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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).$2012-11-16
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    Of course not full dimensional. You can think of any reflection in $\Bbb R^n$, through any affine subspace.2012-11-17
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    What about looking at $f-Id$ , mapping $X$ to $X\subset \mathbb{R}^n$?2012-11-17
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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).2012-11-17
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    @Whuber the fixed point set is a line with a point, in fact it is a circle.2012-11-18
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    @Steve I don't follow: are you claiming that $(0:0:1)$ lies on the circle of fixed points?2012-11-18
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    @whuber: yes, it does. You can see this by lifting your involution to $S^2$, where it is just reflection in the $xy$ plane. The fixed points are a circle, and when pushing down to $RP^2$, we get the fixed points are $RP^1$, which is still just topologically a circle.2012-11-20
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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.2012-11-20

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