bio | website | |
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age | 29 | |
visits | member for | 2 years, 8 months |
seen | 6 hours ago | |
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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 |
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Manifolds admitting flat connections
Related to Q1: mathoverflow.net/questions/91852/… |
Oct 12 |
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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 |
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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 |
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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 |
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Relation between kahler potential and Hermitian metric
If $h$ is the Hermitian form, what is the meaning of $\log h$? |
Jan 4 |
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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 |
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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 |
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Holonomy group of a fiber bundle
Isn't it obvious when using restricted holonomy? |
Jan 3 |
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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 |
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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 |
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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 |
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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 |
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Teaching homology via everyday examples
Arbitrage, no? __ |
Nov 19 |
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Hyper-Complex and quaternionic Kahler Geometry
The link has just been useful, thanks @Vitali. |
Oct 15 |
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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 |
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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 |
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Does this cross-product norm inequality hold?
What is your vector product on $\mathbb{R}^n$? |
Oct 9 |
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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 |
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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. |