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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.

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 $H^2_T(M)$.

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.

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Perhaps you could be more explicit in your question. For instance, what is the action? I can see an obvious action on $\mathbb{P}^1$, but am less sure when there are three homogeneous coordinates. It might save some readers time to clarify that you are talking about complex projective space. Also, why don't you ask your professor? – Mark Grant Oct 25 '12 at 6:21
Do you mean to have the $3$-torus $S^1\times S^1\times S^1$ acting on $\mathbb{CP}^2$? – Mark Grant Oct 25 '12 at 14:49
Did your professor mean that 'triangle' be the orbit space of the torus action on $\mathbb CP^2$? vertices correspondent to the fixed point set and edges correspondent to the orbit with isotropy group $S^1$ and interior correspondent to the principle orbits? – J. GE Oct 25 '12 at 19:16
up vote 2 down vote accepted

It sounds like you're talking about GKM (=Goresky–Kottwitz–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 $$ P = \begin{pmatrix} \ast & \ast & \ast \\ 0 & \ast & \ast \\ 0 & \ast & \ast \end{pmatrix} $$ 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.

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 $H_T^\ast({\text pt}) \cong S(\mathfrak{t}^\ast) \cong \mathbb C[x_1,\ldots,x_n]$.

For more on this, I recommend Julianna Tymoczko nice survey article. Be sure to check out example 4.1, where she computes $H_T^\ast(\mathbb{CP}^2)$ for our $T$ above.

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I must thank you for your help and the article you provided. Meantime please allow me to apologize for those stupid mistakes I made when asking questions – xuxuzhu Oct 27 '12 at 5:37

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