Let $X$ be a smooth projective variety over $F$, and $E/F$ - finite Galois extension. There is an extension of scalars map $CH^\*(X) \to CH^\*(X_E)$. The image lands in the Galois invariant part of $CH^\*(X_E)$, and in the case of rational coefficients, all Galois-invariant cycles are in the image (EDIT: this follows from taking the trace argument).

With integer coefficients Galois-invariant classes don't have to descend. For example, for $CH^1(X) = Pic(X)$ there is an exact sequence: $$ 0 \to Pic(X) \to Pic(X_E)^{Gal(E/F)} \to Br(F), $$ so we can say that the obstruction to descend a cycle lies in a Brauer group.

Are there any known obstructions to descend elements of higher groups $CH^i$ with integer coefficients?

In my case I have a cycle in $CH^*(X_E)^{Gal(E/F)}$ and I want to find out whether or not it's coming from $CH^*(X)$.

(The actual cycle is described in here.)


2 Answers 2


Let us keep the notations from above, and let's write $G:=\mathrm{Gal}(E/F)$. Let me quickly recall the origin of the Brauer obstruction: it really comes from the Hochschild-Serre spectral sequence

$$H^p(G,E^q(X_E,\mathbf{G}_m))\Longrightarrow E^{p+q}(X,\mathbf{G}_m)$$

(I'm writing $E^{\ast}=H^{\ast}_{\mathrm{et}}$ for étale cohomology here, because the system doesn't seem to like too many subscripts.) If we analyze this in low degrees, this gives us the following classical exact sequence (for any $F$-variety $X$):

$$0\to H^1(G,E^0(X_E,\mathbf{G}_m))\to\mathrm{Pic}(X)\to H^0(G,\mathrm{Pic}(X_E))\to H^2(G,E^0(X_E,\mathbf{G}_m))\to\ker\left[\mathrm{Br}(X)\to\mathrm{Br}(X_E)\right]$$ $$\to H^1(G,\mathrm{Pic}(X_E))\to H^3(G,E^0(X_E,\mathbf{G}_m))$$

So we'd like to generalize the sequence above to the situation of $\mathrm{CH}^n(X)=H^n(X,\mathcal{K}_n)$ (where $\mathcal{K}_n$ is the Zariski-sheafification of the presheaf $K_n$). Assume $X$ geometrically regular here. The Gersten resolution of $\mathcal{K}_n$ on $X_E$ is the complex

$$C^{\bullet}(X_E)\colon\quad K_nk(X_E)\to\bigoplus_{x\in X_E^1}K_{n-1}k(x)\to\bigoplus_{x\in X_E^2}K_{n-2}k(x)\to\cdots\to\bigoplus_{x\in X_E^{n-1}}K_1k(x)\to\bigoplus_{x\in X_E^n}K_0k(x)$$

There's a similar complex $C^{\bullet}(X)$ giving the Gersten resolution of $\mathcal{K}_n$ on $X$. We regard the complex $C^{\bullet}(X_E)$ as a $G$-complex; write $\sigma$ for the map

$$C^{n-1}(X_E)=\bigoplus_{x\in X_E^{n-1}}k(x)^{\times}=\bigoplus_{x\in X_E^{n-1}}K_1k(x)\to\bigoplus_{x\in X_E^n}K_0k(x)=Z^n(X_E)$$

of $G$-modules. I want to argue that the kernel of this map is playing the role of $E^0(X_E,\mathbf{G}_m)$ for higher $n$. (When $n=1$, this kernel coincides with $E^0(X_E,\mathbf{G}_m)$.)

Now we might hope for a spectral sequence

$$H^p(G,H^q(C^{\bullet}(X_E)[n]))\Longrightarrow H^{p+q}(C^{\bullet}(X)[n])$$

but of course $K$-theory doesn't quite satisfy Galois descent, so we don't have this convergence ($C^{\bullet}(X)[n]$ is not the homotopy fixed-point complex of $C^{\bullet}(X_E)[n]$). But we're only trying to analyze a very small piece of this spectral sequence — the piece involving $\sigma$. For that, Hilbert Theorem 90 does the work, and we get the following exact sequence:

$$0\to H^1(G,\ker\sigma)\to\mathrm{CH}^n(X)\to H^0(G,\mathrm{CH}^n(X_E))\to H^2(G,\ker\sigma)\to\ker\left[H^2\left(G,C^{n-1}(X_E)\right)\to H^2(G,Z^n(X_E))\right]$$ $$\to H^1(G,\mathrm{CH}^n(X_E))\to H^3(G,\ker\sigma)$$

So we find an obstruction to descending cycles of codimension $n$ in $H^2(G,\ker\sigma)$. Is this the sort of thing you had in mind?

  • $\begingroup$ Thank you, I'll think about that. However, I don't know if it's possible to compute things like H^(G,ker(\sigma)). For example, do you think that one can prove Merkurjev-Suslin result on Chow groups of Severi-Brauer varieties using this approach? $\endgroup$ Jan 15, 2010 at 16:46
  • $\begingroup$ I don't know whether it's possible to compute the obstruction in the particular example you're thinking about, but I do not see a path to find a simpler obstruction. I also don't see a way to use this to prove Merkurjev-Suslin. What did you have in mind? Finally, after I posted this I hunted about for a reference, and I finally found a <a href="arxiv.org/abs/0903.0132">paper</a> that appears to discuss this very result! $\endgroup$ Jan 15, 2010 at 18:56
  • 1
    $\begingroup$ As a test case I'll try to apply this technics to prove that the Chow group of zero cycles on Severi-Brauer in nZ - this is what I called Merkurjev-Suslin theorem. But maybe it won't work out... The argument of Merkurjev and Suslin is an inderect one, invloving BGQ spectral sequence. $\endgroup$ Jan 15, 2010 at 23:12
  • $\begingroup$ Oh, I see. This seems like a pretty good idea. Let me know if it works! $\endgroup$ Jan 18, 2010 at 16:40

Dear Evgeny Shinder,

For CH^2(X) one can find an "obstruction" in a non-ramified cohomology.

Let us write H^i_nr(X, \mu_l^j) for the intersection of kernels of the residu maps d_A in Galois cohomology, where A is running over all discrete valuation rings with field of fractions F(X) containing F. For example, if X is regular, one should take in account all valuations coming from codimension one points, but also the valuations coming from exceptional divisors of various blowing-ups.

If X is a smooth geometrically rational variety defined over a finite field F (i.e. rational over an algebraic closure \bar F of F), one can show that the cokernel of the map CH^2(X)->CH^2(X_{\bar F})^G is isomorphic, up to p-torsion (p=char F), to the group of non-ramified cohomology H^3_nr(X, Q/Z(2)). One can also find examples when this last group is non zero (see http://arxiv.org/abs/1004.1897).

Is it helpful for you?

Best regards,

Alena Pirutka.

  • $\begingroup$ Hi Alena :) Just so you know, if you want your math to display nicely, just write the dollar signs around it; most of the usual Latex works... $\endgroup$ Dec 24, 2010 at 14:54
  • $\begingroup$ Alena, thank you for reply. When I was asking the question I was keeping in mind the vanishing cycle in the Chow group of the middle dimension which was CH^4 in my case. However, by now I realize that my question was not a very good one - the technique I was looking for clearly does not exist in general. Your paper seems very interesting to me and I will take a look at it when I have time. $\endgroup$ Dec 24, 2010 at 19:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.