## New answers tagged 4-manifolds

1

Thank you.
The following gives an answer in the case that all the curves are assume to have genus $g=1$, and $b\geq 2$.
One of the curves, say $C=C_1$ has $C^2>0$. As $p_g=q=0$ we have $\chi(O_X)=1-q+p_g=1$. By Riemann-Roch, we have $\chi(C)=1+ \frac12 (C^2-K\cdot C)$. As $C^2+K\cdot C=2g(C)-2=0$ (by adjunction), we have $\chi(C)=1+C^2$. Also as ...

0

The answer is clearly affirmative for $b=1$, which unfortunately corresponds only to 101 surfaces, including ${\mathbb P}^2$. I doubt that there is a single other case in which it works.
Indeed I do not know a single example with $b>1$ for which your statement holds. It clearly fails for $({\mathbb P}^1)^2$, as mentioned by ulrich, and a similar argument ...

0

If you accept that quantum gravity with matter should be a Topological Quantum Field Theory and that TQFTs probably can't distinguish simply connected homotopy equivalent 4-manifolds, you should come to the conclusion that at least research on quantum gravity would ultimately not depend on the smooth structures of $\mathbb{R}^4$.

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