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From Hirzbuch theorem,

the signature of 4-manifold $\sigma = p_1/3$ with the first Pontryagin class $p_1$.

I know that the $\sigma=p_1/3 =1$ so $p_1=3$ for complex projective space $\mathbb{P}^2$.

  • Is there any 4-manifold $N$ such that it has the minimal first Pontryagin class $p_1=1$ and thus $\sigma=1/3$? What are these $N$?

  • Is there any 3-manifold $M^3$ as a boundary of a 4-manifold $N^4$ such that the first Pontryagin class $p_1=1$ for this whole $M^3= \partial N^4$ extension? (Is this a relative Pontryagin class that can be well-defined?)

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    $\begingroup$ For the first question, the signature $\sigma$ is always an integer. Hence your manifold $N^4$ is closed, then $p_1$ must be divisible by $3$. For the second, the Pontryagin class is a cohomology class, not a relative cohomology class, and hence not naturally an integer. The best way to get an integer is to choose a framing on the boundary which gives a relative Pontryagin class. $\endgroup$
    – Danny Ruberman
    Commented Nov 25, 2023 at 0:09
  • $\begingroup$ "The best way to get an integer is to choose a framing on the boundary which gives a relative Pontryagin class" Do you mean $p_1=1$? Could you write up an answer? $\endgroup$
    – zeta
    Commented Nov 25, 2023 at 1:52
  • $\begingroup$ @DannyRuberman - question --- Are there examples of 4d topological manifold with Spin structure but whose 3d boundary has no Spin structure? $\endgroup$
    – zeta
    Commented Nov 25, 2023 at 17:00
  • $\begingroup$ To answer the second question, a spin structure on an (oriented) manifold induces a spin structure on its boundary. besides, every oriented 3-manifold admits a spin structure. $\endgroup$
    – Danny Ruberman
    Commented Nov 25, 2023 at 21:13

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