It seems to me this come up very often when we talk about group action on (étale) cohomology groups.

For example, let $X$ be a scheme over $\mathrm{spec}\mathbb{Z}$, $\mathcal{F}$ an $\ell$-adic sheaf on $X$, and $V_\ell= H^i(X\otimes \overline{\mathbb{Q}}, \mathcal{F})$. The Galois group of $\overline{\mathbb{Q}}$ of $\mathbb{Q}$ acts, by transport of structure, on $V_\ell$.

For an fancier example, in Deligne's Seminar Bourbaki paper "Formes modulaires et représentations $\ell$-adiques", immediately after definition 3.9 in which a $\mathbb{Q}$-vector space of cohomology of modular curves are defined, it says (in English) "This vector space depends only on the universal elliptic curve (up to isogeny) $f_\infty: E\to M_\infty$ so that, by transport of structure, it is equipped with a left action of the adelic group $\mathrm{GL}_2(\mathbb{A}^f)$". (Here $\mathbb{A}^f$ is the ring of "finite" adeles).

As my knowledge of étale cohomology comes mainly from reading J.S. Milne's online notes with a few glances of Freitag-Kiel, I am pretty lost with the mathematics that happens behind the few words "by transport of structure".

Thank you.

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    $\begingroup$ Maybe it ("ne depend que") should translate as: "It depends only on ...". Cohomology is a functor, so for $\sigma\in Gal(\bar{Q}/Q),$ the automorphism $X\otimes\sigma:X\otimes\bar{Q}\to X\otimes\bar{Q}$ induces an linear isom. on cohom. $\endgroup$
    – shenghao
    May 20, 2011 at 11:38
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    $\begingroup$ To be honest, I've never understood why this phrase is used. As shenghao says, it's just functoriality. If you have a pair $(X,F)$, where $X$ is a space and $F$ is a sheaf over $X$, and a group $G$ acting on the pair, then, for every $g\in G$, we get a map $F\rightarrow g_*g^*F\rightarrow g_*F$, where the first arrow is the adjunction, and the second is induced by the action of $G$ on $F$. Taking cohomology of the composition, you get your action on cohomology (though I'm not sure I haven't just defined its inverse). $\endgroup$ May 20, 2011 at 14:35
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    $\begingroup$ I learned the phrase as a student, when Tate used it and explained what it meant in some course he was teaching. It's nothing fancy, just informal shorthand for standard ways in which we all use natural isomorphisms. There is a wikipedia article about it. $\endgroup$ May 20, 2011 at 14:39
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    $\begingroup$ You have to born French to truly understand "transport de structure". The "structure" is in the sense of Bourbaki. When a ring acts on a vector space on the left, then it acts on the dual vector space on the right by "transport de structure". When a group acts on a vector space on the left, then it can be made to act on the dual on the right or on the left (by inserting an inverse). Only one of these is "transport de structure", but I can never remember which. $\endgroup$
    – mephisto
    May 20, 2011 at 16:19
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    $\begingroup$ Let me try to say it better: Transport of structure simply means if you have this thing and you have some extra structure on this thing and you also have an isomorphism between this thing and another thing then that gives you a way of putting the same kind of structure on that other thing. $\endgroup$ May 21, 2011 at 3:50

2 Answers 2


I am not sure this is adding anything substantial to the comments, but maybe it helps to spell things out. At least this used to confuse me in the past. I found a remark in a recent preprint of Deligne-Flicker helpful: Transport of structures (Bourbaki Ens Ch. IV) is the principle that any isomorphism $Y_1 \to Y_2$ extends to objects constructed from $Y_1$ and $Y_2$. When $Y_1 = Y_2$: symmetries extend. The construction has to be canonical: not involving choices.

Let me now explain why $Aut(\bar k/k)$ acts on $H^i(X \otimes \bar k, \mathbb Q_l)$ (where $X$ is a $k$-scheme and $k \to \bar k$ is a fixed algebraic closure) by transport of structure. Let's first take a step back and use two different algebraic closures $\sigma_i : k \to K_i$. (I learnt this from Tate, in fact I think his global class field theory article in Cassels-Fröhlich.) Then by canonicity of etale cohomology, any isomorphism $K_1 \to K_2$ between $\sigma_1$ and $\sigma_2$ gives us a canonical isomorphism $H^i(X \otimes K_1, \mathbb Q_l)$ and $H^i(X \otimes K_2, \mathbb Q_l)$ of $\mathbb Q_l$-vector spaces. (Never mind that it's easy to say what this map actually is, the point is that we know there has to be one!) In particular, if we take $K_1 = K_2 = \bar k$, we get a canonical $\mathbb Q_l$-linear action of $Aut(\bar k/k)$ on $H^i(X \otimes \bar k, \mathbb Q_l)$. I think the argument is the same when we replace $\mathbb Q_l$ by (the pullback to $X \otimes \bar k$ of) a constructible or lisse $l$-adic sheaf over $X$.

