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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.

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.

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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Xuxu
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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.