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Answer by auniket for Fano plane drawings: embedding PG(2,2) into the real plane

2013-05-14T23:28:12Z

Does this one work?

Interpretation of multiplicity of a point

2013-05-04T19:51:42Z

Let $x$ be a (closed) point on an algebraic variety $X$ (of dimension $n$) defined over an algebraically closed field $k$. What is the multiplicity $mult_x(X)$, and how to compute it?

While having a hard time recently to compute the multiplicity of some surface singularities, I thought it might be useful to have a list of equivalent definitions. These are the ones I know of:

Notations: let $m_x$ be the maximal ideal of $x$ at $X$. For definitions 1 to 3 below assume (a neighborhood of $x$ in) $X$ is embedded in a projective space $\mathbb{P}^N(\mathbb{k})$.

Geometric Definitions:

(Copying from Mumford, Algebraic Geometry I: This is valid only in the case $k = \mathbb{C}$.) For every linear subspace $L$ of dimension $N - n$ such that $x$ is a component of $L \cap X$, define a number $i(x;L \cap X)$ as follows: it is the unique number such that for every sufficiently small neighborhood $U$ of $x$ (in the classical topology), there is a neighborhood $U'$ of $L$ (in the space of $N-n$ dimensional linear subspaces of $\mathbb{P}^N(\mathbb{C})$) such that if $L' \in U'$ and $L'$ intersects $X$ transversally, then $i(x;L \cap X) = |L' \cap X \cap U|$. Then $mult_x(X)$ is the minimum of $i(x;L \cap X)$ as $L$ runs over all $N-n$ dimensional linear subspaces of $\mathbb{P}^N(\mathbb{C})$ for which $x$ is an isolated point of $L \cap X$.
(Corollary of (1), but holds also over positive characteristic - or at least so I think. For my problem this turned out to be the right definition to use) For every linear subspace $L$ of dimension $N - n$ such that $L \cap X$ is discrete, let $s(x;L \cap X)$ be the number of points of intersection (counted with intersection multiplicity) of $L \cap X$ other than $x$. Then $mult_x(X) = \min\lbrace \deg(X) - s(x;L \cap X)\rbrace$ as $L$ runs over all $N-n$ dimensional linear subspaces of $\mathbb{P}^N(k)$ for which $L \cap X$ is discrete.
(From Ramanujam's "On a geometric interpretation of multiplicity") Take a proper birational map $\phi: Y \to X$ such that the pull back of the maximal ideal of $P$ defines an effective Cartier divisor $D$ on $Y$. Then $mult_x(X) = (-1)^{n-1}D^n$.

Algebraic Definitions:

$mult_x(X)$ is $(n-1)!$ times the leading coefficient of the Hilbert–Samuel polynomial of $m_x$.
If $X$ is a hypersurface in a neighbourhood of $x$ defined by a single equation $f$, then $mult_x(X)$ is the integer $q$ such that $f \in m_x^q \setminus m_x^{q+1}$.

What other definitions are out there?

Answer by auniket for jacobian polynomial

2013-02-08T15:41:19Z

Edit: The statement in the next paragraph is wrong! I misunderstood the result of Kaliman: it says that given $(f,g)$ as in the question, there is a polynomial automorphism $\phi$ of $\mathbb{C}^2$ such that each fiber of $\phi \circ (f,g): \mathbb{C}^2 \to \mathbb{C}^2$ is irreducible. So I would assume it is still hard to give a positive answer to the question, but clearly what I wrote below is false.

I would assume the question is quite difficult, since a positive answer would imply the Jacobian conjecture by this result of Kaliman.

Answer by auniket for Normality condition on graded algebra

2013-01-23T22:56:57Z

Hi Isac, One simple criterion can be given which is analogous to Rees' valuations corresponding to ideals. Set $R := \mathbb{C}[x,y]$, define $\nu: R \to \mathbb{N} \cup \infty$ as $\nu(f) := \max\lbrace k: f \in m_k\rbrace$. Then I believe $A$ is normal if $\nu(f^k) = k\nu(f)$ for all $f \in R$. For other criteria, I would look into Kei-ichi Watanabe's articles.

Edit: Here is a general approach.

Claim 1: $A$ is integrally closed iff every homogeneous (with respect to the grading of $A$) element in $\mathbb{C}[x,y,t]$ (i.e. an element of the form $f(x,y)t^k$) which is integral over $A$ is in $A$.

By Claim 1 the integral equation of $ft^k$ (for $f \in R$) over $A$ be of the form $z^d + \sum_{i=0}^d g_ez^{d-e} = 0$ for some $g_e \in m_{ek}$. This proves Claim 2 below.

Claim 2: $A$ is integrally closed iff $\bar m_k := m_k$ for all $k \geq 1$, where

$\bar m_k := \lbrace f \in m_k: f^d + \sum_{i=0}^d g_ez^{d-e} = 0$ for some $d \geq 0$ and $g_1, \ldots, g_d \in m_{ek}\rbrace$.

In the special case that $m_k$ is a monomial ideal for each $k$, it suffices to prove the integral condition only for monomials in $m_k$. And in the special special case that $m_k = m^k$ for some monomial ideal $m$, it follows that your claim is true, i.e. $A$ is integrally closed iff the support of $m^k$ contains all the monomials in the cone spanned by its monomials.

Contracting rational curves on surfaces and getting something non-algebraic

2012-12-30T20:41:58Z

Recently I posted an "announcement" on arxiv where I said something to the effect of "this is the first example we know where contracting (a tree of) rational curves from a non-singular algebraic surface (over $\mathbb{C}$) leads to a (normal) non-algebraic surface." Now that I am writing up the actual paper, I thought that may be I should broaden my knowledge base! So I ask: is there some example of this sort already known?

Remarks:

Grauert constructed (in this article: mathscinet link, springerlink) such non-algebraic surfaces by blowing down curves of genus $\geq 2$ (in Section 4.8, Example d) and remarked that he did not know if it is possible (to construct non-algebraic normal surfaces) from blowing down tori.
An example of Nagata described in a previous question of mine shows that it is indeed possible with tori.
In the second paragraph of the first page of this article (mathscinet link) Artin mentions an example of Hironaka that shows that "in general there are no numerical criteria equivalent with (algebraic) contractibility of a given curve." Does anyone know what is this example?
Happy New Year!

