Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $i : H \to G$ be a subgroup of finite index. The transfer map is a special homomorphism $V(i) : G^\mathrm{ab} \to H^\mathrm{ab}$. The usual ad hoc definition uses a set of representatives of $H$ in $G$ and then you have to check that it is independent from this choice and that it is a homomorphism at all. I think this definition is not enlightening at all (although it is, of course, useful for explicit calculations). A better one uses group homology. Namely, for a $G$-module $A$ there is a natural transformation $A_G \to \mathrm{res}^{G}_{H} A_H$, $[a] \mapsto \sum_{Hg \in H/G} [ga]$, which extends to a natural transformation $H_\*(G;A) \to H_\*(H;\mathrm{res}^{G}_{H} A)$ (usually called corestriction or transfer). Now evaluate at $A = \mathbb{Z}$ and $* = 1$ to get $G^\mathrm{ab} \to H^\mathrm{ab}$. One can then calculate this map using the explicit isomorphisms and homotopy equivalences involved; but now you know by the general theory that it is a well-defined homomorphism.

It also follows directly that the transfer is actually a functor $V : \mathrm{Grp}_{mf} \to \mathrm{Ab}^{\mathrm{op}}$ with object function $G \mapsto G^{\mathrm{ab}}$, where $\mathrm{Grp}_{mf}$ is the category whose objects are groups and whose morphisms are monomorphisms of finite index.

I would like to know if there is an even more "abstract" definition. To be more precise: Is there a categorical characterization of the functor $V$ which only uses the adjunction $\mathrm{Grp} {\longleftarrow \atop \longrightarrow} \mathrm{Ab}$?

Edit: There are many interesting answers so far which give, in fact, very "enlightening" definitions of the transfer. But I would also like to know if there is a pure categorical one, such as the one given by Ralph.

Edit: A very interesting note by Daniel Ferrand is A note on transfer. There a more general statement is proven (even in a topos setting): Let $G$ act freely on a set $X$ such that $X/G$ is finite with at least two elements. Then there is an isomorphism of abelian groups $(\mathrm{Ver},\mathrm{sgn}) : {\mathrm{Aut}_{G}(X)}^{\mathrm{ab}} \cong G^{\mathrm{ab}} \times \mathbb{Z}/2$. It is natural with respect to $G$-isomorphisms. Here again I would like to ask if it is possible to characterize this isomorphism by its properties (instead of writing it down via choices, whose independence has to be shown afterwards).

Proposition 7.1. in this paper includes the interpretation via determinants mentioned by Geoff in his answer, actually something more general: For w.l.o.g. abelian $G$ there is a commutative diagram

$\begin{matrix} {\mathrm{Aut}_{G}(X)}^{\mathrm{ab}} & \cong & \mathrm{Aut}_{\mathbb{Z}G}{\mathbb{Z}X}^{\mathrm{ab}} \\\\ \downarrow & & \downarrow \\\\ G \times \mathbb{Z}/2 & \rightarrow & (\mathbb{Z} G)^{x} \end{matrix} $

Thus we may think of transfer and signature as the embedding the standard units into the group ring.

share|improve this question
5  
Shurely the most 'enlightening' definition is via topology: given a complex X with fundamental group G, and Y the covering space of X corresponding to H, then the transfer map takes a cell in X to its preimage in Y. See Hatcher, for instance, for more details. I find it hard to imagine that a categorical definition could possibly be as enlightening. De gustibus non est disputandum, I suppose. –  HJRW Dec 18 '11 at 20:14
    
In fact, given that transfer extends to all homology groups, you might also like to consider the possibility of a definition using a functor from Grp to the category of graded abelian groups (or something). –  HJRW Dec 18 '11 at 20:18
6  
@Martin : You don't choose a cell in the preimage -- you add up all the cells in the preimage! It is entirely canonical. –  Andy Putman Dec 19 '11 at 0:40
1  
Martin - I realise that you were looking for something more 'categorical' - this is why I left it as a comment rather than as an answer. (Although, as Andy notes, this construction is completely canonical for CW complexes.) I was responding to your complaint that the 'usual' definition is useful but not enlightening. –  HJRW Dec 19 '11 at 11:34
1  
@HW : If you use singular homology like I did in my answer, things are entirely canonical even if your spaces are not equipped with CW complex structures. Though I guess the categorically-minded will just say that I am choosing a particularly nice CW approximation, namely the total singular complex. –  Andy Putman Dec 20 '11 at 4:54
show 4 more comments

