bio | website | maths.adelaide.edu.au/… |
---|---|---|
location | Adelaide | |
age | 57 | |
visits | member for | 4 years |
seen | Mar 24 at 9:25 | |
stats | profile views | 424 |
Mar 19 |
awarded | Yearling |
Mar 19 |
awarded | Yearling |
Dec 20 |
awarded | Good Answer |
Nov 2 |
comment |
looking for an identity for higher jet bundle $J^kM$?
You don't need $M$ to be compact. This is a local fact about the exterior derivative splitting the jet exact sequence. |
Aug 18 |
comment |
Is “Notices of the American Mathematical Society” available before 1995?
There wasn't room in the margins for the proofs. |
Jul 20 |
revised |
$E$ is a holomorphic vector bundle if and only if there is a Dolbeault operator $\bar{\partial}_E$
added 550 characters in body |
Jul 20 |
answered | $E$ is a holomorphic vector bundle if and only if there is a Dolbeault operator $\bar{\partial}_E$ |
Jul 4 |
comment |
Does a free action always induce a diffeomorphism?
Do you have a particular kind of infinite dimensional manifold in mind? Hilbert, Banach, Frechet etc |
Jun 27 |
awarded | Nice Answer |
Jun 8 |
comment |
Research topics restricted to students at top universities?
You should make this an answer instead of a comment. It's excellent. |
Apr 27 |
comment |
Automorphism of a Lie group which preserves a maximal torus is necessarily an inner automorphism?
Assume $G$ compact. If your result was true then every automorphism would be inner. Indeed if $\mu$ was an automorphism you can find $g$ such that $Ad_g \mu (T) = T$. But $Aut(G)/Ad(G)$ is the non-trivial group of automorphisms of the Dynkin diagram. |
Apr 5 |
comment |
A k-form is thought of as measuring the flux through an infinitesimal k-parallelepiped
Can you get hold of a copy of Misner, Thorne and Wheeler's Gravity ? They spend a lot of time explaining this point of view with great pictures. My memory (lost my copy in a postdoctoral move somewhere) is that what is measured is the flux of the $k$-form. |
Mar 19 |
awarded | Yearling |
Feb 21 |
comment |
What is the exterior derivative intuitively?
So the derivative of $\omega \colon X \to \Omega^p(X)$ at $x \in X$ is I guess the tangent map $T_x(\omega) \colon T_x X \to T_{\omega(x)} \Omega^p(X)$. How do you get the $p+1$-form? |
Feb 12 |
comment |
Pseudo-Differentialforms
@Nevermind. Remember the co-ordinate change formula for the integral ? What comes out is the absolute value of the determinant of the jacobian of the co-ordinate change function. If you don't have an orientation this is how things have to transform to get a sensible definition of integral. |
Feb 11 |
comment |
Why does the group act on the right on the principal bundle?
@Ben Webster. It gets even more fun if start worrying about identifying the Lie algebra on the group with left or right invariant vector fields and messing with the definition of connection on your left principal bundle ... |
Feb 8 |
answered | Tools for long-distance collaboration |
Feb 8 |
comment |
group of diffeomorphisms of a manifold
There is an explanation of the Frechet manifold structure on the group of diffeomorphisms of a manifold in Hamilton's paper: projecteuclid.org/… |
Feb 8 |
comment |
From Topological to Smooth and Holomorphic Vector Bundles
@Daniel, @Ricardo. Thanks for the reminders that $\mathbb R$ has lots of different (although diffeomorphic) differentiable structures. Something I had forgotten! |
Feb 8 |
comment |
From Topological to Smooth and Holomorphic Vector Bundles
Ricardo. Yes that was the question. Thanks. You could just call them equal couldn't you in such a case ? |