17

This is very much a question of recent research, solved with varying degrees of generality in the papers listed below. Here is a summary.
The rational map $\mathbb{P}^{2[8]}\dashrightarrow \mathbb{P}^2$ from the Hilbert scheme of $8$ points in $\mathbb{P}^2$ is the map corresponding to an extremal effective divisor on the Hilbert scheme. This suggests ...

15

Already for a finite group $\Gamma$, the only integral closed substacks of $B\Gamma$ are the empty stack and all of $B\Gamma$. So your naive Chow group would be cyclic (typically zero) in each degree. On the other hand, the "correct" Chow group of divisor classes should be the Pontrjagin dual group $\text{Hom}_{\text{Group}}(\Gamma,k^\times)$ (here $k$ is ...

11

In principle, the Chow rings of Hilbert schemes of length $d$ subschemes in $\mathbb{P}^2$ are known (though it may still be a nontrivial task to extract information from the known descriptions). Here are some literature references. (Note that some of these talk about integral cohomology or homology, but because of the Bialynicki-Birula cell structure the ...

10

You don't say what kind of space $X$ and $Y$ are subspaces of. But if they sit in an oriented manifold there's an easy way to define an intersection product in homology. Namely if $M$ is an oriented $d$-manifold then there is a Poincaré duality isomorphism
$$ H_i(M,\mathbf Z) \cong H^{d-i}_c(M,\mathbf Z)$$
between homology and compactly supported cohomology. ...

10

This is called the "bigon criterion". For a discussion, see Section 1.2.4 (and in particular Proposition 1.7) of the "Primer on mapping class groups" by Farb and Margalit.
The Google search "bigon criterion" also finds various references and lecture notes. For example, here is the top hit:
https://math.stackexchange.com/...

9

By compactness, there is a pair of points at the minimal distance. On the other hand, any positive distance can be made smaller (e.g., cf. the proof of Lefschetz theorem in Milnor's "Morse theory": the critical points of the distance function to a complex manifold always have positive index).

9

The Poincaré-Hopf theorem is a consequence of the Lefschetz fixed point theorem if you accept the fairly standard fact that the Euler characteristic of a compact manifold $M$ is equal to the self-intersection of the diagonal $\Delta \subset M\times M$. Here is why :
Since $M$ is compact, the vector field $X$ gives rise to a flow $(\varphi _t)_{t\in\mathbb{R}...

at.algebraic-topology dg.differential-geometry differential-topology intersection-theory fixed-point-theorems

8

### What does taking the graded algebra do to the Grothendieck group, and its relation to the Chow ring?

Does this help?
$K(X) \otimes \mathbb{Q}$ is a graded ring in the sense that it is isomorphic (by the Chern character map) to $A(X) \otimes \mathbb{Q}$, which is graded. I would perhaps prefer to say that it is "gradeable", since the grading isn't very obvious in terms of $K$-theory. The most $K$-theoretic way I know to describe it is that the Adams ...

8

I suppose you want $f$ to be surjective, otherwise $f_*D$ is not defined. Then $f_*D$ is nef: for any curve $C\subset Y$, $\ (f_*D\cdot C)=(D\cdot f^*C)\geq 0$. But it might very well be ample. Consider a smooth quadric $Q\subset \mathbb{P}^3$, and let $f:Q\rightarrow \mathbb{P}^2$ be the projection from a point outside $Q$. Let $D$ be a line contained in $...

8

Note that Nitsure's paper is part of the book FGA Explained. There is a proof of Snapper's lemma in Theorem B.7 of Appendix B ("Basic intersection theory" by Kleiman) in the same book. Kleiman's part of the book can also be found independently on arXiv: 0504020 and contains the relevant appendix.

answered Mar 29 at 20:10

R. van Dobben de Bruyn

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7

There are some special cases where the CSM classes of singular varieties can be computed easily.
For example there is Ehler's formula for $c_{SM}(X)$ where $X$ is any complete toric variety. Let $\Sigma$ be the fan of $X$ with torus orbits $B_\sigma$ for $\sigma \in \Sigma$, then the CSM class of $X$ is given by
$$
c_{SM}(X) = \sum_{\sigma \in \Sigma}[\...

