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For any field $K$ we can define the cyclotomic character $\chi: \operatorname{Gal}(K)\rightarrow GL_1(\hat{\mathbb{Z}})$. For any representation $V$ (I will view this as a module over $\mathbb{Q}[\operatorname{Gal}(K)]$), we may define the $j^{th}$ Tate twist of it to be $V\otimes \chi^j$. I know that I have the general idea of the definition right, but I might have some of the details off. If so, please correct me.

I'm not sure what this is meant for. So as to have a more concrete question, I will ask this in the context of the Tate conjectures. As I understand, one way to state the Tate conjectures is that $A^j(X)\otimes_{\mathbb{Q}} \mathbb{Q}_l$ (where $A^j(X)$ is the image in $H^{2j}(X,\mathbb{Q}_l)$ of the algebraic cycle classes with coefficients in $\mathbb{Q}$) is isomorphic to $(V^j)^G$ where $V^j$ is $H^{2j}(X,\mathbb{Q}_l)(j)$ (the $j^{th}$ Tate twist). Why does the Tate twist appear here? Where does it come from? What's the intuition behind this?

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As far as I've seen, the idea seems to be that you don't merely want to remember certain vector spaces, but you also want to keep certain actions in sight (Frobenius or Galois). –  Matt Aug 16 '11 at 4:37
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1 Answer

up vote 18 down vote accepted

Tate twists have to do with fundamental classes in etale (co)homology.

Usually we talk about etale cohomology, but we can also talk about homology, just as the dual of cohomology.

If $X$ is smooth and projective of dimension $d$, then $H^{2d}(X)$ will be one-dimensional, and transform under $\chi^{-d}$; equivalently, $H_{2d}(X)$ will be one-dimensional, and transform under $\chi^d$.

How do you see this? Of course there is a rigorous proof that you can read, but intuitively, if you think about how fundamental classes are constructed in topology, you can convince yourself that a fundamental class in dimension $d$ will behave like a $d$-fold product of a fundamental class in dimension $1$.

So we reduce to the one-dimensional case.

It is then easy to see that there is a canonical isomorphism $$H_2(\mathbb P^1) \cong H_1(\mathbb A^1\setminus \{0\}).$$ Finally, this last homology is the same as the $\ell$-adic Tate module of $\mathbb A^1 \setminus \{0\}$ (thought of as the multiplicative group), and that Tate module is one-dimensional, with Galois action given by $\chi$.

As for the Tate conjecture:

Imagine that $Y$ is a smooth connected closed subvariety of codimension $j$ of $X$, itself smooth and projective, say, of dimension $d$.

Let's work with homology first, because it's more intuitive.

The map $Y\to X$ will induce a map $H_{2(d-j)}(Y) \to H_{2(d-j)}(X)$. Now since $Y$ has dimension $d-j$, the source of this map is one-dimensional, with Galois action given by $\chi^{d-j}$. So a codimension $j$ cycle gives a Galois-invariant line in $H_{2(d-j)}(X)$ (the image of the above map) which transforms via $\chi^{d-j}$.

Remember that I said $H_{2(d-j)}(X)$ was just the dual of $H^{2(d-j)}(X)$. Now cup product gives a perfect pairing (by Poincare duality) $$H^{2(d-j)}(X) \times H^{2j}(X) \to H^{2d}(X),$$ compatible with Galois. Since $X$ is $d$-dimensional, the Galois action on the one-dimensional space $H^{2d}(X)$ is via $\chi^{-d}$. So we see that as Galois reps., the dual to $H^{2(d-j)}(X)$ can be identified with $H^{2j}(X)\otimes \chi^d$.

Thus our codimension $j$ cycle gives a line in $H^{2j}(X)\otimes \chi^d$ which transforms under Galois via $\chi^{d-j}$, which is the same thing as an invariant line in $H^{2j}(X)\otimes \chi^j$.

So codimension $j$ cycles contribute invariant lines in $H^{2j}(X)(j)$, and the Tate conjecture is that the full space of Galois invariants here is actually spanned by the lines coming from cycles.

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Amazingly clear, as usual. –  Mariano Suárez-Alvarez Aug 16 '11 at 5:33
@Mariano: Dear Mariano, Thanks! Best wishes, –  Matt E Aug 16 '11 at 5:37
Dear Matt, Is the identification of $H^1$ with the Tate module true for any commutative group variety (not necessarily an abelian variety)? (My understanding was that an abelian Galois cover of a comm. group variety needn't be a group scheme in itself.) –  Akhil Mathew Aug 16 '11 at 5:49
(Actually, it can't be true for the affine line, by the "homotopy invariance" of etale cohomology.) –  Akhil Mathew Aug 16 '11 at 13:59
@Akhil: Dear Akhil, I'm probably confused, but I'm not sure about your $\mathbb A^1$ comment. If $\ell$ is different from the residue characteristic, then $\mathbb A^1$ has no $\ell$-torsion, hence has trivial $\ell$-adic Tate module. It also has no $\ell$-power covers, and hence has trivial $\ell$-adic $H^1$ as well. Regards, –  Matt E Aug 17 '11 at 4:31
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