Answer by auniket for Degree of a variety is well-defined

2012-08-01T09:45:02Z

Hi Kiumars, this does not answer your question, since the proof works only when the field is $\mathbb{C}$, but when I was learning it, the most accessible and understandable proof for this case was that of Theorem 5.1 of Mumford's "Algebraic Geometry I: Complex Projective Varieties".

Geometric interpretation of the exact sequence for the cotangent bundle of the projective space

2012-06-02T04:44:47Z

Edit: As Dan Petersen pointed out, this question is a duplicate of a previous one. I would leave it for the moderators to decide if this should be closed. On the other hand, may be this should be left open on the merit of the excellent answers and comments (@Emerton: Thanks!).

I was trying to understand the following exact sequence (for $X := \mathbb{P}^n_k$, where $k$ is an algebraically closed field): $$0 \to \Omega_X \to \mathcal{O}_X(-1)^{n+1} \to \mathcal{O}_X \to 0$$ The explanation (as in the proof of Theorem II.8.13 of Hartshorne) is given by some algebraic formulae, which I am having trouble to digest. I was trying to see in more geometric terms what is going on, and was somewhat successful in the case of the surjection $\mathcal{O}_X(-1)^{n+1} \to \mathcal{O}_X$, namely: we can regard $\mathcal{O}_X(1)$ (respectively $\mathcal{O}_X(-1)$) as the normal bundle $\mathcal{N}$ of (respectively conormal bundle) of $X$ in $Z := \mathbb{P}^{n+1}_k$. Any global section of $\mathcal{O}_X(1)$ therefore induces a map (via evaluation) from $\mathcal{O}_X(-1)$ to $\mathcal{O}_X$. The above surjection comes from taking $n+1$-linearly independent global sections of $\mathcal{O}_X(1)$.

But I do not understand how to interpret the injection $\Omega_X \to \mathcal{O}_X(-1)^{n+1}$. How would someone 'naturally' come up with the algebraic formula?

Answer by auniket for functions satisfying "one-one iff onto"

2012-02-18T01:15:04Z

This addresses the "broader scope" of the question and possibly the comments of Uday on Donu's answer: An injective morphism from an affine algebraic variety over an algebraically closed field to itself is also surjective. Moreover, probably even more surprising is the fact that in the case that the field has characteristic zero (and of course algebraically closed), an injective endomorphism is actually a polynomial automorphism (that is the inverse is also a polynomial map!). See e.g. Chapter 4 of van den Essen's "Polynomial Automorphisms" for proofs of both these statements. Also from the same book: the map $x \mapsto x^3$ from $\mathbb{Q} \to \mathbb{Q}$ shows the necessity of algebraic closedness of the field, and the Frobenius automorphism $x \mapsto x^p$ of an algebraically closed field of characteristic $p > 0$ shows that the second statement is false for positive characteristic. Also, note that both statements are automatically true for proper varieties.

Terminology for the image of the diagonal embedding.

2012-02-18T00:45:44Z

Let $X$ be a topological space equipped with maps into two spaces $\bar X_1$ and $\bar X_2$. Is there a standard notation/terminology for the closure $\bar X$ in $\bar X_1 \times \bar X_2$ of the diagonal map of $X$?

In my case $X$ is an affine algebraic surface (in fact just $\mathbb{C}^2$) which is isomorphic to Zariski open subsets of complete surfaces $\bar X_1$ and $\bar X_2$ and the maps $X \to \bar X_j$'s are the corresponding embeddings.

In a paper I wrote, I used the notation "birational join" for $\bar X$ following Spivakovsky, but the referee does not like it. (S)He suggested something like the "fiber product" $\bar X_1 \times_X \bar X_2$, but that would require the arrows $X \to \bar X_j$ to be reversed. Similarly the ``cofiber product'' requires the arrows $\bar X \to \bar X_j$ to be reversed.

Any suggestions would be much appreciated. Thanks!

Answer by auniket for Algebraic curve cannot suddenly end

2011-11-08T13:14:34Z

Following the idea of Felipe Voloch, I try to give a simple proof based on Puiseux series expansion. Let $C$ be a real algebraic curve at the origin. Look at the Puiseux series expansion (say in terms of $x$) of $C$ near $O$. By assumption one of the branches (over $\mathbb{C}$), call it $C_1$, has the form $$y = a_1x^{r_1} + a_2x^{r_2} + \cdots \quad\quad\quad (1)$$ for $a_i \in \mathbb{R}$. Let $q$ be the least common multiple of the denominators of $r_i$'s. If $q$ is odd, then the branch expands to both sides of the origin and therefore $C_1$ does not end abruptly. So assume $q$ is even. Let $\zeta := e^{2\pi i/q}$. For each $j$, $1 \leq j \leq q$, the complex curve corresponding to $C_1$ has a Puiseux expansion of the form $y = \sum_i a_i \zeta^{jp_i}x^{r_i}$, where $p_i = qr_i$. In particular, taking $j =q/2$ (so that $\zeta^j = -1$), we see that the complex curve corresponding to $C_1$ has an expansion of the form $$y = \sum_i a_i (-1)^{p_i}x^{r_i}. \quad\quad\quad (2)$$ It follows by the minimality of assumption on $q$ that there is $i$ such that $a_i\neq 0$ and $p_i$ is odd , and consequently, $(1)$ and $(2)$ give different real curves, and it follows that $C_1$ does not end abruptly.

PS: The above arguments only show that $C_1$ has at least two end points on the boundary of a small enough disk centered at $O$. But it can not have more than two, because for all $j \not\in \lbrace q/2, q\rbrace$, $\zeta^j$ is non-real, so the corresponding parametrization does not give any real points.

Pathologies of analytic (non-algebraic) varieties.

