Cohomological dimension of a homomorphism - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T02:15:04Z http://mathoverflow.net/feeds/question/89178 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/89178/cohomological-dimension-of-a-homomorphism Cohomological dimension of a homomorphism Mark Grant 2012-02-22T11:29:02Z 2012-02-24T14:31:18Z <p>Let $G$ and $\Gamma$ be discrete groups, and let $\phi\colon\thinspace G\to \Gamma$ be a homomorphism.</p> <p>Define its <em>cohomological dimension</em> $\operatorname{cd}\phi$ to be the least integer $d$ such that $\phi^\ast\colon H^i(\Gamma;M)\to H^i(G;M)$ is the zero homomorphism for all $i>d$ and all $\Gamma$-modules $M$ (where $M$ is regarded as a $G$-module via $\phi$).</p> <p>Given that cohomological dimension of groups is such a well-studied invariant, I would have expected to find references to this relative notion in the literature. Alas, I cannot.</p> <blockquote> <p>Are there any references considering cohomological dimension of homomorphisms?</p> </blockquote> <p>and more specifically</p> <blockquote> <p>Does anyone know an example of a <em>surjective</em> homomorphism $\phi$ as above for which $$\operatorname{cd} \phi &lt; \min\lbrace \operatorname{cd}G, \operatorname{cd} \Gamma \rbrace?$$</p> </blockquote> <p><strong>EDIT:</strong> Thanks to Tom and Ralph's answers, I have been able to prove the following precise statement:</p> <p>Let $$1\to A \to G \stackrel{\phi}{\to} \Gamma \to 1$$ be a central extension, where $H_\ast(A)$ is free and of finite type, and $\Gamma$ is a duality group with $\operatorname{cd}\Gamma = n$. Then $\operatorname{cd}\phi = n$.</p> <p><strong>Proof.</strong> We will show that $0\neq \phi^\ast\colon\thinspace H^n(\Gamma;\mathbb{Z}\Gamma)\to H^n(G;\mathbb{Z}\Gamma)$. This follows from the Lyndon-Hochschild-Serre spectral sequence. Since the action of $\Gamma$ on $A$ is trivial, and $\mathbb{Z}\Gamma$ is a trivial $A$-module, the $E_2$ term has $$H^p(\Gamma;H^q(A;\mathbb{Z}\Gamma))\cong H^p(\Gamma;H^q(A)\otimes\mathbb{Z}\Gamma)$$ in the $(p,q)$-position. Since $\Gamma$ is a duality group, this is zero for $p\neq n$. Hence there are no non-trivial differentials, and the edge homomorphism $$\phi^\ast\colon\thinspace H^n(\Gamma;\mathbb{Z}\Gamma) \to H^n(G;\mathbb{Z}\Gamma)$$ is an isomorphism. $\Box$</p> <p>Tom's answer shows that either centrality or finite type is necessary in the above statement. I haven't accepted it yet because I'm hoping someone will give an example with $\operatorname{cd} G &lt;\infty$.</p> http://mathoverflow.net/questions/89178/cohomological-dimension-of-a-homomorphism/89225#89225 Answer by Tom Goodwillie for Cohomological dimension of a homomorphism Tom Goodwillie 2012-02-22T20:33:57Z 2012-02-23T14:26:06Z <p>How about this? Let $\Gamma$ be free abelian of rank $2$. Poincare duality in the torus identifies $H^2(\Gamma;M)$ with $H_0(\Gamma;M)$, so that in particular there is a natural surjection $M\to H^2(\Gamma;M)$. </p> <p>Let $F$ be the particular $\Gamma$-module $\mathbb Z\Gamma$, free module of rank one for the group ring. A generator of $H^2(\Gamma;F)=H_0(\Gamma;F)=\mathbb Z$ determines an extension of $\Gamma$ by $F$. Call it $G$. </p> <p>$\Gamma$ has cohomological dimension $2$. $G$ has cohomological dimension at least $2$, doesn't it? (EDIT: Yes, just because it has a subgroup $F$ with infinite cd.) But for every $\Gamma$-module the map $H^2(\Gamma;M)\to H^2(G;M)$ is zero, because using that natural surjection $M\to H^2(\Gamma;M)$ it's enough to prove this in the case $M=F$, where it is clear.