Now for your second example, we can play the same game: take two universal elliptic curves $f_{\infty, i} : E_i \to M_\infty$. Any isogeny between them gives us a canonical isomorphism of Deligne's $\mathbb Q$-vector space $W$ (or rather $W_1$, $W_2$). Hence if we take $E_1 = E_2 = E$, ...

Finally, let me consider one more example from Deligne's writing that I found very cryptic in the past. Namely, paragraph 0.2.5 in his article "Valeurs de fonctions et périodes d'intégrales" (http://publications.ias.edu/deligne/paper/379). Basically what he is claiming is the following. Suppose $X$ is a smooth projective variety over a field $k$. For an embedding $\sigma : k \to \mathbb C$ let $H_\sigma := H^*((X \otimes_k \mathbb C)(\mathbb C), \mathbb Q)$, which is a $\mathbb Q$-vector space together with a bigrading on its complexification (by Hodge theory). He says that "by transport of structure", we obtain an isomorphism $F_\infty : H_\sigma \to H_{c \sigma}$ such that moreover $F_\infty \otimes c$ carries $H_\sigma^{pq}$ to $H_{c\sigma}^{pq}$ (where $c$ denotes complex conjugation).

The point is to follow his definitions very carefully: suppose $\sigma : k \to K$ is an embedding into an algebraic closure $K$ of $\mathbb R$ (i.e., abstractly $\mathbb C$, but it's important not to work with just this copy). Then $H_\sigma := H^*((X \otimes_{k, \sigma} K) (K), \mathbb Q)$ and the bigrading is on $H_\sigma \otimes_{\mathbb Q} K = H^*((X \otimes_{k, \sigma} K) (K), K) = \oplus H_\sigma^{pq}$. (Sanity check: for Hodge theory we indeed want to have the same field of definition and coefficients.) Now again take $\sigma_i : k \to K_i$. Any isomorphism $\tau:K_1 \to K_2$ between $\sigma_1$ and $\sigma_2$ gives us a canonical isomorphism $F : H_{\sigma_1} \to H_{\sigma_2}$ of $\mathbb Q$-vector spaces. This in turn gives us a canonical isomorphism $F \otimes \tau : H_{\sigma_1} \otimes K_1 \to H_{\sigma_2} \otimes K_2$, and by canonicity of the Hodge structure it sends $H_{\sigma_1}^{pq}$ to $H_{\sigma_2}^{pq}$. Now specialise to $K_1 = K_2 = \mathbb C$,...

(I find it interesting how Deligne explains in his interview https://www.simonsfoundation.org/science_lives_video/pierre-deligne/ that he is very bad at calculations and therefore needs to work very canonically, otherwise he gets lost. He also says that he learnt about "transport of structure" from reading Bourbaki's "Set theory" as teenager.)

  • $\begingroup$ I was just getting confused reading that same paragraph of Deligne. This cleared it up. Thank you. $\endgroup$
    – Zavosh
    Oct 12, 2014 at 1:34

I'm not completely sure if this is the answer; it just provides an another way of describing the Galois action.

Let $a:X\to\text{Spec }\mathbb Q$ be the structure map. Then $R^ia_*F,$ being an étale sheaf on $\text{Spec }\mathbb Q,$ corresponds to an $\ell$-adic vector space (its stalk at $\text{Spec }\overline{\mathbb Q}$) with a continuous action of $Gal(\overline{\mathbb Q}/\mathbb Q).$ This vector space is nothing else but $H^i=H^i(X_{\overline{\mathbb Q}},F_{\overline{\mathbb Q}}).$ Think of the sheaf $R^ia_*F$ in terms of its espace étalé, and take fiber products: $$ \begin{bmatrix} H^i & \to & H^i & \to & [R^ia_*F] \end{bmatrix} $$ $$ \begin{bmatrix} \downarrow &&&&& \downarrow &&&&& \downarrow\end{bmatrix} $$ $$ \begin{bmatrix} \text{Spec }\overline{\mathbb Q} & \overset{\to}{\sigma} & \text{Spec }\overline{\mathbb Q} & \to & \text{Spec }\mathbb Q.\end{bmatrix} $$ Since the two maps from $\text{Spec }\overline{\mathbb Q}$ to $\text{Spec }\mathbb Q$ are the same, the two pullbacks are both identified with $H^i,$ and the isomorphism between them in the diagram is how $\sigma$ acts. It seems that the $\overline{\mathbb Q}$-structure of $H^i$ is "transported" by $\sigma.$


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