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Consider the affine hyperquadric $Q:=\biggl\{(z_1,...,z_{n+1})\in\mathbb{C}^{n+1}\biggl|\sum_{i=1}^{n+1}z_i^2=1\biggr\}\cong TS^n$.

What is a reasonable Kähler metric for $Q$ (induced by the pullback of the metric from the ambient space $\mathbb{C}^{n+1})$? Furthermore, how do we explicitly calculate the curvature form $\Omega$ on $Q$? Hence, compute the Chern classes of $Q$. Given this, how do we find $\chi(Q,\mathcal{O}_Q)$?

Thanks in advance!

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    $\begingroup$ Since cohomology is a homotopy invariant, $H^*(Q) \cong H^*(S^n)$, with a $\mathbb Z$ in dimensions $0$ and $n$, and $0$ elsewhere. $\endgroup$
    – Arun Debray
    Commented Aug 6, 2018 at 0:28
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    $\begingroup$ The second part of your question doesn't make sense to me; it looks like you're trying to state the Hirzebruch-Riemann-Roch theorem, but this is for compact complex manifolds. $S^n$ is not a compact complex manifold and $Q$ is not a holomorphic vector bundle over it unless I guess $n=2$. Even then, what you would be asking for is a Hermitian metric and the curvature of the corresponding connection (if it was a compact complex manifold). A Kahler metric induces these structures on the tangent bundle, and usually one writes $\chi(X, \mathcal O_X)$ to mean the holomorphic Euler characteristic. $\endgroup$
    – mme
    Commented Aug 6, 2018 at 0:38
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    $\begingroup$ $Q$ is a complex manifold, but not compact. If anything makes sense, it's the left side of the equality, but I think as $Q$ is a Stein manifold (see here) this is the same as the dimension of $H^0(Q;\mathcal O_Q)$, which is the space of holomorphic vector fields on $Q$. I do not know what this space is; it's possible to be infinite-dimensional when the manifold is not compact (as it is in the case of $\Bbb C^n$). $\endgroup$
    – mme
    Commented Aug 6, 2018 at 0:59
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    $\begingroup$ This is simply a guess, since I didn't downvote myself, but my guess is that the downvotes are on the basis of "does not show any research effort". Partly it's that the first question has a trivial answer, noted by Arun. And partly that there are problems with the second part, noted by Mike Miller, that may suggest possibly not thinking things through carefully before asking. To ameliorate this criticism, let me say that I, not being knowledgeable in complex algebraic geometry, wasn't aware of the isomorphism in your display line, so at least I got something out of your question. :-) $\endgroup$
    – Todd Trimble
    Commented Aug 6, 2018 at 14:13
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    $\begingroup$ As suggested by my rejected edit, I really think that "If you downvote"-type pleading doesn't belong here. If you want to talk about the mechanics of the site rather than mathematics, then that belongs on Meta. $\endgroup$
    – LSpice
    Commented Aug 10, 2018 at 12:13

1 Answer 1

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I'm going to answer questions 1 and 2.

  1. You can restrict the Kaehler metric from the ambient affine space to get a Kaehler metric on this hypersurface if you want (there are, of course, many others).

  2. The "curvature form" definition of Chern classes is not usually very useful for calculations. Instead, as you have helpfully pointed out, the quadric is diffeomorphic to the cotangent bundle of $S^n$. This deformation retracts onto the zero section, so you might as well compute the Chern classes of $T(T^*S^n)|_{S^n}$ (living in $H^*(S^n)$). The tangent bundle of $T^*S^n$ restricted to the zero section is $T^*S^n \oplus TS^n$ and, since the $S^n$ is totally real (corresponds to the real locus under your diffeomorphism), this is the complexification of $T^*S^n$. Therefore the Chern classes of the quadric n-fold agree up to a sign with the Pontryagin classes of $S^n$. In particular, you get $c_{n/2}=\pm 1\in Z=H^n(S^n)$ whenever $n/2$ is even (e.g. $S^4$, $S^8$, etc ) and all other Chern classes zero (because they live in cohomology groups which vanish).

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  • $\begingroup$ Thanks so much @JonnyEvans. However, I would like to see how we compute the curvature form. Also, regarding your first answer, do you simply mean the Hermitian metric $h=\delta_{\alpha\overline\beta}dz^{\alpha}\otimes d{\overline z}^{\beta}=dz^{\alpha}\otimes d{\overline z}^{\alpha}$ for $\alpha=1,...,n+1$ restricted to $Q$ by the pullback of the submersion $\theta:Q\hookrightarrow\mathbb{C}^{n+1}$, i.e. $\theta^*h$? $\endgroup$
    – Sergio Charles
    Commented Aug 6, 2018 at 19:23
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    $\begingroup$ That's exactly what I mean (except theta is an immersion, not a submersion). The curvature form is probably not too bad in this case because there is a large symmetry group, but I'm going to abstain from computing it on principle! (There are many useful things you can do with Chern classes, and many useful ways to compute them, but this seems on the face of it to be neither!) $\endgroup$
    – Jonny Evans
    Commented Aug 6, 2018 at 20:08
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    $\begingroup$ (Having said that, I understand what you're going through: I remember when I first learned about the Riemann curvature tensor that I desperately wanted to compute it from the Christoffel symbols in spherical coordinates on S^3, partly as a test of endurance, partly through a misplaced hope that it would give me an aid to understanding curvature in dimension greater than 2. I computed it, but was none the wiser.) $\endgroup$
    – Jonny Evans
    Commented Aug 6, 2018 at 20:13

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