Timeline for Degree of a variety vs degree of its blow-up
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Feb 2, 2021 at 15:24 | answer | added | Pop | timeline score: 3 | |
| Feb 2, 2021 at 9:02 | comment | added | H A Helfgott | True - I had another reason in my head, but I guess it doesn't work. | |
| Feb 2, 2021 at 8:46 | comment | added | Francesco Polizzi | Why should the previous example be symmetrical in $a, \, b$? There is no involution of $\mathbf{P}^2 \times \mathbf{P}^1$ exchanging the two factors. | |
| Feb 2, 2021 at 8:10 | history | edited | H A Helfgott | CC BY-SA 4.0 |
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| Feb 2, 2021 at 7:56 | comment | added | H A Helfgott | @R. van Dobben de Bruyn. You are of course right about $f$ - changed the wording. And the example is nice (though I don't see why it's not symmetrical in $a$ and $b$). | |
| Feb 2, 2021 at 7:35 | history | edited | H A Helfgott | CC BY-SA 4.0 |
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| Feb 2, 2021 at 3:21 | comment | added | R. van Dobben de Bruyn | Also a comment: a blowup in a point gives a morphism $f \colon \mathbf P^n \times \mathbf P^{n-1} \to \mathbf P^n$ such that $f^{-1}(V)$ splits up into the exceptional divisor and $\tilde V$. It's not the image of some morphism $\mathbf P^n \to \mathbf P^n \times \mathbf P^{n-1}$ (which in fact do not exist as every map $\mathbf P^n \to \mathbf P^{n-1}$ is undefined somewhere). | |
| Feb 2, 2021 at 3:13 | comment | added | R. van Dobben de Bruyn | An easy example is $V = \mathbf P^n$ itself for $n = 2$. Then the strict transform $\tilde V$ is given by the equation $x_0y_1=x_1y_0$, corresponding to the divisor $\mathcal O(1,1)$. The cohomology ring of $\mathbf P^2 \times \mathbf P^1$ is $\mathbf Q[h_1,h_2]/(h_1^3,h_2^2)$, and $\mathcal O(1,1)$ gives cohomology class $h_1+h_2$. If $D = ah_1+bh_2$, we get $\deg \tilde V = (h_1+h_2)(ah_1+bh_2)^2 = a^2+2ab$ (the term in $h_1^2h_2$), which is always bigger than $1 = \deg V$ when $a,b > 0$. | |
| Feb 1, 2021 at 23:37 | history | asked | H A Helfgott | CC BY-SA 4.0 |