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It shouldn't be difficult to show that when $C$ is union of lines in generic position in $\mathbb{P}^2_k$, then its normalisation map $f:\tilde C \rightarrow C$ with $\tilde C$ being the disjoint union of corresponding lines induces an equality $\chi(\mathcal{O}_\tilde C)=\chi(f_*\mathcal{O}_\tilde C)$

Is it true (the above equality of holomorphic Euler characteristic) in general for any normalisation map? What could be the conditions providing $\chi(\mathcal{O}_\tilde C)=\chi(f_*\mathcal{O}_\tilde C)$ or more general $\chi(\mathcal{O}_X)=\chi(f_*\mathcal{O}_X)$ with $f:X\rightarrow Y$ ?

I know that when $f:Z\rightarrow X$ is a closed embedding than all cohomologies for $\mathcal{O}_Z$ and $f_*\mathcal{O}_Z$ coincide, in particular the holomorphic euler characteristics are the same.

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    $\begingroup$ I think it's true for every normalization map. In fact, every normalization is an affine morphism, and for an affine morphism of noetherian separated schemes $f : X \rightarrow Y$ and quasicoherent sheaf $\mathcal{F}$ in $X$, $H^i(X, \mathcal{F}) = H^i(Y, f_*\mathcal{F})$ for every $i \geq 0$ (this is exercise III.4.1 of Hartshorne). So your equality is true for all affine morphisms. More generally, I guess you don't want fibres to have higher cohomology (but I'm only guessing) $\endgroup$ – Jane Doé May 3 at 20:48
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    $\begingroup$ @JaneDoé good comment, and I think it would be even better as an answer. $\endgroup$ – KReiser May 4 at 0:18
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    $\begingroup$ @JaneDoé I'm not sure if you were claiming that for $f:X\to Y$ to have $R^i f_\ast\mathcal{F}$ for $i>0$ to be zero (which is what you want for $H^j(X,\mathcal{F})=H^j(Y,f_\ast\mathcal{F})$ to hold) that it's sufficient to have $ H^i(X_y,\mathcal{F}_y)=0$ for all $y\in Y$ and $i>0$. This is not the case. For example, if $f:(\mathbb{A}^2-\{0\})\to \mathbb{A}^2$ is the natural inclusion and $\mathcal{F}$ is the structure sheaf. $\endgroup$ – Alex Youcis May 4 at 23:50
  • $\begingroup$ @JaneDoé I answered the question. I hope you don't mind. $\endgroup$ – Alex Youcis May 6 at 20:25
  • $\begingroup$ @AlexYoucis Of course not, and it looks like I have something to learn from your answer as well! $\endgroup$ – Jane Doé May 6 at 20:34
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$\newcommand{\an}{\mathrm{an}}$As Jane Doé commented above, the key thing is the following:

Theorem: Let $f:X\to Y$ be an affine morphism of schemes. Then, for any quasi-coherent $\mathcal{F}$ we have that there is a natural isomorphism $H^i(X,\mathcal{F})\cong H^i(Y,f_\ast\mathcal{F})$ for all $i\geqslant 0$.

Proof: The Leray spectral sequence for $f$ shows us that we have a spectral sequence

$$E^{p,q}_2=H^p(Y,R^q f_\ast\mathcal{F})\implies H^{p+q}(X,\mathcal{F})$$

But, note that for all $q>0$ we have that $R^q f_\ast\mathcal{F}=0$. Indeed, to show this it suffices to show that for all $y\in Y$ there exists some affine open neighborhood $U$ of $y$ such that $(R^qf_\ast\mathcal{F})\mid_U=0$. But, take $U$ to be affine. Then, as is well-known $(R^qf_\ast\mathcal{F})\mid_U\cong \widetilde{H^q(f^{-1}(U),\mathcal{F})}$. Since $f^{-1}(U)$ is affine, by assumption, this is zero by Serre's affineness criterion. Clearly then the spectral sequence collapses on the second page and we get the desired result. $\blacksquare$

Your literal question is then answered by the fact that normalization morphisms are affine by construction--you literally normalized affine open by affine open.


We actually have the following well-known, but surprisingly difficult result:

Theorem: Let $X$ be a finite type $k$-scheme. Then, the normalization map $\nu:X'\to X$ is finite.

Proof: If you want the most general version you can combine Tag035S and Tag0335. In Vakil this is 9.7.3, but he defers to Eisenbud's book for a proof. $\blacksquare$

The reason that this might be of interest to you is the following:

Theorem: Let $f:X\to Y$ be a finite morphism of schemes (resp. finite morphism of algebraic varieties over $\mathbb{C}$). Then, for any sheaf of abelian groups $\mathcal{F}$ on the etale site of $X$ (resp. any sheaf of abelian groups on $X^\mathrm{an}$, the analytification of $X$) we have a natural isomorphism of cohomology groups $H^i(X,\mathcal{F})\cong H^i(Y,f_\ast\mathcal{F})$ (resp. $H^i(X^\mathrm{an},\mathcal{F}^\mathrm{an})\cong H^i(Y^\mathrm{an},f^\mathrm{an}_\ast\mathcal{F}^\mathrm{an})$).