7

I am sure there are more direct references, but it is fairly easy to prove as well. First of all, the Hilbert scheme $\text{Hilb}_{2m+1}(\mathbb{P}^n)$ is a $\mathbb{P}^5$-bundle over the Grassmannian $G(3,n+1)$. There must be other sources, but one source is Theorem 3.4.1 of Alex Lee's senior thesis.
Alex Lee
The Hilbert Scheme of Curves in $\mathbb{P}^3$...

7

Edit. The odd Bernoulli numbers are zero, of course! So the first attempt below is wrong. The revised examples use complete subvarietes of moduli spaces of curves. These revised examples use nonvanishing of the even Bernoulli numbers.
Revised examples from complete subvarieties of moduli spaces of curves. Let $g\geq 2$ be an integer. Denote by $\mathcal{...

7

Counterexamples. Here are examples showing that each of the properties above can fail. Let $X$ be a smooth cubic surface in $\mathbb{P}^3_k$, where $k$ is a field. Let $H$ be a smooth hyperplane section of $X$. This is a smooth, geometrically connected, projective curve of genus $1$ (a plane cubic).
The Fano scheme of lines on $X$, $F(X)$, is a finite, ...

7

The scroll $S_{(1,k)}$ can be locally parametrized by the map
$$
\begin{array}{cccc}
\phi: & \mathbb{A}^1\times\mathbb{P}^1 & \longrightarrow & \mathbb{P}^{k+2}\\
& (u,[\alpha_0:\alpha_1]) & \mapsto & [\alpha_0 u:\alpha_0:\alpha_1 u^k:\alpha_1 u^{k-1}:\dots:\alpha_1 u:\alpha_1].
\end{array}
$$
Now, consider the Segre embedding
$$
...

ag.algebraic-geometry projective-geometry birational-geometry intersection-theory complete-intersection

6

I've just taught this in my graduate class. Check these notes on intersection theory.
The result you want is contained in Thm. 4.7. Again, you need $2\dim S=\dim X$.

answered Mar 8 '13 at 13:13

Liviu Nicolaescu

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6

The answer is yes. Let $[E^{-1} \rightarrow E^0]$ be a perfect obstruction theory on $M$. After localizing in $M$ we can assume that the map $E^0 \rightarrow \Omega_M$ is induced as $\mathcal{O}_M \otimes \Omega_{\mathbf{A}^n} \rightarrow \Omega_M$ for some map $M \rightarrow \mathbf{A}^n$.(*) As the map on differentials is surjective the relative ...

6

By recursion on $k$.
If $k=1$, this is Bezout's theorem in projective space. Now let $k>1$. We have the Pl\"ucker embedding $\mathbb{G}(k,n) \subset \mathbb{P}(\bigwedge^k \mathbb{C}^n)$. Let $p$ be a general point in $\mathbb{C}^n$. Let $Y_i = \mathbb{P}(\bigwedge^{k-1} \mathbb{C}^n/\langle p\rangle) \cap X_i$.
The variety $Y_i$ is the variety of ...

6

If $C$ is a smooth projective curve of genus $g \geq 3$ and $J(C)$ is the Jacobian of $C,$ then an Abel curve $C \subset J(C)$ is not algebraically equivalent to its image $-C$ under the negation automorphism, even though $C$ is homologically equivalent to $C.$ This was proved by Ceresa in the paper
https://www.jstor.org/stable/2007078
EDIT: Ceresa also ...

