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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 $\mathbb{P}^2$. \mathbb{CP}^2$. Here$\mathbb{S}^1\times\mathbb{S}^1$acts on$\mathbb{P}^2$. \mathbb{CP}^2$. The professor just regards $\mathbb{P}^2$ \mathbb{CP}^2$as a triangle with edges$\mathbb{P}^1$, \mathbb{CP}^1$, 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 $\mathbb{P}^2$ \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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# How to calculate the equivariant cohomology ring of $P^2$?

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 $\mathbb{P}^2$. Here $\mathbb{S}^1\times\mathbb{S}^1$ acts on $\mathbb{P}^2$. The professor just regards $\mathbb{P}^2$ as a triangle with edges $\mathbb{P}^1$, 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 $\mathbb{P}^2$ 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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