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Let $G$ be a finite cyclic group and $X$ a smooth manifold equipped with a trivial $G$-action.

It is known that we can decompose every $G$-equivariant vector bundle with respect to the action:

Every smooth $G$-equivariant vector bundle $V \to X$ can be decomposed as the Whitney sum $V=\bigoplus_{\chi}V_\chi$ of $G$-equivariant vector bundles.

Here $\chi$ runs the character group $X(G):=\mathrm{Hom}(G,\mathbb C^\ast)$ and the $G$-action on $V_\chi$ is given by $g \cdot v = \chi(g)v$ ($g \in G$, $v \in V_\chi$).

I understand the case when $X$ is a point. This is just the eigenspace decomposition.

Question: Why do the eigenspaces vary smoothly so that $V_\chi$ gives a smooth subbundle of $V$?

Notes:

  1. I do not mention above whether the vector bundle is real or complex. Maybe we should assume it to be complex.
  2. If you know a reference containing the proof, please let me know.
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2 Answers 2

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Not a full answer, but as I thought about this for a while, and managed to form something resembling a coherent picture of what's happening, I scribble down the reason, why I think this should be true.

Let $V_x$ be the fiber of $x\in X$, and $V_{x,\chi}\subset V_x$ be the subspace of those vectors $v$ that are acted on according to the character $\chi$, i.e. $v\in V_{x,\chi}$, iff $g\cdot v=\chi(g) v$ for all $g\in G$.

The smoothness of the $G$-action surely means that the inner products $$m_\chi(x):=\langle \chi, V_x\rangle_G=\dim V_{x,\chi}$$ between $\chi$ and the representation of $G$ on the fiber $V_x$ are smooth functions of $x$. Presumably $X$ is connected, so this means that $m_\chi(x)$ is a constant, so at least the dimension of $V_{x,\chi}$ is independent from $x$.

But we can say more. The group algebra $\Bbb{C}G$ acts linearly on the fibers, and as the individual elements of $G$ act smoothly on $V$, so do the elements of the group algebra. Consider the idempotent $$ e_\chi=\frac1{|G|}\sum_{g\in G}\overline{\chi(g)}g\in \Bbb{C}G. $$ At all the fibers $V_x$, we have $V_{x,\chi}=e_\chi(V_x)$. Because this projection $e_\chi$ depends smoothly on $x$. This is because if we denote by $\pi_x$ the representation of $G$ on the fiber $V_x$, then we get a smooth projection $V\to V_\chi$ as $$ e_\chi(x)=\frac1{|G|}\sum_{g\in G}\overline{\chi(g)}\pi_x(g). $$ In place of the image of $e_\chi(x)$ we also get $V_\chi$ as the kernel of $I-e_\chi(x)$.

I think that the claim follows. I'm too ignorant about manifolds/ fiber bundles to call upon a suitable theorem at this point :-(

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  • $\begingroup$ Can you explain why $V_{x,\chi} = e_\chi(V_X)$ holds? $\endgroup$
    – H. Shindoh
    Nov 13, 2013 at 12:10
  • $\begingroup$ That's basic representation theory. It is kind of an averaging argument. Say, if $G$ is of order two, $G=\{1,g\}$, then for all $z\in V_x$ we have that $(z+g\cdot z)/2$ is invariant under $G$, and $(z-g\cdot z)/2$ transforms according to the sign character. Furthermore $$z=\frac{z+g\cdot z}2+\frac{z-g\cdot z}2,$$ the fiber $V_x$ decomposes into a direct sum as prescribed. For cyclic groups this is also equivalent to discrete Fourier analysis. Coming back to this later. A very busy day. $\endgroup$
    – Jyrki Lahtonen
    Nov 13, 2013 at 12:17
  • $\begingroup$ The summands in the previous comment are exactly the $e_\chi\cdot z$ for $\chi$ equal to the trivial (resp. non-trivial) character of $C_2$. For larger cyclic groups you just get more terms. $\endgroup$
    – Jyrki Lahtonen
    Nov 13, 2013 at 12:19
  • $\begingroup$ Do we need $e_\chi = |G|^{-1}\sum_g \overline{\chi(g)}g$? Then it immediately follows from Schur orthogonality that $e_\chi$ is the projection onto $V_\chi$. $\endgroup$
    – H. Shindoh
    Nov 14, 2013 at 9:53
  • $\begingroup$ I think that it is better to regard $V_\chi$ as the kernel of the map $1-e_\chi$ rather than the image of $e_\chi$. The map $1-e_\chi$ is a (smooth) bundle map, and the rank of the kernel is constant. So the kernel forms a subbundle of the original bundle. But this is a small thing. I think that your answer is correct. $\endgroup$
    – H. Shindoh
    Nov 14, 2013 at 9:57
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This is clear on the level of a local trivialisation. If $U$ is any open subset of $X$ such that the pre-image of $U$ under $V\rightarrow X$ is isomorphic to $U\times \mathbb{C}^k$, then the $G$-action on this pre-image is just a $G$-representation on $\mathbb{C}^k$, so locally above $U$ you get the required decomposition. Since the open $U$ overlap and the trivialisations at the overlapping open sets are all compatible, the decomposition into eigenspaces fits together to give you a smooth bundle on the whole manifold.

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  • $\begingroup$ Even within a local trivialisation the $G$-action on the fibers may still vary from point to point. Consider the case $k=2$, where the fiber above point $x$ is a direct sum of two subspace of $\{x\}\times \Bbb{C}^2$, say $M_1(x)$ and $M_2(x)$, belonging to distinct characters $\chi_1$ and $\chi_2$. I don't see why $M_1(x)$ would automatically have to be the same subspace for all $x$? $\endgroup$
    – Jyrki Lahtonen
    Nov 12, 2013 at 14:58
  • $\begingroup$ @JyrkiLahtonen: $G$ acts on the manifold $V$, and since it acts trivially on $X$, it preserves the preimage of any open subset of $X$. So you have an action of $G$ on open sets $U\times \mathbb{C}^k$ for suitable trivialising $U\subseteq X$. But again, $G$ acts trivially on $U\times \{0\}$, so the action on $U\times \mathbb{C}^k$ descends to an action on $(U\times \mathbb{C}^k)/(U\times \{0\}) \cong \mathbb{C}^k$. Or am I oversimplifying something? $\endgroup$
    – Alex B.
    Nov 12, 2013 at 17:04
  • $\begingroup$ I don't know. I am just not sure why you rule out a case like $X=\Bbb{C}^*$, $G=C_2=\langle g\rangle$, $V=X\times\Bbb{C}^2$, and $g$ acts as identity on all sets of the form $\{z\}\times \langle(1,z)\rangle$, and as negation on all sets of the form $\{z\}\times\langle(-z,1)\rangle$. Here I explicitly gave the sub-bundles, so the theorem holds. Furthermore, we can get what you describe by using another diffeomorphism $V\cong U\times \Bbb{C}^2$. In some sense I think that this is what the question is all about (in addition glueing the pieces together), but I may be very wrong. $\endgroup$
    – Jyrki Lahtonen
    Nov 12, 2013 at 17:35
  • $\begingroup$ What I had in mind earlier looks like $g$ acting by $$\frac1{1+z^2}\pmatrix{1-z^2&2z\cr 2z&-1+z^2\cr}$$ on $\{z\}\times\Bbb{C}^2$ with $z\neq\pm i$, so I need $X=\Bbb{C}\setminus\{\pm i\}$ instead. Looks smooth to me. $\endgroup$
    – Jyrki Lahtonen
    Nov 12, 2013 at 17:53
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    $\begingroup$ @AlexB. As Jyrki says, if you take a local trivialisation, then you will get a smooth family of representation. Namely, writing $U \times \mathbb C^k$ for a trivialisation, we can only write $g(z,v)=(z,\pi_z(g)v)$ for the $G$-action. If you want to claim that we can take a trivialisation so that $\pi_z$ is independent of $z$, then the existence of such a trivialisation is exactly my question. $\endgroup$
    – H. Shindoh
    Nov 13, 2013 at 10:44

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