2011-10-26T23:46:30Z

Note: By an "analytic non-algebraic" surface below I mean a two dimensional compact analytic variety $X$ (over $\mathbb{C}$) which is not an algebraic variety.

A property of Nagata's example (see the end of the post for the construction) of a non-algebraic normal analytic surface $X$ is the following:

($\star$) $\quad$ There is a point $P$ on $X$ such that every (compact) algebraic curve $C$ on $X$ passes through $P$.

In a paper I am writing I also constructed (to my surprise) some examples of non-algebraic normal analytic surfaces which have this peculiar property.

Questions: Is this sort of behaviour "normal" for such surfaces? Or, more precisely, if an analytic surface does not satisfy ($\star$), is it necessarily algebraic? How about for higher dimensions?

Nagata's Construction (following Bădescu's book on surfaces): Start with a smooth plane cubic $C$ and a point $P$ on $\mathbb{P}^2$ such that $P - O$ is not a torsion point (where $O$ is any of the inflection points of $C$) on $C$. Let $X_1$ be the blow up of $\mathbb{P}^2$ at $P$, and for each $i \geq 1$, let $X_{i+1}$ be the blow up of $X_i$ at the point of intersection of the strict transform of $C$ and the exceptional divisor on $X_i$. Each blow up decreases the self-intersection number of the strict transform $C_i$ of $C$ by $1$, so that on $X_{10}$ the self-intersection number of $C_{10}$ is $-1$. $X$ is the blow down of $X_{10}$ along $C_{10}$. By some theorems of Grauer and Artin, $X$ is a normal analytic surface.

Ample divisors on projective surfaces

2011-10-17T03:11:48Z

Question: If $X$ is a projective surface and $U$ is an open affine subset of $X$, then is it true that $X \setminus U$ is the support of an (effective) ample divisor on $X$?

Background: I was reading Goodman's paper "Affine open subsets of algebraic varieties and ample divisors" which considers this same question for general varieties. Here is what I understand so far:

If $\dim X = 1$, then the answer to the question is always affirmative. Indeed, if $X$ is a complete curve and $S$ is any finite set of points on $X$, then there is an effective ample Cartier divisor on $X$ which has support $S$ (this is Proposition 5 of the paper, and a straightforward application of the Nakai-Moishezon criterion of ampleness).
For $\dim X = 2$, Theorem 2 of the paper states that the answer is positive if each point of $X\setminus U$ is factorial (i.e. its local ring is a UFD). Actually he proves it only assuming that $X$ is complete (i.e. a priori not necessarily projective) and as a corollary he proves Zariski's theorem that if all the singularities of a complete surface $X$ are contained in an affine open subset, then $X$ is projective.
He presents two examples (of Hironaka and Zariski) where $X$ is a non-singular projective $3$-folds, but $X\setminus U$ is not the support of any ample divisor.
In general he proves (in Theorem 1) that if $X$ is complete then a Zariski open subset $U$ of $X$ is affine iff the complement of (the isomorphic image of $U$) in a blow-up $X'$ of $X$ along a closed subscheme $F$ not meeting $U$ is the support of an effective ample Cartier divisor on $X'$.
For $\dim X \geq 3$, he gives a criterion (in Theorem 3) for when the answer to the question is positive.

As far as I can see, he does not mention anything about the status of the question (i.e. whether if there is a counter-example or not) for general projective surfaces. Therefore I ask it here. I would expect the answer to be negative, but can not think of any examples. For me particularly interesting would be the case when $X$ is normal.

Edit: As the example of Jason Starr in the comment shows: The answer is negative even for normal surfaces (see my comments for an attempt of proof). I wonder what happens if $X$ is rational. In any event, I would gladly accept Jason's answer if he writes one. (And I would also greatly appreciate any answer/remark about the rational case.)

Answer by auniket for Is [mD] very ample if D is ample?

2011-09-30T18:28:55Z

~~At first a question: If I understand correctly, then the rounding down operation depends on your choice of basis for the Néron-Severi group, right?~~

~~So I am assuming you fix a basis $D_1, \ldots, D_k$ of $NS(X) \otimes_\mathbb{Z} \mathbb{R}$ and for each $D := \sum_j r_jD_j$, you define $[mD] := \sum [mr_j]D_j$.~~

~~If this is true, then doesn't your assertion follow from the following geometric fact?~~

~~Let $C$ be a full dimensional cone in $\mathbb{R}^k$ and $K$ be the standard cube of length $2$ in $\mathbb{R}^k$ centered at the origin, i.e.~~

~~$K := \lbrace\sum_{j=1}^k s_je_j: -1 \leq s_j \leq 1$ for all $j$, $1 \leq j \leq k \rbrace$,~~

where $e_1, \ldots, e_k$ are unit vectors along the axes. If $v$ belongs to the interior of a full dimensional cone $C$ in $\mathbb{R}^k$, then $mv + K$ also lies in the interior of $C$ for all sufficiently large $m$.

~~If as your basis you choose ample divisors, then $K$ can be replaced by a cube of length one.~~

Edit 3: This is my 3rd attempt to give an elementary proof. It is essentially the same proof as in Edits 1 and 2, but with some corrections, and hopefully will be clearer. I hope you see that the idea is very simple and geometrically almost obvious. If it seems complicated, then the fault is in my exposition.

Set Up: Let $D_1, \ldots, D_k$ be ample divisors and $D := \sum_j r_jD_j$ for positive real numbers $r_1, \ldots, r_k$. Also, let $D_j = \sum_{i=1}^N a_{ji} C_i$, for irreducible divisors $C_i$ and integers $a_{ji}$. We want to show that $[mD]$ is very ample for large $m$.

In the proof we will use the following fact about finite sums of integral points in a lattice:

Lemma: Let $v_1, \ldots, v_k \in \mathbb{Z}^N$ such that $\mathbb{Z}$-span of $v_j$'s equals $\mathbb{Z}^N$. Let $P$ be the convex hull (over $\mathbb{R}$) of $\lbrace 0, v_1, \ldots, v_k \rbrace$. Then there exists a positive real number $c$ such that for all $n \geq 1$, if $v \in nP \cap \mathbb{Z}^N$ such that the (Euclidean) distance of $v$ from both the origin and the boundary of $nP$ is greater than $c$, then $v$ is in fact an non-negative integral linear combination of $v_1, \ldots, v_k$.

