# The very short version

For any isometry $\sigma$, the fixed point set $\text{Fix}(\sigma)$ is a union of submanifolds. When is the dimension well-defined?

# The long version

In the following, let $M$ always be a Riemannian manifold. The quotes were slightly changed for consistency.

## What I found in the literature

In Klingenberg [1] we can read

1.10.15 Theorem. Let $\sigma\colon M \to M$ be an isometry. Then every connected component of the fixed point set $\text{Fix}(\sigma) := \{p \in M : \sigma(p) = p\}$ is a totally geodesic submanifold.

and Kobayashi [2] writes

Theorem 5.1. Let $\mathfrak S$ be any set of isometries of $M$. Let $\text{Fix}(\mathfrak S)$ be the set of points of $M$ which are left fixed by all elements of $\mathfrak S$. Then each connected component of $\text{Fix}(\mathfrak S)$ is a closed totally geodesic submanifold of $M$.

The latter is at least as strong as the first one, of course. The only concrete example given is in Klingenberg and appears to be just a reflection of the $n$-sphere, where we easily see the components to be of identical dimension. Neither explicitly mentions the possibility of the components having different dimensions.

However, Donnelly and Patodi [3][4] write (emphasis mine)

Let $M$ be a compact Riemannian manifold of dimension $d$ and $\sigma\colon M \to M$ an isometry […]. $\text{Fix}(\sigma)$ is the disjoint union of closed connected submanifolds $N$ of dimension $n$.

This gave me hope that in the compact case all the connected components had the same dimension, but later on [5] they write

Let $\sigma$ be an isometry of the compact Riemannian manifold $M$. The fixed point set of $\sigma$ is the disjoint union of compact connected totally geodesic submanifolds $N$.

and as far as I can tell, they only use $n$ as soon as an $N$ is fixed, so I'm assuming I overinterpreted an implicit dependency of $n$ on $N$.

## What I would like to know

My question is easily phrased as three:

• What are examples where the components of such fixed point sets have different dimensions?
• What conditions can be required of $\mathfrak S \subseteq \text{Aut}(M)$ / $\sigma \in \text{Aut}(M)$ such that $\text{Fix}(\mathfrak S)$ / $\text{Fix}(\sigma)$ is of uniform dimension?
• What conditions can be placed on $M$ such that $\text{Fix}(\mathfrak S)$ / $\text{Fix}(\sigma)$ has a well-defined dimension for all such $\mathfrak S$ / $\sigma$?

In the end, I would like to have as much of a sharp separation between the cases as is possible. The questions above all tackle this same problem from different directions.

## What is trivial

The following repeatedly led to confusion, so I'm saying these right now:

• Yes, if $M$ is not connected, counterexamples are trivial to come by. I'm only interested in the case where $M$ is connected.
• Yes, if $\text{Fix}(\mathfrak S)$ is connected, too, the dimension is well-defined by the standard argument.
• No, the fixed points are not always connected. Think about a reflection of the 1-sphere or a rotation of the 2-sphere.

## What we found out

I'm probably forgetting something but here is what else we know:

• All the standard examples we know and most stuff we came up with was highly symmetrical. This might explain why we failed to find counterexamples.
• Isometries fix geodesics between fixed points as long as the geodesics are unique for their length. This shows for example that on the sphere the only way to obtain a disconnected fixed point set is for it to consist of two antipodal points only.

[1]: Wilhelm Klingenberg, Riemannian Geometry. Page 95 at Google Books.

[2]: Shoshichi Kobayashi, Transformation Groups in Differential Geometry. Referenced at Fixed Points Set of an Isometry.

[3]: Harold Donnelly, Spectrum and fixed point sets of isometries. I. Directly in the beginning.

[4]: Harold Donnelly and V. K. Patodi, Spectrum and fixed point sets of isometries—II. Directly in the beginning.

[5]: Same as [4]. At the start of §2.

For an example of variable dimension of the fixed-point set, take $M=RP^2$ (with a constant curvature metric) and $\sigma$ an isometric reflection. (The projection of a reflection on $S^2$.) Then $Fix(\sigma)$ is a disjoint union of a point and a projective line. You can get more examples like this in higher dimensions. I am not sure about the rest of your questions.

Edit: If $\sigma$ is an isometry of a 2-dimensional orientable connected Riemannian manifold then $Fix(\sigma)$ has constant dimension. The reason is that near an isolated fixed point, $\sigma$ would preserve orientation and near a non-isolated fixed point $\sigma$ would have to reverse orientation. But a homeomorphism of a connected oriented manifold cannot both preserve and reverse orientation.

By working a bit more one can show the following: For every $n$ there exists an $n$-dimensional connected compact Riemannian manifold $M$ and an isometry $\sigma: M\to M$ whose fixed point set contains components of all possible dimensions: $0, 1, 2,..., n-1$.

Here is a construction. First of all, let $M_0,...,M_{n-1}$ be compact connected $n$-dimensional manifolds equipped with involutions $\sigma_0,...,\sigma_{n-1}$ such that $dim(Fix(\sigma_i))=i$. Furthermore, for each $i$ pick points $x_i, y_i\in M_i- Fix(\sigma_i)$ and set $x_i':= \sigma_i(x_i), y_i':= \sigma_i(y_i)$. I am assuming that these choices are generic so that for each $i$ all four points $x_i, x_i', y_i, y_i'$ are distinct. Now, form a connected manifold $M$ by performing a generalized connected sum along small disjoint balls $B_i, B_i', C_i, C_i'$ centered at the points $x_i, x_i', y_i, y_i'$, $i=0,...,n-1$, so that $\sigma_i(B_i)=C_i, \sigma(B_i')=C_i'$.

Namely, remove all these balls from the disjoint union $$\coprod_i M_i$$ ("puncturing each $M_i$") and perform gluing via diffeomorphisms $$f_i: \partial C_i\to \partial B_{i+1}, \partial f_i': C_i'\to \partial B_{i+1}',$$ $i=0, 1,....$. Choose these diffeomorphisms so that $$\sigma_{i+1}\circ f_i= f_i'\circ \sigma_i, i=0, 1,....$$ The result of this gluing is a smooth connected manifold $M$ equipped with a diffeomorphic involution $\sigma$ (whose restriction to each punctured $M_i$ equals $\sigma_i$). Lastly, put a $\sigma$-invariant Riemannian metric on $M$.

• If one wants M to be simply connected, consider the $S^1$ action on $\mathbb{C}P^n$ given in homogeneous coordinates by $z\ast[z_0:...:z_n] = [zz_0: z_1:...:z_n]$. This fixes the point $[1:0:...:0]$ as well as the $\mathbb{C}P^{n-1}$ given by $[0:z_1:...:z_n]$. By having the circle hit more coordinates, you can arrange the fixed point sets to be any pair $\mathbb{C}P^a, \mathbb{C}P^b$ with $a+b = n-1$. Feb 2 '18 at 15:30
• Can you give some idea what “By workig a bit more” entails? I see how to get arbitrary combinations by yours or Jason DeVitos example, but my dimension of $M$ explodes to hard to obtain your claim. Feb 3 '18 at 6:46
• @HermannDöppes: Sure, see the edit. Feb 3 '18 at 10:34
• @MoisheCohen Awesome, thank you very much. Feb 4 '18 at 1:53