The subtlety with (an algebraic phrasing of) the Whitehead conjecture? - MathOverflow most recent 30 from mathoverflow.net 2019-10-20T18:08:12Z https://mathoverflow.net/feeds/question/265770 https://creativecommons.org/licenses/by-sa/4.0/rdf https://mathoverflow.net/q/265770 2 The subtlety with (an algebraic phrasing of) the Whitehead conjecture? HeadingWhiteways https://mathoverflow.net/users/106679 2017-03-28T15:47:19Z 2017-03-30T08:36:06Z <p>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.</p> <p>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 <em>relation module</em>, 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:</p> <blockquote> <p><strong>Whitehead Conjecture</strong>: 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 $R^{\prime}\in\mathbf{r}^{\prime}$.</p> </blockquote> <p>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 $z_1R_1^{\beta^{\prime}}+\cdots +z_nR_n^{\beta^{\prime}}=0_{M(\mathbf{r}^{\prime})}$?</p>