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Let $M$ be a compact oriented Riemannian manifold without boundary, $G$ be a compact Lie group acting smoothly on $M$, and $\mathfrak{g}$ be its Lie algebra.

Let $M_{0}$ be a submanifold of $M$. Let $N$ be the normal bundle of $M_{0}$.

We identify a tubular neighborhood of $M_{0}$ in $M$, with a neighborhood $V$ of $M_{0}$ in $N$. Set $X\in\mathfrak{g}$, $X_{M}$ be the vector field on $M$ corresponding with $X\in\mathfrak{g}$, $\mu^{M}(X)$ is the moment map(see Chapter 7 in 'Heat Kernels and Dirac operators'). Is there a result below:

For $y\in V$, $$X_{M}(y)=-\mu^{M}(X)y.$$ On the left, $y$ is the point in $M$. On the right, $y$ is the tautological section of normal bundle.

Edit: I read this in Bismut's article 'Localization Formulas, Superconnections, and the Index Theorem for Familise' (in the proof No.1 of Theorem 1.3). In Bismut's article, $M_{0}$ is the zeros of $X_{M}$. Here, we didn't have this assumption. I want to know whether the result is also correct in this condition.

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1 Answer 1

No. The result is: $$\alpha_y(X_M(y)) = \langle\mu^{T^*M}(\alpha_y),X\rangle,\quad y\in M,\quad \alpha_y\in T^*_yM,\quad X\in \mathfrak g.$$ The momentum mapping (moment is just wrong since it has nothing to do with torque) goes $\mu^{T^*M}:T^\star M \to \mathfrak g^\star$, and you have to lift the action of $G$ to the cotangent bundle before a momentum mapping can be seen.

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Thanks for your answer. Maybe I didn't catch the key points of your answer. I'm sorry, I make a mistake, the equality is $X_{M}(y)=-\mu^{M}(X)y$. I read this in Bismut's article'Localization Formulas, Superconnections, and the Index Theorem for Familise'(in the proof No.1 of Theorem 1.3). In this article, $M_{0}$ is the zeros of $X_{M}$. Here, we didn't have this assumption. –  Chen Feb 19 '13 at 15:19

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