bio | website | |
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age | 28 | |
visits | member for | 2 years, 2 months |
seen | 4 hours ago | |
stats | profile views | 2,888 |
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.
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. |
Oct 9 |
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Does this cross-product norm inequality hold?
What about the following: Fix $x$ and $z$ and consider the right hand side as a function of $y$; a map $S^2 \to \mathbb{R}$. If we can show that its minimum is larger than the constant on the right, we're done. It's easy to see that the gradient (which for simplicity we can consider to be a vector field using the usual metric on $S^2$) of $y \to || x \times y ||$ is zero at the poles ($\pm x$) and on the equator, and everywhere else points towards the equator. We can assume the vectors $x$ and $z$ are at an angle bigger than $0$ and less than $\pi/2$ (trivial cases)... |
Oct 7 |
awarded | Constituent |
Oct 2 |
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Does every manifold have a flat connection?
@Misha, why not? There is a well-defined two-form with values in the adjoint bundle. |
Oct 2 |
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Does every manifold have a flat connection?
@Peter, there isn't a zero connection. Unless you mean something perculiar? Either way, your conclusion isn't right. |
Oct 2 |
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Does every manifold have a flat connection?
Very close question: mathoverflow.net/questions/91852/… |
Oct 2 |
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Spin manifolds with one parallel spinor
In the Riemannian case, we can assume the manifold is simply connected. The classification given in "Mckenzie Y. Wang, Parallel spinors and parallel forms" shows that the space cannot be irreducible. |