Since $E_3$ was defined as $x_2 = 0, 1 + y_3^2 + z_3^2 = 0$ we have $E_3 \cap E_2 = \emptyset$ in this chart.Similary I got that $E_3 \cap E_2 = \emptyset$ in the other charts (I can add details if necessary).
I called it $E_1$ because it was intersection of the exceptional divisor with the chart $U_1$. Coordinates $U_2$ and $U_3$ are related by $y_3 = y_2x_2^{-1}, z_3 = x_2z, x_2 = x_2$. Stack Exchange network consists of 177 Q&A communities including
But by toric geometry I I should get $3$ copies of $P^1$, with one intersecting the two other transversally, which is clearly not what I did obtain.
By definition . Suppose that η is any other rational n-form, with zeroes minus polesK0 X. Thekeypointisthattheratiof = ω/η isarationalfunction. This is smooth so we don't need to blow-up anymore.
The key trichotomy among compact Riemann surfaces X is whether the canonical divisor has negative degree (so X has genus zero), zero degree (genus one), or positive degree (genus at least 2).
/Filter /FlateDecode By using our site, you acknowledge that you have read and understand our MathOverflow is a question and answer site for professional mathematicians. $$ Q: \ \ \ 1 + y_3^2 + z_3^2 = 0, \ \ \ \ x_2 = 0.$$
Then by checking affine charts of $Y$, $K_Y=f^*K_X+(1+1+2-1)E$ where $E$ is the exceptional divisor of $f$ and $E\cong \mathbb{P}(1,1,2)$. The key trichotomy among compact Riemann surfaces X is whether the canonical divisor has negative degree (so X has genus zero), zero degree (genus one), or positive degree (genus at least 2). and each of these intersections is transversal. 9 0 obj By clicking “Post Your Answer”, you agree to our To subscribe to this RSS feed, copy and paste this URL into your RSS reader. %PDF-1.4 \hat C_2: \ \ \ y_3 = -i, \ \ \ z_3 = 0.$$ The full exceptional divisor is then the union of the three rational curves, $\hat C_1$, $\hat C_2$ and $Q$.It remains to examine how $Q$ intersects $\hat C_1$ and $\hat C_2$. Maybe it was not very clever notation.
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^ovU�5�дj�v�/q�}�:M^�M���2b�Xʃ>��(�MO So $U_2$ and $U_3$ are open charts on different varieties, and it makes no sense to talk about their "intersection" $U_2 \cap U_3$.Instead, we could proceed as follows.
Let P ∈ X be the germ of a smooth variety and a be an R-ideal on X.
Such a divisor is obtained by a weighted blow-up.
Blow up the singular point of $Y$, we get $g: Z\to Y$ and $K_Z\sim g^*K_Y+\frac{1}{2}E'$ since $Y$ is locally a cone over Veronese surface, which is explained well in this question:Also $g^*E=\tilde{E}+E'$, which can be computed by blowing up $\mathbb{P}(1,1,1,2)$ and consider $E$ as a cone over a conic.
We have $E_1 = E_2$ and it is corresponds to two copies of $P^1$ intersecting transversally.
This is precisely the intersection pattern that you were looking for.Thanks for contributing an answer to Mathematics Stack Exchange! Your $U_3$ is an open chart in a blowup of a blowup of $\mathbb A^3$.
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I want to compute $E_3 \cap E_2$. After the first blowup, the exceptional divisor consists of a pair of rational curves intersecting at a point.
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