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If an algebraic variety $X$ over a field characteristic p is given by equations $f_i(x_1,...,x_k) = 0$, we can consider the variety $X^{(p)}$ obtained by applying p-th powers to all the coefficients of all $f_i$'s. Frobenius morphism, as I understand it, is a morphism $X \to X^{(p)}$, given on points as raising all coordinates to p-th power.

Can anyone please explain me, what is the geometric Frobenius, as opposed to the arithmetic one?

EDIT: Thanks to Florian and George for the answers. I understand the difference now. I accepted Florian's answer because he was first and also because I found the last link http://www.math.mcgill.ca/goren/SeminarOnCohomology/Frobenius.pdf he provided especially helpful.

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I do not believe that the argument given for the crucial Corollary 3.2 in the above link is a proof. The link explains a sense in which the pullback and pushforward along the absolute Frobenius are naturally identified (as an adjoint pair) with the identity functor, but why does that formally imply the assertion in Corollary 3.2? It seems one has to really get into how pullback in sheaf cohomology is defined and how the "identification" of the Frobenius pullback/pushforward adjoint pair with the identity is defined. –  BCnrd Jul 4 '10 at 19:05
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2 Answers

up vote 7 down vote accepted

Geometric and arithmetic Frobenius live in a Galois group, they are different from the Frobenius morphism. The Galois group of a finite field of cardinality $q$ has a canonical generator $x \mapsto x^q$; this is the arithmetic Frobenius element. Its inverse, i.e., $x \mapsto x^{1/q}$, is the geometric Frobenius element. The Galois group of a non-archimedean local field (i.e., a finite extension of $\mathbb Q_p$ or $k((x))$ for a finite field $k$) maps surjectively to the Galois group of its residue field (a finite field); an element in the inverse image of an arithmetic/geometric Frobenius is still called arithmetic/geometric Frobenius (but there is no longer a canonical choice).

Finally, I think the reason for the term "geometric" is that for a variety $X/k$ ($k$ a finite field of cardinality $q$), we have a canonical isomorphism $X^{(q)} \cong X$, so the $q$-power Frobenius morphism gives rise to a map $F : X(\bar k) \to X(\bar k)$. The Galois group acts on $X(\bar k)$ as well, and the action of the geometric Frobenius element agrees with $F$.

EDIT: Oops, on $X(\bar k)$ the action of the Frobenius morphism agrees with arithmetic Frobenius, but on the étale cohomology of $X_{\bar k}$ it agrees with geometric Frobenius. Let me try to find a reference...

Here is one (see p.89). The file name seems to indicate that these are Brian Conrad's, but they are not on his web page as far as I can tell, so I hope he doesn't object to the link!

http://math.unice.fr/~dehon/CohEtale-09/Elencj_Etale/CONRAD%20Etale%20Cohomology.pdf

I think I heard that it was Deligne who coined the term "geometric Frobenius element". Deligne's Bourbaki talk in 68/69 doesn't seem to give it a name. (See Jay Pottharst's translation at http://math.bu.edu/people/potthars/writings/deligne-l-adic.pdf, in particular Prop. 4.8.) Deligne mentions SGA 5.XV. I don't have time to check further, I guess it has more on the fact I mentioned but not on the terminology.

http://www.msri.org/publications/books/sga/sga/5/SGA5-page-454.html

Finally see Katz's "Review of l-adic cohomology" in the Motives volumes.

http://books.google.at/books?id=v2CuklFFV5IC&pg=PA26&lpg=PA26&dq=%22geometric+frobenius+element%22&source=bl&ots=QUaysRdc3L&sig=4U_nC8QPWQjdg9RUi1-hHXt1Iec&hl=en&ei=WvstTLzbH8P38Aaj1q2fAw&sa=X&oi=book_result&ct=result&resnum=4&ved=0CBwQ6AEwAw#v=onepage&q=%22geometric%20frobenius%20element%22&f=false

(scroll back one page)

Update: I found some expository notes I couldn't find yesterday. Like Brian Conrad's notes they explain why geometric Frobenius has the same action as the Frobenius morphism on étale cohomology. (They use the terminology of arithmetic/geometric Frobenius morphism though.)

http://www.math.mcgill.ca/goren/SeminarOnCohomology/Frobenius.pdf

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Could you link to a source for those definitions? This is not how I thought the terms were used, but this isn't my field. –  David Speyer Jul 2 '10 at 13:54
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Hi Florian. Curious that the chapter from my never-finished book on something else is there, but that's OK. To summarize the explanation given there, on etale cohomology with constant coefficients for $X_ {\overline{k}}$, pullback by the action of the geometric Frobenius element of the Galois group is proved (not defined!) to coincide with the map induced by the $q$-power endomorphism of the $k$-scheme $X$ (with $|k| = q$). So it recovers the endomorphism induced by a "geometric map", hence the name. [Please replace the set $X(\overline{k})$ above with the scheme $X_ {\overline{k}}$.] –  BCnrd Jul 2 '10 at 15:22
    
Florian, thank you for the answer. I'll think about all that. Actually the Deligne's 68/69 Bourbaki seminar is one of the papers I'm struggling with right now. –  Evgeny Shinder Jul 2 '10 at 15:25
    
I'm glad you are ok with referring to your book draft, Brian! I'm a little confused about your last comment: the natural action of $Gal(\bar k/k)$ on $X_{\bar k}$ doesn't respect the base $\bar k$. Hence it cannot agree with the Frobenius morphism. The action on $X(\bar k)$ of an element of $Gal(\bar k/k)$ that isn't an integer power of Frobenius isn't usually algebraic (this is easily seen, for example, when $X = \mathbb A^1$). (I agree though that one needs $X_{\bar k}$ for étale cohomology, and I edited accordingly. Or maybe you didn't like that I wrote $F$ on $\bar k$-points?) –  fherzig Jul 2 '10 at 17:00
    
