Return to Answer

Let me start being a little nitpicking with the formulation of the question. The fact that $X$ is $\mathbb Q$-factorial does not in itself imply that such $a$ and $b$ exists. One also needs the fact that the Picard number of $X$ is $1$. This is indeed true, but perhaps should be mentioned. In fact, the Picard group of $X$ is $\mathbb Z$ which was silently used in some answers. This is especially not obvious in cases where quotients are taken. Here it is (of course?) trivial, but in my opinion deserves mentioning. One possible way to see it is that $X\setminus \{P\}$ (where $P$ is the vertex) is an $\mathbb A^1$-bundle over $\mathbb P^2$, so its Picard group is $\mathbb Z$ and hence the group of Weil divisors on $X$ is also $\mathbb Z$ (this makes sense as $X$ is normal, because the Veronese embedding is projectively normal). This, by the way, is also a proof that $X$ is $\mathbb Q$-factorial (since it is projective), and that the Picard group is a subgroup which is also isomorphic to $\mathbb Z$.

So, let's get to answering the questions:

First, let us determine $a$ and $b$. These do not require blowing up. Let $H\subset X$ denote a general hyperplane section of $X$, so $H\simeq V\simeq \mathbb P^2$. Then since $V\subset \mathbb P^2$ is the $2$-uple embedding, $H\left|_H\right.$ is (linearly equivalent to the image of) a conic in $\mathbb P^2$. By adjunction $$-3=\deg K_H = \deg \left((K_X+H)\left|_H \right.\right)$$ (notice that $H$ avoids the singular point of $X$, so this is OK). It follows that $$\deg K_X\left|_H\right.= -5.$$

Since the Picard group of $X$ is $\mathbb Z$ (!!) it follows that $2K_X\sim -5H$, i.e., that $a=2$ and $b=-5$ (or any multiple of these values, e.g., $a=-2$ and $b=5$ works as well).

Next, let us determine $d$: From the equation $K_Y\sim f^*K_X + dE$, using adjunction, one obtains that $$(d+1)E\left|_E\right.\sim K_E.\tag{$*$}$$ Since $f$ was obtained as a blow-up of $\mathbb P^6$ at a point, $-E\left|_E\right.$ is linearly equivalent to the restriction of the hyperplane class of the exceptional $\mathbb P^5$ and hence it is a conic on $E$. Conbined with $(*)$ this implies that $$-2(d+1)=-3,$$ that is, that $$d=\frac 12.$$

Let me start being a little nitpicking with the formulation of the question. The fact that $X$ is $\mathbb Q$-factorial does not in itself imply that such $a$ and $b$ exists. One also needs the fact that the Picard number of $X$ is $1$. This is indeed true, but perhaps should be mentioned. In fact, the Picard group of $X$ is $\mathbb Z$ which was silently used in some answers. This is especially not obvious in cases where quotients are taken. Here it is (of course?) trivial, but in my opinion deserves mentioning. One possible way to see it is that $X\setminus \{P\}$ (where $P$ is the vertex) is an $\mathbb A^1$-bundle over $\mathbb P^2$, so its Picard group is $\mathbb Z$ and hence the group of Weil divisors on $X$ is also $\mathbb Z$ (this makes sense as $X$ is normal, because the Veronese embedding is projectively normal). This, by the way, is also a proof that $X$ is $\mathbb Q$-factorial (since it is projective), and that the Picard group is a subgroup which is also isomorphic to $\mathbb Z$.

So, let's get to answering the questions:

First, let us determine $a$ and $b$. These do not require blowing up. Let $H\subset X$ denote a general hyperplane section of $X$, so $H\simeq V\simeq \mathbb P^2$. Then since $V\subset \mathbb P^2$ is the $2$-uple embedding, $H\left|_H\right.$ is (linearly equivalent to the image of) a conic in $\mathbb P^2$. By adjunction $$-3=\deg K_H = \deg \left((K_X+H)\left|_H \right.\right)$$ (notice that $H$ avoids the singular point of $X$, so this is OK). It follows that $$\deg K_X\left|_H\right.= -5.$$

Since the Picard group of $X$ is $\mathbb Z$ (!!) it follows that $2K_X\sim -5H$, i.e., that $a=2$ and $b=-5$ (or any multiple of these values, e.g., $a=-2$ and $b=5$ works as well).