6 Answers

It might be useful here to make the translation between the arguments using covering spaces and those using the Grothendieck construction by noting that there is a well known equivalence for a groupoid $G$ between the category of actions of $G$ on sets and that of covering morphisms of the groupoid $G$. See for example

Higgins, P.J., Notes on categories and groupoids, Mathematical Studies, Volume 32. Van Nostrand Reinhold Co. London (1971); Reprints in Theory and Applications of Categories, No. 7 (2005) pp 1-195. (downloadable)

I have traced this notion of covering morphism back to a 1951 Annals. of Math. paper by P.A. Smith (under the name regular morphism), and the equivalence mentioned above is of course related to the so-called Grothendieck construction, though is was earlier considered by C. Ehresmann.

I believe there are advantages in an exposition of covering spaces using this notion, since a covering map is nicely modelled by a covering morphism.

share|improve this answer
    
Higgins book is fantastic. I recommend it to all. –  Benjamin Steinberg Dec 20 '11 at 17:52
add comment

Edit: As Martin remarked, there is a gap in the proof below. It can be closed by replacing axiom 1 by 1'. However, this isn't very satisfying, as it lowers the categorial flauvor of the characterization. Perhaps one should further investigate, if axiom 1 couldn't be used anyway.

$\hspace{5pt}$1'. If $G=\langle H,x \rangle, n=(G:H)$ and $h \in \cap_{i=0}^{n-1}x^iHx^{-i}$, then $t^G_H(h[G,G])$ is represented $\hspace{10pt}$ $\hspace{12pt}$ by $(hx)^nx^{-n}$.

Futhermore axiom 3 should be

$\hspace{5pt}$3. If $f: G \to G'$ is a homomorphism, $H' \le G, H = f^{-1}(H')$ and $(G:H) = (G':H'),$ $\hspace{5pt}$ $\hspace{12pt}$then the diagram commutes.


As far as I can see, the answers above are all concerned with an explicit construction of the transfer. Here I will go the other direction and characterize the transfer by its properties. Let $V$ denote the usual transfer.

Suppose for each pair $H \le G$ with $(G:H) < \infty$ there is a homomorphism $t^G_H: G_{ab} \to H_{ab}$ satisfying the subsequent properties. Then $t^G_H = V^G_H$.

  1. The composition $G_{ab}\hspace{1pt} \xrightarrow{ t } \hspace{1pt} H_{ab} \hspace{1pt} \xrightarrow{\bar{i}} \hspace{1pt} G_{ab}$ is multiplication by $(G:H)$.

  2. If $H \le K \le G$ then $t^K_H \circ t^G_K = t^G_H$

  3. If $(G:H) = (G':H')$ and $f: G \to G'$ is a homomorphism with $f(H) \le H'$ then the following diagram commutes: $$\begin{array}{ccc} G_{ab} & \xrightarrow{\bar{f}} & G_{ab}' \newline t \downarrow & & \downarrow t' \newline H_{ab} & \xrightarrow[\bar{f}]{} & H_{ab}' \end{array}$$


Proof: a) It's well-known that $V$ satisfies $1.-3.$.

b) By 1., $t^G_H$ and $V^G_H$ agree on $\bar{x}$ for $x \in H$.

c) Suppose $G = \langle H, x \rangle$ and $(G:H) = n$. Let $f: \mathbb{Z} \to G, 1 \mapsto x$. By 1. we have $t: \mathbb{Z} \to n\mathbb{Z}, 1 \to n$. Now 3. implies $t^G_H(\bar{x}) = \bar{x}^n = V^G_H(\bar{x})$. In particular $t^G_G = id|G_{ab} = V^G_G$.

