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Sep
20 |
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Examples of holomorphic Killing vector fields on compact Kahler manifolds
In case it helps you to know, any Killing vector field on a compact Kahler manifold is automatically holomorphic. |
Sep
18 |
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Finding the octonionic analog of the K3 surface, via (almost) hyperkahler geometry?
Yes, I meant $Spin_7$-structure in the sense of $G$-structure, not $Spin_7$-metric (just one of the types). |
Sep
17 |
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Finding the octonionic analog of the K3 surface, via (almost) hyperkahler geometry?
I haven't thought about the details, but are you sure you want almost-hyper-Kahler and not almost-quaternionic-Kahler geometry (or a Spin7-structure) a priori for condition 2 (it might turn out that it reduces further afterwards, as it does for K3)? |
Sep
15 |
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Sobolev Multiplication theorem for Fibre bundles
If $M$ is just a manifold admitting an action, how are you defining a metric on the total space $E(M)$ for your embedding? Also, you seem to be treating $\Gamma(X, E(M))$ as if it is a vector space. |
Aug
27 |
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Asymptotic $\int_M \mathrm{exp}[\mathbf{e}\left(n -\frac{t}{2\pi i}\right)] \left( 1 + \frac{5}{6} \mathbf{e}^2 \right)^{1/2} $ on quintic Calabi-Yau
I got something nearly identical to the RHS by just expanding the integrand as suggested, and I probably made a small mistake somewhere. Given that your manifold is Calabi-Yau, an example of such an $\mathbf{e}$ is a multiple of the Kahler form itself, that's how you're assured there is one. |
Aug
22 |
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When is a conformal class equal to a conformal orbit?
Isn't the first space finite dimensional for $n > 2$ and the second not? |
Jul
21 |
awarded | Yearling |
Jul
17 |
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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 |
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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 |
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Complex symplectic reduction
Shouldn't this be similar to hyper-Kahler reduction? |
Mar
12 |
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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 |
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$Spin(7)$ 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 |
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$Spin(7)$ 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 |
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Tensor calculus on the frame bundle
I guess there is now no need, after Peter's post. |
Mar
1 |
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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 |
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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 |
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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 |
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Octonions product: inversion in the right and identity in the left
@Jjm I think the standard basis does work, I checked. |
Feb
27 |
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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. |