0
$\begingroup$

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.

$\endgroup$

1 Answer 1

0
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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
    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
    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
    Jun 2, 2022 at 17:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.