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Given a compact Kahler manifold $M$ with an $S^1$-action by Kahler isometries, we know that

  1. Its fixed loci $F=M^{S^1}$is a smooth Kahler submanifold.
  2. It splits $F=\sqcup_{\alpha \in A} F_{\alpha}$ into connected components
  3. The tangent space of M at fixed point splits as a complex $S^1$-representation into weight spaces $$T_x M =\oplus_{k\in\mathbb{Z}} H_{k}.$$

These are all classical results, e.g. claim 1. follows from the work of Atiyah-Bott and F. Kirwan (the moment map of the $S^1$-action is a Morse-Bott function). Claims 2. and 3. are immediate.

The question is:

Does the tangent space decomposition from claim 3. induces a split of $T_{F_\alpha} M$ into bundles that correspond to weight-spaces?

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It doesn't give a trivialization (think of the trivial action...), but indeed a splitting. For $y\in F_{\alpha }$, $S^1$ acts on $T_y(M)$; denote by $t_y$ the action of an element $t\in S^{1}$. The coefficients of the characteristic polynomial $\det(X\cdot \mathrm{I}- t_{y})$ are holomorphic functions on $F_{\alpha }$, hence constant; therefore the eigenvalues of $t_y$ and their multiplicities are independent of $y$. It follows that the eigenspaces $E^\chi_y $ of the action of $S^1$ associated to a character $\chi$ of $S^1$ form a subbundle $E^{\chi }$ of $T(M)_{|F_{\alpha }}$, and that $T(M)_{|F_{\alpha }}$ is the direct sum of these subbundles.

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  • $\begingroup$ Does this decomposition of the tangent bundle generalises to a compact Lie group action on a smooth manifold? The decomposition of a tangent space according to characters is just representation theory so generalises easily, but the continuation argument from the answer above uses the holomorphicity... $\endgroup$
    – Filip
    Commented Jul 11, 2022 at 20:42
  • $\begingroup$ Yes, the argument definitely uses the holomorphicity, and also the compactness. I would guess that the result does not hold in the smooth category, though I don't see an obvious counter-example. $\endgroup$
    – abx
    Commented Jul 12, 2022 at 13:14
  • $\begingroup$ Hmm, maybe it still works: given a non-trivial element $t \in S^1,$ the spectrum of $t_y$ consists of a discrete set $\{t^{w_i} \mid w_i \text{ are weights on }y\}$ so one cannot "jump" from one set of weights to an another, while on a path from $y$ to $y'.$ Thus I would still think that the weight decomposition is uniform on a path component $F_\alpha,$ even in the case of smooth actions. And possibly even compactness of $F_\alpha$ is not important. Then the eigenspaces $E_k=ker(t^k \cdot I - t_y)$ are smooth in $y$ hence constitute bundles, for each $k$ that appears as a weight. $\endgroup$
    – Filip
    Commented Jul 12, 2022 at 13:48

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