helpful.
@LeakyNun You can then integrate along a curve in the expected way, much like the physicist's idea. # intersections gives the value of the integral $\int \alpha$ over the curve.
How can I describe a base of $\Bbb R^{\Bbb N}$? Since every vector must a be a linear combination of finitely many vectors, the way I'd construct a "basis" makes $\operatorname{dim} \Bbb R^{\Bbb N} = \Bbb R$, which is wrong (I think...)
@LeakyNun if i want to teach someone i work in $n = 2$ or $3$. If they need to know arbitrary dim i will be going through the multilinear algebra which imo ppl have to know better than the back of their hand (at least i need to)
In my book/course I had no issue doing $\Bbb R^n$. I mean I show examples with $n$ not too large, and eventually I go back and tie it into grad, div, curl with $n=3$.
I don't see what the multilinear algebra is ...
I mean, I defined forms as alternating multilinear guys in the first place, but I don't see why $n$ matters.
Well, I should be careful. I defined everything in terms of $k\times k$ determinants.
@TedShifrin yeah. I'm using facts about consistency and non-singular matrices and I also have that $\mathbf{R}(A)^\perp = \mathbf{N}(A)$, $\mathbf{N}(A^T) = \mathbf{C}(A)^\perp$ and obviously the column-row equivalence of the transpose.
In practice in my course, the only forms on $M\subset\Bbb R^n$ that showed up were restrictions of ambient forms. Although there were a few (hard) exercises—like for orientability—where you have to use partitions of unity to glue together things intrinsically. But hardly any students looked at those.
@TedShifrin i mean ok i just meant going through doing exterior algebra formally w people and connecting that to all your traditional determinant formulas and stuff, but i guess u dont need to do that with newbies but i guess im in a crazy upside down world where the average person i meet is less likely to be cool w expressing things in terms of determinants of minors than in terms of the exterior algebra
@TedShifrin I don't understand how the claim "we know the there exists a system of restriction equations [...] so $b \in \Bbb R^m$ is in $\mathbf{C}(A)$" is necessarily true. I read everything again and still no clue.
actually the grad student who TAed my analysis class way back in freshman year said forms were "pretentious parallelograms" and that i found a little amusing
@TedShifrin when the students then go on and take a course dealing with manifolds, how well do they grapple them having done stuff like your first course?
one thing that's only tangentially related though @Ted, the physicists in my GR class are like linear algebra agnostic, and that is truly a pain when so many GR books go to great lengths to try to avoid spelling out what is actually happening when you do all these index calculations
my book doesnt even tell you that covectors are living in the dual space to vectors, just stuff about them eating each other, and man these grad students were confused as hell when our prof was explaining this without defining anything
@Lucas: So the key thing is that observation that if $W\subset V$, then $V^\perp\subset W^\perp$. I remember thinking it was cool when I figured out this proof.
yeah i mean i always found the best way to be clear on all these big index computations in einstein notation is to be very very clear in your mind about spaces and where everything lives on the tensor algebra and where isomorphisms are being used via the inner product or w.e.
but then the physicists just define everything in terms of components and never tell you anything about where anything lives and blunder through computations like madmen and then they get the right answer seemingly accidentally
I would like to prove that if $\lVert f- f_n\rVert _p\rightarrow 0$ then $f_n \rightarrow f$ in measure. My attempt: let $\delta >0$ and define $D_n = {x\in X: |f_n(x) -f(x)| \ge \delta}$ now assume by contradiction that $\limsup_{n\rightarrow \infty} \mu (D_n) \neq 0$ that is, there is a subsequence $ D_{n_k}$ such that $\lim _{k \rightarrow \infty} D_{n_k} = c > 0$ (might be infinity) . by monotonicity of integral we get $$\int _\mathbb{X} \vert f-f_{n_k} \vert ^p d \mu \ge \int_D_n_k |f-f_n_k|^p d\mu \ge \int_D_n_k \delta^p d\mu = \delta \cdot c $$ since it holds for every $k$ we get th…
Have you worked examples back in Chapter 1, @Lucas? You make the augmented matrix with $b$ on the right, and row reduce to echelon form. Any row of zeroes gives you an equation $c\cdot b = 0$.
i remember in my minimal surfaces class last year we were working through some PDEs u get on the second fundie and no mere mortal can keep track of what's happening on the level of spaces and you just gotta figure out the PDE intuition to work w em
@TedShifrin yes, I remember that. What I mean is that I don't even know where you're seeing a matrix or why you started to talk about constraint equations in the proof.
Imagine Yau's proof of the Calabi conjecture with third- or fourth-order estimates ...
@Lucas: Now you're being silly. We're talking about the column space of $A$ to start with. That's the matrix. And we're thinking of the column space as being characterized by those constraint equations.
@MikeMiller Hi mike, I spend some time on your proof suggestions, here are the things unclear to me: the map $F \colon H \times L_1 \to M$ is such that its differential surjects on the intersection $T_xL_1 \cap T_xL_2$. Why does it implies that $F$ is transverse to $L_2$? I think I misunderstood something here.
@MikeMiller aah sorry, I misunderstood then ahaha. ok I'll elaborate my thoughts here
@MikeMiller So what's unclear is why $F$ is transverse to $L_1$. As far as I understood, the assumption on $H$ is some finite dim vector space of hamiltonian v.f. s.t. $dF$ surjects on the intersection of $T_xL_1 \cap T_xL_2$. Are we adding the directions to make it transverse to $H$ "artificially"?
How so? I'm thinking of the row-reduction algorithm, too. I remember that the parametric description has the ugly non-zero vector if you have a non-zero vector on the other side
That's the main trouble I'm experiencing. The matrix $A$ has row vectors $A_j$ and column vectors $a_j$; I don't understand where the $c_j$ came from. What I think we're trying to do is to make every column space element fit the same homogeneous equation, and this homogeneous equation will be $Cb = 0$.
@TedShifrin I did. My problem was really the proof. I'm not sure if the translator changed it so much that it became unclear or I'm just really really stupid.
I agree. The problem is that (at least for me), it was not obvious that you should get this from $A^Tx = b$, since there could be more facts about the column space to be used, and not the good ol' echelon algorithm.
@Leaky let say we don't know the number of vertices but only the depth, would it be correct to say that the depth has no relation to the number of edges?
Well, then you no longer have the geometric meaning of transpose. It rather means the map on the dual space. But you still have the subspaces, just not the orthogonalities.
The right interpretation then becomes pairing $V^*$ and $V$. To a subspace of $V$ you associate its annhilator in $V^*$.
@MikeMiller I found a group satisfying the requirement we were discussing the other day. Specifically $L=\langle a,b\vert abab^{-1}=bab^{-1}a\rangle$ where $\langle a\rangle$ is not normal in $L$ but $\langle a,bab^{-1}\rangle$ is normal. A subgroup of the symmetric group $S_6$. Allow $a=(12)(56)$ and $b=(135)(246)$, then $bab^{-1}=(12)(34)$ and $bbab^{-1}b^{-1}=(34)(56)$.
