Characterization of the transfer map in group theory

Question

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.

Answer 1

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 $V_G: G^{ab} \to H^{ab}.$.

Answer 2

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 $H$.

Answer 3

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 $$g\mapsto \sum_{x\in X}\tau(g,x).$$

Answer 4

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.

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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: $$\begin{array}{ccc} G_{ab} & \xrightarrow{\bar{f}} & G_{ab}' \newline t \downarrow & & \downarrow t' \newline H_{ab} & \xrightarrow[\bar{f}]{} & H_{ab}' \end{array}$$

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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.

Answer 5

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.

Answer 6

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.