d) We show by induction on $n=(G:H)$ that $t^G_H = V^G_H$ for all $H \le G$. The case $n=1$ was shown in c). Suppose $n>1$ and $t^G_H = V^G_H$ holds for all $H \le G$ with $(G:H) < n$. Let $x \in G$. If $G = \langle H, x \rangle$ then $t^G_H(\bar{x}) = V^G_H(\bar{x})$ by c). So assume $K := \langle H, x \rangle$ is a proper subgroup of $G$. Because of b) we may assume $x \notin H$. Thus $(G:K),(K:H) < n$ and we conclude from 2. and the induction hypothesis and a) that $t^G_H = V^G_H$. q.e.d.

share|improve this answer
    
This is exactly what I was looking for. Can you give me a hint why 3. is true for the transfer map? I can only prove it when $f$ induces a bijection $H/G \to H'/G'$. –  Martin Brandenburg Dec 19 '11 at 8:42
    
I also don't understand the proof of b) at all. And the proof of c) is unclear to me because I don't know which subgroups you take for the application of 3. –  Martin Brandenburg Dec 19 '11 at 9:26
1  
Sorry for replying late - it's a bad time for making errors. Of course, your concerns are reasonable. I undererstimated the problem of evaluating the transfer on elements of the form $h[G,G]$. Concerning 3: I forgot the additional condition $H = f^{-1}(H')$. –  Ralph Dec 20 '11 at 8:01
add comment

Here is another answer. It is in fact equivalent to all the previous answers but is more categorical. Let $X$ be a finite transitive $G$-set and let $\mathcal G=G\ltimes X$ be the corresponding Grothendieck construction. So it is the groupoid with objects $X$ and arrows $(g,x):x\to gx$. The product is $(g,hx)(h,x)=(gh,x)$. It is the groupoid analogue of the covering space of $BG$ associated to the $G$-set $X$.

Now if $H$ is an isotropy group, then $\mathcal G$ is equivalent to $H$ but the choice of equivalence is not unique. This is Martin's complaint. But since any two naturally equivalent functors from a groupoid to an abelian group are the same, there is a CANONICAL functor $\tau\colon \mathcal G\to H^{ab}$ which is just the universal functor from $\mathcal G$ to an abelian group.

The tranfer is the map $$g\mapsto \sum_{x\in X}\tau(g,x).$$

share|improve this answer
    
Actually the transfer is the map on the abelianization induced by the above map. –  Benjamin Steinberg Dec 18 '11 at 21:18
    
Ah I like this point of view! This seems to be the main point of your previous answer applied to the specific situation. However it also just rephrases the usual definition (in a more elegant way!) and does not characterize the transfer. –  Martin Brandenburg Dec 19 '11 at 8:03
    
I'm glad you like this answer. I agree it doesn't characterize the transfer by a universal property. But it makes it clear for categorical reasons why it doesn't depend on the coset reps. –  Benjamin Steinberg Dec 19 '11 at 12:32
    
Maybe the real story is that if one has an etale groupoid $G$ with finitely many objects and a functor $F\colon G\to A$ with $A$ an abelian group, then there is a homomorphism from the inverse monoid of local bisections of $G$ to $A$ by adding up the elements in the local bisection. If a group acts on a finite set it maps into the group of units of the local bisection monoid of the translation groupoid and we get transfer. This still doesn't characterize it but it places transfer in a broader context. –  Benjamin Steinberg Dec 19 '11 at 12:38
add comment

My answer is also not categorical, but it is too long for a comment and I think it sheds light on the nature of the transfer.

I think of the transfer as really being a fact about covering spaces. Let $\pi : X \rightarrow Y$ be a degree $n$ covering map. If $\sigma : \Delta^k \rightarrow Y$ is a singular $k$-simplex on $Y$, then covering space theory provides $n$ different lifts $\tilde{\sigma}_1,\ldots,\tilde{\sigma}_n : \Delta^k \rightarrow X$ of $\sigma$. Define $\tau_k(\sigma)$ to be the singular $k$-chain $\tilde{\sigma}_1 + \cdots+ \tilde{\sigma}_n$ on $X$. This extends by linearity to a map $\tau_k : C_k(Y;R) \rightarrow C_k(X;R)$, where $R$ is any commutative ring and $C_{\ast}(\cdot,R)$ is the abelian group of singular simplices with coefficients in $R$. It is clear that the $\tau_k$ combine together to form a chain map $\tau : C_{\ast}(Y;R) \rightarrow C_{\ast}(X;R)$ that satisfies $$\pi_{\ast}(\tau(x)) = n \cdot x,$$ where $\pi_{\ast} : C_{\ast}(X;R) \rightarrow C_{\ast}(Y;R)$ is the map on singular chains induced by $\pi$. The transfer map $H_{\ast}(Y;R) \rightarrow H_{\ast}(X;R)$ is the map on homology induced by $\tau$.

