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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.

My primary interests are in $G_2$ and $Spin_7$ geometry.


Oct
13
comment Manifolds admitting flat connections
Related to Q1: mathoverflow.net/questions/91852/…
Oct
12
comment Does local reducibility imply global reducibility of universal covering?
You at least need completeness. Otherwise just take any non-rectangular simply connected region in the plane.
Sep
24
awarded  Autobiographer
May
21
comment What is an element of an iterated tangent bundle?
@Ryan, e.g. the second order tangent bundle of $\mathcal{M}$ is the vector bundle over $\mathcal{M}$ whose fibre at $p$ consists of $2$-jets at $p$ of curves through $p$. It has rank $n(n+1)/2$ for $n = dim \mathcal{M}$. Just as for the first order tangent bundle, there is also an algebraic definition if you are so minded.
May
21
comment What is an element of an iterated tangent bundle?
Whatever name you choose, it shouldn't allow confusion with elements of higher-order tangent bundles (as opposed to iterated ones).
Apr
26
comment Relation between kahler potential and Hermitian metric
If $h$ is the Hermitian form, what is the meaning of $\log h$?
Jan
4
comment Metalinear frame bundle on sphere or $\mathbb{C}P^n$
What do you want the complex vector bundle $P$ to be for your three spaces? And, given such a $P$, is there not often a choice of such structure?
Jan
3
comment Holonomy group of a fiber bundle
I agree with Igor; $M = \{\bullet\}$, $N$ any flat space with non-trivial holonomy and $G$ trivial gives a counterexample.
Jan
3
comment Holonomy group of a fiber bundle
Isn't it obvious when using restricted holonomy?
Jan
3
comment Holonomy group of a fiber bundle
@Igor, what is the full holonomy group of $[0,1] \times [0,1]$ with opposite sides identified in the usual way?
Jan
3
comment Holonomy group of a fiber bundle
@Igor, perhaps we are using different definitions but, I do not consider simple connectedness necessary. For example, I consider a flat torus to have trivial holonomy.
Jan
3
comment Holonomy group of a fiber bundle
The reason I suggested the above definition is that it satisfies the properties mentioned; all fibres are isometric. I think "there is no holonomy in the base direction" means $Hol(B)$ is trivial.
Jan
3
comment Holonomy group of a fiber bundle
Do you mean a Riemannian submersion $\pi : M \to B$ whose fundamental tensor fields (as defined in O'Neill's classic paper on Riemannian submersions) $A$ and $T$ both vanish? The vanishing of these fields implies $\pi$ is locally a Riemannian product, as you want. I think your question is about the exactness of $1 \to Hol(F) \to Hol(M) \to Hol(B) \to 1$, as the result follows if $B$ is flat.
Jan
3
comment Teaching homology via everyday examples
Arbitrage, no? __
Nov
19
comment Hyper-Complex and quaternionic Kahler Geometry
The link has just been useful, thanks @Vitali.
Oct
15
comment Para-Complexification of Lie Groups
An almost-paracomplex manifold is a smooth real even-dimensional manifold with a paracomplex structure (an endomorphism defined as above) defined on its tangent bundle, and morphisms are smooth maps preserving it. I can't remember if there is a notion of integrability.
Oct
10
comment Does this cross-product norm inequality hold?
For completeness, I believe I missed a couple of critical points in my reasoning above. However, I think these can be dealt with too.
Oct
9
comment Does this cross-product norm inequality hold?
What is your vector product on $\mathbb{R}^n$?
Oct
9
comment Does this cross-product norm inequality hold?
I seem to have missed the editing window. It should read "the constant on the left". Also, the situation that $x$ and $z$ are orthogonal is not trivial, but does fit in with the rest of what I said.
Oct
9
comment Does this cross-product norm inequality hold?
... Adding the gradient vector fields of $y \to ||x \times y||$ and $y \to ||y \times z||$ together, it is easy to see that the sum is zero at exactly four points: there are maxima where the two 'equators' meet, and minima at the points equidistant from $x$ and $z$ on the great circle containing them. The total function has the same value at each, so just look at the point in between $x$ and $z$. The result follows because $sin 2\theta < 2sin \theta$. A picture would help.