1
$\begingroup$

First we consider the holomorphic Koszul complex on $\mathbb{C}^2$: $$ 0\to \mathcal{O}(\mathbb{C}^2)\overset{\begin{pmatrix}-z_2\\z_1\end{pmatrix}}{\to} \mathcal{O}(\mathbb{C}^2)^{\oplus 2}\overset{(z_1,z_2)}{\to} \mathcal{O}(\mathbb{C}^2)\to \mathbb{C}\to 0. $$ It is well-known that the above complex is exact.

Now we consider the $C^{\infty}$-version of the Koszul complex $$ 0\to C^{\infty}(\mathbb{C}^2)\overset{\begin{pmatrix}-z_2\\z_1\end{pmatrix}}{\to} C^{\infty}(\mathbb{C}^2)^{\oplus 2}\overset{(z_1,z_2)}{\to} C^{\infty}(\mathbb{C}^2)\to \mathbb{C}\to 0. $$ It is clear that the above complex is not exact. For example $\bar{z_1}$ and $\bar{z_2}$ both vanish at the origin but are not in the image of multiplying $z_1$ and $z_2$.

My question is: What is the cohomology of the above complex?

Improve this question
$\endgroup$
2
  • 6
    $\begingroup$ As far as you are concerned there is no difference between $\mathbb{C}^2$ and $\mathbb{R}^4$. In both cases you would be calculating the complex valued functions on them and evaluating at the origin. Hence, the correct Koszul complex to compare with is the one for $\mathbb{R}^4$. This observation would probably give you the answer to your question. $\endgroup$
    – Kapil
    Commented Dec 8, 2021 at 5:51
  • 2
    $\begingroup$ Indeed, @Kapil is absolutely right. Forget for a second about the rightmost term $\mathbb{C}$ (which is usually not included in the Koszul complex anyway). Then your second complex is obtained from your first complex by extending the ground ring from $\mathbb{C}$ to the ring of antiholomorphic functions on $\mathbb{C}^2$, which immediately shows that its homology is the latter ring. $\endgroup$
    – Vladimir Dotsenko
    Commented Dec 8, 2021 at 7:33

1 Answer 1

Reset to default
1
$\begingroup$

As Vladimir mentions in the comment, it is probably easier to formulate this without the $\mathbb{C}$ at the right. However, it is not clear to me that the answer then is exactly what Vladimir claims, i.e., that you get the antiholomorphic functions as cohomology at level $0$.

I am not sure if you intend the statement to be for global sections, as I interpret your notation, but since the involved sheaves $\mathbb{C}$ and $C^\infty$ are acyclic on $\mathbb{C}^2$, and the complexes are exact outside the origin, you might just as well consider the corresponding complexes of stalks at the origin.

By flatness of the smooth functions over the holomorphic functions, it follows that you have an exact complex of sheaves $$0 \to C^\infty_0 \to (C^\infty_0)^{\oplus 2} \to C^\infty_0 \to C^\infty_0/((z_1,z_2)C^\infty_0) \to 0,$$ where $C^\infty_0$ denote the ring of germs of smooth functions at $0$.

The map from the antiholomorphic functions, $\overline{\mathcal{O}_0} \to C^\infty_0/((z_1,z_2)C^\infty_0)$ is injective, since $\overline{\mathcal{O}_0} \cap (z_1,z_2)C^\infty_0 = \\{0\\}$ (which you might verify by reducing to the one-dimensional case, by restricting to complex lines to the origin). However, it is not so clear to me that the map is actually surjective.

Improve this answer
$\endgroup$
3
  • $\begingroup$ What is the answer for the real version? For example, let $R$ be the local ring of smooth functions at $(\mathbb R^2,0)$, and $A$ the local ring of real analytic functions there. The Koszul complex $\DeclareMathOperator\Kos{Kos}\Kos(R;x,y)$ is quasi-isomorphic to the derived tensor product $R\otimes_A^L\mathbb R$, which is the usual tensor product by flatness, but this seems to be complicated? $\endgroup$
    – Z. M
    Commented Jan 16, 2022 at 10:46
  • $\begingroup$ It should follow rather easily from appropriate versions of Taylors formula that the cohomology is just $\mathbb R$ at level $0$. More precisely, at level $0$, it should follow from the fact that any smooth function $f$ may be written as $f(x,y) = f(0,0) + xf_1(x,y)+yf_2(x,y)$ for appropriate smooth functions $f_1,f_2$, and to see that the cohomology vanishes at level $1$, one should be able to use that any smooth function $g(x,y)$ may be written as $g(x,y) = g_1(y) + x g_2(x,y)$ for appropriate smooth functions $g_1,g_2$. $\endgroup$
    – Richard Lärkäng
    Commented Jan 17, 2022 at 15:15
  • $\begingroup$ Thanks. This is called Hadamard's lemma. I forgot this. This vanishing seems also a bit similar to Poincaré's lemma. $\endgroup$
    – Z. M
    Commented Jan 17, 2022 at 15:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .