bio | website | |
---|---|---|
location | ||
age | 30 | |
visits | member for | 3 years, 5 months |
seen | 10 hours ago | |
stats | profile views | 2,962 |
Hello. I completed my PhD in differential geometry in Edinburgh. I have been reading MathOverflow for a year and a half two and a half years three and a half years four and a half years.
My primary interests are in $G_2$ and $Spin_7$ geometry. Mathematics is now a hobby for me.
Jul 21 |
awarded | Yearling |
Jul 17 |
comment |
Oriented volume and determinants: Circularity
Are you really using determinants? Why not define an orientation as the choice of a connected component of the torsor of real frames? |
Jun 5 |
comment |
What is the relation between two Riemannian metrics with the same Riemannian curvatures?
Relevant :mathoverflow.net/questions/100281/… |
Mar 21 |
awarded | Nice Answer |
Mar 12 |
comment |
Complex symplectic reduction
Shouldn't this be similar to hyper-Kahler reduction? |
Mar 12 |
comment |
Bispinors, polyforms, bilinears and supersymmetric manifolds
The first isomorphism is one of vector spaces, not algebras. The second just says that a spinor 'squares' to a form. This squaring map uses a bilinear or sesquilinear form on the space of spinors. There may be several admissible such forms, depending on the dimension or ground field. These give different isomorphisms. There are references but I'm stuck here right now. |
Mar 5 |
comment |
$Spin(7)$ as stabilizer of a $4$-form revisited
Right, thanks, of course, e.g. the volume form on any generic holonomy space. I have never really understood what's particularly special about the forms that do come from spinors vs those that don't, but I guess that's another question. |
Mar 5 |
comment |
$Spin(7)$ as stabilizer of a $4$-form revisited
"and conversely": in general, must every invariant form be the square of some invariant spinor (say, wrt some invariant bilinear/sesquilinear form on complex spinors)? |
Mar 1 |
comment |
Tensor calculus on the frame bundle
I guess there is now no need, after Peter's post. |
Mar 1 |
comment |
Tensor calculus on the frame bundle
The tangent bundle is an associated bundle to the frame bundle, so any tensor on the base can be 'lifted' to a function on the frame bundle with values in a representation that is equivariant under the action of the group. I'm pretty sure you can find this point of view in Sternberg's book Lectures on Differential Geometry, amongst others. Probably in Kobayashi-Nomizu I too. |
Feb 27 |
comment |
Octonions product: inversion in the right and identity in the left
Note that there's nothing special about this basis (and I prefer not to think the octonions have a 'standard' basis at all), and any Cayley frame (i.e. $G_2$-related to this one) will work as well. |
Feb 27 |
comment |
Octonions product: inversion in the right and identity in the left
I would be happy to, but it can be seen just by using the/a multiplication table, such as the one shown here: en.wikipedia.org/wiki/Octonion. You just need to check that $(((((((\mathfrak{i} \ \mathfrak{1}) \ \mathfrak{2}) \ \mathfrak{3}) \ \mathfrak{4}) \ \mathfrak{5}) \ \mathfrak{6}) \ \mathfrak{7})$ gives $\pm \mathfrak{i}$ (where $1, \mathfrak{1}, \ldots, \mathfrak{7}$ is the basis) for each basis member $\mathfrak{i}$, and that when you do it the other way around you get the opposite. It doesn't take too long to check directly. |
Feb 27 |
comment |
Octonions product: inversion in the right and identity in the left
@Jjm I think the standard basis does work, I checked. |
Feb 27 |
comment |
Octonions product: inversion in the right and identity in the left
Doesn't the standard octonion basis do the job? It looks like it to me, unless I'm out by a sign. |
Feb 17 |
comment |
Continuous isometries on Ricci flat compact manifolds
Using a Weitzenboeck identity, one can show that any Killing vector field on a compact Ricci-flat space is parallel. The special cases you mention above are locally irreducible, so have no non-zero parallel vector fields. So it's not about the topology, it's true locally. |
Feb 16 |
comment |
A question about horizontal lifts for an Ehresmann connection
A comment just for completeness: I think Ehresmann defined his connections to always be complete. And as soon as the distribution is complete, the bundle must be a fibre bundle. I guess the fact that 1) is true is just a generalisation of the fact that every homogeneous space is complete. |
Feb 14 |
comment |
dual of the Lie derivative
The Killing-Yano definition you write comes from the decomposition of the covariant derivative on forms into irreducibles. It has three parts. A form is Killing-Yano if the third part vanishes, leaving you with the equation above. |
Jan 29 |
comment |
How many geometric structures on manifolds are there?
Although of course, torsion-freeness is required for the holonomy to be $G$. Note also that any distribution (in the sense of Frobenius) is a structure, and these are studied in many contexts. |
Jan 25 |
comment |
Geodesics and Riemannian submersions
@MarcoGolla, but if a geodesic is horizontal at one point then it is horizontal at all points, no? |
Jan 20 |
comment |
Cohomology of a flat principal connection
The flat connection has a finite holonomy group $H$, and the leaves should be $H$-coverings of the base by the bundle projection. If $H$ is trivial, a leaf is a (parallel) section and $P$ is trivialised by it. |