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Let $f:X\rightarrow X$ be a smooth map of a smooth manifold with $f^2=\operatorname{id}$.

Is the subset $\{x\in X\mid f(x)=x\}$ a smooth submanifold?

I tried to find an argument with the implicit function theorem, but I don't have an answer.

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Not sure yet, but it does not need to be full dimensional. Take the graph $z = xy,$ then the involution $(x,y,xy) \rightarrow (-x,-y, xy).$ –  Will Jagy Nov 16 '12 at 22:44
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What about looking at $f-Id$ , mapping $X$ to $X\subset \mathbb{R}^n$? –  user641 Nov 17 '12 at 0:38
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The connected components of this subset will each be smooth submanifolds, but their dimensions can vary. As such, their union is not a submanifold. As a simple example, consider the vertical reflection $(x:y:z)\to (x:y:-z)$ of the space of one-dimensional linear subspaces of $\mathbb{R}^3$ (that is, $\mathbb{RP}^2$) to itself, expressed in projective coordinates. It's well-defined and smooth. Its fixed points consist of a codimension-1 submanifold $(\cos(\theta):\sin(\theta):0)$ (a circle of all horizontal lines) together with a codimension-2 submanifold $(0:0:1)$ (the vertical line). –  whuber Nov 17 '12 at 23:04
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@Whuber the fixed point set is a line with a point, in fact it is a circle. –  user641 Nov 18 '12 at 2:00
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@Steve It doesn't quite work that way: the z-axis, qua point of the projective plane, also is fixed under this reflection--that's the whole point of the example. To put it another way: an equivalence class (the z-axis) can be fixed without its elements being fixed, so it does not suffice to look only at the fixed points in the covering space. We're not talking about just a single point, either, in general: this example readily generalizes to higher-dimensional projective planes, Grassmannians, flag manifolds, etc., to give a rich set of counterexamples. –  whuber Nov 20 '12 at 8:02
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2 Answers

If $f$ is involutive, so is its derivative in all points: $(T_xf)^2=Id$. And hence, there is a basis $(b_1,..,b_m,..)$ of $T_xX$ in which $T_xf$ becomes of the form $\pmatrix{I_m & 0\\0&-I_k}$ where $I_k$ is the $k\times k$ identity matrix, and $m+k=\dim X$.

So, if $f(x)=x$, then cutting out the correspondent of $\Bbb R^m\cong\langle b_1,..,b_m\rangle \subseteq T_xX\cong \Bbb R^n$ from the local chart $\phi$ will give you charts. Smoothness guarantees that the change of $b_i$'s along $x$ is smooth.

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This argument shows the set is locally a submanifold, but I cannot find a definition of "submanifold" that assures that the entire set will actually be a submanifold: all the definitions I know of are constructed to assure that submanifolds have well-defined dimensions. That's not necessarily the case here. –  whuber Nov 17 '12 at 23:07
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To add to @whuber's comment: there are know topological constraints which will require the fixed-point sets to have globally a constant codimension. But there are also other cases. See sciencedirect.com/science/article/pii/0166864195000887 and references therein. –  Willie Wong Jun 24 '13 at 15:23
    
The requirement that connected components of a manifold (or of a submanifold for that matter) all have the same dimension is in my view a bug in the definition. Bad as the category of smooth manifolds is in many respects, it makes matters worse to artificially rule out even the most elementary categorical constructions (such as coproducts, closure under retracts, etc.) just on the basis of a (relatively unimportant) convention like this. –  user43208 Sep 2 '13 at 0:06
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I can't add a comment to Berci's answer, that is why I post this comment as an answer.

The argument in Berci's answer does not work as stated. (Since the question is a local matter, we can assume that $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$, but I'm not sure that this helps.) The problem is that $f^2=id$ only implies that $(df)_{f(p)}\circ (df)_p = id_{T_pX}$. So you cannot conclude that $(df)_p$ is of the given form, since $(df)_p$ and $(df)_{f(p)}$ are different maps.

A different argument using Riemannian geometry is the following:

Choose a Riemannian metric $g$ on $X$, then $\tilde g= g + (f^* g)$ is a metric for which $f$ is an isometry (because $f^2=id$). The fixed point set of an isometry is easily seen to be a totally geodesic submanifold of $X$.

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At a fixed point $p$, $(df)_p$ and $(df)_{f(p)}$ are the same, so you can conclude that $(df)_p$ is of that form at any fixed point. –  Omar Antolín-Camarena Aug 20 '13 at 16:55
    
Yes. That is true at any fixed point. But not near any fixed point. This is what you would need for the argument above. –  anon Sep 24 '13 at 7:07
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