The above statement (actually a more precise formulation of it) is due to Khovanskii. The proof is very elementary and beautiful, and is in Proposition 2 of this article.

Here starts the proof:

Step 1: Without loss of generality we may assume that $\mathbb{Z}$-span of $D_j$'s equals the $\mathbb{Z}$-span of $C_i$'s. Indeed, it follows from Kleiman's criterion, and finite dimensionality of $N_1(X)$ that for every $m \gg 1$ and $\epsilon := (\epsilon_1, \ldots, \epsilon_N) \in \lbrace 1, 0, -1 \rbrace^N$, $D_{m,\epsilon} := mD_1 + \sum_{i=1}^N\epsilon_i C_i$ is ample. Choosing different values of $\epsilon$ and $m$ and adding $D_{m,\epsilon}$'s to the collection of $D_j$'s, we may ensure that $\mathbb{Z}$-span of $D_j$'s equals the $\mathbb{Z}$-span of $C_i$'s. Moreover, and this is essential, choosing $D_{m,\epsilon}$'s to be sufficiently close to the ray generated by $D_1$, we may ensure that $D$ still lies in the interior of the cone generated by $D_j$'s, i.e. $D = \sum_{j=1}^k r_jD_j$ with each $r_j$ being a positive real number.

Step 2: For each $j$, $1 \leq j \leq k$, let $v_j := (a_{j1}, \ldots, a_{jN}) \in \mathbb{R}^N$, i.e. $v_j$ is the "coordinate" vector of $D_j$ for each $j$ (and therefore $v_j \in \mathbb{Z}^N$ for each $j$). Adding some big multiples of $D_j$'s to the existing collection of $D_j$'s if necessary, we may assume that $v := \sum r_j v_j$ is in the interior of the convex hull $P$ of $0, v_1, \ldots, v_k$.

Step 3: For each $j$, $1 \leq j \leq k$, there exists a positive integer $m_j$ such that $mD_j$ is very ample for all $m \geq m_j$. Indeed, there is $l_j, n_j$ such that $n_jD_j$ is very ample and $mD_j$ is globally generated for all $m \geq l_j$. Setting $m_j := l_j + n_j$ does the job (due to Exercise II.7.5(d) of Hartshorne).

Step 4: There exists a positive integer $m_0$ such that $m_0(D_1 + \cdots +D_k) + \sum s_jD_j$ is very ample for all collections of non-negative integers $s_1, \ldots, s_k$. Indeed, set $m_0 := \max \lbrace m_1, \ldots, m_k \rbrace$ and apply the same exercise of Hartshorne.

Step 5: Let $v, v_1, \ldots, v_k$ and $P$ be as in Step 2. Let $c$ be the constant we get from applying Khovanskii's lemma to $v_1, \ldots, v_k$. Let $v_0 := m_0(v_1 + \cdots + v_k)$, where $m_0$ is as in Step 4. Since $v$ is in the interior of $P$, it follows that if $m$ is sufficiently large, then $[mv] - v_0$ is in the interior of $mP$ and the distance of $[mv] - v_0$ from the origin and the boundary of $mP$ is bigger than $c$. Therefore, Khovanskii's lemma implies that $[mv] - v_0 = \sum a_j v_j$ for non-negative integers $a_j$. Consequently, if $m$ is sufficiently large, then

$$[mD] = m_0(D_1 + \cdots + D_0) + \sum a_j D_j$$

for non-negative integers $a_1, \ldots, a_k$. Step 4 then tells that $[mD]$ is very ample.

Answer by auniket for Pair of curves joining opposite corners of a square must intersect---proof?

2011-10-01T00:08:35Z

How about the following, using the Nested Intervals Theorem (which was in my 2nd year Calculus text) which says the intersection of a nested sequence of closed intervals in $\mathbb{R}$ is non-empty. Here goes the proof:

We construct recursively a nested sequence $I_j := [a_j, b_j]$ of closed intervals for $j \geq 0$. Let $I_0 := [0,1]$. For every $j \geq 0$, construct $I_{j+1}$ as follows: let $m_j$ be the midpoint of $I_j$. If the curves intersect at $t = m_j$, then we are done, so stop the sequence. Otherwise set $I_{j+1}$ to be $[a_j, m_j]$ or $[m_j, b_j]$ depending on whether the curves "switch from left to right" on the first sub-interval or the 2nd (let's say you always make sure that $c_1$ is to the "left" of $c_2$ at $t = a_j$ and to the "right" of $c_2$ at $t = b_j$).

If the sequence is finite, then the curves must intersect, as noted above. So assume the sequence is infinite. The Nested Intervals Theorem and the fact that the length decreases by a factor of 2 at every step implies that $\cap_{j=0}^\infty I_j = \lbrace t\rbrace$ for some $t \in [0,1]$. Then we must have $c_1(t) = c_2(t)$.

Answer by auniket for Examples of seemingly elementary problems that are hard to solve?

2011-09-30T06:45:24Z

The Casas Alvero conjecture: Let $f \in \mathbb{C}[x]$ be a monic polynomial of degree $n$. Suppose that for each $k = 1, \ldots, n-1$, there is a common root of $f$ and $f^{(k)}$. Then $f = (x-a)^n$ for some $a \in \mathbb{C}$. It is known only for the case that $n$ is a prime power or two times a prime power (see for example, this). At some point I thought I proved it :-)

Answer by auniket for Do the elementary properties of mixed volume characterize it uniquely?