6

No.
This is an example with irreducible (and nonsingular) $A,B,C$: Consider $\mathbb{A}^4$ with coordinates $(w,x, y, z)$. Let $B$ be the $(x,y)$-plane (i.e. the set $w = z = 0$), $C$ be the hypersurface $z = xy$, and $A$ be the $(y,z)$-plane. Then $B \cap C$ is the union of the "$x$-axis" and "$y$-axis". Let $V_1$ be "$x$-axis" ...

5

Chern classes of a rank $r$ vector bundle vanish in degree greater that $r$ while Segre classes do not. On the other hand the definition of Segre classes easily generalizes to singular vector bundles such as cones (http://math.stanford.edu/~vakil/245/245class14.pdf). Normal cones are cental objects in deformation theory. For instance the generalization of ...

5

Interesting question. I think the answer is yes, let me try to prove it.
As you noticed, the ideal sheaf sequence shows that $h^1(D)=0$ is equivalent to the fact that $H^0({\mathcal O}_D)$ is 1-dimensional generated by the constant function $1$.
Considering, for every effective decomposition $D=A+B$, the exact sequence
$$
0 \rightarrow {\mathcal O}_B(-A) \...

5

I believe this is a special case of a more general fact; I am not sure of all the signs off the top of my head, but here is the idea.
If $M$ and $N$ are orientable $d$-manifolds, the Künneth theorem gives
$$H^d(M \times N; \mathbb{Q}) \cong \bigoplus_k H^k(M; \mathbb{Q}) \otimes H^{d-k}(N; \mathbb{Q}).$$
To the second factor we first apply Poincare ...

5

You might be interested in the bivariant theories of Fulton and MacPherson.

5

Let $Y\subset\mathbb{P}^n$ be a smooth variety, and let $\epsilon:X = Bl_Y\mathbb{P}^n\rightarrow\mathbb{P}^n$ be the blow-up of $\mathbb{P}^n$ along $Y$.
Let $\widetilde{H}$ be the pull-back of the hyperplane section $H$ of $\mathbb{P}^n$, and $E$ be the exceptional divisor. If $H_Y =H\cdot Y$ we have
$$\widetilde{H}^{h-i}E^i = p^*H_Y^{n-i}\cdot i^*E^{i-1}...

5

If $\frac{2^{n-1}}{n}$ is an integer, that is if $n$ is a power of $2$, then $C$ is actually the set-theoretic complete intersection of $n-1$ quadrics.
This is a theorem of Perron (1941); to see the defining equations see formula (3) of my paper http://www.dima.unige.it/~torrente/RationalNormalCurves.pdf.
Furthermore, for a general degree $n$, Gallarati and ...

5

Regarding your main question, this is done in Cassels, Lectures on elliptic curves, $\S$ 8 (iv) p. 36. We may assume that the common rational point of the quadrics is $(X:Y:Z:T)=(0:0:0:1)$. Then the quadrics have the shape
\begin{align*}
Q_1 & = TL + R\\
Q_2 & = TM + S
\end{align*}
where $L,M$ (resp. $R,S$) are linear (resp. quadratic) in $X,Y,Z$. ...

answered Jul 12 '17 at 11:41

François Brunault

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5

Yes. Put $U:=\widetilde{X}\smallsetminus E=X\smallsetminus W$. By the open-closed exact sequence, $\alpha $ comes from a class in $H^k_c(U)$. But the map $H^k_c(U)\rightarrow H^k(\widetilde{X})$ factors through $H^k(X)$, hence $\alpha =\pi ^*\beta $ for some class $\beta $ in $H^k(X)$. Applying $\pi _*$ gives $\beta =\pi _*\alpha $, hence your formula.

5

No. Because Chow groups lie in the even-dimensional homology, their multiplication structure is commutative. So the natural map is from $\operatorname{Sym}^p A^1$ to $A^p$.
Also no for dimensions reasons. For instance $p=2$, $A$ is a product of two isogenous elliptic curves, $A^1$ is three-dimensional (four if they're CM) but $A^2$ is one-dimensional.

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