The Whitehead conjecture states that if $X$ is a $2$-dimensional aspherical simplicial complex and $Y \subset X$ is a connected sub-complex then $Y$ is aspherical. This can be re-phrased in terms of presentations by considering the presentation complex of $\langle X\mid \mathbf{r}\rangle$; then sub-presentations $\langle X; \mathbf{r}^{\prime}\rangle$, $\mathbf{r}^{\prime}\subset \mathbf{r}$, give sub-complexes.

This presentation setting can be phrased algebraically as follows: Let $G=\langle X\mid \mathbf{r}\rangle$ be a presentation, and let $\alpha: F(X)\rightarrow G$ be the canonical map. So $W^{\alpha}\in G$ for any word $W\in F(X)$. Let $N(\mathbf{r})$ denote the normal closure of $\mathbf{r}$ in $F(X)$, and let $N^{\prime}(\mathbf{r})=[N(\mathbf{r}), N(\mathbf{r})]$ be the derived subgroup. Write $M(\mathbf{r}):=N(\mathbf{r})/N^{\prime}(\mathbf{r})$, and there is a map $\beta: N(\mathbf{r})\rightarrow M(\mathbf{r})$. Then $M(\mathbf{r})$ is a (left, say) $\mathbb{Z}G$-module, the relation module, with action given by $\left(W_1^{\alpha}\pm W_2^{\alpha}\right)\cdot R^{\beta}=\left(W_1RW_1^{-1}W_2R^{\pm1}W_2^{-1}\right)^{\beta}$. Then $\langle X\mid \mathbf{r}\rangle$ is aspherical if and only if the relation module $M(\mathbf{r})$ is freely generated by the images $R^{\beta}$ of relators $R\in\mathbf{r}$. Therefore, the Whitehead conjecture states the following:

Whitehead Conjecture: Suppose $M(\mathbf{r})$ is freely generated as a $\mathbb{Z}G$-module by the images $R^{\beta}$ of relators $R\in\mathbf{r}$. If $\mathbf{r}^{\prime}\subset\mathbf{r}$ then $M(\mathbf{r}^{\prime})$ is freely generated as a $\mathbb{Z}H$-module, $H=\langle X\mid\mathbf{r}^{\prime}\rangle$, by the images $R^{\beta^{\prime}}$ of relators $R^{\prime}\in\mathbf{r}^{\prime}$.

Roughly, my question is: where is the subtlety here? Is it simply the following: there may exist some $z_i\in\mathbb{Z}H\setminus 0$, with $1\leq i\leq n$, which each map to $0\in\mathbb{Z}G$ under the obvious map $\gamma$, where $z_1^{\gamma}R_1^{\beta}+\cdots +z_n^{\gamma}R_n^{\beta}=0_{M(\mathbf{r})}$, but where $z_1R_1^{\beta^{\prime}}+\cdots +z_nR_n^{\beta^{\prime}}=0_{M(\mathbf{r}^{\prime})}$?

  • $\begingroup$ What is $M(R)$? What is $M(R')$? Are they obvious misprint? I.e they must be $\endgroup$ – Alireza Abdollahi Mar 29 '17 at 15:50
  • $\begingroup$ By $M(R)$ I meant $M(\mathbf{r})$, and similarly by $M(R^{\prime})$ I meant $M(\mathbf{r}^{\prime})$. (This is just in the final line, yes?) $\endgroup$ – HeadingWhiteways Mar 30 '17 at 8:35
  • $\begingroup$ You seem to be assuming that $G$ and $H$ have the same generating set $X$, which would correspond geometrically to only looking at subcomplexes with the same $1$-skeleton. $\endgroup$ – Mark Grant Mar 30 '17 at 15:25
  • $\begingroup$ Can you give a reference for your equivalent condition in terms of the relation module? $\endgroup$ – HJRW Mar 31 '17 at 8:57
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    $\begingroup$ @HJRW I believe that it is proven in the Chiswell-Collins-Huebschmann paper "Aspherical group presentations" (Prop. 1.2, although this proposition proves more), but it also is stated in a research announcement of Ivanov "On aspherical presentations of groups". $\endgroup$ – HeadingWhiteways Mar 31 '17 at 16:59

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