14

Note that without any assumption on the singularities the answer to your question is yes. For instance, consider the hypersurface $Y = \{x_0^2+x_1^3+x_2^4\}\subset\mathbb{P}^3$. Then $Sing(Y) = [0:0:0:1]$. The projective tangent cone of $Y$ in the singular point is not reduced. Therefore, the singularity is not ordinary. Let $\pi:X\rightarrow\mathbb{P}^3$ be ...

13

I believe that $(Y,B)$ is always klt for some boundary $B$. In fact by Theorem 5.2 of https://projecteuclid.org/download/pdf_1/euclid.jdg/1090347529, after passing to a truncation of the pluricanonical rings, there is a klt pair of log general type $(Y',B')$ with isomorphic pluricanonical ring (more precisely $R(K_X)^{(a)}\cong R(K_{X'}+B')^{(b)}$ for ...

13

In general, the answer is no. It already fails for a nodal curve. In fancier terms, you can understand the obstruction as follows: if $X$ has a resolution $\tilde X$, and the cohomology of $X$ injects the cohomology of $\tilde X$, then $H^i(X)$ would be pure of weight $i$ as a Galois module/mixed Hodge structure (when over $\mathbb{C}$).

11

As pointed out by Alex in his comment, this is in general not true.
For instance, consider the case $N=d=n=m=2$. Then $|L|$ is the linear system of plane curves of degree $2$ passing through $p_1$ and $p_2$ with multiplicity $2$. Of course there is only one such a curve, namely the line through $p_1$ and $p_2$ counted with multiplicity $2$. In particular, $|...

answered Mar 16 '14 at 0:50

Francesco Polizzi

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11

Put $d:=\deg(X)$. From the exact sequence
$$0\rightarrow \mathcal{O}_X(-d)\rightarrow \Omega ^1_{\mathbb{P}^n|X}\rightarrow \Omega ^1_X\rightarrow 0$$you get an exact sequence $\ 0\rightarrow T_X\rightarrow T_{\mathbb{P}^n|X}\rightarrow \mathcal{O}_X(d)\rightarrow \mathcal{E}xt^1(\Omega ^1_X,\mathcal{O}_X)\rightarrow 0$.
From this you find easily $H^i(X,T_X)=...

ag.algebraic-geometry deformation-theory projective-geometry singularity-theory resolution-of-singularities

11

Somehow that question slipped my radar, sorry!
The truth is that shamefully I'm not able to say much, as I don't have a strong knowledge of resolution of singularities. But at least so far, the flow of information has been from characteristic $p$ to characteristic $0$, or more specifically to mixed characteristic. So maybe I wouldn't be surprised if once ...

10

The answers are the following.
(1) It is well known that the singularity at the vertex of the cone over the Veronese surface is isomorphic to a quotient singularity of type $\frac{1}{2}(1, \, 1, \,1)$, that is the isolated singularity given by the quotient of $\mathbb{C}^3$ by the action of $\mathbb{Z}/2 \mathbb{Z}$ of the form $(x, \,y, \,z) \mapsto (-x, \,...

answered Dec 30 '14 at 15:38

Francesco Polizzi

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10

Consider $A=k[[x^3,x^2y,y^3]]\subset k[[x^3, x^2y, xy^2, y^3]]=B$. $B$ is the integral closure of $A$, $A$ is a hypersurface, but $B$ is not Gorenstein.

9

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 ...

9

The only restriction I see is that $\hat{X}$ must be normal (because $X$ is): if $\phi$ is a rational function on (some affine open subscheme of) $\hat{X}$ which is integral over $\mathscr{O}_\hat{X}$, then $\phi\circ\pi$ is integral over $\mathscr{O}_{X}$, hence in $\mathscr{O}_{X}$. In other words, $\phi$ lies in $\pi_*\mathscr{O}_{X}=\mathscr{O}_\hat{X}$.

answered Mar 26 '15 at 14:59

Laurent Moret-Bailly

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9

It is easy to see that the canonical class of $X'$ is trivial (combine the formula for the canonical class of the blowup of $A^3$ and the adjunction formula). Hence $X'$ is crepant. But a crepant resolution of a surface is unique, so $X = X'$.
By the way, if you look at the exceptional divisor of $X'$, you will note that this is a union of two curves (in ...

8

Background. The threefold compound du Val singularities have been introduced by Miles Reid in the 1980s [R1, R2, R3]. Their geometric description is that a general hyperplane section through the singular point is a du Val singularity. It has been proved by Miles Reid that isolated compound du Val singularities are precisely Gorenstein terminal singularities, ...

8

You gave the definition of normal crossing divisor. The definition of simple normal crossing is the following.
A Weil divisor $D = \sum_{i}D_i \subset X$ on a smooth variety $X$ of dimension $n$ is simple
normal crossing if any component $D_i$ is smooth and for every point $p \in X$ a local equation of $D$ is $x_1\cdot...\cdot x_r$ for independent local ...

8

$\hat{X}$ can be as bad as you want. For example, take your favorite non-Gorenstein variety $\hat{X}$ in $\mathbb{A}^N$. By Noether Lemma there is a finite morphism $\hat{X} \to \mathbb{A}^n =: Y$. Take $X$ to be a resolution of singularities of $\hat{X}$. Then $X \to Y$ is a quasifinite morphism between smooth varieties.

8

Edit. This follows from the Elkik-Fujita Vanishing Theorem. There is a more general vanishing theorem due to Elkik and Fujita. One version of this theorem (where I read the theorem) is Theorem 1.3.1 of the following article.
MR0946243 (89e:14015)
Kawamata, Yujiro; Matsuda, Katsumi; Matsuki, Kenji
Introduction to the minimal model problem. Algebraic ...

