User travis schedler - MathOverflow

Comparison map between de Rham cohomology of analytic and formal neighborhoods of singularities

2012-07-04T23:16:51Z

Suppose that $X$ is a complex algebraic (or complex analytic) variety, and $x \in X$ is a singular point. I am interested in two types of local differential forms at $x$: analytic and formal.

First, let $\mathcal{O}_{X,x}^{\text{an}}$ be the ring of analytic germs of functions at $x$. I am interested in the complex $\Omega_{X,x}^{\text{an}}$ of analytic germs of differential forms, i.e., linear combinations of elements $h_0 dh_1 \wedge \cdots \wedge dh_k$ for $h_0, \ldots, h_k \in \mathcal{O}_{X,x}^{\text{an}}$ , and the corresponding de Rham cohomology $H^\bullet(\Omega_{X,x}^{\text{an}})$ . Explicitly, if $X \subseteq \mathbf{A}^n$ is a subvariety of affine space cut out by equations $f_1, \ldots, f_m$, then this complex is defined as $\Omega_{\mathbf{A}^n,x}^{\text{an}} / (f_1, \ldots, f_m, df_1, \ldots, df_m)$, where we quotient by the ideal in the de Rham differential graded algebra generated by the $f_i$ and $df_i$.

Next, let $\hat {\mathcal{O}}_{X,x}$ be the completion of the local ring of algebraic functions at $x$, i.e., the ring of (not-necessarily convergent) formal power series of functions at $x$. Let $\hat {\Omega}_{X,x}$ be the complex of formal differential forms, i.e., linear combinations of elements $h_0 dh_1 \wedge \cdots \wedge dh_k$ for $h_0, \ldots, h_k \in \hat {\mathcal{O}}_{X,x}$ . For $X \subseteq \mathbf{A}^n$, this is defined as a quotient of $\hat \Omega_{\mathbf{A}^n,x}$ in the same manner as above.

Since one has a canonical inclusion $\mathcal{O}_{X,x}^{\text{an}} \hookrightarrow \hat {\mathcal{O}}_{X,x}$ , one obtains a canonical comparison map

$H^\bullet(\Omega_{X,x}^{\text{an}}) \to H^\bullet(\hat \Omega_{X,x}).$

My question is: When is this map an isomorphism?

I am particularly interested in the case that the LHS is finite-dimensional, e.g., when $X$ has an isolated singularity at $x$ (finite-dimensionality of the LHS then follows from the Theorem of Section 3.17 of Bloom and Herrera's paper ``De Rham Cohomology of an Analytic Space,'' (Invent. Math. 7, 275--296 (1969)).

More details and reformulations:

Under the finite-dimensionality hypothesis, the comparison map is definitely surjective: the RHS is the inverse limit of $H^\bullet(\Omega_{X,x} / \mathfrak{m}_{X,x}^N \cdot \Omega_{X,x})$ , where $\mathfrak{m}_{X,x} \subseteq \mathcal{O}_{X,x}$ is the maximal ideal, and the LHS surjects to each of these (by lifting closed or exact forms modulo $\mathfrak{m}_{X,x}^N$ to closed or exact analytic forms). So both sides are finite-dimensional and the comparison map is surjective.

Thus, under this hypothesis, the question reduces to: When it is true that, if a closed analytic form $\alpha \in \Omega_{X,x}^{\text{an}}$ is the differential of a formal form in $\hat \Omega_{X,x}$, then it is also the differential of an analytic form in $\Omega_{X,x}^{\text{an}}$? (Perhaps, an analytic approximation theorem could be applied to answer this.)

Next, I will restrict this question to the special case that interests me: isolated singularities which are locally complete intersections. In this case, by results of Sections 4 and 5 of Greuel's paper ``Der Gauss-Manin-Zusammenhang isolierter Singularitaeten von vollstaendigen Durchschnitten,'' (Math. Ann. 214, 235--266 (1975)), one has the formula

$H^\bullet(\Omega_{X,x}^{\text{an}}) \cong \mathbf{C}^{\mu_x-\tau_x}[-\operatorname{dim} X],$

where $\mu_x$ is the Milnor number of the singularity at $x$, and the notation above indicates that the de Rham cohomology of the analytic neighborhood of $x$ is concentrated in degree equal to the dimension of $X$. Also, $\tau_x$ is the Tjurina number, which is the dimension of the singularity ring at $x$: explicitly, if $X$ is locally a complete intersection of dimension $n-m$ cut out at $x \in \mathbf{A}^n$ by functions $f_1, \ldots, f_m$, then the singularity ring is the quotient of $\mathcal{O}_{X,x}^{\text{an}}$ by the ideal generated by the $f_i$ together with the determinants of the $(n-m) \times (n-m)$ minors of the Jacobian matrix $(\frac{\partial f_i}{\partial x_j})$. In other words, the Tjurina number here is the dimension of the torsion of the germs of differential forms $\Omega_{X,x}^{\operatorname{dim}(X),\text{an}}$ of degree $\operatorname{dim}(X)$.

In this case, I would only want to know whether the same formula holds for the de Rham cohomology of the formal neighborhood, i.e., that the dimension of $H^\bullet(\hat{\Omega}_{X,x})$ is equal to the Milnor number, and not less.

