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The short: Let

$X= \{(x,y,z) \in \mathbb{C}^* \times \mathbb{C} \times \mathbb{C} \, |\, yz-x\neq 0\}$

Compute $H^*_c(X)$ (say with $\mathbb{C}$-coefficients).

The long: Unless I messed something up, the answer should be

$H_c^3(X) = \mathbb{C}$, $H_c^4(X) = \mathbb{C}^2$, $H_c^5(X) = \mathbb{C}^2$, $H_c^6(X)=\mathbb{C}$, and $0$ otherwise.

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.

Some context: this question is closely related to

Intersection of plus/minus cells in Bialynicki-Birula decomposition

Namely, $X$ is the intersection of the big Bruhat cell and the big opposite Bruhat cell for $SL_3$.

It is also closely related to

Are Kazhdan-Lusztig $R$-polynomials the Poincare polynomials of the corresponding affine varieties

(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).

My convoluted argument: Considering the alternatives $y\neq 0$ and $y= 0$, one obtains a decomposition of $X$ into

$X = (\mathbb{C}^*)^3 \sqcup \mathbb{C}\times \mathbb{C}^*$.

This gives rise to a long exact sequence that puts several restrictions on $H^*(X)$ (but doesn't fully determine it, namely $H_c^3(X), H_c^4(X), H_c^5(X)$ aren't fully determined).

So far this is nice, but now the convoluted bit starts. It is not too hard to see that $H^{*-3}(X)$ 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 $\mathfrak{sl}_3$.

Aside: 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).

Now some standard representation theoretic facts about these $Hom$ spaces combined with the decomposition above yield what I claimed the answer to be.

I would love a simpler/more geometric way of going about this computation!

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1 Answer 1

up vote 6 down vote accepted

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 $H^{BM}_i(Y\cup Z)=\mathbb{C}^2$ 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.

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Ah, perfect! Many thanks! This is exactly the sort of simple answer I was looking for. –  Reladenine Vakalwe Dec 2 '12 at 4:36
    
Reladenine -- welcome! –  algori Dec 2 '12 at 4:48

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