Next, let us determine $d$: From the equation $K_Y\sim f^*K_X + dE$, using adjunction, one obtains that $$(d+1)E\left|_E\right.\sim K_E.\tag{$*$}$$ Since $f$ was obtained as a blow-up of $\mathbb P^6$ at a point, $-E\left|_E\right.$ is linearly equivalent to the restriction of the hyperplane class of the exceptional $\mathbb P^5$ and hence it is a conic on $E$. Combined with $(*)$ this implies that $$-2(d+1)=-3,$$ that is, that $$d=\frac 12.$$

Hm. This is interesting. First of all, an answer is accepted where only one of the three values in question is computed and not the answer where they are all computed and which received the most votes. Second, interestingly, no one noticed that the latter answer could not possibly be correct. The error is in fact minor, but nevertheless it is there. I will make a comment to that answer to explain the details.

Also, let me be a little nitpicking with the formulation of the question. The fact that $X$ is $\mathbb Q$-factorial does not in itself imply that such $a$ and $b$ exists. One also need the fact that the Picard number of $X$ is $1$. This is indeed true, but perhaps should be mentioned. In fact, the Picard group of $X$ is $\mathbb Z$ which was silently used in some answers. This is especially not obvious in cases where quotients are taken. Here it is (of course) trivial, but in my opinion deserves mentioning. The reason is that $X\setminus \{P\}$ (where $P$ is the vertex) is an $\mathbb A^1$-bundle over $\mathbb P^2$, so its Picard group is $\mathbb Z$ and hence the group of Weil divisors on $X$ is also $\mathbb Z$. This, by the way, is also a proof that $X$ is $\mathbb Q$-factorial (since it is projective), and that the Picard group is a subgroup which is also isomorphic to $\mathbb Z$.

Anyway, here is yet another way to approach the original question, not using quotients and perhaps a way which would work more universally. It also seems pretty simple to me.

ANSWER STARTS HERE. BEFORE THIS IS MOSTLY JUST RANTING. FEEL FREE TO SKIP IT.

First, let us determine $a$ and $b$. These do not require blowing up. Let $H\subset X$ denote a general hyperplane section of $X$, so $H\simeq V\simeq \mathbb P^2$. Then since $V\subset \mathbb P^2$ is the $2$-uple embedding, $H\left|_H\right.$ is (linearly equivalent to the image of) a conic in $\mathbb P^2$. By adjunction $$-3=\deg K_H = \deg \left((K_X+H)\left|_H \right.\right)$$ (notice that $H$ avoids the singular point of $X$, so this is OK). It follows that $$\deg K_X\left|_H\right.= -5.$$

Since the Picard group of $X$ is $\mathbb Z$ (!!) it follows that $2K_X\sim -5H$, i.e., that $a=2$ and $b=-5$ (or any multiple of these values, e.g., $a=-2$ and $b=5$ works as well).

Next, let us determine $d$: From the equation $K_Y\sim f^*K_X + dE$, using adjunction, one obtains that $$(d+1)E\left|_E\right.\sim K_E.\tag{$*$}$$ Since $f$ was obtained as a blow-up of $\mathbb P^6$ at a point, $-E\left|_E\right.$ is linearly equivalent to the restriction of the hyperplane class of the exceptional $\mathbb P^5$ and hence it is a conic on $E$. Conbined with $(*)$ this implies that $$-2(d+1)=-3,$$ that is, that $$d=\frac 12.$$

