New answers tagged

0 votes

Is my proof of the equation $d=dx^i\wedge \partial_i$ correct?

Regarding the title Actually the answer can be found in the book that I quoted: In my proof I used the defining properties of the differential, see theorem $2.53$. If we look at the proof of that ...
  • 2,197
0 votes

Reference request: bundle of holomorphic differentials over Teichmüller space

This is a very good question. I ran into the same issue some months ago, and found it very difficult to find a reference. I warmly recommend Bers' 1960 paper "Holomorphic Differentials as ...
  • 4,616
1 vote
Accepted

Generator of $H^1(S^1)$ via integration of a bump $1$-form on $S^1$

There is a map $$ H^1(S^1) \to \mathbb{R} $$ given by $\omega \mapsto \int_{S^1} \omega$. A priori this might be the zero map, but they've proven that it isn't by constructing a closed one-form with ...
  • 26.4k
2 votes
Accepted

Integration of $\mathbb{d}\mathbb{d}^{*}+\mathbb{d}^{*}\mathbb{d}$ on manifolds with boundary

As you know, $d:\Omega^k(V)\to\Omega^{k+1}(V)$. Since $\dim V=p$, we get $\Omega^{p+1}(V)=\{0\}$. Considering that $\iota^*\omega\in\Omega^p(V)$, it follows that $d\iota^*\omega=\iota^*d\omega\in\...
  • 1,261
0 votes

differentiating an expression in differential form

The issue isn't with the mathematics but with the physical implications following from the expression I am working with. For the benefit of future audiences: Note that (some included for completion): $...
0 votes

differentiating an expression in differential form

I think it follows from the fact that $pv = \frac{PV}{m}$ so if you consider the differential of that, you get, using the product rule that $ d(pv) = d(\frac{PV}{m}) = dP \cdot \frac{V}{m} + \frac{P}{...
  • 66
2 votes

How does a trivial canonical bundle for a 2-dimensional complex surface imply the existence of a nowhere vanishing 2-form

You are confused about what it means for the canonical bundle $K_X$ to be trivial. It does not mean that $H^0(X,K_X)=\{0\}$. It means that $K_X\cong\mathcal{O}_X$. In particular, when $\dim_\mathbb{C}...
  • 1,261
1 vote
Accepted

Absolute value in pseudo Riemannian volume form

It's just not true that $\det(A)^2=\det(g)$ in general. Think of flat space $\mathbb{R}^{s,t}$ with signature $(s,t)$, with Cartesian coordinates and $A$ being the identity matrix. I'm not sure how ...
1 vote

Derivative of identity map and moving forms

This discussion is confused. Tildes appear and then disappear. A $1$-form is scalar-valued, unless you specifically speak of a vector-valued $1$-form or, more generally, a $1$-form with values in a ...

Top 50 recent answers are included