# Equivariant tubular neighborhood of an exceptional orbit of a circle action

A pseudofree $$S^1$$-action on a sphere $$S^{2k-1}$$ is a smooth $$S^1$$-action which is free except for finitely many exceptional orbits whose isotropy types $$\Bbb Z_{a_1},\dots,\Bbb Z_{a_n}$$ have pairwise relatively prime orders. Suppose we are given a pseudofree $$S^1$$-action on $$S^5$$. An exceptional orbit in $$S^5$$ with isotropy type $$\Bbb Z_a$$ has an equivariant tubular neighborhood which may be identified with $$\Bbb C\times \Bbb C\times S^1$$ with $$S^1$$-action $$t\cdot (z,w,u)=(t^rz,t^sw,t^au)$$ where $$r, s$$ are relatively prime to $$a$$.

These are asserted in the first two pages of the paper https://www.semanticscholar.org/paper/O(2)-actions-on-the-5-sphere-Fintushel-Stern/8db4650561a2fbcbbff083cd9b636ce0cd5b118f, and I cannot understand the last sentence. How can we assure that an exceptional orbit in $$S^5$$ with isotropy type $$\Bbb Z_a$$ has such an equivariant tubular neighborhood?

• Are you concerned with why there is an equivariant tubular neighborhood? Or why it takes that form? Or both? Sep 27, 2021 at 16:09
• @JasonDeVito I can see that there is an equivariant tubular neighborhood which may be identified with $\Bbb C^2\times S^1$, but I can't see why the action takes the form $t\cdot (z,w,u)=(t^rz,t^sw,t^au)$. Sep 27, 2021 at 20:16

First, a consequence of the slice theorem (see also this link) is that there is a neighborhood of the exceptional $$S^1$$ of the form $$S^1\times_{\mathbb{Z}_a} \mathbb{R}^4$$. Where the $$\mathbb{Z}_a$$ action on $$\mathbb{R}^4$$ is linear, the $$\mathbb{Z}_a$$ action on $$S^1$$ is by right multiplication, and the action of $$S^1$$ on $$S^1\times_{\mathbb{Z}_a} \mathbb{R}^4$$ is just left multiplication on the first factor: $$t\cdot[u,y] = [tu,y]$$.

So, the question is why $$S^1\times_{\mathbb{Z}_a}\mathbb{R}^4$$ is equivariantly diffeomorphic to the action you list on $$\mathbb{C}\times\mathbb{C}\times S^1$$. To answer this, we need to better understand the action of $$\mathbb{Z}_a$$ on $$\mathbb{R}^4$$.

From this MSE answer, all non-trivial irreducible real representations of $$\mathbb{Z}_a$$ are sums of 2-dimensional representations, except when $$a=2$$, where there are also non-trivial 1-dimensional representations. For now, assume $$a\geq 3$$. Then, the action of $$\mathbb{Z}_a$$ on $$\mathbb{R}^4$$ is, up to $$\mathbb{R}$$-linear isomorphism, given by the following action of $$\mathbb{Z}_a$$ on $$\mathbb{C}^2$$: $$\alpha\ast (z,w) = (\alpha^r z, \alpha^s w)$$, where $$\alpha$$ is an $$a$$-th root of $$1$$ (but we do not yet know that $$r$$ and $$s$$ are relatively prime to $$a$$). Thus, we can rewrite $$S^1\times_{\mathbb{Z}_a} \mathbb{R}^4$$ as $$S^1\times_{\mathbb{Z}_a}\mathbb{C}^2$$.

So, why are both $$r$$ and $$s$$ relatively prime to $$a$$? Well, let $$t\in S^1$$ be any $$\gcd(r,a)$$-th root of unity. Then $$t\in \mathbb{Z}_a$$ and thus $$t\cdot [u,z,0] = [tu,z,0] = [u,t^r z, 0] = [u,z,0].$$ In other words, the subgroup $$\mathbb{Z}_{\gcd(r,a)}$$ of $$S^1$$ fixes not only the orbit $$S^1\cdot x$$, but also some of normal neighborhood to it. Since there are no singular orbits by assumption, it follows that $$\gcd(r,a) = 1$$.