Let me start being a little nitpicking with the formulation of the question. The fact that $X$ is $\mathbb Q$-factorial does not in itself imply that such $a$ and $b$ exists. One also needs the fact that the Picard number of $X$ is $1$. This is indeed true, but perhaps should be mentioned. In fact, the Picard group of $X$ is $\mathbb Z$ which was silently used in some answers. This is especially not obvious in cases where quotients are taken. Here it is (of course?) trivial, but in my opinion deserves mentioning. One possible way to see it is that $X\setminus \{P\}$ (where $P$ is the vertex) is an $\mathbb A^1$-bundle over $\mathbb P^2$, so its Picard group is $\mathbb Z$ and hence the group of Weil divisors on $X$ is also $\mathbb Z$ (this makes sense as $X$ is normal, because the Veronese embedding is projectively normal). This, by the way, is also a proof that $X$ is $\mathbb Q$-factorial (since it is projective), and that the Picard group is a subgroup which is also isomorphic to $\mathbb Z$.

So, let's get to answering the questions:

First, let us determine $a$ and $b$. These do not require blowing up. Let $H\subset X$ denote a general hyperplane section of $X$, so $H\simeq V\simeq \mathbb P^2$. Then since $V\subset \mathbb P^2$ is the $2$-uple embedding, $H\left|_H\right.$ is (linearly equivalent to the image of) a conic in $\mathbb P^2$. By adjunction $$-3=\deg K_H = \deg \left((K_X+H)\left|_H \right.\right)$$ (notice that $H$ avoids the singular point of $X$, so this is OK). It follows that $$\deg K_X\left|_H\right.= -5.$$

Since the Picard group of $X$ is $\mathbb Z$ (!!) it follows that $2K_X\sim -5H$, i.e., that $a=2$ and $b=-5$ (or any multiple of these values, e.g., $a=-2$ and $b=5$ works as well).

Next, let us determine $d$: From the equation $K_Y\sim f^*K_X + dE$, using adjunction, one obtains that $$(d+1)E\left|_E\right.\sim K_E.\tag{$*$}$$ Since $f$ was obtained as a blow-up of $\mathbb P^6$ at a point, $-E\left|_E\right.$ is linearly equivalent to the restriction of the hyperplane class of the exceptional $\mathbb P^5$ and hence it is a conic on $E$. Conbined with $(*)$ this implies that $$-2(d+1)=-3,$$ that is, that $$d=\frac 12.$$

Hm. This is interesting. First of all, an answer is accepted where only one of the three values in question is computed and not the answer where they are all computed and which received the most votes. Second, interestingly, no one noticed that the latter answer could not possibly be correct. The error is in fact minor, but nevertheless it is there. I will make a comment to that answer to explain the details.

Also, let me be a little nitpicking with the formulation of the question. The fact that $X$ is $\mathbb Q$-factorial does not in itself imply that such $a$ and $b$ exists. One also need the fact that the Picard number of $X$ is $1$. This is indeed true, but perhaps should be mentioned. In fact, the Picard group of $X$ is $\mathbb Z$ which was silently used in some answers. This is especially not obvious in cases where quotients are taken. Here it is (of course) trivial, but in my opinion deserves mentioning, as $X\setminus \{P\}$ (where $P$ is the vertex) is an $\mathbb A^1$-bundle over $\mathbb P^2$, so its Picard group is $\mathbb Z$ and hence the group of Weil divisors on $X$ is also $\mathbb Z$. This, by the way, is a proof that $X$ is $\mathbb Q$-factorial (since it is projective), and that the Picard group is a subgroup which is also isomorphic to $\mathbb Z$.

Anyway, here is yet another way to approach the original question, not using quotients and perhaps a way which would work more universally. It also seems pretty simple to me.

