In this old question of mine


the presentation of spheres as symplectic group homogeneous spaces was discussed very well, giving $$ S^{4n-1} \simeq Sp(n)/Sp(n-1). $$

Going us back now to the presentation of the spheres most clear in all our minds: $$ S^{2n-1} \simeq SU(n)/SU(n-1) \simeq U(n)/U(n-1). $$ We can recall that this gives to the sphere $S^{2n-1}$ a canaonical $U(1)$-action. Moreover, the invariant space of this $U(1)$-action is the complex projective space, i.e. $$ \mathbb{CP}^{n-1} \simeq S^{2n-1}/U(1). $$

So we can now see what my question is! Does there exist a $U(1)$-action on $S^{4n-1}$ given from its presentation as $Sp(n)/Sp(n-1)$? If yes, then what is the quotient $S^{4n-1}/U(1)$? Maybe it is just $\mathbb{CP}^{2n-1}$?

  • 7
    $\begingroup$ $S^{4n-1}$ has an $Sp(1)$-action which gives the quotient $\mathbb{HP}^{n-1}$. Take $U(1) \subset Sp(1)$ and the quotient is $\mathbb{CP}^{2n-1}$. $\endgroup$ – Michael Albanese May 25 '18 at 14:14

The presentation $\mathbb{S}^{2n-1} = \mathit{U}_n/\mathit{U}_{n-1}$ is tantamount to considering $\mathbb{S}^{2n-1}$ as the unit sphere in $\mathbb{C}^n$ (as $\mathit{U}_n$ is the group of $\mathbb{C}$-linear maps of $\mathbb{C}^n$, acting, say, from the right, preserving the Hermitian inner product and $\mathit{U}_{n-1}$ is the stabilizer of a given vector). Now $\mathbb{C}^n$ has the structure of a $\mathbb{C}$-vector space, so you can quotient out by complex numbers of unit norm, $\mathit{U}_1$, getting, as you say, $\mathbb{P}^{n-1}(\mathbb{C})$ as $\mathbb{S}^{2n-1}/\mathit{U}_1 = \mathit{U}_n/(\mathit{U}_{n-1}\times \mathit{U}_1)$. In other words, the $(2n-1)$-sphere is fibered in circles with base $\mathbb{P}^{n-1}(\mathbb{C})$ (for $n=2$, noting that $\mathit{SU}_2$ is isomorphic to $\mathit{Spin}_3$, this is the classical Hopf fibration $\mathbb{S}^1 \hookrightarrow \mathbb{S}^3 \to \mathbb{S}^2$).

The analogous description of $\mathbb{S}^{4n-1} = \mathit{Sp}_n/\mathit{Sp}_{n-1}$ is to consider $\mathbb{S}^{4n-1}$ as the unit sphere in the space $\mathbb{H}^n$ of quaternionic vectors with the standard Hermitian inner product, so that $\mathit{Sp}_n$ is the group of $\mathbb{H}$-linear maps (acting on the right, say, and where $\mathbb{H}$ acts on the left) preserving the product in question and $\mathit{Sp}_{2n-1}$ is the point stabilizer. This time we can quotient out by quaternions of unit norm, $\mathit{Sp}_1$, and this describes $\mathbb{P}^{n-1}(\mathbb{H})$ as $\mathbb{S}^{4n-1}/\mathit{Sp}_1 = \mathit{Sp}_n/(\mathit{Sp}_{n-1}\times \mathit{Sp}_1)$. In other words, the $(4n-1)$-sphere is fibered in $3$-spheres with base $\mathbb{P}^{n-1}(\mathbb{H})$ (for $n=2$, noting that $\mathit{Sp}_2$ is isomorphic to $\mathit{Spin}_5$, and $\mathit{Sp}_1 \times \mathit{Sp}_1$ to $\mathit{Spin}_3 \times \mathit{Spin}_3 = \mathit{Spin}_4$, this is the Hopf fibration $\mathbb{S}^3 \hookrightarrow \mathbb{S}^7 \to \mathbb{S}^4$).

(Of course you can also quotient $\mathbb{S}^{4n-1}$ by $\mathit{U}_1$ if you want, but I don't see how that can be a question since $\mathbb{S}^{4n-1}$ is already a $\mathbb{S}^{2m-1}$ for $m=2n$ so you already have the answer.)

And before you ask, there is also a Hopf fibration $\mathbb{S}^7 \hookrightarrow \mathbb{S}^{15} \to \mathbb{S}^8$ which makes it look like $\mathbb{S}^8 = \mathbb{P}^1(\mathbb{O})$ should be the quotient of an action of "the group of octonions of unit norm" on the unit sphere in $\mathbb{O}^2$ but, alas, there is no such thing as "the group of octonions of unit norm", and while it makes some sense to think of $\mathit{Spin}_9$ as the "octonionic $U_2$" in the description of $\mathbb{P}^1(\mathbb{O}) = \mathbb{S}^8$ as $\mathit{Spin}_9/\mathit{Spin}_8$, there is no way to decompose $\mathit{Spin}_8$ as the product of two "octonionic $U_1$" factors. Furthermore, while $\mathbb{P}^2(\mathbb{O}) = F_4/\mathit{Spin}_9$ can be defined (where $F_4$ can be thought of as some kind of "octonionic $U_3$"), there does not exist a fibration of $\mathbb{S}^{23}$ with base $\mathbb{P}^2(\mathbb{O})$ and fiber $\mathbb{S}^7$, so the analogies stop there.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.