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I am looking for explicit descriptions of generators of some low-dimensional unstable cobordism groups. For example, $\mathbb CP^2$ embeds into $\mathbb R^7$ by a result of Haefliger. Because it has signature $1$, it is a primitive element in the group of oriented $4$-dimensional submanifolds in $\mathbb R^7$ up to oriented cobordism. By the Pontryagin-Thom construction, this group is isomorphic to $\pi_7(MSO(3))$, which Thom computed to be $\mathbb Z$. Similarly, $\mathbb C P^2$ also generates $\pi_8(MSO(4))$.

I am looking for analogous generators of the unstable group $\pi_8(MU(2))$. Note that $\mathbb CP^2$ seems not to work. A characteristic class computation shows that the normal bundle should have nonvanishing second Chern class, but the normal bundle of $\mathbb CP^2$ in $\mathbb R^8$ has Euler class $0$ by the above, and those two classes should agree. Also by another result of Haefliger, all embeddings of $\mathbb CP^2$ into $\mathbb R^8$ are isotopic, so there cannot be another one with a different normal Euler class.

Hence I am asking if such low-dimensional unstable cobordism groups have ever been computed, and if explicit generators are known?

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    $\begingroup$ I may be getting myself confused here, but is $\pi_8(\mathit{MU}(2))$ really in the unstable range? The usual map $j\colon B\mathrm U(2)\to B\mathrm U$ is an isomorphism on $\mathbb Z$-cohomology in degrees $5$ and below, so $j'\colon \Sigma^{-4}\mathit{MU}(2)\to\mathit{MU}$ is an isomorphism on $\mathbb Z$ cohomology in degrees $5$ and below, so $j'$ should induce an isomorphism on homotopy groups in degrees $4$ and below, if I'm not mistaken? If that is correct, $\pi_8(\mathit{MU}(2))\cong\mathbb Z\oplus\mathbb Z$ generated by $\mathbb{CP}^2$ and $\mathbb{CP}^1\times\mathbb{CP}^1$. $\endgroup$
    – Arun Debray
    Commented Aug 6, 2023 at 13:43
  • $\begingroup$ I believe we are still in the unstable range. From $c(T\mathbb CP^2)=(1+a)^3$ I seem to be getting $c(\nu)=(1-a+a^2)^3=1-3a+6a^2$ if there was an almost complex structure, so $e(\nu)=c_2(\nu)=6a^2$. But maybe I am wrong. I do believe that $\mathbb CP^2$ lives in $\pi_9(S^1MSU(2))$, though. $\endgroup$
    – Sebastian Goette
    Commented Aug 8, 2023 at 7:24
  • $\begingroup$ Sorry, wrong computation. Assuming there was an almost complex structure with $c_1(\nu)=xa$, $c_2(\nu)=0$ for $a\in H^2(\mathbb CP^2)$ a generator with $c(T\mathbb CP^2)=1+3a+3a^2$. Then the total Chern class of $(T\mathbb CP^2\oplus\nu)_{\mathbb R}\otimes_{\mathbb R}\mathbb C$ would be $(1+xa)(1-xa)(1+a)^3(1-a)^3=1-x^2a^2-3a^2$, getting first Pontryagin class $3+x^2\ne 0$. $\endgroup$
    – Sebastian Goette
    Commented Aug 8, 2023 at 9:16
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    $\begingroup$ The group $\pi_8MU(2)$ is in the meta stable range, not in the stable range, so it is an unstable bordism group which fits into an exact sequence of the form $$\pi_9\Omega\Sigma(MU(2)\wedge MU(2))\to\pi_8MU(2)\to\pi_8\Omega\Sigma MU(2)\to\pi_8\Omega\Sigma(MU(2)\wedge MU(2))\simeq\mathbb{Z}.$$ Perhaps this can help in getting a more geometric description of the generators you desire. In particular, the group on the right of the above exact sequence is isomorphic to $\pi_8MU(4)$ through the Thomified map $MU(2)\wedge MU(2)\to MU(4)$ which is again a stable bordism having well known generators. $\endgroup$
    – user51223
    Commented Aug 9, 2023 at 10:39
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    $\begingroup$ The above exact sequence is a part of the $\mathrm{EHP}$-sequence. $\endgroup$
    – user51223
    Commented Aug 9, 2023 at 10:43

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