To recover the classical transfer, let $Y$ be a $K(G,1)$ and $X$ be the cover corresponding to $H$.

share|improve this answer
    
I see that HW wrote a similar thing in the comments while I composed this. I'll leave it up since it gives more details. –  Andy Putman Dec 18 '11 at 20:50
    
Thanks, this is very interesting! Indeed I've already read about this transfer map for covering spaces, but you have also included the missing details. I like this geometric picture. –  Martin Brandenburg Dec 18 '11 at 23:12
add comment

It is not really categorical, so this is maybe more of a comment than an answer, but the way I find easiest to see that transfer really gives a homomorphism (independent of choice of coset representatives, but it's not clear to me that this issue is much easier from this viewpoint) is from a viewpoint which may be due to T. Yoshida, who wrote some papers on "character-theoretic transfer" in the 70s. Given that $[G:H]$ is finite, consider a group homomorphism $\phi: H \to A$ where $A$ is an Abelian group. Let $R$ be the group ring ${\rm GF}(2)[A].$ Consider $\phi$ as a rank $1$-representation of $H$ over $R$. Induce that to a representation from $G \to {\rm GL}_d(R),$ where $d = [G:H]$, and take the determinant of that induced representation. In the case that $A = H/H^{\prime}$, we (implicitly) obtain the homomorphism $V_G: G^{ab} \to H^{ab}.$.

share|improve this answer
    
Oh, that's a nice construction. +1 –  Johannes Hahn Dec 18 '11 at 18:02
    
Very cool, I did NOT know that... –  Igor Rivin Dec 18 '11 at 21:42
add comment

An alternative description along Geoff's line is the following. Let $T$ be a set of coset reps for $H$. Then associated to $T$ is a Krasner-Kaloujnine embedding $$G\hookrightarrow H^{G/H}\rtimes (G/H_G)$$ where $H_G$ is the intersection of the conjugates of $H$. This embedding depends on $T$ only up to an inner automorphism of $H^{G/H}$. The abelianization of the semidirect product above is $H^{ab}\times (G/H_G)^{ab}$ and the restriction of the abelianization map to $G$ yields a homomorphism $$G\hookrightarrow H^{ab}\times (G/H_G)^{ab}\to H^{ab}$$ where the last map is the projection. This induces a homomorphism $G^{ab}\to H^{ab}$ which is the transfer. The independence from $T$ follows the independence of the embedding up to inner automorphism.

share|improve this answer
    
By the way, I would like a categorical interpretation of the Krasner-Kakoujnine embedding. Maybe one needs 2-categories since it is defined up to inner automorphism. –  Benjamin Steinberg Dec 18 '11 at 15:08
2  
The article "Embedding into wreath product and the Yoneda lemma" seems to do what you mentioned... I found it there: kurims.kyoto-u.ac.jp/~kyodo/kokyuroku/contents/pdf/1318-18.pdf –  Sylvain Bonnot Dec 18 '11 at 15:28
    
@Sylvain: Thanks for the reference, this is a very interesting article. It illustrates how natural some ad hoc constructions are when we consider them via (higher) categories. @Benjamin: So the embedding depends on a choice and after all you prove that it is independant from it. But thus is exactly what I would like to avoid. I don't think that your answer (though the content interesting for itsself) answers my question. –  Martin Brandenburg Dec 18 '11 at 18:02
    
Perhaps Sylvain's article gets around the choice (I haven't read it yet)? The choice inner automorphism is uniquely determined by the change in transversals which is why there might be a 2-category explanation. –  Benjamin Steinberg Dec 18 '11 at 18:22
    
Also Geoff's answer has the same defect because one first chooses a basis to define determinant and then shows independence of the basis. This can be disguised using exterior powers but it is still there. –  Benjamin Steinberg Dec 18 '11 at 18:28
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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