And @Don'tdisturb ; Eh, thanks for the compliment? I must admit it’s refreshing to hear, because my boyfriend keeps joking(?) how ugly I am:P Anyhow, there’s no shortage of girls at our mathematical faculty, so I’m surprised there’s not many more of us online - though it is true that the ‘die-hard’ mathematicians seem to be boys in the end, at least from what I see at university and online communities.
I personally find the gender differences slightly fascinating where it concerns mathematics(/science)
@Don'tdisturb I can imagine that someone big like you might not find a lot of satisfaction in the chat, but I've personally learned a lot in the chat and I've been helped a lot with homework exercises
Then if you color the vertex with three colors such that the vertex of the bigger triangle have different colors, the vertex lying on its edges are the color of one of the extremities of this edge, and the rest how you like
@ShaVuklia I'm a bit big, I need to lose some weight. Your intuition is good. Chat can be pleasant, especially with more girls around. :-) I'm sure you got the help you needed.
I've got this space $E$ which fibers over $B$, $\pi:E\to B$ with fiber $\Bbb P^1$ and I want Pontryagin classes of the total space. I can probably get the classes of the base, and I can use that $TE\cong \pi^*(TB)\oplus T\pi$ where $T\pi$ is the vertical tangent bundle or whatever it's called. I know there is no torsion so I have $p(E)=p(B)p(T\pi)$. Now all I need to do is figure out $p(T\pi)$.
I had a previous situation where my space happened to be a projectivized bundle and there I could use that I had a class which restricts to the hyperplane class on each fiber to understand the (in that case Chern) classes of the vertical bundle
@Don'tdisturb Haha! I would have, but I'm not really impressed by the reason why they chose the date for $\pi$ day, because I don't like to approximate mathematical constants, so I skipped:d
@ShaVuklia I wonder if it's fair to let other constants not celebrated like Euler's number, Euler-Mascheroni constant or Catalan's constant. I think you're right to some extent.
@Don'tdisturb Yea, I don't know why $\pi$ is so popular. Maybe because it can be implemented in easy maths/science calculations at high school, so most people know about it? Euler's number comes much later in the curriculum. But that shouldn't be a measure for popularity/celebration, of course.
Yes, Danu, I saw your answer. In general, I can only think of trying to give a difference SES involving the bundle. We might figure out transition functions to see how it's $TF$ twisting along the base. I'll have to ponder.
@TedShifrin So eh... Am I confused or is the Pontryagin class of a complex rank 1 bundle not always trivial (in fact, trivial iff the bundle itself is trivial)?
First was basically a subset of the proof of the spectral theorem on Hilbert spaces, second was proving that the range of a compact operator was closed if and only if the rank was finite, third was proving that the spectrum of a self-adjoint operator was real, fourth was about the space of convergent sequences, and fifth was proving that the set of $4\times 5$ matrices of rank $2$ was a manifold, and what its dimension was
Unfortunately on the last one, I never ended up figuring out why those equations were independent, so once again, on this test, I had to use the flimsy and prob incorrect point from my homework
Yeah, and the part that I was supposed to figure out on my own was one that I only figured out a while after the pset was due, and I didn't remember how that worked, only the nonsense argument about how changing one entry shouldn't affect the others
Hm, my book says that if we take an element $a\in\mathbb Z/n\mathbb Z$, then we can write $a=qn+r$ ($q\in\mathbb Z, 0\leq r<n$). However, I'm confused, because $\mathbb Z/n\mathbb Z$ doesn't consist of integers, yet its elements are entire sets of elements with the property that their remained by division of $n$ is the same?
Why does my book write this $a\in\mathbb Z/n\mathbb Z$ as an integer then?
@ShaVuklia Do you understand why it's the same thing whether they write it as $a$ or $\overline{a}$? (Even though the latter is correct in the context of the group we're talking about)
I think there's a subtle mental shift between writing $\bar a=\bar b$ and $a\equiv b$, where in the former we have equality of sets of integers and in the latter we have equivalence of integers
Because $a$ is an integer and $\bar a$ is an infinite collection of integers (the equivalence class). One of my standard final exam proofs in ring theory was for then to prove that if $\langle a\rangle + I = R$ ($I$ a proper ideal), then $\bar a\in R/I$ was a unit. That problem tested understanding.
@Ted I get that, but I don't get why it's such a big deal as long as you know what you're doing, since the operations are compatible with the equivalence class (which I know you already know)
DogAteMy: In my algebra course, I started just with the latter, and only defined the ring $\Bbb Z_n$ after they were used to working with integers themselves.
@Astyx: Because students do not know what they're doing for the first semester!! :)
I don't think its really necessary to look at how Leibniz or Newton thought about derivatives or how Heaviside thought about dirac deltas and (what would become) weak derivatives
These things are really cool and fun, but a direct presentation is more helpful when learning
