Is there a simple example of an (affine) algebraic variety $X$ over $\mathbb C$ where the $H^*_{dR}(X/\mathbb C) = H^*(\Omega^\bullet_{A/\mathbb C})$ differs from the singular cohomology $H^*_{sing}(X(\mathbb C)^{an},\mathbb C)$?

Such an example has to be singular (by a theorem of Grothendieck), but I am having a hard time finding one. The case $xy = 0$ doesn't work (both cohomology theories give the same answer).


EDIT: I would like a reduced example if possible...

  • 2
    $\begingroup$ $\operatorname{Spec}\mathbb C[X]/(X^2)$ is an example, no? $\endgroup$ Commented Apr 26, 2013 at 19:46
  • $\begingroup$ Right, I was hoping for a reduced one... $\endgroup$
    – Nicolás
    Commented Apr 26, 2013 at 19:50
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    $\begingroup$ @Mariano: For your scheme $D$ we have $H^0_{dR}(D/\mathbb C)=\mathbb C$ and $H^i_{dR}(D/\mathbb C)=0$ for $i\gt0$. $\endgroup$ Commented Sep 9, 2017 at 23:07
  • $\begingroup$ See also this question $\endgroup$ Commented Sep 14, 2017 at 0:54

3 Answers 3


A likely candidate for this would be a non-Du Bois singularity.

Du Bois, following Deligne's ideas, constructed a filtered complex of sheaves with coherent cohomology sheaves that gives a resolution of the constant sheaf $\mathbb C$ for any reduced finite type scheme over $\mathbb C$. This complex agrees with the de Rham complex for smooth varieties and in general its hypercohomology agrees with the singular cohomology of the (underlying topological space of the) scheme.

So, in some sense your question is to see an example when the Du Bois complex is different from the de Rham complex (I know that's not exactly what you are asking, but I think that's the important fact behind this issue).

There is a class of singularities, not surprisingly called Du Bois singularities, with the property that the $0^{\rm th}$ associated graded quotient of the Du Bois complex is quasi-isomorphic to the structure sheaf (this holds for smooth varieties).

So, a good start for finding an example like this is to look at non-Du Bois singularities. For curves being Du Bois is equivalent to semi-normal, so any curve with a cusp is non-Du Bois. I would start there.

In general, Du Bois singularities can be quite varied, but if your $X$ is normal and Gorenstein, then being Du Bois is equivalent to being log canonical. So, you have plenty of examples (For instance, take a normal hypersurface with worse than log canonical singularities). I am sure you will find an example easily.


A very explicit example is given by the cusp $X=\operatorname {Spec}(A)$ where $A=\frac {\mathbb C[\xi,\eta]}{(3\eta^2-2\xi^3)}=\mathbb C[x,y]$.
Since the set of closed points $X(\mathbb C)$ in its classical topology is homeomorphic to $\mathbb C$ we have $H^1_{sing}(X(\mathbb C),\mathbb C)=0$, but we will now show that $H^1_{dR}(X/\mathbb C)\neq0$.

Recall that $3y^2=2x^3$ implies $ydy=x^2dx$ and consider the closed global algebraic differential form $\omega=y^2dy=x^2ydx\in \Omega^1(X)$ .
The claim is that $\omega$ is not a boundary.

Proof of claim
Indeed an arbitrary element $f(x,y)\in A$ can be written uniquely as $$f(x,y)=p(x)+yq(x)\quad (p(t), q(t)\in \mathbb C[t])$$ And an arbitrary element in $\eta\in \Omega^1(X)$ can be writen uniquely as $$\eta =(u(x)+yv(x))dx+w(x)dy \quad (u(x), v(x)),w(x)\in \mathbb C[x])$$

The differential of $f(x,y)=p(x)+yq(x)$ is $$df=p'(x)dx+(q(x)+yq'(x))dy=(p'(x)+x^2 q'(x))dx+q(x)dy$$ and cannot equal $\omega=x^2ydx$.

We have proved that $0\neq [\omega]\in h^1(\Omega ^\bullet(X) )=H^1_{dR}(X/\mathbb C)$ so that $H^1_{dR}(X/\mathbb C)\neq0$ as announced.
(Actually the same sort of computation shows that $H^1_{dR}(X/\mathbb C)=\mathbb C[\omega]$)

Warning (October 1st, 2017)
As explained by Julian Rosen in a comment below, this answer is not correct.
I am very sorry about this mistake and will try to correct it.
Any help from specialists is more than welcome!

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    $\begingroup$ Isn't $\omega=y^2\,dy$ the derivative of $y^3/3$? Also, there is an equality of $1$-forms $2x^3\,dy = 3y^2\,dy=3yx^2\,dx$, and the outer two expressions seem to contradict the uniqueness you assert. $\endgroup$ Commented Sep 14, 2017 at 0:19

Let's look at the cuspidal cubic curve $C: x^3 = y^2$. I claim $H^1_{dR}(C)$ is two-dimensional.

EDIT: Actually $H^1_{dR}(C) = 0$, so that this curve does not give an example of an affine singular variety whose algebraic de Rham cohomology differs from its singular cohomology. This is corroborated by a remark in the beginning of section 3.2 of Huber and Müller-Stach's Periods and Nori Motives (just control-F for "cusp"). Note that the remark is not in their first version, but it is in their sixth which can be found on Prof. Huber's website.

