A cohomology computation request. - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T03:20:51Z http://mathoverflow.net/feeds/question/115141 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/115141/a-cohomology-computation-request A cohomology computation request. Reladenine Vakalwe 2012-12-02T02:53:12Z 2012-12-02T04:07:14Z <p><strong>The short: </strong>Let</p> <p><code>$X= \{(x,y,z) \in \mathbb{C}^* \times \mathbb{C} \times \mathbb{C} \, |\, yz-x\neq 0\}$</code></p> <p>Compute $H^*_c(X)$ (say with $\mathbb{C}$-coefficients).</p> <p><strong>The long: </strong>Unless I messed something up, the answer should be</p> <p>$H_c^3(X) = \mathbb{C}$, $H_c^4(X) = \mathbb{C}^2$, $H_c^5(X) = \mathbb{C}^2$, $H_c^6(X)=\mathbb{C}$, and $0$ otherwise.</p> <p>However, as will become clear I did this through an extremely convoluted argument and I am hoping someone can explain to me a simple way of doing this. </p> <p><b>Some context:</b> this question is closely related to</p> <p><a href="http://mathoverflow.net/questions/111682/intersection-of-plus-minus-cells-in-bialynicki-birula-decomposition" rel="nofollow">http://mathoverflow.net/questions/111682/intersection-of-plus-minus-cells-in-bialynicki-birula-decomposition</a> </p> <p>Namely, $X$ is the intersection of the big Bruhat cell and the big opposite Bruhat cell for $SL_3$.</p> <p>It is also closely related to </p> <p><a href="http://mathoverflow.net/questions/83877/are-kazhdan-lusztig-r-polynomials-the-poincare-polynomials-of-the-corresponding" rel="nofollow">http://mathoverflow.net/questions/83877/are-kazhdan-lusztig-r-polynomials-the-poincare-polynomials-of-the-corresponding</a> </p> <p>(the Hodge-Euler characteristic of $X$ is the $R$-polynomial corresponding to the identity and the longest element in type $A_2$; this is a special case of a general fact about $R$-polynomials).</p> <p><strong>My convoluted argument: </strong> Considering the alternatives $y\neq 0$ and $y= 0$, one obtains a decomposition of $X$ into</p> <p><code>$X = (\mathbb{C}^*)^3 \sqcup \mathbb{C}\times \mathbb{C}^*$</code>.</p> <p>This gives rise to a long exact sequence that puts several restrictions on <code>$H^*(X)$</code> (but doesn't fully determine it, namely $H_c^3(X), H_c^4(X), H_c^5(X)$ aren't fully determined). </p> <p>So far this is nice, but now the convoluted bit starts. It is not too hard to see that <code>$H^{*-3}(X)$</code> equals $Ext^*(\Delta_e, \Delta_{w_0})$ where $\Delta_e$ is the unique simple Verma and $\Delta_{w_0}$ is the unique projective Verma in the principal block of the BGG-category $\mathcal{O}$ of $\mathfrak{sl}_3$. </p> <p><b>Aside:</b> this a special case of a statement connecting extensions of Verma modules with cohomology of intersections of Bruhat cells and opposite Bruhat cells (and also why I am interested in the cohomology of these intersections).</p> <p>Now some standard representation theoretic facts about these $Hom$ spaces combined with the decomposition above yield what I claimed the answer to be.</p> <p>I would love a simpler/more geometric way of going about this computation!</p> http://mathoverflow.net/questions/115141/a-cohomology-computation-request/115143#115143 Answer by algori for A cohomology computation request. algori 2012-12-02T03:36:02Z 2012-12-02T04:07:14Z <p>Reladenine -- your $X$ is the complement in the affine 3-space of the union of two hypersurfaces $Y$ and $Z$, the first given by $x=0$, the second by $yz=x$. The intersection $Y\cap Z$ is the union of two intersecting affine lines. Moreover, both $Y$ and $Z$ are isomorphic to $\mathbb{C}^2$ (note that $Z$ is the graph of a function). So the Borel-Moore homology of $Y\cup Z$ is given by <code>$H^{BM}_i(Y\cup Z)=\mathbb{C}^2$</code> if $i=4,3$ and $\mathbb{C}$ if $i=2$. So, by the Alexander duality $\tilde H^j(X)\cong H^{BM}_{6-1-j}(Y\cup Z)$ where $\tilde H$ stands for the reduced cohomology, $H^i(X)=\mathbb{C}$ if $i=0,3$ and $\mathbb{C}^2$ if $i=1,2$. Applying the Poincar\'e duality $H^i(X)\cong H^{6-i}_c(X)$ one gets the answer you give in your posting.</p>