ANSWER STARTS HERE. BEFORE THIS IS JUST RANTING. FEEL FREE TO SKIP IT

First, let us determine $a$ and $b$. These do not require blowing up. Let $H\subset X$ denote a general hyperplane section of $X$, so $H\simeq V\simeq \mathbb P^2$. Then since $V\subset \mathbb P^2$ is the $2$-uple embedding, $H\left|_H\right.$ is (linearly equivalent to the image of) a conic in $\mathbb P^2$. By adjunction $$-3=\deg K_H = \deg \left((K_X+H)\left|_H \right.\right)$$ (notice that $H$ avoids the singular point of $X$, so this is OK). It follows that $$\deg K_X\left|_H\right.= -5.$$

Since the Picard group of $X$ is $\mathbb Z$ (!!) it follows that $2K_X\sim -5H$, i.e., that $a=2$ and $b=-5$ (or any multiple of these values, e.g., $a=-2$ and $b=5$ works as well).

Next, let us determine $d$: From the equation $K_Y\sim f^*K_X + dE$, using adjunction, one obtains that $$(d+1)E\left|_E\right.\sim K_E.\tag{$*$}$$ Since $f$ was obtained as a blow-up of $\mathbb P^6$ at a point, $-E\left|_E\right.$ is linearly equivalent to the restriction of the hyperplane class of the exceptional $\mathbb P^5$ and hence it is a conic on $E$. Conbined with $(*)$ this implies that $$-2(d+1)=-3,$$ that is, that $$d=\frac 12.$$

Hm. This is interesting. First of all, an answer is accepted where only one of the three values in question is computed and not the answer where they are all computed and which received the most votes. Second, interestingly, no one noticed that the latter answer could not possibly be correct. The error is in fact minor, but nevertheless it is there. I will make a comment to that answer to explain the details.

Also, let me be a little nitpicking with the formulation of the question. The fact that $X$ is $\mathbb Q$-factorial does not in itself imply that such $a$ and $b$ exists. One also need the fact that the Picard number of $X$ is $1$. This is indeed true, but perhaps should be mentioned. In fact, the Picard group of $X$ is $\mathbb Z$ which was silently used in some answers. This is especially not obvious in cases where quotients are taken. Here it is (of course) trivial, but in my opinion deserves mentioning. The reason is that $X\setminus \{P\}$ (where $P$ is the vertex) is an $\mathbb A^1$-bundle over $\mathbb P^2$, so its Picard group is $\mathbb Z$ and hence the group of Weil divisors on $X$ is also $\mathbb Z$. This, by the way, is also a proof that $X$ is $\mathbb Q$-factorial (since it is projective), and that the Picard group is a subgroup which is also isomorphic to $\mathbb Z$.

Anyway, here is yet another way to approach the original question, not using quotients and perhaps a way which would work more universally. It also seems pretty simple to me.

ANSWER STARTS HERE. BEFORE THIS IS MOSTLY JUST RANTING. FEEL FREE TO SKIP IT.

First, let us determine $a$ and $b$. These do not require blowing up. Let $H\subset X$ denote a general hyperplane section of $X$, so $H\simeq V\simeq \mathbb P^2$. Then since $V\subset \mathbb P^2$ is the $2$-uple embedding, $H\left|_H\right.$ is (linearly equivalent to the image of) a conic in $\mathbb P^2$. By adjunction $$-3=\deg K_H = \deg \left((K_X+H)\left|_H \right.\right)$$ (notice that $H$ avoids the singular point of $X$, so this is OK). It follows that $$\deg K_X\left|_H\right.= -5.$$

Since the Picard group of $X$ is $\mathbb Z$ (!!) it follows that $2K_X\sim -5H$, i.e., that $a=2$ and $b=-5$ (or any multiple of these values, e.g., $a=-2$ and $b=5$ works as well).

Next, let us determine $d$: From the equation $K_Y\sim f^*K_X + dE$, using adjunction, one obtains that $$(d+1)E\left|_E\right.\sim K_E.\tag{$*$}$$ Since $f$ was obtained as a blow-up of $\mathbb P^6$ at a point, $-E\left|_E\right.$ is linearly equivalent to the restriction of the hyperplane class of the exceptional $\mathbb P^5$ and hence it is a conic on $E$. Conbined with $(*)$ this implies that $$-2(d+1)=-3,$$ that is, that $$d=\frac 12.$$