</p> <p>EDIT: To spell out this last step, there is a surjective map $M\to H_0(\Gamma;M)$ (for every group $\Gamma$), natural in $M$. For this particular group, there is also a natural isomorphism $H_0(\Gamma;M)\to H^2(\Gamma;M)$. Following this by your natural map $H^2(\Gamma;M)\to H^2(G;M)$, we get a map $M\to H^2(G;M)$, natural in $M$, and we just have to show that it takes every element $x\in M$ to $0$. But there is a map $F\to M$ taking a generator to $x$, so by naturality it is enough to see that you get the zero map when $M$ is $F$.</p> http://mathoverflow.net/questions/89178/cohomological-dimension-of-a-homomorphism/89250#89250 Answer by Ralph for Cohomological dimension of a homomorphism Ralph 2012-02-23T03:38:33Z 2012-02-23T15:21:04Z <p>One more example refering to the second question: Let $$\Gamma = \left\lbrace \left. \begin{pmatrix} 1 &amp; \ast &amp; \ast \newline &amp; 1 &amp; \ast \newline &amp; &amp; 1 \end{pmatrix} \right\vert\ \ast \in \mathbb{Z} \right\rbrace$$ be the group of integral upper triangular matrices with unit diagonal. $\Gamma$ fits into the non-split central extension $$0 \to \mathbb{Z} \to \Gamma \to \mathbb{Z}^2 \to 0$$ that corresponds to a generator $\epsilon \in H^2(\mathbb{Z}^2;\mathbb{Z}) = \mathbb{Z}$. </p> <p><strong>Claim 1:</strong> $cd(\Gamma) = 3$ </p> <p>Since $\mathbb{Z}$ resp. $\mathbb{Z}^2$ has cd $1$ resp. $2$, the LHS spectral sequence $E_2^{ij} = H^i(\mathbb{Z}^2;H^j(\mathbb{Z};M))$ shows $cd(\Gamma) \le 3$. Moreover, by positional reasons $E_\infty^{2,1}=E_2^{2,1}$ and in particular $E_\infty^{2,1} = \mathbb{Z}$ for $M = \mathbb{Z}$. Hence $cd(\Gamma) = 3$. </p> <p><strong>Claim 2:</strong> The inflation map $\text{inf}: H^2(\mathbb{Z}^2;M) \to H^2(\Gamma;M)$ is zero. </p> <p>Since the image of inflation is just $E_\infty^{2,0} = \text{coker}(d_2^{0,1})$, it is sufficient to show that $d_2^{0,1}$ is surjective. Since the action of $\Gamma$ on $M$ is induced by the action of $\mathbb{Z}^2$, it follows that $\mathbb{Z}$ acts trivially on $M$. Futhermore, $\mathbb{Z}^2$ acts trivially on $H^\ast(\mathbb{Z};M)$ because $\mathbb{Z}$ is central. Therefore $E_2^{0,1} = Hom(\mathbb{Z},M)$. </p> <p>Let $\alpha \in Hom(\mathbb{Z},M)$. Then $$d_2^{0,1}: Hom(\mathbb{Z},M) \to H^2(\mathbb{Z}^2;M)$$ is given by $d_2(\alpha)= - \alpha^\ast(\epsilon)$ (well-known formula) where $$\alpha^\ast: H^2(\mathbb{Z}^2;\mathbb{Z}) \to H^2(\mathbb{Z}^2;M)$$ is induced by $\alpha$ on the coefficients. By using a projective resolution or by Poincare duality one easily sees that $$\alpha^\ast : \mathbb{Z} \to M/\lbrace gm-m \mid g \in \mathbb{Z}^2 \rbrace =: \bar{M}$$ is just $\alpha$ composed with the natural projection. Identifying $Hom(\mathbb{Z},M) = M$ now shows that $d_2^{0,1}: M \to \bar{M}$ is the natural projection and hence surjective. </p>