Thanks are due to Julian Rosen for both pointing out a mistake in the derivation defined in the original version as well as realizing the important and subtle point that $\Omega^2_C \neq 0$, being supported at the origin. My sincere apologies for posting prematurely to the (three or more) people who read this answer in the meantime. The derivation as defined originally was not a derivation, but can be replaced by another which serves the same purpose as the original:

$$ \frac{\mathbb C[x,y]}{(x^3-y^2)} \to \frac{\mathbb C[x]}{(x^3)} : x \mapsto x, y\mapsto 1 $$

This derivation still shows that $\eta$ and $x\eta$ are non-zero. The more important point is that the presence of $\Omega^2_C$ means that the forms $x\,dy$ and $x^2\,dy$ are not actually closed, so these forms do not contribute to cohomology as was incorrectly assumed before, because $\Omega^2_C$ is in fact two-dimensional and spanned by $d(x\,dy) = dx\wedge dy$ and $d(x^2\,dy) = 2x\,dx\wedge dy$. These 2-forms are zero everywhere but at the singularity. My assumption that they were closed was the mistake which led me to conclude the cohomology was two-dimensional. The computations and formulas in the post show that the other 1-forms are exact so that $H^1_{dR}(C) = 0$. At any rate this mistake is an instructive one I think, so I'll preserve the post as originally written. Again, the two mistakes of this post are in the definition of the derivation and the incorrect assumption that $x\,dy$ and $x^2\,dy$ are closed forms.

As Georges notes, the singular cohomology of its closed points is trivial.

Regardless of whether it's direct (it's not), we definitely have the sum $$\Omega^1_C = \mathbb C[x]\,dx + \mathbb C[x]y\,dx + \mathbb C[x]\,dy + \mathbb C[x]y\,dy $$

and then because $3x^2\,dx = 2y\,dy$, the last summand is actually unnecessary. The first summand is obviously composed of exact forms, so the only forms in question are things like $x^ny\,dx$ and $x^n\,dy$ where $n\geq 0$. On the other hand,

$$ d\left( \frac{2x^{n+1}y}{2n+5} \right) = x^ny\,dx \quad \text{ for } n > 1 $$ $$d\left( \frac{3 x^ny}{2n+3} \right) = x^n \,dy \quad \text{ for } n > 2$$

For the lower $n$, one needs to divide by $x$'s and $y$'s which is only valid provided $\Omega^1_C$ is torsion-free at the origin (and later we'll see it isn't), so these need to be handled with care. This leaves $y\,dx$, $xy\,dx$, $x\,dy$, and $x^2\,dy$. As $d(xy) = x\,dy + y\,dx$, any $\mathbb C$-linear relation between $y\,dx$ and $x\,dy$ (except for multiples of $y\,dx = -x\,dy$) would show that they are both exact. Similarly, $d(x^2y) = 2xy\,dx + x^2\,dy$ so any $\mathbb C$-linear relation between $xy\,dx$ and $x^2\,dy$ (except for multiples of $2xy\,dx = -x^2\,dy$) would show that they are both exact.

The obvious place to look is the relation $3x^2\,dx = 2y\,dy$, which after multiplying with $y$ gives $$ x^2(3y\,dx - 2x\,dy) = 0 $$ as well as $$ x(3xy\,dx - 2x^2\,dy) = 0 $$ Hence the form $\eta = 3y\,dx - 2x\,dy$ is very relevant: if $\eta = 0$ then we get both linear relations from $\eta = x\eta = 0$ and then we would have shown that the first cohomology group is trivial.

More subtly, it turns out $\eta$ is not zero. What the first formula does show is that it is zero in any localisation of $\Omega^1_C$ away from $(0,0)$. This funny form also maps to zero in the cotangent space at zero, since $\eta \in (x,y)\Omega^1_C$. This implies that any first-order derivation cannot determine whether $\eta \neq 0$.

To see that $\eta \neq 0$ we will use the second-order derivation (the correct derivation is above in the comment) $$ \frac{\mathbb C[x,y]}{(x^3-y^2)} \to \frac{\mathbb C[x,y]}{(x^2,y^2)} : x,y\mapsto 1, \, x^2,y^2 \mapsto 0$$ The universal property of $\Omega^1_C$ then promises that there's a unique map of $\frac{\mathbb C[x,y]}{(x^3-y^2)}$-modules $$ \Omega^1_C \xrightarrow{f} \frac{\mathbb C[x,y]}{(x^2,y^2)} : dx,dy\mapsto 1 $$ Since $f(\eta) = 3y-2x$ and $f(x\eta) = 3xy$ which are both non-zero, we've shown $\eta,x\eta \neq 0$. One way of thinking about $x\eta$ is that it is a Dirac delta supported at $(0,0)$, while $\eta$ is a (weak) derivative of a Dirac delta supported at $(0,0)$.

Now working modulo exact forms, we've shown that $[ydx] = -[xdy]$ and $[x^2\,dy]=-2[xy\,dx]$ in $H^1_{dR}(C)$. I claim these are non-zero and independent.

Following up on David's comment we weight $\frac{\mathbb C[x,y]}{(x^3-y^2)}$ so that $\text{wt }x = 2$ and $\text{wt }y = 3$ and then weight $\Omega_{A/k}$ accordingly in order that $d$ preserves weight (The motivation for this is that we are measuring in units of $t$ under the rationalization map $t \mapsto (t^2,t^3)$).

Because $\eta$ (resp., $x\eta$) is not zero, the weight 5 piece (resp., weight 7) of $\Omega_{C}$ is two-dimensional, spanned by $y\,dx$ and $x\,dy$ (resp., $xy\,dx$ and $x^2\,dy$). On the other hand, the weight 5 piece (resp., weight 7) of $\frac{\mathbb C[x,y]}{(x^3-y^2)}$ is only one-dimensional, spanned by $xy$ (resp., $x^2y$).

We conclude that $H^1_{dR}(C)$ is two-dimensional, supported in weights 5 and 7 and spanned by $[xdy]$ and $[x^2\,dy]$, respectively. See edit comment above.


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