Florian, etale cohomology is functorial in the abstract scheme, so even though the action of ${\rm{Gal}}(\overline{k}/k)$ on $X \otimes_ k \overline{k}$ isn't by $\overline{k}$-morphisms it meaningfully acts on cohomology with constant coefficients, as you know. In particular, the action of the geometric Frobenius in the Galois group induces an endomorphism of that cohomology. This latter endomorphism does equal the endomorphism induced by the $q$-Frobenius of the $k$-scheme $X$ (applied to the $\overline{k}$-pullback). This is proved in my write-up. So I am confused by why you are confused! –  BCnrd Jul 2 '10 at 17:18
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Brion & Kumar ["Frobenius splitting methods in geom. and rep. thy" Birkhauser 2005] call the absolute Frobenius endomorphism the mapping $F_{abs}:X \to X$ which is the identity on $X$ and with comorphism given when $X = \operatorname{Spec}(A)$ is affine by the rule $(f \mapsto F_{abs}^*(f) = f^p):A \to A$.

This is not a morphism "over $k$" since $F^*:A \to A$ is "semilinear" for the Frobenius endomorphism of $k$ (= Frobenius automorphism in Galois group of $k$ if $k$ is perfect).

In Jantzen ["Representations of algebraic groups", 2nd edition] 9.1 and 9.2, he describes the absolute Frobenius map - it is "the same" as the one describe by B&K, except that the codomain is "twisted" to make $F$ a morphism over $k$. For $X = \operatorname{Spec}(A)$ this twisting amounts to: $X^{(p)} = \operatorname{Spec}(A^{(p)})$ where the $k$-algebra $A^{(p)}$ is $A$ as a ring but an element $a \in k$ acts on $A^{(p)}$ as $a^{p^{-1}}= a^{1/p}$ does on $A$.

Geometric and arithmetic Frobenii have meaning only (I believe) when $X$ is "defined over" a finite field; here I'll assume $X$ is defined over $\mathbf{F}_p$. And I'll even suppose $X$ arises by base change to $k$ from the affine $k_0 = \mathbf{F}_p$-scheme $X_0 = \operatorname{Spec}(A_0)$ (otherwise, patch!).

Then $X = \operatorname{Spec}(A)$ where $A = A_0 \otimes_{k_0} k$. The arithmetic Frobenius map on $X$ is the $k_0$-morphism $F_{arith}:X \to X$ whose comorphism is given by $$(f \otimes a \mapsto f \otimes a^p):A = A_0 \otimes_{k_0} k \to A = A_0 \otimes_{k_0} k$$ for $f \in A_0$ and $a \in k$.

Thus the set of $k_0$-points $X_0(k_0)$ is the set of points in $X(k)$ fixed by the arithemtic Frobenius $F_{arith}$; i.e. the action of $F_{arith}$ on points just gives the "usual" action of the Frobenius element of the Galois group on rational points (here I must be supposing $k$ to be perfect...)

The geometric Frobenius of $X$ is the $k$-morphism $F_{geom}:X \to X$ whose comorphism is given by $$(f \otimes a \mapsto f^p \otimes a):A = A_0 \otimes_{k_0} k \to A = A_0 \otimes_{k_0} k.$$ If you pick an embedding $X \subset \mathbf{A}^N$ defined over $k_0$, then $F_{geom}$ is given on $k$-points in these coordinates by $$F_{geom}(x_1,\dots,x_N) = (x_1^p,\dots,x_N^p)$$.

The arithmetic and geometric Frobenius are defined (briefly) in Jantzen (loc. cit.).

Note that $F_{arith} \circ F_{geom} = F_{geom} \circ F_{arith}$ is the "absolute Frobenius" of B&K mentioned above.

Also see Milne's "Lectures on Etale Cohomology" 29.11 for some discussion reconciling the number theorists with their action of the Frobenius automorphism $\phi=(x \mapsto x^p)$ on the Tate group $T_\ell E$ of an elliptic curve defined over $k_0$ and the algebraic geometers with their action of $F_{geom}$ on $H^1(E,\mathbf{Z}_\ell)$.

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These are the definitions I am familiar with. –  David Speyer Jul 2 '10 at 15:21
    
@David, George: for $|k_0| = q$, since absolute $q$-Frobenius induces the identity map on cohomology with constant coeffs (that requires proof, not deep and given in the link in Florian's answer, and probably in some of the other links), the geometric $q$-Frobenius $F_ {\rm{geom}}$ on such cohomology agrees with the inverse of the $q$-Frobenius $F_ {\rm{arith}}$: the $q$-th root map on $k$ when $k$ is perfect (e.g., algebraically closed). One advantage of the latter viewpoint is that it arises from an element of a Galois group when $k = \overline{k}_0$; extremely useful over global fields. –  BCnrd Jul 2 '10 at 15:40
    
@David: I think fherzig's reference to geometric Frobenius element of the Galois group is explained by BCnrd's comment above -- the action of this element coincides with the action of $F_{geom}$ on etale cohomology (with constant coefficients). This is related to the reference to Milne's notes that I mention –  George McNinch Jul 2 '10 at 15:42
    
@Brian: Thanks (our comments "crossed", hence my redundancy...) –  George McNinch Jul 2 '10 at 15:47
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