Proof: By the same Leray spectral sequence argument, it suffices to show that for $q>0$ we have that $R^q f_\ast\mathcal{F}=0$ and $R^q f^\mathrm{an}_\ast\mathcal{F}^\an=0$.

The second case is easier since we have the simpler Hausdorff version of proper base change which tells us that (since $X^\an$ and $Y^\an$ are Hausdorff, $Y^\an$ locally compact, and $f^\an$ universally closed since $f$ is proper) that:

$$(R^q f^\an_\ast\mathcal{F}^\an)_y=H^q(f^{-1}(y),\mathcal{F}^\an)=0$$

where this last equality holds since $f^{-1}(y)$ is a finite discrete set of points.

The first case requires more care. What's true in general is that for any geometric point $\overline{y}$ of $Y$ we have that

$$(R^q f_\ast \mathcal{F})_\overline{y}\cong H^q(X_{(\overline{y})},\mathcal{F})$$

where $X_{(\overline{y})}$ is the so-called Milnor fiber $X\times_Y \mathrm{Spec}(\mathcal{O}_{Y,\overline{y}})$ where $\mathcal{O}_{Y,\overline{y}}$ is the local ring at $\overline{y}$ in the etale topology (i.e. its the strict Henselization of $\mathcal{O}_{Y,y}$ where $y$ is the underlying point of $\overline{y}$). The point though is that since $f$ is finite we have that $X_{(\overline{y})}\to\mathrm{Spec}(\mathcal{O}_{Y,\overline{y}})$ is finite, which means that $X_{(\overline{y})}$ is a finite disjoint union of spectra of strictly Henselian local rings, and thus has trivial etale cohomology. $\blacksquare$

The point of this is that you can now take $\mathcal{F}$ to be any sheaf on the etale site of $X$ or, if $X$ is over $\mathbb{C}$, any sheaf on $X^\an$, not just quasi-coherent ones if your map is finite (e.g. a normalization map).

Two examples:

  • You can take $\mathcal{F}=\mathbb{G}_m$ which can be used to to compute the Picard group of non-normal varieties in terms of their normalizations (e.g. see this).

  • You can compute $H^i(X,\underline{R})$ where $X$ is the projective cuspidal cubic over an algebraically closed fiekd $k$) and $R$ is any finite ring (this cohomology is etale cohomology). Namely, we have that the normalization of $X$ is $\mathbb{P}^1_k$ and so by the above we know that $H^i(\mathbb{P}^1_k,\underline{R})\cong H^i(X,\nu_\ast\underline{R})$ where $\nu$ is the normalization map. But, since $\nu$ is a universal homeomorphism we have that $\nu_\ast\underline{R}=\underline{R}$ which tells you that $H^i(\mathbb{P}^1_k,\underline{R})\cong H^i(X,\underline{R})$.

  • You can take $\mathcal{F}=\underline{R}$, for any ring $R$, on $X^\mathrm{an}$ (assuming that $X$ is finite type over $\mathbb{C}$) to see that $$H^i_\mathrm{sing}(X^\mathrm{an},R)\cong H^i(Y^\mathrm{an},f^\an_\ast\underline{R})$$ which comes up surprisingly often.

NB: The above second example is somewhat perverse. Namely, any universal homeomorphism gives an equivalece of etale sites (e.g. see Tag04DY). This can be used to alternatively compute the result in the second bullet, but this computation is of a different nature, in some sense, than the one presented here.


The last thing I wanted to point out was that, as mentioned in the comments, it is not good enough in general that $H^q(f^{-1}(y),\mathcal{F})=0$ for $q>0$ to imply that we have an isomorphism of cohomology groups $H^i(X,\mathcal{F})\cong H^i(Y,f_\ast\mathcal{F})$.

For example, let $j$ be the inclusion of $X=\mathbb{A}^2-\{0\}$ into $Y=\mathbb{A}^2$. Let us note that $j^{-1}(y)$ is either empty, or a point, for all $y\in Y$ and so evidently $H^q(j^{-1}(Y),\mathcal{F})=0$ for all $q>0$. That said, we claim that it's not necessarily true that $H^i(X,\mathcal{F})\cong H^i(Y,j_\ast\mathcal{F})$.

Indeed, we claim that $j_\ast\mathcal{O}_X=\mathcal{O}_Y$. We have a natural map $\mathcal{O}_X\to j_\ast\mathcal{O}_Y$ coming from restriction. This is actually an isomorphism by algebraic Hartog's lemma (e.g. see Theorem 6.45 of Gortz and Wedhorn). Note then that

$$\dim_k H^1(X,\mathcal{O}_X)=\dim_k H^1(\mathbb{A}^2-\{0\},\mathcal{O}_{\mathbb{A}^2-\{0\}})=\infty$$

but

$$H^1(Y,j_\ast\mathcal{O}_X)=H^1(\mathbb{A}^2,\mathcal{O}_{\mathbb{A}^2})=0$$

and so evidently the two are not isomorphic.

The issue is that $j$ is not affine, of course!

Even if you try this in the analytic category, where it seemed like all you needed was quasi-finiteness, you notice that we applied 'Hausdorff proper base change` and thus actually need properness. But, as ZMT tells us, quasi-finite and proper implies finite (and thus affine) which our map is not! In short, our map is not proper.

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