# Can a point have a nontrivial isometry group?

This question is extremely related to this other question. In fact, a positive answer here directly implies a positive answer there. However, since it is a mathematically different question I decided to ask it separately.

The Freudenthal-Tits Magic Square construction relating exceptional Lie algebras to algebras of infinitesimal isometries of certain "projective planes" over the tensor product of two division algebras

$$\mathfrak{isom}((\mathbb{A}\otimes\mathbb{B})\mathbb{P}^2) = \mathfrak{der}(\mathbb{A}) \oplus \mathfrak{der}(J_3(\mathbb{B})) \oplus (\mathbb{A}_0\otimes J_3(\mathbb{B})_0)$$

can be generalized to projective spaces of arbitrary dimension $n$ for the division rings (and $n\leq2$ for the octonions), yielding

$$\mathfrak{isom}((\mathbb{A}\otimes\mathbb{B})\mathbb{P}^n) = \mathfrak{der}(\mathbb{A}) \oplus \mathfrak{der}(J_{n+1}(\mathbb{B})) \oplus (\mathbb{A}_0\otimes J_{n+1}(\mathbb{B})_0)$$

In particular the right side still makes sense for $n=0$:

$$\mathfrak{isom}((\mathbb{A}\otimes\mathbb{B})\mathbb{P}^0) = \mathfrak{der}(\mathbb{A}) \oplus \mathfrak{der}(\mathbb{B})$$

However, $(\mathbb{A}\otimes\mathbb{B})\mathbb{P}^0$ is just a point.

In his notes on octonions, John Baez says

We can think of $\mathbb{HP}^n$ as the unit sphere in $\mathbb{H}^{n+1}$ with points $x$ and $\alpha x$ identified whenever $\alpha$ is a unit quaternion, and as before, $\mathbb{HP}^n$ inherits a Riemannian metric. The group $\mathrm{Sp}(n+1)$ acts as isometries of $\mathbb{HP}^n$, but this action comes from right multiplication, so

$$\mathrm{Isom} (\mathbb{HP}^n) \cong \mathrm{Sp}(n+1)/\{ \pm 1 \}$$ since not $\mathrm{Sp}(1)$ but only its center $\{\pm 1\}$ acts trivially on $\mathbb{HP}^n$ by right multiplication. At the Lie algebra level, this gives $$\mathfrak{isom}(\mathbb{HP}^n) \cong \mathfrak{sp}(n+1)$$

Does this argument work when $n=0$? How can a point have a nontrivial isometry group?

• The best answer, I think, is that Baez does not offer a mathematical proof, hence, one cannot talk about "his argument"; the claim he makes indeed fails for $n=0$. – Moishe Kohan Aug 23 '16 at 9:52

No, a point cannot have a non-trivial isometry group. There's only a single map $\{*\}\to \{*\}$, the identity (this holds in the category of sets, and hence clearly also in the metric space category).
• Note that his definition of $HP^n$ is different from yours. He starts with the unit sphere in $H^{n+1}$. If one follows Baez' definition then indeed $HP^0$ is a single point. Of course, the group of bijections of a singleton is still trivial. – Moishe Kohan Aug 23 '16 at 9:51
Concerning the quaternion example, there is a fundamental difference between $\mathbb{H}$ and $\mathbb{H}^n$ with $n >1$.
When $n=1$, then every $\mathbb{H}$-linear homotheties $\mathbb{H} \rightarrow \mathbb{H}$ preserve lines (in fact there is only one line). When $n >1$, only the $\mathbb{H}$-linear homotheties $\mathbb{H}^n \rightarrow \mathbb{H}^n$ with real coefficient preserve lines. (Note : if a scalar $\lambda \in \mathbb{H}$ acts on a vector $x$ by $\lambda x$, then a $\mathbb{H}$-linear homothetie is a map $x \mapsto x.a$ where $a \in \mathbb{H}$ is the coefficient. Note that $x \mapsto a.x$ is not linear).
Thus in Baez's notes, when $n=0$, the line $$\mathrm{Isom}(\mathbb{H}P^n) \simeq \mathrm{Sp}(n+1)/\{±1\}$$ should be replaced by $$\mathrm{Isom}(\mathbb{H}P^0) \simeq \mathrm{Sp}(1)/\{ \text{units}\}$$ which is trivial.