# Chromatic orientability of manifolds

If a compact manifold $$M$$ with empty boundary is oriented with respect to all the connective Morava $$K$$-theories $$k(n)_*$$, localized at a prime $$p$$, can one conclude that $$M$$ is orientable with respect to $$p$$-local Brown-Peterson theory $$BP_*$$?

[As an example, Chris Lloyd and I know that the hypothesis holds for the real Grassmanians $$Gr_2(\mathbb R^m)$$ with $$m$$ even (and $$p=2$$).]

Added later: Chris and I have been having fun studying the Morava $$K$$-theory of $$Gr_d(\mathbb R^m)$$ (at $$p=2$$), and among other things, it seems that when $$m$$ is even, all of the spaces $$Gr_d(\mathbb R^m)$$ are $$k(n)$$-orientable for all $$n$$ and $$d$$. So I was just idly pondering what this means.

• That's exactly the sort of question I would pose to Nick Kuhn, if I had thought of it. – Ryan Budney Jul 15 at 19:16
• Your particular example is a complex manifold, indeed a hermitian symmetric space. The orientation double cover (where the parity enters) is $SO(m)/SO(2)\times SO(m-2)$. The stabilizer of a point contains $SO(2)$, which induces a complex structure on the tangent space. Since it is central in the stabilizer, it commutes with the isotropy and thus is canonical. – Ben Wieland Jul 16 at 21:03
• Can you get orientations with respect to the Morava E-theories? Maybe Corollary 3.4 here is useful arxiv.org/pdf/1509.05678.pdf, which says that p-completed BP is a retract of a product of Morava E-theories. – Jeremy Hahn Jul 17 at 17:08
• 1. If a smooth manifold is $E(1)$-orientable, is it $K(n)$ orientable for all $n>1$, since the structure group is the image of $J$? You need to account for the difference between additive and multiplicative structure, but that's not much, is it? Is that just regular orientability? . . . 2. For a simply connected manifold, is orientability the same as equivalence between $M\otimes E$ and $Hom(M,E)$? The latter seems well-suited to combining $K(n)$ into $BP$. But your manifolds are not simply connected. – Ben Wieland Jul 20 at 1:03

If $$p$$ is odd then this is easy but dull. Using the truncation $$k(n) \to HF_p$$, a $$k(n)$$-orientable vector bundle is orientable in the usual sense. On the other hand, $$p$$-locally the Thom spectrum $$MSO$$ has a cell structure with only even cells, so is orientable with respect to any even ring spectrum.
• Right. But the $p=2$ case is really what I care about, as my example suggests. – Nicholas Kuhn Jul 15 at 22:56