That's the $k$ you want (he called it the signature of $\omega$, not sure if that's standard terminology or not). Then you can show using winding number that this gives you the right thing. Then it's easy to show that the difference of two forms is exact iff they have the same signature, and obviously the signature of $k\omega_0$ is $k$, so you're done.
@0celo7 jesus
Right now the guy speaking in our seminar is talking about measures which satisfy differential equations ("in the sense of distributions")
@Daminark In short, there's a pairing $H_{dR}(S^1) \times \pi_1(S^1) \to \Bbb R$ given by taking $(\omega, \gamma)$ and spitting bars about $\int_\gamma \omega$. Every form $\omega$ is completely determined by the value of $(\omega, \gamma_0)$ where $\gamma_0$ is the generating loop in $\pi_1(S^1)$, hence the isomorphism with $\Bbb R$. On the other hand, every homotopy class of loops $\gamma$ is completely determined by $(\omega_0, \gamma)$ where $\omega_0$ is the angle form
Ah I see. In analysis we computed $\pi_1(S^1) = \mathbb{Z}$ using the path lifting business. I don't think I could fully justify out the details at the time as to why you had it but in the smooth case and visually it was clear enough I think, since I had full points
@0celo7 Ah that makes sense. This seminar is very much not a GR one, I'll say that much :P
By the way, how's this for an elementary characterization of a property that distinguishes $\Bbb R^2$ from $\Bbb R^3$? (Without referring to homology/homotopy):
In $\Bbb R^2$, there exists two closed sets $A$ and $B$ whose complements are connected, with $A\cap B$ being a point and $\Bbb R^2\setminus(A\cup B)$ disconnected
@AkivaWeinberger That's just one perspective. That the sapient lifespan has a well-defined start and an endpoint is also very interesting in it's own, and much more than something you'd classify as "poison"
Could someone tell me if this is correct: If $f(x)=0,0<x<1$ and $f(x)=1,1<x<3$ then $lim_{x\to0^+}f(x)=1$,$lim_{x\to0^-}f(x)=0$, $lim_{x\to1^+}f(x)=1$,$lim_{x\to1^-}f(x)=0$,$lim_{x\to3^+}f(x)=0$,$lim_{x\to3^-}f(x)=1$ ?
Let $f\colon\mathbb R^n\to\mathbb R$ be a differentiable function, and $0\in f[\mathbb R^n]$. Assume $\nabla f\neq 0$ on $f^{-1}(0)=M$. Let $a\in\mathbb M$. Then $M$ is an $(n-1)$-dimensional manifold. Show that the tangent space $M_a$ of $M$ at $a$ is given by $$ M_a=\{v_a\in\mathbb R_a^n: v_a\perp\nabla f(a)\}. $$
these were my thoughts: I can see intuitively why this should hold, because our level set will never go in the direction of the gradient, and therefore the only subspace that won’t be part of $M_a$ is the one that lies perpendicular to the gradient. But how can I show this a little bit more rigourously? So I would somehow need to show that $\langle v_a,\nabla f(a)\rangle=0$. I don't really know what coordinate function to use, because $M$ is given implicitly..?
ncatlab.org/nlab/show/simple+object can someone give a simple counterexample to prop 2.1 when the field is not algebraically closed? (An example through quiver representations would be also perfect) Btw, as I understand it, Hom(x,y) is a vector space, not the objects (just a semi-question about the meaning of enriched)
But yeah I saw it as curves in the sense of, you can literally take a curve and take its derivative as one in R^n. That definitely strikes me as the most intuitive one