2011-08-03T05:39:20Z

I think the first three properties do indeed characterize mixed volume. For example, in two dimensions they imply that

$V(A_1, A_2) = \frac{1}{2}(V(A_1 + A_2, A_1 + A_2) - V(A_1, A_1) - V(A_2,A_2))$
$= \frac{1}{2}(Vol(A_1 + A_2) - Vol(A_1) - Vol(A_2)),$

which gives the formula of mixed volume in terms of volume. You can perform the same trick to get in 3 dimensions:

$V(A_1,A_2, A_3) = \frac{1}{6}(Vol(A_1+A_2+A_3) - Vol(A_1+A_2) - Vol(A_2+A_3)$
$- Vol(A_3+A_1) + Vol(A_1) + Vol(A_2) + Vol(A_3))$

In general I believe you get something like:

$V(A_1, \ldots,A_n) = \frac{1}{n!}(Vol(A_1 + \cdots + A_n) - \sum_{i=1}^n Vol(A_1 + \cdots \hat A_i + \cdots + A_n)$
$ + \cdots +(-1)^{n-1}\sum_{i=1}^n Vol(A_i))$

I learned of this from Bernstein's paper that contains his famous result that the number of solutions in $(\mathbb{C}^*)^n$ of $n$ generic Laurent polynomials is precisely the mixed volume of their Newton polytopes.

Answer by auniket for What is the fan of the toric blow-up of $\mathbb{P}^3$ along the union of two intersecting lines?

2011-07-14T13:39:27Z

To find the polytope associated to a toric variety directly you have to realize the variety as the closure of a map from the torus. In this case at least, it is not too hard to get such a description. Let the homogeneous coordinates of $\mathbb{P}^3$ be $[w:x:y:z]$ and the two lines be $C_1 := \lbrace w = x = 0 \rbrace$ and $C_2 := \lbrace w = y = 0 \rbrace$. Then the blow-up $B$ of $\mathbb{P}^3$ along $C_1 \cup C_2$ is the closure of
$\lbrace ([w:x:y:z], [w^2:wx:wy: wz: xy]) : [w:x:y:z] \in \mathbb{P}^3 \setminus (C_1 \cup C_2) \rbrace$
in $\mathbb{P}^3 \times \mathbb{P}^4$. If we identify $\mathbb{C}^3$ with $\mathbb{P}^3 \setminus V(w)$, and write $X, Y, Z$ respectively for $x/w, y/w, z/w$, then $\mathbb{C}^3$ is embedded in $B$ via the map
$(X, Y, Z) \mapsto ([1:X:Y:Z), (1: X: Y: Z: XY])$
Composing with the Segre embedding (and getting rid of duplicate coordinates), we get
$(X, Y, Z) \mapsto [1: X : Y: Z: X^2: XY: XZ: Y^2: YZ: Z^2: X^2Y: XY^2: XYZ]$
Therefore the polytope is the convex hull of the exponents of these monomials. I believe its vertices are (0,0,0), (2,0,0), (0,2,0), (2, 1, 0), (1, 2, 0), (0, 0, 2) and (1,1,1).

PS: There is a detail to be filled: blow-ups along singular subvarieties are not in general normal, so a priori $B$ might not be a normal toric variety (i.e. the polytope is associated not to $B$ but the normalization of $B$). But as David shows in his answer (and probably proved for a general torus invariant subspaces in the article he mentions in the comment), $B$ is indeed normal.

Answer by auniket for Intersection of curves on projective toric surface and some enumerative questions

2011-04-23T05:49:07Z

The answer to your question B is yes and no :) You see: replacing $P$ by $kP$ for any $k \geq 1$ gives rise to the same toric surface (the latter is precisely the $k$-uple Veronese embedding of the former). And, the defining polynomial of any curve will fit (up to translation) in $kP$ for a sufficiently large $P$.

For the phenomenon in your first question, I can give an answer in the case that the underlying field $\mathbb{k}$ is algebraically closed. Here it goes: let $f$ be the Laurent polynomial defining the curve $C$ and $S$ is an edge of the Newton Polygon $N$ of $f$. Let $\nu := (p, q)$ be an outward pointing (with respect to $N$) normal to $S$. For simplicity assume $q > 0$. Then there is a branch of $C$ with degree-wise Puisuex series of the form: $\gamma(t) = (t^p, \sum_{k = 0}^\infty a_k t^{q_k})$, where $q = q_0 > q_1 > \cdots $ are rational numbers with bounded denominators. Now let $\psi_P: X_P \to \mathbb{P}^N$ (where $N := |P \cap \mathbb{Z}^n| - 1$) be the embedding of the toric variety defined by the monomials in $P$, i.e. the restriction of $\psi_P$ to $(\mathbb{k}^*)^n$ is given by: $\psi_P(x) := [x^{\alpha_0}: \cdots : x^{\alpha_N}]$, where $P \cap \mathbb{Z}^n = \lbrace \alpha_0, \ldots \alpha_N \rbrace$. Let $Q$ be the face (i.e. edge or vertex) of $P$ such that $\nu$ is an outer normal to $Q$. W.l.o.g. assume that $Q \cap \mathbb{Z}^n = \lbrace \alpha_0, \ldots, \alpha_q \rbrace$, $q < N$. Then precisely the first $q+1$ coordinates of $ x := \lim_{t \to \infty} \psi_P(\gamma(t))$ are non-zero, i.e. $x$ belongs to the subvariety of $X_P$ determined by $Q$.

Remark: Usually in the books on toric varieties, inner normals are used instead of outer normals. That gives rise to a usual Puiseux series (the exponents being increasing , as opposed to the one in the preceding paragraph). But then the point at infinity (on the curve) is approached as $t \to 0$. I prefer that one approaches the point at infinity as $t$ approaches infinity as well.

Answer by auniket for equations defining a subvariety

2011-03-25T22:03:42Z

Note that $T := \tilde \phi(f^{-1}(Y))$ is a proper Zariski closed subset of $S$. Therefore, for generic $v_1, \ldots, v_s \in V$, $Z(v_1) \cap \cdots \cap Z(v_s) \cap T = \emptyset$. Consequently, in this case $P \cap Y = \emptyset$ and your question boils down to whether $v_1, \ldots, v_s$ generate the ideal of $Y$ (in a neighborhood of $Y$ in $X$), which should in general be false. Am I missing something?

