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comment |
Spin structures on the Grassmannians
One can compute the first Chern class of the complex Grassmannian of $k$-planes in $C^N$ as follows. For the purpose of computing its Chern character, the holomorphic tangent bundle $T$ is a product of the tautological rank $k$ bundle $V$ and (formally) $N−V$. This leads to the formula $c_1(T) = (N−2k)v$ where $v=c_1(V)$ is a generator of $H^2$. So it looks like the Grassmannian is spin iff $N$ is even. |
Dec
4 |
comment |
Is the cotangent bundle to a Kahler manifold hyperkahler?
This answer is pretty complete, but it is worth reading the paper of Calabi in Ann. Ec. Norm. Sup. 12 (1979) for an explicit construction of the HK metric on the cotangent bundle of complex projective space. The precise form of the metric is not obvious, and his approach (subsequently generalized to other HSS's) was to find the Kaehler potential. As in applications of Yau's theorem in the compact case, the HK metric is indeed compatible with the underlying holomorphic symplectic structure. |
Sep
24 |
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6 |
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Aug
2 |
comment |
A geometric interpretation of the Levi-Civita connection?
Yes, but any such section $s$ that passes through $p\in P$ is unique to first order. If we set $a_{ijk} = \Gamma_{ij}^r g_{rk}$ then $s$ is tangent to $P$ at $p$ iff $a_{ijk}+a_{ikj}=0$, which forces the Christoffel symbols to vanish at the point in question. |
Aug
1 |
answered | A geometric interpretation of the Levi-Civita connection? |
Jul
29 |
answered | Do hyperKahler manifolds live in quaternionic-Kahler families? |
Jul
29 |
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Jul
27 |
answered | Diffeomorphism group of the unit sphere of complex n-space |
Jul
27 |
revised |
projection of the co-derivative == co-derivative of the projection ?
added 3 words |
Jul
27 |
answered | projection of the co-derivative == co-derivative of the projection ? |
Jul
25 |
awarded | Editor |