7

It is a good exercise to make the computation in the simplest possible case: $X=\mathbb{C}^2$, $G=\mathbb{Z}/2$ acting by the antipodal involution, so that $Y$ is a quadratic cone, $\tilde{Y}$ obtained by blowing up the vertex of the cone. Locally $\tilde{X} :=X\times _Y\tilde{Y} $ is given in $\mathbb{C}^3$ by the equations $x(y-tx)=y(y-tx)=0$. That is, $\...

7

I assume that you mean $x_1=x_2=0$ in your description of the divisors. I believe that you only really get two different resolutions, but yes, they are both projective varieties. Hodge numbers of birational complex Calabi-Yau varieties are always the same.
You can think of this as a determinantal variety for the matrix $A$ of the form
$$
\Big(\begin{array}{...

7

Here is a proof of $d = \frac{1}{2}$. Then you can argue as Francesco did in point two of his answer.
Consider the action:
$$
\begin{array}{ccc}
\mu_{2}\times\mathbb{A}^{3} & \longrightarrow & \mathbb{A}^{3}\\
(\epsilon,x_{1}, x_{2}, x_3) & \longmapsto & (\epsilon x_{1},\epsilon x_{2}, \epsilon x_3)
\end{array}
$$
The ring of invariants is ...

7

I think the answer is no. In your notation, take $n=5$, $p=1$ and $q=2$. If you consider the blowup at the origin $X\to \mathbb C^2$, then $\mathbb Z_5$ acts on $X$ with two isolated fixed points: at one of the points the action has $p=q=1$ and at the other the action has $p=2$, $q=4$, which is the same as $p=1$, $q=2$. So the blow up introduces an isolated ...

7

I will try to answer the first question only.
As in the remarks, the canonical reference is
Beauville, Narasimhan, Ramanan, Spectral curves and the generalised theta divisor. J. Reine Angew. Math. 398 (1989), 169–179. https://doi.org/10.1515/crll.1989.398.169
First part of your question, about the discriminant locus :
Let $s=(s_1,\cdots,s_N)\in \...

7

Your definition is the usual one. More or less. Probably you should assume that $K_X$ is Cartier or at least $\mathbb Q$-Cartier for the definition to even make sense.
I don't think there is a detailed intro to crepant resolutions. Perhaps because they are rare. It is a strong condition on the singularity that it admits a crepant resolution, so I would be ...

7

I'm assuming that you know "where" in the commensurability class your lattice is. By this, I mean you perhaps have $\Gamma$ as a subgroup of some principal arithmetic lattice $\Lambda$ of known index, e.g., as a subgroup of $\mathrm{PU}(h, \mathcal{O}_k)$ where $h$ is a hermitian form on $k^3$ for an appropriate number field $k$. You then know the ...

ag.algebraic-geometry complex-geometry singularity-theory kahler-manifolds resolution-of-singularities

7

The answer is yes for any reduced equidimensional analytic space. This is the proposition 3.2 (and remark after the proof) page 479 of Variétés polaires II by Bernard Teissier.

ag.algebraic-geometry reference-request complex-geometry singularity-theory resolution-of-singularities

6

Note that $-1 = i^2$. Thus your action is $(x,y)\mapsto (i^2x,iy)$. You can interpret this in the following way. Let $\epsilon\in \mu_4$ and consider tha action given by $(x,y)\mapsto (\epsilon^2 x,\epsilon y)$. The quotient $X = \mathbb{C}^2/\mu_4$ is given by $X = (w^2-uv = 0)\subset\mathbb{A}^3$. This is because as you observed the invariant polynomials ...

6

For what it's worth, one can say the following sort of thing.
Since $Y$ is log terminal so is $(\hat{X}, -\mathrm{Ram})$. This doesn't mean much since in the pair, the boundary has a negative coefficients (ie, the singularities of $\hat{X}$ can be arbitrarily bad). But it does say things like:
if $\hat{X}$ has really bad singularities at some points, ...

6

I am turning Giulia's comment into a CW answer. At over 400 pages, I think that "Travaux de Gabber", published in Astérisque, ought to count as a "comprehensive treatment of [some aspects of ] étale cohomology" using the method of alterations. The style is very SGA, and indeed there is a nonempty intersection among the set of authors.

6

By the Hochster-Roberts Theorem it is Cohen-Macaulay
The canonical sheaf of $\mathbb A^n$ is $G$-invariant (because the elements of $G$ have det=1) and hence it descends to the canonical sheaf of the quotient, which is then a line bundle.

6

Choose five points $P_i$ in general linear position (all choices are
equivalent under ${\rm PGL}_4$, so you might as well put four of them
at the coordinate vectors and the fifth at $(1:1:1:1)$); then
at each $P_i$ the condition of a triple point imposes $1+3+6 = 10$
linear conditions on the space of quintics, which has dimension
${8 \choose 5} = 56$, so you ...

answered Sep 3 '17 at 3:42

Noam D. Elkies

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5

In other words, you want the normalization of $X$ to be smooth, plus a condition on the embedding: the pull back of $\mathcal{O}_X(1)$ to $\tilde{X} $ should be very ample. The latter property depends on the embedding and is not always true, even for curves: see this post and the answers there.

5

I suspect that you will have trouble with trying to "simultaneously" resolve $X$ and $X/G$. We actually run into this issue in the joint paper with Anatoly Libgober arXiv:math/0206241 (although it may not be evident from the paper) and consequently had to settle for working with $\hat X\to\hat Y$ which was $G$-equivariant morphism of smooth varieties which ...

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