For example, let $X := Z(x_0^2x_2 - x_1^3) \subseteq \mathbb{P}^2$. The singular set of $X$ is $Y := \lbrace(0:0:1)\rbrace$ (with respect to homogeneous coordinates $(x_0: x_1: x_2)$ of $\mathbb{P}^2$). Let $L$ (resp. $V$) be the linear system with basis $x_0, x_1, x_2$ (resp. $x_0, x_1$). Then $S = \mathbb{P}^1$ and $T = \lbrace(0:1)\rbrace$. Therefore, if we take $v_1 := a_0x_0 + a_1x_1$ with $a_1 \neq 0$, then $Z(v_1) \cap T = \emptyset$ and consequently $P \cap Y = \emptyset$. Let $U$ be the affine neighborhood of $Y$ in $\mathbb{P}^2$ with coordinates $u_0 := x_0/x_2$ and $u_1 := x_1/x_2$. Then ideal of $Y$ on $U \cap X$ is $\mathcal{I} := \langle u_0, u_1 \rangle$ and the ideal generated by $v_1$ is $\mathcal{J} := \langle a_0u_0 + a_1u_1 \rangle$. Since the ideal in $\mathbb{C}[x,y]$ generated by $a_0u_0 + a_1u_1$ and $u_0^2 - u_1^3$ does not equal the ideal generated by $u_0$ and $u_1$, it follows that $\mathcal{I} \neq \mathcal{J}$.

Edit: The heuristics in the first paragraph remains valid in the case that $X$ is normal. Below I give an explicit example where $X$ is a normal surface. I don't know anything about secant varieties to comment about the validity of the statement in that case.

Let $X$ be the weighted projective space $\mathbb{P}^2(1, 1, 2)$. We view $X$ as the toric surface corresponding to the polygon $\mathcal{P}$ which is the triangle in $\mathbb{R}^2$ with vertices $(0,0)$, $(2,0)$ and $(0,4)$. Let me draw $\mathcal{P} \cap \mathbb{Z}^2$.

 
| | | | |
x-o-o-o-o-
| | | | |
x-o-o-o-o-
| | | | |
x-x-o-o-o-
| | | | |
x-x-o-o-o-
| | | | |
x-x-x-o-o-

Here I marked the integral points which belong to $\mathcal{P}$ by 'x' and the others by 'o' (the coordinates of the point at the bottom-left corner being $(0,0)$). Since $|\mathcal{P} \cap \mathbb{Z}^2| = 9$, it follows that $X$ is isomorphic to a subvariety of $\mathbb{P}^8$. Denote the homogeneous coordinates of $\mathbb{P}^8$ by $z_\alpha$ for all $\alpha \in \mathcal{P} \cap \mathbb{Z}^2$. Then the equations of $X$ in $\mathbb{P}^8$ determined by relations between $x_1^{\alpha_1}x_2^{\alpha_2}$ for all $\alpha := (\alpha_1, \alpha_2) \in \mathcal{P} \cap \mathbb{Z}^2$.

Let $L$ be the linear system with basis $\lbrace z_\alpha \rbrace$ and $V$ be the subspace of $L$ with basis $\lbrace z_\alpha : \alpha \neq (2,0) \rbrace$. Then you can check that $Y := BS(V)$ (as a set) consists of the only singular point of $X$ and the blow-up $\tilde X$ of $X$ along $Y$ is non-singular. Moreover, $f^{-1}(Y)$ is a curve. Finally, $S$ is the toric surface corresponding to the polygon

 
     | | | | |
     x-o-o-o-o-
     | | | | |
     x-o-o-o-o-
Q := | | | | |
     x-x-o-o-o-
     | | | | |
     x-x-o-o-o-
     | | | | |
     x-x-o-o-o-

It follows that $S$ is non-singular, and $\dim (\tilde \phi(f^{-1}(Y))) \leq 1$. Therefore, for generic $v_1, v_2 \in V$, $Z(v_1) \cap Z(v_2) \cap \tilde \phi(f^{-1}(Y)) = \emptyset$, and consequently, $P \cap Y = \emptyset$. We claim that there is a neighborhood $U$ of $Y$ such that the ideal of $Y$ on $U$ can not be generated by $2$ elements.

Indeed, let $U := X \setminus Z(z_{(2,0)})$. Then $U \cong \text{Spec}~ \mathbb{C}[x^{-1}, x^{-1}y, x^{-1}y^2] \cong \text{Spec}~ (\mathbb{C}[u,v,w]/\langle uw - v^2 \rangle)$ and $Y = Z(u,v,w) \subseteq U$. Since $uw - v^2$ is a homogeneous polynomial of degree $2$, the ideal generated by $u$, $v$ and $w$ in $\mathbb{C}[u,v,w]$ does not equal the ideal generated by $uw-v^2$, $g_1$ and $g_2$ for all $g_1, g_2 \in \mathbb{C}[u,v,w]$. This proves the claim and completes the counter example.

Are there non-projective normal surfaces which are rational?

2011-01-05T06:26:43Z

Every non-singular complete surface is projective. On the other hand, there are non-projective complete surfaces (see e.g. Excercise II.7.13 of Hartshorne) - and there are such examples where the surface is also normal (see e.g. this ). All the examples I have seen of complete normal non-projective surfaces are non-rational. Hence the question: are there (complete) rational non-projective normal surfaces?

Edit: I just saw a previous question which asked for examples of normal non-projective varieties. So I guess this is a sub-question of that one.

Can a curve intersect a given curve only at given points?

2010-11-30T11:21:54Z

Clearly the question in the title has a positive answer for analytic (or smooth, or continuous ...) curves, but what about the algebraic category? More specifically, given an irreducible polynomial curve $X \subseteq \mathbb{CP}^2$ and points $P_1, \ldots, P_k$ on $X$, when can we find another curve $Y$ (defined by a polynomial) such that the $Y$ intersects $X$ only at $P_1, \ldots, P_k$?

