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?

  • $\begingroup$ Are you concerned with why there is an equivariant tubular neighborhood? Or why it takes that form? Or both? $\endgroup$ Sep 27, 2021 at 16:09
  • $\begingroup$ @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 $t\cdot (z,w,u)=(t^rz,t^sw,t^au)$. $\endgroup$
    – user302934
    Sep 27, 2021 at 20:16

1 Answer 1


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 $$\psi(t\cdot[ u,z,w]) = \psi[tu,z,w] = (t^r u^r \overline{z}, t^s u^s \overline{w}, t^a u^a) = t\cdot (u^r \overline{z}, u^s \overline{w}, u^a) = t\cdot \psi([u,z,w]).$$


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