@Axoren I disagree with one of your post in Meta. I hope don't disturb you, and I am saying then I am sorry. My opinion is that such OP tried help (do a comparison between the activity of such OP versus a moderator that also answered the question) to a younger student. Notice that this site democratizes the access to mathematics to all students. I believe that this message from me to you isn't 100% right, thus I am saying I am sorry, but I wanted said it. You are welcome.
One of my former colleagues allows his students to do test corrections, so he has to do two rounds of grading. Since I post solutions (unless it's senior-level courses), I absolve myself of such options.
Well, @0celo, OK ... pick topics you know you didn't learn and sit down with a decent book (I won't even insist on mine) and do a few problems. You can also look at one or two of my YouTube lectures if you want to get some intuition or see examples.
When you write $\sum dx_{j_1}\wedge dx_{j_2}$, I write down all possible terms. If $n=2$, I get $dx_1\wedge dx_1+dx_1\wedge dx_2+dx_2\wedge dx_1+dx_2\wedge dx_2$... which is a longwinded way of writing $0$.
If you indicate $\sum_{j_1<j_2}$, then I write down just $dx_1\wedge dx_2$.
But, in general, there are $\binom nk$ terms when I use increasing $j_\ell$'s.
Just check out the dedicated chat room---you can find the link on the page. I'm not doing anything different from what you meant, I just don't know at all what the terminology is, as far as I can tell.
On occasions, it's convenient to sum over all multiindices, @Balarka. Note that even if $I$ and $J$ are increasing, when I concatenate, $IJ$ is very unlikely to be. So even my definition of wedge came along before the increasing convention.
That's another reason why notation like $A_{(i} B_{j)}$ for (anti?)symmetrization is annouying: It's not clear whether one includes a division by $k!$ in it
what you'll get is (anti)symmetric either way, of course, but it doesn't necessarily map an antisymmetric thing back to itself. it depends on how it'bs being used, and that's tiresome
What I was saying is this: If you thought 3 was an eigenvalue, but then you solved for the nullspace of $A-3I$ and got only the trivial solution, that would tell you $3$ cannot in fact be an eigenvalue.
Yeah, too many things wrong to sort out and save yourself.
So here's an interesting conundrum to mull over, @Balarka. Invariance of domain tells us that a space-filling curve cannot be an injective mapping. How badly non-injective must it be?
@Balarka: In one of the constructions of a space-filling curve, there are only countably many points hit more than once and none gets hit more than 4 times.
Yes, Andy in his thesis commented that you can repeatedly divide the unit square into $k$ squares (Hilbert is $k=4$, Peano is $k=9$) and for any $k$ there are a countable number of points with more than one preimage and always at most 4.
@BalarkaSen So this is a specific definition for Linear Spaces, but if we don't utilize the linearity of the space and equip it with a group action, does that group action necessarily have the same property as the one defined with the linearity of the set in mind?
I don't know if I asked that question very well.
@Brody Glad I could help I guess, lol. I think that might actually have been the result of a tangent off of a question I asked.
Let me think of rewording that question, @BalarkaSen
I'm not 100% on selecting which mapping preserves algebraic structure for arbitrary groups and sets, but after the other night, I'm fairly confident with linear spaces and now I understand the necessary pieces for it.
@Axoren It's natural in the sense that it's the "largest possible" group that can act on $\Bbb R^n$. If $G$ acts on $\Bbb R^n$, then multiplying by $g \in G$ gives a linear map. A linear map is literally a matrix, so this realizes $G$ as a subgroup (I lie here a bit) $\text{GL}_n(\Bbb R)$.
It actually gives a map $G \to \text{GL}_n(\Bbb R)$, but whatever.
Nah, the point is you can go as far as representing every element of $G$ as a nxn matrix, but you might have a group action for which two different elements are represented by the same matrix.
Aka, the map $G \to \text{GL}_n(\Bbb R)$ might not be injective
