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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.

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    $\begingroup$ That's exactly the sort of question I would pose to Nick Kuhn, if I had thought of it. $\endgroup$
    – Ryan Budney
    Commented Jul 15, 2021 at 19:16
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    $\begingroup$ 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. $\endgroup$
    – Ben Wieland
    Commented Jul 16, 2021 at 21:03
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    $\begingroup$ 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. $\endgroup$
    – Jeremy Hahn
    Commented Jul 17, 2021 at 17:08
  • $\begingroup$ 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. $\endgroup$
    – Ben Wieland
    Commented Jul 20, 2021 at 1:03

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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.

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    $\begingroup$ Right. But the $p=2$ case is really what I care about, as my example suggests. $\endgroup$
    – Nicholas Kuhn
    Commented Jul 15, 2021 at 22:56

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