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Let $M$ be a projective manifold of dimension $n$ such that $T^*M$ is algebraically completely integrable system. Let $\mu: T^*M \to \mathbb{C}^n$ be the moment map. There is a natural $\mathbb{C}^*$ action on $T^*M$, namely, $c.(p, v)= (p, cv)$, where $c \in \mathbb{C}^*$ and $v$ is a co-tangent vector at $p$.

Question: Does there exist a $\mathbb{C}^*$ action on $\mathbb{C}^n$ such that the moment map $\mu$ is $\mathbb{C}^*$ equivariant ?

I was thinking in the following way: any functions $f$ on the $T^*M$ can be thought as an element of $\oplus_{i=1}^{\infty}H^0(M, S^nTM)$, where $S^nTM$ denotes the $n-$th symmetric power of $TM$. If $f$ is represented by $(s_1, s_2, ....)$ then define $c.f= (cs_1, cs_2,..., )$, where $c \in \mathbb{C}^*$, which induces a $\mathbb{C}^*$ action on $\mathbb{C}^n$. On the other hand, on $T^*X$, there is a natural $\mathbb{C}^*$ action, anmely $c.(p, v)=(p, cv)$, where $p \in X$ and $v$ is a cotangent vector of $X$ at $p$. It seems the moment is the $\mathbb{C}^*$-equivariant with this action.

Please correct me if I am wrong.

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You'll only have $c\cdot f =(cs_1,\dots)$ if the components of $\mu$ are vector fields (thought of as functions on $T^*M$). In this case, the integrable system is induced by an action of $\mathbb{C}^n$ on $M$, so you'll have this equivariance if and only if this is the case.

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  • $\begingroup$ I guess for an algebraically completely integrable system there are $n$ functions $f_1, f_2,..., f_n$ on $T^*X$ which are in involution and the moment map is defined by these functions. The functions are given by $f_i=(s^i_1, s^i_2, ...)$. Please correct me if I am wrong. $\endgroup$
    – LAPRAS
    Commented Jun 2, 2022 at 16:16
  • $\begingroup$ May the appropriate $\mathbb{C}^*$ action will be $c.f_i=(cs^i_1, c^2s^i_2,...)$. $\endgroup$
    – LAPRAS
    Commented Jun 2, 2022 at 16:27
  • $\begingroup$ I think this will be the correct approach. A section of $S^kTX$ gives regular function on $T^*X$ which is a homogenius polynomial of degree $k$ in each fiber. Now the $\mathbb{C}^*$ action on the cotangent bundle namely $c(p, v)= (p, cv)$ gives us an action on each component of $\mathbb{C}^n$ as $s(cv)= c^ks(v)$, which is clearly $\mathbb{C}^*$ equivariant. $\endgroup$
    – LAPRAS
    Commented Jun 2, 2022 at 17:41

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