I find the question to be nontrivial even for $k = 1$. Here are some observations for $k = 1$ case:

If $P$ is a point on $X$ with multiplicity $\deg X - 1$, then a tangent of $X$ through $P$ intersects $X$ only at $P$ (by Bezout's theorem).
If $X$ is a rational curve and $X \setminus {P} \cong \mathbb{C}$, then there is a curve $Y$ such that $X \cap Y = {P}$.
Let $X$ be a non-singular cubic. Give it a group structure such that the origin is an inflection point. Then for all $P \in X$, there exists $Y$ such that $Y \cap X = {P}$ iff $P$ is a torsion point in the group.

If $X$ (of degree $d$) is non-singular at $P$, then the most direct approach for finding a $Y$ of degree $e$ intersecting $X$ only at $P$ seemed to blow it up $de$ times and look for the conditions under which $Y$ goes through each of the points on $X$ in the $i$-th infinitesimal neighborhood of $P$, $0 \leq i \leq de - 1$. But the conditions on the coefficients of the polynomial defining $Y$ did not appear very tractable.

Edit: I would like to make a correction to observation 3. This is what I know about a non-singular cubic curve $X$: If $P$ is an inflection point, then there is a curve $Y$ such that $Y \cap X = P$ (take $Y$ to be the tangent of $X$ at $P$). If $P$ is a non-torsion point (for the group structure on $X$ for which the origin is an inflection point), then there is no such $Y$. I don't know what happens for torsion points.

Answer by auniket for CM for radical ideal

2010-11-26T14:28:02Z

Yes. From Eisenbud's Commutative Algebra: a ring $S$ is Cohen-Macaulay iff all the maximal ideal $m$ of $S$ satisfies codim($m$) = depth($m$). Now, the maximal ideals of $R/J$ are the same as $R/I$ and their depths and codimensions are the same as well.

Any implemented algorithm to compute the closure of an affine variety in a product of projective spaces?

2010-08-30T06:16:57Z

Let $I$ be an ideal of $k[x_1, \ldots, x_m, y_1, \ldots, y_n]$, $k$ being a field. Does any of the computer algebra systems implement any algorithm to calculate the generators of the 'bi-homogenization' $\tilde I$ of $I$ with respect to $x$ and $y$ variables?

(Recall that the 'bi-homogenization' of a polynomial $f = \sum a_{\alpha, \beta} x^\alpha y^\beta$ is by definition $\tilde f := \sum a_{\alpha, \beta} x^\alpha y^\beta x_0^{d - |\alpha|} y_0^{e- |\beta|}$, where $x_0$ and $y_0$ are two new variables, $d := \deg_x(f)$ and $e := \deg_y(f)$. Then $\tilde I := ${$\tilde f: f \in I$}.)

My motivation is geometric: to find the closure $\overset{-}{V}$ of a subvariety $V$ of $k^{m+n}$ in $\mathbb{P}^m \times \mathbb{P}^n$. Of course I could as well calculate the Segre embedding of $\overset{-}{V}$ in $\mathbb{P}^{mn + m +n}$, but I would like to have something computationally less expensive.

I can think of an algorithm which involves introducing $n$ (or $m$, whichever is the smaller) new variables $t_1, \ldots, t_n$ and computing the monomial basis of an ideal $J$ in $k[x,y,t]$, where $J$ is to be constructed from $I$. But I was wondering if someone had already implemented some (possibly better) algorithm which would do this job.

Answer by auniket for Jokes in the sense of Littlewood: examples?

2010-09-25T08:16:22Z

You can't prove something exists just by computing the probability of its existence - right? The first application of the "probabilistic arguments" in Combinatorics I have encountered was this; took me a long time to get it.

Tensor product of a line bundle with a large multiple of another positive line bundle also positive?

2010-09-05T21:31:25Z

Let $X$ be a complex manifold and $\mathcal{L}$ be a positive line bundle on $X$. If $E$ is any other line bundle on $X$, then is it true that for all sufficiently large $m$, $\mathcal{L}^m \otimes E$ is also positive?

When $X$ is compact, the answer is positive, and it follows by a standard compactness argument if you start with the definition that $\mathcal{L}$ is positive iff the Chern class $\omega$ of $\mathcal{L}$ satisfies: $\omega(x; v, Iv) > 0$ for all $x \in X$ and $v \in T_{\mathbb{R}, x}(X)$ (the real tangent space of $X$ at $x$) and $I: T_{\mathbb{R}, x}(X) \to T_{\mathbb{R}, x}(X)$ is the map induced by multiplication by $i$.

So my real question is: is the above question true when $X$ is not compact? What if $X$ is an affine algebraic variety?

Appropriate journal to publish a determinantal inequality

2010-08-29T20:41:56Z

I have recently made the following observation:

Let $v_i := (v_{i1}, v_{i2})$, $1 \leq i \leq k$, be ~~non-zero~~ positive elements of $\mathbb{Q}^2$ such that no two of them are proportional. Let $M$ be the $k \times k$ matrix whose entries are $m_{ij} := \max${$v_{ik}/v_{jk}: 1 \leq k \leq 2$}. Then $\det M \neq 0$.

The above statement is equivalent to the basic case of a result I recently discovered about pull back of divisors under a birational mapping of algebraic surfaces. I was going to include it as a part of another paper, then noticed the equivalent statement stated above and found it a bit amusing. My question is: is it worthwhile to try to publish it in a journal (as an example of an application of algebraic geometry to derive an arithmetic inequality), and if it is, then which journal(s)?

It is of course also very much possible that it is already known, or has a trivial proof (or counterexample!) - anything along those directions would also be appreciated.

Edit: Let me elaborate a bit about the geometric statement. In the 'other' paper, I define, for two algebraic varieties $X \subseteq Y$, something called "linking number at infinity" (with respect to $X$) of two divisors with support in $Y \setminus X$. I can show that when $Y$ is a surface, (under some additional conditions) the matrix of linking numbers at infinity of the divisors with support in $Y \subseteq X$ is non-singular. In a special (toric) case, the matrix of linking numbers takes the form of $M$ defined above. So the question is if the result about non-singularity of the matrix and its corresponding implication(s) are publishable anywhere.