When $$a=2$$, the same conclusions are true. Here, up to equivalence, the action on $$\mathbb{R}^4$$ is given by $$\alpha \cdot ( y_1,y_2, y_3, y_3) = ((-1)^{\beta_1} y_1, (-1)^{\beta_2} y_2, (-1)^{\beta_3} y_3, (-1)^{\beta_4} y_4)$$. But note that the $$\mathbb{Z}_2$$ action on $$S^1\times \mathbb{R}^4$$ is orientation preserving since the quotient is diffeomorphic to $$S^1\times \mathbb{R}^4$$, which is orientable. This implies that an even number of the $$\beta_i$$ are even. But then this action is equivalent to $$\mathbb{Z}_2$$ action on $$\mathbb{C}^2$$ of the form $$\alpha\ast(z,w) = (\alpha^r z, \alpha^s w)$$ where $$r=s= 0$$ if all $$\beta_i$$ are even, $$r=0,s=1$$ if precisely two $$\beta_i$$ are even, and $$r=s=1$$ if all $$\beta_i$$ are odd. Since $$a=2$$ and $$r,s\in\{0,1\}$$, the $$\gcd$$ condition is obvious.

Ok, so, in summary, regardless of the value of $$a$$, we know that there is a neighborhood of the form $$S^1\times_{\mathbb{Z}_a}\times \mathbb{C^2}$$, where $$\mathbb{Z}_a$$ acts on $$S^1\times \mathbb{C}^2$$ as $$\alpha\ast(u,z,w) = (\alpha u, \alpha^r z, \alpha^s w)$$ with $$r,s$$ relatively prime to $$a$$, and where $$t\in S^1$$ acts by left multiplication: $$t\cdot[u,z,w] = [tu,z,w]$$.

As this point we are ready to write down an equivariant diffeomorphism from $$S^1\times_{\mathbb{Z}_a}\mathbb{C}^2$$ to $$\mathbb{C}\times \mathbb{C}\times S^1$$ with your action. Define $$\psi:S^1\times_{\mathbb{Z}_a}\mathbb{C}^2\rightarrow \mathbb{C}\times\mathbb{C}\times S^1$$ by $$\psi[u,z,w] = (u^{r}\overline{z},u^{s}\overline{w}, u^a)$$.

We claim that $$\psi$$ is well defined in the sense that if $$\alpha\in \mathbb{Z}_a$$, then $$\psi([u\alpha, \alpha^r z, \alpha^s w]) = \psi([u,z,w])$$. Let's compute: $$\psi([u\alpha,\alpha^r z, \alpha^s w]) = (u\alpha)^{r} \alpha^{-r} \overline{z}, (u\alpha)^{s} \alpha^{-s} \overline{w}, (u\alpha)^a) = (u^{r} \overline{z}, u^{s} \overline{w}, u^a) = \psi([u,z,w]).$$

We next claim $$\psi$$ is injective. Indeed, if $$\psi([u_1,z_1,w_1]) = \psi([u_2,z_2,w_2])$$, so $$(u_1^{r} \overline{z}_1, u_1^{s} \overline{w}_1, u_1^a) = (u_2^{r} \overline{z}_2, u_2^{s} \overline{w}_2, u_2^a),$$ then the last coordinate gives $$u_1^a = u_2^a$$, so $$u_1 = \alpha u_2$$ for some $$\alpha \in \mathbb{Z}_a$$.

Now, the first coordinate equality simplifies to $$z_1 = \alpha^r z_2$$ and the second coordinate equality simplifies to $$w_1 = \alpha^s w_2$$. In other words, $$(u_1, z_1 w_1) = (u_2 \alpha, \alpha^r z_2, \alpha^2 w_1)$$, so $$[u_1,z_1,w_1] = [ u_2,z_2,w_2]$$. That is, $$\psi$$ is injective.

We next claim that $$\psi$$ is surjective. Given (z',w',u'), select $$u$$ to be an $$a$$-th root of $$u'$$, set $$z = u^r \overline{z}'$$, and $$w = u^s \overline{w}'$$. Then $$\psi([u,z,w]) = (z',w',u')$$.

We also note that $$\psi$$ is obviously smooth. Then inverse $$\psi^{-1}$$ exists by the above work. Moreover, since $$u'\in S^1$$, we can locally smoothly pick an $$a$$-th root of unity. So, by the surjectivity calculation, it follows that $$\psi^{-1}$$ is smooth Thus, $$\psi$$ is a diffeomorphism.

The last thing to check is that $$\psi$$ is equivariant. But $$\psi(t\cdot[ u,z,w]) = \psi[tu,z,w] = (t^r u^r \overline{z}, t^s u^s \overline{w}, t^a u^a) = t\cdot (u^r \overline{z}, u^s \overline{w}, u^a) = t\cdot \psi([u,z,w]).$$