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Let $G$ be a group $H\leq G$ a subgroup of finite index. Further, let ${\mathcal E}^G_H$ denote the class of those short exact sequences of $G$-modules (over some fixed base ring) which split when regarded as sequences of $H$-modules. Then $(G\text{-mod},{\mathcal E}^G_H)$ is a Frobenius category.

Does anybody know if it is true that $(G\text{-mod},{\mathcal E}^G_H)$ is not Frobenius if $(G:H)=\infty$?

To show this, it would be enough to construct for and pair $(G,H)$ with $(G:H)=\infty$ an $H$-module $M$ such that the canonical $H$-split map

$\text{Ind}^G_H M\hookrightarrow\text{Coind}^G_H\text{Res}^G_H\text{Ind}^G_H M$

is not $G$-split. For example, if $M$ is the trivial $H$-module, this map becomes

${\mathbb Z}[G/H]\hookrightarrow\text{Hom}_{{\mathbb Z}H}({\mathbb Z}G,{\mathbb Z}[G/H])$

$g_0 H\longmapsto (g\mapsto g g_0 H)$.

Unfortunately, I'm not able to proof that this map is not $G$-split.

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If you take $G$ infinite and $H=(1)$ trivial, you are asking whether the category $G$-mod, with the class of short exact sequences which split in the category of abelian groups, is Frobenius. In that category $\mathbb ZG$ is projective, yet it is not injective---indeed, the relative $\mathrm{Ext}$-groups $\mathrm{Ext}_{\mathbb ZG,\mathbb Z}^p(\mathbb Z,\mathbb ZG)$ are just the usual $\mathrm{Ext}$'s, and they are not zero in general.

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  • $\begingroup$ Thank you for your answer! Do you also know an example working for any H and G? Somehow it would be nice to see that $(G\text{-mod},{\mathcal E}^G_H)$ is Frobenius if and only if $(G:H)<\infty$). $\endgroup$
    – Hanno
    Commented Jan 22, 2010 at 16:01
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The representation $\mathrm{Hom}_{H}(\mathbb{Z}G,\mathbb{Z}[G/H])$ is the $\mathbb Z$-valued functions on the $G$-set $G\times_HG/H$ such that the each fiber of the natural map $G\times_HG/H\to G/H$ is only has finitely many non-zero entries.

Now, what would a map of this representation to the trivial one look like? It would have to send the characteristic function of at least one element $\delta_x$ to a non-zero value. But, take a transversal $t:G/H\to G$ and consider $\sum_{k\in G/H}t(k)\cdot \delta_x$. This is a perfectly good element of $\mathrm{Hom}_{H}(\mathbb{Z}G,\mathbb{Z}[G/H])$, but it is sent to $\infty$ by our coinvariant. Thus, the coinvariants are trivial. The coinvariants of $\mathbb{Z}[G/H]$ are not trivial, and any inclusion which is not injective on coinvariants does not split.

So indeed this exact structure is Frobenius if and only if the index is finite.

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  • $\begingroup$ Thank you, Ben! One question concerning your identification of $\text{Hom}_H({\mathbb Z}G,{\mathbb Z}[G/H])$ with the ${\mathbb Z}$-valued functions on $G\times_H G/H$: In the former, the image of some element in $G$ has only finitely many nonzero coefficients in ${\mathbb Z}[G/H]$, whereas you impose no finiteness conditions on the functions on $G\times_H G/H$. How does this fit together? $\endgroup$
    – Hanno
    Commented Jan 22, 2010 at 21:28
  • $\begingroup$ Ben, yet another question: How do you know from the infinity of the G-orbits that the coinvariants are trivial? $\endgroup$
    – Hanno
    Commented Jan 22, 2010 at 22:46
  • $\begingroup$ With your first comment, you're right that I made a mistake (fixed above), but it makes no difference to the argument. On the second point, I've tried to clarify. $\endgroup$
    – Ben Webster ♦
    Commented Jan 22, 2010 at 23:44
  • $\begingroup$ It's not clear to me why there has to exist a characteristic function on which the given map does not vanish. Further, what's the formal argument to show that there's no integer value $\sum_{k\in G/H} t(k)\cdot\delta_x$ could be sent to? (One could try to split off a large finite sum to see that it's value equals a large multiple of $\delta_x$'s value -- but this doesn't help, because the remaining sum/function could have an equally large value. Reminds me a bit of the proof that ${\mathbb Z}^{{\mathbb N}}$ is not free.) $\endgroup$
    – Hanno
    Commented Jan 23, 2010 at 8:35

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