@TedShifrin why would one expect that the second extension of the cauchy completion of the rationals would ever be algebraically closed? it just feels like a gift from god to me
@CaptainBohemian: Yes, multiplication by $i$ is a 90º rotation, geometrically. We're used to what $i$ means when we work with $\Bbb C$, but with vector bundles there's no well-defined coordinate system as you move around.
@TedShifrin is it the case that if there is well-defined coordinate system when you multiply $i$ as you move thrroughout a manifold, that means the manifols has a complex structure?
Not a well-defined coordinate system, @CaptainBohemian — that would require a trivial tangent bundle. But a well-defined, smooth notion of $i$, yes. There's a bit more to get a complex manifold. There is an integrability condition that will give you holomorphic charts rather than just smooth ones.
@Balarka: That was so weak I didn't even deem it a pun.
they have a midterm coming up in a week. We're doing polar coordinates now. After the midterm we start complex numbers and then eventually linear algebra. We'll see how they do on the exam.
Another question, on the RHS of the main equation there, I'm sure $f(p)vg + g(p)vf \not\in \mathbb{R}$, since $vg$ and $vg$ are just multiplication of linear maps
Ohhh so the linear map $v : C^{\infty}(M) \to \mathbb{R}$ is a derivation at $p$ if it satisfies $v(f\cdot g(x)) = v(f(x)\cdot (g(x)) = f(p)\cdot v(x)\cdot g(x)+g(p)\cdot v(x) \cdot f(x)$?
Basicaly, you can identify vector fields on $\Bbb R^3$ either as either 1-forms or 2-forms (which is a coincidence that can only happen in 3 dimensions, the important thing is the Hodge-$*$ operator here)
for differential forms you have a derivative which turns out to generalize grad curl and div under this identifications
@TedShifrin It's like Ted has gone. But I want to reply the last message of Ted. I think I can imagine the definition of complex structure from that of spin structure: yes. a global frame should not usually be required for these structures. but it's like for a noncompact spacetime manifold the exitence of a spin structure entails a global frame.
I am trying to factor $N$ by using Dixon's factorization method, so I am looking at the equation:
$$a^2\equiv b(\mod{N})$$
If I be able to find $b$ that is a perfect square, I will be able to factor $N$.
But while looking at the values of $f(x)$ I notice something very interesting
$$f(x)\equi...
I am trying to factor $N$ by using Dixon's factorization method, so I am looking at the equation:
$$a^2\equiv b(\mod{N})$$
If I be able to find $b$ that is a perfect square, I will be able to factor $N$.
But while looking at the values of $f(x)$ I notice something very interesting
$$f(x)\equi...
