Virtual Class Exercises

1. We know \(S^\bullet\) is a graded \(\mathcal O_X\)-algebra generated in degree \(1\) since \(C\) is a cone over \(X\). This means there is a surjective \(\mathcal O_X\)-algebra homomorphism \(\Sym S^1\twoheadrightarrow S^\bullet\), hence \(S^\bullet \cong \Sym S^1/\mathcal I\) for some ideal sheaf \(\mathcal I \subseteq \Sym S^1\), hence \(C\to A(C)\) is a closed embedding.
2. Suppose \(E = \GlSpec \Sym \mathcal E\) is some other abelian cone such that there exists a closed embedding \(\Sym (\mathcal E)\twoheadrightarrow S^\bullet\). Then in particular \(\Sym(\mathcal E)\) surjects onto \(S^1\), and hence we can lift this to a surjection \(\Sym(\mathcal E) \twoheadrightarrow \Sym (S^1)\). This corresponds to a composition of closed embeddings \(C\hookrightarrow A(C) \hookrightarrow E\) after applying \(\GlSpec(-)\).

The following is a statement of the lemma:

If \(X\hookrightarrow Y\) is a Koszul-regular embedding, then \(C_{X/Y} = N_{X/Y}\) and these cones are vector bundles. If moreover \(Y\) has pure dimension then the rank of \(N_{X/Y}\) is the codimension of \(X\) in \(Y\).

Here \(X\) and \(Y\) are algebraic stacks. By Remark 1.2.3 in the text, Koszul-regularity of \(X\hookrightarrow Y\) is fpqc-local on \(Y\), so we may assume \(X\) and \(Y\) are affine schemes, reducing the problem to showing that \[\Sym(I/I^2) \twoheadrightarrow \bigoplus_{i=0}^\infty I^i/I^{i+1} \] is surjective, where \(I\) is the ideal sheaf of \(X = \Spec (R/I)\) in \(Y = \Spec (R)\). However, by Stacks Section 062D, “Koszul regular sequences”, a Koszul-regular ideal is a quasi-regular ideal, and hence by definition \[(R/I)[x_1,...,x_n] \cong \bigoplus_{i=0}^\infty I^i/I^{i+1}.\] In particular this means \(I/I^2\) is free, and thus \(\Sym(I/I^2) \cong (R/I)[x_1,...,x_n]\).

The statement of Lemma 1.2.4 is

Suppose we have a cartesian diagram of schemes of algebraic stacks

where vertical maps are closed embeddings. Then there is a closed embedding \(C_{X'/Y'}\hookrightarrow C_{X/Y}\times_X X'\).

Suppose then that we’re in this situation. Let \(\mathcal I\) be the defining ideal of \(X\) in \(\mathcal O_Y\) and \(\mathcal I'\) the defining ideal of \(X'\) in \(Y'\). Taking the exact sequence \[0 \to \mathcal I \to \mathcal O_Y \to \iota_*\mathcal O_X\to 0\] and pulling back to \(Y'\) gives us \[f^*\mathcal I\to \mathcal O_{Y'}\to \iota'_*\mathcal O_{X'} \to 0\] using the natural identifications \(f^*\mathcal O_Y = \mathcal O_{Y'}\) and \(f^*\iota_*\mathcal O_X = \iota'_*\mathcal O_{X'}\) (see Vakil 16.3.9). This means that \(f^*\mathcal I\) maps onto \(\mathcal I'\) in \(\mathcal O_Y\), i.e. \(f^*\mathcal I\twoheadrightarrow \mathcal I'\). This surjection also holds if we pull back to \(X'\) , so we get a surjection of graded algebras \[\bigoplus_{n\geq 0}g^*(\mathcal I^n/\mathcal I^{n+1}) \twoheadrightarrow \bigoplus_{n\geq 0}\mathcal I'^n/\mathcal I'^{n+1}.\] Applying \(\Spec(-)\) then gives us \[C_{X'/Y'}\hookrightarrow C_{X/Y}\times_{X}X'.\]

Let \(I = (x)\subset k[x]/(x^2) \). Then \[\Sym I/I^2 \cong k[x],\] so \(N_{X/Y} \cong \mathbb A^1\). But \[\bigoplus_{n\geq 0} I^n/I^{n+1}\cong k\oplus k\cdot x \cong k[x]/(x^2),\] simply because \(I^2 = 0\) in \(\mathcal O_Y\). So \(C_{X/Y}\cong Y \not\cong \mathbb A^1 \cong N_{X/Y}\).

