How to calculate the equivariant cohomology ring of $P^2$? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T04:09:42Z http://mathoverflow.net/feeds/question/110608 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110608/how-to-calculate-the-equivariant-cohomology-ring-of-p2 How to calculate the equivariant cohomology ring of $P^2$? xuxuzhu 2012-10-25T02:27:21Z 2012-10-25T18:21:07Z <p>It is well known that Kirwan's injection theorem gives an ring injection from $H^{\ast}_T(M)$ to $H^{\ast}_T(M^T)$ which is induced by the inclusion $M^T \to M$, where $T$ is a torus acting on manifold $M$ and $M^T$ is the fixed point set of this torus action. </p> <p>I came across a problem when my professor tried to use Kirwan's injection theorem to explore the ring structure of $\mathbb{CP}^2$. Here $\mathbb{S}^1\times\mathbb{S}^1$ acts on $\mathbb{CP}^2$. The professor just regards $\mathbb{CP}^2$ as a triangle with edges $\mathbb{CP}^1$, with orthogonal axis $u$ and $v$. Then he said on each vertex there is a polynomial since $H^{\ast}_T(M^T)=H^{\ast}(M^T)\otimes\mathbb{C}[u,v]$. Suppose the triangle is put with two orthogonal edges parallel to the axis $u$ and $v$. Then for the two vertex on the edge of $u$ direction, set $u=0$ to obtain the relations between coefficients. For the case $\ast=2$, each vertex has a polynomial of the form $au+bv$. So there would be 6 unknowns with 3 equations, which gives the rank of $H^2_T(M^T)$ to be 3, same for $H^2_T(M)$.</p> <p>Now my questions are: Firstly, how should I understand the view of $\mathbb{CP}^2$ as a triangle sitting in the orthogonal coordinate system, and why the $u$ and $v$ here coincident with the coordinate axis? Secondly, what is the intepretation of setting $u=0$ when we are trying to find the structure of the cohomology ring? Hope someone can help me with those questions.</p> http://mathoverflow.net/questions/110608/how-to-calculate-the-equivariant-cohomology-ring-of-p2/110694#110694 Answer by Faisal for How to calculate the equivariant cohomology ring of $P^2$? Faisal 2012-10-25T18:16:00Z 2012-10-25T18:21:07Z <p>It sounds like you're talking about GKM (=Goreskyâ€“Kottwitzâ€“MacPherson) theory, in which case it's better to think of tori as complex tori, i.e. as products of copies of $\mathbb C^\times$ and not of $S^1$. The triangle to which you're referring is the so-called moment graph of $\mathbb{CP}^2 = SL_3(\mathbb C)/P$, where <code>$$P = \begin{pmatrix} \ast &amp; \ast &amp; \ast \\ 0 &amp; \ast &amp; \ast \\ 0 &amp; \ast &amp; \ast \end{pmatrix}$$</code> and $T=\mathbb C^\times \times \mathbb C^\times$ is the diagonal subgroup of $SL_3$ acting by left multiplication. The vertices of the moment graph are the $T$-fixed points, of which there are three in this case. Two vertices are connected by an edge if and only if there's a one-dimensional $T$-orbit whose closure contains the corresponding fixed points. The closure of such an orbit is a copy of $\mathbb{CP}^1$, so that might explain why your professor labeled the edges as such. But anyway, the moment graph already comes with a useful labeling and a direction, though let me not say more about this here.</p> <p>GKM theory provides a combinatorial description of $H_T^\ast(M)$ in terms of the moment graph of $M$, and it appears that this is what your professor was using. (Here $M$ refers to a projective variety on which a complex torus $T$ is acting in some "nice" fashion. If $M=G/P$ is a generalized flag variety then the action of a maximal torus $T\subset P$ of $G$ is "nice" enough for GKM theory.) One is also provided with an isomorphism $$H^\ast(M) = \frac{H_T^\ast(M)}{\mathfrak{m} H_T^\ast(M)},$$ where $\mathfrak m$ is the augmentation ideal $(x_1,\ldots,x_n)$ in $H_T^\ast({\text pt}) \cong S(\mathfrak{t}^\ast) \cong \mathbb C[x_1,\ldots,x_n]$.</p> <p>For more on this, I recommend Julianna Tymoczko <a href="http://arxiv.org/abs/math/0503369" rel="nofollow">nice survey article</a>. Be sure to check out example 4.1, where she computes $H_T^\ast(\mathbb{CP}^2)$ for our $T$ above.</p>