[Readers who are tired of reading can stop here---I will give one more alternative formulation:]

Alternatively, one can work with the de Rham complex modulo torsion, $\tilde{\Omega}_{X,x}^{\text{an}}$ , obtained from $\Omega^{\text{an}}_{X,x}$ by modding by the torsion submodule over $\mathcal{O}_{X,x}^{\text{an}}$ . This is equivalent to working with germs of forms modulo those forms that become zero when restricted to the smooth locus, i.e., whose representatives on open neighborhoods of $x$ have zero restriction to smooth open subsets. In this case, Greuel's formula (still for an isolated singularity at $x$ which is locally a complete intersection) remains the same,

$H^\bullet(\tilde{\Omega}_{X,x}^{\text{an}}) \cong \mathbf{C}^{\mu_x-\tau_x}[-\operatorname{dim} X].$

In the alternative formulation, I would like to know again if the same formula holds replacing analytic germs of forms mod torsion, $\tilde{\Omega}_{X,x}^{\text{an}}$ , by formal forms mod torsion. It follows from Greuel's paper that, still assuming $x$ is an isolated singularity which is locally a complete intersection, the two questions are equivalent.

Answer by travis schedler for Comparison map between de Rham cohomology of analytic and formal neighborhoods of singularities

2012-12-07T21:24:48Z

For the case where x is an isolated singularity (which is probably a necessary hypothesis), we positively answered this question in arXiv:1211.1883, Theorem 4.45.

Answer by travis schedler for Groups of matrices that preserve several quadratic forms

2012-07-06T10:58:20Z

Let $V$ be the vector space you begin with. As you probably know, transformations $T \in \operatorname{End}(V)$ that preserve a symmetric form $(-,-)$ of full rank are called orthogonal, and the group of these transformations is denoted $O(V)$ (let me work with transformations instead of matrices).

Now, given any other form $\langle -,- \rangle$, there must exist a transformation $A \in \operatorname{End}(V)$ such that $\langle v,w \rangle = (Av, w)$, since $(-,-)$ had full rank. More precisely, one can view forms as linear transformations $V \to V^*$ via the map $v \mapsto (v, -)$ and similarly $v \mapsto \langle v, - \rangle$, and full rank ones are invertible, so we can obtain $A$ by composing $\langle -,- \rangle$ with the inverse of $(-,-)$. This obtains the desired transformation $A$.

Thus you are asking for the subgroup of $O(V)$ which also preserves $\langle v, w \rangle = (Av, w)$. This is nothing but the subgroup of $O(V)$ of transformations which commute with $A$. Indeed, if $B$ is orthogonal, then $\langle Bv, Bw \rangle = (ABv, Bw) = (B^{-1} AB v, w)$, which equals $\langle v,w \rangle$ for all $v$ and $w$ if and only if $A=B^{-1}AB$.

Of course this generalizes to the setting where you have your original nondegenerate symmetric form and $k$ other forms $v,w \mapsto (A_i v, w)$: then you are interested in the subgroup of $O(V)$ of transformations commuting with all $A_i$.

Computing this group is a standard exercise in linear algebra. As pointed out by the next author, returning to the case where $k=1$ and $A=A_1$, one can restrict to the generalized eigenspaces of $A$, which are each preserved by all $B$ in the desired group, and ask that $B$ commute with $A$ on each of those. For example, if you are working over the field of real numbers and your first form is an inner product (i.e., positive-definite), and the transformation $A$ is diagonalizable over the complex numbers (i.e., the generalized eigenspaces are all actual eigenspaces), then, up to conjugation, your group is a direct product of $O(V_\lambda)$ for the real eigenspaces $V_\lambda$ along with $U(V_{\lambda,\bar \lambda})$ for the complex nonreal pairs of eigenvalues $\lambda, \bar \lambda$ (where $V_{\lambda, \bar \lambda} \subseteq V$ has the property that its complexification is the sum of complex eigenspaces of $\lambda$ and $\bar \lambda$, and the group $U(V_{\lambda,\bar \lambda})$ is the unitary group of $V_{\lambda, \bar \lambda}$ equipped with a complex structure given by $A$, which makes the original inner product into a Hermitian one with respect to this complex structure).

Answer by travis schedler for Is the singular locus ideal preserved by all derivations?

2012-06-03T14:14:11Z

The fact that the set-theoretic singular locus is preserved is true only in characteristic zero: for example if you take the curve ($x^p = y^2$) in $\mathbb{A}^2$, then the ideal $(x, y)$ is not preserved by the derivation $d/dx$ in characteristic $p$. On the other hand, the Jacobian ideal in this case, $(y)$, is preserved.

In characteristic zero one can prove that the set-theoretic singular locus is preserved by exponentiating derivations: given a derivation $D$ of $\mathcal{O}_X$ one can consider $e^{t D}$, a derivation of $\mathcal{O}_X[[t]]$, and the composition of this with a character $\mathcal{O}_X \to k$ corresponding to a singular point gives a character $\mathcal{O}_X((t)) \to k$ also with the same dimension of tangent space, hence corresponding to a singular point. Therefore the singular locus is preserved, since the set-theoretic singular locus is the intersection of all such kernels.

This theorem is due to, I believe, Seidenberg: it is the corollary to Theorem 12 in "Differential ideals in rings of finitely generated type" in AJM 1967.

Comment by travis schedler

2012-07-21T08:15:36Z

Not immediately obvious to me, but one can prove that it implies n-CY by noticing that the condition is exactly what is needed to have a cyclically supersymmetric superpotential as in the sense of, e.g., Bocklandt-Schedler-Wemyss, and then one can see that the corresponding complex is indeed a projective bimodule resolution of your algebra A, just as in the case where all the $a_{ij}$ are one.

Comment by travis schedler

2012-07-06T14:45:36Z

When you say "positive definite" I assume for this you need the field to be a subfield of the real numbers, since otherwise this notion is not well-defined for quadratic forms (which is the context the poster used). One could also generalize positive-definiteness to subfields of the complex numbers if one replaces quadratic forms by Hermitian forms.