Let \(I = (y^2 + x^3 - x^2)\) and \(J = (x,y)\subset k[x,y]/I\). The minimal generating set of \(J\) is \(x,y\) so \(N_{X/Y}\) is two dimensional and is isomorphic to \(\mathbb A^2\). However since \(y^2 = x^2(1-x)\), \(J^2\) is generated by \(x^2, xy\). Looking at the morphism \[(k[x,y]/I)[A,B]\to \bigoplus_{n\geq 0}J^n/J^{n+1}\] defined by \(A\mapsto x\), \(B \mapsto y\), we see the kernel is generated by \(A^2 - (1-x)B^2\), so \(C_{X/Y} \cong \)

Part (a) Example 5 demonstrates that a space can carry multiple virtual classes. If \(X = \mathbb P^1\) then \(X\) is the zero section of the zero rank vector bundle on \(X\), producing the virtual class \([X]\). The space \(X\) can also be realized as the twisted cubic given by \(\mathbb P^1_{tu} \hookrightarrow \mathbb P^3_{xyzw}\), \(t\mapsto [t^3, t^2u, tu^2, u^3]\). The image of this embedding is cut out by \((xz - y^2, yw - z^2, xw - yz)\), meaning it is the vanishing of a section of \(\mathcal O_{\mathbb P^3}(2)^{\oplus 3}\). The associated virtual class is \(8[\pt]\), and this is what we must justify. Let \(s\) be the section above whose zero locus is \(X\) and let \(E\) be the vector bundle \(\mathcal O_{\mathbb P^3}(2)^{\oplus 3}\). Then for \(Y = \mathbb P^3\) we have the fiber diagram

with Gysin pullback \(i^!_E\) (or maybe denoted \(0^!_E\)) is given by the composition \[A_*(Y) \xrightarrow{\sigma}A_*(C_{X/Y})\xrightarrow{j_*}A_*(j^*N_{Y/E}) \xrightarrow{(p^*)^{-1}}A_{*-d}(X)\] where

• \(\sigma\) sends the class \([V]\) corresponding to a pure dimensional subvariety \(V\subset Y\) to \([C_{V\cap X/Y}]\)
• \(j\) is the composition \(C_{X/Y}\hookrightarrow C_{Y/E}\times_Y X \xrightarrow{\sim} N_{Y/E}\times_Y X = i^*N_{Y/E} \) (see Exercise 1.3) and \(j_*\) is proper pushforward
• \((p^*)^{-1}\) is the inverse of the flat pullback map given by \(p:i^*N_{Y/E}\to X\).

With this data, \([X]^{vir} = i_E^!([Y])\). Because we’re in the nice regular embedding into a vector bundle case, it satisfies the equation \[i_*[X]^{vir} = c_{top}(E)\cap [Y]\] by Fulton’s Intersection Theory Section 14.1. Since \(Y = \mathbb P^3\) in this case and \(E = \mathcal O_{\mathbb P^3}(2)^{\oplus 3}\), we have \[i_*[X]^{vir} = c_1(\mathcal O_{\mathbb P^3}(2))^{\cap 3} \cap [Y] = 8[pt].\]

Part (b) We can of course realize \([X]\) as a virtual class by taking the identity embedding \(X\hookrightarrow X\), replacing \(Y\) with \(X\) and \(E\) with the trivial vector bundle (also \(X\)). Then nothing is happening and all the maps above are simply identity maps on \(A_*(X)\). Alternatively, we just use the Fulton formula and get that \([X]^{vir} = c_{top}(E)\cap [X]\), but since \(E\) is the zero vector bundle, its top Chern class is the identity operator.

I think we need \(X\) to be pure dimensional so the fundamental class is pure dimensional…otherwise something wacky happens with the normal cone construction, but I’m not exactly sure where it breaks.

Part (a) We need to check that for every \(y \in Y\) with \(x = s(y)\in E\) that \(Y_x\) is geometrically regular over \(k(x)\). Since \(E\) is a vectory bundle, \(Y_x \cong \mathbb A^d_{k(x)}\), which is indeed regular over \(k(x)\).

Part (b) The composition \(p\circ s\) is the identity \(\id_Y\).

Part (c)