Consider the real projective variety on $\mathbb{R}P^n \times \mathbb{R}P^n$ defined by the bihomogeneous polynomial $x_0y_0+...+x_ny_n$, i.e. $V=\{([x_0:...:x_n],[y_0:...:y_n])|x_0y_0+...+x_ny_n=0\}$.

Since $V$ is nonsingular I can consider it as a (real) smooth manifold of dimension $n$. I am interested to know whether $V$ is orientable as a manifold or not, yet I am not sure how to check this myself.

I have tried barehand-ly constructing an orientation for each affine piece $V\cap U_i=V\cap \{x_i \neq 0\}$ and fitting them together, but to determine whether they can indeed be patched was difficult. I have also tried computing Jacobian of transition functions between the affine pieces, for if they can be made to be all positive, $V$ is orientable, however I found it difficult to modify the parametrization on each piece suitably to compare the signs.

I would appreciate if someone can provide me a way to determine the orientability of $V$. Thank you in advance.

  • $\begingroup$ If someone comments who knows this writes; delete this. I am not a expert: isn't the original calculation of homology/cohomology what this compute's. From Wikipedia: "The Betti numbers of the manifold are the rank of the free part of the homology group, and the non-orientable cycles are described by the torsion part. ". I think I still have the book, somewhere, that gives a cohomology/homology calculation via differential forms. In the first chapter :) In other words, the failure of ddw=0 ; as I recall. it's been a long time and I didn't study it thoroughly. $\endgroup$ – rrogers Nov 8 '16 at 21:56
  • $\begingroup$ @rrogers I believe you are describing the de Rham cohomology, the manifold is orientable iff the top de Rham cohomology group is $\mathbb{R}$, but how I can calculate the de Rham cohomology? $\endgroup$ – Chi Cheuk Tsang Nov 9 '16 at 0:35
  • $\begingroup$ Found the book: "From Calculus to Cohomology" ... Ib Madsen and Jorgen Tornehave page 15. I will look at it; I should be able to say "review" but my memory is really poor. $\endgroup$ – rrogers Nov 9 '16 at 15:41
  • $\begingroup$ $\mathbb{R}P^n$ is orientable iff the $\mathbb{Z}/2$ action $S^n\rightarrow S^n$ is orientable. Similarly here, your subvariety can be given by taking the quotient of some $M\subset S^n\times S^n$ by $\mathbb{Z}/2\times\mathbb{Z}/2$. It suffices to check whether the $\mathbb{Z}/2\times\mathbb{Z}/2$ action is orientable. The computation might be a bit confusing, but should be doable. $\endgroup$ – DCT Nov 13 '16 at 20:35
  • $\begingroup$ You could try: www1.maths.leeds.ac.uk/~rb/DG34ch7.pdf which says " A level hypersurface in Rn+1 given by the inverse image of a regular value of a smooth function f has canonical orientation given by grad f /|grad f|." Ignoring your "projective" condition your problem fits this. It seems that your problem is represented by a vector X and the Y is in the orthogonal plane to X. Normalizing to |X|=1 you have a sphere with the orthogonal plane either on the surface or through the center. In other words you have the "inner product". $\endgroup$ – rrogers Nov 14 '16 at 14:17

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