Hi all,

Given a variety $X$ over the real numbers, we can consider the singular cohomology of the space $X(\mathbb{C})$, with coefficients in $\mathbb{Q}$, say. The action of complex conjugation on $X(\mathbb{C})$ induces an action on these cohomology groups. How do you compute/describe this action in concrete examples? For example, what happens in the case of $H^1$ of a curve of genus $g$, or even in the supposedly trivial case where $X = Spec(\mathbb{R})$? Other examples would also be interesting, or general hints on how to understand the action.

  • 2
    $\begingroup$ The case $X=Spec(\mathbb R)$ is indeed trivial: the complex conjugation $c$ acts on $H^0=\mathbb Q$ trivially: it must have trace 1, by Lefschetz trace formula. A related example in dim. 0, but a bit more interesting, is that we compact $Spec(\mathbb R)\coprod Spec(\mathbb R)$ with $Spec(\mathbb R[T]/(T^2+1)).$ They only have $H^0,$ which is 2-dimensional, and $c$ acts on the first one trivially, and on the 2nd one by switching the two components. $\endgroup$ – shenghao Apr 7 '11 at 16:27

Suppose that $X$ is smooth and projective. Then Hodge theory gives a decomposition of $H^i(X(\mathbb C),\mathbb C)$ into the direct sum of $H^{p,q}$'s (with $p + q = i$).
Now there are two complex conjugations acting on $H^i$ with $\mathbb C$ coefficients: the "trivial action" just coming from conjugating the coefficients, which I'll denote by $c$ (and which is conjugate-linear as an automorphism of $H^i(X(\mathbb C),\mathbb C)$ by its very definition), and the non-trivial one coming from the action of complex conjugation on $X(\mathbb C)$ itself (which is $\mathbb C$-linear), which I'll denote by $Fr_{\infty}$ (the Frobenius at $\infty$).

It is a familiar fact from beginning Hodge theory that $c$ interchanges $H^{p,q}$ and $H^{q,p}$. What you can check is that $Fr_{\infty}$ also interchanges $H^{p,q}$ and $H^{q,p}$.

Another fact is that if $X$ is geometrically connected, then $Fr_{\infty}$ acts on the top dimensional cohomology as multiplication by $(-1)^d$, if $X$ is $d$-dimensional. (This is a special case of the fact that the top-dimensional etale cohomology of a geometrically connected smooth projective $d$-dimensional variety is always the $-d$th Tate twist.)

These two facts taken together serve to establish most of the claims in David Speyer's answer, for example.

What can be hard to work out in general is what $Fr_{\infty}$ does on $H^{p,p}$. Just as an example, if $X$ is Spec $\mathbb R$, a single point, then the cohomology is just a one-dimensonal $H^{0,0}$, and $Fr_{\infty}$ acts trivially. If we then take a pair of such points, we get a two-dimensional $H^{0,0}$, again with $Fr_{\infty}$ acting trivially.

On the other hand, if $X$ is Spec $\mathbb R[x]/(x^2 + 1)$, so that $X$ is a pair of complex conjugate points, then the cohomology of the complex points is again a two-dimensional $H^{0,0}$, but with $Fr_{\infty}$ acting with one $+1$ and one $-1$ eigenspace (because it switches the two points).

One can make this example a bit more interesting by blowing up $\mathbb P^2$ at (a) a pair of points each defined over $\mathbb R$, and (b) a pair of complex conjugate points.

In case (a) one gets $X$ whose complex points have a three-dimensional $H^{1,1}$ with $Fr_{\infty}$ acting by $-1$, while in case (b) one again gets a three-dimensional $H^{1,1}$, but now the eigenvalues of $Fr_{\infty}$ are $(1,-1,-1)$. (In each case the $H^{1,1}$ is spanned by the fundamental class of a line in $\mathbb P^2$ together with the fundamental classes of the two exceptional divisors. In case (a) each of these fundamental classes is defined over $\mathbb R$, and so each contributes an eigenvalue of $-1 = (-1)^1$ --- here the exponent $1$ is because these are fundamental classes of curves --- while in case (b) we see that $Fr_{\infty}$ is switching the two exceptional divisors.)

  • $\begingroup$ Dear Mariano, You're welcome. Best wishes, Matt $\endgroup$ – Emerton Apr 8 '11 at 6:40

Even the case of curves is surprisingly technical. The two references which I refer to when I need the details are Gross and Harris and Vinnikov.

To summarize: If $X$ is a compact curve defined over $\mathbb{R}$, then complex conjugation acts by $-1$ on $H^2(X)$. The rational cohomology $H^1(X, \mathbb{Q})$ splits into a $+1$ and a $-1$ eigenspace, each $g$-dimensional and each Lagrangian for the symplectic form. Call them $H^1_+(X, \mathbb{Q})$ and $H^1_{-}(X, \mathbb{Q})$. Let $H^1_{\pm}(X, \mathbb{Z}) = H^1_{\pm}(X, \mathbb{Q}) \cap H^1(X, \mathbb{Z})$.

If $X(\mathbb{R})$ has $k$ connected components, with $k \geq 1$, then $$H^1(X, \mathbb{Z})/ \left( H^1_{+}(X, \mathbb{Z}) \oplus H^1_{-}(X, \mathbb{Z}) \right) \cong (\mathbb{Z}/2)^{g+1-k}.$$ (In particular, $k \leq g+1$, which is an interesting fact in itself.) When $X(\mathbb{R})$ is empty, so $k=0$, there are two different topologies which can occur.

If you want more detail than that, I recommend the above papers.

  • $\begingroup$ +1 especially for the reference to Gross-Harris. There is a lot of material in there that should be classical but is otherwise very hard to find. Note also that your "interesting act" is (a version of) en.wikipedia.org/wiki/Harnack%27s_curve_theorem. $\endgroup$ – Pete L. Clark Apr 7 '11 at 17:17
  • $\begingroup$ "Interesting fact", I meant to say. $\endgroup$ – Pete L. Clark Apr 7 '11 at 22:19
  • 4
    $\begingroup$ A lot of what's in this Gross-Harris paper (of which I am also a fan!) has been revisited in the recent work of Kirsten Wickelgren (math.harvard.edu/~kwickelg/papers/delta2real.pdf) who interprets them in the following way: a real point gives (up to conjugacy) a section from Gal(C/R) to the arithmetic fundamental group of X/R, and it turns out that the conjugacy classes of such sections exactly match the components of X(R) -- even if you only use the 2-nilpotent quotient of the fundamental group, which reduces everything to questions about cup product. $\endgroup$ – JSE Apr 9 '11 at 13:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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