Answer by auniket for Applications of compactness

2010-06-11T16:02:28Z

Gordan's lemma is another application of "compact + discrete => finite", but it is one of the basic building blocks of the theory of toric varieties.

In some sense the theory of division by polynomials (i.e. the Gröbner basis theory) is a manifestation of compactness, e.g. Dickson's lemma can be stated and proved as an application of compactness, see e.g. the wikipedia entry. I guess this is an example of the principle mentioned in the answer of Michael Greinecker.

Answer by auniket for Knuth's intuition that Goldbach might be unprovable

2010-06-11T01:39:48Z

Again from combinatorics: The strengthened finite Ramsey theorem . It was proved by Paris and Harrington that it is true, but not provable in Peano arithmetic. Wikipedia says: "This was the first "natural" example of a true statement about the integers that could be stated in the language of arithmetic, but not proved in Peano arithmetic".

Can Cantor set be the zero set of a continuous function?

2010-05-09T17:57:46Z

More generally, can the zero set $V(f)$ of a continuous function $f : \mathbb{R} \to \mathbb{R}$ be nowhere dense and uncountable? What if $f$ is smooth?

About two days ago I discovered that in this proof I am working on, I have implicitly assumed that $V(f)$ has to be countable if it is nowhere dense - hence this question.

Is the absolutely continuous image of a nowhere dense set is also nowhere dense?

2010-05-10T04:27:16Z

Let $f: [a, b] \subseteq \mathbb{R} \to \mathbb{R}$ be an absolutely continuous map. Does $f$ map a nowhere dense subset of $[a, b]$ to a nowhere dense set?

Remarks:

The answer is "no" if $f$ is only assumed to be continuous and almost everywhere differentiable, e.g. take the Cantor function .
If $f$ is assumed to be $C^1$, then the answer is yes - a nice proof can be found at this page. Essentially the same proof works if it is assumed that $f$ is not differentiable at at most countably many points. Edit: I would like to retract the previous sentence. Now I don't see why it should be true.

Comment by

2013-05-06T19:56:01Z

Hi Oleg, the comment of Steven Landsburg on this question: mathoverflow.net/questions/129673/… implies that Samuel's book should have a reference to at least the first question of yours.

Comment by

2013-05-05T12:38:04Z

@Steven: Thanks!

Comment by

2013-04-27T01:07:02Z

Hi Philip, this article (of mine) gives a necessary and sufficient criterion for algebraicity in a special case: arxiv.org/abs/1301.0126 PS: I myself am interested in your Question 1, and I don't know of any other reference other than Grauert's original article, which is in German and therefore I can't read :(

Comment by

2013-02-14T20:28:24Z

Is $a_{I,m}$ by definition $\lim_{k\to \infty} \sigma_{I,m}(k)/kh_I(k)$? If not, what is it?

Comment by

2013-01-24T17:44:43Z

Hi Isac, A look at Chapter 5 of people.reed.edu/~iswanson/book/SwansonHuneke.pdf might help.

Comment by

2012-12-31T13:33:34Z

@Jason: OK, I admit it wasn't such a good choice of words :) Would you prefer if I change "something non-algebraic" to "non-algebraic surfaces"?

Comment by

2012-12-30T21:41:36Z

@Angelo: Thanks! But I knew of these (perhaps should have mentioned them in the question) and they do not contain (and as far as I can see, do not shed any light on the construction of) any such examples.

Comment by

2012-12-12T20:26:24Z

Hi Francesco, the projection from a generic $p \in L$ may not be birational! The correct answer seems to be that either $L$ is a component of $C$, or there are hyperplanes $H_1, \ldots, H_m$ containing $L$ and curves $C_j \subseteq H_j$ of degree $d_j$ such that $1 \leq m \leq n$, $d_1 + \cdots d_m = d$ and $C = C_1 \cup \cdots \cup C_m$.

Comment by

2012-12-09T18:29:42Z

Oops! I just voted up and destroyed the magic of 27! I am really sorry, now if I downvote it becomes 26!

Comment by

2012-11-13T23:44:54Z

Oops! I can see that I was stupid. I will cross the answer out until I have something more useful to say.

Comment by

2012-06-02T19:38:28Z

Thanks! This is pretty close to what I wanted.

Comment by

2012-03-27T05:49:33Z

Yeah! Careless mistake ...

Comment by

2012-03-27T05:32:36Z

$F$ is proper in a neighborhood of $F^{-1}(c)$ means that there is an open set $V$ containing $c$ such that $F$ restricted to $F^{-1}(V)$ is proper (i.e. for every compact subset $Z$ of $V$, $F^{-1}(Z)$ is also proper). E.g. $F := \mathbb{C}^2 \to \mathbb{C}^2$ be defined by $u = x^2y - x + y$ and $v = xy$ (where $(u,v)$ are the coordinates in the 'target'). Then $F$ is not proper at $F^{-1}(c)$ for every $c$ on the line $v = 1$.

Comment by

2012-03-27T04:52:32Z

Answer to 1) should be always true, I think. However, for 2), you need some sort of properness condition. More precisely, if $F$ is not proper in a neighborhood of $F^{-1}(c)$, then there is a curve $\gamma(t)$ such that as $t$ goes to infinity, $\gamma(t)$ goes to infinity and $F(\gamma(t))$ goes to $c$. Since there can be only finitely many isolated solutions of $F(x) = c$,your condition (2) will be violated.

Comment by

2012-03-16T06:23:36Z

No - any $m\times n$ matrix can be expressed as $BAB$ for some $m\times n$ matrix $B$ and $n\times m$ matrix $A$. In fact for each $B$ you can choose $A$ so that $BAB=B$ (to see it note that you can express any $B$ as $EJF$, where $E,F$ are invertible and $J$ is a block matrix with 4 blocks such that the left most block is a $p \times p$ identity matrix for some $p\leq\min(m,n)$ and the other blocks are zero matrices. Consequently $BAB=EJA'JF$, where $A'=FAE$. Choosing $A$ so that $A'$ has the "same" form as $J$ (of course the zero blocks have to have different dimensions) gives the result.