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Jan
26
revised Show $\mathbb{R}^2$ with $dr^2 + \sinh^2r d \theta^2$ is isometric to the Poincare disc with $g = \frac{4(dx^2 + dy^2)}{(1-x^2-y^2)^2}$
Forgot the fraction.
Jan
26
suggested approved edit on Show $\mathbb{R}^2$ with $dr^2 + \sinh^2r d \theta^2$ is isometric to the Poincare disc with $g = \frac{4(dx^2 + dy^2)}{(1-x^2-y^2)^2}$
Jan
26
accepted Show $\mathbb{R}^2$ with $dr^2 + \sinh^2r d \theta^2$ is isometric to the Poincare disc with $g = \frac{4(dx^2 + dy^2)}{(1-x^2-y^2)^2}$
Jan
26
comment Show $\mathbb{R}^2$ with $dr^2 + \sinh^2r d \theta^2$ is isometric to the Poincare disc with $g = \frac{4(dx^2 + dy^2)}{(1-x^2-y^2)^2}$
Actually, I just used your map, found $dx$ and $dy$, and then showed the expression for $g$ reduces to $dr^2 + \sinh^2{r} d\theta^2$. Also, I think you should've plugged in $\frac{1}{2}\sech^2{(r/2)}\sin{\theta}$ as the second coordinate in your calculation of $g_{rr}$, right?.
Jan
26
comment Show $\mathbb{R}^2$ with $dr^2 + \sinh^2r d \theta^2$ is isometric to the Poincare disc with $g = \frac{4(dx^2 + dy^2)}{(1-x^2-y^2)^2}$
How did you get $g_{\theta \theta} = \sinh^2{r}$? What did you plug in as the vector coordinates for the argument for the quadratic form?
Jan
25
asked Show $\mathbb{R}^2$ with $dr^2 + \sinh^2r d \theta^2$ is isometric to the Poincare disc with $g = \frac{4(dx^2 + dy^2)}{(1-x^2-y^2)^2}$
Dec
2
comment Are maps inducing the same cohomology homomorphisms homotopic?
How would you show the statement in your first sentence, though? Also, by $H^{\ast}$ you mean the cohomology group, right?
Dec
2
comment $\alpha \wedge \beta = 0$ iff $\beta = \alpha \wedge \gamma$
@Peter, I don't think you can just write $\beta = \sum \beta_J \alpha_{j_1} \wedge \alpha_{j_2} \wedge \dotsc \wedge \alpha_{j_{p+1}}$ with $\alpha_{j_1} = \alpha$, because it seems that in this way you're already assuming what you're trying to prove, i.e. that $\beta = \alpha \wedge \gamma$, no? I mean, $\alpha$ itself is a $1$-form, so a linear combination of $\mathrm{d}x_i$, so why would you be able to write $\beta$ like that?
Nov
2
comment Product of manifolds & orientability
After seeing your original answer, I figured this would be the way to go about it, but thanks for the edit and the confirmation I was on the right track.
Nov
1
comment Product of manifolds & orientability
Could you maybe explain the "a little computation shows that $\{W_i\}$ with each $\phi_i$ correspondingly changed is a positive atlas of $M$" part a bit? What exactly is $\phi_i$ in your proof and what and how would you be computing?
Oct
11
comment Find the interior product of a basic p-form $\alpha = dx_1 \wedge dx_2 \wedge \ldots \wedge dx_p$ a and a vector field $X$
@MichaelAlbanese, thanks. So I'm guessing the first property is only true for functions $f$ then, right? That is, we don't have $i_X d \alpha = X(\alpha)$ when $\alpha$ is a $p$-form for $p > 1$. Because I guess this is what was throwing me off, as I was assuming that property did hold for any $p$-form.
Oct
11
revised Find the interior product of a basic p-form $\alpha = dx_1 \wedge dx_2 \wedge \ldots \wedge dx_p$ a and a vector field $X$
added 556 characters in body
Oct
11
asked Find the interior product of a basic p-form $\alpha = dx_1 \wedge dx_2 \wedge \ldots \wedge dx_p$ a and a vector field $X$
Jun
2
awarded  Popular Question
Mar
12
accepted Show that $\lim_{r\to 0} \frac{1}{r^2}\int_{C_{r}}f(z)dz = 2 \pi i\frac{\partial f}{\partial \bar{z}}(z_0)$
Mar
12
comment Show that $\lim_{r\to 0} \frac{1}{r^2}\int_{C_{r}}f(z)dz = 2 \pi i\frac{\partial f}{\partial \bar{z}}(z_0)$
@Dr.MV, you could have a $C^1$ function that does not satisfy the Cauchy-Riemann equations, and I don't see anything in those notes that would contradict that.
Mar
12
comment Show that $\lim_{r\to 0} \frac{1}{r^2}\int_{C_{r}}f(z)dz = 2 \pi i\frac{\partial f}{\partial \bar{z}}(z_0)$
@Dr.MV, no, but that's not true, as Umberto already mentioned in the comments to his answer (and as the book I'm using would also agree). $C^1$ is not the same as holomorphic.
Mar
12
comment Show that $\lim_{r\to 0} \frac{1}{r^2}\int_{C_{r}}f(z)dz = 2 \pi i\frac{\partial f}{\partial \bar{z}}(z_0)$
@UmbertoP., thanks for that example, and I don't know why I was thinking that would be the case in the first place.
Mar
12
comment Show that $\lim_{r\to 0} \frac{1}{r^2}\int_{C_{r}}f(z)dz = 2 \pi i\frac{\partial f}{\partial \bar{z}}(z_0)$
@Dr.MV, you're absolutely right. I thought the question required me to work without the assumption that $f$ is $C^1$ for that part, and I made that mistake because I also immediately jumped to the conclusion that $C^1$ implies complex differentiable. But assuming $f$ is $C^1$, that follows immediately from the definition of $\frac{\partial f}{\partial \bar{z}}$, since the function will then also satisfy the Cauchy-Riemann equations.
Mar
11
comment Show that $\lim_{r\to 0} \frac{1}{r^2}\int_{C_{r}}f(z)dz = 2 \pi i\frac{\partial f}{\partial \bar{z}}(z_0)$
@Dr.MV, is there another, easier way to show the above is true without Morera's Theorem? Also, Morera's Theorem doesn't really address that, because here we have an additional factor of $\frac{1}{r^2}$ and the limit, no? And thanks for the upvote :)