NOTE: scroll down to read my latest edit first if you're reading this for the first time :)

My aim is to calculate the de Rham cohomology of the variety $U = \text{Spec} \ A$, where: $$A = \frac{\mathbb{C}[U,V,W]}{\langle V^2 - UW\rangle} $$

So far I have defined:

$$\Omega^1_{A/\mathbb{C}} := \frac{\Omega^1_{\mathbb{C}[U,V,W]/ \mathbb{C}}}{\langle d(V^2-UW)\rangle } \ \ , \qquad \Omega^p_{A/\mathbb{C}} := \wedge^p \Omega^1_{A/\mathbb{C}} $$


$$ d^0(f+I) := \frac{ \partial f}{ \partial U} \,dU + \frac{ \partial f}{ \partial V} dV + \frac{ \partial f}{ \partial W} \, dW + \langle d(V^2-UW)\rangle $$ (letting $ \ I = \ \langle V^2 - UV\rangle \ \ $)

Now, assuming what I have above is ok, we have $ \langle d(V^2-UW) \rangle \ = \langle 2V\, dV - W\,dU - U\,dW \rangle $

So ($U$ affine etc) : $$\small{ H^0_{dR}(U) \cong \ker d^0 = \lbrace f+I : \left(\frac{ \partial f}{ \partial U} + I \right) dU + \left(\frac{ \partial f}{ \partial V} + I \right) dV + \left(\frac{ \partial f}{ \partial W} + I \right) dW \in \langle 2V \,dV - W\,dU - U\,dW \rangle \rbrace }$$

Here is where I get stuck. I know we should end up with $H^0_{dR}(U) \cong \mathbb{C} $, but these maps are pretty confusing, especially now as our generators are no longer independent thanks to the new relation we've introduced.

My questions are:

1) is what i've done above correct so far?

2) if it is, then how would one go about showing $H^0_{dR}(U) \cong \mathbb{C} $ continuing what i've done above? (this can be just a hint in the right direction if you want)


I've looked through the suggested MO thread, and commented below on it. It provides a good proof of (2) using a more general method. I would still like to get at it through this calculation if that's possible.


Also I tried this earlier; is this be correct for the ring $\mathbb{C}[x,y]/\langle xy \rangle $ ?:

Suppose $\frac{ \partial f}{ \partial x} dx + \frac{ \partial f}{ \partial y} dy \in \langle ydx + xdy \rangle $ (working in $ \Omega^1_{\mathbb{C}[x,y]} $)

Now $ \exists g \in \mathbb{C[x,y]} \ \ \frac{ \partial f}{ \partial x} = yg(x,y) \ , \frac{ \partial f}{ \partial y} = xg(x,y)$

Choose $G_x(x,y) \ , \ G_y(x,y)$ such that $\frac{ \partial G_x}{ \partial x} = g \ , \ \frac{ \partial G_y}{ \partial y} = g$.

Now we see $f = yG_x + H_1(y) $ and $ f = xG_y + H_2(x) $ for some $H_1,H_2$

without loss of generality then $G_x$ is divisible by $x$ and similarly $y | G_y$. Then we see $H_1(y) = f = H_2(x) $ (since $xy=0$), so $H_1,H_2$ (and hence also $f$) constant.

I realize I've argued this fairly loosely and it could do with more rigour.


David below has pointed out that this calculation will not give me what I want. I would still like to know a sketch of the method that should be used to find kernels and images of linear maps like this, because it's this which I find tough. I guess the situation I am thinking about is:

$ \phi : \ V \rightarrow W$, where $V,W$ are infinite dimensional $\mathbb{C}$-vector spaces and in general do not have a basis, (but the case where $V$ has a basis is also important) and where we know all the relations imposed on $V,W$.

Is there a standard method to find kernels and/or images in this case?

(this will get the bounty)

  • 3
    $\begingroup$ Here's a proof that works in a much greater generality than you are mathoverflow.net/questions/75329/…, and is actually surprisingly simple. $\endgroup$ Aug 12 '14 at 22:55
  • $\begingroup$ thanks, i'll have to look through it tomorrow morning. this is a good excuse for me to get up to date with the more general definitions $\endgroup$
    – JC574
    Aug 12 '14 at 23:06
  • $\begingroup$ Although this proof is very nice, and solves the problem, I'd like to be able to do it by simply solving the kernel of the map above. Is this feasible? It's more the method I'm interested in, If i just needed to prove the result I could use the comparison to the singular cohomology or something $\endgroup$
    – JC574
    Aug 13 '14 at 16:36
  • 1
    $\begingroup$ General advise: Calculating with cosets makes life complicated. Cosets provide only one construction of quotient rings. Quotient rings should be interpreted via their universal property. $A$ is the universal complex algebra containing elements $u,v,w$ such that $v^2=uw$. You may calculate with $u,v,w$. It follows that $\Omega^1_A$ is the universal $A$-module containing elements $d(u),d(v),d(w)$ such that $2 v d(v) = u d(w) + w d(u)$. Now verify that $\mathbb{C} \to A \xrightarrow{d} \Omega^1_A$ is exact, using a $\mathbb{C}$-basis of $A$. $\endgroup$ Aug 13 '14 at 17:24
  • 2
    $\begingroup$ For a singular variety like yours, computing de rham cohomology in this naive way is not the morally right thing to do; in particular, it need not compute actual topological cohomology. See mathoverflow.net/questions/128853 for some examples, and footnote 9 in numdam.org/item?id=PMIHES_1966__29__95_0 for the appropriate fix. Of course, the fact that it isn't the right thing to do doesn't mean it isn't fun to do it. $\endgroup$ Aug 15 '14 at 20:53

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